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kuku
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1. Homework Statement
The beam ABCD with the cross section has pin connections at B and C and is subjected to a uniformly distributed load w. If the allowable bending stress is 150MPa and the allowable shear stress is 80 Mpa. ( Hint : the horizontal reaction forces at A and D are both zero).
the picture of beam is :
https://fbcdn-sphotos-f-a.akamaihd...._=1425149738_3735a63d312d409ebc2ebe6405dca32c
Homework Equations
Bending stress = My/I
Shear stress = VQ/It
The Attempt at a Solution
I tried in this way but cannot get the answer. answer : w = 57.1 kn/m
6 Rd-5w*3=0
6Rd = 15w
Rd = 2.5w
Ra=2.5 w
then i draw shear force diagram to find Vmax.
V max = 2500w N
I also find neutral axis at 165mm
Moment of Inertia I = 1/12 * 300^3*10 + 2 [ 1/12*15^3*70+15*170*157.5^2]
= 1.491075*10^-4 m^4
t= 0.01 m
Q = y'A' = 150*10*75+170*15*157.5
= 5.14125*10^-4 m^3
So, we have 80*10^6 = 2500w * 5.14125*10^-4 / 1.491075*10^-4 *0.01
w = 92.8 kN
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For bending moment stress
I draw bending moment diagram to find max bending moment = 4.375 kN.m
150*10^6 = (4.375w * 165/1000) / 1.491075*10^-4
w = 30.98 kNCan anyone tell me what's wrong that I can't get the solution w = 57.1 kN/m
Thank you