Night School 10 Session 7

AISC Night School March 29, 2016

Steel Design 2: Selected Topics Session 7: Bracketed, Stepped, and Tapered Columns

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There’s always a solution in steel!

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Copyright © 2016 American Institute of Steel Construction 7.1



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Course Description Session 7: Bracketed, Stepped, and Tapered Columns

March 29, 2016 This session will investigate the behavior of bracketed, stepped, and tapered columns based on elastic buckling analysis. The design of columns with load introduced along the length versus load introduced at the ends will be investigated. Bracketed and stepped columns will be compared and an approach to strength determination will be explained. Tapered columns will be briefly considered to establish a simple starting point for design.





Learning Objectives • Gain an under understan standing ding of elast elastic ic bucklin buckling g load and effec effective tive length factors • Determi Determine ne effectiv effective e length length factors factors for for columns columns with with load applied along their length • Observe Observe the the influenc influence e of a reduced reduced moment moment of inertia inertia along along a portion of the length in a stepped column • Apply Apply concep concepts ts of steppe stepped d column column design design to the the design design of of tapered columns


Steel Design 2: Selected Topics Topics based on AISC 360-10 Specification for Structural Steel Buildings Lesson 7 – Bracketed, Stepped, and Tapered Tapered Columns Presented by Louis F. Geschwindner, Ph.D., P.E. Emeritus Professor at Penn State University Former Vice-President at AISC


8




Steel Design 2: Selected Topics based on AISC 360-10 Specification for Structural Steel Buildings Night School 10 Lesson 7 Bracketed, stepped, and tapered columns


7.9

Lesson 7 • Column Column desi design gn appea appears rs to be be based based on the determination of effective length factors • Effective Effective length length factors factors are really really just a short-hand way of determining elastic buckling load • This lesson lesson will will start start with with a discussion discussion of elastic buckling load and effective length factors There’s always a solution in steel!

7.10




Lesson 7 • It will follow up with determination of effective length factors for columns with load applied along their length • This will be followed by discussion of bracketed columns and stepped columns • Tapered columns will be viewed with the goal of establishing a starting point for their design There’s always a solution in steel!

7.11

Elastic Buckling Analysis • Determine the elastic buckling load for a perfect column P

L y x

•Assumptions •Perfectly elastic •Perfectly straight •Constant cross section •Pin ends Equilibrium at a point on a free body in the displaced configuration

P

x There’s always a solution in steel!

= Py 7.12





•From the principles of mechanics using small displacement theory

d2y

L

dx 2

y x

x

=−

EI

•Combining and rearranging terms

d2y

P

dx 2

+

P

y = 0 EI


7.13

Elastic Buckling Analysis • Determine the elastic buckling load for a perfect column •Define

P

k 2 =

P EI

L y

•The result is the differential equation of the column

x

d2y P

dx 2


+ k2y = 0 7.14





•Solution

y = A sin kx + B cos kx L y

•From boundary conditions

x

B = 0 A sin kL = 0

P


7.15


•Thus, and L

y

sin kL = 0

kL = nπ

x

•The shape of the deflected column is P

y = A sin


nπx L

7.16





•Remembering that

k2 = L y

P EI

and kL = nπ

thus,

x

2

k =

P

P EI

=

n 2 π2 L2


7.17


•So the final solution for the differential equation is L

y

P =

n 2 π2 EI L2

x

P

which has a minimum value when

n = 1


7.18





L

•Thus, we have the well known Euler Equation for the elastic buckling load:

y x

P e =

π2 EI L2

P


7.19

Elastic Buckling Analysis • Our design approach is assisted by comparing the real column/structure to the perfect, Euler Column • The elastic buckling load for a real column, as part of a real structure with real imperfections can be thought of as

Pcr = P e × (modification factor) There’s always a solution in steel!

7.20




Elastic Buckling Analysis • If the modification factor is defined as: modification factor =

1 2 K exact

• The elastic buckling load for a real column would be given by: P cr =

π2 EI ( K exact L)2


7.21

Elastic Buckling Analysis • Every approach proposed to determine the effective length factor, K , is really an attempt to determine the exact effective length factor, K exact , such that the exact elastic buckling load may be determined, without the need to actually resort to an elastic buckling analysis.


7.22




E3. Flexural Buckling • For nonslender element members in uniform compression Pn = Fcr Ag When

When

F y F e F y F e

E3-1

F   Fcr =  0.658 F  F y   y

≤ 2.25

> 2.25

e

E3-2

E3-3

Fcr = 0.877 F e


7.23

E3. Flexural Buckling • For nonslender element members in uniform compression F y = 36 ksi



F y



F Fcr = 0.658 e  F y





E3-2

Eq. E3-2

Fcr = 0.877 F e Eq. E3-3 Inelastic Buckling

F e = Elastic Buckling


π2 E

 KL     r 

E3-3

E3-4 2

7.24




E3. Flexural Buckling • F e is the elastic buckling stress, what we usually call the Euler buckling stress F e =

π2 E

 KL     r 

2

E3-4

• This is derived from what we call the “perfect column” or the “Euler column” as we just saw


7.25

E3. Flexural Buckling However, the use of K is not the only way to determine the elastic buckling stress We are used to starting with K and determining F e


But we could also start with F e and determine K

7.26




Elastic Buckling Analysis • An elastic buckling analysis is the solution to an eigenvalue problem • There are many ways to solve that problem, usually involving a computer solution • For all elastic buckling analysis solutions in this lesson I have used the computer program GT STRUDL There’s always a solution in steel!

7.27

Example 1 • Determine the buckling load for the given frame using an elastic buckling analysis P

P

All members W8x24 I x = 82.7 in.4

t f

0 . 0 1

This structure will buckle in a sidesway buckling mode at a critical load P e = 232 kips

A = 7.08 in.2 r x = 3.42 in. 20.0 ft

From this we determine the elastic buckling stress as F e =


232 7.08

= 32.8 ksi 7.28




Example 1 Using stress determine K

F e =

32.8

KL r

93.4

π2 E

 KL     r 

π2 E

=

32.8

= 32.8 ksi

2

= 93.4

 3.42   10 (12 )  = 2.66  

K = 93.4  Using force determine K

P e =

π2 EI

( KL )

K =

2

π EI L

P e

=

π

29,000 (82.7 )

10(12)

232

= 2.66


7.29

Example 1 • Determine the nominal strength of the columns in this frame P

From our buckling analysis

P

All members W8x24 I x = 82.7 in.4 2

A = 7.08 in. r x = 3.42 in. 20.0 ft

F e = 32.8 ksi

t f

0 . 0 1

F y F e

=

50 32.8

= 1.52 < 2.25

F 50     Fcr =  0.658 F  F y =  0.65832.8  50 = 26.4 ksi       P n = 26.4( 7.08) = 187 kips y e

Note that we determined the nominal strength without using effective length There’s always a solution in steel!

7.30




Example 1

F cr

The critical stress is determined without first obtaining the slenderness ratio

F e

32.8 26.4


7.31

Elastic Buckling Analysis • So it appears that for this structure, K = 2.66. • But for this same structure with different loading, the buckling load is P e = 460 kips. P K = All members W8x24 I x = 82.7 in.4

π EI L

P e

t f

0 . 0 1

A = 7.08 in.2

=

π

29,000 (82.7 )

10(12)

460

= 3.42 in. r x 20.0 ft


= 1.89 ≠ 2.66 7.32




Elastic Buckling Analysis • Now consider how load location might impact the buckling strength of a two story structure P P The frame is permitted to sway sideways, L = 10 ft

P

P

All members are W8x24

10.0 ft

10.0 ft

20.0 ft


7.33

Elastic Buckling Analysis Both stories loaded

P

P

P

P

P cr = 140 kips K upper =

π

29,000 (82.7 )

10(12)

140

10.0 ft

= 3.43 K lower =

π

29,000 (82.7 )

10(12)

2 (140 )

10.0 ft

= 2.42 20.0 ft

Sidesway Permitted There’s always a solution in steel!

7.34




Elastic Buckling Analysis Top story loaded

P

P

P cr = 242 kips K upper =

π

29,000 (82.7 )

10(12)

242

10.0 ft

= 2.61 K lower =

π

29,000 (82.7 )

10(12)

242

10.0 ft

= 2.61 20.0 ft


7.35

Elastic Buckling Analysis Bottom story loaded P cr = 290 kips K upper =

π

29,000 (82.7 )

10(12)

0

P

P

10.0 ft

=? K lower =

π

29,000 (82.7 )

10(12)

290

10.0 ft

= 2.38 20.0 ft


7.36




Elastic Buckling Analysis P

In our earlier derivation, we took the minimum buckling load represented by n = 1. For higher modes of buckling, the integer n will increase. L

Thus, for n = 2, the mode shape will be a full sine wave and the buckling load will be 4 times greater.

P cr =

P

4π 2 EI L2

y = A sin

2πx L


7.37


L

Thus, for n = 3, the mode shape will be 1½ sine waves and the buckling load will be 9 times greater.

P cr =

9π2 EI L2

y = A sin

3πx

L

P


7.38





L

And, for n = 4, the mode shape will be 2 sine waves and the buckling load will be 16 times greater.

P cr =

16π2 EI L2

y = A sin

4πx

L

P


7.39


But all this depends on how we define L For this column to buckle in any of these other modes, it must be braced appropriately. L

P

Thus, if it is braced at the 1/3 points it buckles in the mode defined for n = 3 when the column length is given as the full length as shown

P cr =

9π2 EI L2


y = A sin

3πx L 7.40





However, if the length is defined as shown here, n = 1 because it buckles in a half sine wave over that length

L

So the important point here, and particularly for our discussion to come for columns loaded along their length, is how do you define length.

P cr =

P

π2 EI L2

y = A sin

πx L


7.41

Elastic Buckling Analysis • Now consider how the point of load application might impact buckling P

W16x77 P t f

5 4 = L

P

x

For buckling about the x -axis

P


7.42




Elastic Buckling Analysis • If we are then to determine the effective length factor it too will vary with point of load application For x = 21 ft, P cr = 2050 kips K =

=

π EI L

P cr

π

29,000 (1110 )

45(12)

2050

= 0.729


7.43

Elastic Buckling Analysis • The bracketed and stepped columns we are going to be looking at will have load applied at the top (roof load) and also at an intermediate point (crane rail load) • Now look at how application of two loads on a column will influence the buckling load


7.44




Elastic Buckling Analysis • If equal loads are applied at the top and at an intermediate point P 2

P 2

W16x77 x

L = 45 ft

For buckling about the x-axis

P


7.45

Elastic Buckling Analysis • The effective length factor starts at 1.0 when all load is applied at the top, the Euler Column For x = 21 ft, P cr = 1442 kips K =

=

π EI L

P cr

π

29,000 (1110 )

45(12)

1442

= 0.869


7.46




Example 2 • Determine if the given bracketed column will support the applied loads P 1 t f

5 t f

8

P 2 H

The column is a W16x77 braced out of the plane at the roof truss, the crane rail girder and at 16 ft o/c below that Section Properties

t f

A = 22.6 in.2

2 3

r x = 7.00 in. 4

I x = 1110 in.

r y = 2.47 in.


7.47

Example 2 • Determine if the given bracketed column will support the applied loads P 1

Use Association of Iron and Steel Engineers Technical Report No. 13

t f

5 t f

8

P 2 H

t f

2 3

Two load cases will be considered Case 2: DL, LL, Crane Load Case 3: DL, Crane Vertical, Wind See Design Guide 7


7.48




Example 2 • Results of a second-order analysis for ASD (DG 7) P 1

– Case 2

t f

5

50 ft-kips

t f

8

P 1 = 31 kips P 2 = 50 kips

P 2

35 ft-kips

125 ft-kips

H

t f

P total = 81 kips

2 3

Second-order moments

121 ft-kips

7.49



t f

– Case 3

5 t f

P 1 = 33 kips

8

117 ft-kips

H

P 2 = 37.5 kips t f

P total = 70.5 kips


P 2

2 3

110 ft-kips


7.50




Example 2 • To determine column strength we must determine either the elastic buckling load or the equivalent effective length factor. • Design Guide 7 includes tables of effective length factors based on the paper by Agrawal and Stafeij • AISE Report 13 includes similar tables


7.51

Example 2 • Note that we have two load cases. This will result in different buckling loads – Case 2

P 1

P 1 = 31 kips P 2 = 50 kips

t f

5 t f

8

P total = 81 kips

P 2 H

– Case 3 P 1 = 33 kips P 2 = 37.5 kips

t f

2 3

P total = 70.5 kips There’s always a solution in steel!

7.52




Example 2 • To determine the elastic buckling load of our structure. Assume a fixed base and a rotationally rigid top with permitted sway P

P1 = 0.38P

P1 = 0.47 P t f

t f

5

5 t f

t f

P2 = 0.61P

8

P2 = 0.53P

8

L

P cr = 1090 kips

t f

t f

2 3

2 3

K = 1

P cr = 1179 kips

P

Case 2 There’s always a solution in steel!

P cr = 1166 kips

Case 3 7.53

(See: Anderson and Woodward)

Example 2 • Consider Case 2 P 1 = 31 kips

Effective length factors

K top =

50 ft-kips

P 2 = 50 kips 35 ft-kips

125 ft-kips

= 1.55

P total = 81 kips K bot =

P cr = 1179 kips

π2 ( 29,000 )(1110 ) 2  31  ( 45 (12 ) ) (1179 )    81 

π2 ( 29,000 )(1110 )

 ( 45 (12 ) ) (1179 )  2

81 

  81 

= 0.961 121 ft-kips



Note that in both cases we are using the full length, L = 45 ft 7.54




Example 2 • Consider Case 2 P 1 = 31 kips

We could have obtained F e directly from P cr

50 ft-kips

 31    81  = 20.0 ksi

1179 

P 2 = 50 kips

F e upper =

22.6

35 ft-kips

125 ft-kips

P total = 81 kips

 81    81  = 52.2 ksi

1179  F e lower =

P cr = 1179 kips

22.6

However, in our case the other axis is going to control so we will not be taking advantage of these. 7.55

121 ft-kips



Example 2 • Determine the nominal compressive strength of the lower segment KL r x KL r y

=

=

0.961( 45)(12) 7.00 1.0 (16)(12) 2.47

= 74.1 < 4.71

= 77.7 < 4.71

29,000 50

29,000 50

= 113

= 113

For compression, y-axis controls There’s always a solution in steel!

7.56




Example 2 • Determine the nominal compressive strength of the lower segment F e =

π2 E

 KL   r   

F  Fcr =  0.658 F   y e

2

=

π 2 ( 29,000 )

( 77.7 )

2

= 47.4 ksi

This is the y -axis, for the x -axis we had F e = 52.2 ksi

50    47.4  F y =  0.658  ( 50) = 32.2 ksi    

Pn = 32.2 ( 22.6 ) = 728 kips

P n Ω = 728 1.67 = 436 kips


7.57

Example 2 • Determine the flexural strength for bending about the x -axis of the lower segment • Unbraced length is Lb = 16 ft • W16x77 has compact flange and web • From Manual Table 3-2 L p = 8.72 ft M p

Ω

Lr = 27.8 ft BF Ω = 7.34 ft-kips/ft

= 374 ft-kips

M r

Ω


= 234 ft-kips 7.58




Example 2 • Available moment strength  BF  −  ( Lb − Lp ) Ω Ω  Ω  = 374 − 7.34(16.0 − 8.72) = 321 ft-kips

M n

=

• Interaction

M p

P r P c

=

81 436

= 0.186 < 0.2 ∴ use H1-1b

Pr 2 Pc 81 2 ( 436 )

+

125 321

+

M r M c

≤ 1.0

H1-1b

= 0.093 + 0.389 = 0.482 ≤ 1.0


7.59

Example 2 • Consider Case 3

Effective length factors

P 1 = 33 kips P 2 = 37.5 kips

K top = 117 ft-kips

π2 ( 29,000 )(1110 ) 2  33  ( 45 (12 ) ) (1166 )    70.5 

= 1.41

P total = 70.5 kips K bot =

P cr = 1166 kips

π2 ( 29,000 )(1110 )

 ( 45 (12 ) ) (1166 )  2

70.5 

  70.5 

= 0.967 110 ft-kips



7.60




Example 2 • Consider Case 3

We could have obtained F e directly from P cr

P 1 = 33 kips

 33    70.5  = 24.1 ksi

1166 

P 2 = 37.5 kips

F e upper =

117 ft-kips

P total = 70.5 kips

22.6

 70.5    70.5  = 51.6 ksi

1166  F e lower =

P cr = 1166 kips

22.6

110 ft-kips



7.61

Example 2 • Determine the nominal compressive strength of the upper segment KL r x KL r y

=

=

1.41( 45)(12) 7.00

1.0 ( 8.0 )(12) 2.47

= 109 < 4.71

= 38.9 < 4.71

29,000 50 29,000 50

= 113

= 113

For compression, the x -axis controls There’s always a solution in steel!

7.62




Example 2 • Determine the nominal compressive strength of the upper segment F e =

π2 E

 KL   r   

F  Fcr =  0.658 F   y e

2

=

π2 ( 29,000 )

(109 )

2

= 24.1 ksi

Note that this is the same F e = 24.1 ksi that we had just calculated, without using K

50    24.1  F y =  0.658  ( 50) = 21.0 ksi    

Pn = 21.0 ( 22.6 ) = 475 kips

P n Ω = 475 1.67 = 284 kips


7.63

Example 2 • For bending about the x -axis, Lb = 8.0 ft Lb = 8.0 < L p = 8.72 ft

• Interaction

P r P c

=

33 284

2 Pc 2 ( 284 )

+

117 374

M n

Ω

=

M p

Ω

= 374 ft-kips

= 0.116 < 0.2 ∴ use H1-1b

Pr 33

∴

+

M r M c

≤ 1.0

= 0.058 + 0.313 = 0.371 ≤ 1.0


7.64




Elastic Buckling Analysis • In order to consider stepped columns, we must investigate the influence of reducing the moment of inertia along a portion of the column length • We will consider the same pin-pin column that was the basis for our consideration of load applied at various points along the length There’s always a solution in steel!

7.65

Elastic Buckling Analysis • The W16x77 used earlier will be the base of this column. The upper portion will have ½ the moment of inertia P

P

x

I top = 555 in.4

45 ft I bottom = 1110 in.4

P

P


7.66




Elastic Buckling Analysis • Note that when the column is a full height with I = 1110 in.4 the elastic buckling load is the same as it was for our previous study of load placement P cr = 1090 kips • When I is cut in half for the full length, the elastic buckling load is also cut in half P cr = 545 kips There’s always a solution in steel!

7.67

Elastic Buckling Analysis • If we then determine the effective length factor it too will vary as we alter the moment of inertia For x = 21 ft, P cr = 738 kips K =

=

π EI L

P cr

π

29,000 (1110 )

45(12)

738

= 1.22

Note that for this plot we have used, I = 1110 in.4 in our determination of K There’s always a solution in steel!

7.68




Elastic Buckling Analysis • When it comes to design, we will want to use the properties of the individual segments • We must also account for the magnitude of force in each segment • So consider the stepped column with loads applied at the top and the step


7.69

Elastic Buckling Analysis • Define

γ=

Pcr

( P1 + P2 )

=

P cr

P total

π2 EI T K = 2 L γ ( P 1) 2 T

so that

P 1

π2 EI B K = 2 L γ ( P1 + P 2 ) 2 B

I T

P 2

Take the ratio π2 EI T L2 γ ( P1 )

L

I B

2

K T

2

KB

=

2

π EI B 2 L γ ( P1 + P 2 )


I T

=

( P 1 ) I B

( P1 + P 2 )

=

I T  P1 + P 2 



IB 

P 1

 

7.70




Elastic Buckling Analysis • Solve for K T in terms of K B

 P 2  1 +  P K T I T  P1 + P 2  1   =  = KB I B  P1  I B I T

P 1

P 2

L

1.0 2.0 3.0 4.0 5.44 I B I T


7.71

Example 3 • Determine if the stepped column given will be adequate to support the loading shown P 1

Upper segment W12x35c, L = 13 ft

t f

5 t f

8

A = 10.3 in.2 P 2

I x = 285 in.4 5.25 in. r x =

H

r y = 1.54 in. t f

Lower segment W24x62c, L = 32 ft

2 3


A = 18.2 in.2

S x = 131 in.3

I x = 1550 in.4

r ts = 1.75 in.

9.23 in. r x =

J = 1.71 in.4

r y = 1.38 in.

ho = 23.1 in. c = 1.0

7.72





Case 2

t f

53.3 ft-kips

5

P 1 = 31 kips P 2 = 50 kips

t f

P 2

8

60.4 ft-kips

H

P total = 81 kips

t f

2 3

127 ft-kips


7.73

Example 3 • As for Example 2, we will assume a fixed base and a rotationally rigid top with permitted sway P

P1 = 0.38P t f

5 t f

8

P2 = 0.62P

K top =

π2 ( 29,000)( 285)

( 45 (12 ) ) ( 916) 

31    81 

2

H

= 0.896 4

I x = 1550 in.

(full length)

L

t f

P cr = 1520 kips

2 3

K = 1

K bot =

P cr = 916 kips

π2 ( 29,000)(1550 )

 ( 45 (12 )) ( 916 )  2

81    81 

= 1.29 P


7.74




Example 3 • Determine the nominal compressive strength of the upper segment KL r x

=

KL r y

0.896 ( 45)(12)

=

5.25 1.0 ( 8)(12) 1.54

= 92.2 < 4.71

= 62.3 < 4.71

29,000 50

29,000 50

= 113

= 113

For compression, the x -axis controls There’s always a solution in steel!

7.75

Example 3 • Determine the nominal compressive strength of the upper segment, assume Q=1 π ( 29,000 ) π E 2

2

F e =

 KL   r   

F  Fcr =  0.658 F   y e

2

=

( 92.2 )

2

= 33.7 ksi

50     F y =  0.65833.7  ( 50) = 26.9 ksi    


7.76




Example 3 • Check column element slenderness, Section E7.2(a) using F cr for Q = 1 h tw

= 36.2 < 1.49

E F cr

= 1.49

29,000 26.9

= 48.9

• So the element slenderness does not impact column strength Pn = 26.9 (10.3) = 277 kips

P n Ω = 277 1.67 = 166 kips


7.77

Example 3 • Determine the flexural strength for bending about the x -axis of the upper segment • Unbraced length is Lb = 8 ft • W12x35 has compact flange and web • From Manual Table 3-2 L p = 5.44 ft M p

Ω


= 128 ft-kips

M r

Ω


= 79.6 ft-kips 7.78




Example 3 • Available moment strength  BF  −  ( Lb − Lp ) Ω Ω  Ω  = 128 − 4.34 ( 8.0 − 5.44) = 117 ft-kips

M n

=

M p

P r

• Interaction

P c

=

31 166

= 0.187 < 0.2 ∴ use H1-1b

Pr 2 Pc 31 2 (166 )

+

53.3 117

+

M r M c

≤ 1.0

H1-1b

= 0.093 + 0.456 = 0.549 ≤ 1.0


7.79

Example 3 • Determine the nominal compressive strength of the lower segment KL r x KL r y

=

=

1.29 ( 45)(12) 9.23 1.0 (16)(12) 1.38

= 75.5 < 4.71

= 139 > 4.71

29,000 50

29,000 50

= 113

= 113

For compression, the y -axis controls There’s always a solution in steel!

7.80




Example 3 • Determine the nominal compressive strength of the lower segment, assume Q=1 π ( 29,000 ) π E 2

2

F e =

 KL   r   

2

=

(139 )

2

= 14.8 ksi

Fcr = 0.877F e = 0.877 (14.8) = 13.0 ksi


7.81

Example 3 • Check column element slenderness, Section E7.2(a) using F cr for Q = 1 h tw

= 50.1 < 1.49

E F cr

= 1.49

29,000 13.0

= 70.4

• So the element slenderness does not impact column strength Pn = 13.0 (18.2 ) = 237 kips

P n Ω = 237 1.67 = 142 kips


7.82




Example 3 • Determine the flexural strength for bending about the x -axis of the lower segment • Unbraced length is Lb = 16 ft • W24x62 has compact flange and web • From Manual Table 3-2 L p = 4.87 ft M p

Ω


= 382 ft-kips

M r

Ω

= 229 ft-kips


7.83

Example 3 • Since Lb > Lr , determine flexural strength from Eqs. F2-3 and F2-4 M n = Fcr S x =

=

Cb π2 ES x

 Lb     r ts 

2

1 + 0.078

1.0π2 ( 29, 000 )(131)

 16 (12 )     1.75 

2

Jc  Lb 



2



S x ho  r ts 

(1.71)(1.0 )  16 (12 )  1 + 0.078   131( 23.1)  1.75 

= 3850 in.-kips  321 ft-kips There’s always a solution in steel!

M n

Ω

=

321 1.67

2

= 192 ft-kips 7.84




Example 3 • Interaction P r P c

=

81 142

= 0.570 > 0.2 ∴ use H1-1a

Pr Pc

+

8 M r 9 M c

≤ 1.0

H1-1a

8  127  +   = 0.570 + 0.588 = 1.16 > 1.0 142 9  192  81


7.85

Example 3 • The result is that the upper segment will work fine for this loading condition • But, the lower segment will not be adequate for this loading condition • Increasing the size of the lower segment will – Change the second-order analysis results – Change the elastic buckling load There’s always a solution in steel!

7.86




Elastic Buckling Analysis • Now consider how we might treat a tapered member. • We will only consider members loaded at their ends (no intermediate loads) • Remembering that the ultimate goal is to determine the elastic buckling load, based on Design Guide 25 we will determine a modified moment of inertia and hold K = 1 There’s always a solution in steel!

7.87

Elastic Buckling Analysis • If we are determining an effective length factor K =

π

EI

L

P e

• If instead we determine an effective moment of inertia, using K = 1.0 2

I =

( KL ) P e π

2

E


=

L2 P e 2 π

E 7.88




Elastic Buckling Analysis • Define the tapered member We will base our discussion of the tapered column on the properties at the top and the bottom. These will generally be referred to as the large end and small end although it really does not matter which end is up.

L

x

We will then define properties at another point, x from the small end.


7.89

Elastic Buckling Analysis • If our software permits us to model the taper, we can do an elastic buckling analysis of this member • Consider these properties

L

x

b f = 8.00 in.

hlarge = 36.0 in.

t f = 0.500 in.

L = 360 in.

t w = 0.188 in.

I small = 776 in.4

hsmall = 18.0 in.


I large = 3400 in.4

7.90




Elastic Buckling Analysis • An elastic buckling analysis yields P e = 3730 kips

• By our previous approach we might want to determine the effective length factors so that K small =

=

π2 EI small

π2 EI large

K large =

L2 P e

π2 ( 29,000)( 776 )

L2 P e

π2 ( 29,000 )( 3400 )

=

2

( 360) ( 3730)

= 0.678

2

( 360) ( 3730 )

= 1.42


7.91

Elastic Buckling Analysis • However, there is another approach that will prove more useful in the end • If we set K = 1.0 we can then determine an equivalent moment of inertia that will give us the same elastic buckling load From K =

π2 EI L2 P e

we can determine 2

2

(1.0 ) (360 ) (3730 ) = = 1690 in.4 I ′ = 2 2 π E π ( 29,000 ) K 2 L2 P e


7.92




Elastic Buckling Analysis • With a bit of calculating of section properties, we can determine that this moment of inertia will exist on this specific column when 2 3  d   0.188d ′ I = 2  4  + 0.25   + = 1690 in.4  2 12   

d = 26.1 in. x = 162 in. = 0.45L There’s always a solution in steel!

7.93

Elastic Buckling Analysis • A study presented in DG 25 shows that the distance x can be determined, as a function of the moment of inertia at each end, which leads to prediction of the elastic buckling load with very good accuracy  I  x = 0.5L  small   I large    There’s always a solution in steel!

0.0732

7.94




Elastic Buckling Analysis • As we have seen in our previous examples, once we know the elastic buckling load, we can determine the elastic buckling stress, F e, and then proceed to determine the nominal compressive strength.


7.95

Example 4 • Determine the nominal compressive strength for this column first without considering the influence of slender elements

L x

b f = 8.00 in.

Asmall = 11.4 in.2

t f = 0.500 in.

Alarge = 14.8 in.2

t w = 0.188 in.

I small = 776 in.4

hsmall = 18.0 in.

I large = 3400 in.4

hlarge = 36.0 in.

L = 360 in.


7.96




Example 4 • We already know that the elastic buckling load is P e = 3730 kips • So, at the top, with Q = 1 (ignoring element slenderness) F e =

P e Alarge

=

3730

F y

= 252 ksi

14.8

F  Fcr =  0.658 F  

F e

=

50 252

= 0.198 < 2.25 L

50     F y =  0.658252  ( 50) = 46.0 ksi    

y e

P n = 46.0 (14.8) = 681 kips There’s always a solution in steel!

7.97

Example 4 • Check element slenderness • Flange

b f 2t f h t w

= =

8.0 2 ( 0.5 ) 36 0.188

= 8.0

= 191

k c =

4 h t w

λ r = 0.64

=

4 191

kc E F y

= 0.289 < 0.35

= 0.64

0.35( 29,000) 50

= 9.12

So the flange is not slender

• Web λ r = 1.49

E F y

= 1.49

29,000 50

= 35.9 < 191


So the web is slender

7.98




Example 4 • Determine effective width be = 1.92 ( 0.188 )

29,000  0.34 1 − 46.0  191

29,000  46

 = 8.66 in. 

E7-17

• Effective area Ae = 2 ( 8.0 ( 0.5 ) ) + 0.188 ( 8.66 ) = 9.63 in.

2

• Slender element reduction factor Q=

Ae A g

=

9.63 14.8

= 0.651


7.99

Example 4 • Determine nominal strength at column top QF y F e

=

0.651( 50 ) 252

QF  Fcr = Q  0.658 F  

y

e

= 0.129 < 2.25

  F y = 0.651( 0.6580.129 ) ( 50) = 30.8 ksi  

P n = 14.8 ( 30.8 ) = 456 kips


For the column top

7.100




Example 4 • So, at the bottom, with Q = 1 (ignoring element slenderness) F e =

P e Asmall

=

3730 11.4

= 327 ksi

F  Fcr =  0.658 F   y e

F y F e

=

50 327

= 0.153 < 2.25

50     F y =  0.658327  ( 50) = 46.9 ksi    

L

P n = 46.9 (11.4 ) = 535 kips


7.101

Example 4 • Check element slenderness • Flange

b f 2t f h t w

= =

8.0 2 ( 0.5 ) 18 0.188

= 8.0

k c =

4 h t w

=

4 95.7

= 0.409

= 95.7 λ r = 0.64 kc E = 0.64 0.409( 29,000) = 9.86 50

F y

So the flange is not slender

• Web λ r = 1.49

E F y

= 1.49

29,000 50

= 35.9 < 95.7


So the web is slender

7.102




Example 4 • Determine effective width be = 1.92 ( 0.188 )

29,000  0.34 1 − 46.9  95.7

29,000 

 = 8.18 in.

E7-17

46.9 

• Effective area Ae = 2 ( 8.0 ( 0.5 ) ) + 0.188 ( 8.18) = 9.54 in.

2

• Slender element reduction factor Q=

Ae A g

=

9.54 11.4

= 0.837


7.103

Example 4 • Determine nominal strength at bottom QF y F e

=

0.837( 50 ) 327

QF  Fcr = Q  0.658 F  

y

e

= 0.128 < 2.25

  F y = 0.837 ( 0.6580.128 ) ( 50) = 39.7 ksi  

P n = 11.4 ( 39.7 ) = 453 kips


For the column bottom

7.104




Example 4 • Example summary Column Top

Column Bottom

Q

P n kips

Q

P n kips

1.0

681

1.0

535

0.651

456

0.837

453

The smaller section has the lower strength, but not by as much when we include element slenderness. There may be other locations along the column length that might control. Based on what we have calculated, we don’t know. There’s always a solution in steel!

7.105

Example 4 • If we were to take this example to its proper conclusion and determine the final nominal compressive strength we would need to address – The flanges at the top are slender – The flanges at the bottom are not slender – The web at the top is slender – The web at the bottom is less slender There’s always a solution in steel!

7.106




Summary • We looked at the effect of placing half the load at the top and half along the member – Any other division of the applied load will give different elastic buckling loads

• We have illustrated the impact of a 2:1 ratio for moment of inertia – Any other ratio will result in different elastic buckling loads. There’s always a solution in steel!

7.107

Summary • We have worked with pin end columns in our parameter studies – Other boundary conditions will give different results

• Combining all these variables we see that there are an infinite number of possibilities • Then we looked at a tapered member and see yet another approach There’s always a solution in steel!

7.108




Conclusion • Recognizing the complexities associated with using the effective length factor should raise the question; – Isn't there a better way?

• There is, it is the Direct Analysis Method given in Chapter C of the Specification. – With it, you always use K = 1 and you let the analysis take care of things for you (See Design Guide 28) There’s always a solution in steel!

7.109

References •

Agrawal, K. M., and Stafiej, H. A., “Calculation of Effective Lengths of Stepped Columns,” Engineering Journal , Vol. 17, No.4, 1980, AISC, Chicago

•

Anderson, J. P., and Woodward, J. H., “Calculation of Effective Lengths and Effective Slenderness Ratios of Stepped Columns,” Engineering Journal , Vol. 9, No.4, 1972, AISC, Chicago

•

Association of Iron and Steel Engineers, Technical Report No. 13, Guide for the Design and Construction of Mill Buildings, AISE, Pittsburgh, PA 2003

•

Vasquez, J., and Riddell, R., “A Simple Stepped-Column Buckling Model and Computer Algorithm,” Engineering Journal , Vol. 48, No.1, 2011, AISC, Chicago


7.110




Lesson 8 • Our next and final lesson will address column base plates • Base plates are required to distribute the column load to the concrete foundation • We will look at base plates that transfer compression only • Those that transfer tension • And those that also must transfer moment There’s always a solution in steel!

7.111

Thank You

American Institute of Steel Construction One East Wacker Drive Chicago, IL 60601 There’s always a solution in steel!

7.112




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Night School 10 Session 7

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