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Theory of Structures - Engineering

Q1:

For permissible shear stress fs, the torque transmitted by a thin tube of mean diameter D and wall thickness t, is

A

B

C πD2t fs

D

ANS:A -

T = Fs*J/R.

where, J = π*(Do^4-Di^4)/32.
=π/32*(Do^2+Di^2)(Do+Di)(Do-Di).

= " * "* 2D *2t (where D=(Do+Di)/2, t=(Do-Di)/2).
=π*D*t/8 *[(Do-Di)^2+2*Do*Di].
=π*D*t/8*[ 4t^2 + 2D^2].
As the first term has a higher power of t , that can be neglected so;
=π*D^3*t/4.

Now Torque T=fs*J/R.
=π*D^2*t*fs/2.