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For Circular X-section beam, show that maximum shear stress is 4/3 times of the average shear stress.
1 Answer
written 6.4 years ago by | • modified 6.4 years ago |
a=2R
b2=√R2−y2
b2=4(R2−y2) (i)
Aˉy=∫day
Now, I=πR44
Aˉy=∫Rybdy∗y
b2=4(R2−y2)
2b.db=4∗(−2y.dy)
b.db=−4y.dy
y.dy=−b.db4
Aˉy=∫0bb(−b.db4)
Aˉy=b312
q11=FIb∗Aˉy
=FIb∗b312=Fb212I
F12I∗4∗(R2−y2) from (i)
q will be minimum (zero) when y=R
q will be maximum when y=0
qmax=F3πR44(R2−02)=434πR2
qmax=43qaug