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Example - Tension-Field Design

A36 steel plate girder - 40 ft. simple span. Assume full lateral support of the compression

flange. Section: 60" x 1/4" web and 19" x 1" flange; Af= 19 in2, Aw = 15 in

2. The loading for

the girder, as well as the shear and moment diagram, are shown below.

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Determine the moment capacity of the plate girder.

3180=M>73321.=)12

136ksi)()(0.9570)(n.6i(0.9)(1285=M

0.9570=161.8]-][240300(0.789)+1200

0.789[-1=R

0.7895=19

15=

tb

th=a

ni1285.6=31

39,853=y/I=S

ni39,853=).5(2)(19)(30+))(19)(112

1(2)(+))(60

4

1)(

12

1(=I

=36

29,000

0.38=10.8240=

60=

t

h

ulk3

n

PG

ff

wcw

3

maxxx

4233

x

pf

f

w

Assume that bearing stiffeners are used at Points A, B, C and D.

Shear in Region BC

Investigate the need for stiffeners in the middle portion of the girder (Region BC) where the

shear force is quite low.

If no stiffeners are used, then kv = 5.

Also, note that h/tw = 240 > 1.37 86.9F

Ek

y

v = .

Hence, the shear strength with no stiffeners is given as:

24=V>31.8=V0.9=V

kips-35.35=V

(36)(240)

000)(5)(1.51)(29,15.5)(0.6)(36)(

F)t

h(

1.51EkA0.6F=V

in15.5)

4

1((62)A

kunn

n

2

y

2

w

vwyn

2

w

max

=

==

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Therefore, the web shear strength is adequate and no stiffeners are needed in Region BC.

Shear in Regions AB and CD

Tension-field used to compute transverse stiffener lay-out, except for first panel where tension-

field is not permitted.

For the first panel, tension-field is not permitted. Hence, solve for spacing required using Sect.

G2.2.

First Stiffener Spacing

0.830(36)(15.5)(0.9)(0.6)

250C

in15.5)4

1

(62)(A

A0.6F

VC

VCA0.6FV

v

2

w

wy

uv

uvwyn

=

==

=

0.830

t

h

E/FK1.10C

w

yv

v =

in22.50)(0.3744)(6a

0.374435.68

5a/h

40.68(a/h)

55K

40.6829,000

36

1.10

)(0.83)(240K

2v

2

v

=

=

+=

=

Hence, select a = 22. This value will provide

kips250V254.6V

42.19K

u

k

n

v

=>=

=

For the remaining shear panels, then tension-field action may be used to locate the transverse

intermediate stiffeners.

Assume a = 41

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adequate.is41"aso,242.7V244.5V

67.24212

22/4250

244.57)(0.9)(271.V

271.7(0.6833)11.15

0.331710.331715.5)(0.6)(36)(V

0.3317(36)(240)

0)(15.71)1.51(29,00C

154.136

,000)(15.71)(29

1.37240t

h

15.71(0.6833)

55k

0.683360

41

h

a

k

u

k

n

k

n

k

2n

2v

w

2v

==>=

=

=

==

=

+

+=

==

=>=

=+=

==

k

uftkV

A spread sheet program was used to solve for the remaining stiffener spacings.

ix uV a nV

0

22

63

111

250

242.7

229.0

213.0

22

41

48

58

254.6

244.5

230.8

213.1

The stiffener layout above requires three transverse intermediate stiffeners, in addition to the two

bearing stiffeners in the end region of the beam.

0 22 43 111 168

22 41 48 57

If an additional stiffener is added, then the spacing can be made more uniform:

0 20 57 94 131 168

20 37 37 37 37

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Stiffener Design

Single Plate Transverse Stiffeners

Assume that single plate transverse stiffeners are used. Base the design on the largest

stiffener spacings, and use for all transverse stiffeners between Points A and B.

Calculation for 20" Stiffener Spacing

Observe thatwt

his between the limits: 220.8 < ,9.274240

t

h

w

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Calculation for 37" stiffener spacing

0.3833=(36))(240

0)(18.15)1.51(29,00

=C

165.6=36

,000)(18.15)(291.37>240=h/t

18.15=)(0.617

5+5=k

0.617=60

37=a/h

2v

2

Assume that the ratio of Vr/Vc is equal to 1.0, so

2st

2

st

in2.211.1253.33A

])(18)(1/4-0.3833)(1)-)(14)(60)(1/4[(0.15)(2.(1)A

=

Also, we must compute Ist:

As before, the 7" x 5/8" stiffener plate appears to be adequate.

The stiffener plate must also satisfy local buckling requirements:

Hence use 7" x 5/8" transverse stiffeners.

43

st

2st

in71.5=))(78

5)(

3

1(=I

in4.375=)8

5((7)=A

43

st

2

in2.64=(4.574))(37)(1/4I

4.574=2-

)(0.617

2.5=j

15.9=36

29,0000.56