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For Circular X-section beam, show that maximum shear stress is 4/3 times of the average shear stress.
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a=2R

b2=R2y2

b2=4(R2y2) (i)

Aˉy=day

Now, I=πR44

Aˉy=Rybdyy

b2=4(R2y2)

2b.db=4(2y.dy)

b.db=4y.dy

y.dy=b.db4

Aˉy=0bb(b.db4)

Aˉy=b312

q11=FIbAˉy

=FIbb312=Fb212I

F12I4(R2y2) from (i)

q will be minimum (zero) when y=R

q will be maximum when y=0

qmax=F3πR44(R202)=434πR2

qmax=43qaug

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