Question 5.13: A vertical pole consisting of a circular tube of outer diame...
A vertical pole consisting of a circular tube of outer diameter d_{2} = 4.0 in. and inner diameter d_{1} = 3.2 in. is loaded by a horizontal force P = 1500 lb (Fig. 5-36a).
(a) Determine the maximum shear stress in the pole.
(b) For the same load P and the same maximum shear stress, what is the diameter d_{0} of a solid circular pole (Fig. 5-36b)?
(a) Maximum shear stress. For the pole having a hollow circular cross section (Fig. 5-36a), we use Eq. (5-44) with the shear force V replaced by the load P and the cross-sectional area A replaced by the expression \pi\left(r_{2}^{2}-r_{1}^{2}\right); thus,
\tau_{\max }=\frac{V Q}{I b}=\frac{4 V}{3 A}\left(\frac{r_{2}^{2}+r_{2} r_{1}+r_{1}^{2}}{r_{2}^{2}+r_{1}^{2}}\right) (5-44)
\tau_{\max }=\frac{4 P}{3 \pi}\left(\frac{r_{2}^{2}+r_{2} r_{1}+r_{1}^{2}}{r_{2}^{4}-r_{1}^{4}}\right) (a)
Next, we substitute numerical values, namely,
P = 1500 lb r_{2}=d_{2} / 2=2.0 \text { in. } \quad r_{1}=d_{1} / 2=1.6 \text { in. }
and obtain
\tau_{\max }=658 psiwhich is the maximum shear stress in the pole.
(b) Diameter of solid circular pole. For the pole having a solid circular cross section (Fig. 5-36b), we use Eq. (5-42) with V replaced by P and r replaced by d_{0} /2:
\tau_{\max }=\frac{V Q}{I b}=\frac{V\left(2 r^{3} / 3\right)}{\left(\pi r^{4} / 4\right)(2 r)}=\frac{4 V}{3 \pi r^{2}}=\frac{4 V}{3 A} (5-42)
\tau_{\max }=\frac{4 P}{3 \pi\left(d_{0} / 2\right)^{2}} (b)
Solving for d_{0}, we obtain
d_{0}^{2}=\frac{16 P}{3 \pi \tau_{\max }}=\frac{16(1500 lb )}{3 \pi(658 psi )}=3.87 in. ^{2}from which we get
d_{0}=1.97 in .In this particular example, the solid circular pole has a diameter approximately one-half that of the tubular pole.
Note: Shear stresses rarely govern the design of either circular or rectangular beams made of metals such as steel and aluminum. In these kinds of materials, the allowable shear stress is usually in the range 25 to 50% of the allowable tensile stress. In the case of the tubular pole in this example, the maximum shear stress is only 658 psi. In contrast, the maximum bending stress obtained from the flexure formula is 9700 psi for a relatively short pole of length 24 in. Thus, as the load increases, the allowable tensile stress will be reached long before the allowable shear stress is reached.
The situation is quite different for materials that are weak in shear, such as wood. For a typical wood beam, the allowable stress in horizontal shear is in the range 4 to 10% of the allowable bending stress. Consequently, even though the maximum shear stress is relatively low in value, it sometimes governs the design.