STRC13 Bridges Steel Members 0715

Structural Engineering Review Course

Bridges

Bridges: Steel Members Structural Engineering Review Course

STRC ©2015 ProfessionalPublications,Inc.

©2015 Professional Publications, Inc.

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Bridges: Steel Members

Lesson Overview •

Design of Steel Beams for Flexure

•

Design of Steel Beams for Shear

•

Design of Shear Studs

•

Design of Stiffeners

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Learning Objectives You will learn •

•

•

•

how to determine the plastic bending strength of composite beams

•

how to determine transformed section properties of composite beams when to use the different values for ultimate and service level decisions

•

how to calculate shear strength of cross‐section and shear connectors •

how to determine flexural strength of plate girders for non‐composite condition according to •

lateral‐torsional buckling

•

flange local buckling

how to determine the shear strength of plate girders, depending upon web stability how to design and space stiffeners

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Prerequisite Knowledge and Skills You should already be familiar with •

•

•

elementary mechanics of geometric cross‐sections •

moment of inertia

•

section modulus

•

radius of gyration

•

•

plastic bending strength of cross‐ sections causes of lateral‐torsional buckling and local buckling of steel beams

AASHTO design moments and shears AISC cross‐section properties of rolled steel shapes

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Referenced Codes and Standards AASHTO LRFD Bridge Design Specifications (AASHTO, 2012)

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Composite Beam Construction steel beam usually rolled steel, wide flange section

Since concrete slab only contributes in compression, composite construction is •

concrete slab has longitudinal reinforcing steel

•

very efficient for positive bending (bottom of beam in tension) not efficient for negative bending (top of beam in tension)

shear connectors required to ensure that steel and concrete will act as a composite cross‐ section

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AASHTO Design Procedure AASHTO App. C6 contains flow charts explaining the process for design of steel members. •

print, laminate, and put in a 3‐ring binder

•

useful for understanding how AASHTO code is organized

•

reference while flipping through AASHTO code (saves time on exam)

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Flexure Design Procedure All of the following must be checked. •

cross‐section proportion limits (AASHTO Sec. 6.10.2)

•

constructability (AASHTO Sec. 6.10.3)

•

service limit state (AASHTO Sec. 6.10.4)

•

fatigue and fracture limit state (AASHTO Sec. 6.10.5)

•

strength limit state (AASHTO Sec. 6.10.6)

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Flexure Design Procedure strength limit state •

resistance factor   1.0

•

design composite with effective flange width = tributary width of beam

•

design procedure •

•

composite sections in positive flexure per AASHTO 6.10.6.2.2 composite sections in negative flexure and noncomposite sections (for constructability checks) per AASHTO 6.10.6.2.3

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AASHTO Flexure Design Procedure composite sections in positive flexure composite section is compact if all of the following are true •

flange fy ≤ 70 ksi AASHTO Eq. 6.10.2.1.1‐1

•

no longitudinal stiffener, and

•

satisfies web slenderness limit,

AASHTO Eq. 6.10.6.2.2‐1

otherwise, section is noncompact



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AASHTO Flexure Design Procedure compact composite sections in positive flexure •

design per AASHTO 6.10.7.1 & 6.10.7.3

•

at strength limit state,

•

•

AASHTO Eq. 6.10.7.1.1‐1

is typically = 0 therefore,



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AASHTO Flexure Design Procedure compact composite sections in positive flexure ductility requirement AASHTO Eq. 6.10.7.3‐1

Dp = distance from top of concrete deck to neutral axis of composite section at plastic moment

Dt = total depth of composite section



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AASHTO Flexure Design Procedure compact composite sections in positive flexure nominal flexural resistance •

Otherwise, AASHTO Eq. 6.10.7.1.2‐2

•

= distance from top of concrete deck to neutral axis of composite section at plastic moment

Dt

= total depth of composite section

Mn

= nominal flexural resistance

Mp

= plastic moment of composite section

If Dp ≤ 0.1Dt, then Mn = Mp. AASHTO Eq. 6.10.7.1.2‐1

•

Dp

In a continuous span, Mn ≤ 1.3RhMy AASHTO Eq. 6.10.7.1.2‐3

My = yield moment Rh



= hybrid factor

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Plastic Design of Composite Beams •

•

•

•

bending strength of compact composite beams in positive flexure

Figure 8.9 Determination of Mp

uses plastic bending strength of cross‐section assume all material is at ultimate strength •

concrete in compression at 0.85f c’

•

steel in tension at Fy

•

steel in compression at Fy

concrete in tension assumed to have no strength STRC ©2015 ProfessionalPublications,Inc.


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Plastic Moment, M Assuming the plastic neutral axis (PNA) is within the concrete slab, determine the location/depth of the PNA.


The PNA is located where the total steel force is balanced by the portion of the concrete acting in compression.



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Plastic Moment, M •

•

•

Calculate force developed by each component of cross‐section (slab, flange, web, etc.).


Component force is located at centroid of component. Determine distance between component forces and the plastic neutral axis.



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Plastic Moment, M The internal moment will be Figure 8.9 Determination of Mp

Plastic mechanics do not require the cross‐sectional properties used for linear elastic mechanics.



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Poll Question: Plastic Bending Strength Which of the following items is related to determining the plastic bending strength of a cross‐section? (A) centroid of cross‐section (B) modular ratio (C) transformed cross‐section (D) steel strength



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Poll Question: Plastic Bending Strength Which of the following items is related to determining the plastic bending strength of a cross‐section? (A) centroid of cross‐section (B) modular ratio (C) transformed cross‐section (D) steel strength The answer is (D).



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Example: Plastic Strength Example 8.26

Assume the member is compact.



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Example: Plastic Strength



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When PNA is Located in Steel •

This is an uncommon condition, but sometimes the steel cannot be fully balanced by the concrete in compression.

•

In this condition, the top portions of the steel will be in compression.

•

It usually requires an iterative process to locate the PNA.

•

•

•

If there is no better option, assume the location of the PNA.

•

A tabulated solution helps to organize the calculation.

The correct PNA will result in zero‐sum net axial force along the beam. Once the location of the PNA is known, the internal moment is calculated by summing each of the components force by the lever arm.



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Example: Added Plate at Bottom Flange Using an iterative approach, determine the narrowest width of a 1 in thick plate to be added to the bottom of the beam from Ex. 8.26 that will increase the section’s plastic moment by at least 30%. Assume the plate width can be ordered in increments of 2 in. The W36 × 194 girder has the following properties.

The plastic moment capacity is 5313 ft‐kips.



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Example: Added Plate at Bottom Flange If the beam needs to be 30% stronger, the new plastic bending moment will be M p , new  1.30 M p , old

 in   1.30  5313 ft-kips  12  ft    82,884 in-kips



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Example: Added Plate at Bottom Flange For an iterative approach, assume that the neutral axis is in the top flange of the beam. Assume that the added plate’s width will be 8.0 in. Since the neutral axis will likely be in the top flange, allow for two lines of calculation for the top flange. The lever arm in the table is the distance from the centroid of the item to the bottom of the cross‐section.



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Example: Added Plate at Bottom Flange Solve for depth in tension and compression. Solve for the force in all members other than the top flange.

The top flange compression is CTF Fbdy Fb

y

x   1.26

F bxy The top flange tension is TTF  F bd y

b = width; d = depth



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Example: Added Plate at Bottom Flange The concrete force is Ps  bs ts 0.85  n 0.85  f c  96 in 8i    4.5 kips  2  in  

2941kips

 kips   33.98  in 50 The web force is Pw  t w DFy   0.765in  1300 kips in 2   kips   The bottom flange force is Pt  bt f Fy  12.12  in 1.26  in 50 2  764 kips in   kips   The plate force is Pp  bpt p Fy   8in 1 in  50 2  400 kips in  

The sum of the web force, bottom flange force, and plate force is     T

1300 kips

764 kips

400 kips

2464 kips STRC ©2015 ProfessionalPublications,Inc.


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Example: Added Plate at Bottom Flange Set the total compression equal to the total tension and solve for x. Ps  Fyb 1.26   x 

T

x

Fybx

Ps  T  1.26 Fyb

2 2941 kips  2464 kips  1.26 kips   50 12.12 in    2  in    2  1.02 in

Use a table to calculate the values.



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Example: Added Plate at Bottom Flange width

depth

area

stress

(in)

(in2)

(kips/in2)

(kips)

(in)

(in‐kips)

‐3.83

−2 9 4 1

4 1 .5

−122,052

‐50

‐1 4 6

37.38

−5457

(in) concrete steel

96 ‐

8 ‐

768

‐

‐

‐

‐

force

lever

moment

‐

top flange (Comp.)

12.12

0.24

2 .9 1

top flange (Tens.)

12.12

1.02

12.36

web

0.765

33.98

25.99

50

13 00

19.25

25,025

bottom flange

1 2 .1 2

1 .2 6

15.27

50

7 64

1 .6 3

1245

plate total

8

1

8

50

50

6 18

40 0

4 5 .5



36.75

0.5 −5

22,712

200 −78,327

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Example: Added Plate at Bottom Flange This provides a moment of 78,327 in‐kips, which does not exceed the required increase in plastic bending strength. Increase the width of the plate to 10 in and update the table.



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Example: Added Plate at Bottom Flange width

depth

area

stress

(in)

(in2)

(kips/in2)

(kips)

(in)

moment (in‐kips)

−3.83

‐2 9 3 8

4 1 .5

−121,927

−194

37.34

−7244

(in) concrete steel

96 ‐

8 ‐

768

‐

‐

‐

‐

force

‐

compression of flange

1 2 .1 2

0 .3 2

3 .8 8

tension of flange

1 2 .1 2

0 .9 4

1 1 .3 9

web

0.765

33.98

26.0

50

13 00

bottom flange

1 2 .1 2

1 .2 6

15.3

50

763.2

plate

10

total

1

10

‐50

50

50

5 6 9 .5

5 00

45.5



lever

0.5 1 .7

3 6 .7 1

20,906

19.25

25,025

1 .6 3

1 244

25 0 −81,746

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Example: Added Plate at Bottom Flange This wider plate provides a moment of 81,746 in‐kips, which does not exceed the required increase in plastic bending strength. Since the value so close, stop iterating and assume that 12 in is acceptable. Use 1 in by 12 in grade 50 steel plate. Note that by adding 12 in2 of steel to the bottom, we made a significant increase in the bending strength. Adding bottom plates to composite beams is a very efficient way to increase the strength.



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AASHTO Flexural Design Procedure all other situations Use elastic mechanics to determine bending strength. •

top and bottom flange checked separately in terms of stresses (not moment)

•

stresses calculated as M/S (S varies for different loads) •

for beam dead load and weight of deck, S = S of beam

•

for superimposed dead loads, S = long term composite property

•

for live loads, S = short term composite property



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Elastic Cross‐Sections Convert concrete slab to an “equivalent” steel area. Ignore concrete that will be in tension.

Figure 8.1 Composite Section Properties



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Modular Ratio •

•

•

Converting concrete to “equivalent” steel requires the area of the concrete to be reduced by the ratio of the modulus of elasticity of the two materials.

Figure 8.1 Composite Section Properties

This ratio is the modular ratio, n. Concrete modulus is a function of concrete compressive strength.

•

Use n for short term (live load)

•

Use 3n for long term (dead load)



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Transformed Section •

assume neutral axis will be in steel – uses the full amount of concrete

•

locate centroid of cross‐section •

if located in steel, then all concrete in compression

•

if located in concrete, iterative process usually needed (Assume the neutral axis is slightly lower than what was calculated when the assumption was that full slab would be in compression. Correct location given when the only area of concrete considered is above calculated neutral axis.)

•

•

Calculate the moment of inertia, I, about the centroidal axis. Calculate Stc, Sts and Sbs, the section moduli for top of concrete, top of steel, and bottom of steel. STRC ©2015 ProfessionalPublications,Inc.


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AASHTO Flexural Design Procedure noncompact composite sections in positive flexure •

•

design per AASHTO 6.10.7.2 & 6.10.7.3 at strength limit state, compression flange state satisfies AASHTO Eq. 6.10.7.2.1‐1

ϕf = resistance factor for flexure

fbu = flange stress calculated without consideration of flange lateral bending Fnc = nominal flexural resistance of compression flange



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AASHTO Flexural Design Procedure noncompact composite sections in positive flexure Tension flange satisfies AASHTO Eq. 6.10.7.2.1‐2

ft = flange lateral bending stress Fnt = nominal flexural resistance of tension flange typically,

= 0, so

\ STRC ©2015 ProfessionalPublications,Inc.


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AASHTO Flexural Design Procedure noncompact composite sections in positive flexure nominal flexural resistance of compression flange taken as AASHTO Eq. 6.10.7.2.2‐1

Rb = web load‐shedding factor Rh = hybrid factor nominal flexural resistance of tension flange taken as AASHTO Eq. 6.10.7.2.2‐2



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AASHTO Flexural Design Procedure noncompact composite sections in positive flexure •

•

maximum longitud inal compressive stress in the concrete deck at the strength limit may not exceed 0.6fc’ ductility requirement same as for compact composite sections in positive flexure



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AASHTO Flexure Design Procedure composite sections in negative flexure and noncomposite sections •

•

designed per AASHTO 6.10.8

•

compression flange: choose the smaller of •

•

stress to cause local buckling of flange

•

yielding of tension flange,

Fnt  Rh F yt

tension flange governs only for non‐ symmetric plate girders (where the tension flange significantly smaller than the compression flange)

stress to cause lateral‐torsional buckling of beam cross‐section



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AASHTO Flexure Design Procedure composite sections in negative flexure and noncomposite sections hybrid plate girders •

use different materials for the flange and the web

•

not common in modern design and construction

Rh = hybrid factor Rh = 1 if the plate girder is not a hybrid design



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•

local buckling of compression flange slenderness of compression flange  f 

b fc 2t fc

AASHTO Eq. 6.10.8.2.2‐3

•

slenderness limit for a compact compression flange  pf  0.38

•

slenderness limit for a noncompact flange  pf  0.56



E Fyc

E Fyc

AASHTO Eq. 6.10.8.2.2‐4

AASHTO Eq. 6.10.8.2.2‐5

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•

The flange does not have a local buckling problem if  f If the flange does not have a local buckling problem

Fnc  Rb R F h yc •

AASHTO Eq. 6.10.8.2.2‐1

If the flange does have a local buckling problem

  

  

Fnc   1   1  •

  pf

 F yr

 Rh F yc 

f

  ych b  R R F rf pf   pf 

AASHTO Eq. 6.10.8.2.2‐2

Rb is the web load‐shedding factor; usually 1.0 (see AASHTO Sec. 6.10.1.10.2)



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Lateral‐Torsional Buckling limits on unbraced length, Lb L p  1.0rt

E Fyc

AASHTO Eq. 6.10.8.2.3‐4

Lr   rt

E Fyr

AASHTO Eq. 6.10.8.2.3‐5

effective radius of gyration for lateral torsional buckling AASHTO Eq. 6.10.8.2.3‐9



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Lateral‐Torsional Buckling •

•

When Lp  Lb  Lr, inelastic buckling occurs and

  

  

Fnc b C 1  1 

•

•

AASHTO Eq. 6.10.8.2.3‐1

F When Lb  Lp, no buckling occurs and Fnc  Rb R h yc

F 

y r

  Rh F yc 

  ychb  R R F r L p L  L  L

b p

AASHTO Eq. 6.10.8.2.3‐2

When Lb  Lr, elastic buckling occurs and Fcr 

Cb Rb 2 E

 Lb     rt 

2

AASHTO Eq. 6.10.8.2.3‐8

F ≤F cr

y STRC ©2015 ProfessionalPublications,Inc.


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Example: Flexural Strength of Plate Girder Determine the available flexural capacity for the welded plate girder of grade A36 steel shown. Lateral support of the compression flange is provided at Lb = 10 ft centers and Cb = 1.0.



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Example: Flexural Strength of Plate Girder Determine the available flexural capacity for the welded plate girder of grade A36 steel shown. Lateral support of the compression flange is provided at Lb = 10 ft centers and Cb = 1.0.

Since the cross‐section is symmetrical, the tension flange will not govern. Since the flanges and web are all A36, it is not a hybrid beam, so Rh = 1.0. The web load‐shedding factor, Rb, is also 1.0. Check the strength of the plate girder based on the compression flange. Check the compression flange for local buckling.

f 

b fc 2t fc





18 in

 9.0

 2 1 in 50

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Bridges


Example: Flexural Strength of Plate Girder Compare to  pf  0.38

Determine the strength based on lateral‐ 29,000

E  0.38 Fyc

36

torsional buckling.

kips

in kips

2

rt 

in 2

 10.8 Since f <  pf , the flange will not locally buckle and

FncFLB   Fyc  36

kips in

2



b fc

  1  D t  12  1     c w     3   b fct fc      18 in

 in    1  26 in 0.3125 12   1           3   18in 1in

 4.84 in



AASHTO Eq. 6.10.8.2.3‐9

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Example: Flexural Strength of Plate Girder Use rt to find the lengths that define the various types of failure. 29,000

in 2 kips

1.0  (4.84 in) L p  1.0rt

E Fyc

36

 12

36 12

in

ft 11.4 

ft

 (4.84 in)

E  Fyc

in 2

in

29,000

Lr   rt

kips

kips

in kips in

2

2

36.0 ft

ft STRC ©2015 ProfessionalPublications,Inc.


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Example: Flexural Strength of Plate Girder Since the unbraced length, Lb, is 10 ft and is less than Lp, lateral‐torsional buckling will not occur.

FncFLB   Fyc  36

kips in 2

Since neither type of buckling will occur, the strength is based on the elastic mechanics.

Ix  Sx 

bd 3 h 3b t  

 18in  54 in   

3

52 in

12

3

18  in

0.312 in

12

I x 28,940 in 4   1072 in 3 c 27 in

1072 in  36 3

So the bending strength is M x  S x Fnc 



in 12 ft

kips 



in 2   3216 ft-kips



 28,940 in 4

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Bridges


Example: Effect of Unbraced Length For a non‐composite plate girder of grade 50 steel, determine the value of Fnc(LTB) for the following three different unbraced lengths for a plate girder cross‐section. The top and bottom flanges are 1 in by 12 in plate. The web is ⁄ in by 96 in plate. Assume Cb, Rb, and Rh are 1.0. 1) 60 in 2) 120 in 3) 240 in



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Example: Effect of Unbraced Length For a non‐composite plate girder of grade 50 steel, determine the value of Fnc(LTB) for the following three different unbraced lengths for a plate girder cross‐section. The top and bottom flanges are 1 in by 12 in plate. The web is ⁄ in by 96 in plate. Assume Cb, Rb, and Rh are 1.0. 1) 60 in 2) 120 in 3) 240 in

(1) Solve. I oy 

t f bf 3 12

tw hw 6



 1in  12 in    17.0 in rt 

I oy Af

 144 in 4

12

Af  t f b f 





3

1.0 12 

0.625  in

 48 in

6

2



144 in 4 17.0 in

2

 2.91 in

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Example: Effect of Unbraced Length Calculate limits for slenderness. Lp  1.1rt

(2) Lp will be the same value as before. I oy  144 in 4

E Fy

A f  17.0 in 2 29,000

 1.1  2.91in 50

rt  2.91 in

kips

in kips

2

Calculate limits for slenderness

in 2

 77.1 in

Lr   rt

Since Lb is given as 60 in, Lb  Lp. The value of Fnc(LTB) is Fy = 50 ksi.

29,000 50

kips

in kips in

2

2

 220.2 in



E    2.91 in  Fy

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Example: Effect of Unbraced Length Lb  Lr but the unbraced length of 120 in exceeds Lp, so the value of Fnc(LTB) must be calculated as   

  

Fnc b C  1   1 

Lb L p 

Fyr

  Rh F yc 



r

R R F  

L p L

      0.7  50   1.0  1  1       1.0   50    45.0 kips/in 2

yc hb

  120in 69.8in  kips in        50      1.0 1.0 kips in    220.2in 69.8 in    2   in    kips   2



2

57

57

– Steel


Bridges


Example: Effect of Unbraced Length (3) Lr will be the same value as before. Since Lb  Lr the value of Fcr must be calculated using Fcr 

Cb Rb 2 E

 Lb  r   t 

2

1.0 1.0  

 

 2  29,000

 240 in     2.91 

kips  2



in 

2

 42.1 kips/in 2



58

58

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Bridges


Shear Connectors •

•

•

•

to create a composite action between the steel and concrete, there must be a shear connector installed physical fastener that can resist horizontal shear welded to the steel beam before concrete is cast

•

per AASHTO •

•

pitch of fasteners is governed by fatigue requirements total quantity of fasteners is governed by ultimate strength requirements

channel connectors oriented to scoop concrete under deflection



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59

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Bridges


Shear Connectors: Fatigue Strength •

•

Fatigue shear resistance, Zr, is based upon service level fatigue loading. For infinite cycles of life AASHTO Sec. 6.10.10.2‐1

•

When fatigue II (finite life) is needed, the fatigue shear resistance is changed to AASHTO Sec. 6.10.10.2‐2 AASHTO Sec. 6.10.10.2‐3



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Bridges


Shear Connectors: Shear Flow •

•

determine shear flow on beam from transformed cross‐section •

need statical moment,

•

need shear flow,

pitch, p, or spacing, of connectors



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Bridges


Example: Shear Connector Spacing Example 8.28



62

62

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Bridges


Example: Shear Connector Spacing Example 8.28



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63

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Bridges


Example: Shear Connector Spacing



64

64

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Bridges


Example: Shear Connector Spacing



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65

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Bridges


Ultimate Strength of Connectors •

•

Shear connectors also need to be checked that they can provide shear transfer when the beam is loaded to plastic capacity. This is not a fatigue limit state so the full strength for the connectors can be used.

•

Assume all connectors fail in unison.

•

ϕ factor according to AASHTO is

•

•

shear strength, Qn, of welded stud

Design for the force required to fail either the steel or the concrete using the lesser of



66

66

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Bridges


Poll Question: Connector Layout Which of the patterns shows the correct layout of channel connectors for a composite beam with a moment diagram as shown? (A) pattern A (B) pattern B (C) pattern C (D) pattern D



67

67

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Bridges


Poll Question: Connector Layout Which of the patterns shows the correct layout of channel connectors for a composite beam with a moment diagram as shown? (A) pattern A (B) pattern B (C) pattern C (D) pattern D The answer is (A).



68

68

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Bridges


Example: Ultimate Shear Connector Strength Example 8.29



69

69

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Bridges


Example: Ultimate Shear Connector Strength Example 8.29



70

70

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Bridges


Example: Ultimate Shear Connector Strength



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71

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Bridges


AASHTO Shear Design Procedure •

•

ϕ = 1.0 for shear At the strength limit state, straight and curved web panels must satisfy Vu  vVn

AASHTO Eq. 6.10.9.1‐1

AASHTO Sec. C.6.10.9.1 flowchart for determining the shear resistance of I‐sections STRC ©2015 ProfessionalPublications,Inc.


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Bridges


Shear Resistance Check whether the web is stiffened or unstiffened. Where D = depth of web of the plate girder and do = distance between transverse (vertical) stiffeners, •

For sections without longitudinal stiffeners, if d0 ≤ 3D, the web is stiffened.

•

For sections with longitudinal stiffeners, if d0 ≤ 1.5D, the web is stiffened.

•

Otherwise, the web is unstiffened.



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73

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Bridges


Shear Strength (Unstiffened Webs) •

shear strength is based on steel web of cross‐section (ignore concrete slab) AASHTO Eq. 6.10.9.2‐1

•

•

web considered to be distance between inside faces of flanges, D steel yields in shear at a lower stress than it yields in tension AASHTO Eq. 6.10.9.2‐2



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Bridges



For stout webs, D  1.12 Ek tw Fyw

AASHTO Eq. 6.10.9.3.2‐4

Shear coefficient is C = 1.0 Shear buckling coefficient k = 5.0 •

For moderately slender webs, 1.12

The shear coefficient is C 

1.12

D    tw 

2

Ek D Ek   1.40 Fw wt yw F

AASHTO Eq. 6.10.9.3.2‐5

Ek Fw

Shear buckling coefficient k = 5.0 STRC ©2015 ProfessionalPublications,Inc.


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Bridges



For slender webs,

D Ek  1.40 tw Fyw

   1.57 Shear coefficient is C   2  D t   w 

    Ek    F    yw   

AASHTO Eq. 6.10.9.3.2‐6

Shear buckling coefficient k = 5.0



76

76

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Bridges


Example: Shear Strength (Unstiffened Web) Example 8.27

Assume that the web is stout.



77

77

– Steel


Bridges


Example: Shear Strength (Unstiffened Web)



78

78

– Steel


Bridges


Example: Shear Strength (Unstiffened Web)



79

79

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Bridges


Example: Shear Strength (Unstiffened Web) For the welded plate girder shown, the yield strength of the steel is 36 ksi and the web is unstiffened. Determine the available shear capacity for the welded plate girder.



80

80

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Bridges


Example: Shear Strength (Unstiffened Web) For the welded plate girder shown, the

Determine the slenderness of the web.

yield strength of the steel is 36 ksi and the web is unstiffened. Determine the available shear capacity for the welded plate girder.

D 52 in   166 tw 0.313 in For an unstiffened web, k = 5.



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81

– Steel


Bridges


Example: Shear Strength (Unstiffened Web)  29,000 kips  5  Ek in 2   The lower limit for web slenderness is 1.12  1.12  71 kips Fyw 36 in 2

kips    29,000  5  Ek in 2   1.40   89 The upper limit for web slenderness is 1.40 kips Fyw 36 in 2



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82

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Bridges


Example: Shear Strength (Unstiffened Web) Since the web is more slender than the upper limit, elastic shear buckling of the web will cause failure.      1.57   Ek  C  2   F    D    yw  t    w     1.57    29,000    166  2   36   

Vcr CVp

C 

F Dt  0.58 yw w



kips     0.229  0.58  36   52 in 0.313 in in 2    77.8 kips

AASHTO uses a reduction factor of 1.0 for shear, so the available shear capacity is 77.8 kips.   5   in   kips  2  in  kips  2

 0.229 STRC ©2015 ProfessionalPublications,Inc.


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– Steel


Bridges


Tension Field Action •

•

•

Tension field action happens when a web is able to develop a post‐buckling tension strut. Truss action combines tension strut in web, compression in the stiffener, tension in the bottom flange and compression in the top flange. tension field action •

requires the use of transverse (vertical) stiffeners for the web

•

is not allowed in end panel (a panel closest to the end of the beam)

•

can significantly increase shear strength of a plate girder



84

84

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Bridges


Poll Question: Tension Field Action Which of the panels in the plate girder CANNOT use tension field action? (A) panel A (B) panel B (C) panel C (D) all panels can use tension field action



85

85

– Steel


Bridges


Poll Question: Tension Field Action Which of the panels in the plate girder CANNOT use tension field action? (A) panel A (B) panel B (C) panel C (D) all panels can use tension field action The answer is (A).



86

86

– Steel


Bridges


Shear Strength (Stiffened Web) Interior Panels For beams where

2 Dtw

b fct

b t ft fc 

f

  0.87 1  C     2.5 , Vn  V p  C  2   do    1     D  

   0.87 1  C Otherwise, Vn  V p  C   2 do  do   1  D  D    

      

AASHTO Eq. 6.10.9.3.2‐8



AASHTO Eq. 6.10.9.3.2‐1 and Eq. 6.10.9.3.2‐2

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87

– Steel


Bridges


Shear Strength (Stiffened Web) Interior Panels AASHTO Eq. 6.10.9.3.2‐3

•

•

C equations are the same as those for unstiffened webs, except with k, the shear‐ buckling coefficient, taken as AASHTO Eq. 6.10.9.3.2‐7



88

88

– Steel


Bridges


Shear Strength (Stiffened Web) End Panels nominal resistance of web end panel,

AASHTO Eq. 10.9.3.3‐1

AASHTO Eq. 10.9.3.3‐2

C

= ratio of shear‐buckling resistance to shear yield strength

Vcr

= shear‐buckling resistance

Vp

= plastic shear force

transverse stiffener spacing for all end panels ≤ 1.5D



89

89

– Steel


Bridges


Example: Shear Strength (Stiffened Web) For the welded plate girder shown, the yield strength of the steel is 36 ksi. Transverse stiffeners are spaced 100 in apart. Determine the available shear capacity for an interior panel.



90

90

– Steel


Bridges


Example: Shear Strength (Stiffened Web) For the welded plate girder shown, the

Since the section does not have

yield strength of the steel is 36 ksi. Transverse stiffeners are spaced 100 in apart. Determine the available shear capacity for an interior panel.

longitudinal stiffeners, the web is stiffened if d0 ≤ 3D. d 0  3D 100 in  3 52 in

 156 in The web is stiffened.



91

91

– Steel


Bridges


Example: Shear Strength (Stiffened Web) Determine the available shear capacity for an interior panel. The web slenderness in the previous example was 166. Since the transverse stiffeners are 100 in apart, determine k.

k  5

5

5

 2 5 2  d o   100 in   D  52 in   

 6.352



92

92

– Steel


Bridges


Example: Shear Strength (Stiffened Web)  29,000 kips  6.532  Ek in 2   The lower limit for web slenderness is 1.12  1.12  80 kips Fyw 36 in 2

kips   29,000   6.532  Ek in 2  The upper limit for web slenderness is 1.40  1.40   100

Fyw



36

kips in 2

93

93

– Steel


Bridges


Example: Shear Strength (Stiffened Web) Since the web is more slender than the upper limit, elastic shear buckling of the web will cause failure.

  kips      29,000 6.532    2       1.57 in   1.57  Ek   0.291 C       2  2  kips Fyw   166       D  36     2     in  t    w    

V p  0.58 F Dt yw

w

kips     0.58   36  2  52in  0.313in in  

 339.3 kips



94

94

– Steel


Bridges


Example: Shear Strength (Stiffened Web) Determine which equation should be used for the tension field action. 2 Dtw

b fct

fc

b t tf

 tf

2 52in 0.313    in  0.90  2.5 18in 1 in    18  in  1 in

Use AASHTO Eq. 6.10.9.3.2‐2 for the shear strength.     0.87 1  C    Vn V p  C    339.3kips 2    do   1     D  

     0.291    

  0.87 1 0.291  195.4kips 2  100 in  1    52 in  

AASHTO uses a reduction factor of 1.0 for shear, so the available shear capacity is 195.4 kips.



95

95

– Steel


Bridges


Stiffener Design bearing stiffener stiffener installed directly above a support

transverse stiffener stiffener installed perpendicular to the main axis of the plate girder (since plate girder is usually a horizontal beam, the transverse stiffeners are usually vertical)

longitudinal stiffener stiffener installed parallel to the main axis of the plate girder (longitudinal stiffeners are not commonly used)



96

96

– Steel


Bridges


Transverse Stiffener Design transverse stiffeners •

•

may be installed on both faces of a plate girder may be installed on only a single face of a plate girder (especially for the fascia girder, where the external face of the girder is intended to have a minimalist aesthetic design)

•

increase the resistance for web buckling by shortening the panel length

•

designed per AASHTO 6.10.11.1



97

97

– Steel


Bridges


Transverse Stiffener Design transverse stiffeners (continued) The width, bt, of each projecting stiffener element must satisfy both of the following. AASHTO Eq. 6.10.11.1.2‐1

AASHTO Eq. 6.10.11.1.2‐2

bf = full width of widest compression flange within field section under consideration [I‐sections] = full width of widest top flange within field section under consideration [tub sections]

tp = thickness of projecting stiffener element

(Limit of bf/4 does not apply for closed box sections.)



98

98

– Steel


Bridges


Transverse Stiffener Design transverse stiffeners (continued)

In which

For transverse stiffeners adjacent to web panels in which neither panel supports a shear force, Vu, larger than the factored shear buckling resistance, φvVcr, the moment of inertia, It, of the transverse stiffener must satisfy the smaller of

It ≥ It1

AASHTO Eq. 6.10.11.1.3‐1

It ≥ It2

AASHTO Eq. 6.10.11.1.3‐2

AASHTO Eq. 6.10.11.1.3‐3 AASHTO Eq. 6.10.11.1.3‐4

AASHTO Eq. 6.10.11.1.3‐5 AASHTO Eq. 6.10.11.1.3‐6

AASHTO Eq. 6.10.11.1.3‐7 AASHTO Eq. 6.10.11.1.3‐8 STRC ©2015 ProfessionalPublications,Inc.


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– Steel


Bridges


Transverse Stiffener Design transverse stiffeners (continued) For panels with tension field action, if It2 > It1, AASHTO Eq. 6.10.11.1.3‐9

Otherwise,

AASHTO Eq. 6.10.11.1.3‐10

Vn = smaller of nominal combined buckling and tension‐field shear resistances of adjacent web panels



100

100

– Steel


Bridges


Transverse Stiffener Design transverse stiffeners (continued) Transverse stiffeners used in web panels with longitudinal stiffeners must also satisfy AASHTO Eq. 6.10.11.1.3‐11

bt = projecting width of transverse stiffener bℓ = projecting width of longitudinal stiffener lℓ = moment of inertia of longitudinal stiffener



101

101

– Steel


Bridges


Bearing Stiffener Design •

required to support concentrated load or reaction

•

design per AASHTO Sec. 6.10.11.2

•

width of each projecting stiffener element must satisfy

AASHTO Eq. 6.10.11.2.2‐1

Fys = specified minimum yield strength of stiffener tp = thickness of projecting stiffener element



102

102

– Steel


Bridges


Bearing Stiffener Design factored bearing resistance for fitted ends of bearing stiffeners AASHTO Eq. 6.10.11.2.3‐1

ϕb = resistance factor for bearing

Apn = area of projecting elements of stiffener outside the web‐to‐ flange fillet welds, but not beyond the edge of the flange

nominal bearing resistance for the fitted ends of bearing stiffeners



  1.4 Apn Fys

sb n

AASHTO Eq. 6.10.11.2.3‐2

Fys = specified minimum yield strength of stiffener



103

103

– Steel


Bridges


Bearing Stiffener Design factored axial resistance, Pr •

designed as a column per AASHTO 6.9.2.1

•

radius of gyration calculated about mid‐thickness of web

•

effective length = 0.75D

•

for stiffeners bolted to web, effective column section = stiffener elements only

•

•

for welded stiffeners, effective column section = stiffeners + portion of web (centrally located web strip extending no more than 9tw on either side of stiffeners/outer projecting elements) for exclusions for hybrid girders, see AASHTO Sec. 6.10.11.2.4b



104

104

– Steel


Bridges


Example: Bearing Stiffener Two pairs of bearing stiffeners are welded on each side of a 5/16 in thick web 10 in apart. Each stiffener is ½ in by 8 in. All steel is grade 50. What is the bearing resistance for the fitted ends of the bearing stiffeners?



105

105

– Steel


Bridges


Example: Bearing Stiffener Two pairs of bearing stiffeners are welded

The area of the projecting stiffener

on each side of a 5/16 in thick web 10 in apart. Each stiffener is ½ in by 8 in. All steel is grade 50. What is the bearing resistance for the fitted ends of the bearing stiffeners?

elements is Apn   4 0.5in  8in 

16in

2

The nominal bearing resistance is

 Rsb n  1.4 ApnFys 

1.4 16in

2

  50  

kips 



in 2 

 1120 kips Factored bearing resistance is

 Rsb r  b Rsb n 

1 1120 kips

 1120 kips



106

106

– Steel


Bridges


Learning Objectives You have learned •

•

•

•

how to determine the plastic bending strength of composite beams

•

how to determine transformed section properties of composite beams when to use the different values for ultimate and service level decisions

•

how to calculate shear strength of cross‐section and shear connectors •

how to determine flexural strength of plate girders for non‐composite condition according to •

lateral‐torsional buckling

•

flange local buckling

how to determine the shear strength of plate girders, depending upon web stability how to design and space stiffeners



107

107

– Steel


Bridges


Lesson Overview Design of Steel Beams for Flexure Design of Steel Beams for Shear Design of Shear Studs Design of Stiffeners



108

108

– Steel

STRC13 Bridges Steel Members 0715

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