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Answer:

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The three suspender bars AB, CD, and EF are made of A-36 steel and have equal cross-sectional areas of 500 mm2. Determine the average normal stress in each bar if the rigid beam is subjected to a force of P = 70kN , let d = 2.4 m , L = 4m  

Answer :

Stress = force / area

for Bar 1 = (40.03 * 1000) / 500  = 80.06 MPa

Bar 2 = ( 23.33 * 1000 ) / 500 = 46.66 MPa

for Bar 3 = ( 5.33 * 1000 ) / 500 = 10.66 MPa

Explanation:

Given data:

Type of steel = A-36

cross-sectional area = 500 mm^2

Calculate the average normal stress in each bar

we have to  make some assumptions

assume forces in AB, CD, EF  to be p1,p2,p3  respectively

∑ Fy = 0 ; p1 + p2 + p3 = 70kN  ----------  ( 1 )

∑ Mc = 0 ; P1 * d - p * d/2 - p3 * d = 0

where d = 2.4 hence ; p1 - p3 = 35 -------- ( 2 )

Take ; Tan∅

Tan∅  = MN / 2d = OP/d

i.e. s1 - 2s2 - s3 = 0

[tex]\frac{P1L}{AE} - \frac{2P2}{AE} + \frac{P3L}{AE} = 0[/tex]

L , E and A are the same hence

P1 - 2p2 + p3 = 0 ----- ( 3 )

Next resolve the following equations

p1 = 40.03 kN,  p2 = 23.33 kN, p3 = 5.33 kN

Stress = force / area

for Bar 1 = (40.03 * 1000) / 500  = 80.06 MPa

Bar 2 = ( 23.33 * 1000 ) / 500 = 46.66 MPa

for Bar 3 = ( 5.33 * 1000 ) / 500 = 10.66 MPa