Reinforced Concrete Mechanics Design Solutions

Chapter 3 3-1

What is the significance of the “critical stress”? (a)

with respect to the structure of the concrete?

A continuous pattern of mortar cracks begins to form. As a result there are few undamaged portions to carry load and the stress-strain curve is highly nonlinear. (b)

with respect to spiral reinforcement?

At the critical stress the lateral strain begins to increase rapidly. This causes the concrete core within the spiral to expand, stretching the spiral. The tension in the spiral is equilibrated by a radial compression in the core. This in turn, biaxially compresses the core, and thus strengthens it. (c)

with respect to strength under sustained loads?

When concrete is subjected to sustained loads greater than the critical stress, it will eventually fail.

3-2

A group of 45 tests on a given type of concrete had a mean strength of 4780 psi and a standard deviation of 525 psi. Does this concrete satisfy the requirements of ACI Code Section 5.3.2 for 4000-psi concrete?

From Eq. 3-3a:

Using (for design)

From Eq. 3-3b:

Using (for design)

Because both of these exceed 4000 psi, the concrete satisfies the requirements of ACI Code Section 5.3.2 for 4000 psi concrete.

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3-1

3-3

The concrete containing Type I cement in a structure is cured for 3 days at 70° F followed by 6 days at 40° F. Use the maturity concept to estimate its strength as a fraction of the 28-day strength under standard curing.

Note: C 

5  F  32  , so 70° F = 21.1° C and 40° F = 4.4° C 9

From Eq. 3-6: n

M   (Ti  10)(ti ) i 1

 (21.1  10)(3)  (4.4  10)(6)  180 C days From Fig. 3-8 the compressive strength will be between 0.60 and 0.70 times the 28-day strength under standard curing conditions.

3-4

3-5

Use Fig. 3-12a to estimate the compressive strength 2 for bi-axially loaded concrete subject to: (a)  (b)

1 = 0.0, 2 = fc'

(c)

1 = 0.5 fc' in compression, 2 = 1.2 fc'

1 = 0.75 ft' in tension, 2 = 0.5 fc'

The concrete in the core of a spiral is subjected to a uniform confining stress 3 of 750 psi. What will the compressive strength, 1 be? Assume .

From Eq. 3-16:


3-2

3-6

3-7

What factors affect the shrinkage of concrete? (a)

Relative humidity. Shrinkage increases as the relative humidity decreases, reaching a maximum at RH ≤ 40%.

(b)

The fraction of the total volume made up of paste. As this fraction increases, shrinkage increases.

(c)

The modulus of elasticity of the aggregate. As this increases, shrinkage decreases.

(d)

The water/cement ratio. As the water content increases, the aggregate fraction decreases, causing an increase in shrinkage.

(e)

The fineness of the cement. Shrinkage increases for finely ground cement that has more surface area to attract and absorb water.

(f)

The effective thickness or volume to surface ratio. As this ratio increases, the shrinkage occurs more slowly and the total shrinkage is likely reduced.

(g)

Exposure to carbon dioxide tends to increase shrinkage.

What factors affect the creep of concrete? (a)

The ratio of sustained stress to the strength of the concrete. The creep coefficient, , is roughly constant up to a stress of 0.5 fc', but increases above that value.

(b)

The humidity of the environment. The amount of creep decreases as the RH increases above 40%.

(c)

As the effective thickness or volume to surface ratio increases, the rate at which creep develops decreases.

(d)

Concretes with a high paste content creep more that concretes with a large aggregate fraction because only the paste creeps.


3-3

3-8

A structure is made from concrete containing Type 1 cement. The average ambient relative humidity is 70 percent. The concrete was moist-cured for 7 days. fc' = 4000 psi. (a)

Compute the unrestrained shrinkage strain of a rectangular beam with crosssectional dimensions of 8 in. x 20 in. at 2 years after the concrete was placed.

1. Compute the humidity modification factor from Eq. (3-30a):

2. Use Eq. (3-31) to compute the volume/surface area ratio modification factor:

( ⁄

)

(

)

⁄

3. Use Eq. (3-29) to compute the ultimate shrinkage strain: (

)

4. Use Eq. (3-28) to compute the shrinkage strain after 2 years:

(

)

(b)

(

)

Compute the stress dependent (creep) strain in the concrete of a 20 in. x 20 in. x 12 ft column at age 3 years. A compression load of 400 kips was applied to the column at 30 days.

1. Compute the ultimate shrinkage strain coefficient,

⁄

, using Eqs. (3-36)-(3-39).

,

Where:

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3-4

2. Compute the creep coefficient for the time since loading,

(

3. Compute the total stress-dependent strain,

, using Eq. (3-35).

), using Eqs. (3-5), (3-18), and (3-35).

First, calculate the creep strain since the load was applied:

(

)

(

√

√ ( ) ( )

)

Then, calculate the initial strain when the load is applied: ( )

(

)

( )

( )

( )

√

( )

√

( ) ( )

( ) Thus, (

)

( )

(

)


3-5

Chapter 4 4-1

Figure P4-1 shows a simply supported beam and the cross-section at midspan. The beam supports a uniform service (unfactored) dead load consisting of its own weight plus 1.4 kips/ft and a uniform service (unfactored) live load of 1.5 kip/ft. The concrete strength is 3500 psi, and the yield strength of the reinforcement is 60,000 psi. The concrete is normal-weight concrete. Use load and strength reduction factors from ACI Code Sections 9.2 and 9.3. For the midspan section shown in part (b) of Fig. P4-1, compute and show that it exceeds .

1. Calculate the dead load of the beam. Weight/ft =

24  12  0.15  0.3 kips/ft 144

2. Compute the factored moment, M u : Factored load/ft: wu = 1.2(0.30 + 1.40) + 1.6(1.50) = 4.44 k/ft M u  wu

2

8  4.44  202 8  222 kip-ft

3. Compute the nominal moment capacity of the beam, M n and the strength reduction factor,  . Tension steel area: As = 3 No. 9 bars = 3 1.00 in.2 = 3.00 in.2 Compute the depth of the equivalent rectangular stress block,  , assuming that tension steel is yielding. From equilibrium (using Eq. (4-14)):

  c 1

A f 3.00  60000 s y   5.04 in. 0.85 f ' b 0.85  3500  12 c

For fc'  3500 psi, 1  0.85 . Therefore, c  

1

 5.04

0.85

 5.93 in.

Check whether tension steel is yielding:  d c  21.5  5.93     0.003  0.00788 Using Eq.(4-18)      s t  c  cu  5.93  Thus,  > 0.002 and the steel is yielding ( f s  f y ). s

 

Compute the nominal moment strength, using Eq. (4-21): 5.04   3.00  60000   21.5    2    M  A f d     285 kip-ft n s y 2 12000 Since,  t  0.00788  0.005 the section is clearly tension-controlled and  =0.9. Then,  M n  0.9  285 kip-ft  256 kip-ft. Clearly,  M n  M u © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write 4-1 to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

4-2

A cantilever beam shown in Fig. P4-2. The beam supports a uniform service (unfactored) dead load of 1 kip/ft plus its own dead load and it supports a concentrated service (unfactored) live load of 12 kips as shown. The concrete is normal-weight concrete with psi and the steel is Grade 60. Use load and strength-reduction factors form ACI Code Section 9.2 and 9.3. For the end section shown in part (b) of Fig. P4-2, compute and show it exceeds .

1. Calculate the dead load of the beam. Weight/ft =

30  18  0.15  0.563 kips/ft 144

2. Compute the factored moment, M u . Factored distributed load/ft: wu = 1.2(0.563 + 1.0) = 1.88 k/ft Factored live load is a concentrated load: Pu  1.6 12  19.2 kips



M u   wu

2







2  Pu   1   1.88  102 2  19.2  9   267 kip-ft

3. Compute the nominal moment capacity of the beam, M n and the strength reduction factor,  . Tension steel area: As = 6 No. 8 bars = 6  0.79 in.2 =4.74 in.2 Compute the depth of the equivalent rectangular stress block,  , assuming that tension steel is yielding. From equilibrium (using Eq. (4-14)): A f 4.74  60000 s y   2.79 in. 1 0.85 f ' b 0.85  4000  30 c ' For fc  4000 psi, 1  0.85 . Therefore, c    2.79  3.28 in. 1 0.85

  c

Check whether tension steel is yielding:  d c  15.5  3.28     0.003  0.011 > 0.0021 Using. Eq.(4-18)      s t  c  cu  3.28  Thus,  > 0.002 and the steel is yielding ( f s  f y ). s

 

Compute the nominal moment strength, using Eq. (4-21):

2.79   4.74  60000  15.5    2    M  A f d     334 kip-ft n s y 2 12000 Since,  t  0.011  0.005 the section is clearly tension-controlled and  =0.9. Then,  M n  0.9  334  301 kip-ft  267 kip-ft. Clearly,  M n  M u © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write 4-2 to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

4-3

(a)

Compare for singly reinforced rectangular beams having the following properties. Use strength reduction factors from ACI Code Sections 9.2 and 9.3.

Beam

b

d

No.

(in.)

(in.)

1 2 3 4 5

12 12 12 12 12

22 22 22 22 33

Bars 3 No. 7 2 No. 9 plus 1 No. 8 3 No. 7 3 No. 7 3 No. 7

f c'

fy

(psi) 4,000 4,000 4,000 6,000 4,000

(psi) 60,000 60,000 80,000 60,000 60,000

Beam No.1 Compute the depth of the equivalent rectangular stress block,  , assuming that tension steel is yielding.

, 1  0.85 . Therefore,

For (

)

(

)

(

⁄

⁄

)

Thus,  > 0.002 and the steel is yielding ( f s  f y ). s Since,  t  0.005 the section is tension-controlled and  =0.9.

(

(

)

)

-

For Beam 1, Beam No.2

Compute the depth of the equivalent rectangular stress block,  , assuming that tension steel is yielding. (

For

)

, 1  0.85 . Therefore,

⁄

⁄

© 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write 4-3 to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

(

)

(

)

(

)

Thus,  > 0.002 and the steel is yielding ( f s  f y ). s Since,  t  0.005 the section is clearly tension-controlled and  =0.9.

(

(

)

)

(

)

-


Compute the depth of the equivalent rectangular stress block,  , assuming that tension steel is yielding.

, 1  0.85 . Therefore,

For (

)

(

)

(

⁄

⁄

)

Thus,  > 0.002 and the steel is yielding ( f s  f y ). s Since,  t  0.005 the section is clearly tension-controlled and  =0.9.

(

(

)

)

-



For (

, )

(

)

. Therefore, (

⁄

⁄

)



(

(

)

)

-



, 1  0.85 . Therefore,

For (

)

(

)

(

⁄

⁄

)


( For Beam 5, (b)

(

)

)

Taking beam 1 as the reference point, discuss the effects of changing and d on . (Note that each beam has the same properties as beam 1 except for the italicized quantity.)

Beam No. 1 2 3 4 5

M

n

(kip-ft) 167 250 219 171 257


Effect of As (Beams 1 and 2)

An increase of 55% in As (from 1.80 to 2.79 in.2) caused an increase of 50% in  M n . Increasing the tension steel area causes a proportional increase in the strength of the section, with a loss of ductility. Note that in this case, the strength reduction factor was 0.9 for both sections. Effect of f y (Beams 1 and 3) An increase of 33% in f y caused an increase of 31% in  M n . Increasing the steel yield strength has essentially the same effect as increasing the tension steel area. Effect of f c' (Beams 1 and 4) An increase of 50% in f c' caused an increase of 2% in  M n . Changing the concrete strength has approximately no impact on moment strength, relative to changes in the tension steel area and steel yield strength. Effect of d (Beams 1 and 5) An increase of 50% in d caused an increase of 54% in  M n . Increasing the effective flexural depth of the section increases the section moment strength (without decreasing the section ductility).

(c)

What is the most effective way of increasing effective way?

? What is the least

Disregarding any other effects of increasing d , As or f y such as changes in cost, etc., the most effective way to increase  M n is to increase the effective flexural depth of the section, d , followed by increasing f y and As . Note that increasing f y and As too much may make the beam over-reinforced and thus will result in a decrease in ductility. The least effective way of increasing  M n is to increase f c' . Note that increasing f c' will cause a significant increase in curvature at failure.


4-4

A 12-ft-long cantilever supports its own dead load plus an additional uniform service (unfactored) dead load of 0.5 kip/ft. The beam is made from normal-weight 4000-psi concrete and has in., in., and in. It is reinforced with four No. 7 Grade-60 bars. Compute the maximum service (unfactored) concentrated live load that can be applied at 1ft from the free end of the cantilever. Use load and strength –reduction factors from ACI Code Sections 9.2 and 9.3. Also check .

1. Compute the nominal moment capacity of the beam, M n and the strength reduction factor,  . Tension steel area: As = 4 No. 7 bars = 4  0.60 in.2 =2.40 in.2 Compute the depth of the equivalent rectangular stress block,  , assuming that tension steel is yielding. From equilibrium (using Eq. (4-14)): A f 2.4  60000 s y   c   2.65 in. 1 0.85 f ' b 0.85  4000  16 c For fc'  4000 psi, 1  0.85 . Therefore, c    2.65  3.1 in.  0.85 1

Check whether tension steel is yielding:  d c  15.5  3.1     0.003  0.012 Using Eq.(4-18)      s t  c  cu  3.1  Thus,  > 0.002 and the steel is yielding ( f s  f y ). s

 

Compute the nominal moment strength, using Eq. (4-21): 2.65   2.4  60000  15.5    2     M  A f d     170 kip-ft n s y 2 12000 Since,  t  0.012  0.005 the section is clearly tension-controlled and,  M n  0.9 170 kip-ft = 153 kip-ft 2. Compute Live Load Set M u   M n  153 kip-ft 16  18  0.15  0.3 kips/ft 144 Factored dead load = 1.2  0.3  0.5  0.96 kips/ft

Weight/ft of beam =

Factored dead load moment = wl 2 2  0.96 122 2  69.1 kip-ft Therefore the maximum factored live load moment is: 153 kip-ft – 69.1 kip-ft = 83.9 kip-ft Maximum factored load at 1 ft from the tip = 83.9 kip-ft / 11 ft = 7.63 kips Maximum concentrated service live load = 7.63 kips / 1.6 = 4.77 kips © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written 4-7 permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

3. Check of As ,min The section is subjected to positive bending and tension is at the bottom of this section, so we should use bw in Eq. (4-11). Also, 3 f c' is equal to 189 psi, so use 200 psi in the numerator: As ,min 

200 200 2 bw d   16  15.5  0.82 in. < As (o.k.) fy 60,000


4-5

Compute psi and

and check psi.

for the beam shown in Fig. P4-5. Use

1. Compute the nominal moment capacity of the beam, M n and the strength reduction factor,  . Tension steel area: As = 6 No. 8 bars = 6  0.79 in.2 =4.74 in.2 The tension reinforcement for this section is provided in two layers, where the distance from the tension edge to the centroid of the total tension reinforcement is given as d  19 in. Assuming that the depth of the Whitney stress block is less than or equal to the thickness of the compression flange,   h f and that the tension steel is yielding,  s   y , using Eq. (4-16):











A f 4.74  60000 s y   1.55 in.  h f  6 in. (o.k.) ' 0.85 f b e 0.85  4500  48 c

For fc'  4500 psi, 1  0.825 . Therefore, c    1.55  1.88 in.  0.825 1

Comparing the calculated depth to the neutral axis, c , to the values for d and d t , it is clear that the tension steel strain,  s , easily exceeds the yield strain (0.00207) and the strain at the level of

the extreme layer of tension reinforcement,  t , exceeds the limit for tension-controlled sections (0.005). Thus,  =0.9 and we can use Eq. (4-21) to calculate M n :

1.55   4.74  60000  19    2    M  A f d     432 kip-ft n s y 2 12000  M n  0.9  432 kip-ft = 389 kip-ft 2. Check of As ,min The section is subjected to positive bending and tension is at the bottom of this section, so we should use bw in Eq. (4-11). Also, 3 f c' is equal to 201 psi, so use 3 f c' in the numerator: As ,min 

3 f c' fy

bw d 

201  12  19  0.76 in.2 < As (o.k.) 60,000


4-6

Compute psi and

and check psi.


1. Compute the nominal moment capacity of the beam, M n and the strength reduction factor,  . Tension steel area: As = 6 No. 8 bars = 6  0.79 in.2 =4.74 in.2 The tension reinforcement for this section is provided in two layers, where the distance from the tension edge to the centroid of the total tension reinforcement is given as d  18.5 in. Assuming that the depth of the Whitney stress block is less than or equal to the thickness of the compression flange,   h f and that the tension steel is yielding,  s   y , using Eq. (4-16):











A f 4.74  60000 s y   4.18 in.  h f  5 in. (o.k.) 0.85 f ' b e 0.85  4000  20 c


Check whether tension steel is yielding:  d c  18.5  4.95     0.003  0.0082 Using Eq.(4-18)    s  c  cu  4.95  Thus,  > 0.002 and it is clear that the steel is yielding in both layers of reinforcement. s It is also clear that the section is tension-controlled (  =0.9), but just for illustration the value of

 can be calculated as: t  dt  c   19.5  4.92      0.003  0.0089  cu 4.92    c 

  t

We can use Eq. (4-21) to calculate M n :

4.18   4.74  60000  18.5    2    M  A f d     389 kip-ft n s y 2 12000  M n  0.9  389 kip-ft = 350 kip-ft 2. Check of As ,min The section is subjected to positive bending and tension is at the bottom of this section, so we should use bw in Eq. (4-11). Also, 3 f c' is equal to 190 psi, so use 200 psi in the numerator: As ,min 

200 200 2 bw d   12  18.5  0.74 in. < As (o.k.) fy 60,000


4-7

Compute the negative-moment capacity, shown in Fig. P4-7. Use psi and

, and check psi.

for the beam

1. Calculation of  M n This section is subjected to negative bending, so tension will develop in the top flange and the compression zone is at the bottom of the section. ACI Code Section 10.6.6 requires that a portion of the tension reinforcement be distributed in the flange, so assuming that the No. 6 bars in the 2 flange are part of the tension reinforcement: As  6  0.44  2.64 in. The depth of the Whitney stress block can be calculated using Eq. (4-16), using b  12 in., since the compression zone is at the bottom of the section:

, 1  0.85 . Therefore,

For (

)

(

⁄

⁄

)





The steel is yielding  s   y  0.00207 and it is tension-controlled  t  0.005 so  = 0.9. (

)

(

)

2. Check of As ,min The flanged portion of the beam section is in tension because the beam is subjected to negative bending. Therefore, the value of As ,min will depend on whether the beam is statically determinate. Assuming that the beam is part of a continuous, statically indeterminate floor system, the minimum tension reinforcement should be calculated using bw in Eq. (4-11). Also, 3 f c' is equal to 189 psi, so use 200 psi in the numerator: (

)

However, for a statically determinate beam, bw should be replaced by the smaller of

2bw   24 in. or be . Given that be is 48 in. for this beam section, (

)


4-8

For the beam shown in Fig. P4-8, (a)

psi and

psi.

Compute the effective flange width at midspan.

The limits given in ACI Code Section 8.12 for determining the effective compression flange, be , for a flanged section that is part of a continuous floor system are:

    4   be   bw  2(8h f )  b  2(clear trans. distance)/2  w    Assuming that the columns are 18 in. 18 in. , the longitudinal span is approximated as:  18 in.   ft  22.5 ft  21 ft    12 in.  ft    12 in.    8.5 ft The clear transverse distance for the 9 ft.-6 in. span is: 9.5 ft    12 in.  ft   1  12 in. 18 in.  and for the 11 ft. span is: 11 ft     9.75 ft 2  12 in. 12 in.  ft ft   So, the average clear transverse distance is 9.125 ft

The effective compression flange can now be computed as:   22.5 ft 12 in./ft   67.5 in.   4     be   12 in.  2  8  6 in.  108 in.    12 in.  2  9.125 ft 12 in./ft   /2=122 in.     The first limit governs for this section, so be  67.5 in.

(b)

Compute for the positive- and negative-moment regions and check for both sections. At the supports, the bottom bars are in one layer; at midspan, the No. 8 bars are in the bottom, the No. 7 bars in a second layer.

Positive moment region 1. Calculation of  M n Tension steel area: As = 3 No. 8 bars + 2 No. 7 bars = 3  0.79  2  0.60  3.57 in.2 © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write 4-12 to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

The tension reinforcement for this section is provided in two layers. Assuming the section will include a No. 3 or No. 4 stirrup, it is reasonable to assume that the distance from the extreme tension edge of the section to the centroid of the lowest layer of steel is approximately 2.5 in. Thus the distance from the top of the section to the extreme layer of tension reinforcement, d t , can be calculated to be: dt  21 in. – 2.5 in. =18.5 in.

The minimum spacing required between layers of reinforcement is 1 in. (ACI Code Section 7.6.2). Thus the spacing between the centers of the layers is approximately 2 in. So the distance from the tension edge to the centroid of the total tension reinforcement is:

 3  0.79   2.5   2  0.60   4.5 3.57

 3.17 in.

Therefore, the effective flexural depth, d , is:

d  21 in. – 3.17 in. =17.8 in. Assuming that the depth of the Whitney stress block is less than or equal to the thickness of the compression flange,   h f  6 in. and that the tension steel is yielding,  s   y ; using Eq.(4-









16) we have: A f 3.57  60000 s y    1.07 in.  h f  6 in. (o.k.) ' 0.85  3500  67.5 0.85 f b e c


Comparing the calculated depth to the neutral axis, c , to the values for d and d t , it is clear that the tension steel strain,  s , easily exceeds the yield strain (0.00207) and the strain at the level of

the extreme layer of tension reinforcement,  t , exceeds the limit for tension-controlled sections (0.005). Thus,  =0.9 and we can use Eq. (4-21) to calculate M n :

1.07   3.57  60000  17.8   2    M  A f d     308 kip-ft n s y 2 12000  M n  0.9  308 kip-ft = 277 kip-ft



Check of As ,min : The section is subjected to positive bending and tension is at the bottom of this section, so we should use bw in Eq. (4-11). 3 f c' is equal to 177 psi, so use 200 psi in the numerator.


As ,min 

200 200 2 bw d   12  17.8  0.71 in. < As (o.k.) fy 60,000

Negative moment region The tension and compression reinforcement for this section is provided in single layers. Assuming the section will include a No. 3 or No. 4 stirrup, it is reasonable to assume that the distance from the extreme tension or compression edge of the section to the centroid of the tension or compression layer of steel is approximately 2.5 in. 2 As = 7 No. 7 bars = 7  0.60  4.2 in. , d  18.5 in.

As'  = 2 No. 8 bars = 2  0.79  1.58 in.2 , d '  2.5 in. Because this is a doubly reinforced section, we will initially assume the tension steel is yielding and use the trial and error procedure described in Section 4-7 to find the neutral axis depth, c. Try c  d 4  4.5 in.  c  d'   4.5  2.5    cu     0.003  0.00133  4.5   c  f s'  E s  s'  29,000 ksi  0.00133  38.6 ksi  f y

 s'  

C 's 

As'





f s'

 0.85 f c'



  1.58 in. 38.6 ksi  2.98 ksi   56.3 kips 2

Cc  0.85 fc'b1c  0.85  3.5 ksi 12 in.  0.85  4.5 in.=137 kips T  As f y  4.20 in.2  60 ksi  252 kips Because T  Cc  C 's , we should increase c for the second trial. Try c  5.9 in.

 s'  0.00173



f s'  50.2 ksi  f y



C 's  74.6 kips Cc  179 kips T  254 kips  Cc  Cs  254 kips

With section equilibrium established, we must confirm the assumption that the tension steel is yielding.  d c  18.5  5.9     0.003  0.0064 using Eq.(4-18)     s  c  cu  5.9 





Thus, the steel is yielding   0.00207 and it is a tension-controlled section  t   s  0.0102 . s So, using   1c  0.85  5.9 in.  5.0 in. , use Eq. (4-21) to calculate M n .

  M  Cc  d    C 's d  d '  179 kips  16 in.  74.6 kips  16 in. n 2 





M  2865 k-in.  1195 k-in  4060 k-in  338 k-ft n © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write 4-14 to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

 M n  0.9  338 kip-ft = 304 kip-ft Check of As ,min : The flanged portion of the beam section is in tension and the value of As ,min will depend on the use of that beam. Since the beam is part of a continuous, statically indeterminate floor system, the minimum tension reinforcement should be calculated using bw in Eq. (4-11). Also, 3 f c' is equal to 177 psi, so use 200 psi in the numerator. As ,min 

200 200 2 bw d   12  18.5  0.74 in. < As (o.k.) fy 60,000


4-9

Compute psi and (a)

and check psi, and


the reinforcement is six No. 8 bars.

1. Compute the nominal moment capacity of the beam, M n and the strength reduction factor,  . Tension steel area: As = 6 No. 8 bars = 6  0.79 in.2 = 4.74 in.2 Assuming that the depth of the Whitney stress block is less than or equal to the thickness of the top flange,   5 in. and that the tension steel is yielding,  s   y , using Eq. (4-16) with





b  30 in. : ( , 1  0.85 . Therefore,

For (

)

(

)

(

⁄

)

⁄

)

Thus,  > 0.002 and the steel is yielding ( f s  f y ). s Since,  t  0.005 the section is tension-controlled and  =0.9. We can use Eq. (4-21) to calculate M n : (

)

(

)

2. Check of As ,min The flanged portion of the beam section is in tension and the value of As ,min will depend on the use of that beam. Assuming that the beam is part of a continuous, statically indeterminate floor system, the minimum tension reinforcement should be calculated using bw  2  5  10 in. in Eq. (4-11). Also, 3 f c' is equal to 189 psi, so use 200 psi in the numerator: As ,min 

200 200 2 bw d   10  32.5  1.08 in. < As (o.k.) fy 60,000

However, for a statically determinate beam, bw should be replaced by the smaller of

2bw   20 in. or be . Given that be is 30 in. for this beam section, As ,min 

200 200 2 bw d   20  32.5  2.17 in. < As (o.k.) fy 60,000


(b)

the reinforcement is nine No. 8 bars.

1. Compute the nominal moment capacity of the beam, M n and the strength reduction factor,  . Tension steel area: As = 9 No. 8 bars = 9  0.79 in.2 =7.11 in.2 Assuming that the depth of the Whitney stress block is less than or equal to the thickness of the compression flange,   h f  5 in. and that the tension steel is yielding,  s   y , using Eq. (4-









16) with b  30 in. : ( , 1  0.85 . Therefore,

For (

)

(

)

(

⁄

)

⁄

)

Thus,  > 0.002 and the steel is yielding ( f s  f y ). s Since,  t  0.005 the section is tension-controlled and  =0.9. We can use Eq. (4-21) to calculate M n :

(

)

(

)

2. Check of As ,min

As ,min is the same as in part (a).


4-10

Compute psi and

and check psi.


1. Compute the nominal moment capacity of the beam, M n and the strength reduction factor,  . Tension steel area: As = 8 No. 7 bars = 8  0.60 in.2 =4.8 in.2 Tension will develop in the bottom flange and the compression zone is at the top of the section. Thus, assuming that the tension steel is yielding,  s   y , in Eq. (4-16) we should use





b  2  6  12 in. and we find the depth of the Whitney stress block as:



A f 4.8  60000 s y   5.65 in. ' 0.85 f b 0.85  5000  12 c

For fc'  5000 psi, 1  0.80 . Therefore, c  

1

 5.65

0.80

 7.06 in.

Check whether tension steel is yielding:  d c  23.5  7.06     0.003  0.007 using Eq.(4-18)     t     s 7.06   c  cu  Thus,  > 0.002 and the steel is yielding ( f s  f y ). s Since,  t  0.005 the section is tension-controlled and  =0.9. We can use Eq. (4-21) to calculate M n :

5.65   4.8  60000   23.5    2    M  A f d     496 kip-ft n s y 2 12000  M n  0.9  496 kip-ft = 446 kip-ft 2. Check of As ,min The flanged portion of the beam section is in tension and the value of As ,min will depend on the use of that beam. Assuming that the beam is part of a continuous, statically indeterminate floor system, the minimum tension reinforcement should be calculated using bw  2  6  12 in. in Eq. (4-11). Also, note that 3 f c' is equal to 212 psi: As ,min 

212 212 2 bw d   12  23.5  1.00 in. < As (o.k.) fy 60,000

However, for a statically determined beam, bw should be replaced by the smaller of

2bw   24 in. or be . Given that be is 42 in. for this beam section, As ,min 

212 212 bw d   24  23.5  1.99 in.2 < As (o.k.) fy 60,000


4-11

(a)

Compute

for the three beams shown in Fig. P4-11. In each case, psi and ksi,

Beam No. 1 Tension steel area: As = 6 No. 9 bars = 6 1.00 in.2 =6.00 in.2 The tension reinforcement for this section is provided in two layers. Assuming the section will include a No. 3 or No. 4 stirrup, it is reasonable to assume that the distance from the extreme tension edge of the section to the centroid of the lowest layer of steel is approximately 2.5 in. Thus the distance from the top of the section to the extreme layer of tension reinforcement, d t , can be calculated to be:

dt  36 in. – 2.5 in. =33.5 in. The effective flexural depth, d , is given as : d  32.5 in.





Assuming that the tension steel is yielding,  s   y , using Eq. (4-16):

For

, (

)

⁄

. Therefore, (

⁄

)

Thus,  > 0.002 and the steel is yielding ( f s  f y ). s Also,  t  0.005 , the section is tension-controlled and  =0.9. We can use Eq. (4-21) to calculate M n :

(

)

(

)

Beam No. 2 Tension steel area: As = 6 No. 9 bars = 6 1.00 in.2 =6.00 in.2 Compression steel area: A s' = 2 No. 9 bars = 2 1.00 in.2 =2.00 in.2 As was discussed for beam No. 1, d  32 in., dt  33.5 in. and d ' is given as d '  2.5 in. Because this is a doubly reinforced section, we will initially assume the tension steel is yielding and use the trial and error procedure described in Section 4-7 to find the neutral axis depth, c. © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write 4-19 to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Try c  d 4  8 in. (

(

)

) (

(

)

(

) )

Because T  Cc  C 's , we should decrease c for the second trial. Try (

)

With section equilibrium established, we must confirm the assumption that the tension steel is yielding. (

)

(

)





Clearly, the steel is yielding   0.00207 and it is a tension-controlled section. s , use Eq. (4-21) to calculate M n .

So, using *

(

)

(

)+

[

(

)

(

)]

Beam No. 3 Tension steel area: As = 6 No. 9 bars = 6 1.00 in.2 =6.00 in.2 Compression steel area: A s' = 4 No. 9 bars = 4 1.00 in.2 =4.00 in.2 As was discussed for beam No. 1, d  32.5 in., and dt  33.5 in. The compression reinforcement for this beam section is provided in two layers and d ' is given as 3.5 in. Because this is a doubly reinforced section, we will the same procedure as for beam No. 2 (assuming that the tension steel is yielding). The depth of the neutral axis for this section should be smaller compared with beam section No. 2, since the compression reinforcement is increased for this section. Try

(Note that both layers of the compression steel will be in the compression zone)

(

)


Because T  Cc  C 's , we should decrease c for the second trial. Try (

)


)

(

)





Clearly, the steel is yielding   0.00207 and it is a tension-controlled section. s , use Eq. (4-21) to calculate M n .

So, using *

(b)

(

)

(

)+

[

(

)

(

)]

From the results of part (a), comment on whether adding compression reinforcement is a cost-effective way of increasing the strength, , of a beam.

Comparing the values of  M n for the three beams, it is clear that for a given amount of tension reinforcement, the addition of compression steel has little effect on the nominal moment capacity, as long as the tension steel yields in the beam without compression reinforcement. As a result, adding compression reinforcement in not a cost effective way of increasing the nominal moment capacity of a beam. However, adding compression reinforcement improves the ductility and might be necessary when large amounts of tension reinforcement are used to change the behavior from compression controlled to tension controlled.


4-12

Compute

for the beam shown in Fig. P4-12. Use psi and psi. Does the compression steel yield in this beam at nominal strength?

As = 6 No. 8 bars = 6  0.79 in.2 = 4.74 in.2 , d  25 in.  2.5 in.  22.5 in. As' = 2 No. 7 bars = 2  0.60 in.2 =1.2 in.2 , d '  2.5 in. Because this is a doubly reinforced section, we will initially assume the tension steel is yielding and use the trial and error procedure described in Section 4-7 to find the neutral axis depth, c. ⁄ Try For psi, . Thus, Since the depth of the Whitney stress block is less than 5.0 in. ,   5.0 in. , the width of the compression zone is constant and equal to 10 in., i.e. b  10 in. (

(

)

) (

(

)

(

) )

, we should increase c for the second trial.

Because Try (

(

)

) (

(

)

(

) )

Since , the width of the compression zone is not constant. Using a similar reasoning as in the case of flanged sections, where the depth of the Whitney stress block is in the web of the section, the compression force can be calculated from the following equations (refer to Fig. S4-12): (

)(

)


)

(

)





Thus, the tension steel is yielding   0.00207 and it is a tension-controlled section. s © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write 4-22 to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

Summing the moments about the level of the tension reinforcement: (

[ [

) (

( )

) (

(

)] )

(

)]

The strain in the compression steel at nominal moment capacity is 0.00185, the compression steel has not yielded at nominal strength.


bw 0.85f'c ht

a f's

d

h

f fs=fy b

(assumed)

a) total beam section and stress distribution

bw ht

a/2

a

d'

Cs Ccw

d

h

T1 F b b) Part 1: web of section and corresponding internal forces

bw ht

(a+ht)/2

a

Ccf d

h

T2 F b c) Part 2: overhanging flanges and corresponding internal forces

Fig. S4-12.1 Beam section and internal forces for the case of   ht . © 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write 4-24 to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.


Chapter 5 5-1

Give three reasons for the minimum cover requirements in the ACI code.

- To ensure enough concrete is present to develop the reinforcement. - To protect reinforcement from corrosive agents. - To insulate reinforcement in case of fire. Under what circumstances are larger covers used? - Highly corrosive environments. - Situations where abrasion to the concrete surface may result in a reduction in cover provided. Note: See Section 5.3 “Concrete Cover and Bar Spacing” for further discussion.

5-2

Give three reasons for using compression reinforcement in beams.

- To reduce long-term deflections (i.e. creep). - It tends to lead to a more ductile failure mode. - Some compression reinforcement is always required for fabrication of rebar cages.

5-1


5-3

Design a rectangular beam section, i.e. select b, d, h, and the required tension reinforcement, at midspan for a 22 ft-span simply supported rectangular beam that supports its own dead load, a superimposed service dead load of 1.25 kip/ft, and a uniform service load of 2 kip/ft. Use the procedure in Section 5.3 for the design of beam sections where the dimensions are unknown. Use and fy = 60 ksi.

Step 1: Estimate the dead load due to self-weight of the beam: Method 1:

DL  0.10 to 0.15  (SDL  LL) DL  0.10 to 0.15  (1.25 k/ft  2 k/ft ) DL  325 lb/ft to 490 lb/ft

Method 2:

/18  h  /12, so estimate h  22 in b  0.8h  18 in DL 

bh lb 18 in  22 in lb lb  150 3   150 3  413 2 2 2 2 144 in /ft ft 144 in /ft ft ft

Therefore, DL = 410 lb/ft seems like a good first estimate of the weight of the beam.

Step 2: Compute the total factored load and factored design moment, Mu: From ACI-08 Chapter 9: wu  1.2 wD  1.6 wL , or 1.4 wD wu  1.2  (0.41 k/ft  1.25 k/ft)  1.6  (2 k/ft), or 1.4  (0.41 k/ft  1.25 k/ft) wu  5.20 k/ft, or 2.35 k/ft

For this simply supported span, w 2 5.20 k/ft  (22 ft)2 Mu  u   315 k-ft  3775 k-in 8 8 Step 3: Select ρ and the corresponding R-factor: Assume the desirable strain diagram shown in Fig. 5-27b, which leads to   0.9 . From Eq. (5-19):  f ' 0.825  4,500 psi Note: β1 = 0.825 for f’c = 4,500 psi  1 c   0.0154 4 fy 4  60,000 psi From Eq. (5-21): f 0.0154  60,000 psi  y   0.205 f 'c 4,500 psi From Eq. (5-22): R   f 'c (1  0.59)  0.205  4.5 ksi  (1  0.59  0.205)  0.811 ksi (Note that this R-factor is reasonable based on values given in Table A-3)

5-2


Step 4: Select section dimensions, b and h: From Eq. (5-23): M 3775 k-in bd 2  u   5170 in 3  R 0.9  0.811 ksi Since no column dimensions are given which control the width of the beam, the designer can assume any reasonable α value. Here we assume α = 0.7 1

1

 M 3  3775 k-in 3 d  u     19.5 in  0.7  0.9  0.811 ksi    R  h  d  2.5 in  22 in b   d  0.7 19.5 in  13.65 in  14 in

Note that both h and b are rounded up to the nearest even inch value for constructability.

Step 5: Determine As and select the reinforcing bars: First, go back and recalculate the weight of the beam with the final selected dimensions: bh lb 14 in  22 in lb lb DL   150 3   150 3  321 2 2 2 2 144 in /ft ft 144 in /ft ft ft So, wu  5.09 k/ft and M u  308 k-ft  3695 k-in Calculate the required area of steel, assuming jd  0.9d : From Eq. (5-16): Mu 3695 k-in As    3.90 in 2 a  0.9  60 ksi   0.9  19.5 in    fy  d   2  With this estimate, iterate once to have a better estimate of the lever arm jd . From Eq. (5-17): As f y 3.90 in 2  60 ksi a    4.37 in 0.85 f 'c b 0.85  4.5 ksi 14 in From Eq. (5-16): Mu 3695 k-in As    3.95 in 2 a 4.37 in      f y  d   0.9  60 ksi  19.5 in  2 2     No further iterations are necessary, since the estimated lever arm was very reasonable. Select 4#9 bars as bottom reinforcement at the critical section of the beam. As  4 Ab  4 1.0 in 2  4.0 in 2  3.95 in 2 OK Step 6: Required Checks: 1) Use minimum bar spacing and cover values, or Table A-5, to verify that b = 14 in is sufficiently wide for 4 #9 bars to be placed in a single layer. OK

5-3


2) Use ACI-08 Eq. (10-3) to check that the reinforcement provided is more than the minimum required. For f’c = 4,500 psi, 3 f 'c  201  200 so use 3 f 'c As ,min 

3 f 'c fy

bw d 

3 4500  14 in  19.5 in  0.92 in 2  4.0 in 2 60,000 psi

OK

3) Calculate the strain in the extreme layer of tension steel to verify that assuming   0.9 is valid. From Eq. (5-17): As f y 4.0 in 2  60 ksi a    4.48 in 0.85 f 'c b 0.85  4.5 ksi 14 in We know that 1  0.825 (see above). a 4.48 in c   5.43 in 1 0.825 From Eq. (4-18): d -c 19.5 in - 5.43 in OK t   cu   0.003  0.0078  0.005 c 5.43 in Therefore the designer is permitted to use   0.9 for this beam design. 4) Finally, verify that the nominal flexural strength is sufficient for the applied loads. From Eq. (5-15): a 4.48 in     M n   As f y  d    0.9  4.0 in 2  60 ksi  19.5 in   3725 k-in 2 2     M n  310 k-ft  308 k-ft  M u Therefore, this design is sufficient without being too conservative. Note that other combinations of b, h, and As may also be correct if different assumptions were made by the designer. If all checks listed in Step 6 are satisfied, without being unreasonably conservative, the design may be considered adequate.

14in.

22in. 4 #9 bars

2.5in. Fig. S5-4 Cross-section of final design at mid-span

5-4


5-4

The rectangular beam shown in Fig. P5-4 carries its own dead load (you must guess values for b and h) plus an additional uniform service load of 0.5 kip/ft and a uniform service live load of 1.5 kip/ft. The dead load acts on the entire beam, of course, but the live load can act on parts of the span. Three possible loading cases are shown in Fig. P5-4. Use load and strength reduction factors from ACI Code sections 9.2 and 9.3. a)

Draw factored bending-moment diagrams for the three loading cases shown and superimpose them to draw a bending-moment envelope.

Begin by estimating the dead load due to self-weight of the beam: Method 1:

DL  0.10 to 0.15  (SDL  LL) DL  0.10 to 0.15  (0.5 k/ft  1.5 k/ft ) DL  200 lb/ft to 300 lb/ft

Method 2:

/18  h  /12, so estimate h  22 in b  0.8h  18 in DL 

bh lb 18 in  22 in lb lb  150 3   150 3  413 2 2 2 2 144 in /ft ft 144in /ft ft ft

We select DL = 350 lb/ft as a first estimate of the weight of the beam. So, using wu  1.2wD  1.6wL from ACI 318-08, Chapter 9, the bending-moment envelope is as follows:

Fig. S5-5a Bending moment envelope

5-5


b)

Design a rectangular beam section for the maximum positive bending moment between the supports, selecting b, d, h, and the reinforcing bars. Use the procedure in Section 5.3 for the design of beam sections where the dimensions are unknown. Use f’c = 5000 psi and fy = 60 ksi.

It is necessary to design this beam section for both negative and positive bending. The need for a practical design makes it reasonable to assume that the outer dimensions of the beam will be constant along the length, and that these dimensions will be controlled by the design of the section subjected to the largest absolute value of moment. As seen in Part (a), the largest expected moment is a positive moment of 226 kip-ft (2715 k-in). Therefore, it is reasonable to begin the beam design by designing the beam at this location. Step 1: Select ρ and the corresponding R-factor: Assume the strain diagram shown in Fig. 5-27b, which leads to   0.9 . From Eq. (5-19):  f ' 0.80  5,000 psi Note: β1 =0.80 for f’c = 5,000 psi  1 c   0.0167 4 fy 4  60,000 psi From Eq. (5-21): f 0.0167  60,000 psi  y   0.20 f 'c 5,000 psi From Eq. (5-22): R   f 'c (1  0.59)  0.20  5 ksi  (1  0.59  0.20)  0.882 ksi (Note that this R-factor is reasonable based on values given in Table A-3) Step 2: Select section dimensions, b and h: From Eq. (5-23): M 2715 k-in bd 2  u   3420 in 3  R 0.9  0.882 ksi Since no column dimensions are given which control the width of the beam, the designer can assume any reasonable α value. Here we assume α = 0.7 1

 M 3  2715 k-in  d  u     0.7  0.9  0.882 ksi    R  h  d  2.5 in  20 in b   d  0.7 17.5 in  12.5 in  14 in

1

3

 17.0 in  17.5 in

Note that both h and b are rounded up to the nearest even inch value for constructability. Step 3: Determine As and select the reinforcing bars: First, go back and recalculate the weight of the beam with the final selected dimensions: bh lb 14 in  20 in lb lb DL   150 3   150 3  292 2 2 2 2 144 in /ft ft 144 in /ft ft ft So, using pattern loading again, the maximum positive moment is: M u  222 k-ft  2670 k-in

5-6


Now calculate the required area of steel, assuming jd  0.9d : From Eq. (5-16): Mu 2670 k-in As    3.14 in 2 a  0.9  60 ksi   0.9  17.5 in    fy  d   2  With this estimate, iterate once to have a better estimate of the lever arm jd From Eq. (5-17): As f y 3.14 in 2  60 ksi a    3.16 in 0.85 f 'c b 0.85  5 ksi 14 in From Eq. (5-16): Mu 2670 k-in As    3.11 in 2 a 3.16 in     f y  d   0.9  60 ksi  17.5 in 2 2     No further iterations are necessary, since the estimated lever arm was very reasonable. Select 4#8 bars as bottom reinforcement at the critical section of the beam. OK As  4 Ab  4  0.79 in 2  3.16 in 2  3.11 in 2 Step 4: Required Checks: 1) Use minimum bar spacing and cover values, or Table A-5, to verify that b = 14 in is sufficiently wide for 4 #8 bars to be placed in a single layer. OK 2) Use ACI-08 Eq. (10-3) to check that the reinforcement provided is more than the minimum required. For f’c = 5,000 psi, 3 f 'c  212  200 so use 3 f 'c As ,min 

3 f 'c fy

bw d 

3 5000  14 in  17.5 in  0.87 in 2  3.16 in 2 OK 60,000 psi

3) Calculate the strain in the extreme layer of tension steel to verify that assuming   0.9 is valid. From Eq. (5-17): As f y 3.16 in 2  60 ksi a    3.19 in 0.85 f 'c b 0.85  5 ksi 14 in We know that 1  0.80 (see above). a 3.19 in c   3.99 in 1 0.80 From Eq. (4-18): d -c 17.5 in - 3.99 in t   cu   0.003  0.0102  0.005 c 3.99 in

OK

Therefore the designer is permitted to use   0.9 for this beam design.

5-7


4) Finally, verify that the nominal flexural strength is sufficient for the applied loads. From Eq. (5-15): a 3.19 in     M n   As f y  d    0.9  3.16 in 2  60 ksi  17.5 in   2715 k-in 2 2     M n  226 k-ft  222 k-ft  M u Therefore, this design for positive bending is sufficient without being too conservative. c)

Using the beam section from part (b), design flexural reinforcement for the maximum negative moment over the roller support.

Since the outer dimensions are selected, the design for negative bending follows the method for designing a rectangular section where the section dimensions are known. The maximum expected negative moment, considering pattern loading, is Determine As and select the reinforcing bars for negative bending: Assume : From Eq. (5-16): ( ) ( ) With this estimate, iterate once to have a better estimate of the lever arm jd From Eq. (5-17):

From Eq. (5-16): (

)

(

)

No further iterations are necessary since the solution has essentially converged. Select 2#9 bars as top reinforcement at the critical section of the beam.

Required Checks: 1) Use minimum bar spacing and cover values, or Table A-5, to verify that b = 14 in is sufficiently wide for 2 #9 bars to be placed in a single layer. OK 2) Use ACI-08 Eq. (10-3) to check that the reinforcement provided is more than the minimum required. For f’c = 5,000 psi, √ so use √ As ,min 

3 f 'c fy

bw d 

3 5000  14 in  17.5 in  0.87 in 2  2.0 in 2 60,000 psi

5-8

OK


3) Calculate the strain in the extreme layer of tension steel to verify that assuming   0.9 is valid. From Eq. (5-17): As f y 2.0 in 2  60 ksi a    2.02 in 0.85 f 'c b 0.85  5 ksi 14 in We know that 1  0.80 (see above). a 2.02 in c   2.53 in 1 0.80 From Eq. (4-18): d -c 17.5 in - 2.53 in t   cu   0.003  0.0178  0.005 c 2.53 in OK Therefore the designer is permitted to use   0.9 for this beam design. 4) Finally, verify that the nominal flexural strength is sufficient for the applied loads. From Eq. (5-15): a 2.02 in     M n   As f y  d    0.9  2.0 in 2  60 ksi  17.5 in   1780 k-in 2 2    Therefore, this design is sufficient without being too conservative. The cross-sections of the beam design at maximum positive and negative bending moments is shown below.

14in.

14in. 2.5in. 20in.

2 #9 bars

20in.

4 #8 bars

2.5in. Fig. S5-5b Cross-sections of final designs for positive and negative bending respectively

Note that other combinations of b, h, and As may also be correct if different assumptions were made by the designer. If all checks listed in Steps 4 and 6 are satisfied, without being unreasonably conservative, the design may be considered adequate.

5-9


5-5

Design three rectangular beam sections, i.e. select b and d and the tension steel area As, to resist a factored design moment, Mu = 260 kip-ft. For all three cases select a section with b = 0.5d and use f’c = 4000 psi and fy = 60 ksi. Start your design by assuming that εt = 0.0075 (as was done in Section 5.3)

a)

The equations presented in section 5.3 initially assumed that εt = 0.0075. So, while no changes need to be made, their derivation will briefly be shown to easy comparisons with the solutions to parts (b) and (c).  cu 0.003 c d   d  0.286d  cu   s 0.003  0.0075 Cc  0.85 f 'c 1bc  0.85 f 'c 1b  0.286d  0.241 f 'c bd Now, enforce equilibrium: Cc  T

0.241 f 'c bd  As f y Thus we have an expression for the initial ρ value when εt = 0.0075 is assumed, from Eq. 5-19: 0.241 f 'c 0.24  0.85  4,000 psi initial    0.0136 fy 60,000 psi From Eq. (5-21): f 0.0136  60,000 psi  y   0.204 f 'c 4,000 psi From Eq. (5-22): R   f 'c (1  0.59)  0.204  4 ksi  (1  0.59  0.204)  0.718 ksi (Note that this R-factor is reasonable based on values given in Table A-3) From Eq. (5-23): M 3120 k-in bd 2  u   4830 in 3 Note that with εt = 0.0075,   0.9  R 0.9  0.718 ksi We are told to assume α = 0.5 1

 M 3  3120 k-in  d  u     0.5  0.9  0.718 ksi    R  b   d  0.5  21.5 in  10.8 in  12 in

1

3

 21.3 in  21.5 in

Note that both d and b are rounded up so that h and b both result in even inch values for constructability. Now, determine As and select the reinforcing bars, assuming jd  0.9d : From Eq. (5-16): Mu 3120 k-in As    2.99 in 2 a  0.9  60 ksi   0.9  21.5 in    fy  d   2 

With this estimate, iterate once to have a better estimate of the lever arm jd .

5-10


From Eq. (5-17): As f y 2.99 in 2  60 ksi a    4.40 in 0.85 f 'c b 0.85  4 ksi 12 in From Eq. (5-16): Mu 3120 k-in As    3.00 in 2 a 4.4 in     f y  d   0.9  60 ksi   21.5 in 2 2    No further iterations are necessary, since the estimated lever arm was very reasonable. Select 3#9 bars as bottom reinforcement at the critical section of the beam. OK As  3 Ab  3 1.0 in 2  3.0 in 2  3.0 in 2 Now do the required checks: 1) Use minimum bar spacing and cover values, or Table A-5, to verify that b = 12 in is sufficiently wide for 3 #9 bars to be placed in a single layer. OK 2) Use ACI-08 Eq. (10-3) to check that the reinforcement provided is more than the minimum required. For f’c = 4,000 psi, 3 f 'c  190  200 so use 200. As ,min 

200 psi 200 psi bw d   12 in  21.5 in  0.86 in 2  3.0 in 2 OK fy 60,000 psi

3) Calculate the strain in the extreme layer of tension steel to verify that assuming   0.9 is valid. From Eq. (5-17): As f y 3.0 in 2  60 ksi a    4.41 in 0.85 f 'c b 0.85  4 ksi 12 in We know that 1  0.85 . a 4.41 in c   5.19 in 1 0.85 From Eq. (4-18): d -c 21.5 in - 5.19 in OK t   cu   0.003  0.0094  0.005 c 5.19 in Therefore the designer is permitted to use   0.9 for this beam design. 4) Finally, verify that the nominal flexural strength is sufficient for the applied loads. From Eq. (5-15): a 4.41 in     M n   As f y  d    0.9  3.0 in 2  60 ksi   21.5 in   3125 k-in 2 2     M n  260 k-ft  260 k-ft  M u Therefore, this design is sufficient without being too conservative.

b)

Start your design by assuming that εt = 0.005

5-11


We need to re-derive an expression for ρ, using the same approach used in part (a).  cu 0.003 c d  d  0.375d  cu   s 0.003  0.005 Cc  0.85 f 'c 1bc  0.85 f 'c 1b  0.375d  0.319  1 f 'c bd Now, enforce equilibrium: Cc  T

0.3191 f 'c bd  As f y Thus we have an expression for the initial ρ value when εt = 0.005 is assumed: 0.3191 f 'c 0.319  0.85  4,000 psi initial    0.0181 fy

60,000 psi

From Eq. (5-21): f 0.0181 60,000 psi  y   0.271 f 'c 4,000 psi From Eq. (5-22): R   f 'c (1  0.59)  0.271 4 ksi  (1  0.59  0.271)  0.911 ksi (Note that this R-factor is reasonable based on values given in Table A-3) From Eq. (5-23): M 3120 k-in bd 2  u   3805 in 3 Note that with εt = 0.005,   0.9  R 0.9  0.911 ksi We are told to assume α = 0.5 1

 M 3  3120 k-in  d  u     R 0.5  0.9  0.911 ksi     h  d  2.5 in  22 in b   d  0.5 19.5 in  10 in

1

3

 19.7 in  19.5 in

Note that both h and b are rounded to even inch values for constructability. Normally these values would be rounded up, but since the estimate for d is so much nearer to 19.5 in than 21.5 in, it is rounded down in this solution. Adequate strength of the section will still be achieved by selection of an appropriate As value. Now, determine As and select the reinforcing bars, assuming jd  0.9d : From Eq. (5-16): Mu 3120 k-in As    3.29 in 2 a  0.9  60 ksi   0.9  19.5 in    fy  d   2  With this estimate, iterate once to have a better estimate of the lever arm jd . From Eq. (5-17): As f y 3.29 in 2  60 ksi a    5.81 in 0.85 f 'c b 0.85  4 ksi 10 in From Eq. (5-16): Mu 3120 k-in As    3.48 in 2 a 5.81 in     f y  d   0.9  60 ksi  19.5 in 2 2     Iterate once more to have a better estimate of the lever arm jd .

5-12


From Eq. (5-17): As f y 3.48 in 2  60 ksi a    6.14 in 0.85 f 'c b 0.85  4 ksi 10 in From Eq. (5-16): Mu 3120 k-in As    3.52 in 2 a 6.14 in     f y  d   0.9  60 ksi  19.5 in 2 2    No further iterations are necessary, since the solution has converged. Select 3#10 bars as bottom reinforcement at the critical section of the beam. OK As  3 Ab  3 1.27 in 2  3.81 in 2  3.52 in 2

Now do the required checks: 1) Use minimum bar spacing and cover values, or Table A-5, to verify that b = 10 in is sufficiently wide for 3 #10 bars to be placed in a single layer. NOT OK So, either different bars must be selected, or the beam must be widened. Here we choose to widen the beam so that b = 12 in. 2) Use ACI-08 Eq. (10-3) to check that the reinforcement provided is more than the minimum required. For f’c = 4,000 psi, 3 f 'c  190  200 so use 200. As ,min 


3) Calculate the strain in the extreme layer of tension steel to verify that assuming   0.9 is valid. From Eq. (5-17): As f y 3.81 in 2  60 ksi a    5.60 in 0.85 f 'c b 0.85  4 ksi 12 in We know that 1  0.85 . a 5.60 in c   6.59 in 1 0.85 From Eq. (4-18): d -c 19.5 in - 6.59 in OK t   cu   0.003  0.0059  0.005 c 6.59 in Therefore the designer is permitted to use   0.9 for this beam design. 4) Finally, verify that the nominal flexural strength is sufficient for the applied loads. From Eq. (5-15): a 5.60 in     M n   As f y  d    0.9  3.81 in 2  60 ksi  19.5 in   3435 k-in 2 2     M n  286 k-ft  260 k-ft  M u

5-13


Therefore, this design is sufficient. Since  M n is 10% larger than M u , this design is perhaps too conservative. Options for optimizing this design might be selecting different bars, or slightly resizing the element. Notice that the design in part (a) was more efficient, so different initial assumptions can lead to different designs. Start your design by assuming that εt = 0.0035. You will probably need to add compression reinforcement to make this a tension-controlled section.

c)

We need to re-derive an expression for ρ, using the same approach used in part (a).

Now, enforce equilibrium: Cc  T

Thus we have an expression for the initial ρ value when εt = 0.0035 is assumed:

From Eq. (5-21):

From Eq. (5-22): ( ) ( ) (Note that this R-factor is reasonable based on values given in Table A-3) Before we can estimate the dimensions of the beam with Eq. (5-23), we need to determine  . Although we began by assuming that εt = 0.0035, tension controlled sections are so desirable that we will ensure that the section is tension controlled. Compression reinforcement might be required. Set . From Eq. (5-23):

We are told to assume α = 0.5. Also, since we had difficulty placing all the steel required in part (b) due to insufficient beam width, and part (c) requires a higher reinforcement ratio to limit the tensile strains, we will begin by assuming that two layers of reinforcement will be required. ⁄

⁄

(

)

(

)

Recall that we use 3.5 in. instead of 2.5 in. to account for the effect the second layer of steel has on the location of the centroid of the tension reinforcement.

5-14


Note that both h and b are rounded to even inch values for constructability. Now, Determine As and select the reinforcing bars, assuming jd  0.9d : From Eq. (5-16): (

(

)

)

With this estimate, iterate once to have a better estimate of the lever arm jd . From Eq. (5-17):

From Eq. (5-16): (

)

(

)

Iterate once more to have a better estimate of the lever arm jd . From Eq. (5-17):

From Eq. (5-16): (

)

(

)

No further iterations are necessary, since the solution has converged. Select 5#8 bars as bottom reinforcement at the critical section of the beam. OK Check whether the section is tensioned controlled:

c

a

1



6.97 in  8.2 in 0.85

Compression steel must be added to have a tension-controlled section. Try adding 2 #8 bars in the compression zone, which is approximately Try (

)

(

) (

(

)

(

) )

5-15

, with


Try (

)

Now do the required checks: 1) Use minimum bar spacing and cover values, or Table A-5, to verify that b = 10 in is sufficiently wide for 3 #8 bars to be placed in a single layer. OK 2) Use ACI-08 Eq. (10-3) to check that the reinforcement provided is more than the minimum required. For f’c = 4,000 psi, 3 f 'c  190  200 so use 200. As ,min 


3) Calculate the strain in the extreme layer of tension steel to verify that assuming From Eq. (4-18):

Ok,

is valid.

.

4) Finally, verify that the nominal flexural strength is sufficient for the applied loads. From Eq. (5-15): *

(

) -

(

)+

[

(

)

(

)]

-

Therefore, this design is sufficient. Again, since  M n is 10% larger than M u , this design is perhaps too conservative. Options for optimizing this design might be selecting different bars, or slightly resizing the element. Notice that the design in part (a) was more efficient, so different initial assumptions can lead to different designs.

5-16


d)

Compare and discuss your three section designs. 12in.

12in.

10in. 2 #8 bars

24in.

2.5in.

22in.

22in.

3.5in. 2.5in. 3 #10 bars 5 #8 bars Fig S5-6 Cross sections of section designs from parts (a), (b), and (c), respectively 3 #9 bars

Note that the area of reinforcement provided is higher in the third design compared to either of the first two, even though the section size is smaller. The design for part (c) also required compression steel to ensure a tension controlled section, whereas the sections in parts (a) and (b) were tension controlled as singly-reinforced sections.

5-17


5-6

You are to design a rectangular beam section to resist a negative bending moment of 275 kip-ft. Architectural requirements will limit your beam dimensions to a width of 12 in. and a total depth of 18 in. Using those maximum permissible dimensions, select reinforcement to provide the required moment strength following the ACI Code provisions for the strength reduction factor, . Use and .

Begin by trying to design the section as singly reinforced, with one layer of tension steel. Set

(

(

)

)

Either 5 #9 bars of 6 #8. To fit within the 12 in. beam width, 2 layers of bars are required. Try 5 #9 bars:

Check whether the tension steel has yielded, and whether the section is tension controlled.

Add compression steel so that

; try 3 #8 bars so that

By iteration, (

)

Check whether the section is tension controlled.

Calculate the nominal moment capacity: *

(

)

(

)+

[

(

5-18

)

(

)]


Now do the required checks: 1) Use minimum bar spacing and cover values, or Table A-5, to verify that b = 12 in. is sufficiently wide for 3 #9 bars to be placed in a single layer. OK 2) Use ACI-08 Eq. (10-3) to check that the reinforcement provided is more than the minimum required. For f’c = 5,000 psi, √ so use 212. √

Already verified that

is OK, and calculated the nominal moment capacity.

5-19


5-7

For column line 2, use the ACI Moment Coefficients given in ACI Code section 8.3.3 to determine the maximum positive and negative factored moments at the support faces for columns A-2 and B-2, and at the midspan of an exterior span and the interior span.

First, confirm that the ACI moment coefficients can be used. a) Two or more spans b) Longer span length within 20% of shorter c) Loads are uniformly distributed d) Unfactored LL load does not exceed 3 times DL e) Members are prismatic

OK OK OK OK OK

Now, estimate the dead load supported by column line 2: Dead load from beam, per foot of beam: 12 in  18 in lb lb  150 3  225 2 ft ft 144 in 2 ft Dead load from slab, per foot of beam:  12 ft  11 ft   6 in   2  lb lb   150 3  863 ft ft 12 in ft Superimposed dead load, per foot of beam: 12 ft  11 ft lb lb  20 2  230 2 ft ft Can the live load be reduced? From Eq. (5-3): At column A-2 and midspan of exterior beam:    15  15 2 2 Lr  L 0.25    0.050 k/ft  0.25    0.041 k/ft K A 2  11.5 ft  30 ft    LL T   At column B-2:    15  15 2 2 Lr  L 0.25    0.050 k/ft  0.25    0.034 k/ft K LL AT  2  11.5 ft  55 ft    At midspan of interior span:    15  15 2 2 Lr  L 0.25    0.050 k/ft  0.25    0.044 k/ft K LL AT  2  11.5 ft  25 ft    Live load, per foot of beam: 12 ft  11 ft lb lb At column A-2 and midspan of exterior beam:  41 2  472 2 ft ft 12 ft  11 ft lb lb At column B-2:  34 2  391 2 ft ft 12 ft  11 ft lb lb At midspan of interior span:  44 2  506 2 ft ft

5-20


So, using wu  1.2DL  1.6LL , the factored loads are: At column A-2 and midspan of exterior beam: wu  1.2   0.225 k/ft  0.863 k/ft  0.230 k/ft   1.6   0.472 k/ft   2.34 k/ft At column B-2:

wu  1.2   0.225 k/ft  0.863 k/ft  0.230 k/ft   1.6   0.391 k/ft   2.21 k/ft At midspan of interior span: wu  1.2   0.225 k/ft  0.863 k/ft  0.230 k/ft   1.6   0.506 k/ft   2.39 k/ft Calculate the clear span length, At column A-2: At midspan of exterior span:

n

: n

 30 ft -16 in  28.67 ft  30 ft -16 in  28.67 ft

At column B-2 (exterior):

n

 30 ft  25 ft

At column B-2 (interior):

n

At midspan of interior span:

n

n

-16 in  26.17 ft  2   30 ft  25 ft  -16 in  26.17 ft 2

 25 ft -16 in  23.67 ft

Design Moments using ACI Moment Coefficients from section 8.3.3: 2 2.34 k/ft   28.67 ft  w 2 At column A-2: Mu   u n    120 k-ft 16 16 2 w 2 2.34 k/ft   28.67 ft  At midspan of exterior span: Mu  u n   137 k-ft 14 14 2 2.21 k/ft   26.17 ft  w 2 At column B-2 (exterior): Mu   u n    151 k-ft 10 10 2 2.21 k/ft   26.17 ft  w 2 At column B-2 (interior): Mu   u n    138 k-ft 11 11 2 w 2 2.39 k/ft   23.67 ft  At midspan of interior span: Mu  u n   83.7 k-ft 16 16

5-21


5-8

Repeat problem 5-7, but use structural analysis software to determine the maximum positive and negative moments described in problem 5-7. The assumed beam, slab and column dimensions are given in the figure. Assume 12 ft story heights above and below this floor level. You must use appropriate live load patterns to maximize the various factored moments. Use a table to compare the answers from Problems 5-7 and 5-8.

From Problem 5-7, the factored loads are: At column A-2 and midspan of exterior beam: At column B-2: At midspan of interior span:

wu  2.34 k/ft wu  2.21 k/ft wu  2.39 k/ft

Properties of elements used for model: Column: Ag  16 in 16 in  256 in 2 I g  16 in  / 12  5,460 in 4 4

Beam (Use properties of the web as an approximation of the cracked properties): Ag  12 in  24 in  288 in 2 I g  12 in   24 in  / 12  13,800 in 4 3

We assume story heights of 12 ft above and below the continuous beam being modeled, and include columns, fixed at their ends, in our model. Then we apply the appropriate load combinations following Example 5-2, and the following design moments result: Location At column A-2: At midspan of exterior span: At column B-2 (exterior): At column B-2 (interior): At midspan of interior span:

ACI Design Moment (From Problem 5-7) -120 kip-ft 137 kip-ft -151 kip-ft -138 kip-ft 83.7 kip-ft

5-22

Design Moment (From software) -135 kip-ft 93.0 kip-ft -159 kip-ft -114 kip-ft 55.0 kip-ft


5-9

Repeat problems 5-7 and 5-8 for column line 1.


OK OK OK OK OK

Now, estimate the dead weight supported by column line 1: Dead weight from beam, per foot of beam: 12 in  18 in lb lb  150 3  225 2 ft ft 144 in 2 ft Dead weight from slab, per foot of beam: 6 in  6.5 ft lb lb  150 3  488 ft ft 12 in ft Superimposed dead load, per foot of beam: lb lb 6.5 ft  20 2  130 ft ft Can the live load be reduced? From Eq. (5-3): At column A-1 and midspan of exterior beam:    15  15 2 2 Lr  L 0.25    0.050 k/ft  0.25    0.052 k/ft K LL AT  2  6 ft  30 ft    since Lr ≥ LL, no reduction is possible, so use LL = 0.050 k/ft2 At column B-1:    15  15 2 2 Lr  L 0.25    0.050 k/ft  0.25    0.042 k/ft K LL AT  2  6 ft  55 ft    At midspan of interior span:    15  15 2 2 Lr  L 0.25    0.050 k/ft  0.25    0.056 k/ft K A 2  6 ft  25 ft    LL T   2 since Lr ≥ LL, no reduction is possible, so use LL = 0.050 k/ft Live load, per foot of beam: At column A-1 and midspan of exterior beam: At column B-1: At midspan of interior span:

lb lb  325 ft 2 ft lb lb 6.5 ft  42 2  273 ft ft lb lb 6.5 ft  50 2  325 ft ft 6.5 ft  50

5-23


So, using wu  1.2DL  1.6LL , the factored loads are: At column A-1 and midspan of exterior beam: wu  1.2   0.225 k/ft  0.488 k/ft  0.130 k/ft   1.6   0.325 k/ft   1.53 k/ft At column B-1:

wu  1.2   0.225 k/ft  0.488 k/ft  0.130 k/ft   1.6   0.273 k/ft   1.45 k/ft At midspan of interior span: wu  1.2   0.225 k/ft  0.488 k/ft  0.130 k/ft   1.6   0.325 k/ft   1.53 k/ft Calculate the clear span length, At column A-1: At midspan of exterior span:

n

: n

 30 ft -16 in  28.67 ft  30 ft -16 in  28.67 ft

At column B-1 (exterior):

n

 30 ft  25 ft

At column B-1 (interior):

n


n

n

-16 in  26.17 ft  2   30 ft  25 ft  -16 in  26.17 ft 2

 25 ft -16 in  23.67 ft

Design Moments using ACI Moment Coefficients from section 8.3.3: 2 1.53 k/ft   28.67 ft  w 2 At column A-1: Mu   u n    78.6 k-ft 16 16 2 w 2 1.53 k/ft   28.67 ft  At midspan of exterior span: Mu  u n   89.8 k-ft 14 14 2 1.45 k/ft   26.17 ft  w 2 At column B-1 (exterior): Mu   u n    99.3 k-ft 10 10 2 1.45 k/ft   26.17 ft  w 2 At column B-1 (interior): Mu   u n    90.3 k-ft 11 11 2 w 2 1.53 k/ft   23.67 ft  At midspan of interior span: Mu  u n   53.6 k-ft 16 16 Now, assemble a model using structural analysis software. Use the following properties: Column: Ag  16 in 16 in  256 in 2 I g  16 in  / 12  5,460 in 4 4

Beam (Use properties of the web as an approximation of the cracked properties): Ag  12 in  24 in  288 in 2

I cr  12 in   24 in  /12  13,800 in 4 3

5-24


We assume story heights of 12 ft above and below the continuous beam being modeled, and include columns, fixed at their ends, in our model. Then we apply the appropriate load combinations following Example 5-2, and the following design moments result:

Location At column A-1: At midspan of exterior span: At column B-1 (exterior): At column B-1 (interior): At midspan of interior span:

ACI Design Moment -78.6 kip-ft 89.8 kip-ft -99.3 kip-ft -90.3 kip-ft 53.6 kip-ft

5-25

Design Moment (From software) -88.8 kip-ft 60.9 kip-ft -103 kip-ft -74.7 kip-ft 35.5 kip-ft


5-10

Repeat problems 5-7 and 5-8 for the beam m-n-o-p in Fig. P5-7. Be sure to comment on the factored design moment at the face of the spandrel beam support at point m.


OK OK OK OK OK

Now, estimate the dead weight supported by beam m-n-o-p: Dead weight from beam, per foot of beam: 12 in  18 in lb lb  150 3  225 2 ft ft 144 in 2 ft Dead weight from slab, per foot of beam: 6 in  12 ft lb lb  150 3  900 in ft ft 12 ft Superimposed dead load, per foot of beam: lb lb 12 ft  20 2  240 ft ft Can the live load be reduced? From Eq. (5-3): At m and midspan of exterior beam:    15  15 2 2 Lr  L 0.25    0.050 k/ft  0.25    0.040 k/ft K LL AT  2  12 ft  30 ft    At n:    15  15 2 2 Lr  L 0.25    0.050 k/ft  0.25    0.033 k/ft K LL AT  2  12 ft  55 ft    At midspan of interior span:    15  15 2 2 Lr  L 0.25    0.050 k/ft  0.25    0.043 k/ft K LL AT  2  12 ft  25 ft    Live load, per foot of beam: At m and midspan of exterior beam: At n: At midspan of interior span:

lb lb  480 2 ft ft lb lb 12 ft  33 2  396 ft ft lb lb 12 ft  43 2  516 ft ft 12 ft  40

5-26


So, using wu  1.2DL  1.6LL , the factored loads are: At m and midspan of exterior beam: wu  1.2   0.225 k/ft  0.900 k/ft  0.240 k/ft   1.6   0.480 k/ft   2.41 k/ft At n:

wu  1.2   0.225 k/ft  0.900 k/ft  0.240 k/ft   1.6   0.396 k/ft   2.27 k/ft At midspan of interior span: wu  1.2   0.225 k/ft  0.900 k/ft  0.240 k/ft   1.6   0.516 k/ft   2.46 k/ft Calculate the clear span length, At m: At midspan of exterior span:

n

: n n

At n (exterior):

n

At n (interior):

n


n

 30 ft -12 in  29 ft  30 ft -12 in  29 ft

-12 in  26.5 ft  2   30 ft  25 ft  -12 in  26.5 ft 2  30 ft  25 ft

 25 ft -12 in  24 ft

Design Moments using ACI Moment Coefficients from section 8.3.3: 2 2.41 k/ft   29 ft  w 2 At m: Mu   u n    84.5 k-ft 24 24 2 w 2 2.41 k/ft   29 ft  At midspan of exterior span: Mu  u n   145 k-ft 14 14 2 2.27 k/ft   26.5 ft  w 2 At n (exterior): Mu   u n    159 k-ft 10 10 2 2.27 k/ft   26.5 ft  w 2 At n (interior): Mu   u n    145 k-ft 11 11 2 w 2 2.46 k/ft   24 ft  At midspan of interior span: Mu  u n   88.6 k-ft 16 16 Now, assemble a model using structural analysis software. Use the following properties: Beam (Use properties of the web as an approximation of the cracked properties): Ag  12 in  24 in  288 in 2 I g  12 in   24 in  / 12  13,800 in 4 3

Here we follow the recommendations from chapter 5 of the text, and assume that the beam is pinned at m and supported by rollers (which are free to rotate) at n, o, and p. This neglects the relatively small amount of moment transferred into the supporting beams due to their torsional rigidity.

5-27


Now apply the appropriate load combinations following Example 5-2, and the following design moments result: Location At m: At midspan of exterior span: At n (exterior): At n (interior): At midspan of interior span:

ACI Design Moment -84.5 kip-ft 145 kip-ft -159 kip-ft -145 kip-ft 88.6 kip-ft

Design Moment (From software) -14.6 kip-ft 184 kip-ft -195 kip-ft -200 kip-ft 23.5 kip-ft

The software model assumes that there is no torsional rigidity supplied by the supporting beams. Therefore, the moment that is predicted by the software at m is only due to the fact that the beam is offset from the centerline of the supporting spandrel beam. While neglecting the torsional rigidity of the spandrel beams is not realistic, it is also unlikely that the spandrel beam is torsionally rigid enough to result in a moment as high as the ACI Design Moments.

5-28


5-11

Repeat problems 5-7 and 5-8 for the one-way slab strip shown in Fig. P5-7. For this problem, find the factored design moments at all the points, a through i, indicated in Fig. P5-7.


OK OK OK OK OK

Now, estimate the dead load supported by slab strip a-i: Dead load from slab, per foot of slab: 6 in  12 in lb lb  150 3  75 2 ft ft 144 in 2 ft Superimposed dead load, per foot of slab: lb lb 1 ft  20 2  20 ft ft Live load, per foot of slab: lb lb 1 ft  50 2  50 ft ft This live load cannot be reduced due to the very small influence area of the slab strip. So, using wu  1.2DL  1.6LL , the factored loads are:

wu  1.2   0.075 k/ft  0.020 k/ft   1.6   0.050 k/ft   0.194 k/ft

Calculate the clear span lengths, n n

n

:

 12 ft -12 in  11 ft  11 ft -12 in  10 ft

Also, assemble a model using structural analysis software. Use the following cracked properties: Beam: Ag  12 in  6 in  72 in 2 I cr  0.5I g  0.5 12 in   6 in  / 12  108 in 4 3

Here, we follow the recommendations from chapter 5 of the text, and assume that the slab strip is pinned at a, and supported by rollers (which are free to rotate) at c, e, g, and i. Note that although i is a point of geometrical symmetry, it cannot be modeled as fixed, since the pattern loads are not necessarily symmetrical. Also note that pinning these supports neglects the relatively small amount of moment transferred into the supporting beams due to their torsional rigidity. Once the model is constructed, apply the appropriate load combinations following Example 5-2.

5-29


Design Moments using ACI Moment Coefficients from section 8.3.3, compared to design moments resulting from software model: Location a b c d e f g h i

ACI Moment Coefficient -1/24 1/14 -1/11 1/16 -1/11 1/16 -1/11 1/16 -1/11

ACI Design Moment -0.978 kip-ft 1.68 kip-ft -2.13 kip-ft 1.47 kip-ft -1.76 kip-ft 1.21 kip-ft -1.76 kip-ft 1.21 kip-ft -1.76 kip-ft

5-30

Design Moment (From Software) 0.423 kip-ft 2.14 kip-ft -2.86 kip-ft 1.29kip-ft -1.99 kip-ft 1.29 kip-ft -1.99 kip-ft 1.24 kip-ft -1.94 kip-ft


5-12

Use structural analysis software to find the maximum factored moments for the girder on column line C. Find the maximum factored positive moments at o and y, and the maximum factored negative moments at columns C-1, C-2, and C-3.

Estimate the loads supported by the beam along column line C: Distributed dead load from beam, per foot of beam: 12 in  24 in lb lb  150 3  300 2 ft ft 144 in 2 ft The remaining loads are transferred to the girder as a point load at mid-span. Loads transferring from the adjacent interior span will transfer directly to the girder, whereas loads from the adjacent exterior span must be amplified by 15% according to the ACI shear coefficient. Dead load from beams and slab, applied at the mid-point of the girder between C-1 and C-2:      12 in  18 in   25 ft  6 in   6 in 12 ft   25 ft  6 in    150 lb  13.5 k  144 in 2  2  2 ft 3 12 in  12 in 12 in   2 ft ft ft       ft    12 in  18 in  30 ft 6 in  6 in 12 ft  30 ft 6 in   lb 1.15        ×150 3  18.8 k  144 in 2  in in in  2   2  ft 12 12 12 ft  ft ft    ft 2   18.8 k  13.5 k  32.3 k

Dead load from beams and slab, applied at the mid-point of the girder at the interior spans:      12 in  18 in   25 ft  6 in   6 in 11 ft   25 ft  6 in    150 lb  12.6 k  144 in 2  2  2 ft 3 12 in  12 in 12 in   2 ft ft ft     ft     12 in  18 in  30 ft 6 in  6 in 11 ft  30 ft 6 in   lb  1.15        150 3  17.5 k  144 in 2  in in in  2   2  ft 12 12 12 ft  ft ft    ft 2  

12.6 k  17.5 k  30.1 k

Superimposed dead load, applied at the mid-point of the girder at the exterior spans:  lb  25 ft   12 ft   2    20 ft 2  3.0 k     lb  30 ft   1.15  12 ft      20 ft 3  4.14 k 2    3.00 k  4.14 k  7.14 k

Superimposed dead load, applied at the mid-point of the girder at the interior spans:  lb  25 ft   11 ft   2    20 ft 2  2.75 k     lb  30 ft   1.15  11 ft    20 3  3.80 k   ft  2  

5-31


2.75 k  3.80 k  6.55 k Live load, applied as a point load: For negative moment at C-1 and positive moment at o:    15  15 2 2 Lr  L  0.25    50 lb/ft  0.25    33 lb/ft A 24 ft  55 ft    I    25 ft    30 ft  LLo  33 lb/ft 2    12 ft  1.15    12 ft   11.8 k    2   2  

For negative moment at C-2:    15  15 2 2 Lr  L  0.25    50 lb/ft  0.25    27.4 lb/ft AI  46 ft  55 ft     25 ft    30 ft  LLo  27.4 lb/ft 2    12 ft  1.15    12 ft   9.78 k    2   2    25 ft    30 ft  LLy  27.4 lb/ft 2    11 ft  1.15    11 ft   8.97 k    2   2  

For positive moment at y:    15  15 2 2 Lr  L  0.25    50 lb/ft  0.25    34 lb/ft AI  22 ft  55 ft     25 ft    30 ft  LLy  34 lb/ft 2    11 ft  1.15    11 ft   11.1 k    2   2   For negative moment at C-3:    15  15 2 2 Lr  L  0.25    50 lb/ft  0.25    27.7 lb/ft A 44 ft  55 ft    I    25 ft    30 ft  LLy  27.7 lb/ft 2    11 ft  1.15    11 ft   9.06 k    2   2  

Calculate the clear span length, Between C-1 and C-2: Between C-2 and C-3:

n

: n n

 24 ft -16 in  22.67 ft  22 ft -16 in  20.67 ft

Now assemble a model using structural analysis software. Use the following cracked properties: Column: Ag  16 in 16 in  256 in 2 I g  16 in  / 12  5,460 in 4 4

Beam (Use properties of the web as an approximation of the cracked properties): Ag  12 in  24 in  288 in 2

I cr  12 in   24 in  /12  13,800 in 4 3

5-32


We assume story heights of 12 ft above and below the continuous beam being modeled, and include columns, fixed at their ends, in our model. Then we apply the appropriate load combinations following Example 5-2, and the following design moments result: Location

Design Moment (From Software) -121 kip-ft

Negative at C-1 Positive at o

236 kip-ft

Negative at C-2 (o-side)

-224 kip-ft

Negative at C-2 (y-side)

-198 kip-ft

Positive at y

175 kip-ft

Negative at C-3

-164 kip-ft

5-33


5-13

Assume the maximum factored positive moment near midspan of the floor beam between columns A-2 and B-2 is 60 kip-ft. Using the beam dimensions given in Fig. P5-7, determine the required area of tension reinforcement to satisfy all the ACI Code requirements for strength and minimum reinforcement area. Select bars and provide a sketch of your final section design.

The maximum expected positive moment is given as M u  60 k-ft  720 k-in Determine the effective flange width: By ACI code section 8.12:

b f  30 ft / 4  7.5 ft  90 in Check that:  2  8  6 in  12 in  108 in  b f  11 ft -12 in  2  12 in  132 in  2

OK

Determine As and select the reinforcing bars: Assume jd  0.95d : From Eq. (5-16): Mu 720 k-in As    0.65 in 2 a  0.9  60 ksi   0.95  21.5 in    fy  d   2  With this estimate, iterate once to have a better estimate of the lever arm jd From Eq. (5-17): As f y 0.65 in 2  60 ksi a    0.13 in 0.85 f 'c b 0.85  4 ksi  90 in From Eq. (5-16): Mu 720 k-in As    0.62 in 2 a 0.13 in     f y  d   0.9  60 ksi   21.5 in 2 2    This seems like a very small area of required steel, so check minimum steel requirement before selecting bars. From ACI-08 Eq. (10-3): For f’c = 4,000 psi, 3  f 'c  189  200 so use 200 As ,min 

3 f 'c fy

bw d 

200  12 in  21.5 in  0.86 in 2 60,000 psi

The minimum steel requirement will govern here. Select 3 #5 bars. As  3 Ab  3  0.31 in 2  0.93 in 2  0.86 in 2

5-34

OK


Required Checks: 1) Use minimum bar spacing and cover values, or Table A-5, to verify that b = 12 in is sufficiently wide for 3 #5 bars to be placed in a single layer. OK 2) Use ACI-08 Eq. (10-3) to check that the reinforcement provided is more than the minimum required. Since this requirement governed our bar selection, this is satisfied by default. OK 3) Calculate the strain in the extreme layer of tension steel to verify that assuming   0.9 is valid. From Eq. (5-17): As f y 0.93 in 2  60 ksi a    0.182 in 0.85 f 'c b 0.85  4 ksi  90 in We know that 1  0.85 for f’c = 4,000 psi a 0.182 in c   0.214 in 1 0.85 From Eq. (4-18): d -c 21.5 in - 0.214 in OK t    cu   0.003  0.297  0.005 c 0.214 in Therefore the designer is permitted to use   0.9 for this beam design. 4) Finally, verify that the nominal flexural strength is sufficient for the applied loads. From Eq. (5-15): a 0.182 in     M n   As f y  d    0.9  0.93 in 2  60 ksi   21.5 in    1075 k-in 2 2     M n  89.6 k-ft  60 k-ft  M u Therefore, this design is sufficient. It seems too conservative, but that is because the minimum area requirement governed bar selection.

90in.

24in.

2.5in.

3 #5 bars

Fig. S5-13 Cross-section of final design for positive bending region Note that other selections As may also be correct. If all checks are satisfied, without being unreasonably conservative, the design may be considered adequate.

5-35


5-14

Assume the maximum factored negative moment at the face of column B-2 for the floor beam along column line 2 is -120 kip-ft. Using the beam and slab dimensions given in Fig. P5-7, determine the required area of tension reinforcement to satisfy all the ACI Code requirements for strength and minimum reinforcement area. Select bars and provide a sketch of your final section design.

The maximum expected negative moment is given as M u  120 k-ft  1440 k-in Although this is a T-section, there is no need to determine the effective flange width since the compression zone is located in the web of the beam. Determine As and select the reinforcing bars: Assume jd  0.9d : From Eq. (5-16): Mu 1440 k-in As    1.38 in 2 a  0.9  60 ksi   0.9  21.5 in    fy  d   2  With this estimate, iterate once to have a better estimate of the lever arm jd From Eq. (5-17): As f y 1.38 in 2  60 ksi a    2.03 in 0.85 f 'c b 0.85  4 ksi 12 in From Eq. (5-16): Mu 1440 k-in As    1.30 in 2 a 2.03 in      f y  d   0.9  60 ksi   21.5 in 2 2    Select 3 #5 bars over the web, and 4 #3 bars in the flange. This will result in “part” of the reinforcement (about 1/3) being spread into the flange, and most (2/3) remaining over the web. OK As  3 Ab1  4 Ab 2  3  0.31 in 2  4  0.11 in 2  1.37 in 2  1.30 in 2 Required Checks: 1) Use minimum bar spacing and cover values, or Table A-5, to verify that b = 12 in is sufficiently wide for 3 #5 bars to be placed in a single layer. OK 2) Use ACI-08 Eq. (10-3) to check that the reinforcement provided is more than the minimum required. From ACI-08 Eq. (10-3): For f’c = 4,000 psi, 3 f 'c  190  200 so use 200 psi As ,min 

3 f 'c fy

bw d 

200 psi  12 in  21.5 in  0.86 in 2  1.37 in 2 OK 60,000 psi

5-36


3) Calculate the strain in the extreme layer of tension steel to verify that assuming   0.9 is valid. From Eq. (5-17): As f y 1.37 in 2  60 ksi a    2.01 in 0.85 f 'c b 0.85  4 ksi 12 in We know that 1  0.85 for f’c = 4,000 psi a 2.01 in c   2.36 in 1 0.85 From Eq. (4-18): d -c 21.5 in - 2.36 in OK t    cu   0.003  0.024  0.005 c 2.36 in Therefore the designer is permitted to use   0.9 for this beam design. 4) Finally, verify that the nominal flexural strength is sufficient for the applied loads. From Eq. (5-15): a 2.01 in     M n   As f y  d    0.9  1.37 in 2  60 ksi   21.5 in   1515 k-in 2 2     M n  126 k-ft  120 k-ft  M u Therefore, this design is sufficient.

90in. 3 #5 bars and 4 #3 bars

12in.

24in.

Fig. S5-14 Cross-section of final design for negative bending region

Note that other selections of As may also be correct. If all checks are satisfied, without being unreasonably conservative, the design may be considered adequate. Also note that all flange reinforcement that is considered to contribute to the negative bending capacity of this section is placed within two flange depths of the web.

5-37


5-15

Assume the maximum factored negative moment at support n of the floor beam m-n-o-p is -150 kip-ft. Using the design procedure for singly reinforced beam sections given in Section 5-3 (design of beams when section dimensions are not known), determine the beam dimensions and select the required area of tension reinforcement to satisfy all the ACI Code requirements for strength and minimum reinforcement area. Select bars and provide a sketch of your final section design.

The maximum expected negative moment is given as M u  150 k-ft  1800 k-in Although this is a T-section, there is no need to determine the effective flange width since the compression zone is located in the web of the beam. Step 1: Select ρ and the corresponding R-factor: Assume the desirable strain diagram shown in Fig. 5-27b, which leads to   0.9 . From Eq. (5-19):  f ' 0.85  4,000 psi Note: β1 =0.85 for f’c = 4,000 psi  1 c   0.0142 4 fy 4  60,000 psi From Eq. (5-21): f 0.0142  60,000 psi  y   0.2125 f 'c 4,000 psi From Eq. (5-22): R   f 'c (1  0.59)  0.2125  4 ksi  (1  0.59  0.2125)  0.743 ksi (Note that this R-factor is reasonable based on values given in Table A-3) Step 2: Select section dimensions, b and h: From Eq. (5-23): M 1800 k-in bd 2  u   2690 in 3  R 0.9  0.743 ksi Since no column dimensions are given which control the width of the beam, the designer can assume any reasonable α value. Here we assume α = 0.5 1

 M 3  1800 k-in  d  u     0.5  0.9  0.734 ksi    R  h  d  2.5 in  20 in b   d  0.5 17.5 in  8.75 in  10 in

1

3

 17.6 in  17.5 in

Note that both h and b are rounded to the nearest even inch value for constructability. Determine As and select the reinforcing bars: Assume jd  0.9d : From Eq. (5-16): Mu 1800 k-in As    2.12 in 2 a  0.9  60 ksi   0.9  17.5 in    fy  d   2 

5-38


With this estimate, iterate once to have a better estimate of the lever arm jd From Eq. (5-17): As f y 2.12 in 2  60 ksi a    3.74 in 0.85 f 'c b 0.85  4 ksi 10 in From Eq. (5-16): Mu 1800 k-in As    2.13 in 2 a 3.74 in     f y  d   0.9  60 ksi  17.5 in 2 2    Since I am not satisfied that any of the possible bar combinations that fulfill this requirement are not too conservative, I will choose to widen the beam slightly to help get a more efficient design. Select b = 12 in. Now re-determine As and select the reinforcing bars: Assume jd  0.9d : From Eq. (5-16): Mu 1800 k-in As    2.12 in 2 a  0.9  60 ksi   0.9  17.5 in    fy  d   2  With this estimate, iterate once to have a better estimate of the lever arm jd From Eq. (5-17): As f y 2.12 in 2  60 ksi a    3.12 in 0.85 f 'c b 0.85  4 ksi 12 in From Eq. (5-16): Mu 1800 k-in As    2.09 in 2 a 3.12 in      f y  d   0.9  60 ksi  17.5 in  2 2     Select 3 #6 bars over the web, and 4 #4 bars in the flange. This will result in “part” of the reinforcement (about 1/3) being spread into the flange, and most (2/3) remaining over the web. As  3 Ab1  4 Ab 2  3  0.44 in 2  4  0.2 in 2  2.12 in 2  2.09 in 2 OK Required Checks: 1) Use minimum bar spacing and cover values, or Table A-5, to verify that b = 12 in is sufficiently wide for 3 #6 bars to be placed in a single layer. OK 2) Use ACI-08 Eq. (10-3) to check that the reinforcement provided is more than the minimum required. From ACI-08 Eq. (10-3): For f’c = 4,000 psi, 3 f 'c  190  200 so use 200 As ,min 

3 f 'c fy

bw d 

200 psi  12 in  17.5 in  0.70 in 2  2.12 in 2 OK 60,000 psi

5-39


3) Calculate the strain in the extreme layer of tension steel to verify that assuming   0.9 is valid. From Eq. (5-17): As f y 2.12 in 2  60 ksi a    3.12 in 0.85 f 'c b 0.85  4 ksi 12 in We know that 1  0.85 for f’c = 4,000 psi a 3.12 in c   3.67 in 1 0.85 From Eq. (4-18): d -c 17.5 in - 3.67 in OK t    cu   0.003  0.011  0.005 c 3.67 in Therefore the designer is permitted to use   0.9 for this beam design. 4) Finally, verify that the nominal flexural strength is sufficient for the applied loads. From Eq. (5-15): a 3.12 in     M n   As f y  d    0.9  2.12 in 2  60 ksi  17.5 in   1820 k-in 2 2     M n  152 k-ft  150 k-ft  M u Therefore, this design is sufficient without being too conservative.


6in. 20in.

12in.

12in. Fig. S5-15 Cross-section of final design for negative bending region

Note that other selections As may also be correct. If all checks are satisfied, without being unreasonably conservative, the design may be considered adequate. Also note that all flange reinforcement that is considered to contribute to the negative bending capacity of this section is placed within two flange depths of the web.

5-40


5-16

Assume the maximum factored negative moment at the face of column C-2 for the girder along column line C is -250 kip-ft. Using the design procedure given in section 5-4 for the design of doubly reinforced sections, determine the beam dimensions and select the required areas of tension and compression reinforcement to satisfy all the ACI Code requirements for strength and minimum reinforcement area. Select bars and provide a sketch of your final section design.

The maximum expected negative moment is given as M u  250 k-ft  3000 k-in Although this is a T-section, there is no need to determine the effective flange width since the compression zone is located in the web of the beam. Select ρ and the corresponding R-factor: Assume the desirable strain diagram shown in Fig. 5-27b, which leads to   0.9 . From Eq. (5-25): 0.361 f 'c 0.36  0.85  4,000 psi Note: β1 =0.85 for f’c = 4,000 psi    0.0204 fy 60,000 psi From Eq. (5-21): f 0.0204  60,000 psi  y   0.306 f 'c 4,000 psi From Eq. (5-22): R   f 'c (1  0.59)  0.306  4 ksi  (1  0.59  0.306)  1.0 ksi (Note that this R-factor is reasonable based on values given in Table A-3) Select section dimensions, b and h: From Eq. (5-23): M 3000 k-in bd 2  u   3330 in 3  R 0.9  1.0 ksi Since no column dimensions are given which control the width of the beam, the designer can assume any reasonable α value. Here we assume α = 0.65 1

1

 M 3  3000 k-in 3 d  u     17.2 in  17.5 in  0.65  0.9  1.0 ksi    R  h  d  2.5 in  20 in b   d  0.65 17.5 in  11.4 in  12 in

Note that both h and b are rounded to the nearest even inch value for constructability. Determine As and select the reinforcing bars: Assume jd  0.9d : From Eq. (5-16): Mu 3000 k-in As    3.53 in 2 a  0.9  60 ksi   0.9  17.5 in    fy  d   2 

5-41


With this estimate, iterate once to have a better estimate of the lever arm jd From Eq. (5-17): As f y 3.53 in 2  60 ksi a    5.2 in 0.85 f 'c b 0.85  4 ksi  12 in From Eq. (5-16): Mu 3000 k-in As    3.73 in 2 a 5.2 in     f y  d   0.9  60 ksi  17.5 in 2 2    Select 3 #9 bars over the web, and 4 #4 bars in the flange. This will result in “part” of the reinforcement (about 1/5) being spread into the flange, and most (4/5) remaining over the web. As  3 Ab1  4 Ab 2  3 1.0 in 2  4  0.2 in 2  3.80 in 2  3.73 in 2 OK Also, select compression reinforcement such that the area of compression reinforcement (A’s) is greater than one half of the tension steel. Here we select 3 #8 bars so that A’s = 2.37 in2. Required Checks: 1) Use minimum bar spacing and cover values, or Table A-5, to verify that b = 12 in is sufficiently wide for 3 #9 bars to be placed in a single layer. OK 2) Use ACI-08 Eq. (10-3) to check that the reinforcement provided is more than the minimum required. From ACI-08 Eq. (10-3): For f’c = 4,000 psi, 3 f 'c  190  200 so use 200 As ,min 

3 f 'c fy

bw d 

200  12 in  17.5 in  0.70 in 2  3.80 in 2 OK 60,000 psi

3) Calculate the strain in the extreme layer of tension steel to verify that assuming   0.9 is valid. First, an iterative procedure must be used to determine the depth of the neutral axis. Following the procedure shown in Chapter 4, c = 4.2 in. From Eq. (4-18): d -c 17.5 in - 4.2 in OK t    cu   0.003  0.010  0.005 c 4.2 in Therefore the designer is permitted to use   0.9 for this beam design. 4) Finally, verify that the nominal flexural strength is sufficient for the applied loads. From Eq. (5-15): As a result of the iterations used to determine c, the following forces were determined: T  As f y  3.8 in 2  60 ksi  228 k Cc  1c0.85 f 'c bw  0.85  4.2 in  0.85  4 ksi 12 in  145 k

 cd'  4.2 in  2.5 in  2 C 's     0.003Es As     0.003  29,000 ksi  2.37 in  83 k c 4.2 in    

5-42


So, the nominal moment capacity of the section is:    a  M n   Cc  d    C 's  d  d ' 2    

 

 M n  0.9  145 kip  17.5 in 

 0.85  4.2 in   83 k  17.5 in  2.5 in    2  

  M n  3170 k-in  264 k-ft  250 k-ft  M u

Therefore, this design is sufficient without being too conservative.


20in.

12in.

2.5in.

3 #8 bars

12in. Fig. S5-16 Cross-section of final design for negative bending region

Note that other selections of b, h, and As may also be correct. If all checks are satisfied, without being unreasonably conservative, the design may be considered adequate. Also note that all flange reinforcement that is considered to contribute to the negative bending capacity of this section is placed within two flange depths of the web.

5-43


5-17 point

For the one-was slab shown in Fig. P5-7, assume the maximum negative moment at support c is -3.3 kip-ft/ft, and the maximum factored positive moment at midspan b is 2.4 kip-ft/ft. (a)

Using the given slab thickness of 6 in, determine the required reinforcement size and spacing at both of these locations to satisfy ACI Code flexural strength requirements. Be sure to check the ACI Code requirements for minimum flexural reinforcement in slabs.

Negative Moment M u  3.3 k-ft  39.6 k-in : Assume 0.75 in of cover will be provided. This results in a d = 5 in. From Eq. (5-16): Mu 39.6 k-in 2 As ft ft    0.16 in ft a  0.9  60 ksi   0.9  5 in  ft   fy  d   2  With this estimate, iterate once to have a better estimate of the lever arm jd From Eq. (5-17): As f y 0.16 in 2  60 ksi a    0.24 in 0.85 f 'c b 0.85  4 ksi 12 in From Eq. (5-16): Mu 39.6 k-in 2 As ft ft    0.15 in ft a 0.24 in  ft    f y  d   0.9  60 ksi   5 in  2 2    Minimum reinforcement: 2 As ,min  0.0018bh  0.0018  12 in  6 in  0.13 in ft ft Strength requirements govern here. Maximum reinforcement spacing is limited to 3h or 18 in, which is the same value for this 6 in deep slab. Also, we must check reinforcement spacing for crack control. Since fy and cc are the same here as in Example 5-7, the maximum spacing for crack control is 12 in. This governs. If we select #3 bars at 8 in, the result is: 2 2 12 in As  0.11 in 2   0.165 in  0.15 in ft ft ft 8 in

OK

Check that strength is satisfied: a 0.24 in     M n   As f y  d    0.9  0.165 in 2  60 ksi   5 in   43.5 k-in 2 2    OK  M n  3.62 k-ft ft  3.3 k-ft ft  M u

5-44


Remember to calculate the strain in the extreme layer of tension steel to verify that assuming   0.9 is valid. d -c 5 in - 0.24 / 0.85 in OK t    cu   0.003  0.050  0.005 c 0.24 / 0.85 in Therefore the designer is permitted to use   0.9 for this beam design.

Positive Moment M u  2.4 k-ft  28.8 k-in : Assume 0.75 in of cover will be provided. This results in a d = 5 in. From Eq. (5-16): Mu 28.8 k-in 2 As ft ft    0.12 in ft a  0.9  60 ksi   0.9  5 in  ft   fy  d   2  With this estimate, iterate once to have a better estimate of the lever arm jd From Eq. (5-17): As f y 0.12 in 2  60 ksi a    0.17 in 0.85 f 'c b 0.85  4 ksi 12 in From Eq. (5-16): Mu 28.8 k-in 2 As ft ft    0.11 in ft a 0.17 in  ft    f y  d   0.9  60 ksi   5 in  2 2    Minimum reinforcement: 2 As ,min  0.0018bh  0.0018  12 in  6 in  0.13 in ft ft Minimum requirements govern here. Maximum reinforcement spacing is limited to 3  h or 18 in, which is the same value for this 6 in deep slab. Also, we must check reinforcement spacing for crack control. Since fy and cc are the same here as in Example 5-7, the maximum spacing for crack control is 12 in. This governs. If we select #3 bars at 10 in, the result is: 2 2 12 in As  0.11 in 2   0.132 in  0.13 in ft ft ft 10 in

OK

Check that strength is satisfied: a 0.17 in     M n   As f y  d    0.9  0.132 in 2  60 ksi   5 in   35.0 k-in 2 2    OK  M n  2.9 k-ft ft  2.4 k-ft ft  M u

5-45


Remember to calculate the strain in the extreme layer of tension steel to verify that assuming   0.9 is valid. d -c 5 in - 0.17 / 0.85 in OK t    cu   0.003  0.072  0.005 c 0.17 / 0.85 in Therefore the designer is permitted to use   0.9 for this beam design. (b)

At both locations, determine the required bar size and spacing to be provided in the transverse direction to satisfy ACI Code section 7.12.2 requirements for minimum shrinkage and temperature reinforcement.

Since the positive flexural region was controlled by temperature and shrinkage reinforcement, the reinforcement specified there would suffice in the transverse direction at all locations. So, use #3 bars at 10 in. Placement near the top or bottom of the slab makes no difference here, so specify that bars are to be placed wherever is easiest. (c)

For both locations provide a sketch of the final design of the slab section.

#3@8in.

#3@10in.

6in.

6in. 12in.

12in.

Fig. S5-17 Cross-section of final design for negative and positive bending regions, respectively Note that other selections of As may also be correct. If all checks are satisfied, without being unreasonably conservative, the design may be considered adequate.

5-46


Chapter 6

(b)

-17.7

Assuming the beam is uncracked, show the direction of the principal tensile stresses at middepth at points A, B, and C.

A

(c)

10

0.2 -9.8

V, kips

16.0

For the rectangular beam shown in Fig P6-1: (a) Draw a shear force diagram.

7.3

6-1

B

C

On a drawing of the beam sketch the inclined cracks that would develop at A, B, and C.

A

B

6-1

C


6-2, 6-3, 6-4 and 6-5 (

6-2

Compute for the cross sections shown in Figs. P6-2, P6-3, P6-4 and P6-5. In each case use and )

√

(

6-3

)

(

]⁄

) )

√

(

⁄

√ [(

⁄ (

)⁄ ]⁄ )

)

√ ⁄ ( 6-6

⁄ )⁄

√ [(

⁄ (

6-5

)⁄ ]⁄ )

√

6-4

⁄

√ [(

⁄ (

(

√ [(

)

⁄ )⁄ ]⁄

)

ACI Sec. 11.4.5.1 sets the maximum spacing of vertical stirrups at ⁄ . Explain why. Every inclined crack must be crossed by at least one stirrup. The assumed horizontal projection of the crack is d so every crack will be crossed by at least one stirrup.

6-2


6-7

Figure P6-7 shows a simply supported beam. The beam has No. 3 Grade 40 doubleleg stirrups with and 4 No.8 Grade 60 longitudinal bars with . The plastic truss model for the beam is shown in the figure. Assuming that the stirrups are all loaded to . (a)

Use the method of joints to compute the forces in each panel of the compression and tension chords and plot them. The force in member ⁄ . is

Assume all stirrups yield. As f yt  8.8 kips Joint L11 :

Vertical force in stirrup = 8.8 kips

V  0 at Joint L

11

gives vertical component in strut U12  L11 as 8.8

kips downward (compression in U12  L11 ). Horizontal component in U12  L11 

12 in.  8.8 kips = 5.28 kips acting 20 in.

to the left on Joint L11 . Joint U12 :

Each inclined strut has a vertical component of 8.8 kips, therefore 6 struts are needed to equilibrate 52.8 kips force at Joint U12 as shown.

Joint U 8 :

Total downward load is 26.4 kips applied load plus 8.8 kips in stirrup

U8  L8 , therefore 4 struts are needed to equilibrate 35.2 kips force at

U 8 as shown. Forces in Inclined Struts in Web

Member

U12  L11 U12  L10 U12  L9 U11  L8 U10  L7 U 9  L6 U8  L5

U8  L4 U8  L3 U8  L2

Vertical Component (kips) 8.8

Horizontal Projection (in.) 12

Horizontal Component (kips) 5.28

8.8

24

10.56

8.8

36

15.84

8.8

36

15.84

8.8

30

13.20

8.8

24

10.56

8.8

18

7.92

8.8

24

10.56

8.8

30

13.20

8.8

36

15.84

6-3


U 7  L1 Member (cont’d)

U 6  L0 U 5  L0

8.8 Vertical Component (kips) 8.8

36 Horizontal Projection (in.) 36

15.84 Horizontal Component (kips) 15.84

8.8

30

13.20

8.8

24

10.56

8.8

18

7.92

8.8

12

5.28

8.8

6

2.64

U 4  L0 U 3  L0 U 2  L0 U1  L0 Compute Forces in Lower Chord

Maximum moment = 52.8  8  26.4  4  316.8 ft-kips Force in L11  L13 

 H at Joint L

11

M 316.8 12   190.0 kips (in tension) jd 20

gives force in L10  L11  190.0  5.28  184.7 kips

Force in Lower Chords

Member

L11  L13 L10  L11 L9  L10 L8  L9 L7  L8 L6  L7 L5  L6 L4  L5 L3  L4 L2  L3

L1  L2 L0  L1

Horizontal Force in Strut (kips) -

Joint -

L11 L10 L9 L8

L7 L6 L5 L4 L3 L2 L1 L0

5.28

184.7

10.56

174.2

15.84

158.3

15.84

142.5

13.20

129.3

10.56

118.7

7.92

110.8

10.56

100.2

13.20

87.0

15.84

71.2

15.84

55.4

15.84 +13.20

6-4

Lower Chord Force (kips) 190.0 tension


+10.56

Member (cont’d)

+7.92 Horizontal Force in Strut (kips) +5.28

Joint

Lower Chord Force (kips)

+2.64 Total:

55.4

The force in L0  L1 is 55.4 kips. This must be anchored in Joint L0 . The sum of the horizontal forces in the struts at Joint L0 is 55.4 kips and

 H  0 at Joint L

0

.

Forces in Upper Chords: Joint U1 : There is no force in the compression chord to the right of U1 .

Member

Joint

Horizontal Force in Struts (kips)

Lower Chord Force (kips)

U1  U 2

U1

2.64

U 2  U3

U2

5.28

2.64 compression 7.92

U3  U 4

U3

7.92

15.84

U 4  U5

U4

10.56

26.4

U5  U6

U5

13.20

39.8

U6  U7

U6

15.84

55.4

U 7  U8

U7

15.84

71.3

U8  U9

U8

15.84 +13.20 +10.56 +7.92

118.8

U 9  U10

U9

10.56

129.4

U10  U11

U10

13.20

142.6

U11  U12

U11

15.84

158.4

Compression at Midspan

U12

15.84 +10.56

6-5


+5.28

Plot of Forces in Upper Chords:

Plot of Forces in Lower Chords:

6-6

190.0


(b)

⁄ on the diagram from part (a) and compare the bar forces Plot from the truss model to those computed from ⁄ . The bar forces in the lower chords provided by As f s  M / jd appear to be less than the values provided by the truss model throughout the span except at the midspan where the two values match.

(c)

Compute the compression stress in the diagonal member 11)). The beam width, , is 12 in. Slope of L1  L7 : tan   20 / 36  0.556    29.05

f cd 

V  1  52800  1   tan     0.556    518 psi bw jd  tan   12  20  0.556 

6-7

(see Eq. (6-


The beam shown in Fig. P6-8 supports the unfactored loads shown. The dead load includes the weight of the beam. Draw shearing force diagram for

wDu  1.2 1.4  1.68 k/ft wLu  1.6 1.5  2.4 k/ft (1)

factored dead and live load on the entire beam.

At the left support:

Vu 

1.68  2.4   25  51.0 kips 2

Vu , kips

- 51.0

Case (1)

(2)

factored dead load on the entire beam plus factored live load on the left haft-span.

1.68 

252 12.52  2.4  2 2  28.5 kips 25

At the right support:

Vu 

At the right support:

Vu  1.68  25  2.4 12.5  28.5  43.5 kips

At the midspan:

Vu  43.5  1.68  2.4 12.5  7.5 kips

Vu , kips

(3)

-28.5

Case (2)

-7.5

(a)

51.0

6-8

factored dead load on the entire beam plus factored live load on the right haft-span.

Due to the asymmetry of loadings Case (3) to loadings Case (2), the shear diagram of Case (3) is asymmetric to that of Case (2).

6-8


- 43.5

7.5

28.5

Case (3)

Vu , kips

Superimpose the diagram to get a shear force envelope. Compare the shear at midspan to that from Eq. 6-26.

Design stirrups. Use .

, kips

- 68.0

  0.75

10.0



- 10.0

Vu

and No.3 double-leg stirrups with

68.0

(c)

- 51.0

-7.5

Vu , kips

wLu 2.4  25   7.5 kips 8 8

51.0

Eq. 6-26: Vu (midspan) 

7.5

(b)

Are stirrups required?

Vc  2 4500 12 17.5 /1000  28.2 

Vu



 68.0 kips

 Stirrups are required. Check stirrups anchorage Use No. 3 Grade 40 stirrups. ACI code Sec. 12.13.2.1. allows these to be anchored by a hook around a top bar. Maximum spacing d/2 = 8.75 in.

Vu  6 fcbwd  82.1 kips  Maximum spacing of d / 4 is not required.  Av f y 0.22  40000   14.6 in. (note 0.75 fc  50.3 psi  50 psi, use 50.3 psi ) 50.3 12 0.75 f cbw

6-9


 Maximum spacing smax  8.5 in.

Compute spacing to resist shear forces  At d from supports:

Vu



s

 68 

 68  10  17.5  61.2 12.5 12 

Av f yt d Vu /   Vc



kips

0.22  40 17.5  4.5 in.  smax  8.5 in.  61.4  27.4 

 Use s = 4.0 in. 

Change this to 6 in. where

This occurs at x 





0.22  40 17.5  27.4  53.1 kips. 6

Vu





0.22  40 17.5  27.4  46.7 kips. 8

68  46.7  12.5 12   55.1 in. from end. 68  10

Terminate stirrups where






68  53.1  12.5 12   38.5 in. from end. 68  10

Change this to 8 in. where


Vu

Vu





Vc 28.2   14.1 kips. 2 2

68  14.1  12.5 12  = 139 in. 68  10

Compute the number of stirrups From the center of the support, use 1 @ 2 in.

38.5  2  9.1 , use 10 @ 4 in. 4 55.1   2  10  4  Required no. of stirrups at the spacing of 6 in.:  2.2 , use 3 @ 6 in. 6 139   2  10  4  3  6  Required number of stirrups at the spacing of 8 in.:  9.9 , use 8 Required number of stirrups at the spacing of 4 in.:

10 @ 8 in.

 Provide No.3 U stirrups Grade 40 steel. Starting from the center of the support, use 1 @ 2 in., 10 @ 4 in., 3 @ 6 in., and 10 @ 8 in. from each end.

6-10


(Note that Vs (max) 

6-9

0.22  40 17.5  38.5 kips < 8 f c bw d  113 kips , OK) 4

The beam shown in Fig. P6-9 supports the unfactored loads shown in the figure. The dead load includes the weight of the beam. (a)

Draw shearing force diagrams for (1) factored dead and live load on the entire length of beam. (2) factored dead load on the entire beam plus factored live load between B and C. (3) factored dead load on the entire beam plus factored live load between A and B and between C and D.

Loadings (2) and (3) will give the maximum positive and negative shears at B. wDu  2.4 k/ft

wLu  2.4 k/ft wu  4.8 k/ft

6-11


28.8

(b)

-43.8 -29.4

-10.2

28.8

-36.3

14.4

21.3

(a3)

2.1

(a2)

28.2

-5.4

(a1)

33.0

V, kips

Draw the factored shear force envelope. The shear at B should be the factored dead load shear plus or minus the shear from Eq. 6-26.

Eq. 6-26: Vu (midspan) 

2.4 16  4.8 kips 8

6-12

28.8

2.1

33.0

Envelope from Eq. 6-26

28.8

4.8-2.7 = 2.1

-43.8

-10.2

Envelope from Part (a)

33.0


-43.8

-4.8-2.7 = -7.5

Eq. 6-26 is correct for s simple beam without overhangs. It is an approximation for all other cases. However, we shall assume it is close enough and use it for the rest of this example.

(c)

Design stirrups. Use

and

.

Are stirrups required? ⁄

√

 Stirrups are required. Check anchorage Use No. 3 stirrups. These can be anchored by a hook around a top bar. Maximum spacing d/2 = 10.8 in. √

 Maximum spacing of d / 4 is not required.

Av f y 50bw



0.22  40000  14.7 in. (note 50 12

√

 Maximum spacing smax  10.8 in. Compute spacing to resist shear forces – Part AB

6-13

)




At d from support A:

Vu

 44 

s

0.22  40  21.5  86 in.  smax  10.8 in.  34.8  32.6



 44  2.8  21.5  34.8 8 12 

kips

 Use s = 10 in. 

Terminate stirrups where:

This occurs at:

 Part A-B: Provide No.3 stirrups Grade 40 steel. Starting from A use 1 @ 5 in., 7 @ 10 in. Compute spacing to resist shear forces – Part BC  At d from support C

Vu



 58.4 

 58.4  10   21.5  47.6

kips

8 12 0.22  40  21.5 s  12.6 in.  smax  10.8 in.  47.6  32.6   Use s = 10 in. 

Terminate stirrups where:

This occurs at:

 Part BC: Provide No.3 stirrups Grade 40 steel. Starting from C use 1 @ 5 in., 9 @ 10 in.

Compute spacing to resist shear forces – Part CD  At d from support C

6-14


 Can terminate stirrups where

Vu





Vc . However, place minimum shear reinforcement 2

throughout part CD.

 Part CD: Use No. 3 U stirrups, Grade 40. Starting from C use 1 @ 5 in., 7 @ 10 in.

6-15


Fig. P6-10 shows an interior span of a continuous beam. The shears at the ends are ⁄ . The shear at midspan is from Eq. (6-26). (a)

Draw a shear force envelope.

wDu  1.2 1.5  1.8 k/ft wLu  1.6 1.8  2.88 k/ft

wu  4.68 k/ft End shears = 4.68  22 / 2  51.5 kips Midspan shears = 2.88  22 / 8  7.92 kips , kips

(b)

Design stirrups using

- 68.7

  0.75

10.6



- 10.6

Vu

68.7

6-10

and

.


Vc  2 4000 12 17.5 /1000  26.6 kips  Stirrups are required. Check anchorage Use No. 3 stirrups. These can be anchored by a bend around a top bar. Maximum spacing d/2 = 8.8 in.

Vu  6 f cbwd  79.8 kip  Maximum spacing of d / 4 is not required.  Av f y 0.22  40000   14.7 in. (note 0.75 fc  47.4 psi < 50 psi, use 50 psi ) 50bw 50 12  Maximum spacing smax  8.8 in. Computing spacing to resist shear forces 

At d from face of column:

Vu

 68.7 

s

0.22  40 17.5  4.5 in.  61.0  26.6 



 68.7  10.6  17.5  61.0 1112 

kips

6-16


 Use s = 4 in. 

Changing stirrup spacing to 6 in. where






0.22  40 17.5  26.6  52.3 kips 6

Vu





0.22  40 17.5  26.6  45.9 kips 8

68.7  45.9  1112   51.8 in. 68.7  10.6

Terminate stirrups when






68.7  52.3  1112   37.3 in. 68.7  10.6

Changing stirrup spacing to 8 in. where


Vu

Vu





Vc 26.6   13.3 kips 2 2

68.7  13.3  1112   126 in. 68.7  10.6

 Use No.3 Grade 40 U stirrups. Starting from face of column at each end, use 1 @ 2 in., 9@ 4 in., 3 @ 6in., and 9 @ 8in.

6-17


6-11

Design shear reinforcement for the C1-C2 span of the girder designed in Example 56 (final section given in Fig. 5-32). From the structural analysis discussed in Example 5-6, the factored design end shears for this girder are 28.9 kips at the face of column C1 and 39.2 kips at the face of column C2. Use and . Are stirrups required? ⁄ √ Stirrups are required.  Check anchorage Select No. 3 stirrups. These can be anchored by a bend around a top bar. Maximum spacing ⁄ √

 Maximum spacing of ⁄ is not required. √

(

)

 Maximum spacing Compute the spacing of stirrups required to resist shear forces It is desirable to have the most efficient design of stirrups possible, so stirrup size and spacing should be adjusted over the span length whenever possible to account for the changing shear demand. Likewise, it is advantageous to use the factored shear demand at from the face of the column rather than directly at the face of the column to design stirrups. However, for a girder loaded like this one, the majority of shear comes from the point load at midspan. Therefore, it is reasonable to select only two different stirrup spacing, one for each half of the girder, based on the factored shear demand at the face of the columns. 

At the exterior column face (C1):

⁄  The maximum spacing controls, so select s = 8 in. 

At the interior column face (C2):

⁄  The strength requirement controls, so select s = 6 in.

 Use No.3 Grade 40 U stirrups. Starting from face of C2 column, use s = 6 in. until past the point load, then use s = 8 in.

6-18


Fig. P6-12 shows a rigid frame and the factored loads acting on the frame. The 7-kip horizontal load can act from the left or the right. and . (a)

Design stirrups in the beam 34.9 27.2

Wind from right

Vu , kips 3.85

CL

-3.85 CL

Wind from left

-27.2 -34.9

46 44 . 5 38 . 1 .0

6-12

Vu



, kips

CL 5.1 d from face of column face of column

-5.1

Are stirrups required? Assume d = 20 – 2.5 = 17.5 in. √ ⁄  Stirrups are required. Maximum spacing Use No. 3 d/2 = 8.8 in. √

 Maximum spacing of d / 4 is not required. √  Maximum spacing smax = 8.8 in.

(

6-19

√

)


Compute spacing required to resist shear 

At d from face of column,

Vu



 38 kip

⁄ 

Terminate stirrups when:

This occurs at:

 Provide No. 3, Grade 40 U stirrups. Starting at face of column from each end, use 1 @ 4 in., 11@ 8 in. (b)

Are stirrups required in the columns? If so, design the stirrups for the columns.

Maximum shear is Vu  8 kips in leeward column. )√

( (

(

)

)√

(


 Stirrups are not required. Use minimum amount of shear reinforcement.

6-20

)


Chapter 7 7-1

A cantilever beam 8 ft long and 18 in wide supports its own dead load plus a concentrated load located 6 in from the end of the beam and 4.5 in away from the vertical axis of the beam. The concentrated load is 15 kips dead load and 20 kips live load. Design reinforcement for flexure, shear, and torsion. Use fy = 60,000 psi for all steel and f’c = 3750 psi.

Design for moment: Since wide sections are desirable for torsion, estimate that α = 0.8, so d  18 in / 0.8  22.5 in . Therefore, the dead load due to the weight of the beam is: (22.5 in  2.5 in) 18 in DL   0.15 k/ft 3  0.469 k/ft 144 in 2 /ft 2 So, the moment demand due to factored loads is:

0.469 k/ft  8 ft 

2

 1.2  15 k  1.6  20 k   7.5 ft  393 k-ft  4720 k-in 2 Select ρ and corresponding R-factor, assuming the desirable strain diagram shown in Fig. 5-27b, which leads to   0.9 .  f ' 0.85  3,750 psi  1 c   0.0133 Note: β1 = 0.85 for f’c = 3,750 psi 4 fy 4  60,000 psi M u  1.2 

 fy

0.0133  60,000 psi  0.213 f 'c 3,750 psi R   f 'c (1  0.59)  0.213  3.75 ksi  (1  0.59  0.213)  0.698 ksi (Note that this R-factor is reasonable based on values given in Table A-3)





Now select h with b = 18 in: M 4720 k-in bd 2  u   7500 in 3  R 0.9  0.698 ksi Since the width of the beam is given as 18 in, we can directly solve for a reasonable d value. Mu 4720 k-in d   20.4 in  21.5 in  bR 0.9  18 in  0.698 ksi h  d  2.5 in  24 in b  18 in

Note that both h and b are rounded up to the nearest even inch value for constructability. Finally, determine As required for resisting this applied moment by first going back and recalculating the weight of the beam with the final selected dimensions: bh k 18 in  24 in k k DL   0.15 3   0.15 3  0.45 2 2 2 2 144 in /ft ft 144 in /ft ft ft So M u  393 k-ft  4720 k-in Calculate the required area of steel, assuming jd  0.9d : Mu 4720 k-in As    4.52 in 2 a  0.9  60 ksi   0.9  21.5 in    fy  d   2 

7-1


With this estimate, iterate once to have a better estimate of the lever arm jd . As f y 4.52 in 2  60 ksi a    4.73 in 0.85 f 'c b 0.85  3.75 ksi 18 in Mu 4720 k-in As    4.57 in 2 a 4.73 in     f y  d   0.9  60 ksi   21.5 in 2 2    No further iterations are necessary, since the estimated lever arm was very reasonable. Use b = 18 in, h = 24 in, and As = 4.57 in2. Bars will be selected later. Use ACI-08 Eq. (10-3) to check that the reinforcement provided is more than the minimum required. For f’c = 3,750 psi, 3 f 'c  184  200 so use 200 psi

As ,min 

200 200 psi bw d   18 in  21.5 in  1.29 in 2  4.57 in 2 fy 60,000 psi

OK

Also, calculate the strain in the extreme layer of tension steel to verify that assuming   0.9 is valid. As f y 4.57 in 2  60 ksi a    4.78 in 0.85 f 'c b 0.85  3.75 ksi 18 in We know that 1  0.85 . a 4.78 in c   5.62 in 1 0.85

d -c 21.5 in - 5.62 in   cu   0.003  0.0085  0.005 c 5.62 in Therefore the designer is permitted to use   0.9 for this beam design.

t 

OK

Now design for torsion and shear: At d = 21.5 in away from the face, the factored torsion is: 4.5 in Tu  1.2  15 k  1.6  20 k    18.8 k-ft 12 in/ft At d = 21.5 in away from the face, the factored shear is: 21.5 Vu  1.2 15 k  1.2  (8 ft  ft)  0.45 k/ft  1.6  20 k  53.4 k 12 Should torsion be considered in this design? Acp  18 in  24 in  432 in 2

Pcp  (18 in  24 in)  2  84 in From Eq. (7-18b):  Acp 2 Tth   f 'c   Pcp 



 432 in 2    0.75  1.0  3,750 psi    84 in  

7-2



2

   102,000 lb-in  8.5 k-ft  


Since Tu  Tth , torsion must be considered. Also, since we are dealing with a statically determinate system, we have a case of equilibrium torsion, in which the full factored torsion must be sustained by our beam. Therefore, the design torsion cannot be reduced. Check whether the section dimensions are sufficient to withstand the combined stresses due to shear and torsion. From Eq. (7-33): 2

2

 Vu   Tu Ph   V     c  8 f 'c     2   bw d   1.7 Aoh   bw d 

Assuming #4 stirrups, Ph  2   24 in  2 1.5 in  2  0.25 in  18 in  2 1.5 in  2  0.25 in   70 in

Aoh   24 in  2 1.5 in  2  0.25 in   18 in  2 1.5 in  2  0.25 in   297 in 2

2  53.4 k    18.8 k-ft  70 in   18 in  21.5 in   2 2   1.7  297 in 





2

     0.75   2 3,750 psi 18 in  21.5 in  8 3,750 psi     18 in  21.5 in   

0.174 ksi  0.459 ksi

The section is sufficiently large.

Now determine the area of stirrups required to resist Vu: Vc  2 f 'c bw d  2 3,750 psi 18 in  21.5 in  47.4 k Vu

53.4 k  47.4 k  23.8 k  0.75 Av V 23.8 k in 2  s   0.018 s f yt d 60 ksi  21.5 in in

Vs 

 Vc 

Determine the additional area of stirrups required to resist Tu: T 18.8 k-ft Tn  u   25.1 k-ft 0.75 0.75 From Eq. (7-24), using Ao  0.85 Aoh : At Tn 25.1 k-ft  12 in/ft   s 2 Ao f yt cot  2  0.85  297 in 2  60 ksi  cot 45

 

 0.0099

So, evaluate the total required area of stirrups: in 2 in 2 in 2 For strength: 0.018  2  0.0099  0.0378 in in in 2 50bw 50  18 in in Minimum required:   0.015 f yt 60,000 psi in The strength requirement governs here.

7-3

in 2 in


Now determine stirrup size and spacing: 12 in   smax   Ph / 8  70 / 8  8.75 in d / 2  10.75 in (for shear)  2  0.11 in 2  5.82 in in 2 0.0378 in 2  0.20 in 2 If we select #4 stirrups, s   10.6 in in 2 0.0378 in Although it is a tight spacing, select to use closed #3 stirrups at 5 in spacing.

If we select #3 stirrups, s 

Determine the need for longitudinal reinforcement resisting torsion:  in 2   A   f yt  2 2 Al   t  Ph  cot    0.00990 For strength:    70 in  1.0  1.0  0.693 in in  s   f yl    Minimum required:

Al ,min 

5 f 'c Acp f yl



f yt At Ph s f yl

At 25bw in 2 in in 2 , we must use 0.022  0.022   0.0075 in s in f yt in 2

Since

Al ,min 

5 3,750 psi  432 in 2 in 2  0.022  70 in 1.0  0.66 in 2 60 ksi in

Use Al  0.693 in 2 Longitudinal bars are required in the corner of each stirrup. Also, longitudinal bars must be spaced no more than 12 in apart around the perimeter of the section, so a bar is needed in the middle of each face. So 8 bars are required. Area / bar  0.693 in 2 / 8 bars  0.0866 in 2 /bar

Select #3 bars along bottom and sides of the cantilevered section. The reinforcement required along the top of the beam for resisting moment and torsion is: As  4.56 in 2  3  0.0866 in 2  4.82 in 2 Select 5#8 bars along the top of the beam such that As  5 Ab  5 1.0 in 2  5 in 2  4.82 in 2 .

7-4


7-2

Explain why the torsion in the edge beam A-B in Fig. 7-21a is called “equilibrium torsion,” while the torsion in the edge beam A1-B1 in Fig. P7-3 is called “compatibility torsion.”

If the edge beam A-B in Fig. 7-21c did not resist torsion, the beam would rotate, uninhibited, about its longitudinal axis and fail to resist the action of load P. Essentially, the torsional resistance of the beam is required for equilibrium to be satisfied. On the contrary, if beam A1-B1 in Fig. P7-3 did not resist torsion, the beam would rotate only slightly before the floor’s weight and superimposed loads would be redistributed to other elements, thereby satisfying equilibrium through the redundancy of the system. The torsion in A1B1 only arises from the need to maintain compatibility of deformations between the ends of the joists and the twisting of the edge beam.

7-3

The two parts of this problem refer to the floor plan shown in Fig. P7-3. Assume that the entire floor system is constructed with normal-weight concrete that has a compressive strength, f’c = 4,500 psi. Also, assume that the longitudinal steel has a yield strength of fy = 60 ksi and that the transverse steel has a yield strength of fyt = 40 ksi. a)

Design the spandrel beam between columns B1 and C1 for bending, shear, and torsion. Check all of the appropriate ACI Code requirements for strength, minimum reinforcement area, and reinforcement spacing are satisfied.

Step 1: Determine Mu, Vu, and Tu: Dead weight from beam, per foot of beam: 12 in  18 in lb lb  150 3  225 2 ft ft 144 in 2 ft If we assume that the slab dimensions given in the figure are measured center to center of the beams, we need to account for half the width of the beam in addition the span length given when calculating loads. Therefore the dead weight from slab, per foot of beam, is: 6 in  6.5 ft lb lb  150 3  488 ft ft 12 in ft Superimposed dead load, per foot of beam: lb lb 6.5 ft  20 2  130 ft ft To simplify the design, we will not reduce the live load. Therefore, the live load per foot of beam: lb lb 6.5 ft  50 2  325 ft ft So, using wu  1.2DL  1.6LL , the factored load is, per foot of beam:

wu  1.2   0.225 k/ft  0.488 k/ft  0.130 k/ft   1.6   0.325 k/ft   1.53 k/ft

7-5


With that, we can determine the design moments using ACI Moment Coefficients: At columns: (

)

(

At midspan: ( )

(

)

)

Note that n is taken as the average of the adjoining clear spans for calculating the negative moment at the columns. The design shear is: At midspan:

At d away from the column face: ⁄

(

) ⁄

(

)

Note that n is taken as the clear span length of the interior span when calculating the shear acting on the face of the support. Do not reduce this value to the n used for calculating the negative moment at column B-1. To calculate the design torsion, we first have to determine the moment and shear that the slab is applying to the edge of the beam. wu  1.2   0.075 k/ft  0.020 k/ft   1.6   0.050 k/ft   0.194 k/ft n

 12 ft -12 in  11 ft

0.194 k/ft  11 ft  w 2 Mu   u n    0.978 k-ft 24 24 w 0.194 k/ft 11 ft Vu  u n    1.07 k 2 2 Therefore, the torsion applied to our beam by the slab is: 6 in t  0.978 k-ft  1.07 k   1.51 k-ft/ft 12 in/ft 2

And thus our design torsion, at d away from the ends of our beam, is: ( ) ( )

7-6


Step 2: Determining the area of longitudinal steel required for flexure. At the column:

Refer to Table A-3. For f’c = 4,500 psi, and R = 229 psi, column,

. So, at

the

At midspan:

Refer to Table A-3. There are no R values below 190 psi, which corresponds to the √

√

minimum reinforcement ratio. Therefore,

Step 3: Determine whether torsion must be considered in the design of this beam. Begin by determining the dimensions of the beam section active in torsion, and calculate the threshold torsion.  height of beam below slab  18 in , where f is the length of flange active in torsion. f  4h f  24 in   Therefore, Acp  24 in 12 in  6 in 18 in  396 in 2

Pcp  24 in  12 in  18 in  18 in  6 in  30 in  108 in  Acp 2 Tth   f 'c   Pcp 



 396 in 2    0.75  1.0  4,500 psi    108 in  



2

   73,000 lb-in  6.09 k-ft  

Since Tu  Tth , torsion must be considered in this design. Step 4: Since the torsion resisted by this edge beam is not required to maintain equilibrium of the structure, we have a case of compatibility torsion. Therefore we can likely reduce our Tu to the following:  396 in 2 2   Acp 2    292,000 lb-in  24.4 k-ft Tu ,comp   4 f 'c   0.75  4  1.0  4,500 psi   Pcp   108 in      Unfortunately, since Tu  Tu ,comp , we cannot reduce our design torsion. Since our design torsion is





not being reduced, no redistribution of design moments is required in the adjoining slab. Step 5: Determine whether the section is large enough to resist the combined actions of shear and torsion. First assume that a closed #4 stirrup will be used in the web of this beam. Aoh   24 in  2 1.5 in  0.5 in   12 in  2 1.5 in  0.5 in   174 in 2

Ph  2   24 in  2 1.5 in  0.5 in  12 in  2 1.5 in  0.5 in   58 in

7-7


From Eq. (7-33): √( √(

) )

( (

)

(

)

(

√ ) √

√

The section is sufficiently large. Step 6: Determine the area of stirrups required to resist Vu: Vc  2 f 'c bw d  2 4,500 psi 12 in  21.5 in  34.6 k

Step 7: Determine the area of stirrups required to resist Tu: From Eq. (7-24), using Ao  0.85 Aoh : (

)

Step 8: Evaluate the total required area of stirrups, and select spacing: For strength:

Minimum required: 0.75 4500  bw 50.31 psi  12 in in 2   0.015 f yt 40,000 psi in The strength requirement governs here. Now determine stirrup size and spacing: 12 in   smax   Ph / 8  58 in / 8  7.25 in d / 2  10.75 in (for shear)  If we select #3 stirrups,

If we select #4 stirrups,

7-8

)


Select closed #3 stirrups at 5 in. spacing at the ends of the beam. It is also possible that stirrups are not required along the full length of this beam. For torsion, determine where Tu  Tth /  . By similar triangles, this occurs at 66.5 in. away from the face of the column. However, torsional reinforcement must be continued for (d  bt )  (21.5 in  12 in)  33.5 in past this theoretical point. Therefore, no torsional reinforcement is required beyond 100 in. away from the face of the column. Since Vu  Vc / 2 at this point, no shear reinforcement is required beyond this point either. Final design of transverse reinforcement: Using closed #3 stirrups, One stirrup at 2.5 in from either column face, followed by stirrups spaced at 5 in. until beyond 100 in. from the face of the column.

Step 9: Finalize the design of the longitudinal reinforcement: Determine the need for longitudinal reinforcement resisting torsion: For strength:

Minimum required: 5 f 'c Acp At f yt Al ,min   Ph f yl s f yl in 2 in 2 25bw in 2 A  Since  t   0 , we must use 0.0075   0.0075 in in f yt in  s ,min

5 4,500 psi  396 in 2 in 2 2 Al ,min   0.0075  58 in   1.92 in 2 60 ksi in 3 Use Al  1.92 in 2 Longitudinal bars are required in the corner of each stirrup. Also, longitudinal bars must be spaced no more than 12 in apart around the perimeter of the section, so a bar is needed in the middle of each vertical face. So 6 bars are required. Area / bar  1.92 in 2 / 6 bars  0.32 in 2 / bar

-At the columns, use #6 bars in the bottom corners and halfway up the vertical face of the beam. ( )⁄ As top reinforcement, ⁄ , so specify 3 #7 bars along the top of the section. -At midspan, use #6 bars in the top corners and halfway up the vertical face of the beam. As ( )⁄ bottom reinforcement, ⁄ , so specify 3 #7 bars along the bottom of the section.

7-9


b)

Design the spandrel beam between columns A1 and A2 for bending, shear, and torsion. Check that all of the appropriate ACI Code requirements for strength, minimum reinforcement area, and reinforcement spacing are satisfied.

Step 1: Determine Mu, Vu, and Tu: Dead weight from spandrel beam, per foot of beam: 12 in  24 in lb  150 lb 3  300 2 ft ft 144 in 2 ft Superimposed dead load, per foot of spandrel beam: 1 ft  20 lb 2  20 lb/ft ft To simplify the design, we will not reduce the live load. Therefore, the live load per foot of beam: 1 ft  50 lb 2  50 lb/ft ft Dead weight from joist, applied as a point load at midspan: 12 in  18 in  150 lb 3  14.5 ft  3.26 k 2 ft in 144 ft 2 Dead weight from slab, applied as a point load at midspan: 6 in  12 ft  150 lb 3  14.5 ft  13.1 k ft 12 in ft Superimposed dead load, applied as a point load at midspan: 12 ft 14.5 ft  20 lb 2  3.48 k ft Since the live load is not reduced, the live load per foot of beam is: 12 ft 14.5 ft  50 lb 2  8.7 k ft So, using wu  1.2DL  1.6LL , the factored load is:

wu  1.2   0.300 k/ft  0.02 k/ft   1.6   0.05 k/ft   0.464 k/ft , per foot of beam

wu , po int  1.2   3.26 k  13.1 k  3.48 k   1.6  8.7 k   37.7 k , applied as a point load n

 24 ft -16 in  22.67 ft

With that, we can determine the design moments, but structural analysis software must be used since the ACI Moment Coefficients cannot be applied when not all loads are distributed. Input the structural model and applied loads using the appropriate pattern loading for the live loads. M u  102 k-ft At column A-1: M u  128 k-ft At midspan: M u  139 k-ft At column A-2: The design shear at d away from the supports is: At column A-1: w   2d  wu , po int 0.464 k/ft   22.67 ft  2  21.5 in  37.7 k Vu  u n     23.3 k 2 2 2 2 Vu  0 k At midspan: Vu  1.15  Vu , A1  26.8 k At column A-2:

7-10


To calculate the design torsion, we first have to determine the moment and shear that the joist is applying to the edge of the spandrel beam. wu  1.2   0.225 k/ft  0.900 k/ft  0.240 k/ft   1.6   0.600 k/ft   2.60 k/ft n

 30 ft -12 in  29 ft

2.6 k/ft   29 ft  w 2 Mu   u n    91.1 k-ft 24 24 Vu  wu , po int  37.7 k 2

Thus our design torsion, at d away from the ends of our beam, is: 6 in 6 in Tu  M u  Vu   91.1 k-ft  37.7 k   110 k-ft 12 in/ft 12 in/ft Step 2: Determining the area of longitudinal steel required for flexure. Mu 102 k-ft  12 in/ft At column A-1: R   245 psi 2 2  bd 0.9  12 in   21.5 in  Refer to Table A-3. For f’c = 4,500 psi, and R = 245 psi,   0.0042 . So, at column A-1, As  bd  0.0042 12 in  21.5 in  1.08 in 2 . Mu 128 k-ft  12 in/ft At midspan: R   308 psi 2 2  bd 0.9  12 in   21.5 in  Refer to Table A-3. For f’c = 4,500 psi, and R = 308 psi,   0.0054 . So, at midspan, As  bd  0.0054 12 in  21.5 in  1.39 in 2 . Mu 139 k-ft  12 in/ft At column A-2: R   334 psi 2 2  bd 0.9  12 in   21.5 in  Refer to Table A-3. For f’c = 4,500 psi, and R = 334 psi,   0.0058 . So, at column A-2, As  bd  0.0058 12 in  21.5 in  1.50 in 2 .





 396 in 2    0.75  1.0  4,500 psi    108 in  

Since Tu  Tth , torsion must be considered in this design.

7-11



2

   73,000 lb-in  6.09 k-ft  


Step 4: Since the torsion resisted by this edge beam is not required to maintain equilibrium of the structure, we have a case of compatibility torsion. Therefore we can likely reduce our Tu to the following:  396 in 2 2   Acp 2    292,000 lb-in  24.4 k-ft Tu ,comp   4 f 'c   0.75  4  1.0  4,500 psi   Pcp   108 in     





Since Tu  Tu ,comp , we can reduce our design torsion to 24.4 k-ft. Since our design torsion for the spandrel beam is being reduced, it is necessary to redistribute the design moments for the joist that frames into the spandrel beam. See chapter 7 for further discussion of why this is required. Step 5: Determine whether the section is large enough to resist the combined actions of shear and torsion. First assume that a closed #4 stirrup will be used in the web of this beam. Aoh   24 in  2 1.5 in  0.5 in   12 in  2 1.5 in  0.5 in   174 in 2

Ph  2   24 in  2 1.5 in  0.5 in  12 in  2 1.5 in  0.5 in   58 in From Eq. (7-33): 2

2

 Vu   Tu Ph   V     c  8 f 'c     2   bw d   1.7 Aoh   bw d  2

   2 4,500 psi  12 in  21.5 in  25.6 k    24.4 k-ft  58 in    0.75   8 4,500 psi    12 in  21.5 in   2   2  12 in  21.5 in   1.7  174 in     2





0.344 ksi  0.503 ksi

The section is sufficiently large. Step 6: Determine the area of stirrups required to resist Vu: Vc  2 f 'c bw d  2 4,500 psi 12 in  21.5 in  34.6 k At column A-1: At column A-2:

Vs  0 k V 25.6 k Vs  u  Vc   34.6 k  0 k  0.75

Step 7: Determine the area of stirrups required to resist Tu: T 24.4 k-ft Tn  u   32.5 k-ft 0.75 0.75 From Eq. (7-24), using Ao  0.85 Aoh : At Tn 32.5 k-ft  12 in/ft   s 2 Ao f yt cot  2  0.85  174 in 2  40 ksi  cot 45

 

 0.0330

in 2 in

Step 8: Evaluate the total required area of stirrups, and select spacing: in 2 in 2 For strength: 2  0.0330  0.066 in in Minimum required:

0.75 f c'  bw

f yt The strength requirement governs here.



50.3 psi  12 in in 2  0.015 40,000 psi in

7-12


Now determine stirrup size and spacing: 12 in   smax   Ph / 8  58 in / 8  7.25 in d / 2  10.75 in (for shear)  2  0.2 in 2  6.1 in in 2 0.066 in 2  0.31 in 2 If we select #5 stirrups, s   9.4 in in 2 0.066 in Select closed #4 stirrups at 6.0 in spacing at the ends of the beam. While in some cases it is possible that stirrups are not required along the full length of the beam, the torsion in this case is constant along the length of the beam, as is the shear. Therefore, closed #4 stirrups are required along the full length of the beam.

If we select #4 stirrups, s 

Step 9: Finalize the design of the longitudinal reinforcement: Determine the need for longitudinal reinforcement resisting torsion:  in 2  2  At   f yt  2 2 A  P cot   0.0330 For strength:    58 in   1.0  1.28 in l  s  h  f  in  3    yl  

Al ,min 

Minimum required:

5 f 'c Acp f yl



f yt At Ph s f yl

in 2 in 2 25bw in 2 A  Since  t   0.0330 , we must use 0.0330   0.0075 in in f yt in  s ,min

Al ,min

5 4,500 psi  396 in 2 in 2 2   0.0330  58 in   0.94 in 2 60 ksi in 3

Use Al  1.28 in 2 Longitudinal bars are required in the corner of each stirrup. Also, longitudinal bars must be spaced no more than 12 in apart around the perimeter of the section, so a bar is needed in the middle of each vertical face. So 6 bars are required. Area / bar  1.28 in 2 / 6 bars  0.21 in 2 / bar

-At column A-1, use #5 bars in the bottom corners and halfway up the vertical face of the beam. As top reinforcement, As / bar  1.08 in 2  2  0.21 in 2 / 3  1.50 in 2 / 3  0.50 in 2 , so specify 3 #7





bars along the top of the section. -At midspan, use #5 bars in the top corners and halfway up the vertical face of the beam. As bottom reinforcement, As / bar  1.39 in 2  2  0.21 in 2 / 3  1.81 in 2 / 3  0.60 in 2 , so specify 3 #8





bars along the bottom of the section. -At column A-2, use #5 bars in the bottom corners and halfway up the vertical face of the beam. As top reinforcement, As / bar  1.50 in 2  2  0.21 in 2 / 3  1.92 in 2 / 3  0.64 in 2 , so specify 3 #8





bars along the top of the section.

7-13


7-4

The two parts of this problem refer to the floor plan shown in Fig. P7-3. Assume that the entire floor system is constructed with sand light-weight concrete that has a compressive strength, f’c = 4,000 psi. Also assume that the longitudinal steel has a yield strength of fy = 60 ksi and that the transverse steel has a yield strength of fyt = 60 ksi. a)

Design the spandrel beam between columns B1 and C1 for bending, shear, and torsion. Check all of the appropriate ACI Code requirements for strength, minimum reinforcement area, and reinforcement spacing are satisfied.

Step 1: Determine Mu, Vu, and Tu: Note that this problem is very similar to the previous problem. One noticeable change is the use of lightweight concrete, which will affect the dead weight used in the calculation of the design loads. Although no guidance is given on the density of the sand-lightweight concrete used in this problem, any reasonable assumption would be acceptable. Here a density of 120 lb/ft 3 is assumed. Dead weight from beam, per foot of beam: 12 in  18 in lb lb  120 3  180 2 ft ft 144 in 2 ft Dead weight from slab, per foot of beam: 6 in  6.5 ft lb lb  120 3  390 in ft ft 12 ft Superimposed dead load, per foot of beam: lb lb 6.5 ft  20 2  130 ft ft To simplify the design, we will not reduce the live load. Therefore, the live load per foot of beam: lb lb 6.5 ft  50 2  325 ft ft So, using wu  1.2DL  1.6LL , the factored load is, per foot of beam:

wu  1.2   0.180 k/ft  0.390 k/ft  0.130 k/ft   1.6   0.325 k/ft   1.36 k/ft

With that, we can determine the design moments using ACI Moment Coefficients: At columns: (

At midspan: ( )

)

(

(

)

)

Note that n is taken as the average of the adjoining clear spans for calculating the negative moment at the columns.

7-14


The design shear is: At midspan:

At d away from the column face: ⁄

(

) ⁄

(

)

Note that n is taken as the clear span length of the interior span when calculating the shear acting on the face of the support. Do not reduce this value to the n used for calculating the negative moment at column B-1. To calculate the design torsion, we first have to determine the moment and shear that the slab is applying to the edge of the beam. wu  1.2   0.060 k/ft  0.020 k/ft   1.6   0.050 k/ft   0.176 k/ft n

 12 ft -12 in  11 ft

0.176 k/ft  11 ft  wu n 2   0.887 k-ft 24 24 w 0.176 k/ft 11 ft Vu  u n    0.968 k 2 2 Therefore, the torsion applied to our beam by the slab is: 6 in t  0.887 k-ft  0.968 k   1.37 k-ft/ft 12 in/ft And thus our design torsion, at d away from the ends of our beam, is: ( ) ( ) 2

Mu  

Step 2: Determining the area of longitudinal steel required for flexure. At the column:

Refer to Table A-3. For f’c = 4,000 psi and R = 204 psi,

. So, at the column,

At midspan:

Refer to Table A-3. There are no R values below 190 psi, which corresponds to the minimum reinforcement ratio. Therefore,

7-15






 396 in 2    0.75  0.85  4,000 psi    108 in  



2

   58,500 lb-in  4.88 k-ft  

Since Tu  Tth , torsion must be considered in this design. Step 4: Since the torsion resisted by this edge beam is not required to maintain equilibrium of the structure, we have a case of compatibility torsion. Therefore we can likely reduce our Tu to the following:  396 in 2 2   Acp 2    234,000 lb-in  19.5 k-ft Tu ,comp   4 f 'c   0.75  4  0.85  4,000 psi   Pcp   108 in      Unfortunately, since Tu  Tu ,comp , we cannot reduce our design torsion. Since our design torsion is





not being reduced, no redistribution of design moments is required in the adjoining slab. Step 5: Determine whether the section is large enough to resist the combined actions of shear and torsion. First assume that a closed #4 stirrup will be used in the web of this beam. Aoh   24 in  2 1.5 in  0.5 in   12 in  2 1.5 in  0.5 in   174 in 2

Ph  2   24 in  2 1.5 in  0.5 in  12 in  2 1.5 in  0.5 in   58 in

From Eq. (7-33): √( √(

) )

( (

)

(

)

(

The section is sufficiently large. Step 6: Determine the area of stirrups required to resist Vu: √ √

7-16

√ ) √

√

)


Step 7: Determine the area of stirrups required to resist Tu: From Eq. (7-24), using Ao  0.85 Aoh : (

)


Minimum required: √

The strength requirement governs here. Now determine stirrup size and spacing: 12 in   smax   Ph / 8  58 in / 8  7.25 in d / 2  10.75 in (for shear)  If we select #3 stirrups,

If we select #4 stirrups,

Select closed #3 stirrups at 7 in. spacing at the ends of the beam. It is also possible that stirrups are not required along the full length of this beam. For torsion, determine where Tu  Tth /  . By similar triangles, this occurs at 75 in. away from the face of the column. However, torsional reinforcement must be continued for (d  bt )  (21.5 in  12 in)  33.5 in past this theoretical point. Therefore, no torsional reinforcement is required beyond 108.5 in. away from the face of the column. Since Vu  Vc / 2 at this point, no shear reinforcement is required beyond this point either. Final design of transverse reinforcement: Using closed #3 stirrups, One stirrup at 3.5 in from either column face, followed by stirrups spaced at 7 in. until beyond 109 in. from the face of the column.

7-17


Step 9: Finalize the design of the longitudinal reinforcement: Determine the need for longitudinal reinforcement resisting torsion: For strength:

Minimum required: 5 f 'c Acp At f yt Al ,min   Ph f yl s f yl in 2 in 2 25bw in 2 A  Since  t   0 , we must use 0.0075   0.0075 in in f yt in  s ,min

5 4,000 psi  396 in 2 in 2  0.0050  58 in  1.0  1.80 in 2 60 ksi in Use Al  1.80 in 2 Al ,min 

Longitudinal bars are required in the corner of each stirrup. Also, longitudinal bars must be spaced no more than 12 in apart around the perimeter of the section, so a bar is needed in the middle of each vertical face. So 6 bars are required. Area / bar  1.80 in 2 / 6 bars  0.30 in 2 / bar

-At the columns, use #5 bars in the bottom corners and halfway up the vertical face of the beam. ( )⁄ As top reinforcement, ⁄ , so specify 3 #7 bars along the top of the section. -At midspan, use #5 bars in the top corners and halfway up the vertical face of the beam. As ( )⁄ bottom reinforcement, ⁄ , so specify 3 #7 bars along the bottom of the section.

7-18


b)

Design the spandrel beam between columns A1 and A2 for bending, shear, and torsion. Check that all of the appropriate ACI Code requirements for strength, minimum reinforcement area, and reinforcement spacing are satisfied.

Step 1: Determine Mu, Vu, and Tu: Again, assume the density of the concrete is 120 lb/ft 3 . Dead weight from beam, per foot of beam: 12 in  24 in lb lb  120 3  240 2 ft ft 144 in 2 ft Superimposed dead load, per foot of spandrel beam: 1 ft  20 lb 2  20 lb/ft ft To simplify the design, we will not reduce the live load. Therefore, the live load per foot of beam: 1 ft  50 lb 2  50 lb/ft ft Dead weight from joist, applied as a point load at midspan: 12 in  18 in lb  120 3  14.5 ft  2.61 k 2 ft 144 in 2 ft Dead weight from slab, applied as a point load at midspan: 6 in  12 ft lb  120 3  14.5 ft  10.4 k in ft 12 ft Superimposed dead load, applied as a point load at midspan: lb 12 ft  14.5 ft  20 2  3.48 k ft To simplify the design, we will not reduce the live load. Therefore, the live load per foot of beam: lb 12 ft  14.5 ft  50 2  8.7 k ft So, using wu  1.2DL  1.6LL , the factored load is:

wu  1.2   0.240 k/ft  0.02 k/ft   1.6  50 k/ft   0.392 k/ft , per foot of beam

wu , po int  1.2   2.61 k  10.4 k  3.48 k   1.6  8.7 k   33.7 k , applied as a point load

 24 ft -16 in  22.67 ft With that, we can determine the design moments, but structural analysis software must be used since the ACI Moment Coefficients cannot be applied when not all loads are distributed. M u  90.4 k-ft At column A-1: M u  114 k-ft At midspan: M u  123 k-ft At column B-1: The design shear at d away from the supports is: At column A-1: w   2d  wu , po int 0.392 k/ft   22.67 ft  2  21.5 in  33.7 k Vu  u n     20.6 k 2 2 2 2 Vu  0 k At midspan: Vu  1.15  Vu , A1  23.7 k At column B-1: n

7-19


To calculate the design torsion, we first have to determine the moment and shear that the beam is applying to the edge of the beam. wu  1.2   0.180 k/ft  0.720 k/ft  0.240 k/ft   1.6   0.600 k/ft   2.33 k/ft n

 30 ft -12 in  29 ft

2.33 k/ft   29 ft  w 2 Mu   u n    81.6 k-ft 24 24 Vu  wu , po int  33.7 k 2

Thus our design torsion, at d away from the ends of our beam, is: 6 in 6 in Tu  M u  Vu   81.6 k-ft  33.7 k   98.5 k-ft 12 in/ft 12 in/ft Step 2: Determining the area of longitudinal steel required for flexure. Mu 90.4 k-ft  12 in/ft At column A-1: R   217 psi 2 2  bd 0.9  12 in   21.5 in  Refer to Table A-3. For f’c = 4,000 psi, and R = 217 psi,   0.0037 . So, at column A-1, As  bd  0.0037 12 in  21.5 in  0.95 in 2 . Mu 114 k-ft  12 in/ft At midspan: R   274 psi 2 2  bd 0.9  12 in   21.5 in  Refer to Table A-3. For f’c = 4,000 psi, and R = 274 psi,   0.0048 . So, at midspan, As  bd  0.0048 12 in  21.5 in  1.24 in 2 . Mu 123 k-ft  12 in/ft At column A-2: R   296 psi 2 2  bd 0.9  12 in   21.5 in  Refer to Table A-3. For f’c = 4,000 psi, and R = 296 psi,   0.0052 . So, at column A-2, As  bd  0.0052 12 in  21.5 in  1.34 in 2 .





 396 in 2    0.75  0.85  4,000 psi    108 in  

Since Tu  Tth , torsion must be considered in this design.

7-20



2

   58,500 lb-in  4.88 k-ft  


Step 4: Since the torsion resisted by this edge beam is not required to maintain equilibrium of the structure, we have a case of compatibility torsion. Therefore we can likely reduce our Tu to the following:  396 in 2 2   Acp 2    234,000 lb-in  19.5 k-ft Tu ,comp   4 f 'c   0.75  4  0.85  4,000 psi    Pcp   108 in     





Since Tu  Tu ,comp , we can reduce our design torsion to 19.5 k-ft. Since our design torsion for the spandrel beam is being reduced, it is necessary to redistribute the design moments for the joist that frames into the spandrel beam. See chapter 7 for further discussion of why this is required. Step 5: Determine whether the section is large enough to resist the combined actions of shear and torsion. First assume that a closed #4 stirrup will be used in the web of this beam. Aoh   24 in  2 1.5 in  0.5 in   12 in  2 1.5 in  0.5 in   174 in 2

Ph  2   24 in  2 1.5 in  0.5 in  12 in  2 1.5 in  0.5 in   58 in From Eq. (7-33): √( √(

) )

( (

)

(

)

(

√ ) √

√

The section is sufficiently large. Step 6: Determine the area of stirrups required to resist Vu: √ √ At column A-1:

At column A-2:

Step 7: Determine the area of stirrups required to resist Tu: T 19.5 k-ft Tn  u   26.0 k-ft 0.75 0.75 From Eq. (7-24), using Ao  0.85 Aoh : At Tn 26.0 k-ft  12 in/ft   s 2 Ao f yt cot  2  0.85  174 in 2  60 ksi  cot 45

 

7-21

 0.0176

in 2 in

)



Minimum required:

50bw 50  12 in in 2   0.010 f yt 60,000 psi in

The strength requirement governs here. Now determine stirrup size and spacing: 12 in   smax   Ph / 8  58 in / 8  7.25 in d / 2  10.75 in (for shear)  If we select #3 stirrups, If we select #4 stirrups,

Select closed #3 stirrups at 5 in. spacing for the full length of the beam. While in some cases it is possible that stirrups are not required along the full length of the beam, the torsion in this case is constant along the length of the beam, as is the shear. Therefore, closed #3 stirrups are required along the full length of the beam. Final design of transverse reinforcement: Using closed #3 stirrups, One stirrup at 2.5 in from either column face, followed by stirrups spaced at 5 in.

Step 9: Finalize the design of the longitudinal reinforcement: Determine the need for longitudinal reinforcement resisting torsion:  in 2   At   f yt  2 2 A  P cot   0.0176 For strength:    58 in  1.0  1.0  1.02 in l  s  h  f  in     yl   Minimum required:

Al ,min 

5 f 'c  Acp f yl



f yt At Ph s f yl

in 2 in 2 25bw in 2 A  Since  t   0.0176 , we must use 0.0176   0.0050 in in f yt in  s ,min

Al ,min

5 4,000 psi  396 in 2 in 2   0.0176  58 in  1.0  1.07 in 2 60 ksi in

Use Al  1.07 in 2

7-22


Longitudinal bars are required in the corner of each stirrup. Also, longitudinal bars must be spaced no more than 12 in apart around the perimeter of the section, so a bar is needed in the middle of each vertical face. So 6 bars are required. Area / bar  1.07 in 2 / 6 bars  0.18 in 2 / bar

-At column A-1, use #4 bars in the bottom corners and halfway up the vertical face of the beam. As top reinforcement, As / bar  0.95 in 2  2  0.18 in 2 / 3  1.31 in 2 / 3  0.437 in 2 , so specify 3





#6 bars along the top of the section. -At midspan, use #4 bars in the top corners and halfway up the vertical face of the beam. As bottom reinforcement, As / bar  1.24 in 2  2  0.18 in 2 / 3  1.60 in 2 / 3  0.53 in 2 , so specify 3





#7 bars along the bottom of the section. -At column A-2, use #4 bars in the bottom corners and halfway up the vertical face of the beam. As top reinforcement, As / bar  1.34 in 2  2  0.18 in 2 / 3  1.7 in 2 / 3  0.57 in 2 , so specify 3 #7





bars along the top of the section.

7-23


Chapter 8 8-1

Figure P8-1 shows a cantilever beam with containing three No. 7 bars that are anchored in the column by standard hooks. (normalweight) and . If the steel is stressed to at the face of the column, can these bars: (a)

be anchored by hooks into the column? The clear cover to the side of the hook is 2 ¾ in. The clear cover to the bar extension beyond the bend is 2 in. The joint is enclosed by ties at 6 in. o.c.

Except for the side cover of 2 ¾ in., a clear cover of 2 in. is assumed throughout the beam and joint. See Fig. S8-1 below. 18 in.

48 in.

3 #7 2 in. (clear cover)

7.5 in.

2 in. (clear cover) 2 in.

6 in.

18 in.

#3, double legs Side cover = 2¾ in. for columns and 2 in. for beams; Beam width = 12 in.

Fig. S8-1 1. Check if a 90 standard hook can be used  Length of the hook tail including the bend: 12db  3db  0.5db  15.5db  15.5  0.875 in.  13.6 in.  Available vertical room for the hook tail: 18 in.  2  2 in.  14 in.  13.6 in. , OK. 2. Required development length of 90 standard hooks:  Basic development length of 90 standard hooks: 0.02 e f y 0.02  1 60000 db   0.875  14.8 in. hb   f c 1 5000 Note:

Coating factor  e  1 for uncoated reinforcement (assumed) Lightweight-aggregate-concrete factor   1 for normal-weight concrete

8-1


 Applicable factors: ACI Code Section 12.5.3(a) applies (factor of 0.7) because: bar size smaller than bar No. 11  side cover: 2 3/4 in.  2 1/2 in.  tail cover: 2 in.  2 in.  ACI Code Section 12.5.3(b) does not apply because: tie spacing: 6 in.  3db  2.6 in. ACI Code Section 12.5.3(d) does not apply because f y  60000 psi

 dh  hb  0.7  14.8 in.  0.7  10.4 in.  Minimum development length dh ,min  max(8db ,6 in.)  max(8  0.875 in., 6 in.)  7 in.  10.4 in. , OK. 3. Available horizontal room for 90 standard hook: 18 in.  2 in.  16 in.  10.4 in. , OK. Conclusion: The bars can be anchored in the column by using 90˚ standard hook. (b)

be developed in the beam? The bar ends 2 in. from the end of the beam. The beam has No. 3 double-leg stirrups at 7.5 in.

1. Available room in the beam for straight anchorage: 48 in.  2 in.  46 in. 2. Required development length of straight anchorage conforming to ACI Code Section 12.2.2.:  Determine which formulas to use in the Table 8-1 12 in.  2  2 in.  3  0.875 in.   2.7 in.  2db  1.75 in. Clear spacing: 2  Clear cover: 2 in.  db  0.875 in. Bar #7    e t f y 1.3  1 60000  hb  db   0.875  48.3 in.  46 in. , NG. 20 f c 20  1 5000 Bar-location factor  t  1.3 is used because the bars are top reinforcement with a depth of fresh concrete below: 18 in.  2 in.  0.875 in.  15.1 in.  12 in. Per ACI Code Section 12.2.2, the bars cannot be developed in the beam by using straight anchorage. This problem can be solved by reducing the bar size. However, ACI Code Section 12.2.3 allows another method to calculate the development length which is likely to yield a shorter development length. Let try this. Note:

3. Required development length of straight bars conforming to ACI Code Section 12.2.3.  Bar-size factor  s  1 for #7 bars

8-2


 Bar-spacing factor cb One-half center-to-center spacing of the bars: 12 in.  2  2 in.  0.875 in. 0.5   1.78 in. 2 Smallest distance from beam surface to centers of bars: 0.875 in. 2 in.   2.44 in. 2 cb  min(1.78 in.,2.44 in.)  1.78 in.  Transverse reinforcement index 2 40 Atr 40   2  0.11 in.  Ktr    0.39 in. sn 7.5 in.  3  Required development length per ACI Code Section 12.2.3 3 f y  t e s 3  60000 1.3  1 1 db   0.875  29.2 in.<46 in. d  40 f c cb  Ktr 40  1 5000 1.78  0.39 0.875 db

cb  Ktr  2.48  2.5,OK. ) db Per ACI Code 12.2.3, the bars can be developed in the beam by using straight anchorage. (Note that

8-2

Give two reasons why the tension development length is longer than the compression development length. 1. A bar stressed in compression transfers some of its force to the concrete by bearing on the end of the bars. 2. A bar stressed in tension transfers its tensile stress into concrete. As a result the concrete is cracked and “in-and-out” bond stresses exist. In such a case there are localized bond stresses which are several times greater the average bond stress. A bar stressed in compression transfers its stress into concrete which is compressed and hence un-cracked. There are no “in-and-out” bond stresses in such a case.

8-3

Why do bar spacing and cover to the surface of the bar affect bond strength? The lugs on deformed bars transfer forces to the concrete. The radial component of these lug forces causes a tensile stress in an annulus of concrete around the bar. The thicker the wall of this annulus the lower the tensile stresses are in it. The thickness of the wall is governed by the minimum distance to the surface of the concrete or to the next bar. Thus, the larger the cover and bar spacing are, the larger the bond stresses can be developed.

8-3


8-4

A simply supported rectangular beam with and and No. 3 Grade 40 stirrups at , spans 14 ft and supports a total factored uniform load of 6.5 kips/ft, including its own dead load. It is built of 4000 psi light-weight concrete and contains 2 No. 10 Grade 60 bars that extend 5 in. past the centers of the supports at each end. Does this beam satisfy ACI Section 12.11.3? If not, what is the largest size bars which can be used? The question asks to check if

d



Mn  Vu

a

at the support.

1. Calculation of d .  Bar-location factor  t  1 for bottom bars  Coating factor  e  1 for uncoated reinforcement (assumed)  Bar-size factor  s  1 for #10 bars  Lightweight-aggregate-concrete factor for light-weight concrete  Bar-spacing factor cb  min 2.5 in.,0.5  14 in.  2  2.5 in.  2.5 in. 

Transverse reinforcement index K tr . ( )



Calculate the development length d



3 fy

 t e s

40 f c cb  Ktr db

(Note that

2. Calculation of

 db 

3  60000 1 1 1 1.27  50.2 in. 2.5  0.55 40  0.75 4000 1.27

cb  Ktr  2.4  2.5,OK. ) db

Mn  Vu

a

.

( ⁄ ) 6.5 k/ft  14 ft Vu   45.5 kips 2



8-5

d



Mn  Vu

a

(

⁄ )

So, the use of 2 #10 bars satisfies ACI Code Section 12.11.3.

Why do ACI Section 12.10.3 and 12.12.3 require that bars extend off points?

past their cut-

Inclined cracking due to shear increases the tension in the flexural reinforcement at all points except the points of maximum moments. As a result, the tensile force in the flexural reinforcement computed from the moment at a given section actually exists at a point about 0.75d to d from that point in the direction of decreasing moment.

8-4


8-6

Why does ACI Section 12.10.2 define “points within the span where adjacent reinforcement terminates” as critical sections for development of reinforcement in flexural members? If flexural reinforcement is cut off according to the moment diagram, the flexural cut-off point is the point in the beam where the remaining steel not cut off is just adequate for the moment if stressed to f y . Due to the effect of shear, this point actually occurs about d farther away and hence this point is critical.

A rectangular beam with cross section , , and supports a total factored load of 3.9 kips/ft, including its own dead load. The beam is simply supported with a 22-ft span. It is reinforced with 6 No. 6 Grade 60 bars, two of which are cut off between midspan and the support and four of which extend 10 in. past the centers of the supports. . The beam has No. 3 stirrups satisfying ACI Code Sections 11.4.5 and 11.4.6. CL

h = 24 in.

21.5 in. in. d =21.7

cut-off point

4 bars #6

10 in. in. 12

6 bars #6 11 ft.

ℓd = 28.5 in.

ℓd = 28.5 in.

øMn = 163 160 k-ft. k-ft

flexural cut-off point cut-off point 12 10 in.

35.6 37 in.in.

factored moment diagram M

M = 236 k-ft.

236 k-ft. k-ft øMn = 238

160 k-ft M = 163 k-ft.

8-7

21.7in. in. 21.5

Fig. S8-2 ( ), where x is Plot to scale the factored moment diagram. ( ) the distance from the support and is the span. 3.9 11 ft Maximum moment at the midspan: M  x  / 2    22 ft  11 ft   236 k-ft 2 The factored moment diagram is shown in Fig. S8-2. (a)

8-5


(b)

Plot a resisting moment diagram and locate the cut-off point for the two cutoff bars.

1. Moment capacity of beam sections with 6 #6 bars 6  0.44 in.2   60000 psi As f y  a   3.3 in. 0.85 f cb 0.85  4000 psi  14 in. ( ) ( ⁄ ) ( a 3.3 in. c   3.9 in. ( 1  0.85 for fc  4000 psi ) 1 0.85 (

)

2. Moment capacity of beam sections with 4 #6 bars As f y  4  0.44 in.2   60000 psi  2.2 in. a  0.85 f cb 0.85  4000 psi  14 in. ( ) ( ⁄ ) ( a 2.2 in. c   2.6 in. ( 1  0.85 for fc  4000 psi ) 1 0.85 (

⁄ )

⁄ )

)

3. Development length for straight bars. ACI Code Section 12.2.2 will be used to determine the development length.  Minimum clear spacing between bars: 14 in.  2  2.3 in.  5  0.75 in.  1.1 in.  db  0.75 in. 5  Since the shear reinforcement is provided such that ACI Code Sections 11.5.4 and 11.5.5.3 will be satisfied, shear reinforcement is not less than the minimum code requirement.  Bars used are #6  e t f y 1 1 60000  d db   0.75  28.5 in. 25 f c 25  1 4000 4. Locate the cut-off point of the 2 #6 bars (see Fig. S8-2.)  Determine the flexural cut-off point of the 2 #6 bars. This point is the intersection of the factored moment diagram and the line of moment capacity  M n (4#6) , i.e.: M ( x)   M n (4#6) (

)

8-6




Extend the flexural cut-off point a distance d toward the left support to take into account the shear effect. The distance from the left support to the cut-off point is

The resisting moment diagram is shown in Fig. S8-2.

8-8

Why does ACI Code Section 12.10.5 require extra stirrups at bar cut-off points in some cases? A severe discontinuity in longitudinal bar stresses exists in the region of a cut-off point in a zone of flexural tension. This causes a reduction in the inclined cracking load in that region. To compensate, more stirrups are required.

The beam shown in Fig. P8-9 is built of 4000 psi normal-weight concrete and Grade 60 steel. The effective depth The beam supports a total factored uniform load of 5.25 kips/ft, including its own dead load. The frame is not part of the lateral load-resisting system for the building. Use Figs. A-1 to A-4 to select cut-off points in Problems 8-9 to 8-11. 8-9

Select cut-off points for span A-B based on the following requirements: (a) Cut off two No. 6 positive moment bars when no longer needed at each end. Extend the remaining bars into the columns. 1. Development length for bottom #6 and #7 bars From the effective depth of 18.5 in, assume a concrete cover of 2.5 in. to centers of the bars.  Clear spacing: 

Shear reinforcement: Av 0.22 in.2 50bw 50  12 A    0.028 in.2 /in.   v     0.015 in.2 /in. s 8 in. f yt 40000  s min (Note that 0.75 fc  47.4 psi  50 psi, use 50 psi )

The development length is computed following Case 1 d

 #6, bottom  

d

 #7, bottom  

 e t f y

25 f c

 e t f y 20 f c

db  db 

1 1 60000 25  1 4000 1 1 60000 20  1 4000

 0.75  28.5 in.  0.875  41.5 in.

2. Locate the cutoff point for the #6 bars After the 2 #6 bars are cutoff, the remaining #7 bars will provide an of , which is 0.577 times at midspan. Therefore, the #6 bars can be cutoff when the moment demand is less than 0.577 at midspan. From Fig. A-2, this occurs at from the face of the exterior support. The actual cutoff has to be the larger of and beyond the theoretical cutoff point.

8-7


and , so the actual cutoff is the face of the exterior support.

from

3. Verify that the distance from midspan to the actual cutoff point is larger than the development length.

Therefore, the cutoff for the 2 #6 bars is 34 in. from the face of the exterior support. 4. Do the same calculations for the interior support (B) for this exterior span: After the 2 #6 bars are cutoff, the remaining #7 bars will provide , which is 0.577 times at midspan. Therefore, the #6 bars can be cutoff when the moment demand is less than 0.577 at midspan. From Fig. A-2, this occurs at from the face of the interior support. The actual cutoff has to be the larger of and beyond the theoretical cutoff point. and , so the actual cutoff is from the face of the interior support. Verify that the distance from midspan to the actual cutoff point is larger than the development length.

Therefore, the cutoff for the 2 #6 bars can be 36 in. from the face of the interior support. To limit the potential for errors on the jobsite, given the very small difference between exterior and interior support cutoff locations, we will select the more conservative value for both supports of the exterior span. Cutoff 2 #6 bars at 34 in. from the face of the columns in the exterior span. 5. Verify that enough bars are extended into the column: ( ) ( ) This requirement is satisfied. Also, specify that the #7 bars extend 6 in. into the columns. 6. With the #7 bars extended 6 in. into the columns, verify that the #7 bars are sufficiently developed. To check this, ensure that the continuous #7 bars extend more than past the theoretical cutoff point of the #6 bars. ( ) This requirement is satisfied. 7. Check if

d



Mn  Vu

(

a

at the positive moment point of inflection near end A.

⁄ )

(

8-8

⁄ )


From Fig. A-2, the positive moment point of inflection closest to end A is 0.1 n  2 ft  24 in. 5.25 k/ft  20 ft Vu   5.25k/ft  2 ft  42 kips 2 ( ) { }

Mn  a at the positive moment point of inflection near end B Vu Everything remains the same as for end A, except the point of inflection. From Fig. A-2, the positive moment point of inflection closest to end B is M 0.104 n  2.08 ft. The change is not significant, so d  n  a is ensured. Vu

8. Check if

d



9. Check whether ACI Code Section 12.10.5.1 is satisfied. ⁄ ⁄ ⁄ √ √ (

)

(

)

Therefore, we must add stirrups near the point of termination for the #6 bars. Additional stirrup area required is:

If we keep the same #3 U-shaped stirrups, . Therefore, Specify a stirrups spacing of 3 in. within d of the #6 bar termination point. (b)

Extend all negative moment bars past the negative moment point of inflection before cutting them off.

1. Development length for # 6 and #7 top bars Note that the clear spacing of top #7 bars is: The development length is computed following Case 1, with  t  1.3 for top bars. d

 #6, top  

d

 #7, top  

 e t f y

25 f c

 e t f y 20 f c

db  db 

1 1.3  60000 25  1 4000 1 1.3  60000 20  1 4000

8-9

 0.75  37.0 in.  0.875  54.0 in.


2. Determine the cut-off point conforming to reinforcement continuity for negative reinforcement near column A  Since all reinforcement is required to extend past the point of inflection, the A (face) requirement of greater than s to extend past the point of inflection is 3 automatically satisfied.  All reinforcement has to extend past the inflection point a length: {

} ⁄  Negative moment point of inflection from face of column A is 0.164 n  3.28 ft = 39.4 in.  Total length from face of column A to the reinforcement cut-off point is: ( ) Choose 58 in. 3. Determine the cut-off point conforming to reinforcement continuity for negative reinforcement near column B  Since all reinforcement is required to extend past the point of inflection, the A (face) requirement of greater than s to extend past the point of inflection is 3 automatically satisfied.  All reinforcement has to extend past the inflection point a length: {

} ⁄  Negative moment point of inflection from face of column B is 0.24 n  4.8 ft = 57.6 in.  Total length from face of column B to the reinforcement cut-off point is: ( ) , so choose 78 in. (c)

Check the anchorage of the negative moment bars at A and modify the bar size if necessary.

Reinforcement continuity and structural integrity require that all negative moment reinforcement be fully anchored into column A. Note that the development length of straight #6 bars is 37.1 in., which far exceeds the available column width of 18 in. A hook must be used for anchorage.  Development length for a standard 90º hook for #6 bars: 0.02 e f y 0.02  1 60000 db   0.75  14.2 in. hb   f c 1 4000  Applicable factors. The problem does not give sufficient information to decide if reduction factors per ACI Code Section 12.5.3 shall apply. To be conservative, assume that no reduction of hb will be made.  dh  hb  14.2 in.  Minimum development length

8-10


,min  max(8db ,6 in.)  max(8  0.75 in., 6 in.)  6 in.  14.2 in. , OK.  Available horizontal room for the 90º standard hook: 18 in.  2.4 in.  15.6 in.  14.2 in., OK Extend the hook past the column face and to the other side of the column. dh

15 in.

58 in.

78 in.

21 in. 6 in.

34 in.

34 in.

6 in.

Fig. S8-3 8-10

Repeat Problem 8-9(a) and (b) for span B-C (a)

Cut off two No. 6 positive moment bars when no longer needed at each end. Extend the remaining bars into the columns. 1. Development length for bottom #6 and #7 bars From the effective depth of 18.5 in, assume a concrete cover of 2.5 in. to centers of the bars.  Clear spacing: 

Shear reinforcement: Av 0.22 in.2 50bw 50  12 A    0.028 in.2 /in.   v     0.015 in.2 /in. s 8 in. s f 40000  min yt (Note that 0.75 fc  47.4 psi  50 psi, use 50 psi )

The development length is computed following Case 1 d

 #6, bottom  

d

 #7, bottom  

 e t f y

25 f c

 e t f y 20 f c

db  db 

1 1 60000 25  1 4000 1 1 60000 20  1 4000

 0.75  28.5 in.  0.875  41.5 in.

2. Locate the cutoff point for the #6 bars. Both supports are interior supports. After the 2 #6 bars are cutoff, the remaining #7 bars will provide , which is 0.577 times at midspan. Therefore, the #6 bars can be cutoff when the moment demand is less than 0.577 at midspan. From Fig. A-1, this occurs at from the face of the interior support. The actual cutoff has to be the larger of and beyond the theoretical cutoff point. and , so the actual cutoff is from the face of the interior support. 3. Verify that the distance from midspan to the actual cutoff point is larger than the development length.

8-11


Therefore, the cutoff for the 2 #6 bars can be 52 in. from the face of the support. 4. Verify that enough bars are extended into the column: ( ) ( ) This requirement is satisfied. Also, specify that the #7 bars extend 6 in. into the columns. 5. With the #7 bars extended 6 in. into the columns, verify that the #7 bars are sufficiently developed. To check this, ensure that the continuous #7 bars extend more than past the theoretical cutoff point of the #6 bars. ( ) This requirement is satisfied. M 6. Check if d  n  a at the positive moment point of inflection. Vu

(

⁄ )

(

⁄ )

From Fig. A-1, the positive moment point of inflection closest to end A is

(

{

)

}

7. Check whether ACI Code Section 12.10.5.1 is satisfied. ⁄ ⁄ ⁄ √ √ (

)

(

)


If we keep the same #3 U-shaped stirrups, . Therefore, Specify a stirrup spacing of 4 in. within d of the #6 bar termination point. (b)

Extend all negative moment bars past the negative moment point of inflection before cutting them off.

8-12


1. Similar to Problem 8-9(b), extend all negative reinforcement past the negative moment point of inflection a length of 18.5 in. 2. Negative moment point of inflection from face of column B and C is 0.24 n  5.04 ft = 60.5 in. 3. Total length from face of column B to the reinforcement cut-off point is ( ) Select 80 in. See Fig. S8-4 for reinforcement details. 80 in.

80 in.

21 in. 6 in.

52 in.

52 in.

6 in.

Fig. S8-4 8-11

Assume the beam is constructed with 4000 psi light-weight concrete. Select cut-off points for span A-B based on the following requirements: (a)

Extend all negative moment bars at A past the negative moment point of inflection.

1. Development length for #6 top bars Note that the clear spacing of top #6 bars is: The development length is computed following Case 1, with  t  1.3 for top bars.

(

)

√

√

2. Determine the cut-off point conforming to reinforcement continuity for negative reinforcement near column A  Since all reinforcement is required to extend past the point of inflection, the A (face) requirement of greater than s to extend past the point of inflection is 3 automatically satisfied.  All reinforcement has to extend past the inflection point a length: {

} ⁄  Negative moment point of inflection from face of column A is 0.164 n  3.28 ft = 39.4 in.  Total length from face of column A to the reinforcement cut-off point is: ( ) Choose 58 in.

8-13


(b)

Cut off the two No. 6 positive moment bars when no longer needed at each end. Extend the remaining bars into the columns.

1. Development length for bottom #6 and #7 bars From the effective depth of 18.5 in, assume a concrete cover of 2.5 in. to centers of the bars.  Clear spacing: 

Shear reinforcement: Av 0.22 in.2 50bw 50  12 A    0.028 in.2 /in.   v     0.015 in.2 /in. s 8 in. s f 40000  min yt (Note that 0.75 fc  47.4 psi  50 psi, use 50 psi )

The development length is computed following Case 1

(

)

(

)

√

√

√

√

2. Locate the cutoff point for the #6 bars After the 2 #6 bars are cutoff, the remaining #7 bars will provide an of , which is 0.577 times at midspan. Therefore, the #6 bars can be cutoff when the moment demand is less than 0.577 at midspan. From Fig. A-2, this occurs at from the face of the exterior support. The actual cutoff has to be the larger of and beyond the theoretical cutoff point. and , so the actual cutoff is from the face of the exterior support. 3. Verify that the distance from midspan to the actual cutoff point is larger than the development length.

Therefore, the cutoff for the 2 #6 bars is 34 in. from the face of the exterior support. 4. Do the same calculations for the interior support (B) for this exterior span: After the 2 #6 bars are cutoff, the remaining #7 bars will provide , which is 0.577 times at midspan. Therefore, the #6 bars can be cutoff when the moment demand is less than 0.577 at midspan. From Fig. A-2, this occurs at from the face of the interior support. The actual cutoff has to be the larger of and beyond the theoretical cutoff point. and , so the actual cutoff is from the face of the interior support. Verify that the distance from midspan to the actual cutoff point is larger than the development length.

8-14


Therefore, the cutoff for the 2 #6 bars can be 36 in. from the face of the interior support. To limit the potential for errors on the jobsite, given the very small difference between exterior and interior support cutoff locations, we will select the more conservative value for both supports of the exterior span. Cutoff 2 #6 bars at 34 in. from the face of the columns in the exterior span. 5. Verify that enough bars are extended into the column: ( ) ( ) This requirement is satisfied. Also, specify that the #7 bars extend 6 in. into the columns. 6. With the #7 bars extended 6 in. into the columns, verify that the #7 bars are sufficiently developed. To check this, ensure that the continuous #7 bars extend more than past the theoretical cutoff point of the #6 bars. ( ) This requirement is satisfied. 7. Check if

d



Mn  Vu

(

a

at the positive moment point of inflection near end A.

⁄ )

(

⁄ )

From Fig. A-2, the positive moment point of inflection closest to end A is 0.1 n  2 ft  24 in. 5.25 k/ft  20 ft Vu   5.25k/ft  2 ft  42 kips 2 ( ) { }

Mn  a at the positive moment point of inflection near end B Vu Everything remains the same as for end A, except the point of inflection. From Fig. A-2, the positive moment point of inflection closest to end B is M 0.104 n  2.08 ft. The change is not significant, so d  n  a is ensured. Vu 9. Check whether ACI Code Section 12.10.5.1 is satisfied. ⁄ ⁄ ⁄ √ √ 8. Check if

d

(



)

(

)

8-15



If we keep the same #3 U-shaped stirrups, . Therefore, Specify a stirrup spacing of 3 in. within d of the #6 bar termination point. (c)

Cut off two of the negative moment bars at B when no longer needed. Extend the remaining bars past the point of inflection.

1.

Calculate the development length of a #7 bar: (

)

√

√

2.

Cut-off points for the remaining 3 #7 bars: These bars need to be extended past the negative point of flexure. Problem 8-9(b) shows that the required length is 76.2 in. Use 78 in. as before.

3.

Cut-off points for 2 #7 bars:  Ratio of remaining reinforcement area after cutting off the 2 #7 bars is 3/5 = 0.6 Moment at x  2 #7 bars can be cut off at the location x where  0.6 . This Maximum moment occurs at x  0.08 n  1.6 ft.=19.2 in. from B  To consider the shear effect, extend 2 #7 bars a distance d past the flexural cutoff point away from B. The distance from the face of column B to the cut-off point is ( ) , which is not OK. An option is to specify a length of 72 in. However, to simplify construction, another option is to extend all the bars a distance of 78 in. as in Item 1. Select a cutoff point of 78 in. from the column face for all 5 #7 bars. No bars are terminated within a flexural tension zone, so there is no need to check shear at the cut off point. 58 58 in. in.

78 in. 4#3#3stirrups @ 5 in.@(extra) 3 in.

4 #6

4#3#3stirrups @ 5 in.(extra) @ 3 in.

5 #7 12 21in. in.

A 6 in. 18 in.

2 #7

2 #6 Cut-off points for bars #6 240 in.

8-16

B 6 in. 18 in.


Fig. S8-5

8-17


Chapter 9 9-1

Explain the differences in appearance of flexural cracks, shear cracks, and torsional cracks.

Flexural cracks are approximately vertical cracks extending from the tensile face of the member towards the level of the zero-strain axis (neutral axis). Shear cracks are inclined to the axis of the element. Most often these cracks start at a flexural crack on the tensile face of the member and extend diagonally through the member toward the point of maximum moment. Torsional cracks spiral around the member and, for pure torsion, are roughly 45 deg. on all faces. For combined torsion and shear, the cracks tend to be pronounced on the face where the direct shear stress and the shear stresses due to torsion add, and less pronounced (or even absent) on the opposite face, where the stresses counteract.

9-2

Why is it necessary to limit the width of cracks?

Cracks are of concern for three main reasons:  appearance: wide cracks are unsightly and sometimes lead to a concern by owners and occupants;  leakage of air or fluids through the crack: crack control is important in the design of liquidretaining structures;  corrosion of reinforcement: traditionally corrosion of reinforcement has been related to crack width, but more recent studies suggest that the factors governing the eventual development of corrosion are independent of the crack width, although the period of time required for corrosion to start is a function of crack width.

9-3

Does the beam shown in Fig. P9-3 satisfy the ACI Code crack control provisions (Section 10.6.4)? .

 40,000   40,000  ACI Code Eq. (10-4) : s  15    2.5cc  12    fs   fs  using cc  1.5 in.  (3 / 8) in.  1.875 in. f s  2 f y  2  60,000 psi=40,000 psi , 3 3 (The value for f s can be also calculated as the stress in the reinforcement closest to the tension face at service load based on the unfactored moment.) So,

 40,000   40,000  s  15     2.5  1.875  10.3 in.< 12    12 in. OK. 40,000    40,000 

 

  1  12   2  1.5  3  1.128    8 2     2.37 in. < 10.3 in. From Fig. P9-3, bar spacing  3 Thus, the beam satisfies ACI Code Section 10.6.4.

9-1


9-4

Compute the maximum spacing of No. 5 bars in a one-way slab with 1 in. of clear cover that will satisfy the ACI Code crack-control provisions (Section 10.6.4). .  40,000   40,000  ACI Code Eq. (10-4) : s  15    2.5cc  12    fs   fs  using cc  1.0 in.. f s  2 f y  2  60,000 psi=40,000 psi 3 3 So,  40,000   40,000  s  15     2.5  1.0  12.5 in.> 12    12 in.  40,000   40,000  Thus, the maximum spacing for the No. 5 bars that satisfies ACI Code Section 10.6.4 is 12 in.

9-5 and 9-6 For the cross sections shown in Figs. P9-5 and P9-6, compute: (a) the gross moment of inertia, ; (b) the location of the neutral axis of the cracked section and ; and (c) for . The beams have a 1.5 in. of clear cover and No. 3 stirrups. The concrete strength is ; . 9-5

(a) Ig 

the gross moment of inertia,

;

16  24  18432 in.4 (ignoring the effect of reinforcement for simplicity) 12 3

(b) the location of the neutral axis of the cracked section and ; The distance from the extreme tension edge of the section to the centroid of the lowest 1   layer of steel is 1.5 in.+ 3 in.+ 8 in.   2.38 in. Assuming that the spacing between the 8 8 2   centers of the layers is 2 in., find:  4  0.79  2.38   2  0.79  4.38 Centroid =  3.05 in. above bottom and d  20.95 in. 4.74 E 29  106 psi n s   8.04 Ec 57000 4000 psi

 

 

Transformed area of steel = 8.04   6  0.79  38.11 in.2 Compute location of neutral axis Let depth of neutral axis be c and sum moments about the neutral axis to zero. Part Compression zone Tension steel

Area, in.2

y , in.

Ay ,in.3

16c

c 2

8c 2

38.11

 c  20.95

38.11c  798.4

Ay  8c  38.11c  798.4  0 and c  7.89 in. Thus, the neutral axis is 7.89 in. below the top of the beam 2

9-2


Compute Icr Part Compression zone Tension steel

Area, in.2

y , in.

I own

Ay 2 ,in.4

126.2

3.94

654.9

1959.1

38.11

-13.06



6500 9114 in.4

I cr  (c) for First calculate  M n for the beam: A f 4.74  60000 s y   c   5.23 in. and c  5.23  6.15 in. 1 0.85 ' 0.85  4000  16 0.85 f b c  d c  20.95  6.15        0.003  0.00722  0.002 O.K. s  c  cu  6.15  and   0.005    0.90 t 5.23   4.74  60000   20.95   a 2    M  A f d     5214 kips-in. n s y 2 1000 and  M n  0.9  5214  4693 kips-in. -

Thus, M cr 

fr I g

 7.5  



4000 psi  18432 in.4

lbs 12 in.  1000 kips So, using Eq. (9-10a) find yt

(

9-6

(a)

)

[

) ]

(

the gross moment of inertia,

Part Web Flanges

3888  9 in. 432 I g  22032 in.4 ytop 

(

 728.6 kips-in.

Area, in.2 288 144   432

)

[

(

) ]

;

ytop , in.

Aytop

I own , in.4

12 3

3456 432   3888

13824 432 Ig 

ybtm  15 in.

9-3

Ay 2 ,in.4 2592 5184 22032 in.4


(b)

the location of the neutral axis of the cracked section and E 29  106 psi n s   8.04 Ec 57000 4000 psi

;

Transformed area of steel = 8.04   4  0.79  25.4 in.2 d  24 in.  2.5 in.  21.5 in. Assume that the neutral axis is less than 6 in. Compute location of neutral axis. Part Compression zone

Area, in.2

y , in.

Ay ,in.3

36c

c 2

18c 2

nAs

25.4

 c  21.5

25.4c  546

Ay  18c  25.4c  546  0 and c  4.85 in. Thus, the neutral axis is 4.85 in. below the top of the beam (i.e. the compression zone is rectangular). 2

Compute Icr Part Compression zone

Area, in.2

y , in.

I own

Ay 2 ,in.4

174.6

2.42

342

1022.5

nAs

25.4

-16.65



7041

I cr 

8405 in.4

(c) for First calculate  M n for the beam: A f 3.16  60000 s y a c   1.55 in. and c  1.55  1.82 in. 1 0.85 ' 0.85  4000  36 0.85 f b c The steel is yielding and it is a tension-controlled section   0.90  .

1.55   3.16  60000   21.5   a 2    M  A f d     3930 kips-in. n s y 2 1000 and  M n  0.9  3930  3537 kips-in. Thus, M cr 

fr I g

-

 7.5  



4000 psi  22032 in.4

lbs 15 in.  1000 kips So, using Eq. (9-10a) find yt

(

)

[

(

) ]

(

 697 kips-in.

)

9-4

[

(

) ]


9-7

Why are deflections limited in design?

Deflections are limited for several reasons. 1. Deflections greater than 250 of the span are visible and may be unsightly. 2. Excessive deflections may cause cracking of partitions, malfunctioning of doors and windows and similar damage to non-structural elements. 3. Excessive deflections may interfere with the use of the structure, particularly if the structure supports machinery that must be carefully aligned. 4. Ponding of water on deflected roofs may overload the roofs. 5. Very large deflections may damage structural members and change the load path.

9-8

A simply supported beam with the cross section shown in Fig. P9-5 has a span of 25 ft and supports an unfactored dead load of 1.5 kips/ft, including its own self-weight plus an unfactored live load of 1.5 kips/ft. The concrete strength is 4500 psi. Compute (a) the immediate dead load deflection. (b) the immediate dead-plus-live load deflection. (c) the deflection occurring after partitions are installed. Assume that the partitions are installed one month after shoring for the beam is removed and assume that 20 percent of the live load is sustained.

(a) the immediate dead load deflection. From question 9-5 we found: I g  18432 in.4 , M cr  728.6 kips-in.

n

Es 29  106 psi   7.58 Ec 57000 4500 psi

Transformed area of steel = 7.58   6  0.79   35.93 in.2 Compute location of neutral axis Let depth of neutral axis be c and sum moments about the neutral axis to zero. Part Compression zone Tension steel

Area, in.2

y , in.

Ay ,in.3

16c

c 2

8c 2

35.93

 c  20.95

35.93c  752.7

Ay  8c2  35.93c  752.7  0 and c  7.71 in. Thus, the neutral axis is 7.71 in. below the top of the beam Compute Icr Part Compression zone Tension steel

Area, in.2

y , in.

I own

Ay 2 ,in.4

123.4

3.85

611

1829

35.93

-13.24



6298 8738 in.4

I cr 

9-5


Unfactored dead load moment: M DL 

   117.2 kips-ft  1406 kips-in.

1.5  kip ft   252 ft 2

8 Thus, M a  M cr  cracked section and need to calculate I eff

3 3   M 3    728.6 3   M cr   728.6  4 cr Ie   18432  1    I g  1     I cr      8738  10087 in. M M 1406 1406        a      a  

This is a simply supported beam with distributed loading, so using deflection Case 1 from Fig. 913, the immediate dead load deflection can be calculated as: 2 2 5 M pos 5 1406  1000   25  12   iD     0.342 in. 48 EI 48 3.824  106  10087 (b) the immediate dead-plus-live load deflection. Unfactored dead plus live load moment: 3.0  kip ft   252 ft 2  in.  M D L   12    2812 kips-in. 8  ft  3   728.6 3   728.6  4 Ie   18432  1      8738  8907 in. 2812  2812      So using again the deflection Case 1 from Fig. 9-13, the immediate dead-plus-live-load deflection can be calculated as:

 

5 2812  1000   25  12   iD+L    0.774 in. 48 3.824  106  8907 2

(c)

the deflection occurring after partitions are installed. Assume that the partitions are installed one month after shoring for the beam is removed and assume that 20 percent of the live load is sustained. The deflection occurring after the partitions are installed can be calculated from Eq. (9-14):

  iL    t0 ,   iD   iLS The immediate dead load deflection, iD , was found from part (a) to be 0.342 in. However, after the live load has been applied and the beam has cracked, the deflection due to dead load will be increased by an amount equal to the ratio of the I e values used in part (a) and (b). Thus the immediate dead load deflection on the structure which has been loaded to D+L will be calculated and used in Eq. (9-14). 

iD 

0.342 

10087 =0.387 in. 8907

The immediate live load deflection,  iL , is found as: iL  i,L+D  iD  0.774 in.  0.387 in.  0.387 in. Twenty percent of this results from sustained live loads, so: iLS  0.20  0.387 in.  0.077 in.

9-6


Since the beam has no compression reinforcement, and with the partitions installed 1 month after removal of the shoring, (

)

The deflection occurring after the partitions are installed is found as:

9-9

Repeat Problem 9-8 for the beam section in Fig. P9-6.

(a) the immediate dead load deflection. From question 9-6 we found: , , , and The unfactored dead load moment: M DL 

   117.2 kips-ft  1406 kips-in.

1.5  kip ft   252 ft 2

8 Thus, M a  M cr  cracked section and need to calculate I eff (

)

[

(

) ]

(

)

[

(

) ]

This is a simply supported beam with distributed loading, so using deflection Case 1 from Fig. 913, the immediate dead load deflection can be calculated as: ( )

(b) the immediate dead-plus-live load deflection. Unfactored dead plus live load moment: 3.0  kip ft   252 ft 2  in.  M D L   12    2812 kips-in. 8  ft 

 

(

)

[

(

) ]

So using the deflection Case 1 from Fig. 9-13, the immediate dead-plus-live-load deflection can be calculated as: ( )

(c)

the deflection occurring after partitions are installed. Assume that the partitions are installed one month after shoring for the beam is removed and assume that 20 percent of the live load is sustained. The deflection occurring after the partitions are installed can be calculated from Eq. (9-14):

  iL    t0 ,   iD   iLS The immediate dead load deflection, iD , was found from part (a) to be 0.342 in. However, after the live load has been applied and the beam has cracked, the deflection due to dead load will be

9-7


increased by an amount equal to the ratio of the I e values used in part (a) and (b). Thus the immediate dead load deflection on the structure which has been loaded to D+L will be calculated and used in Eq. (9-14).

The immediate live load deflection,  iL , is found as:

Twenty percent of this results from sustained live loads, so:

Since the beam has no compression reinforcement, and with the partitions installed 1 month after removal of the shoring, (

)


9-10

The beam shown in Fig. P9-10 is made of 4000-psi concrete and supports unfactored dead and live loads of 1 kip/ft and 1.1 kips/ft. Compute:

(a) the immediate dead-load deflection. Compute Ig for the T-section (ignore the effect of the reinforcement for simplicity): Assume flange width = effective flange width from ACI Code Section 8.12.2. = 72 in. E 29  106 psi n s   8.04 Ec 57000 4000 psi Part Web Flanges

Area, in.2 240 360 A  600

ytop , in.

Aytop

10 3

2400 1080 Ay  3480

I own , in.4 8000 1080 Ig 

Ay 2 ,in.4 4234 2822 16136 in.4

3480  5.8 in. ybtm  14.2 in. 600 Compute Icr at the left end: The positive-moment reinforcement is not developed for compression at the support and will therefore be neglected. The section is a rectangular section, with 3 No. 7 bars , d  17.5 in. ,and the following properties: As  1.80 in.2 , nAs  14.47 in.2 ytop 



1.80  0.0086, n  0.069 12  17.5

Using Eq. (9-3): k  2 n    n    n  0.309 and c  kd  5.41 in. 2

9-8


Part Compression zone

Area, in.2

y , in.

I own

Ay 2 ,in.4

64.9

2.71

158

477

nAs

14.47

5.41  17.5  12.1



2119

I cr 

2754 in.4

Compute Icr at the right end:

As  3.16 in.2 , nAs  25.41 in.2



3.16  0.0150, n  0.121 12  17.5

Using Eq. (9-3): k  2 n    n    n  0.386 and c  kd  6.75 in. 2


Area, in.2

y , in.

I own

Ay 2 ,in.4

81

3.37

307.5

920

nAs

25.41

 6.75  17.5  10.75



2936

I cr 

4163 in.4

Compute Icr at the midspan:

As  2.18 in.2 , nAs  17.53 in.2 Assuming that c is less than 6 in. Part Compression zone

Area, in.2

y , in.

Ay ,in.3

72c

c 2

36c 2

nAs

17.53

 c  17.5

17.53c  307

Ay  36c2  17.53c  307  0 and c  2.7 in.


Area, in.2

y , in.

I own

Ay 2 ,in.4

194

1.35

118

354

nAs

17.53

 2.7  17.5  14.8



3840

I cr 

4312 in.4

The moment at the end and the midspan can be calculated using the ACI Moment Coefficients

9-9


Unfactored dead load moments w 2n 1 2    1  22.67   32.1 kips-ft 16 16 w 2n 1 2 Positive moment at midspan =   1  22.67   36.7 kips-ft 14 14 w 2 1 2 Negative moment at right end =  n    1  22.67   51.4 kips-ft 10 10

Negative moment at left end = 

Cracking moments f r  7.5 4000 psi  474 psi 474  16136 Ends: M cr   110 kips-ft 5.8  12,000 474  16136 Midspan: M cr   45 kips-ft 14.2  12,000 Compute the average effective moment of inertia for both ends and the midspan, M a  M cr and thus the sections are uncracked under dead loads and I eff  I g  16136 in.4 Compute immediate dead-load deflection This is an exterior beam span with a column as the exterior support. Therefore, the deflection can be calculated using row 3 of Table 9-2, which requires a combination of Cases 2 and 8 from Fig. 9-13. 4 1 1000   24  12  w 4 12  =0.0533 in. (down) For deflection Case 2:   185EI 185  3.6  106  16136









M 2  32.1 12000    24  12  For deflection Case 8:    =  0.0343 in. (up) 16 EI 16  3.6  106  16136 2





Thus, iD  0.0533  0.0343  0.019 in. (b) the immediate dead-plus-live load deflection. Unfactored dead plus live-load moments w 2 1 2 Negative moment at left end =  n    1  1.1   22.67   67.4 kips-ft 16 16 2 w n 1 2 Positive moment at midspan =   1  1.1   22.67   77.1 kips-ft 14 14 w 2 1 2 Negative moment at right end =  n    1  1.1   22.67   107.9 kips-ft 10 10 Compute the average effective moment of inertia left end: M a  M cr  I e  I g  16136 in.4 3   45 3   45  4 midspan: M a  M cr  I e   16136  1    4312  6663 in.    77.1    77.1  

9-10


right end: M a  M cr  I e  I g  16136 in.4 Thus, the weighted average value of I eff using Eq. (9-11a) is:

I e( avg )  0.70  6663  0.30 16136  9505 in.4 Immediate plus live-load deflection Using the same procedure as in part (a),





2.1 1000   24  12  w 4 12 for deflection Case 2:    =0.190 in. (down) 185EI 185  3.6  106  9505



4



M 2  67.4  12000    24  12   =  0.122 in. (up) 16 EI 16  3.6  106  9505 2

for deflection Case 8:  





Thus, iD+L  0.190  0.122  0.068 in. (c)

the deflection occurring after partitions are installed. Assume that the partitions are installed two months after the shoring is removed and assumed that 15 percent of the live load is sustained.

The deflection occurring after the partitions are installed can be calculated from Eq. (9-14):   iL    t0 ,   iD   iLS The immediate dead load deflection, iD , was found from part (a) to be 0.019 in. However, after the live load has been applied and the beam has cracked, the deflection due to dead load will be increased by an amount equal to the ratio of the I e values used in part (a) and (b). Thus the immediate dead load deflection on the structure which has been loaded to D+L will be calculated and used in Eq. (9-14). 

iD 

0.019 

16136 =0.032 in. 9505

The immediate live load deflection,  iL , is found as: iL  i,L+D  iD  0.068 in.  0.032 in.  0.036 in. Fifteen percent of this results from sustained live loads, so:

Since the beam has no compression reinforcement at midspan, and with the partitions installed 2 months after removal of the shoring, (

)


9-11


Chapter 10 10-1

A five-span one-way slab is supported on 12-in.-wide beam with center-to-center spacing of 16 ft. The slab carries a superimposed dead load of 10 psf and a live load of 60 psf. Using and , design the slab. Draw a cross section showing the reinforcement. Use Fig. A-5 to locate the bar cut-off points.

slab design strip

Plan

n

 16 ft.  12 in.  15 ft. in. 12 ft

Estimate slab thickness Assume partitions are not sensitive to deflections. Will require recheck if sensitivity is established later. Table A-9: End bay: Min h 

 15 12

 7.50 in. 24 Interior bay: Min h  n  15 12  6.43 in. 28 28 Note that slab thickness was chosen on the basis of deflection control, since flexure and shear probably won’t govern the design (this will be checked later). n

24

Try h  7.0 in. (this may need to be checked for deflections in the end span). Assuming a cover of 0.75 in. and No. 4 bars as the slab reinforcement, d  7.0  0.75  0.5  6.0 in. 2





Compute factored loads Considering a 1-ft wide strip of slab: 7.0  150  87.5 psf Slab self weight: wDs  12 Superimposed dead load: wDi  10 psf Total dead load: wD  87.5  10  97.5 psf

10-1


Live load: Factored load: Load per foot along design strip wL  3wD , so we can use the ACI Moment coefficients for the calculation of the positive and the negative moments.

Thickness for flexure The maximum value for M u is at the first interior support since appropriate moment coefficient from ACI Code Section 8.3.3, ( )

n

 15 ft. throughout. Using the

For a reinforcing ratio of   0.01 , which is a reasonable upper limit for a slab, the reinforcing index can be found from Eq. (5-21), 0.01 60000   0.15 4000 From Eq. (5-22) calculate the flexural resistance factor, R. R  0.15  4000  1  0.59  0.15  547 psi Using this value of R, the required value of bd 2 can be determined using Eq. (5-23a), assuming that   0.9 (will check it later).

For

√

,

Actual d  6 in.  will be less

Therefore, the minimum d to keep   0.01 is than 0.01 (O.K. for flexure). Thickness for shear

The max shear Vu is at the exterior face of the first interior support. Using the appropriate shear coefficient from ACI Code Section 8.3.3, ( √

)

( √

)

OK

So, use a 7 in. slab. Flexural reinforcement Max M u  4.8 kips-ft/ft

a   Assuming that  d   0.95d  and  s   y , find the required reinforcement for a 1-ft wide strip 2   of slab. (

)

(

)

10-2


Iterate to find the depth of the compression stress block and recompute the value of the required reinforcement:

Since the depth to the neutral axis, c, is less than 3 8 of d , the section is tension controlled



s



  y ,  0.9 .

The minimum reinforcement required by ACI Code Section 10.5.4, is As ,min  0.0018bh  0.0018 12  7  0.15 in.2 /ft The maximum spacing of the bars is, by ACI Code Section 7.6.5, 3h  21 in. smax     18 in  Select No. 4 bars at 12 in. ⁄

⁄

Temperature and shrinkage steel as required by ACI Code Section 7.12.2, 5h  35 in. As ,min  0.0018bh  0.15 in.2 /ft and smax     18 in  So provide No. 4 bars at 16 in.  As

ft







0.2 in.2  12 in.  0.15 in.2 ft 16 in.

The flexural reinforcement for the supports and the midspan for all the spans is calculated in the following table.

10-3


Calculation of reinforcement required in the slab.

 ft.

1.

n

2.

wu

15.0

2 n

47.9

47.9

15.0 47.9

47.9

47.9

47.9

3.

Moment Coef.

1 24

1 14

1 10

1 11

1 16

1 11

1 16

4.

M u  kips-ft/ft 

2.0

3.4

4.8

4.4

3.0

4.4

3.0

5.

As reqd. in.2 /ft

0.08

0.13

0.19

0.12

0.17

0.12

6.

As,min in.2 /ft

0.15

0.15

0.15

0.15

0.15

0.15

7.

Reinforcement

#4 @ 16 in.

#4 @ 16 in.

#4 @ 12 in.

#4 @ 16 in.

#4 @ 12 in.

#4 @ 16 in.

8.

As provided in.2 /ft

0.15

0.15

0.2

0.15

0.2

0.15













Fig. S10-1.1 shows a cross-section of the slab showing the reinforcement. The bar cut-off points were located using Fig. A-5(c).

12”

12”

16”

Fig. S10-1.1 Slab reinforcement detailing.

10-4


10-2

A four-span one-way slab is supported on 12-in.-wide beams with center-to-center spacing of 14, 16, 16, and 14 ft. The slab carries a superimposed dead load of 20 psf and a live load of 100 psf. Design the slab, using and . Select bar cut-off points using Fig. A-5 and draw a cross-section showing the reinforcement.

slab design strip

Plan  14 ft.  12 in.  13 ft. in. 12 ft  15 ft. Long span clear length: n  16 ft.  12 in. in. 12 ft 13 ft.  15 ft. Average clear span length for first interior support: n,avg  Short span clear length:

n

2

 14 ft.

Estimate slab thickness Assume partitions are not sensitive to deflections. Will require recheck if sensitivity is established later. Table A-9: End bay: Min h 

 13 12

 6.50 in. 24 Interior bay: Min h  n  15 12  6.43 in. 28 28 Note that slab thickness is chosen on basis of deflection control, since flexure and shear probably won’t govern the design (will be checked later). n

24

Try h  6.5 in. Assuming

cover of 0.75 d  6.5  0.75  0.5  5.5 in. 2



a



in.

and

No.

10-5

4

bars

as

the

slab

reinforcement,


Compute factored loads Considering a 1-ft wide strip of slab: 6.5  150  81.25 psf Slab self weight: wDs  12 Superimposed dead load: wDi  20 psf Total dead load: wD  81.25  20  101.25 psf Live load: Factored load: Load per foot along design strip = wL  3wD , so we can use the ACI Moment coefficients for the calculation of the positive and the

negative moments (ACI Code Section 8.3.3). Thickness for flexure The maximum moment will occur at either: (1) exterior face of the first interior support, or (2) face of the middle support For negative moments at the face of an interior support, ACI Code Section 8.0 defines n as the average of the clear spans of the two adjacent spans. Using the appropriate moment coefficient from ACI Code Section 8.3.3, ( ) (

)

For a reinforcing ratio of   0.01 , which is a reasonable upper limit for a slab, the reinforcing index can be found from Eq. (5-21), 0.01 60000   0.171 3500 From Eq. (5-22) calculate the flexural resistance factor, R. R  0.171 3500  1  0.59  0.171  538 psi Using this value of R, the required value of bd 2 can be determined using Eq. (5-23a), assuming that   0.9 (will check it later).

For

,

√

i.e., min d to keep   0.01 is flexure).

Actual d  5.5 in.  will be less than 0.01 (O.K. for

Thickness for shear The max shear Vu will occur in one of the two locations discussed for the maximum moments. Using the appropriate shear coefficient from ACI Code Section 8.3.3,

10-6










Vc  0.75 2 fc' bw d  0.75 2  3500  12  5.5  5860 lbs/ft  ok for shear Flexural reinforcement -

Max

a   Assuming that  d   0.95d  and  s   y , find the required reinforcement for a 1-ft wide strip 2   of slab. (

(

)

)

Iterate to find the depth of the compression stress block and recompute the value of the required reinforcement:

Since the depth to the neutral axis, c, is less than 3 8 of d , the section is tension controlled



s



  y ,  0.9 .

The minimum reinforcement required by ACI Code Section 10.5.4, is As ,min  0.0018bh  0.0018 12  6.5  0.14 in.2 /ft The maximum spacing of the bars is, by ACI Code Section 7.6.5, 3h  19.5 in. smax     18 in  Select No. 4 bars at 9 in. ⁄

⁄

Temperature and shrinkage steel as required by ACI Code Section 7.12.2, 5h  32.5 in. As ,min  0.0018bh  0.14 in.2 /ft and smax     18 in  So provide No. 4 bars at 16 in. 

As

ft







0.2 in.2  12 in.  0.15 in.2 ft 16 in.

The flexural reinforcement for the supports and the midspan for all the spans is calculated in the following table.

10-7


Calculation of reinforcement required in the slab.

1.

n

13.0

13.0

14.0

15.0

15.0

15.0

55.3

63.5

63.5

63.5

2.

wu

 kips-ft 

47.7

47.7

3.

Moment Coef.

1 24

1 14

1 10

1 11

1 16

1 11

1 16

4.

M u  kips-ft/ft 

2.0

3.4

5.5

5.0

4.0

5.8

4.0

5.

As reqd. in.2 /ft

0.09

0.14

0.23

0.17

0.25

0.17

6.

As,min in.2 /ft

0.14

0.14

0.14

0.14

0.14

0.14

7.

Reinforcement

#4 @ 16 in.

#4 @ 16 in.

#4 @ 9 in.

#4 @ 11 in.

#4 @ 9 in.

#4 @ 11 in.

8.

As provided in.2 /ft

0.15

0.15

0.27

0.18

0.27

0.18

2 n









 

Fig. S10-2.1 shows a cross-section of the slab showing the reinforcement. The bar cut-off points were located using Fig. A-5(c).

9”

11”

16”

Fig. S10-2.1 Slab reinforcement detailing.

10-8


10-3

A three span continuous beam supports 6-in.-thick one-way slabs that span 20 ft center-to-center of beams. The beams have clear spans, face-to-face of 16-in.-square columns, of 27, 30, and 27 ft. The floor supports ceiling, ductwork, and lighting fixtures weighing a total of 8 psf, ceramic floor tile weighting 16 psf, partitions equivalent to a uniform deal load of 20 psf, and a live load of 100 psf. Design the beam, using . Use for flexural reinforcement and for shear reinforcement. Calculate cut-off points, extending all reinforcement past points of inflection. Draw an elevation view of the beam and enough cross-sections to summarize the design.

1. Compute the trial factored load on the beam (a) Dead load 6 in.  75 psf 12 in. ft. ceiling, tile, partitions: 8 psf  16 psf  20 psf  44 psf 150 pcf 

slab self-weight:

The beam size is not known at this stage, so it must be estimated for preliminary design purposes. Once the size of the beam has been established, the factored load will be corrected and then used in subsequent calculations. The beam size will be estimated in step 2. (b) Live Load The ASCE/SEI 7-10 recommendations allow live-load reductions based on tributary area multiplied by a live-load element factor,  LL  2 , to convert the tributary area to an influence area. 

Positive moment at span AB and negative moment at exterior support, A.

AT  15 ft.  27 ft.  405 ft.2

 15 L  Lo  0.25   K LL AT 

   15   100   0.25    100  0.777  77.7 psf > 0.5 100 psf  2  405   

10-9


Note that L shall not be less than 0.50 Lo for members supporting one floor (O.K.) 

Negative moment at interior support, B.

AT  15 ft.   27  30 ft.  855 ft.2



 15 L  Lo  0.25   K LL AT 

  15    100   0.25    100  0.613  61.3 psf > 0.5 100 psf (O.K.)  2  855   

Positive moment at span BC. AT  15 ft.  30 ft.  450 ft.2  15 L  Lo  0.25   K LL AT 

   15   100   0.25    100  0.75  75.0 psf > 0.5 100 psf (O.K.)  2  450   

The size of the beam will be chosen on the basis of negative moment at the first interior support. For this location, the factored load on the beam, not including the beam stem below the slab, is: wu  1.2   75  44  1.6  61.3  241 psf The tributary width for the beam is 15 ft and the factored load from the slab per foot of beam is 241 psf 15 ft.  3,615 lbs  3.6 kips ft ft Two approximate methods can be used to estimate the weight of the beam stem: (a) the factored dead load of the stem is taken as 12 to 20 percent of the other factored loads on the beam. This gives 0.43 to 0.72 kip/ft. (b) the overall depth of beam h is taken to be 1/18 to 1/12 of the larger span, , and bw is taken to be 0.5h . This gives the overall h as 20 to 30 in., with the stem extending 14 to 24 in. below the slab, and gives bw as 10 to 15 in. The factored load of such sizes ranges from 0.17 to 0.45 kip/ft. As a first trial, assume the factored weight of the stem to be 0.50 kip/ft. Then, total trial load per foot  3.6  0.5  4.1 kip/ft 2. Estimate the size of the beam stem (a)

Calculate the minimum depth based on deflections.

ACI Table 9.5(a) (Table A-9) gives the minimum depths, unless deflections are checked. For partitions flexible enough to undergo some deflection, minimum depth for beam BC is 16 in. hmin   28.3 ft , where  the span center-to-center of supports  27 ft  18.5 12 in. ft 28.3  12  18.4 in. Thus, hmin  18.5

10-10


(b) Determine the minimum depth based on the negative moment at the exterior face of the first interior support. The beam fits the requirements in ACI Code Section 8.3.3 and can use the moment and shear coefficients. For the support at B, w 2 4.1  28.5 Mu  u n   333 kips-ft 10 10 27 ft  30 ft  28.5 ft (the average of the two adjacent spans). where n  2 2

Using the procedure developed in Chapter 5 for the design of singly reinforced beam sections, the reinforcement ratio that will result in a tension-controlled section can be estimated from Eq. (5-18) as,  f ' 0.825  4.5   initial   1 c   0.0155 4 fy 4  60 For this reinforcement ratio, use Eq. (5-21), to find the reinforcing index, 0.0155  60   0.207 4.5 From Eq. (5-22) calculate the flexural resistance factor, R. R  0.207  4500  1  0.59  0.207   818 psi Using this value of R, the required value of bd 2 can be determined using Eq. (5-23a), assuming that   0.9 (will check it later). 333 bd 2   12,000  5430 in.3 0.9  818 Since columns are 16 in., try a 14 or 16 in. wide stem. Let’s try b  14 in. 5428  19.7 in.  d  21.5 in. Then, d  14 With one layer of steel at supports, h  21.5  2.5  24 in. (O.K. for deflections). So, try a 14-in. wide-by-24-in. beam. (c)

Check the shear capacity of the beam

Maximum shear is at the exterior face at support B, 1.15  wu n 1.15  4.1 27 Vu ,max    63.6 kips 2 2 From ACI Code Section 11.2.1.1, Vc  2 fc' bw d  2 1 4500  14  21.5  1  40.4 kips 1000 ACI Code Section 11.4.7.9 sets the maximum nominal Vs is









Vs  8 fc' bw d  8  4500 14  21.5  1  162 kips 1000

Thus, Vn  0.75   40.4  161  151 kips (O.K. for shear)

10-11


(d)

Use :

Summary b  14 in. h  24 in. 18 in. below slab  d  21.5 in.,assuming one layer of steel

3. Compute the dead load of the stem, and recompute the total load per foot.

18  14  0.26 kip/ft 144 Corrected total factored load for 1st internal support moment:  3.6  1.2   0.26  3.9 kip/ft Since this is less that the 4.1 kip/ft used earlier to estimate the beam size, the section chosen will be adequate. Weight per foot of the stem below slab  0.15 

Factored Total dead load: wD  1.2  119 15  1  0.26  2.5 kips/ft 1000





Factored total loads: (a) Positive moment at span AB and negative moment at exterior support, A. wu  2.5  1.6  77.7 15  1  4.4 kips/ft 1000 (b) Negative moment at interior support, B. wu  2.5  1.6  61.3 15  1  4.0 kips/ft 1000 (c) Positive moment at span BC. wu  2.5  1.6  75.0 15  1  4.3 kips/ft 1000













4. Calculate the beam flange width for positive-moment regions From ACI Code Section 8.12.2,

 0.25  be    

n

 based on the shorter span for simplicity   0.25   27 12  in.  81 in.   bw  2 8  6   14  2 8  6   110 in.   15 12   180 in. 

Therefore, the effective flange width is 81 in. and shown in Fig. S10-3.1

Fig. S10-3.1 Beam cross-section

10-12


5. Can we use the ACI Code Moment Coefficients?    

Ratio of successive spans = 30 ft. / 27 ft. = 1.11<1.20 O.K. ; Live load/ft 0.115   0.6  3.0 O.K. ; Dead load/ft 2.5 There are more than two spans; The loads are uniformly distributed.

Thus, we can use the ACI Code Moment Coefficients 6. Compute the beam moments

n,

ft.

wu , kip/ft.

wu

2 n ,kips-ft

Cm

M u  Cm wu

2 n ,kips-ft

27.0

27.0

28.5

30.0

4.4

4.4

4.0

4.3

3210

3210

3250

3870

1 16

1 14

 1 and  1 10 11

1 16

-200

230

-325

242

7. Design the flexural reinforcement (a) Max negative moment (first interior support) Max M u_,max  325 kips-ft Because the beam acts as a rectangular beam with compression in the web, we can assume that a    d  2  0.90d  .For  s   y and  =0.9 , the required reinforcement for that section is,   M u  12,000 325  12,000 As    3.73 in.2 a  0.90  60,000   0.90  21.5    fy d   2  Iterate to find the depth of the compression stress block and recompute the value of the required As f y 3.73  60000 reinforcement: a    4.18 in., c  4.18  5.07 in. ' 0.825 0.85 fc b 0.85  4500  14

10-13


Since the depth to the neutral axis, c, is less than 3 8 of d , clearly the section is tension





controlled  s   y ,  0.9 , and 325  12,000  3.72 in.2 4.18   0.90  60,000   21.5  2   The other negative moment sections have a lower design moment, so it will be conservative to use the ratio of As obtained here to quickly determine the area of tension steel required at Mu those other locations. That ratio is As ,red 



As 3.72 in.2   0.0114 in.2 /kips-ft M u 325 kips-ft



(eq. A)

(b) Max positive moment Max M u,max  242 kips-ft Because the beam acts as a T-shape beam with compression in the top flange, assume that the a   compression zone is rectangular, i.e. a  h f  6 in. , use  d   0.95d  .For  s   y and  =0.9 , 2   the required reinforcement that section is, As 

M u  12,000 242  12,000   2.63 in.2 a  0.90  60,000   0.95  21.5    fy d   2 

Iterate to find the depth of the compression stress block and recompute the value of the required reinforcement: As f y 2.63  60000 a   0.51 in., c  0.51  0.62 in. ' 0.825 0.85 fc be 0.85  4500  81





The section is tension controlled  s   y ,  0.9 , and doing one iteration for the negative moment section results in As  2.53 in.2 The other positive moment section has a lower design moment, so it will be conservative to use the ratio of As obtained here to quickly determine the area of tension steel required at those Mu other locations. That ratio is



As 2.53 in.2   0.0104 in.2 /kips-ft M u 242 kips-ft



(eq. B)

10-14


(c) Calculate the minimum reinforcement From ACI Code Section 10.5.1, As ,min 

3 f c' fy

bw d and 

200bw d . fy

For 4500 psi concrete, 3 fc'  201 psi , thus As ,min 

3 4500 14  21.5=1.0 in.2 60,000

(d) Calculate the area of steel and select the bars. The remaining calculations are done in the following table.

M u ,kips-ft As coef. Eq. A and B

-200 0.0114

230 0.0104

-325 0.0114

242 0.0104

As ( red .) ,in.2

2.28

2.39

3.70

2.52

As  As.min

Yes

Yes

Bars selected

4 No. 7

4 No. 7

As provided, in.2

2.40

2.40

Yes 2 No. 8 and 4 No. 7 3.98

Yes 3 No. 7 and 2 No.6 2.68



Yes



Yes

bw   14 in.  bmin

Note that in the negative moment regions some of the bars can be placed in the slab besides the beam and it is not necessary to check whether they will fit into the web width. 8. Check the distribution of the reinforcement (a) Positive moment region From ACI Code Section 10.6.4, the maximum bar spacing is  40,000   40,000  s  15    2.5cc  12   , where  fs   fs  f s  2 f y  40,000 psi and cc  1.5 in. cover  0.375 in. stirrups  1.875 in. 3 Thus, s  15  2.5 1.875  10.3 in.  12 in.

10-15


14   2  1.5  0.375  7  8   3.1 in. <10.3 in. OK.  Bar spacing  3

It was also clear that the bar spacing is smaller than 10.3 in., since there are four bars and bw  14 in.

(b) Negative moment region ACI Code Section 10.6.6 says “part” of the negative moment steel shall be distributed over a width equal to the smaller of the effective flange width (81 in.) and n 10  34.2 in. At the interior supports, there are 6 top bars. Place the two No. 8 bars at the corners of the stirrups, two No. 7 bars over the beam web, and the other two No. 7 bars in the slab. Within a width of 34.2 in. we must place six bars. These cannot be further apart than 10.3 in. (as calculated in part a). We shall arbitrarily place two bars at 5 in. outside the web of the beam. ACI Code Section 10.6.6 requires “some” longitudinal reinforcement in the slab outside this band. We shall assume that the shrinkage and temperature steel in the slab will satisfy this requirement. 9. Design the shear reinforcement The shear force diagrams are calculated in the following table and shown at the bottom of the table. The shear coefficients for the supports are from ACI Code Section 8.3.3 and the coefficient fro the midspan of the beam is based on Eq. (6-26).

n ,ft

wu ,kips/ft wLu ,kips/ft Cv at support and midpsan wu n 2 wLu n Vu , kips Vn  Vu 

27 4.4 1.9 1.0

0.125

59.4 59.4 79.2

51.3 6.4 8.5

30 4.3 1.8 1.15

1.0

59.4

64.5

68.3 91.1

10-16

64.5 86.0

0.125

1.0 64.5

54 6.7 8.9

64.5 86.0


(a) Exterior end of beam AB Because the beam is supported by a column, the critical section is located at d away from the face of the support. Equation for Vu  : Vu   kips   5.24x  ft   79.2

 21.5 in.    79.2  69.8 kips Vu   @ d from A   5.24    12 in.  ft   V Vc  2 1 4500  14  21.5  1  40.4 kips , c  20.2 kips 1000 2 Vc Vu  20.2 kips , stirrups are required. Because  69.8 kips exceeds  2







Try No. 3 Grade 40 stirrups double-leg stirrups with a 90o hook Av  2  0.11  0.22 in.2 

Max spacing:

smax

Vu





 Vc  4 f c' bw d and from ACI Code Section 11.4.5.1,

d   10.75 in.    2  10.75 in.   24 in.  

 To satisfy the minimum stirrup requirement in ACI Code Section 11.4.6.3, the stirrup spacing must be, Av f yt 0.22  40,000 smax    12.6 in. 50bw 50  14 Note that 0.75 4500  50.3 psi > 50 psi, so use 50 psi in ACI Code Eq. (11-13) . Thus, use 10.5 in. as maximum stirrup spacing. 

The spacing required to support the shear force at the support is,  0.22  40  21.5 s  6.45 in.-say 6 in. on centers 69.8  40.4

We can change the stirrup spacing to 10.5 in. when

Vu





0.22  40  21.5  40.4  58.4 kips 10.5

This occurs at about 4 ft from face of support A. V V We can stop the stirrups when u  c  x  11.3 ft from face of support A.  2 Place the first stirrup at 3 in. from support A, then 9 stirrups at 6 in. and 9 stirrups at 10.5 in.

10-17


(b) Interior end of beam AB Equation for Vu  : Vu   kips   6.12x  ft   91.1

 21.5 in.    91.1  80.1 kips > Vc ,stirrups required Vu   @ d from B   6.12   in.  12  2 ft   The spacing required to at this point is,  0.22  40  21.5 s  4.8 in.-say 4.5 in. on centers 80.1  40.4 V Change the stirrup spacing to 10.5 in. when u  58.4 kips .  This occurs at 5.3 ft from face of support B. We can stop the stirrups at 11.6 ft from face of support B. Place the first stirrup at 2 in. from support B, then 15 stirrups at 4.5 in. and 9 stirrups at 10.5 in. (c) Ends of beam BC Equation for Vu  : Vu   kips   5.14x  ft   86.0

 21.5 in.    86  76.8 kips > Vc ,stirrups required Vu   @ d from C   5.14    12 in.  2 ft   The spacing required to at this point is,  0.22  40  21.5 s  5.2 in.-say 5.0 in. on centers 76.8  40.4 V Change the stirrup spacing to 10.5 in. when u  58.4 kips .  This occurs at 5.4 ft from face of support. We can stop the stirrups at 12.8 ft from face of support. Place the first stirrup at 2.5 in. from support B, then 14 stirrups at 4.5 in. and 9 stirrups at 10.5 in.

10. Bar cutoffs

d  21.5 in. (a) Detailing requirements: 12db  12 in. for No. 8 bar n

16 Thus, d exceeds 12db and

n

16

 20.25 in. for 27 ft span, 22.5 in. for 30 ft span

for AB span, while

n

16

governs for span BC.

The bottom and top bars have clear spacing and cover of at least d b and are enclosed by at least minimum stirrups. Therefore, this is Case 1 in Table 8-1 (ACI Code Section 12.2.2).

10-18


d



f y t e 20 f c'

db 

60,000  1.0  1.0  db 20  1 4500

33.5 in. for No. 6    44.7db  39.1 in. for No. 7  44.7 in. for No. 8  

(b) Cutoffs for bottom steel Span AB 4 No. 7-Extend 2 full length into each support, cut off the other two at the positive moment point of inflection so that extra stirrups are not required. Exterior end: From Fig. A-2, inflection point at  0.10  27 12  32.4 in. from face of column. Rule 3-a - Extend d  21.5 in. past the flexural cutoff point, i.e. 32.4 in.-21.5 in = 10.9 in. from face of column at A. Say 10 in. Rule 4-a - Distance from midspan to cutoff point greater than d . Rule 1-b - This is an interior beam with open stirrups. Since this is a discontinuous end use 90 deg. standard hooks on 2 No. 7 bars. Rule 4-d – At the inflection point, the remaining steel is two No. 7, As  1.2 in.2 Thus,

1.2  60  1.57 in. 0.85  4.5  12 1.57   M n  1.2  60   21.5  =1490 kip-in=124 kip-ft 2   The shear at 32.4 in. from the exterior end is,  32.4  Vu  5.24     79.2  65 kips  12  a  21.5 in. M 1490  21.5  44.4 in. Thus, n  a  Vu 65 This exceeds d - therefore, OK. a

Interior end: From Fig. A-2, inflection point at  0.104  27 12  33.7 in. from face of column Rule 3-a - Extend bars to 33.7 in.- 21.5 in. = 12.2 in. from face of column. Use 10 in. to match other end. Rule 4-a - Satisfied Rule 1-b - This is an interior beam with open stirrups. Rule 1-b applies. Lap splice 2#7 bars from the exterior span with 2 No. 7 bars from the interior span with a Class A tension lap splice  1.0  39.1 in.  3 ft - 2 in. Rule 4-d M n  1490 kip-in=124 kip-ft The shear at 33.7 in. from the exterior end is,  33.7  Vu  6.12     91.1  73.9 kips  12  a  21.5 in. M 1490  21.5  41.7 in. Thus, n  a  Vu 73.9 This exceeds d - therefore, OK.

10-19


Span BC 2 No. 8 and 3 No. 7 at midspan – Extend 2 No. 7 into supports. Cutoff 2 No. 8 and 1 No. 7 bars at the positive moment point of inflection so that extra stirrups are not required. Inflection point at  0.146  30 12  52.6 in. from face of column n  22.5 in. past the flexural cutoff point, i.e. 52.6 in.-22.5 in. = 30.1 in. 16 from face of column. Say 30 in. Rule 4-a - Satisfied. Rule 1-b – Lap splice 2#7 bars 3 ft - 2 in at support Rule 4-d M n  2650 kip-in=221 kip-ft The shear at 52.6 in. from the exterior end is,  52.6  Vu  5.14     86  63.5 kips  12  a  21.5 in. M 2650  21.5  63.3 in. Thus, n  a  Vu 63.5 This exceeds d - therefore, OK.

Rule 3-a - Extend

Cutoffs for top steel Span AB, exterior end, 4 No. 7 0.02  1.0  60,000 Use standard hook, dh    0.75  13.4 in. 1.0  4500 Set tail cover = 2 in., then dh  0.7 13.4  9.4 in. . This is OK in a 16 in. column.

Negative moment point of inflection at  0.164  27 12  53.1 in. Rule 3-b - Extend d to 53.1+21.5 = 74.6 in. Cutoff at 75 in. = 6 ft-3 in. from face of column. Rule 4-b - Since 75 in. > 39.1 in. Rule 2 is satisfied. Span AB, interior end, 2 No. 8 and 4 No. 7 bars Point of inflection at  0.24  27 12  77.8 in. Rule 3-b - Extend d to 77.8+21.5 = 99.3 in. Say 8 ft.-4 in. from face of column Rule 4-b – Since 100 in. > 44.7 in. Rule 2 is satisfied. Span BC, 2 No. 8 and 4 No. 7 bars Negative moment point of inflection at  0.24  30 12   86.4 in. Rule 3-b – Extend to 108.9 in. Say 9 ft- 1 in, from face of column Rule 4-b – OK Since all cutoffs are past points of inflection, they are not in zones of flexural tension, therefore extra stirrups are not needed. Provide 2 No. 4 top bars as stirrup support, lab splice with negative moment steel.

10-20


Fig. S10-3.2 Beam reinforcing detailing.

10-21


10-4

Repeat Problem 10-3, but cut off up to 50 percent of the negative- and positivemoment bars in each span where they are no longer required.

Bat cutoff for positive moment steel in AB span Flexural reinforcement: 4#7 bars Extend two bars full length and into each support, cutoff the other two where they no longer required  50% of M u  . From Fig. A-2, flexural cutoff point is at 0.21 (support A), and 0.22

n

n

 0.21  27 12   68 in. from exterior end

 0.22   27 12   71 in. from interior end (support B).

Rule 3-a: Extend d  21.5 in. past the flexural cutoff point 68 in.  21.5 in.  46.5 in. from exterior end 71 in.  21.5 in.  49.5 in. from interior end Rule 4-a: Distance between midspan to cutoff point = 46.5 115.5   13.5  12 in.    in.    in. 49.5  112.5  115.5 for #7 bar  39.1 in.    in. 112.5 Bar cutoff for negative moment steel d

Flexural reinforcement: 2#8 and 4#7 bars AB span, interior end From Fig. A-2, flexural cutoff point is at 0.104 n  0.104   27 12   34 in. Rule 3-b: Extend d  21.5 in. past the flexural cutoff point 34 in.+ 21.5 in.  55.5 in. Rule 4-b: d for #7 top bar  39.1 in.< 55.5 in. Therefore, use 55.5 in. for cutoff point. BC span From Fig. A-1, flexural cutoff point is at 0.1 n  0.1  30 12   36 in. Rule 3-b: Extend d  22.5 in. past the flexural cutoff point 36 in.+ 22.5 in.  58.5 in. Rule 4-b: d for #7 top bar  39.1 in.< 58.5 in. Therefore, use 58.5 in. for cutoff point.

10-22


Chapter 11 11-1

The column shown in Fig. P11-1 is made of 4000 psi concrete and Grade-60 steel. a)

Compute the theoretical capacity of the column for pure axial load.

First compute the gross area of the section, and the area of steel in the column. Ag  18 in 18 in  324 in 2

Ast  6 Ab  6 1.0 in 2  6 in 2 Now we can compute the theoretical capacity of the column. Po  0.85 f 'c  Ag  Ast   f y A st  0.85  4 ksi  324 in 2  6 in 2  60 ksi  6 in 2  1440 k



b)



Compute the maximum permissible  Pn for the column.

 Pn,max  0.80 Po  0.80  0.65 1440 k  749 k Note that the factors 0.80 and 0.65 would change to 0.85 and 0.75 for spiral columns.

11-2

Why does a spiral improve the behavior of a column?

As any column is loaded, and thus shortened, the concrete will expand laterally. When this expansion occurs, transverse reinforcement is engaged and will tend to react against any further expansion of the concrete within the core. This results in a state of tri-axial compression within the core of the column, which significantly improves both strength and ductility. The circular spiral is much more effective than tied reinforcement at confining this expansion for two reasons. First, spirals are often spaced more closely together than tied stirrups, so the confinement is more uniformly applied to the core. Second, the confinement stresses are transformed directly into hoop stresses in the spiral, which is a much more efficient mechanism for reacting to the core’s expansion than the straight legs of tied stirrups can provide (see section 11-2).

11-3

Why are tension splices required in some columns?

Even when columns are subjected to axial loads, reinforcing bars can still often be stressed in tension when moments are concurrently applied to the section. When this is the case, tension splices (either Class A or Class B) are required for those bars expected to be resisting tension. However, since it is most practical from a construction standpoint to use the same length of lap splices on all bars within a section, all splices should be specified as tension splices when some of the bars are expected to be in tension.

11-1


11-4

Compute the balanced axial load and moment capacity of the column shown in Fig. P11-1. Use and .

Assume bending around an axis parallel to the two layers of steel. To calculate the balanced point of the interaction diagram, set the extreme compression fiber strain to  c  0.003 and the extreme steel tensile strain to  y  f y / E  60 ksi / 29,000 ksi  0.00207 . Begin by calculating the depth of the compression zone: c c d  c  y

c

0.003 1.128 in    18 in  1.5 in  0.375 in   0.003   0.00207   2 

c  9.21 in a  1c  0.85  9.21 in  7.83 in Now calculate the strain and stress in each layer of steel:  s1  0.00207 f s1  60 ksi

Fig. S11-3 9.21 in  1.5 in  0.375 in  1.128 in / 2  cd'  s2   c   0.003  0.00221 c 9.21 in 60 ksi  f s 2  smaller of   E s 2  29,000 ksi  0.00221  64.1 ksi Calculate the force in the concrete and in each layer of steel: Cc  0.85 f 'c ab  0.85  4 ksi  7.83 in 18 in  479 k

Fs1  As1 f s1  3 in 2  60 ksi  180 k

Fs 2  As 2  f s 2  0.85 f 'c   3 in 2   60 ksi  0.85  4 ksi   170 k

Now we can calculate the nominal axial load and moment at the balanced point: Pn  Cc  Fs 2  Fs1  479 k  170 k  180 k  469 k

h a h  h  M n  Cc     Fs1   d   Fs 2   d '  2 2 2  2   18 in 7.83 in   18 in   18 in  M n  479 k     180 k    15.56 in   170 k    2.44 in   2   2  2   2  M n  4730 k-in  394 k-ft

And, since  s1   y , the section is compression controlled, and   0.65 .  Pn  0.65  469 k  305 k  M n  0.65  394 k-ft  256 k-ft

11-2


11-5

For the column shown in Fig. P11-5, use a strain-compatibility solution to compute five points on the interaction diagram corresponding to points 1 to 5 in Fig. 11-22. Plot the interaction diagram. Use and .

Begin by calculating the depth and area of each layer of steel, and other constants. d1  18 in  1.5 in  0.375 in  1.128 in / 2  15.56 in

A1  3 Ab  3 1.0 in 2  3.0 in 2 d2  h / 2  18 in / 2  9.0 in

A2  2 Ab  2 1.0 in 2  2.0 in 2

d3  1.5 in  0.375 in  1.128 in / 2  2.44 in

A3  3 Ab  3 1.0 in 2  3.0 in 2

Ag  hb  18 in 18 in  324 in 2

Ast  8 Ab  8 1.0 in 2  8.0 in 2

1  0.80

Point 1: Pure Axial Load: Po  0.85 f 'c ( Ag  Ast )  f y Ast  0.85  5 ksi  (324 in 2  8 in 2 )  60 ksi  8 in 2  1830 k Note that this 0.85 is k3, not β1. Since the column is compression controlled,   0.65 .  Pn  0.65 1830 k  1190 k

 Pn,max  0.80  0.65 1830 k  949 k  M n  0.65  0 k-ft  0 k-ft

Point 2: Zero tension on one face: Begin by calculating the depth of the compression zone: c  h  18 in a  1c  0.80 18 in  14.4 in Now calculate the strain and stress in each layer of steel: c  d1 18 in  15.56 in  s1   c   0.003  0.00041 c 18 in 60 ksi  f s1  smaller of   E s1  29,000 ksi  0.00041  11.9 ksi

c  d2 18 in  9 in  c   0.003  0.00150 c 18 in 60 ksi  f s 2  smaller of   E s 2  29,000 ksi  0.00150  43.5 ksi

 s2 

c  d3 18 in  2.44 in  c   0.003  0.00259 c 18 in 60 ksi  f s 3  smaller of   E s 3  29,000 ksi  0.00259  75.1 ksi

 s3 

Calculate the force in the concrete and in each layer of steel: Cc  0.85 f 'c ab  0.85  5 ksi 14.4 in 18 in  1100 k

Fs1  As1  f s1  0.85 f 'c   3 in 2  11.9 ksi  0.85  5 ksi   23.0 k

11-3


Fs 2  As 2  f s 2  0.85 f 'c   2 in 2   43.5 ksi  0.85  5 ksi   78.5 k and, since d3  a ,

Fs 3  As 3 f s 3  3 in 2  60 ksi  180 k Now we can calculate the nominal axial load and moment: Pn  Cc  Fs1  Fs 2  Fs 3  1100 k  23.0 k  78.5 k  180 k  1380 k

h a h  h  h  M n  Cc     Fs1   d1   Fs 2   d 2   Fs1   d3  2 2 2 2 2         14.4 in   M n  1100 k   9 in   23.0 k  9 in  15.56 in   78.5 k  9 in  9 in   180 k   9 in  2.44 in  2   M n  3010 k-in  251 k-ft And, since  s1 is in compression, the section is compression controlled, and   0.65 .  Pn  0.65 1380 k  897 k  M n  0.65  251 k-ft  163 k-ft

Point 3: Balanced Point: Begin by calculating the depth of the compression zone: c 0.003 c  d1   15.56 in  9.21 in  c  y 0.003   0.00207  a  1c  0.80  9.21 in  7.37 in

Now calculate the strain and stress in each layer of steel: c  d1 9.21 in  15.56 in  s1   c   0.003  0.00207 c 9.21 in 60 ksi  f s1  smaller of   E s1  29,000 ksi  0.00207  60 ksi

c  d2 9.21 in  9 in  c   0.003  0.00007 c 9.21 in 60 ksi  f s 2  smaller of   E s 2  29,000 ksi  0.00007  1.98 ksi

 s2 

c  d3 9.21 in  2.44 in  c   0.003  0.00221 c 9.21 in 60 ksi  f s 3  smaller of   E s 3  29,000 ksi  0.00221  64.1 ksi

 s3 

Calculate the force in the concrete and in each layer of steel: Cc  0.85 f 'c ab  0.85  5 ksi  7.37 in 18 in  564 k

Fs1  As1 f s1  3 in 2  60 ksi  180 k and, since d 2  a ,

Fs 2  As 2 f s 2  2 in 2 1.98 ksi  3.96 k

Fs 3  As 3  f s 3  0.85 f 'c   3 in 2   60 ksi  0.85  5 ksi   167 k

11-4


Now we can calculate the nominal axial load and moment at the balanced point. Pn  Cc  Fs1  Fs 2  Fs 3  564 k  180 k  3.96 k  167 k  555 k

h a h  h  h  M n  Cc     Fs1   d1   Fs 2   d 2   Fs1   d3  2 2 2 2 2         7.37 in   M n  564 k   9 in   180 k   9 in  15.56 in   3.96 k  9 in  9 in   167 k   9 in  2.44 in  2   M n  5280 k-in  440 k-ft

And, since  s1   y , the section is compression controlled, and   0.65 .  Pn  0.65  546 k  355 k  M n  0.65  440 k-ft  286 k-ft

Point 4: Tension control limit Begin by calculating the depth of the compression zone: c 0.003 c  d1   15.56 in  5.84 in  c  s1 0.003   0.005 a  1c  0.80  5.84 in  4.67 in

Now calculate the strain and stress in each layer of steel: c  d1 5.84 in  15.56 in  s1   c   0.003  0.00500 c 5.84 in 60 ksi  f s1  smaller of   E s1  29,000 ksi  0.00500  145 ksi

c  d2 5.84 in  9 in  c   0.003  0.00162 c 5.84 in 60 ksi  f s 2  smaller of   E s 2  29,000 ksi  0.00162  47.1 ksi

 s2 

c  d3 5.84 in  2.44 in  c   0.003  0.00175 c 5.84 in 60 ksi  f s 2  smaller of   E s 2  29,000 ksi  0.00175  50.8 ksi

 s3 

Calculate the force in the concrete and in each layer of steel: Cc  0.85 f 'c ab  0.85  5 ksi  4.67 in 18 in  357 k

Fs1  As1 f s1  3 in 2  60 ksi  180 k Fs 2  As 2 f s 2  2 in 2  47.1 ksi  94.2 k

Fs 3  As 3  f s 3  0.85 f 'c   3 in 2   50.8 ksi  0.85  5 ksi   140 k

Now we can calculate the nominal axial load and moment at the balanced point. Pn  Cc  Fs1  Fs 2  Fs 3  357 k  180 k  94.2 k  140 k  223 k

h a h  h  h  M n  Cc     Fs1   d1   Fs 2   d 2   Fs1   d3  2 2 2  2  2 

11-5


4.67 in   M n  357 k   9 in   180 k   9 in  15.56 in   94.2 k  9 in  9 in   140 k   9 in  2.44 in  2   M n  4480 k-in  373 k-ft

And, since  s1  0.005 , the section is tension controlled, and   0.9 .  Pn  0.9  223 k  201 k  M n  0.9  373 k-ft  336 k-ft Point 5: Pure Tension Pn  Ast f y  8in 2  60 ksi  480 k M n  0 k-ft

And, since the section is in pure tension, it is tension controlled, and   0.9 .  Pn  0.9  480 k  432 k  M n  0.9  0 k-ft  0 k-ft 2000

Axial Load (kip)

1500

1000

500

0 0

100

200

300

Moment (k-ft) -500

Fig. S11-4

11-6

400

500


11-6

Use the interaction diagrams in Appendix A to compute the maximum moment, Mu, that can be supported by the column shown in Fig. P11-1 if (Use and ): (a)

Pu = 583 kips.

Calculate the parameters required to use the interaction diagrams: A 6  1.0 in 2  g  st   0.0185 Ag 18 in  18 in

d  d ' 18 in  1.5 in  0.375 in  1.128 in / 2   1.5 in  0.375 in  1.128 in / 2    0.729 h 18 in  Pn Pu 583 k    1.80 ksi bh bh 18 in 18 in



Now go to the interaction diagrams: Mn  0.46 ksi for   0.60 From Fig. A-7a: bh2 Mn  0.50 ksi for   0.75 From Fig. A-7b: bh2 Mn  0.49 ksi By interpolation for   0.729 , bh2 Finally, we can compute the maximum moment carried by this section:

M u   M n  0.49bh2  0.49 18 in  18 in   2860 k-in  238 k-ft 2

(b)

Pu = 130 kips.


d  d ' 18 in  1.5 in  0.375 in  1.128 in / 2   1.5 in  0.375 in  1.128 in / 2    0.729 h 18 in  Pn Pu 130 k    0.40 ksi bh bh 18 in 18 in




11-7


M u   M n  0.54bh2  0.54 18 in  18 in   3150 k-in  262 k-ft (c) e = 4 in. 2




d  d ' 18 in  1.5 in  0.375 in  1.128 in / 2   1.5 in  0.375 in  1.128 in / 2    0.729 h 18 in e 4 in   0.222 h 18 in


M u   M n  0.44bh2  0.44 18 in  18 in   2570 k-in  214 k-ft 2

11-7

Use the interaction diagrams in Appendix A to select tied-column cross sections to support the loads given in the accompanying list. In each case, use and . Design the ties and calculate the required splice lengths, assuming that the bars extending up from the column below are the same diameter as in the column you have designed. Draw a typical cross section of the column showing the bars and ties. (a)

Pu = 390 kips, Mu = 220 k-ft, square column with bars in two faces.

First estimate the size of the section required. For the first iteration, assume  g  0.02 . Ag ,trial 

Pu

0.40  f 'c  f y  g 



390 k  188 in 2  13.7 in 13.7 in 0.40   4 ksi  60 ksi  0.02 

So try a 14 in square column, assuming #3 stirrups and #8 bars. Then calculate the parameters required to use the interaction diagrams, and reference them to select an appropriate reinforcement ratio: 14 in  2  1.5 in  0.375 in  0.5 in    0.66 14 in  Pn Pu 390 k    1.99 ksi bh bh 14 in 14 in

11-8


Mn bh

2



M u 220 k-ft  12 in/ft   0.962 ksi 2 bh2 14 in  14 in 

From Figs. A-6a and A-6b, the section would require , so a larger section is required. Therefore, we will try a square column with 18 in sides. Re-calculate the parameters required to use the interaction diagrams, and reference them to select an appropriate reinforcement ratio: 18 in  2  1.5 in  0.375 in  0.5 in    0.736 18 in  Pn Pu 390 k    1.20 ksi bh bh 18 in 18 in  M n M u 220 k-ft  12 in/ft  2   0.453 ksi 2 bh2 bh 18 in  18 in  From Fig. A-6a:  g  0.021 for   0.60 From Fig. A-6b:  g  0.016 for   0.75 By interpolation for   0.736 ,  g  0.017 Finally, we can compute the area of steel required to reinforce this section: As ,required  0.017bh  0.017 18 in 18 in  5.51 in 2 Now select the bars: fits in an 18 in section 8#8 bars  6.32 in 2 2 fits in an 18 in section 10#7 bars  6.00 in 2 fits in an 18 in section 14#6 bars  6.16 in Select #7 bars, placing 5 along the top and bottom faces. As indicated on the interaction diagrams, f s  0.5 f y in the extreme tensile layer of steel. Therefore, Class B tension splices are required. From Table A-13, the splice length must be 54 in. As transverse reinforcement we are permitted to use #3 bars since the longitudinal bars are not larger than #10s. Now select the vertical spacing of the ties:  16d b  16  0.875 in  14 in  s   48dbt  48  0.375 in  18 in min(b, h)  min(18 in,18 in)  18 in  Select #3 stirrups spaced at 14 in o.c.

11-9


Fig. S11-7a (b)

Pu = 710 kips, Mu = 50 k-ft, square column with bars in four faces.

First estimate the size of the section required. For the first iteration, set  g  0.02 : Ag ,trial 

Pu

0.40  f 'c  f y  g 



710 k  341 in 2  18.5 in 18.5 in 0.40   4 ksi  60 ksi  0.02 

So try a 20 in square column, assuming #3 stirrups and #8 bars. Then calculate the parameters required to use the interaction diagrams, and reference them to select an appropriate reinforcement ratio: 20 in  2  1.5 in  0.375 in  0.5 in    0.76 20 in  Pn Pu 710 k    1.78 ksi bh bh 20 in  20 in  M n M u 50 k-ft  12 in/ft  2   0.075 ksi 2 bh2 bh 20 in   20 in  From Figs. A-9a and A-9b, the section would require  g

0.01 , so a smaller section

would be desirable. Try a square column with 18 in sides. Re-calculate the parameters required to use the interaction diagrams, and reference them to select an appropriate reinforcement ratio: 18 in  2  1.5 in  0.375 in  0.5 in    0.736 18 in  Pn Pu 710k    2.19 ksi bh bh 18 in 18 in  M n M u 50 k-ft  12 in/ft  2   0.103 ksi 2 bh2 bh 18 in  18 in  From Fig. A-9a:  g  0.016 for   0.60 From Fig. A-9b:  g  0.015 for   0.75 By interpolation for   0.736 ,  g  0.015

11-10


Finally, we can compute the area of steel required to reinforce this section: As ,required  0.015bh  0.015 18 in 18 in  4.86 in 2 Now select the bars: 12#6 bars  5.28 in 2 fits in an 18 in. section Select #6 bars, placing 4 along each face. From the interaction diagrams, the extreme tensile layer of steel is under compression. Therefore, from Table A-13, a splice of length 0.83  23 in  19.1 in  20 in is required.

As transverse reinforcement, we have selected #3 bars since the longitudinal bars are not larger than #10 bars. Now select the vertical spacing of the ties: 16d b  16  0.75 in  12 in   s   48dbt  48  0.375 in  18 in min(b, h)  min(18 in,18 in)  18 in  Select #3 stirrups spaced at 12 in o.c.

Fig. S11-7b

(c)

Pu = 200 kips, Mu = 240 k-ft, square column with bars in four faces.

First estimate the size of the section required. For the first iteration, set  g  0.02 : (

)

(

)

However, since the moment is relatively high, try a 16 in square column, assuming #3 stirrups and #8 bars. Then calculate the parameters required to use the interaction diagrams, and reference them to select an appropriate reinforcement ratio:

11-11




16 in  2  1.5 in  0.375 in  0.5 in  16 in

Mn bh

2



 0.70


From Figs. A-9a and A-9b, the section would require a very high required. Try a square column with 18 in sides.

, so a larger section is

Re-calculate the parameters required to use the interaction diagrams, and reference them to select an appropriate reinforcement ratio: 18 in  2  1.5 in  0.375 in  0.5 in    0.736 18 in

Mn bh

2




From A-9a: for   0.60 From A-9b: for   0.75 By interpolation for   0.736 , Finally, we can compute the area of steel required to reinforce this section:

Now select the bars: 8#8 bars  6.32 in 2 fits in an 18 in section Select #8 bars, placing 3 along the each face. From the interaction diagrams, f s  0.5 f y in the extreme tensile layer of steel. Therefore, Class B tension splices are required. From Table A-13, the splice length must be 62 in. As transverse reinforcement, we have selected #3 bars since the longitudinal bars are not larger than #10 bars. Now select the vertical spacing of the ties: 16d b  16  1.00 in  16 in   s   48dbt  48  0.375 in  18 in min(b, h)  min(18 in,18 in)  18 in  So we must use #3 stirrups spaced at 16 in o.c.

11-12


Fig. S11-7c

11-8

Use the interaction diagrams in Appendix A to select spiral-column cross sections to support the loads given in the accompanying list. In each case, use and . Design the spirals and calculate the required splice lengths. Draw a typical cross section of the column showing the bars and spiral. (a)

Pu = 600 kips, Mu = 65 k-ft.

First estimate the size of the section required. For the first iteration, set  g  0.02 . Ag ,trial 

Pu

0.50  f 'c  f y  g 



600 k  194 in 2  15.7 in dia. 0.50   5 ksi  60 ksi  0.02 

Try a 16 in diameter column, assuming #3 spiral and #8 bars. Then calculate the parameters required to use the interaction diagrams, and reference them to select an appropriate reinforcement ratio: 16 in  2  1.5 in  0.375 in  0.5 in    0.703 16 in  Pn Pu 600 k    2.99 ksi Ag Ag 201 in 2

Mn hAg



M u 65 k-ft  12 in/ft   0.243 ksi hAg 16 in  201 in 2

From Fig. A-13a:  g  0.015 for   0.60 From Fig. A-13b:  g  0.012 for   0.75 By interpolation for   0.703 ,  g  0.013 Finally, we can compute the area of steel required to reinforce this section:

11-13


16 in   2.61 in 2 d2  0.013     0.013    4 4 2

As ,required

Now select the bars: 6#6 bars  2.64 in 2 Select #6 bars, spaced evenly around the column. From the interaction diagrams, the extreme tensile layer of steel is under compression. Therefore, from Table A-13, a splice of length 0.75  23 in  17.3 in  18 in is required. As transverse reinforcement, we have selected a #3 spiral since the longitudinal bars are not larger than #10 bars. Now select the pitch of the spiral: 3 in   2  d sp 2 f yt    0.375 in   60 ksi s  0.45D f '  A / A  1  0.45  13 in  5 ksi  201 in 2 / 134 in 2  1  1.81 in c c  g ch     Use a 16 in diameter column reinforced with 6 #6 bars. Use 18 in lap splices, and #3 spirals with a pitch of 1.75 in as transverse reinforcement. (b) Pu = 200 kips, Mu = 150 k-ft. First estimate the size of the section required. For the first iteration, set  g  0.02 : ( ) ( ) Since there is also a moment applied, try an 18 in diameter column, assuming #3 spiral and #8 bars. Then calculate the parameters required to use the interaction diagrams, and reference them to select an appropriate reinforcement ratio: 18 in  2  1.5 in  0.375 in  0.5 in    0.736 18 in

 Pn Ag

Mn hAg



Pu  Ag



200  18    2

2

 0.79 ksi

M u 150 k-ft  12 in/ft   0.394 ksi hAg 18 in  254 in 2

From Fig. A-13a:  g  0.016 for   0.60 From Fig. A-13b:  g  0.012 for   0.75 By interpolation for   0.74 ,  g  0.012 Finally, we can compute the area of steel required to reinforce this section:

18 in   3.05 in 2 d2  0.012     0.012    4 4 2

As ,required

11-14


Now select the bars: 8#6 bars  3.52 in 2 6#7 bars  3.6 in 2 Select #6 bars, spaced evenly around the column. From the interaction diagrams, the extreme tensile layer of steel is in tension, but only requires a Class A splice as long as not all of the bars are spliced at the same location. Since, in reality, it is likely that all bars will be spliced in the same plane, specify a Class B splice regardless. Therefore, from Table A-13, a splice of length 33 in is required. As transverse reinforcement, we have selected a #3 spiral since the longitudinal bars are not larger than #10 bars. Now select the pitch of the spiral: 3 in   2  d sp 2 f yt    0.375 in   60 ksi s  0.45D f '  A / A  1  0.45  15 in  5 ksi  254 in 2 / 177 in 2  1  1.81 in c c  g ch     Use an 18 in diameter column reinforced with 8 #6 bars. Use 33 in lap splices, and #3 spirals with a pitch of 1.75 in as transverse reinforcement.

11-9

Design a cross section and reinforcement to supports Pu = 450 kips, Mux = 100 k-ft, and Muy = 130 k-ft. Use and .

Although the strain compatibility method (shown in example 11-5) is the most theoretically correct method for designing columns for biaxial loading, it is seldom used in design. Here we will use two more common methods for designing a column, the equivalent eccentricity method and the Bresler reciprocal load method. Any method outlined in section 11-7 is appropriate for the solution to this problem. Equivalent eccentricity method: First select the dimensions of the trial section, assuming   0.015 : Pu 450 k Ag ,trial    230 in 2  15.2 in by 15.2 in 0.40  f 'c  f y  g  0.40   4 ksi  60 ksi  0.015  Since biaxial moments are also applied to the section, select an 18 in square section. Assuming #3 ties and #9 bars: 18 in  2  1.5 in  0.375 in  0.56 in    0.73 18 in Now compute ex, ey, and eox. M uy 130 k-ft  12 in/ft ex    3.47 in Pu 450 k M 100 k-ft  12 in/ft ey  ux   2.67 in Pu 450 k

11-15


Since our trial section is square,

ex



ey

x

y

Pu 450 k   0.347  0.4 f 'c Ag 4 ksi  18 in 2

 Pu   f y  40,000 psi       f ' A c g    100,000 psi     60,000 psi  40,000 psi  450 k     0.5    0.847 2    100,000 psi 4 ksi  18 in       ey x 0.847  2.67 in  18 in eox  ex   3.47 in   5.73 in 18 in y

   0.5 

Therefore, we can design our column for: Pu  450 k , and

M oy  Pu eox  450 k  5.73 in  2580 k-in  215 k-ft

Now use the interaction diagrams in Appendix A to determine  g .

 Pn



Ag

Mn hAg

Pu 450 k   1.39 ksi Ag 324 in 2



M oy hAg



2580 k-in  0.442 ksi 18 in  324 in 2

Since   0.73  0.75 we can use Fig A-9b without interpolating for this design. So use  g  0.023 Finally, we can compute the area of steel required to reinforce this section, and select the bars: Ast   g Ag  0.023  324 in 2  7.45 in 2 Select 8 #9 bars, with 3 bars along each face. Select ties and splice lengths as appropriate. We can check this solution using the Bresler reciprocal load method. Remember that we have an 18 in square column reinforced with 8 #9 bars. A 8 in 2  g  st   0.0247 Ag 324 in 2



18 in  2  1.5 in  0.375 in  0.56 in 

 0.73 18 in Compute  Pnx , the factored load capacity corresponding to ex and  g . ex 

M uy Pu



130 k-ft  12 in/ft e 3.47 in  3.47 in , and x   0.193 . 450 k 18 in x

From Fig. A-9b,

 Pnx  1.9 ksi bh

Therefore  Pnx  616 k Compute  Pny , the factored load capacity corresponding to ey and  g .

11-16


ey 

M ux 100 k-ft  12 in/ft e 2.67 in   2.67 in , and x   0.148 . Pu 450 k 18 in x

 Pnx

 2.2 ksi bh Therefore  Pny  713 k From Fig. A-9b,

From equ. 11-7 calculate Pno :





Pno   k3 f 'c   Ag  Ast   f y Ast   0.85  4 ksi  324 in 2  8 in 2  60 ksi  8 in 2  1560 k 1 1 1 1 1 1 1 1        Pu  Pn  Pnx  Pny  Pno 616 k 713 k 0.65  1560 k

Pu  490 k , so our section is sufficient for the loading defined here.

11-17


Chapter 12 12-1

A hinged end column 18-ft tall supports unfactored loads of 100 kips dead load and 60 kips live load. These loads are applied at an eccentricity of 3 in. at bottom and 5 in. at the top. Both eccentricities are on the same side of the centerline of the column. Design a tied column with at least three bars per face using and . Factored loads and moments Pu  1.2 100 kips +1.6  60 kips  216 kips M1 / M 2  0.6 (Note that the column is bent in single curvature)

Estimate column size Assume  g  0.015 Ag (trial ) =

Pu 216 kips   110 in.2 0.4  f c  f y  g  0.4   4 ksi + 60 ksi  0.015

Choose a column cross section of 12 in. x 12 in. Check whether the column is slender k u 1.0  (18  12) in.   60 r (0.3  12 in.) M  34  12  1   34  12  0.6  26.8  M2  M  k u  60 > 34  12  1   26.8 r  M2  The column is quite slender, increase column size to 16 in. x 16 in.

Are moments greater than the minimum? M 2,min  Pu (0.6  0.03h) =216 kips   0.6 in.  0.03 16 in.  233 k-in. < M 2  use M 2

Compute EI

EI 

0.4 Ec I g 1  d

Ec  57,000 4000  3605 103 psi

16 in. 

4

 5460 in.4 12 factored dead load 1.2 100 kips d    0.56 all factored load 216 kips Ig

12-1


EI 

0.4   3605  103 psi   5460 in.4 1  0.56

 5.05  109 lb-in.2

Magnified moment M c   ns M 2

 ns 

Cm  1.0 P 1 u  Pc

Cm  0.6  0.4 Pc 

 ns 

 2 EI

k 

2



M1  0.6  0.4  0.6  0.84 M2

 2  5.05  109 lb-in.2  1.0  18  12  in. 2

2

 1070  103 lb =1070 kips

0.84  1.15 216 kips 1 0.75  1070 kips

Select reinforcement Assume # 8 bars for longitudinal reinforcement, # 3 bars for the ties, and a clear concrete cover of 1.5 in. 16 in.  2  1.5 in.  2  0.375 in.  1 in.   0.70 16 in.  Pn Pu 216 kips    0.84 ksi bh bh (16 in.)2

Assume E-type reinforcement Fig.A-9a shows Fig.A-9b shows  Ast  0.01Ag  2.56 in.2



. Use

.



Use 6 bars # 6 Ast  2.64 in.2 for the column of 16 in.  16 in.

12-2

Repeat Problem 12-1, but with the top eccentricity to the right of the centerline and the bottom eccentricity to the left. Factored loads and moments Pu  216 kips M1 / M 2  0.6 (Note that the column is bent in double curvature)

12-2


Estimate column size Assume  g  0.015 Ag  110 in.2

Choose a column cross section of 14 in. x 14 in. Check if the column is slender k u 1.0  (18  12) in.   51.4 r (0.3  14 in.) M  34  12  1   34  12  (0.6)  41.2  40 (40 governs)  M2  M  k u  51.4 > 34  12  1   40 r  M2  The column is slender.

Are moments greater than the minimum? M 2,min  Pu (0.6  0.03h) =216 kips   0.6 in.  0.03 14 in.  220 k-in. < M 2  use M 2

Compute EI

EI 

0.4 Ec I g 1  d

Ec  3605 103 psi Ig

14 in. 

4

12 d  0.56

EI 

 3200 in.4

0.4   3605  103 psi   3200 in.4 1  0.56

 2.96  109 lb-in.2


 ns 

Cm  1.0 P 1 u  Pc

Cm  0.6  0.4 Pcr 

 ns 

 2 EI

k 

2



M1  0.6  0.4  (0.6)  0.36 M2

 2  2.96  109 lb-in.2  1.0  18  12  in. 2

2

 626  103 lb =626 kips

0.36  0.67  1, use  ns  1 216 kips 1 0.75  626 kips

12-3


Select reinforcement Assume # 8 bars for longitudinal reinforcement, # 3 bars for ties, and a clear concrete cover of 1.5 in. 14 in.  2  1.5 in.  2  0.375 in.  1 in.   0.66 14 in.

 Pn bh



Pu 216 kips   1.10 ksi bh (14 in.)2

Assume R-type reinforcement Fig.A-9a shows Fig.A-9b shows

Use 8 #6 bars (

12-3

) for the column of 14 in.  14 in.

Figure P12-3 shows an exterior column in a multistory frame. The dimensions are center-to-center of the joints. The beams are 12 in. wide by 18 in. in over-all depth. The floor slab is 6 in. thick. The building includes a service core which resists the majority of the lateral loads. Use f c = 5000 psi and f y = 60,000 psi. The loads and moments on column AB are: Factored dead load: Axial force = 260 kips Moment at top = 60 kip-ft Moment at bottom = -80 kip-ft Factored live load: Axial force = 200 kips Moment at top = 50 kip-ft Moment at bottom = -75 kip-ft Design column A-B using a square cross section with at least three bars per face. Since the building has a service core which resists the majority of the lateral loads, the frame is braced, or non-sway. Factored loads and moments

|

|

The column is in double curvature. ⁄ ⁄ 12-4


22 ft

C 12 ft A

24 ft

B 12 ft D

Fig. P12-3

Estimate column size Assume  g  0.015 ( ) ( ) Choose a column cross section of 16 in. x 16 in. Check if the column is slender From Table 12-2, k  0.90 u   24  12  in.  18 in.  270 in.

k

u

r



0.9  270 in.  50.6 (0.3  16 in.) (

k

)

(

)

 50.6 > 42.5 r Column is slender. u

Are moments greater than the minimum? ( )  use M 2 12-5

(

)


Compute k Assume the columns CA and BD also have a cross section of 16 in. x 16 in.

I  column   0.7 I g  column   0.7 

16 in. 12

4

 3820 in.4

I  beam   0.35I g (beam)  0.35   2I web   0.7 

12 in.  18 in.

3

12

 4080 in.4

 EI  3820 Ec  26.5Ec     c CA 12  12 

 EI  3820 Ec  13.3Ec     c  AB 12  24   EI  4080 Ec   15.5Ec      b  beams 12  22 

26.5Ec  13.3Ec  2.57    bottom joint  15.5Ec From nomograph, read k  0.87

  top joint  

Compute EI

EI 

Ec I g 2.5(1   d )

Ec  57,000 5000  4030 103 psi I g  5460 in.4

260  0.57 460 4030  103 psi  5460 in.4 EI   5.61 109 lb-in.2 2.5  (1  0.57)

d 


 ns 

Cm  1.0 P 1 u  Pc

Cm  0.6  0.4 Pc 

 ns 

 2 EI

k

u



2



M1  0.6  0.4  (0.71)  0.32 M2

 2  5.61 109 lb-in.2 

0.87  270 in.

2

 1000  103 lb =1000 kips

0.32  0.83  1, use  ns  1 460 kips 1 0.75  1000 kips

12-6


Select reinforcement Assume # 8 bars for longitudinal reinforcement, # 3 bars for ties, and a clear concrete cover of 1.5 in. 16 in.  2  1.5 in.  2  0.375 in.  1 in.   0.70 16 in.

Assume E-type reinforcement Fig. A-10a yields Fig. A-10b yields

) for the column of 16 in.  16 in.

Use 8 #8 bars (

12-4

Use the ACI moment-magnifier method to redesign the columns in the main floor of Example 12-3 assuming that the floor-to-floor height of the first story is 16 ft 0 in. rather than 18 ft 0 in. Also, assume the lateral wind forces are 15 percent larger than those used in Example 12-3. The floor plan and elevation is shown in Fig. S12-1 The following design shall be performed for a typical interior frame, for instance the fram along line 2. The design of the columns will follow the following steps: A. B. C. D. E. F.

Calculate loads Calculate the beam and column properties and modulus of elasticity Select preliminary column size Check with gravity load case Check with gravity plus wind load case Finalize the design of column reinforcement

A. Calculate loads for a typical interior frame Dead load from roof Distributed loads on beams 18 in.  24 in.  0.150 k/ft 3  6 in.  DL    0.150 k/ft 3  0.025 k/ft 2   10  144 in 2 /ft 2 12 in/ft   1.45 k/ft Distributed load on girders (beams supporting other beams)  18 in.   18 in.  30 in  DL    0.150  0.6 k/ft   0.025   2 2   12 in./ft   144 in /ft  Concentrated on the interior columns (

)

12-7


6

20'

5

20'

4 a) Plan

20' 1

1 3

20'

N 2

20'

1 32' A

30' B

32' C

D

Roof 11' 6" 5 th floor 11' 6" 4 th floor 11' 6" 3 th floor

b) Section 1-1 11' 6"

2 th floor

16' 0" Ground floor 11' 6"

Slab thickness: 6"; Column size: 18" x 18"; Beam size: 18" x 30"

Fig. S12-1

12-8


Concentrated on the exterior columns (

)

Dead load from other floors Distributed loads on beams 18 in.  24 in.  0.150 k/ft 3  6 in.  DL    0.150 k/ft 3  0.020 k/ft 2   10  144 in 2 /ft 2 12 in/ft   1.40 k/ft Distributed load on girders (beams supporting other beams)  18 in.   18 in.  30 in  DL    0.150  0.593 k/ft   0.020   2 2   12 in./ft   144 in /ft  Concentrated on the interior columns (

)

Concentrated on the exterior columns (

)

For each column of 11.5 ft, a weight of 3.88 kips is added for the column self-weight. For each column of 16.0 ft, a weight of 5.40 kips is added for the column self-weight.

Live load from roof Concentrated live load on the interior columns (

)

Concentrated live load on the exterior columns (

)

Live load from other floors Concentrated live load on the interior columns (

)

Concentrated live load on the exterior columns (

)

B. Calculate the beam and column properties and modulus of elasticity Column: A  18 in. 18 in.  324 in.2

I c  0.7 I g

18 in.  0.7  12

4

 6120 in 4 12-9


Beam: Effective flange width 18 in.  8  6 in.  66 in.

A  6  66  18  24  828 in 2

I b  0.7 I g  beam-web   0.7 

18 in.   30 in.

3

 28350 in.4 12 Note that the selection of rigid end zones follows Example 12-3. Modulus of elasticity E  57000 4000  3600 103 psi

C. Select preliminary column size Columns are sized based on the gravity load 1.2D  1.6L  0.5Lr  . In this load combination, live load can be reduced. From separate analyses of dead load, live load from the roof floor, and live load from the other floors, the axial loads of the columns in the ground floor are shown in the Table P12-1. The axial load in the exterior column from the live load from the other floors then can be reduced with a reduction factor as follows: 15 0.25   0.46 4   4  16.75  20  The axial load in the interior column from the live load from the other floors then can be reduced with a reduction factor as follows: 15 0.25   0.40 4   4  31 20  The calculation of reduced axial live load and factored load is shown in Table P12-1. Table P12-1 Exterior column 285 10.0 107 48.8 425

All unit are in kips Dead load Live load from the roof floor Live load from the other floors Reduced live load from the other floors Total factored load

Interior column 487 18.5 198 79.2 720

Assume  g  0.015 Exterior column (

)

(

)

(

) 12-10

(

)


Interior column (

)

(

)

(

)

(

)

Based on the result of Example 12-3, select a column cross section of 18 in. 18 in. ( Ag  324 in.2 ) for both exterior and interior columns. D. Check with gravity load case 1.2D  1.6L  0.5Lr 1. Is the story being designed sway or non-sway? In order to answer this question, we need P  to calculate the stability index Q  u oh .In order to have the terms from the same Vus  c analysis, we need to analyze a frame with an arbitrary lateral load of 20 kips applied at the 2nd floor level in conjunction with the factored dead load and live load as shown in Fig. S12-2. In order to take into account the live load reduction, an average of live load reduction factor for exterior and interior columns (0.43) is multiplied with the live load factor 1.6, yielding 0.69. Therefore, the load combination used is: 1.2D  0.69L  0.5Lr  20 k lateral load . A structural analysis gives the following results: oh  0.127 in. Pu  414  722  723  418  2277 kips c  16 ft  192 in. 2277 kips  0.127 in. Q   0.075  0.05 20 kips  192 in.

The first story is a sway story. Note that the Pu  2277 kips does not differ significantly from ( ) 2. Are the columns slender? k u 1.2  (192  30) in.   36  22 r (0.3  18 in.) The columns are slender. 3. Compute the factored axial loads and moments from a first-order frame analysis. As explained in Example 12-3, the unfactored moments for exterior columns can be determined based on the live load pattern shown in Fig. S12-3 while those for interior columns based on the live load pattern shown Fig. S12-4. After a structural analysis is made, live load reduction factors will be applied. All results and calculation are shown in Table P12-2.

12-11


20 kips

Factored dead load and live load plus arbitrary lateral load to evaluate stability index, Q

Fig. S12-2

All span loaded with live load

Fig. S12-3 12-12


Staggered live load pattern

Fig. S12-4 Table P12-2 (Forces in kips Moments in k-ft) PD

Exterior Column 285

Interior Column 487

PL  reduced 

48.8

79.9

PLr

10

18.5

M D  top 

37.0

-4.9

M D  bottom 

-36.2

5.5

M L  top, reduced 

20.9  0.67  14

18.2  0.46  8.4

M L  bottom, reduced 

20.5  0.67  13.7

2.8  0.46  1.3

The factored load on exterior and interior columns are as follows: Exterior column

M  top   1.2  37  1.6 14  66.8 k-ft  M 2 M  bottom  1.2   36.2  1.6   13.7   65.4 k-ft  M1

12-13


Interior column

M  top   1.2   4.9   1.6   8.4   19.3 k-ft  M 2 M  bottom   1.2   5.5  1.6  1.3  8.68 k-ft  M1 4. Find  ns for the exterior and interior column? Cm  ns   1.0 P 1 u  Pc Exterior column M 65.4 Cm  0.6  0.4 1  0.6  0.4  ( )  0.21 M2 66.8 Pc 

 2 EI

k 

2

EI 

0.2 Ec I g  Es I se 1   dns I g  8750 in.4 Ec  3600 ksi

I se  150 in.4 (

)

EI  5.88 106 kip-in 2 Pc 

 2 EI

k 

2



 2  5.88  106 kip-in 2 

1.0 162 in.

2

 2210 kips

M c  66.8 k-ft

Interior column

Cm  0.6  0.4 (

M1 8.68  0.6  0.4  ( )  0.42 M2 19.3

)

Since dns (int) does not change significantly, EI and Pc will remain essentially the same.

M c  67.8 k-ft

12-14


5. Check initial column sections for gravity load case Exterior column

e 1.9   0.11 h 18  g  0.015 Fig. A-9b yields

Interior column

 g  0.015

Because reading from the graph may not be accurate given the two values are so close, we need to select reinforcement for the column and check its capacity against the demand. Select 8 bars #8, Example 12-3 shows , OK. E. Check with gravity plus wind load case 1.2D  1.6W  0.5  L  Lr  Wind loads are given in Fig. S12-5

Pu  oh Vus  c Similar to the gravity load case, we need to do one single analysis with the wind load plus gravity load case. To take into account of the live load reduction, an average live load reduction factor of 0.43 will be multiplied with the live load factor 0.5, yielding 0.22. Therefore, the load combination becomes 1.2D  1.6W  0.215Lr  0.5L . A structural analysis yields the following results: 1. Calculation of the stability index Q 

oh  0.357 in. Pu  352  635  633  385  2005 kips c  16 ft  192 in. Vus  4.46  3  6.61  8.49  32.9 kips

2005 kips  0.357 in.  0.113  0.05 32.9 kips  192 in. 1 1 s    1.13 1  Q 1  0.113 The first story is a sway story. Q 

12-15


4.46 kips

6.61 kips

6.61 kips

6.61 kips

8.49 kips

Wind load

Fig. S12-5

2. Factored axial loads and moments A structural analysis of the frame subjected to the wind load (without the load factor) yields the following results, as shown in Table P12-3. Table P12-3 Forces in kips Moments in k-ft PW

Exterior Column

Interior Column

10.4

0.5

MW  top 

46.2

65

MW  bottom 

-46

-64.6

Exterior column

( ) M 2  M 2ns   s M 2 s M 2ns  1.2M D  0.5M L  1.2  37  0.5 14  51.4 k-ft M 2 s  1.6MW  1.6  46.2  73.9 k-ft M 2  51.4  1.13  73.9  135 k-ft

Interior column 12-16


( ) M 2  M 2ns   s M 2 s M 2ns  1.2M D  0.5M L  1.2  4.9  0.5  8.4  10.1 k-ft M 2 s  1.6MW  1.6  65  104 k-ft M 2  10.1  1.13 104  128 k-ft Note that the Pu  2005 kips does not differ significantly from

Pu  2   387  632   2038 kips

3. Check column sections for axial loads and moments Exterior column

 g  0.01 Fig. A-9b yields

At this point, we can select 8 bars #6 ( Ast  3.52 in.2 ,  g  0.011 ) for the exterior columns. Interior column

 g  0.02

Conclusion: The cross section of exterior and interior columns is 18 in. x 18 in. Use 8 bars #6 for exterior columns, and 8 bars #8 for interior columns.

12-17


Chapter 13 13-1 Compute for the edge beam shown in Fig. P13-1. The concrete for the slab and beam was placed in one pour. Because the slab and the beam have the same elastic modulus, Eq. (13-9) reduces to  f 

Ib

Is 1. Compute I b . The cross section of the beam is shown in Fig. S13-1.1 and I b is computed for the shaded area. 13 in. 7 in.

20 in. 45

16 in.

≤ 28 in.

Fig. S13-1.1 Section through edge of slab.

Area, in.2 320 91   411

Part Web Flanges

ytop 

4

ytop , in.

Aytop

I own , in.

10 3.5

3200 319   3519

10,670 372 Ib 

Ay 2 ,in.4 663 2331 14,030 in.4

3519  8.56 in. 411

I g  14,030 in.4 2. Compute I s . The cross section of the slab is shown in Fig. S13-1.2 and I s is computed for the shaded portion of the slab. Is 

(108  8)  73  3316 in.4 12

Fig. S13-1.2 Edge beam.

13-1


3. Compute  f .

 f  Ib I  14,030 3316  4.23 s

13-2

Compute the column-strip and middle-strip moments in the long-span direction for an interior panel of the flat-slab shown in Fig. 13-25. Assume the slab is 6 in. thick, the design live load is 40 psf and the superimposed dead load is 5 psf for ceiling, flooring, and so on, plus 25 psf for the partitions. The columns are 10 in.  12 in., as shown in Fig. 13-25.

1. Compute the factored load.

6  qu  1.2    150  5  25   1.6  40  190 psf  12  Note: if the local building code allows a live-load reduction, the 40-psf live load could be multiplied by the appropriate reduction factor. 2. Compute the static moment in the long span of the slab.

n 2

12  13.5 ft 12  13.2 ft  14.5 

The column strip extends the smaller of 2 4 or 1 4 on each side of the column centerline (ACI Code Section 13.2.1). Thus, the column strip extends 13.2 4  3.3 ft on each side of column centerlines. The total width of the column strip is 6.6 ft. Each half-middle strip extends from the edge of the column strip to the centerline of the panel. The total width of two half-middle strips is 13.2  6.6  6.6 ft The static moment M o can be calculated from Eq. (13-5), Mo 

qu

2 2 n

8







190  13.2 13.52  1  57 kip-ft 1000 8

3. Divide M o into negative and positive moments. From ACI Code Section 13.6.3.2, for an interior span, the total moment is divided as follows: Negative moment = 0.65M o  0.65  57  37 kip-ft Positive moment = 0.35M o  0.35  57  20 kip-ft 4. Divide the moments between the column and middle strips. Negative moments From Table 13-3 for  f 1

2

1

 0 (since there are no beam between the columns).

Column-strip negative moment = 0.75   37   27.8 kip-ft Middle-strip negative moment = 0.25   37   9.3 kip-ft

13-2


Half of the middle-strip negative moment, -4.7 kip-ft, goes to each of the adjacent half-middle strip. Because the adjacent bays have the same width, 2 , a similar moment will be assigned to the other half of each middle strip so that the total middle-strip negative moment is 9.3 kip-ft. Positive moments From Table 13-4 for  f 1 2 1  0 , Column-strip positive moment = 0.60  20  12 kip-ft Middle-strip positive moment = 0.40  20  8 kip-ft

13-3

Use the direct-design method to compute the moments for the column-strip and middle-strip spanning perpendicular to the edge of the exterior bay of the flat-plate shown in Fig. P13-3. Assume the slab is 7.5 in. thick and supports a superimposed dead load of 25 psf and a live load of 50 psf. There is no edge beam. The columns are all 18 in. square.

1. Compute the factored load.

 7.5  qu  1.2    150  25   1.6  50  223 psf  12  Note: if the local building code allows a live-load reduction, the 50-psf live load could be multiplied by the appropriate factor. 2. Compute the static moment for the span perpendicular to the edge of the exterior bay 18  18.5 ft n  20  12 2

 19 ft

The column strip extends the smaller of 2 4 or 1 4 on each side of the column centerline (ACI Code Section 13.2.1). Thus, the column strip extends 19 4  4.75 ft on each side of column centerlines. The total width of the column strip is 9.5ft. Each half-middle strip extends from the edge of the column strip to the centerline of the panel. The total width of two half-middle strips is 19  9.5  9.5 ft The static moment M o can be calculated from Eq. (13-5), Mo 

qu

2 2 n

8







223  19.0  18.52  1  181 kip-ft 1000 8

3. Divide M o into negative and positive moments. From ACI Code Section 13.6.3.3, for a “slab without beams between interior supports and without edge beam”, the total moment is divided as follows: Interior negative moment = 0.70M o  0.70 181  127 kip-ft Positive moment = 0.52M o  0.52 181  94 kip-ft Exterior negative moment = 0.26M o  0.26 181  47 kip-ft

13-3


4. Divide the moments between the column and middle strips. Interior negative moments From Table 13-3 for  f 1 2

1

 0 (since there are no beam between the columns),

Interior column-strip negative moment = 0.75   127   95 kip-ft Interior middle-strip negative moment = 0.25   127   32 kip-ft Half of the middle-strip negative moment, -16 kip-ft, goes to each of the adjacent half-middle strip. Because the adjacent bays have the same width, 2 ,a similar moment will be assigned to the other half of each middle strip so that the total middle-strip negative moment is 32 kip-ft. Positive moments From Table 13-4 for  f

2

1

 0,

Column-strip positive moment = 0.60  94  56 kip-ft Middle-strip positive moment = 0.40  94  38 kip-ft Exterior negative moment From Table 13-5, for  f 1

2

1

 0 (since there is no beam parallel to

1

) and for t  0 (since

there is no edge beam), Exterior column-strip negative moment = 1.0  47  47 kip-ft Exterior middle-strip negative moment = 0  47  0 kip-ft

13-4

For the slab configuration and loading conditions in P13-3, use the direct-design method to compute moments for the edge-column strip and the middle strip spanning parallel to the edge of the slab.

1. Compute the factored loads.

 7.5  qu  1.2    150  25   1.6  50  223 psf  12  Note: if the local building code allows a live-load reduction, the 50-psf live load could be multiplied by the appropriate factor. 2. Compute the static moment for the span parallel to the edge of the slab. 18  17.5 ft n  19  12 For the definition of 2 refer to Fig. 13-22 in the textbook. 20 9 Edge frame: 2,e    10.8 ft 2 12

13-4


Interior frame:

 20 ft

2,i

Generally, the column strip extends the smaller of

4 on each side of the column 19 9 centerline (ACI Code Section 13.2.1).Thus; the width of the edge-column strip is   5.5 ft 4 12 2

4 or

1

The half-middle strip extends from the edge of the column strip to the centerline of the panel. The total width of two half-middle strips is 20  9.5  10.5 ft . The static moment M o can be calculated from Eq. (13-5). Edge frame: M o 

qu

Interior frame: M o 

2 2,e n

8 qu



2 2,i n

8









223  10.8  17.52  1  92.2 kip-ft 1000 8 

223  20  17.52  1  171 kip-ft 1000 8

3. Divide M o into negative and positive moments. From ACI Code Section 13.6.3.2, for the edge frame, the total moment is divided as follows: Negative moment = 0.65M o,e  0.65  92.2  60 kip-ft Positive moment = 0.35M o,e  0.35  92.2  32 kip-ft For the interior frame, the total moment is divided as follows: Negative moment = 0.65M o,i  0.65 171  111 kip-ft Positive moment = 0.35M o,i  0.35 171  59.9 kip-ft

4. Divide the moments between the edge-column and middle strips. Exterior negative moment From Table 13-3, for  f 1

2

1

 0 (since there is no beam between the columns),

Edge column-strip negative moment = 0.875  60  52.5 kip-ft 1 Middle-strip negative moment =   0.25  60  0.25  111 =21.4 kip-ft 2 1  Note that   0.25  0.875   1.0 2  Positive moments From Table 13-4 for  f

2

1

 0,

Edge column-strip positive moment = 0.80  32  25.6 kip-ft 1 Middle-strip positive moment =   0.40  32  0.40  59.9  =18.4 kip-ft 2

13-5


1  Note that   0.40  0.80   1.0 2  

13-6


13-5

A 7-in. thick flat-plate slab with spans of 20 ft in each direction is supported on 16 in.  16 in. columns. The average effective depth is 5.6 in. Assume the slab supports its own dead load, plus 25 psf superimposed dead load and 40 psf live load. The concrete strength is 4000 psf. Check two-way shear at a typical interior support. Assume unbalanced moments are negligible.

1. Compute the factored uniform load. (

)

2. Check one-way shear. One-way shear is critical at a distance from the face of the column. Thus, the critical sections for one-way shear are A-A and B-B in Fig. S13-5.1. The loaded areas causing shear on these sections are cross hatched. Their outer boundaries are lines of symmetry on which Vu  0 . We will only check the shear for section A-A, since the check for section B-B is the same.

6 6 6

6

Fig. S13-5.1 Critical section for one-way shear at interior column.

(a) Compute Vu at section A-A.

13-7


(

)

(b) Compute Vc for one-way shear. Because there is no shear reinforcement, we have Vn  V c and from Eq. (13-27), (

√

)

)⁄

( √

Thus, the slab is OK for one-way shear.

3. Check two-way shear Punching shear is critical on a rectangular section located at d 2 away from the column face, as shown in Fig. S13-5.2. The critical perimeter is 21.6 in. by 21.6 in. The average d value for determining the shear strength of the slab is d  5.6 in.

Fig. S13-5.2 Critical section for two-way shear at interior column.

(a) Compute Vu on the critical perimeter for two-way shear. (

)

(b) Compute Vc for the critical section.

13-8


The length of the critical perimeter is bo  2  21.6  21.6  86.4 in. Now, Vc is to be taken as the smallest of the following. From Eq. (13-24),





Vc  4 fc' bo d  4 1 4000  86.4  5.6  1  122 kips 1000 For Eq. (13-25),   1.0 (since column is square). Therefore,





 4 Vc   2    f c' bo d   2  4   1 4000  86.4  5.6  1  184 kips 1000  

For Eq. (13-26),  s  40 for this interior column. Therefore,  d   40  5.6  Vc   s  2   fc' bo d    2   1 4000  86.4  5.6  1  140 kips 1000  86.4   bo 





Therefore, the smallest values is Vc  122 kips , so Vc  0.75 122  91.5 kips > Vu and the slab is OK in two-way shear.

13-6

Assume the slab described in Problem 13-5 is supported on 10 in.  24 in. columns. Check two-way shear at a typical interior support. Assume unbalanced moments are negligible.

1. Compute the factored uniform load. (

)

2. See the solution to problem 13-5 for one-way shear calculations. 3. Check two-way shear Punching shear is critical on a rectangular section located at d 2 away from the column face, as shown in Fig. S13-6.2. The critical perimeter is 29.6 in. by 15.6 in. The average d value for determining the shear strength of the slab is d  5.6 in. (a) Compute Vu on the critical perimeter for two-way shear. (

)

(b) Compute Vc for the critical section. The length of the critical perimeter is bo  2  29.6  15.6  90.4 in. Now, Vc is to be taken as the smallest of the following. From Eq. (13-24),





Vc  4 fc' bo d  4 1 4000  90.4  5.6  1  128 kips 1000

13-9


Fig. S13-6.2 Critical section for two-way shear at interior column. 24  2.4 (since column is 10 in. by 24 in.). Therefore, 10  4 4   1 Vc   2    fc' bo d   2    1 4000  90.4  5.6  1000  117 kips  2.4    

For Eq. (13-25),  





For Eq. (13-26),  s  40 for this interior column. Therefore,  d   40  5.6  Vc   s  2   fc' bo d    2   1 4000  90.4  5.6  1  143 kips 1000 b 90.4    o 





Therefore, the smallest values is Vc  117 kips , so Vc  0.75 117  88 kips > Vu and the slab is OK in two-way shear.

13-7

The slab shown in Fig. P13-7 supports a superimposed dead load of 25 psf and a live load of 60 psf. The slab extends 4 in. past the exterior face of the column to support an exterior wall that weighs 400 lbs/ft of length of wall. The story-to-story height is 9.5 ft. Use 4500-psi concrete and Grade-60 reinforcement. (a)



Select slab thickness.

Determine the thickness to limit deflections. From Table 13-1, the minimum thicknesses of the four typical slab panels are as follows: Panel 1-2-A-B (corner; treat as exterior), and panels 2-3-A-B and 1-2-B-C (exterior)

13-10


Maximum

n

  20 12  16  224 in.

Minimum h 

n

30



224  7.47 in. 30

Panel 2-3-B-C (interior) Maximum n  224 in. 224 Minimum h  n   6.78 in. 33 33 Try h  8.0 in. 

Check the thickness for shear. We should check the shear at columns A2 and B2

The tributary area for column A2 is cross-hatched in Fig. S 13-7.1 The factored uniform load can be calculated as: 8  qu  1.2    150  25   1.6  60  246 psf  12  Note that if the area of any of the panels exceeded 400 ft 2 , it would be possible to reduce the live load before factoring it.

Fig. S 13-7.1 Initial critical shear perimeters and tributary areas for column A2. The critical shear perimeter is located at d 2 away from the interior column face and 4 in. from the exterior column face, as shown in Fig. S 13-7.1. In the following calculation for the factored shear force transmitted to column A2, the shear force multiplier of 1.15 required for the first interior support will be applied directly to the appropriate tributary lengths. Then, davg  8  0.75  0.5  6.75 in. (assuming 3 in. clear cover and No. 4 bars as slab 4

reinforcement). bo  22.75  2  23.38  69.5 in.

 12   22.75  23.38    Vu  246   9  1.15  9  10        1.2  400  9 1.15  9    60700 lbs  61 kips 12   144   

13-11


From Eq.(13-25), 16   1 16  4  2    6.0  4 (does not govern)   From Eq. (13-26),  s  30 , for an exterior slab-column connection  sd   30  6.75   2    2   4.91  4.0 (does not govern)    bo   69.5

Thus, using Eq. (13-24):





Vc   4 fc' bo d  0.75  4 1 4500  69.5  6.75  11000  94.4 kips > Vu Note: Vu Vc  0.65  0.75

For this ratio, ACI Code Section permits modification of  f for moment transfer about an axis parallel to the edge of the slab. With that information and because this ratio is below 0.8, the slab thickness at this connection should be sufficient for checking shear and moment transfer about an axis perpendicular to the edge of the column. Shear check for column B2 follows the same procedure as for column A2. Thus, use an 8-in. slab. Final shear checks will be made in part (c) after completing the flexural design of the slab. (b)

Use the direct design method to compute moments, and then design the reinforcement for the column and middle strips associated with column line 2.

Because there is no edge beams,  f  0 

Compute moments in the slab strip along column line 2 A2 1 (ft) n (ft) 2 (ft) qu (ksf)

Mo 

qu

2 2 n

(kip-ft) 8 Moment Coef. Moments (kip-ft)

-0.26 -51

B2

C2

20.0 18.67 18.0 0.25

20.0 18.67 18.0 0.25

196

196

0.52 +102

-0.70 -137

13-12

-0.65 -127

0.35 +69

-0.65 -127




Compute moments in the slab strip along column line 1 A1 1 (ft) n (ft) 2 (ft) qu (ksf)

Mo 

qu

2 2 n

(kip-ft) 8 Moment Coef. Moments (kip-ft) Wall load (kip/ft) q  2 Wall M o  wall n 8 Moments from wall (kip-ft)

-0.26 -28

B1

C1

20.0 18.67 10.0 0.25

20.0 18.67 10.0 0.25

109

109

0.52 +57 0.48

-0.70 -76

-0.65 -71

21 -5.5

0.35 +38 0.48

-0.65 -71

21

11

-15

-14

+7

-14

 Distribute the negative and positive moments to the column and middle strips and design the reinforcement. In each panel, the column strip extends 0.25  min  1 , 2   0.25  18 12  54 in. on each side of the column lines. The total width of the column strip is 2  54 in.= 108 in.  9 ft . The width of the middle strip is 9 ft. The edge strip has a width of 54 in.  12 in.  66 in.  5.5 ft . Place the steel in the long direction close to the surface of the slab. Try No. 4 bars. Thus, d  8  0.75  0.25  7.0 in. Compute trial As required at the section of maximum moment (column strip at B2). The largest M u is 102.3 kip-ft. Assuming that

 jd  0.95 ,

102.3  12,000  3.42 in.2 0.9  60,000  0.95  7.0 Compute a and check whether the section is tension controlled: 3.42  60,000 a  0.82 in. 0.85  4500   5.5  12  As (trial) 

0.82  1.00 in. 0.825 Clearly, the section is tension-controlled; therefore,   0.9 . 0.82 Compute the value of jd : jd  7.0   6.59 in. 2 Assuming that a is constant for all sections (conservative assumption), compute a constant for computing As : c

As (in.2 ) 

M u  12,000  0.0337 M u (kip-ft) (Eq. A) 0.9  60,000  6.59

13-13


The values of As required in the following table are computed from Eq. (A). From ACI Code Section 13.3.1, As ,min  0.0018bh for Grade-60 reinforcement. Maximum bar spacing is 2h (ACI Code Section 13.3.2), but not more than 18 in. (ACI Code Section 7.12.2.2). Therefore maximum is 16 in. Edge column strip: As,min  0.0018   5.5 12   8  0.95 in.2 5.5  12  4.1 16 Therefore, the minimum number of bars is 5. Other strips:

Minimum number of bar spaces 

As,min  0.0018   9 12   8  1.56 in.2

The minimum number of bars is 8.

Division of moment to column and middle strip: north-south strips Edge Column Strip

Middle Strip

Column Strip

Middle Strip

Column Strip

Strip Width, ft

9.0

9.0

9.0

9.0

5.5

Exterior Negative Moments

A1 -28 1.0

0.0

-28

0.0

Slab moment (kip-ft) Moment Coef. Distributed moments to strips Wall moment (kip-ft) Total strip moment (kip-ft)

0.0

A2 -51 1.0

0.0

0.0

A3 -51 1.0

0.0

-51

0.0

0.0

-51

-5.5 -33.5 1.13

0.0 0.0

-51 1.72

0.0 0.0

-51 1.56

Minimum As (in. ) Selected Steel

0.95 6 #4

1.56 9 #4

1.56 9 #4

1.56 9 #4

1.56 9 #4

As (in.2 ) provided

1.20

1.80

1.80

1.80

1.80

2

Required As (in. ) 2

End Span Positive Moments


57 0.6

0.4

0.2

102 0.6

0.2

0.2

102 0.6

34.2

22.8

20.4

61.2

20.4

20.4

61.2

11 45.2

43.2

61.2

40.8

61.2

2

1.52

1.45

2.06

1.37

2.06

2

0.95 8 #4 1.60

1.56 10 #4 2.00

1.56 10 #4 2.00*

1.56 10 #4 2.00

1.56 10 #4 2.00*

Required As (in. ) Minimum As (in. ) Selected Steel

As (in.2 ) provided

13-14


First Interior Negative Moments


B1 -76 0.75

0.25

-57

-19

0.125

B2 -137 0.75

0.125

0.125

B3 -137 0.75

-17.1

-102.3

-17.1

-17.1

-102.3

-15 -72

-36.1

-102.3

-34.2

-102.3

2

2.43

1.22

3.45

1.15

3.45

2

0.95 8 #5 2.48

1.56 9 #4 1.80

1.56 11 #5 3.41*

1.56 9 #4 1.80

1.56 11 #5 3.41*


As (in.2 ) provided Interior Positive Moments

Slab moment (kip-ft) Moment Coef. Distributed moments to strips Wall moment (kip-ft) Total strip moment (kipft)

38 0.6

0.4

0.2

69 0.6

0.2

0.2

69 0.6

22.8

15.2

13.8

41.1

13.8

13.8

41.1

7 29.8

29

41.1

27.6

41.1

2

1.00

0.98

1.36

1.04

1.36

2

0.95 6 #4 1.20

1.56 9 #4 1.80

1.56 9 #4 1.80

1.56 9 #4 1.80

1.56 9 #4 1.80


As (in.2 ) provided Interior Negative Moments


C1 -71 0.75

0.25

-53.2

-17.8

0.125

C2 -127 0.75

0.125

0.125

C3 -127 -0.75

-15.9

-95.2

-15.9

-15.9

-95.2

-14 67.2

-33.7

-95.2

-31.8

-95.2

2

2.26

1.14

3.21

1.07

3.21

2

0.95 8 #5 2.48

1.56 9 #4 1.80

1.56 11 #5 3.41

1.56 9 #4 1.80

1.56 11 #5 3.41


As (in.2 ) provided * As(provided) < As(required) is o.k. because adjacent positive moment regions are over-designed and some moment redistribution can occur in ductile slabs. (c)

Check two-way shear and moment transfer at columns A2 and B2. Neglect unbalanced moments about column line 2.

Column A2

13-15


The critical perimeter is at d 2 from the face of the column, where d is the average depth. At all exterior ends, the reinforcement is No. 4 bars and davg  6.75 in. The shortest perimeter results from the section shown in Fig. S13-7.2 and the perimeter dimensions are, b 1  20 in.  d 2  23.38 in. b 2  16 in.  d  22.75 in. For moments about the z  z axis, 2   23.38  5.69   23.38 2 y AB   7.86 in. 2  23.38  5.69    22.75  5.69  Therefore, cAB  7.86 in. and cCD  15.52 in. For moments about the w  w axis, 22.75 cCB  cAD   11.38 in. 2 As calculated in part (a), Vu  61 kips . For slabs designed by the direct-design method, the moment transferred from the slab to the column axis z-z is 0.3M o , and using the moments calculated from part (b), 0.3  M o  0.3 196  58.8 kip-ft (acting about the centroid of the shear perimeter).

Fig. S 13-7.2 Critical section- ColumnA2

From part (b), we found that the unbalanced moment due to the wall moments is 7 kip-ft and assuming that the loads acts at 2 in. from the edge of the slab,  23.38  2.0  7.86  13.52 in. from centroid The total moment to be transferred is, 13.52 M z  z  58.8  7   50.9 kip-ft 12

13-16


Note that the unbalanced moment about column line 2  M w w  is neglected as stated in the problem.

From Eq. (13-32), calculate the fraction of moment transferred by flexure, 1 1 f    0.60 2 2 1 b1 b2 1  23.38 22.75 3 3 ACI Code Section 13.5.3.3 allows  f to be increased to 1.0 if Vu Vc and the resulting  is less than 0.375b within a width of c2  3h centered in the column. From part (a), Vu Vc  0.65 . Therefore, take  f  1.0 and check the reinforcement required.

Width effective for flexure  c2  3h  16  3  8  40 in. Effective width  c2  2ct  c2  2c1 (second expression governs) Effective width  16  2 16  48 in. Assume that  jd  0.95d  , with d  7.0 in. 50.9  12,000  1.70 in.2 0.9  60,000  0.95  7.0 The steel provided is 9 No. 4 in a column strip width of 9 ft  108 in. , or roughly 13.5 on centers. The bars within the 40 in. effective width can be used for the moment transfer. Place four As 

column-strip bars into this region and add 5 No. 4 bars, giving As  1.8 in.2 in the effective width. Recompute As , 1.80  60,000 a  0.71 in. 0.85  4500  40 50.9  12,000  1.70 kip-ft (steel chosen OK) 0.71   0.9  60,000   7  2   The reinforcement ratio is, A 1.8  s   0.0064 bd 40  7 and from Eq. (4-24), 0.85  0.825  4500  0.003  b     0.0311 , 60,000  0.003  0.00207  As (in.2 ) 

and thus, 0.375b  0.0117  0.0064 and we can use  f  1.0 . As a result, it is not necessary to transfer any of the moment about z-z axis by eccentric shear stresses. Column B2 The critical perimeter is shown in Fig. S 13-7.3 and the centroidal axes pass through the centers of the sides. bo  2  22.75  22.75  91 in.

13-17


The factored shear force transmitted to column B2 is,   22.75  22.75   Vu  246   9  1.15  9 10 1.15  10       101500 lbs  102 kips 144   

Fig. S 13-7.3 Critical section- Column B2 From Eq.(13-25),  4  2    6.0  4 (does not govern)   From Eq. (13-26),  s  40 , for an interior slab-column connection  sd   40  6.75   2    2   4.97  4.0 (does not govern)  b 91   o  

Thus, using Eq. (13-24):





Vc   4 fc' bo d  0.75  4 1 4500  91 6.75  11000  124 kips > Vu Since, Vu Vc  0.82  0.4 no adjustment will be permitted in the ratio of unbalanced moment resisted by eccentric shear. The moment about x-x axis to be transferred comes from part (b) and is the difference between the negative moments on the two sides of column B2, i.e. M u , x  x  137  127  10 kip-ft . From Eq. (13-32), calculate the fraction of moment transferred by flexure (x-x axis), 1 1 f    0.6 2 2 1 b1 b2 1  1 3 3 The torsional moment of inertia can be calculated from Eq. (13-34), 2  b d 3 db3  b  J c  2  1  1   2  b2 d   1  12  2  12 Where d  6.75 in. and b1  b2  22.75 in. Thus, J c  54150 in.4 By inspection, the reinforcement that is already in the slab is adequate for moment transfer.

13-18


From Eq. (13-30) and neglecting unbalanced moment about column line 2 (i.e. about axis y-y), M u (shear transfer)    M u  1  0.6 10  4 kip-ft  48,000 lb-in. Then,

 Muc Jc



48,000  11.38  12.1 psi 45,150

So, 102,000 124,000  12.1  166 psi  12.1 psi=178 psi  c   202 psi 91 6.75 91 6.75 Thus, the shear is OK at this column.

u (max) 

13-8

Refer to the slab shown in Fig. P13-7 and the loadings and material strengths given in Problem 13-7.

(a) Select slab thickness. Problems 13-7 and 13-8 refer to the same flat-slab. As a result, the thickness of the slab was chosen in part (a) of problem 13-7. Thus, use an 8 in. thick slab. (b)

Use the direct design method to compute moments, and then design the reinforcement for the column and middle strips associated with column line A.

Because there is no edge beams,  f  0 

Compute moments in the slab strip along column line A A1 A2 (ft) 18.0 1 (ft) 16.67 n 11.0 2 (ft) qu (ksf) 0.25 2 q 96 M o  u 2 n (kip-ft) 8 Moment Coef. -0.26 0.52 -0.70 -0.65 Moments (kip-ft) -25 +50 -67 -62 Wall load (kip/ft) 0.48 16.7 q  2 Wall M o  wall n 8 Moments from wall -4.3 8.7 -11.7 -10.9 (kip-ft)



Compute moments in the slab strip along column line B B1 B2 (ft) 18.0 1 (ft) 16.67 n 20.0 2 (ft) qu (ksf) 0.25

13-19

A3 18.0 16.67 11.0 0.25 96 0.35 +34 0.48 16.7

-0.65 -62

+5.8

-10.9

B3 18.0 16.67 20.0 0.25


Mo 

qu

2 2 n

174 174 (kip-ft) 8 Moment Coef. -0.26 0.52 -0.70 -0.65 0.35 -0.65 Moments (kip-ft) -45 +90 -122 -113 +61 -113  Distribute the negative and positive moments to the column and middle strips and design the reinforcement. In each panel, the column strip extends 0.25  min  1 , 2   0.25  18 12  54 in. on each side of the column lines. The total width of the column strip is 2  54 in.= 108 in.  9 ft . The width of the middle strip is 9 ft. The edge strip has a width of 54 in.  12 in.  66 in.  5.5 ft . Place the steel in the long direction close to the surface of the slab. Try No. 4 bars. Thus, d  8  0.75  0.25  7.0 in. Compute trial As required at the section of maximum moment (column strip at B2). The largest M u is 91.5 kip-ft. Assuming that

 jd  0.95 ,

Compute a and check whether the section is tension controlled:

The section is tension-controlled; therefore,   0.9 . ⁄ Compute the value of jd : Assuming that a is constant for all sections (conservative assumption), compute a constant for computing As (Eq. A):

The values of As required in the following table are computed from Eq. (A). From ACI Code Section 13.3.1, As ,min  0.0018bh for Grade-60 reinforcement. Maximum bar spacing is 2h (ACI Code Section 13.3.2), but not more than 18 in. (ACI Code Section 7.12.2.2). Therefore maximum is 16 in. Edge column strip: As,min  0.0018   5.5 12   8  0.95 in.2 5.5  12  4.1 16 Therefore, the minimum number of bars is 5.

Minimum number of bar spaces 

Other strips: As,min  0.0018   9 12   8  1.56 in.2 The minimum number of bars is 8.

13-20


Division of moment to column and middle strip: east-west strips Edge Column Strip

Middle Strip

Column Strip

Strip Width, ft

9.0

9.0

9.0

Exterior Negative Moments

A1 -25 1.0

0.0

0.0

B1 -45 1.0

-25

0.0

0.0

-45


-4.3 -29.3

0.0

-45

2

0.99

0.0

1.52


2

0.95 5 #4

1.56 8 #4

1.56 8 #4

As (in.2 ) provided

1.00

1.60

1.60

Required As (in. )

End Span Positive Moments


50 0.6

0.4

0.2

90 0.6

30

20

18

54

8.7 38.7

38

54

2

1.31

1.28

1.82

2

0.95 7 #4 1.40

1.56 8 #4 1.60

1.56 10 #4 2.00


As (in.2 ) provided First Interior Negative Moments


A2 -67 0.75

0.25

0.125

B2 -122 0.75

-50

-17

-15

-91.5

-11.7 -51.7

-32

-91.5

2

1.75

1.08

3.09

2

0.95 9 #4 1.80

1.56 8 #4 1.60

1.56 10 #5 3.10


As (in.2 ) provided

13-21


Interior Positive Moments

Slab moment (kip-ft) Moment Coef. Distributed moments to strips Wall moment (kip-ft) Total strip moment (kip-ft) 2

Required As (in. ) 2


As (in.2 ) provided

13-9

34 0.6

0.4

0.2

61 0.6

20.4

13.6

12.2

36.6

5.8 26.2 0.89

25.8 0.87

36.6 1.24

0.95 5 #4 1.00

1.56 8 #4 1.60

1.56 8 #4 1.60

For the corner column (A1) in Fig. P13-7 and the loadings and material strengths given in Problem 13-7, select a slab thickness to satisfy ACI Code strength requirements for two-way shear and moment transfer, and deflection control. (a)

Make the check for moment transfer in only one principal direction (use the more critical direction).

First determine the thickness to limit deflections: From Table 13-1, the minimum thicknesses of the corner panel is (treat it as an exterior column): in   Maximum n   20 ft  12   16 in  224 in ft   224 in Minimum h  n   7.47 in 30 30 Like Problem 13-8, try h  8.0 in Now check whether this thickness is OK for shear at column A1, considering moment transfer in the more critical direction: Note that if the area of any of the panels exceeded 400 ft 2 , it would be possible to reduce the live load before factoring it, but this is not the case for column A1. Therefore the factored uniform load acting on the corner panel is:    8 in  lb qu  1.2    150 3  25 psf   1.6  60 psf  246 psf ft  12 in   ft  Although the shear force acting on the first interior column face is amplified by 1.15, the ACI Code does not permit reducing the tributary shear area for an exterior column to below 0.5 in either principal direction. Also note that the critical shear perimeter is located at d 2 away from the interior column faces. Therefore,

davg  8 in  0.75 in  0.5 in  6.75 in

13-22


bo  2  16 in  4 in  6.75 in / 2   46.8 in       23.4 in 2      26100 lb  26.1 k Vu  246 psf  10 ft  1 ft    9 ft  1 ft    in 2      144 2    ft     Now calculate the ultimate shear stress given this Vu and moment transfer in the more critical direction. V  M c vu  u  v u bo d Jc where: 1 1 v 1 1  0.4 2 2 1 b1 / b2 1 1 3 3 b1  b2  23.4 in

c  16 in cAB 

d  b1 

2

2  b1d  b2 d 



6.75 in   23.4 in 

2

4  23.4 in  6.75 in 

 5.85 in

2  b d 3 db 3 b   J c   1  1  b1d  1  cAB    b2 dcAB 2 12 2    12 2  23.4 in   6.75 in 3 6.75 in   23.4 in 3  23.4 in   Jc     23.4 in  6.75 in    5.85 in   12 12  2   

 23.4 in  6.75 in   5.85 in  J c  18,600 in 4 And, the more critical moment transfer axis is along column line 1, which is presented in the solution to Problem 13-7. M u  28 k-ft

Finally, we can calculate the maximum combined shear stress considering moment transfer in the more critical direction as follows: in 0.4   28 k-ft  12 16 in Vu  v M u c 26.1 k ft vu      0.198 ksi  198 psi bo d Jc 46.8 in  6.75 in 18,600 in 4 Now compare compute vn to check whether this level of shear stress is acceptable.    4 f c'  0.75  4  1  4500 psi  201 psi    4 4    vn     2   f c'  0.75   2    4500 psi  302 psi  1        d    s  2  f c'  0.75   20  6.75 in  2   4500 psi  246 psi   bo  46.8 in  

13-23

2


Therefore, OK  vn  201 psi  198 psi  vu (b) Make the check for moment transfer in both principal directions, but permit a 20 percent increase in the maximum permissible shear stress calculated at the corner of the critical shear perimeter. From part (a),

     23.4 in 2 Vu  246 psf  10 ft  1 ft    9 ft  1 ft    in 2   144   ft 2  

     26100 lb  26.1 k   

M u1  28 k-ft

Following the same procedure shown in Problem 13-7, the moment in the E-W direction is: M u 2  25 k-ft

Now we can calculate the maximum combined shear stress considering moment transfer in both principal directions as follows: V  M c  M c vu  u  v u1  v u 2 bo d Jc Jc 26.1 k  46.8 in  6.75 in vu  0.301 ksi  301 psi vu 

in in  16 in 0.4   25 k-ft  12 16 in ft ft  18,600 in 4 18,600 in 4

0.4   28 k-ft   12

Now compare compute vn to check whether this level of shear stress is acceptable.    4 f c'  0.75  4  1  4500 psi  201 psi    4 4    vn     2   f c'  0.75   2    4500 psi  302 psi  1      d    s  2  f c'  0.75   20  6.75 in  2   4500 psi  246 psi   bo  46.8 in   Even allowing for a 20% increase in the maximum permissible shear stress, a slab thickness of 8 in does not seem to be deep enough to satisfy this shear check. Repeat with h = 9.5 in. davg  9.5 in  0.75 in  0.5 in  8.25 in

bo  2  16 in  4 in  8.25 in / 2   48.3 in      24.1 in 2  Vu  246 psf  10 ft  1 ft    9 ft  1 ft    in 2   144   ft 2   b1  b2  24.1 in

cAB 

d  b1 

2

2  b1d  b2 d 



8.25 in   24.1 in 

2

4  24.1 in  8.25 in 

 6.03 in

13-24

     26100 lb  26.1 k   


2  b d 3 db 3 b   J c   1  1  b1d  1  cAB    b2 dcAB 2 12 2    12 2  24.1 in   8.25 in 3 8.25 in   24.1 in 3  24.1 in   Jc     24.1 in  8.25 in    6.03 in   12 12  2   

 24.1 in  8.25 in   6.03 in  J c  25, 200 in 4 So, now the shear stress demand is: V  M c  M c vu  u  v u1  v u 2 bo d Jc Jc in in 0.4   28 k-ft   12  16 in 0.4   25 k-ft  12 16 in 26.1 k ft ft vu    48.3 in  8.25 in 25, 200 in 4 25, 200 in 4 vu  0.227 ksi  227 psi If a 9.5 in deep slab is used, then 227 psi  vu  1.2 vn  241 psi .

(c)

Check one-way shear for a critical diagonal section across the corner near the corner column.

Assume that the critical section for one-way shear at this corner column occurs at a 45 degree angle from either of the exterior edges, at a distance, d, from the most interior corner. If we assume that d = 6.75 in, as in part (a), the length of the critical section is as follows:  2   4 in  16 in   2  6.75 in   70 in Therefore,

     35.0 in 2  Vu  246 psf  10 ft  1 ft    9 ft  1 ft    in 2   144   ft 2  

     25000 lb  25.0 k   

The capacity of the section is: Vn  Vc  6.75 in  70.0 in  0.75 1.0  2  4,500 psi  6.75 in  70.0 in  47,500 lb  47.5 k Therefore, a slab with h = 8 in will not fail in one-way shear along this diagonal failure plane.

13-25

2


13-10 For the slab system shown in Fig. P13-7, assume the slab has four equal spans in the north-south direction and three equal spans in the east-west direction. Use the loading and material strengths given in Problem 13-7 and assume a slab thickness of 7.5 in. (a)

Use an equivalent-frame method to analyze the factored design moments along column line 2 and compare the results with the moments used in part (b) of Problem 13-7.

For this solution, structural analysis software is used to model an equivalent frame used to represent the column line in question. Before this can be done, an appropriate equivalent frame must be defined, which is done here following recommendations from the text. Column properties: Ag  16 in 16 in  256 in 2 16 in  16 in 

3

Ig 

12 I e  I g  5460 in 4

 5460 in 4

Note that a full 9.5 ft of column will be modeled above and below the slab. These elements will be fixed at their ends. Beam properties: In the positive bending regions and the negative bending regions near interior columns:   0.5 ,   0.5 bh3  2 h3 0.5  18 ft  12 in/ft   7.5 in     3800 in 4 12 12 12 I e   I g  0.5I g  1900 in 4 3

Ig 

In the negative bending regions near exterior columns (assumed to be 0.2    0.2 ,   0.33

1

ft long):

bh3  2 h3 0.2  18 ft 12 in/ft   7.5 in  Ig     1520 in 4 12 12 12 I e   I g  0.33I g  500 in 4 3

Loads to be applied:    7.5 in lb lb  lb k qD  18 ft    150 3  25 2   2140  2.14 in ft ft  ft ft  12  ft   lb k  lb  qL  18 ft   60 2   1080  1.08 ft ft  ft  use qU  1.2qD  1.6qL with appropriate pattern loading schemes for each location

13-26


Location

Equivalent Frame Analysis with Direct Design Method Software-Based Model (From Problem 13-7) Column A-2 -62.8 k-ft -51 k-ft Midspan A-B 80.3 k-ft 102 k-ft Exterior of col. B-2 -156 k-ft -137 k-ft Interior of col. B-2 -130 k-ft -127 k-ft Midspan B-C 64.5 k-ft 69 k-ft Column C-2 -126 k-ft -127 k-ft Note that the remaining locations along column line 2 are symmetrical across column C-2. Some discussion should be presented in the solution, in addition to this table, commenting on the relative precision of the two approaches, noting that neither is necessarily accurate nor “correct”, although they are hopefully close to reality. It might be noted that similar discrepancies show up when the results of a software analysis of a frame system are compared with the results from using an ACI moment coefficient approach.

(b)

Use an equivalent-frame method to analyze the factored design moments along column line A and compare the results with the moments used in part (b) of Problem 13-8.

For this solution, structural analysis software is used to model an equivalent frame used to represent the column line in question. Before this can be done, an appropriate equivalent frame must be defined, which is done here following recommendations from the text. Column properties: Ag  16 in 16 in  256 in 2 16 in  16 in 

3

Ig 

12 I e  I g  5460 in 4

 5460 in 4

Note that a full 9.5 ft of column will be modeled above and below the slab. These elements will be fixed at their ends. Beam properties: In the positive bending regions and the negative bending regions near interior columns:   0.5 ,   0.5 ( )

In the negative bending regions near exterior columns (assumed to be 0.2    0.2 ,   0.33 ( )

13-27

1

ft long):


Loads to be applied: (

)(

(

)(

) )

use qU  1.2qD  1.6qL with appropriate pattern loading schemes for each location Location

Equivalent Frame Analysis with Direct Design Method Software-Based Model (From Problem 13-7) Column A-1 -52.5 k-ft -29.3 k-ft Midspan A1-A2 46.9 k-ft 58.7 k-ft Exterior of col. A-2 -89.6 k-ft -78.7 k-ft Interior of col. A-2 -78.8 k-ft -72.9 k-ft Midspan A2-A3 39.4 k-ft 39.8 k-ft Note that the remaining locations along column line A are symmetrical across Midspan A2-A3. Some discussion should be presented in the solution, in addition to this table, commenting on the relative precision of the two approaches. It should be mentioned that neither is necessarily accurate nor “correct”, although they are hopefully close to reality. It might be noted that similar discrepancies show up when the results of a software analysis of a frame system are compared with the results from using an ACI moment coefficient approach.

13-11 For the same floor system described in Problem 13-10 and the loading and material strengths given in Problem 13-7, assume the slab thickness has been selected to be 6.5 in. (a)

For a typical interior floor panel, calculate the immediate deflection due to live load and compare to the limit given in ACI Code Table 9.5 (b).

This solution will follow the approach shown in Example 13-16. Step 1: Compute the immediate deflection of an interior column strip, which should be taken in the N-S direction, as these are the longer span column strips. Take the span between columns B2 and C-2 as the interior span in question. First compute Ma. The loads we must consider for deflection calculations are: 6.5 in lb  150 3  25 psf  106 psf Dead load: 1.0   Dead Load   in ft 12 ft Service load: 1.0   Dead  Live   106 psf  60 psf  166 psf Construction load: 2.0  Slab Dead Load   2  81.3 psf   163 psf Therefore, cracking will be governed by the service load. Since we know that the moments calculated in Problem 13-7b are based on an area load of 246 psf, we can take Ma to be 166/246 = 0.675 times the column strip moments calculated in Problem 13-7b, as follows: Negative moment at B-2 and C-2: 0.675   95.2 k-ft   64.3 k-ft Positive moment at midspan: 0.675  41.1 k-ft  27.7 k-ft

13-28


Now compute Mcr:

in 3   6.5 in  1 12 ft M cr    383,000 lb-in  31.9 k-ft yt 3.25 in It is a safe bet that the slab will be cracked in both the positive and negative moment regions regardless, due to temperature and shrinkage effects. fr I g

7.5 4500 psi  9 ft  12

Next, we have to compute Icr and Ie. For simplicity, we will assume that the reinforcement selected in Problem 13-7 is used in this slab as well. Negative moment region: 11 0.31 in 2   0.00486 in 9 ft  12  6.5 in ft 29,000,000 psi n  7.58 57,000 4500 psi  n  0.0368 k  2 n    n    n  2  0.0368   0.0368  0.0368  0.237 2

2

1 3 2 I cr  b  kd   Ast n  d  kd  3 1 in 3 2   9 ft  12   0.237  5.4 in   3.41 in 2  7.58  5.4 in  0.237  5.4 in   514 in 4 3 ft 3

M   31.9 k-ft  4 I e  I cr   I g  I cr   cr   514 in 4  2470 in 4  514 in 4    753 in M 64.3 k-ft    a  Positive moment region: I e  I g  2470 in 4



3



So, the weighted average value of Ie is: I e( average)  0.7 I em  0.15  I e1  I e 2   0.7  2470 in 4  0.15  753 in 4  753 in 4   1950 in 4 Now we can calculate the immediate deflection due to live load. Using the same logic described in Example 13-16, assume that 67.5% of the loads will be carried by the column strip. wL  60 psf 18 ft  0.675  0.729 k/ft  60.8 lb/in wD  106 psf 18 ft  0.675  1.29 k/ft  107 lb/in 4

 L (column strip max)

in   60.8 lb/in   20 ft  12  wL 4 ft    0.0048   0.0048   0.130 in EI 57,000 4500 psi  1950in 4 4

 D (column strip max)

in   107 lb/in   20 ft  12  4 w ft    0.0026  D  0.0026   0.124 in EI 57,000 4500 psi  1950in 4

Step 2: Compute the immediate deflection of an interior middle strip, which should be taken in the E-W direction, as these are the shorter span middle strips. Take an interior span between column lines 2 and 3.

13-29


First compute Ma. The loads we must consider for deflection calculations are: 6.5 in lb  150 3  25 psf  106 psf Dead load: 1.0   Dead Load   in ft 12 ft Service load: 1.0   Dead  Live   106 psf  60 psf  166 psf Construction load: 2.0  Slab Dead Load   2  81.3 psf   163 psf Therefore, cracking will be governed by the service load. Since we know that the moments calculated in Problem 13-7b are based on an area load of 246 psf, we can take Ma to be 166/246 = 0.675 times the slab strip moments calculated using the loading described in Problem 13-7b. Negative moment at B-2 and C-2: 0.675   28.3 k-ft   19.1 k-ft Positive moment at midspan: 0.675  24.4 k-ft  16.5 k-ft Now compute Mcr:

M cr 

fr I g

7.5 4500 psi 11 ft 12

in 3   6.5 in  1 12 ft  468,000 lb-in  39.0 k-ft

 yt 3.25 in It is unlikely that the middle strip will be significantly cracked given that the moments which control the cracking state are so far below the cracking moment for the strip section. Therefore, we can assume that: Negative moment region: I e  I g  3020 in 4 Positive moment region: I e  I g  3020 in 4 And therefore the weighted average value of Ie is obviously: I e( average)  3020 in 4 Now we can calculate the immediate deflection due to live load. Using the same logic described in Example 13-16, assume that 32.5% of the loads will be carried by the middle strip. wL  60 psf  20 ft  0.325  0.390 k/ft  32.5 lb/in wD  106 psf  20 ft  0.325  0.689 k/ft  57.4 lb/in 4

 L (middle strip max)

in   32.5 lb/in  18 ft  12  4 w ft    0.0048  L  0.0048   0.029 in EI 57,000 4500 psi  3020in 4 4

 D (middle strip max)

in   57.4 lb/in  18 ft  12  4 w ft    0.0026  D  0.0026   0.028 in EI 57,000 4500 psi  3020in 4

Step 3: Compute the maximum immediate total deflection in the panel due to the live load, and compare against allowable deflections according to ACI Code limits.  L(max)  0.130 in  0.029 in  0.159 in

13-30


ACI Code Limit  / 360 

(b)

20 ft  12

in ft  0.667 in

We are OK. 360 For the same floor panel, calculate the total deflection after the attachment of partitions and compare to the limit given in ACI Code Table 9.5(b) for partitions that are not likely to be damaged by long term deflections. Assume that 85 percent of the dead load is acting when the partitions are attached to the structure and assume that 25 percent of the live load will be sustained for a period of one year.

We will assume that the total expected deflection after the attachment of partitions is a sum of the following: 1. Dead load Assuming that 85% of the dead load is already acting when the partitions are attached, consider 15% of the immediate deflection due to dead load affects the partitions. Dead immediate  0.1 (0.124 in  0.028 in)  0.015 in

2. Live load Assume that the total instantaneous deflection due to live load could occur at any time once the partitions are attached, so include the full instantaneous deflection due to live load. Live immediate  0.159 in

3. Long term deflections Assume that the full dead load acts plus 25% of the live load over the course of a year. Use the suggested deflection coefficient of 3.0 to adjust for long-term amplification effects. Long term  3.0   0.124 in  0.028 in  0.25  0.159 in   0.575 in

Therefore, the total expected deflection felt by the partitions is:  partitions   Dead immediate   Live immediate   Long term  partitions  0.015 in  0.159 in  0.575 in  0.749 in

Compare this to the ACI limit for slab systems supporting partitions which are not likely to be damaged by large deflections: ACI Code Limit  / 240 

20 ft  12 240

in ft  1.0 in

We are OK.

13-12 Repeat the questions in Problem 13-11 for the following panels. (a)

An exterior panel along the west side of the floor system.

13-31


This solution will follow the same lines as the solution for Problem 13-11. The column strip in question will still be taken along the N-S direction. The major change will lie in the definition of moments acting in the column and middle strips.

(b)

An exterior panel along the north side of the floor system.

This solution will follow the same lines as the solution for Problem 13-11. The column strip in question will still be taken along the N-S direction. The major change will again lie in the definition of moments acting in the column and middle strips. (c)

A corner panel.

This solution will again follow the same lines as the solution for Problem 13-11. The column strip in question will still be taken along the N-S direction. The major change will still lie in the definition of moments acting in the column and middle strips.

13-32


Chapter 14 14-1

For the slab panel shown in Fig. P14-1, use the yield-line method to determine the minimum value of the area load, , at the formation of the critical yield-line mechanism.

We may need to investigate two yield-line mechanisms to find the minimum qf. 1. Select first plastic mechanism.

9 ft.

II

18 ft.

I A

B

L1 22ftft. LL11==20 Mechanism No. 1

2. Assume a virtual displacement equal to

along the positive yield line from A to B.

3. Compute the internal work. For slab segment I the internal work is: (

(

)

)

Similarly, the internal work for slab segment II is: (

( )

)

Therefore, the sum of the internal work is: ∑

(

)

(

)

4. Compute the external work. The two end regions outside of the points A and B are essentially two half-pyramids that can be combined into a single pyramid with a base area 18 ft x 2 x 20 ft. The central region between points A and B has a triangular cross-section and extends over a length of 20 ft x (1 – 2 ). Thus, the external work is: [

(

14-1

)

]

(

)


5. Equate the external and internal work. Set EW = ∑

and solve for qf.

(

)

(

)

6. Solve for the minimum value of qf. The solution table below starts with decreases .

= 0.5 and slowly

Numerator, kips

Denominator, ft2

qf, ksf

0.50

57.4

120

0.478

0.48

59.1

122

0.482

0.46

60.8

125

0.488

From this table the minimum value of qf is 0.478 ksf and it occurs for a value of 0.50. Because of this result, we should investigate a second possible yield-line mechanism. 7. Select a second possible plastic yield-line mechanism.

II D

I

L2 = 18 ft.

C  L2 11 ftft 10

20 22 ft ft. Mechanism No. 2

8. Assume a virtual displacement of

along positive yield line C – D.


(

)

)


)

( )


(

)

(

)

10. Compute the external work. The two end regions outside of the points C and D are essentially two half-pyramids that can be combined into a single pyramid with a base area 20 ft x 2 x 18 ft.

14-2


The central region between points C and D has a triangular cross-section and extends over a length of 18 ft x (1 – 2 ). Thus, the external work is: (

[

)

11. Equate the external and internal work. Set EW = ∑ (

]

(

)

and solve for qf. )

(

)

12. Solve for the minimum value of qf. The solution table below starts with decreases .

= 0.5 and slowly

Numerator, kips

Denominator, ft2

qf, ksf

0.50

57.2

120

0.477

0.48

57.9

122

0.475

0.46

58.7

125

0.47

0.44

59.6

127

0.469

0.42

60.6

130

0.468

0.4

61.6

132

0.467

0.38

62.8

134

0.467

From this table the minimum value of qf is 0.467 ksf and it occurs for an value of approximately 0.47. In this case it was important for us to investigate this second possible yield-line mechanism.

14-3


14-2


1. Select plastic yield-line mechanism. free edge

10 ft. B II

A L1 = 20 ft.

 L1

I

20 ft. Assumed yield-line mechanism


along the positive yield line from A to B.


(

)

)


(

)

)

Therefore, the sum of the internal work is: (

∑

)

(

)

4. Compute the external work. The end region below the point A is essentially a half-pyramid with a base area of 20 ft x x 20 ft. The region between points A and B has a triangular cross-section and extends over a length of 20 ft x (1 – ). Thus, the external work is: (

[

5. Equate the external and internal work. Set EW = ∑ (

)

]

and solve for qf. )

(

)

14-4

(

)


6. Solve for the minimum value of qf. The solution table below starts with increments of 0.05.

= 0.75 and increase

Numerator, kips

Denominator, ft2

qf, ksf

0.75

34.7

150

0.231

0.80

33.6

147

0.229

0.85

32.7

143

0.228

0.90

31.9

140

0.228

0.95

31.2

137

0.228

1.00

30.5

133

0.229

From this table the minimum value of qf is 0.228 ksf and it occurs for a 0.90.

14-5

value of approximately

in


14-3


1. Select plastic yield-line mechanism. F

LIII

X C III II

D

18 ft I

Y

A

B

E

24 ft Assumed yield-line mechanism

This is a complex mechanism with many potential variables. We will select values for some of the distances shown in this figure as an initial trial. A few points should be noted regarding this mechanism. As stated in the text, a positive yield line between two adjacent plate segments will extend to or toward the intersection of the axes of rotation for those two plate segments. Thus, the positive yield line A-D starts at the intersection between the axes of rotation for segments I and II (at point A). Similarly, the other two positive yield lines branching out from point D aim toward the points E (intersection of axes of rotation for segments I and III) and F (intersection of axes of rotation for segments II and III). Experience with this particular mechanism indicates that the minimum value of qf will occur when the point D is relatively close to the corner column support. 2. Select trial values for the assumed mechanism. The following values were selected for the mechanism shown above. B-E = 24 ft

C-F = 18 ft

Y = 15 ft

X = 20 ft

With these values selected, geometry can be used to determine the additional values shown on the following figure. It should be noted that the three lines extending from the slab to line E-F (the axes of rotation for segment III) are all perpendicular to that line. 3. Assume a virtual displacement equal to

I 



15 ft

;  II 

occurs at the point D.



20 ft

, and  III 

14-6



4.80 ft


F 24 .5'

4.5 ' 4.1 ' 4.8 0'

17.1'

III

D 18 ft

4. 11 '

4.0 9'

C

26 .9'

II I 12.9'

A

B

E


4. Compute the internal work. For the three slab segments the internal work is:

IW ( I )  (m p  mn )  24 ft   I

k-ft   24 ft   (22.4 k)   ft 15 ft IW ( II )  (m p  mn ) 18 ft   II  (6  8)

k-ft  18 ft   (12.6 k)   k 20 ft IW ( III )  m p  LIII   III  (6  8)

6

k-ft   8.60 ft   (10.8 k)   k 4.80 ft

Therefore, the sum of the internal work is:

 IW   45.8 k   

5. Compute the external work. Rather than calculate a displaced volume, we will determine the magnitude of load acting on each plate segment and then multiply that load times the displacement at the centroid of the segment. If we did this directly for the defined segments, it would be difficult to find the centroids of each segment. Thus, for plate segment I we will calculate the work done by the distributed load acting on the triangular segment A-D-E, and then subtract to work for the triangular piece outside the actual slab panel. So, for segment I:

 (12.9 15)     EW ( I )   A(total)   A(outside)    q f 3 3  1 1 12.9 15  1 EW ( I )    48 15    24 12.9   qf  3 2 3  2  (75.6 ft 2 )  q f  

14-7


Similarly, for segment II:

1 1 17.1 20  1 EW ( II )    36  20   18 17.1   qf  3 2 3  2  (76.1 ft 2 )  q f   And for segment III, where some additional small triangles are used outside the slab panel:

1 1 1 4.11 4.80 1 4.09 4.80 EW ( III )  [  60  4.80    26.9  4.11    24.5  4.09   2 3 2 3 2 3 1 4.11 4.80 1 4.09 4.80   3.10  4.11    5.5  4.09  ] q f   2 3 2 3 EW ( III )  (13.0 ft 2 )  q f   Thus, the sum of external work is:

 EW  (165 ft

2

)  q f 

6. Setting the sum of the internal work equal to the sum of the external work gives:

q f (trial) 

45.8 kips  0.278 ksf 165 ft 2

7. Estimate for the minimum value of qf. Rather than doing several trials, we will assume that the selected trial mechanism is reasonable and should give an answer within ten percent of the true minimum value for qf. Thus, we can say:

q f (min)  0.9  0.278  0.250 ksf

14-8


14-4


1. Select a plastic yield-line mechanism. Try the mechanism shown below. Note that there is a single variable for this mechanism, the value of . free edge A

II

L1 = 18 ft.

 L1

I

 L2 L2 = 24 ft. Assumed yield-line mechanism


occurs at the point A. (

)


(

))

(

(

))

[

(

) ]


(

)


) (

[

(

[

)

)

]

]

4. Compute the external work. The deflected shape is essentially a half-pyramid, with a combined base area (18 ft x 24 ft)/2. Thus, the external work is: [

]

5. Equate the external and internal work. Set EW = ∑ ( ) (

14-9

and solve for qf. )


6. Solve for the minimum value of qf. Because the denominator is constant, we need to minimize the numerator. It is expected that ≈ 0.5. Numerator, kips

Denominator, ft2

qf, ksf

0.48

29.0

72.0

0.403

0.50

28.5

72.0

0.396

0.52

28.1

72.0

0.390

0.54

27.7

72.0

0.385

0.56

27.5

72.0

0.382

0.58

27.4

72.0

0.381

0.60

27.5

72.0

0.382

From this table the minimum value of qf is 0.381 ksf and it occurs for a

14-10

value of 0.58.


14-5

The slab panel shown in Fig. P14-5 is 7.5 in. thick and is made of normal weight concrete. The slab will need to support a superimposed dead load of 25 psf and a live load of 60 psf. Assume that the slab will be designed with isotropic top and bottom reinforcement. (a)

Find the critical yield-line mechanism and then use the appropriate load factors and strength-reduction factors to determine the required sum of the nominal negative- and positive-moment capacities ( ) in units of - ⁄ .

1. Determine the total factored load to be resisted. Slab weight is (7.5/12) x 150 = 94 psf; Thus, the total dead load = 94 + 25 = 119 psf. The live load can be reduced based on the panel area (20 x 26 = 520 ft2). Note: KLL = 1.0. For this area, the reduced live load is equal to 54.5 psf. So, the total factored load, qu = 1.2 x 119 + 1.6 x 54.5 = 230 psf = 0.230 ksf 2. Select a possible plastic yield-line mechanism. A couple items should be noted. First, in text Example 14-4, the value of was close to 0.5, so having a virtual displacement of only at a single point (point A) will give a relatively good answer. Also, because of the fixed supports at the support edges for plate segments I and II, we would like to make the rotations of those plate segments smaller than for plate segments III and IV. Thus, will be greater than 0.5.  x 26 ft

III A II

20 ft

IV

 x 20 ft

I



I 



at point A.

, and  II 



  20 ft   26 ft    III  , and  IV  (1   )  20 ft (1   )  26 ft

14-11


3. Compute the internal work. For the four slab segments the internal work is:

IW ( I )  (mn  m p )  26 ft   I 

1.3(mn  m p )



IW ( II )  (mn  m p )  20 ft   II 

0.77(mn  m p )





1.3m p

IW ( III )  m p  26 ft   III  IW ( IV )  m p  20 ft   IV



 1  0.77m p   1 

From these, the sum of the internal work is: ∑

(

[

)

]

At some point in the design process need to make a decision on the ratio between the flexural strength of the bottom reinforcement, mp, vs. the strength of the top reinforcement, mn. Reading ahead to problem 14.5(b), we are asked to assume that . Thus, the internal work can be expressed as a function of mp. ∑

[

(

)

]

4. Compute the external work. In previous problems we have solved for the distributed load, qf, at the development of a failure mechanism. In this case, qf is equivalent to qu divided by the strength reduction factor, , which will be assume to be equal to 0.9. Thus,

EW 

qu    0.230 ksf 20 ft  26 ft    (173 ft 2 )    (44.2 k)      3 0.9

5. Equate the external and internal work. [

(

)

[

6. Solve for the minimum required value of increase in increments of 0.05.

0.50 0.55 0.60 0.65

]

]

. The solution table below starts with

= 0.5 and

Required , kip-ft/ft 2.97 3.07 3.13 3.11

From this table the minimum required value of kip-ft/ft.

is 3.13 kip-ft/ft. Assuming that

14-12

,


(b)

Start with approximately equal to , and then select top and bottom isotropic reinforcement to provide the required combined flexural strength calculated in part (a). Be sure to check reinforcement spacing requirements given in ACI Code Section 13.3.2.

1. From part (a), and required moment strength. Assume suffice, use

kip-ft/ft. Select top and bottom bars to provide the and . Also, assuming that #4 bars will

For negative bending, (

)

(

(

)

)

Using #4 bars, Using #3 bars, For positive bending, (

)

(

(

)

)

Using #4 bars, Using #3 bars, 2. Check reinforcement detailing requirements and select reinforcement: Minimum reinforcement area: bottom reinforcement required for strength purposes. Maximum spacing:

{

}

. This is easily satisfied by the top and

To satisfy the strength requirements from step 1

and this maximum spacing requirement, select #3 bars for both top and bottom reinforcement. Bars shall be spaced at 7.5 in. and 11 in. along the top and bottom faces of the slab, respectively, in both orthogonal directions. Clear cover requirements of 0.75 in. shall be enforced. Proper anchorage shall be provided for all slab reinforcement.

14-13


Chapter 15 15-1

Design wall footings to be supported 3 ft below grade for the following conditions. ⁄ and normal weight concrete for both problems. Assume a soil density of Service dead load is 6 kips/ft, service live load is 8 kips/ft. Wall is 12 in. thick. Allowable soil pressure, , is 4000 psf. and .

The following design is done for a 1 ft length of footing. Although not included for simplicity, the “per foot” notation applies to all units in the following solution. 1) Estimate size of footing and factored net pressure The bottom face of the footing is 3 ft below the ground surface. Assume the thickness of the footing is 12 in. The soil density is 120 lb/ft 3 Allowable net soil pressure:

qn  4 ksf  (1.0 ft  0.15 k/ft 3  2.0 ft  0.12 k/ft 3 )  3.61 ksf Required footing area:

Areq. 

6 k 8 k  3.88 ft 2 3.61 ksf

 Try 4 ft wide strip footing. Factored net pressure:

qnu 

1.2  6 k  1.6  8 k  5 ksf 4 ft 1 ft

2) Check one-way shear Assume a concrete cover of 3 in., and #8 bars being used.

d  12 in.  3 in.  0.5 in.  8.5 in.

The distance the footing extends to each side from the column face:  48 in.  12 in. / 2  18 in.

 18 in.  8.5 in.  Vu  5 ksf    1 ft  3.96 k  12 in/ft  Vc   2 fcbwd  0.75  2 3500 12  8.5  9050 lb  9.05 k Since Vc is significantly greater than Vu , we reduce the footing thickness to the minimum allowable by ACI Section 15.7, i.e.: 6 in. + bar dia. + 3 in. cover. = 10 in.

d  10 in.  3 in.  0.5 in.  6.5 in.  18 in.  6.5 in.  Vu  5 ksf    1 ft  4.79 k  12 in/ft 

√  Use 10 in. thick footing

√

15-1


3) Design reinforcement

M u   5 ksf 1ft 

1.5 ft  

2

 5.63 k-ft 2 Assume j  0.90 5.63 k-ft 12 in/ft As   0.21 in.2 0.9  60 ksi  (0.9  6.5 in) As ,min  0.0018bh  0.0018 12 in. 10 in.  0.22 in.2 Use #4 bars at with a spacing of 10 in. ( As  0.24 in.2 )

d  10 in.  3 in.  0.25 in.  6.75 in. 0.24 in.2  60 ksi a  0.4 in. 0.85  3.5 psi  12 in. t 

6.75   0.4 / 0.85 d c  cm   0.003  0.040  0.005,  0.9 c 0.4 / 0.85

 M n  0.9  0.24 in.2  60 ksi   6.75 in.  0.4 in./2   84.9 k-in.  7.07 k-ft  M u

4) Check development Bar spacing exceeds 2db and cover exceeds d b  This is Case 2 development.  e t f y 1 1 60000 psi db   0.5 in.  20 in.  18 in.  3 in., NG. d  25 f c 25  1 3500 psi We shall use 90º standard hooks to anchor the bars. The required development length for a 90º standard hook is: 0.02 e f y 0.02  1 60000 psi db   0.5 in.  10 in. < 18 in.  3 in., OK. d   f c 1 3500 psi 5) Temperature requirement for longitudinal reinforcement:

As , req.  0.0018bh  0.0018  48 in. 10 in.  0.86 in.2  Use 5 #4 bars for longitudinal reinforcement.

Use 10 in. thick by 4 ft. wide footing with #4 bars at 10 in. o.c. with 90º hooks at both ends in the transverse direction, and 5 #4 bars in the longitudinal direction.

15-2

Design wall footings to be supported 3 ft below grade for the following conditions. ⁄ and normal weight concrete for both problems. Assume a soil density of Service dead load is 15 kips/ft, service live load is 8 kips/ft. Wall is 16 in. thick. Allowable soil pressure, , is 6000 psf. and .

1) Estimate size of footing and factored net pressure The bottom face of the footing is 3 ft below the ground surface. Assume the thickness of the footing is16 in. The soil density is 120 lb/ft 3

15-2


Allowable net soil pressure: ( Required footing area:

⁄

⁄

)

 Try 4.5 ft wide strip footing. Factored net pressure:

2) Check one-way shear Assume a concrete cover of 3 in., and #8 bars being used.

d  16 in.  3 in.  0.5 in.  12.5 in.

The distance the footing extends to each side from the column face:  54 in.  16 in. / 2  19 in.

 19 in.  12.5 in.  Vu  6.84 ksf    1 ft  3.71 k 12 in/ft   Vc   2 fcbwd  0.75  2 3000 12 12.5  12.3 k Since Vc is significantly greater than Vu , we reduce the footing thickness to 12 in.:

d  12 in.  3 in.  0.5 in.  8.5 in.  19 in.  8.5 in.  Vu  6.84 ksf    1 ft  5.99 k  12 in/ft  √  Use 12 in. thick footing

√


M u   6.84 ksf 1ft 

1.583 ft  

2

 8.58 k-ft 2 Assume j  0.90 8.58 k-ft 12 in/ft As   0.25 in.2 0.9  60 ksi  (0.9  8.5 in) As ,min  0.0018bh  0.0018 12 in. 12 in.  0.26 in.2 Use #5 bars at with a spacing of 12 in. ( As  0.31 in.2 )

d  12 in.  3 in.  0.31 in.  8.7 in. a

t 

0.31 in.2  60 ksi  0.61 in. 0.85  3.0 psi  12 in.

8.7   0.61/ 0.85 d c  cm   0.003  0.033  0.005,  0.9 c 0.61/ 0.85

 M n  0.9  0.31 in.2  60 ksi  8.7 in.  0.61/ 2 in.  141 k-in.  11.7 k-ft  M u

15-3


4) Check development Bar spacing exceeds 2db and cover exceeds d b  This is Case 2 development.  e t f y 1 1 60000 psi db   0.625 in.  27.4 in.  19 in.  3 in., NG. d  25 f c 25  1 3000 psi We shall use 90º standard hooks to anchor the bars. The required development length for a 90º standard hook is: 0.02 e f y 0.02  1 60000 psi db   0.625 in.  14 in. < 19 in.  3 in., OK. hb   f c 1 3000 psi 5) Temperature requirement for longitudinal reinforcement:

As , req.  0.0018bh  0.0018  54 in. 12 in.  1.17 in.2  Use 6 #4 bars for longitudinal reinforcement.

Use 12 in. thick by 4 ft – 6 in. wide footing with #5 bars at 12 in. o.c. with 90º hooks at both ends in the transverse direction, and 6 #4 bars in the longitudinal direction.

15-3

Design a square spread footing for the following conditions: ⁄ . Service dead load is 350 kips, service live load is 275 kips. Soil density is Allowable soil pressure is 4500 psf. Column is 18 in. square. (normal weight) and . Place bottom of footing at 5 ft below floor level.

1) Estimate size of footing and factored net pressure Estimate thickness between 18 and 36 in., say 30 in. Assume the floor thickness is 6 in. Allowable net soil pressure:

qn  4.5 ksf  (2.5 ft  0.5 ft)  0.15 k/ft 3  2.0 ft  0.13 k/ft 3  3.79 ksf Required footing area:

Areq. 

350 k  275 k  165 ft 2 3.79 ksf

 Try 13 ft square footing. Factored net soil pressure:

qnu 

1.2  350 k  1.6  275 k  5.09 ksf 169 ft 2

2) Check thickness for two-way shear Assume a concrete cover of 3 in., and #8 bars being used. Average effective depth:

d  30 in.  3 in.  1 in.  26 in. 2   18 in.  26 in.   2 Vu  5.09 ksf  169 ft      792 k  12 in./ft   

15-4


Vc is taken as the smaller of the following:  4  4  Vc   2    f cbw d   2    f cbw d  6 f cbw d c  1   (Note  c is the ratio of the column dimensions)

  d 40  26 in.  Vc   2  s   f cbw d   2    f b d  7.9 f cbw d bo   4  44 in.  c w   (Note  s  40 for a column placed at the center of its footing)

Vc  4 fcbwd √

√

 Use 30 in. thick footing.

3) Check one-way shear The distance the footing extends to each side from the column face: 156 in.  18 in. / 2  69 in. [

]

√

√


M u   5.09 ksf 13 ft 

 5.75 ft  

2 Assume j  0.90

2

 1095 k-ft

Use 18 #7 ( As  10.8 in.2 )

(

)

5) Check development Bar spacing exceeds 2db and cover exceeds d b  This is a Case 2 development.  e t f y 1 1 60000 psi db   0.875 in.  35.5 in.<69 in.  3 in., OK. d  25 f c 25  1 3500 psi Use 13 ft square footing, 30 in. thick with 18 #7 bars each way. 6) Design footing-column joint

Factored load at base of column:

15-5


1.2  350  1.6  275  860 kips Allowable bearing on footing: 0.65  0.85  3.5 18 18  2  1250 k

Allowable bearing on column (assume column f c  4000 psi )

0.65  0.85  4 18 18  716 k 860  716 Area of dowels required   3.7 in.2 0.65  60 Depending on how the column reinforcement is selected, the dowel bars will be selected accordingly for construction simplicity.

15-4

Design a rectangular spread footing for the following conditions. Service dead loads are: axial = 350 k, moment = 80 k-ft; service live loads are: axial = 250 k, moment = 100 k-ft. The moments are acting about the strong axis of ⁄ . Allowable soil pressure is 5500 psf. the column section. Soil density is Place the bottom of the foundation at 4.5 ft below the basement floor. Assume the basement floor is 6 in. thick and supports a total service load of 80 psf. Assume the column section is 24 in. x 16 in., is constructed with 5000 psi normal-weight concrete and contains 6 No. eight bars (Grade 60) placed to give maximum bending resistance about the strong axis of the column section. Assume the footing will use and .

1) Select footing dimensions based on allowable soil pressure. Assume a footing thickness of 3 ft. The net permissible bearing pressure is: ( ) Consider only the axial loads to establish the width of the footing. ⁄ Use axial loads and moments to establish the length of the footing. Assume a linear distribution of soil pressure as shown in Fig. 15-5(b). Thus,

Set

and

.

(

Solving for the positive root, Set 2) Calculate factored soil pressure: Factored loads are:

With these values, the factored soil pressures are:

15-6

)


3) Check footing thickness for two-way shear: Dimension of the critical perimeter are:

Therefore, (

)

(

)

Calculate the two-way shear capacity, with Vc taken as the smaller of: ) √

(

) √

(

√

(Note  c is the ratio of the column dimensions) ) √

(

) √

(

√

(Note  s  40 for a column placed at the center of its footing)

Vc  4 fcbwd √  Use 36 in. thick footing.

√

4) Check for combined transfer of shear and moment: √ ⁄ (

) (

√

( )

(

)

) √

√

5) Check for one-way shear: ( √

⁄

) √

6) Design flexural reinforcement for long direction:

15-7

(

)


(

)

Assume jd = 0.95d, and use

 Use 15 #7 (

(

)

)

7) Check development: The available length is 72 in., so this is ok.

15-8


Chapter 17 17-1

The deep beam shown in Fig. P17-1 supports a factored load of 1450 kips. The beam and columns are 24 in. wide. Draw a truss model neglecting the effects of stirrups and the dead load of the wall. Check the strength of the nodes and struts, and design the tension tie. Use (normal weight concrete) and . 69

36

1450 k

1290 k 644 k B

C

Node 1

21.1 10.5

Node 2

55.3°

35.8º

A

Node 3

D 644 k

1290 k 30

24 243

Note: Length units are in inches.

Fig. S17-1 1. Compute Pu /  and reactions To include  in calculations use Pu /   1450 kips/0.75  1930 kips Sum moments about the center of the support at A, assuming the member acts as a simple beam and all loads and reactions act along the axes of the columns.

17-1


69  18  15  1930 kips  644 kips 243  12  15 RA  1930 kips  644 kips  1286 kips  1290 kips

RD 

2. Select size of beam Assume RA is the maximum value of Vu /  , and set this equal to ( 6 to 8 ) f cbd . For b  24 in. , solve to find d values between 142 and 106 in. Try an overall beam height h of 10 ft. and assume that d is approximately 9 ft. = 108 in. 3. Isolate D-regions Because the distance between the load and the reactions is less than 2h at both ends, the beam consists of two D-regions, one of each side of the load. 4. Draw a strut-and-tie model and establish base dimensions of the nodes Use a truss with two inclined struts and a tie, as shown below. Compute f cu Node 1 (CCC node): fcu  0.85n fc  0.85 1.0  4000 psi  3400 psi Node 2 and 3 (CCT nodes): fcu  0.85n fc  0.85  0.8  4000 psi  2720 psi Strut 1-2 (assume sufficient web reinforcement is used to satisfy Eq. 17-10): fcu  0.85s fc  0.85  0.75  4000 psi  2550 psi Select the sizes of the nodes assuming they are hydrostatic nodes with fcu  2550 psi Node 1 Divide the load into two components equal to the two reactions. Width of left part of node  1290 k /(2.55 ksi  24 in.) = 21.1 in. Similarly, the width of the right part of the node = 10.5 in. Total node width  21.1  10.5  31.6 in. < 36 in. , OK. Node 2 Width of node = 21.1 in. < 30 in. (assume reaction is at center of column) Node 3 Width of node = 10.5 in. < 24 in. (assume reaction is at center of column) 5. Establish geometry of truss and forces in struts and tie: Assume that dv  h  2 ft  10 ft.  2 ft. = 8 ft. = 96 in. Then, tan ( left part of span ) = 96 in./(69  18  15) in.; so,  ( left )=53.1 Total span length = 243  12  15  216 in. Then, tan ( right part of span )  96 in./(216  72)in.; so, ( right)  33.7

17-2


Strut 1-2 Axial force = 1290 kips/sin53.1  1610 kips Horizontal component = 1610 kips  cos53.1  967 kips  force in Tie 2-3

Node 2 (Note that  was included in force calculations) Node height = force in the Tie 2-3/( fcu  b)  967 k /(2.55 ksi  24 in.) Node height = 15.8 in.; this is also the height of Node 1. At this point we must check the assumed value for d v Revised: dv  h  2  (15.8 in./2)=120 in.  15.8 in. = 104.2 in. (should recycle) Assume: dv  104 in. With this value:  (left)= 55.3 and  (right) = 35.8 Axial forces: Strut 1-2 = 1570 kips and Tie 2-3 = 893 kip Heights of nodes 1 and 2 = 14.6 in., and dv  120 in.  2  (14.6 in./2)  105.4 in. , OK. Strut 1-3 Axial force: 644 kips/sin35.8  1100 kips Horizontal component: 1100 kips  cos35.8  892 kips ( approx. 893 kips, OK.) 6. Select reinforcement for Tie 2-3 (Note that  was included in force calculations) As  893 k / 60 ksi = 14.9 in.2

Use 16 #9 bars ( As  16.0 in.2 ) Provide 4 layers of 4 No. 9 hooked bars. These must be anchored into the column at each end with anchorage starting where the centroid of the tie first meets the inclined struts at each end of the beam. The length required for a 90° standard hook is 21.4 in. for a No. 9 bar, which is less than the dimension of each support of column. Thus, this should be OK. The centroid of the bars should be at about the mid-height of the nodal zones at each end, i.e. about 7.5 in. above the bottom of the beam. 7. Minimum web reinforcement to control cracking inclined struts Because  s  0.75 was used for the inclined struts in both the left and right spans, minimum web reinforcement that satisfies Eq. 17-10 is required throughout the length of the beam. For a deep beam, it is a good practice to also satisfy the reinforcement requirements in ACI sections 11.8.4 and 11.8.5 Part of the beam to the left of the concentrated load: For horizontal reinforcement, try #5 bars on the front and back faces at a vertical spacing of 12 in. 2  0.31 in.2 h   0.0022  0.0015 (OK.) 12 in.  24 in.  h  90  55.3  34.7 For vertical reinforcement, try #5 two-legged stirrups at a spacing of 10 in.

17-3


2  0.31 in.2  0.0026  0.0025 (OK.) 10 in.  24 in.  v  55.3

v 

Checking Eq. 17-10: A  bsi sin  i   i sin  i si

 (0.0022)  (sin 34.7 )  (0.0026)  (sin 55.3 )  0.0034  0.003,OK. Part of the beam to the right of the concentrated load: Use the same reinforcement. Here  v  35.8 and  h  90  35.8  54.2

  sin  i

i

 0.00174  0.00151  0.00325  0.003 (OK.)

Therefore, throughout the span use 2 #5 horizontal bars at a vertical spacing of 12 in. and use #5 two-legged stirrups as a spacing of 10 in.

17-2

Repeat Problem 17-1, but include the dead load of the wall. Assume that stirrups crossing the lines AB and CD have a capacity ∑ equal to one-third or more of the shear due to the column load. This problem is solved similarly to Problem 17-1. The factored dead load of the wall can be added to the factored load applied on top of the beam. In this case, the vertical reinforcement is designed such that the dead load of the wall portion below the lines AB and CD can be transferred to the level at least above those lines.

17-3

Design a corbel to support a factored vertical load of 120 kips acting at 5 in. from the face of a column. You should include a horizontal load equal to 20 percent of the factored vertical load. The column and corbel are 14 in. wide. The concrete in the column and corbel was cast monolithically. Use 5000-psi normal-weight concrete and .

Interface transfer shear stress is limited to the smallest of 0.2 fc  1000 psi 480  0.08 fc  880 psi and 1600 psi The transfer shear stress is taken as 880 psi. 120,000 Compute minimum effective depth: d   13 in. 0.75  880  14 Try d  16 in. and h  18 in. Factored shear = 120 kips = 120,000 lbs. Normal force  0.2 120 kips  24 kips  24000 lbs. Moment  120  5  24  2  648 k-in. (Note: 2-in. moment arm assumed between horizontal load and horizontal reinforcement)

17-4


Shear Friction Steel V 120 Avf  u   1.90 in.2  f y 0.75  1.4 1.0   60 (Note that we have used   1.4 for monolithic concrete and   1 for normal-weight concrete) Flexural Steel 648  1.0 in.2 0.75  60  (0.9 16) 1.0  60 a  1 in. 0.85  5  14 648 Af   0.93 in.2 0.75  60(16  1/ 2) Af 

Direct Tension Steel 24 An   0.53 in.2 0.75  60 Area of Tension Tie will be taken as the largest of: Af  An  (0.93  0.53)  1.5 in.2 2 Avf / 2  An  2 1.90/ 3  0.53  1.8 in.2

Asc ,min 

0.04 f c 0.04  5 bw d   14  16  0.75 in.2 fy 60

 Use 3 #7 bars (or 2 #9), As  1.80 in.2 (or 2.00 in.2 ) Horizontal Stirrups Area required: Avf / 3  0.63 in.2 Use 2 #4 closed stirrups, As  0.80 in.2 Anchor tension tie into column Use 90° standard hooks to anchor the tension tie. Use #7 bars. Also assume that the ties are inside the column cage; therefore a modification factor of 0.7 is multiplied to the basic development length per ACI Code Section 12.5.3(a): 0.02 e f y 0.02  1 60000 0.7 dh  0.7  db  0.7   0.875  10.4 in.  f c 1 5000 Assume a column size of 14 in. 14 in. , #8 bars used for column longitudinal reinforcement, and a concrete cover of 2.5 in. to the longitudinal bars. The available space for anchorage is: 14 in.  2.5 in.  0.5 in.  12 in.  10.4 in. , OK. Fig. S17-2 shows the detailed reinforcement of the corbel.

17-5


9 in.

3 #7 welded to the angle plate

9 in.

2 #4 closed ties 2 #4 bars

14 in.

8 in.

Fig. S17-2

17-4

Repeat Problem 17-3, but with a factored vertical load of 100 kips and a factored horizontal load of 40 kips. The problem is solved similar to Problem 17-3.

17-5

Figure P17-5 shows the dapped support region of a simple beam. The factored vertical reaction is 100 kips, and you should include a horizontal reaction of 20 kips. Use normal-weight concrete with and reinforcement with (weldable). The beam is 16 in. wide. (a) Isolate the D-region. (b) Draw a truss to support the reaction. (c) Detail the reinforcement.

The factored load on the dap is 100 kips. To include  in the design we shall design the dap for Vn  100/ 0.75  133 kips In addition, design the dap for a horizontal force of H n  0.2 133  26.6 kips (a) Isolate the D-region. Fig S17-3(a) shows the D-region.

17-6


(a) Truss model

2.5

D-region

14

4

17 47.9°

147

D

C

4

6

8

67.2°

147 E

203 G

42.2°

3.5

14.5

15

19

133

133

133

A

16

10

2.5

15

9

F

10.5

120 B

10.5 30

(b) Reinforcement details

17 4 #5 U-stirrups @ 3 in.

4 #5 closed double-leg stirrups

4 #5 U-stirrups @ 2 in. top reinforcement

2 #4 U bars @ 3.5 in.

B

F 5 # 7 with left ends welded to the angle plate

D

2 #5 U bars @ 3.8 in.

A

2 #8 U bar

C

E

G bottom reinforcement

22 48

Note: Force units are in kips, length units are in inches.

17-7


(b) Draw a truss to support the reaction. 1. Select a strut-and-tie model There are two basic types of truss which can be drawn: - Vertical stirrups at end of deeper part - Inclined stirrups in the end We shall use the former detail. Fig. S17-3(a) shows the assumed truss model. 2. Locate nodes at A, B, and C  V B  0  NBC  133 kips As ,required  133/ 60  2.22 in.2

Use 4 #5 closed double leg stirrups ( As  2.48 in.2 ). Place them as shown in Fig. S17-3(b) because these must be as close to the end of the deeper part of the beam as possible. The centroid of this group is at 1.5  (2  0.625)  0.75  3.5 in. from the end of the deeper section. Assume 2.5 in. from the bottom of the beam to the centroid of the steel at C, 2 in. at A, and assume the centroid of the compression force B-F is 2.5 in. below the top of the beam. 4. Compute the strut and tie forces Solve Join at A 10.5 in. tan    1.11    47.9 9.5 in. N AB  133/ sin 47.9  179 kips () 9.5 N AD   133  26.6  147 kips    10.5 Solve Joint at B V  0  N BC  133 kips    H  0  N BF 

9.5  133  120 kips    10.5

Locate node at D Strut CD is axially compressed by the horizontal tensile force in AD and the vertical tensile force in DE. The horizontal distance from C to D is 147  14.5  16.0 in. 133 The axial force in CD is 198 kips. The horizontal distance EG is 30 in.  3.5 in.  16 in.  10.5 in. Solve Joint at E Horizontal component in EF 

10.5  133  55.9 kips 25

Axial force in EF = 144 kips Force in EG = 147 + 55.9 = 203

17-8


Joint F Horizontal force to the right of F = 176 kips As a check on the calculations, cut a section parallel to EF and sum moments about F to calculate the force in EG 133  36 in. +26.6 10.5 = TEF  25 TEF  203 kips  OK 5. Compute strut widths and see if they will fit fcu  0.85s fc  0.85  0.75  5000  3190 psi (Note that the struts are assumed to have confinement steel, hence  s  0.75 ) Strut AB Assume the concrete outside the ties spalls at B. Remaining thickness of strut  16  2 1.5  13 in. 179 Width   4.32 in. 3.19  13 Strut BF 120 Width   2.89 in. 3.19  13 Strut CD Use full thickness. 198 Width   3.88 in. 3.19  16 Strut BE Width 

144  2.82 in. 3.19  16

Struts having these approximate widths are shown in the drawing. 6. Select reinforcement for the ties Tie BC We have already chosen 4 #5 double leg closed stirrups in two groups of two stirrups. Ties DE and FG Use 4 #5 double leg stirrups with 135 hooks spaced at 3 in. o.c. for tie DE and 4 #5 double-leg stirrups with 135 hooks spaced at 2-in. for tie FG. Ties AD As  147 / 60  2.45 in.2 Use 5 #7 bars. Weld these to the angle at A. We need to extend these bars past D a distance equal to the development length. These bars are considered top bars, from Table A-6, d  55.2  48.3 in. So, extend the bars 50 in past D.

17-9


Tie CE As  2.45 in.2 Provide 2 #8 U bars spaced 1 in. clear above the bottom steel. Lap splice them with the tension reinforcement in the bottom of the beam using a splice length of

7. Design confinement reinforcement: Strut AB. Use only horizontal steel. Try 2 #4 U bars spaced equally at 3.5 in. o.c. A 2  0.20 in.2 h  sh   0.0071 bsv 16 in.  3.5 in.

 h  47.9 h sin  h  0.0053  0.0030 , OK. Strut CD. Use only horizontal steel. Try 2 #5 U bars spaced equally at 3.83 in. o.c.

h 

Ash 2  0.31 in.2   0.0101 bsv 16 in.  3.83 in.

 h  42.2 h sin  h  0.0068  0.0030 , OK.

17-10


Chapter 18 18-1

Check moment and shear strength at the base of the structural wall shown in Fig. P181. Also, show that the given horizontal and vertical reinforcement satisfies all of the ACI Code requirements regarding minimum reinforcement percentage and maximum spacing. The given lateral loads are equivalent wind forces, considering both direct lateral forces and the effects of any torsion. Use a load factor of 1.6 for the wind load effects. The given vertical loads represent dead loads, and you can assume that the vertical live loads are equal to 60 percent of the dead loads. Assume the wall is constructed with normal-weight concrete that has a compressive strength of 4500 psi. Assume all of the steel is Grade-60.

1. Calculate factored loads at base of wall.

M u (base)  1.6  (7 k 15 ft  12 k  26 ft  18 k  37 ft  22 k  48 ft  20 k  59 ft)  5310 kip-ft Vu (base)  1.6   7  12  18  22  20  kips  126 kips Nu ,min (base)  0.9  230 k  207 kips 2. Calculate flexural strength. The total area of vertical wall reinforcement is:

Ast  2  0.20 in.2 

w

s

 0.40 in.2 

240 in.  8.00 in.2 12 in.

From Eq. (18-26a), the percentage of vertical (longitudinal) reinforcement is:

Ast 8.00 in.2   0.00333 h  w 10 in.  240 in.

 

From Eq. (18-26b), the vertical reinforcement index is:

 

fy fc

 0.00333 

60 ksi  0.0444 4.5 ksi

From Eq. (18-27), the axial load index is:



Nu h

w

 f c



207 kips  0.0192 10 in.  240 in.  4.5 ksi

For 4500 psi concrete, 1 = 0.825. Then, from Eq. (18-28) the depth to the neutral axis is:

    c   0.851  2 

w

0.0192  0.0444     240 in.  19.3 in.  0.85  0.825  2  0.0444 

This is very small compared to d (taken as 0.8 ℓw), so this is clearly a tension-controlled section and  = 0.9. From Eq. (18-25a), the tension force in the vertical reinforcement is:

 T  Ast f y  

w

c   240  19.3  2   8.00 in.  60 ksi    441 kips  240  w 

18-1


And, from Eq. (18-29) the nominal moment strength at the base of the wall is:

   c   240 in.   240 in.  19.3 in.  M n  T  w   Nu  w  441 k    207 k    2  2     2   2   53, 000 k-in.  22,800 k-in.  75,800 k-in.  6320 k-ft Using the strength reduction factor, , the design strength is:

 M n  0.9  6320  5690 k-ft  M u (o.k.)

3. Check shear strength. Because the wall is subjected to compression, we are permitted to use Eq. (18-41) to determine the concrete contribution to shear strength. For this calculation we will assume d = 0.8 x ℓw = 0.8 x 240 = 192 in., as permitted in ACI Code Section 11.9.4.

Vc  2 f c  hd  2 1 4500 psi 10 in. 192 in.  258, 000 lbs  258 kips Using  = 0.75 for shear, the design strength contribution from the concrete is:

Vc  0.75  258  193 kips  Vu

Thus, no horizontal reinforcement is required for shear strength. However, because Vu exceeds one-half of Vc, the reinforcement requirements in ACI Code Section 11.9.9 must be satisfied. Horizontal reinforcement: From Eq. (18-45a), the percentage of horizontal reinforcement is:

t 

Av ,horiz h  s2



2  0.20 in.2  0.0025  0.0025 (o.k.) 10 in. 16 in.

The maximum center-to-center spacing for the horizontal reinforcement is the smallest of ℓw/5 (48 in.), 3h (30 in.) and 18 in. Thus, the spacing of 16 in. for the horizontal reinforcement is ok. Vertical reinforcement: Because the wall aspect ratio, hw/ℓw = 59ft/20ft = 2.95, exceeds 2.5, the minimum required percentage of vertical reinforcement is 0.0025. From step 2, ℓ = 0.00333, which exceeds the minimum value. The maximum center-to-center spacing for the vertical reinforcement is the smallest of ℓw/3 (80 in.), 3h (30 in.) and 18 in. Thus, the provided spacing of 12 in. for the vertical reinforcement is ok.

18-2


18-2

Design a uniform distribution of vertical and horizontal reinforcement for the structural wall shown in Fig. P18-2. Your design must satisfy all of the ACI Code requirements, as well as the requirements for minimum reinforcement percentage and minimum spacing. The given lateral loads are equivalent wind forces, considering both direct lateral forces and the effects of any torsion. Use a load factor of 1.6 for the wind load effects. Assume the wall is constructed with normal-weight concrete that has a compressive strength of 4000 psi. Assume all of the steel is Grade-60.

1. Calculate factored loads at base of wall.

M u (base)  1.6  (75 k 12 ft  100 k  24 ft)  5280 kip-ft

Vu (base)  1.6   75  100  kips  280 kips

Nu ,min (base)  0.9  (70  40) kips  99.0 kips 2. Flexural design. Based on the results from Problem 18-1 (a wall with similar design base moment and similar dimensions), select a trial value for ℓ = 0.0035. Then, from Eq. (18-26b),

 

fy f c

 0.0035 

60 ksi  0.0525 4 ksi

From Eq. (18-27), the axial load index is:



Nu h

w

 f c



99.0 kips  0.0103 10 in.  240 in.  4 ksi

For 4000 psi concrete, 1 = 0.85. Then, from Eq. (18-28) the depth to the neutral axis is:

    c   0.851  2 

w

0.0103  0.0525     240 in.  18.2 in.  0.85  0.85  2  0.0525 

This is very small compared to d (taken as 0.8 ℓw), so this is clearly a tension-controlled section and  = 0.9. The total area of vertical wall reinforcement is:

Ast    h 

w

 0.0035 10 in.  240 in.  8.40 in.2

From Eq. (18-25a), the tension force in the vertical reinforcement is:

 T  Ast f y  

w

c   240  18.2  2   8.40 in.  60 ksi    466 kips  240  w 

And, from Eq. (18-29) the nominal moment strength at the base of the wall is:

   c   240 in.   240 in.  18.2 in.  M n  T  w   Nu  w  466 k    99.0 k    2  2     2   2   55,900 k-in.  11, 000 k-in.  66,900 k-in.  5570 k-ft Using the strength reduction factor, , the design strength is:

 M n  0.9  5570  5020 k-ft  M u (not o.k.)

18-3


For a wall with relatively low axial load, the nominal moment strength should increase approximately linearly with increases in the percentage of vertical reinforcement. Thus,

 (req'd.)   (trial) 

Mu 5280  0.0035   0.00368  M n (trial) 5020

Round this up a little and try ℓ = 0.0038. Then, redo the calculations to find:

  0.0570, c  19.3 in., Ast  9.12 in.2 , T  503 kips, and  M n  5350 k-ft  M u (o.k.) Select vertical reinforcement as: No. 4 bars at 10 in. spacing in each face (EF), results in ℓ = 0.00400 No. 5 bars at 16 in. spacing in each face (EF), results in ℓ = 0.00388 Either selection will work. To use less total bars, select No. 5 bars at 16 in. spacing, EF. It is good practice to replace the pair of No. 5 bars at the edges of the wall with a pair of No. 6 bars. 3. Shear Design. The aspect ratio for this wall is, hw/ℓw = 24/20 = 1.20. Thus, this is a short wall and the shear strength contribution from the concrete is probably given by Eq. (18-43). Using d = 0.8 x ℓw = 192 in., results in:

Vc  3.3 f c hd 

Nu d 4 w

99, 000 lbs 192 in. 4  240 in.  (401, 000  19,800) lbs  421, 000 lbs  421 kips  3.3 1 4000 psi 10 in. 192 in. 

Before accepting this value, we will check the value of Vc from Eq. (18-44). For this flexural-shear strength equation, we need to evaluate the ratio of Mu/Vu at the critical section above the base of the wall, as defined in Fig. 18-19. For this wall, ℓw/2 = 10 ft, governs. At that section the factored moment is,

M u (crit. sect.)  M u (base)  Vu (base)

w

2  5280 kip-ft  280 kip 10 ft  2480 kip-ft

Thus, the ratio of Mu/Vu = 2480/280 = 8.86 ft. Using this value, the denominator in the second term of Eq. (18-43) is,

Mu  w  8.86 ft  10 ft  1.14 ft Vu 2 Because this is a negative number, Eq. (18-44) is not valid for this wall. So, using  = 0.75 and the value for Vc from Eq. (18-43):

Vc  0.75  421 kips  316 kips  Vu

18-4


Thus, no horizontal reinforcement is required for shear strength. However, because Vu exceeds one-half of Vc, the reinforcement requirements in ACI Code Section 11.9.9 must be satisfied. Horizontal reinforcement: Use 2 No. 4 bars at 16 in. spacing in each face. Then, from Eq. (18-45a), the percentage of horizontal reinforcement is:

t 

Av ,horiz h  s2



2  0.20 in.2  0.0025  0.0025 (o.k.) 10 in. 16 in.

The maximum center-to-center spacing for the horizontal reinforcement is the smallest of ℓw/5 (48 in.), 3h (30 in.) and 18 in. Thus, the provided spacing of 16 in. for the horizontal reinforcement is ok. Vertical reinforcement: Because t = 0.0025, the minimum value for ℓ in Eq. (18-46) is 0.0025. From the flexural design, the provided value for ℓ = 0.00388, so it is o.k. Also, the selected spacing of 16 in. is less than the smallest of ℓw/3 (80 in.), 3h (30 in.) and 18 in. (o.k.)

18-3

The structural wall shown in Fig. P18-3 is subjected to gravity loads ( and ) and an equivalent, static, lateral earthquake load, ) (torsion effects included). Check the moment strength at the base of the structural wall assuming that the distance, , from the base of the structure to the lateral force, , is equal to two-thirds of the wall height. Use a capacity-design approach to check the shear strength of the wall and assume the distance, , is equal to one-half of the wall height for this check. Also, show that the given horizontal and vertical reinforcement in the web of the structural wall satisfies all of the ACI Code requirements regarding minimum reinforcement percentage and maximum spacing. Assume the wall is constructed with normal-weight concrete that has a compressive strength of 5000 psi. Assume all of the steel is Grade-60. Use: for the flexural strength check for the capacity-based shear strength check

1. Calculate factored loads for flexural strength check at base of wall.

M u (base)  1.0 150 k  2 3  60 ft  6000 kip-ft Nu ,min (base)  0.9 150 k  135 kips

2. Flexural strength. For a boundary element in tension, use Eq. (18-30) to find:

T  As f y  12  0.79 in.2  60 ksi  569 kips Assuming that the depth of the compression stress block does not exceed the size of the boundary element, use Eq. (18-32) to find:

a

T  Nu 569 k  135 k   8.28 in. (  20 in., o.k.) 0.85 f c b 0.85  5 ksi  20 in.

18-5


Assume d = ℓw – 20in./2 = 180 – 10 = 170 in. Then, from Eq. (18-33), the nominal moment strength is:

a   M n  T  d    Nu  2  

w

a 2 

8.28 in.    180 in.  8.28 in.   569 k  170 in.    135 k   2  2     (94, 400  11, 600)k-in.  8830 kip-ft With a = 8.28 in., it is clear that this is a tension-controlled section, and thus  = 0.9. So,

 M n  0.9  8830  7950 kip-ft

Mu

The wall is substantially over-designed in flexure and we should reduce the steel in the boundary element to reduce the shear required to develop the flexural strength. One possible redesign is to use eight No. 9 bars in each boundary element. This leads to the final result of Mn = 6860 kip-ft. 3. Design shear (determine capacity-based design shear using original flexural design). Assume that the probable axial load is:

N pr  N D  N L  150 k  100 k  250 kips With this axial load the wall moment strength will be reevaluated and referred to as the probable moment strength, Mpr. First, the depth of the compression stress block is,

a

T  Nu 569 k  250 k   9.64 in. 0.85 f c b 0.85  5 ksi  20 in.

This is larger than calculated previously, but it is still clear that the tension steel in the boundary element will be yielding. With this value of a, the moment strength is:

a   M pr  T  d    N pr  2  

w

a 2 

 94, 000 k-in.  21,300 k-in.  115, 000 kip-in.  9610 kip-ft With this moment and assuming that x = 0.5 x 60 ft = 30 ft, the capacity-based design shear is,

Vu (cap-based) 

M pr 0.5hw



9610 k-ft  320 kips 0.5  60 ft

4. Check shear strength. For this wall, the value of Acv in Eq. (18-48) is:

Acv  h 

w

 10 in. 180 in.  1800 in.2

Because this is a slender wall, c = 2.0. Eq. (18-45b) will be used to determine t for the distributed horizontal reinforcement.

18-6


t 

Av ,horiz h  s2



2  0.20 in.2  0.0025 10 in. 16 in.

Using the values calculated here, the nominal shear strength of the wall from Eq. (11-48) is,



Vn  Acv  c  f c  t f y





 1800 in.2 2 1 5000 psi  0.0025  60, 000 psi



 1800 in.2 141 psi  150 psi   524, 000 lbs  524 kips Using  = 0.75 for shear,

Vn  0.75  524 k  393 kips  Vu (cap-based) (o.k.)

The vertical and horizontal steel percentages in the web of the wall (both 0.0025) and the bar spacing (16 in. both horizontal and vertical) satisfy the requirements of ACI Code Section 11.9.9, which are applicable for this wall.

18-4

The structural wall shown in Fig. P18-4 is subjected to gravity loads ( and ) and a single, equivalent, static, lateral earthquake load ( ) at the top of the wall (at roof level). For the given uniform distribution of vertical and horizontal reinforcement, check the moment strength at the base of the wall and use a capacity-deisgn approach to check the shear strength of the wall. Also, show that the given horizontal and vertical reinforcement in the structural wall satisfies all of the ACI Code requirements regarding minimum reinforcement percentage and maximum spacing. Assume the wall is constructed with normal-weight concrete that has a compressive strength of 4000 psi. Assume all of the steel is Grade-60.

1. Calculate factored loads for flexural strength check at base of wall.

M u (base)  1.0  220 k 15 ft  3300 kip-ft Nu ,min (base)  0.9  80 k  72 kips

2. Flexural strength. From Eq. (18-454a) the vertical reinforcement percentage is:

2  0.31 in.2     0.00388 h  s1 10 in. 16 in. Av ,vert

From Eq. (18-26b) the reinforcement ratio for the vertical reinforcement is:



fy f c

 0.00388 

60 ksi  0.0582 4 ksi

From Eq. (18-27) the axial load ratio is:

Nu

 h

f

w c



72 kips  0.00750 10 in.  240 in.  4 ksi

18-7


For 4000 psi concrete, 1 = 0.85. Then, from Eq. (18-28), the depth to the neutral axis is:

    c   0.851  2 

w

0.0075  0.0582     240 in.  18.8 in.  0.85  0.85  2  0.0582 

This is very small compared to d (0.8ℓw = 192 in.), so this is clearly a tension-controlled section and  = 0.9. For the longitudinal steel,

Ast  2  0.31 in.2 

240 in.  9.30 in.2 16 in.

Then, from Eq. (18-25a), the flexural tension force is:

 T  Ast f y  

w

c   240 in.  18.8 in.  2   9.30 in.  60 ksi    514 kips 240 in.   w 

Finally, the nominal moment capacity is found using Eq. (18-29):

   c  M n  T  w   Nu  w   2   2   240 in.   240 in.  18.8 in.   514 k     72 k   2  2     (61, 700  7960)k-in.  5810 kip-ft Applying the strength factor, the design flexural strength is,

 M n  0.9  5810  5230 kip-ft

Mu

The wall is substantially over-designed in flexure and we should reduce the vertical reinforcement in the wall to reduce the shear required to develop the flexural strength. One possible redesign is to use two No. 4 bars at a 16 in. spacing in each face. For this reinforcement ℓ = 0.0025 (the minimum value) and the design flexural strength, Mn = 3670 kip-ft. 3. Design shear (determine capacity-based design shear using original flexural design). Assume that the probable axial load is:

N pr  N D  N L  80 k  40 k  120 kips With this axial load, the axial load ratio is:



120 kips  0.0125 10 in.  240 in.  4 ksi

And, the depth to the neutral axis is:

0.0125  0.0581   c  240 in.  20.2 in.  0.85  0.85  2  0.0581 

18-8


Then, the flexural tension force is:

 T  Ast f y  

w

c   240 in.  20.2 in.  2   9.30 in.  60 ksi    511 kips 240 in.   w 

Finally, the probably moment capacity is:

 240 in.   240 in.  20.2 in.  M pr  511 k     120 k   2  2     (61,300  13, 200)k-in.  6210 kip-ft

With this probable moment strength, the capacity-based design shear is,

Vu (cap-based) 

M pr hw



6210 k-ft  414 kips 15 ft

4. Check shear strength. For this wall, Eq. (18-45b) will be used to determine t for the distributed horizontal reinforcement.

2  0.20 in.2 t    0.0025 h  s2 10 in. 16 in. Av ,horiz

The effective wall shear area from Eq. (18-48) is:

Acv  h 

w

 10 in.  240 in.  2400 in.2

Because this is a squat wall (hw/ℓw = 0.75), c = 3.0. Thus, the nominal shear strength of the wall from Eq. (11-48) is,



Vn  Acv  c  f c  t f y





 2400 in.2 3 1 4000 psi  0.0025  60, 000 psi



 2400 in.2 190 psi  150 psi   815, 000 lbs  815 kips Using  = 0.75 for shear,

Vn  0.75  815 k  612 kips  Vu (cap-based) (o.k.)

The vertical and horizontal steel percentages in the web of the wall (both ≥ 0.0025) and the bar spacing (16 in. both horizontal and vertical) satisfy the requirements of ACI Code Section 11.9.9, which are applicable for this wall.

18-5

Design a uniform distribution of vertical and horizontal reinforcement for the first story of the structural wall shown in Fig. P18-5. The given horizontal loads are strength-level static equivalent earthquake forces, so use a load factor of 1.0. The vertical dead and live loads are from the tributary area adjacent to the wall. Assume the given lateral loads include the direct shear force and any torsional effects that need to be considered.

18-9


Use a compressive strength of 4000 psi and Grade 60 reinforcement. Your final design should satisfy all ACI Code requirements for minimum reinforcement percentages and maximum spacing. Try using

( )

( )

. Use 2#4 bars at 16 in. both ways. ⁄ (

) {

Start with flexural design. Calculate

}

:

(

)

(

)

⁄

(

Calculate

)

: (

)

(

)

Therefore, in. on-center are OK.

(

)

-

, and the two curtains of #4 vertical bars spaced at 16

Continue with the shear design. Calculate ( ) based on the probable moment capacity of the wall and assuming that the lateral force resultant is located at . (

)

(

)

(

)

⁄

(

)

(

18-10

)


(

Calculate

⁄

)

:

⁄

( (

√

)

⁄

(

)

) (

)

√

Therefore, (

)

The capacity and demand are approximately equal, so the horizontal reinforcement is adequate.

Final design: Vertical Reinforcement: Horizontal Reinforcement:

2 curtains of #4 bars spaced at 16 in. on-center. 2 curtains of #4 bars spaced at 16 in. on-center.

18-11


Chapter 19 19-1

Use Eq. (19-28) to check the need for specially confined boundary elements in the structural wall described in Problem 18-3. Assume the design displacement, , at the top of the wall is 0.6 ft. Use all of the dimensions, loading information, and material properties given in Problem 18-3. If a specially confined boundary element is required, define the required vertical and horizontal dimensions for the boundary element.

1. Building drift ratio. The ratio, u/hw, represents the building drift ratio due to the design earthquake for the building in question. We are given a value for the design displacement, u, at the top of the building, so the building drift ratio is:

Building drift ratio 

u hw



0.6 ft  0.010 (1%) 60 ft

2. Limit for neutral axis depth. Eq. (19-27) gives a limiting value for the neutral axis depth. For larger values, the wall boundary elements must be confined.

c(limit) 

w

600  u hw 



15 ft 12 in./ft  30.0 in. 600  0.010

3. Calculated neutral axis depth. The depth of the neutral axis for this check should come from the calculation of the probable moment strength of the wall, which was calculated in step 3 of Problem 18-3. In that calculation the depth of the equivalent stress block, a, was found to be 9.64 in. Then, using 1 = 0.80 for 5000 psi concrete, the corresponding neutral axis depth is:

c

a

1



9.64 in.  12.1 in.  c(limit) 0.80

Therefore, special confinement reinforcement is not required in the wall boundary elements.

19-1


19-2

Use Eq. (19-29) to check the need for specially confined boundary elements in the structural wall described in Problem 18-3. Use all of the dimensions, loading information, and material properties given in Problem 18-3. If a specially confined boundary element is required, define the required vertical and horizontal dimensions for the boundary element.

1. Factored loads at base of wall. From step 1 of Problem 18-3:

M u (base)  6000 kip-ft  72, 000 kip-in. Nu ,min (base)  135 kips 2. Wall section properties. To determine the gross area and moment of inertia of the wall, divide the wall into rectangular pieces representing the web and the two boundary elements. Then,

A  A(web)  2 A(boundary)  10 in. 140 in.  2   20 in.  2200 in.2 2

And,

I    I i  Ai yi2   i

 20  203  10 1403  0  2  20  20  802  12  12 

 7.43 106 in.4 Also,

y

w

2



180 in.  90 in. 2

3. Combined stress in edge of wall at base of wall. Eq. (19-28) will be used to find the combined stress at the edge of the wall at its base. If that stress exceeds the limit of 0.2fc', then we will need use special confinement reinforcement in the wall boundary and continue that reinforcement up the wall until we reach a section where the combined stress in the edge of the wall is less than 0.15fc'. Using the values from steps 1 and 2:

Nu M u y  A I 135 k 72, 000 k-in.  90 in.   2 2200 in. 7.43 106 in.4  0.061 ksi  0.872 ksi  0.934 ksi  0.2  5 ksi  1.0 ksi

fc 

Therefore, special confinement reinforcement is not required in the wall boundary elements.

19-2

Reinforced Concrete Mechanics Design Solutions

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