Question 5.6: A vertical post 2.5-meters high must support a lateral load ...
A vertical post 2.5-meters high must support a lateral load P = 12 kN at its upper end (Fig. 5-20). Two plans are proposed—a solid wood post and a hollow aluminum tube.
(a) What is the minimum required diameter d_{1} of the wood post if the allowable bending stress in the wood is 15 MPa?
(b) What is the minimum required outer diameter d_{2} of the aluminum tube if its wall thickness is to be one-eighth of the outer diameter and the allowable bending stress in the aluminum is 50 MPa?
Maximum bending moment. The maximum moment occurs at the base of the post and is equal to the load P times the height h; thus,
M_{\max } = Ph = (12 kN)(2.5 m) = 30 kN·m
(a) Wood post. The required section modulus S_{1} for the wood post (see Eqs. 5-19b and 5-24) is
S=\frac{\pi d^{3}}{32} (5-19b)
S=\frac{M_{\max }}{\sigma_{\text {allow }}} (5-24)
S_{1}=\frac{\pi d_{1}^{3}}{32}=\frac{M_{\max }}{\sigma_{\text {allow }}}=\frac{30 kN \cdot m }{15 MPa }=0.0020 m ^{3}=2 \times 10^{6} mm ^{3}Solving for the diameter, we get
d_{1}=273 mmThe diameter selected for the wood post must be equal to or larger than 273 mm if the allowable stress is not to be exceeded.
(b) Aluminum tube. To determine the section modulus S_{2} for the tube, we first must find the moment of inertia I_{2} of the cross section. The wall thickness of the tube is d_{2}/8, and therefore the inner diameter is d_{2} – d_{2}/4, or 0.75d_{2}. Thus, the moment of inertia (see Eq. 5-19a) is
I=\frac{\pi d^{4}}{64} (5-19a)
I_{2}=\frac{\pi}{64}\left[d_{2}^{4}-\left(0.75 d_{2}\right)^{4}\right]=0.03356 d_{2}^{4}The section modulus of the tube is now obtained from Eq. (5-17) as follows:
S_{2}=\frac{I_{2}}{c} (5-17)
=\frac{0.03356 d_{2}^{4}}{d_{2} / 2}=0.06712 d_{2}^{3}The required section modulus is obtained from Eq. (5-24):
S_{2}=\frac{M_{\max }}{\sigma_{\text {allow }}}=\frac{30 kN \cdot m }{50 MPa }=0.0006 m ^{3}=600 \times 10^{3} mm ^{3}By equating the two preceding expressions for the section modulus, we can solve for the required outer diameter:
d_{2}=\left(\frac{600 \times 10^{3} mm ^{3}}{0.06712}\right)^{1 / 3}=208 mmThe corresponding inner diameter is 0.75(208 mm), or 156 mm.