8/3/2019 Timoshenko Problem

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A simply supported beam of 2ft. span in subjected to a load of

60,000 lb at its centre the cross-section is show in Fig. Find the

maximum normal and shearing stresses at the junction of theflange and web.

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P = 60,000 lb

R2 30,000 lbR1 30,000 lb

1" 1"

+30,000

-30,000

+30,000 lb-ft

+

+

SFD

BMD

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C1 = 6"

C2 = 6"

275.05

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4

4

13

4

.2

31.286

04.11923.4804.119

)(04.119

23.48023.48

inI

I

symmetrytodueinII

inI AN

43

2

4

2

.1

43

1

23.4812

)5.10)(5.0(

04.1192

75.05)75.5(18.0

18.012

)75.0)(5(

inI

inI

inI

AN

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PsiI

MC

f tlbM

lbV

27.754433.286

612000,30

formulebyStressFlexural

000,30

000,30

1max

max

max

Psix

x

x

6601

)75.06(6

20.7544

6

20.7544

75.06

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The flexural stress profile is

max -7544.20

(6-0.75)"

6"

-7544.20

x6"

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5.6255.25

5.25/2

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Psi

bI

VQ

NA

NA

V

17.5871

2

25.525.55.0625.575.05

5.031.286

000,30

formulashearbyStressShear

psi

JunctionAt

xy

4424

2

25.575.05

5.031.286

000,30

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4424 Psi

5871.17 Psi

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Then the principal stress is

.52.5519

.8820

44242

6601

2

6601

0

22

max

1

2

2

1

2

2

1

Anspsi

Anspsi

As y

xy

yxyx

It is seen that (1) the principal stress at the junction between flange

and web is larger than the tensile stress in the most remote fiber (1 >

max) and therefore should be considered in design.

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Find the span length (l) such that the maximum Principal stress

at the junction between flange and the web will be equal to the

maximum flexural stress in the extreme fiber.

I

Pl

I

P

If

VQ

I

Pl

I

Pl

I

Pl

I

Plc

I

MC

xy

x

1.21)5.0(

625.575.3

2

312.112

5.10

)1(5.1

6

44

max

max

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.8.39;1580;2.44528.0

5.12.445430.0656.0

(2)and(1)Equating

)2(2.445430.0656.0

0

22

22

22

22

1

2

2

1

inlll

I

Pl

I

Pl

I

Pl

I

Pl

I

Pl

I

Pl

I

Pl

y

xy

yxyx