Question 12.7: A sign is supported by a pipe which has 100 mm outer diamete...
A sign is supported by a pipe which has 100 mm outer diameter and 80 mm inside diameter as shown in Figure 12.15. The sign is 2 m long and 0.75 m high, and its lower edge is 3 m above the ground where it is supported by the pipe. Calculate the maximum shear stresses at points A, B and C. The wind pressure on the sign is 1.5 kPa.
Total wind load acting on the sign is
P = ( Pressure ) × ( Projected area)
=1.5\left(10^3\right) \times(1.95 \times 0.75) N =2.194 \times 10^3 N
Let us refer to Figure 12.16:
It is clear from the figure that this wind load produces bending and twisting of the pipe.
Bending of the pipe can be clearly seen in Figure 12.17. This bending takes place in such a way that B – C lies on the neutral surface. Therefore, stresses at points B and C are due to torsional shear stress only
\tau_{ B }=\tau_{ C }=\frac{2.25\left(10^6\right)(100 / 2)}{(\pi / 32)\left(100^4-80^4\right)} \frac{ N }{ mm ^2}=19.41 MPa
Clearly, stress at A will be due to the combined effect. So, maximum shear stress at point A is
\tau_{ A }=\sqrt{\left\lgroup \frac{\sigma_{x x}}{2} \right\rgroup^2+\tau_{x y}^2}
where
\left|\sigma_{x x}\right|=\left|\frac{M\left(d_0 / 2\right)}{(\pi / 64)\left(d_{ o }^4-d_{ i }^4\right)}\right|=\left|\frac{7404.75\left(10^3\right)(100 / 2)}{(\pi / 64)\left(100^4-80^4\right)}\right| N / mm ^2
or \sigma_{x x}=127.75 MPa
\text { and } \tau_{x y}=19.41 MPa \text {, so }\tau_{ A }=\sqrt{63.88^2+19.41^2}=66.76 MPa
Hence shear stress developed at points A, B and C are
\tau_{ A }=66.76 MPa , \quad \tau_{ B }=\tau_{ C }=19.41 MPa
