Question 11.3: A brass bar AB projecting from the side of a large machine i...
A brass bar AB projecting from the side of a large machine is loaded at end B by a force P = 1500 lb acting with an eccentricity e = 0.45 in. (Fig. 11-25). The bar has a rectangular cross section with height h = 1.2 in. and width b = 0.6 in.
What is the longest permissible length L_{\max} of the bar if the deflection at the end is limited to 0.12 in.? (For the brass, use E = 16 × 10^{6} psi.)
Critical load. We will model this bar as a slender column that is fixed at end A and free at end B. Therefore, the critical load (see Fig. 11-19b) is
P_{cr}=\frac{\pi^{2}EI}{4L^{2}} (a)
The moment of inertia for the axis about which bending occurs is
I=\frac{hb^{3}}{12}=\frac{(1.2 in.)(0.6 in.)^{3}}{12}=0.02160 in.^{4}Therefore, the expression for the critical load becomes
P_{cr}=\frac{\pi^{2}(16,000,000 psi)(0.02160 in.^{4})}{}=\frac{852,700 lb-in.^{2}}{L^{2}} (b)
in which P_{cr} has units of pounds and L has units of inches.
Deflection. The deflection at the end of the bar is given by Eq. (11-54), which applies to a fixed-free column as well as a pinned-end column:
\delta=e\left[\sec\left(\frac{\pi}{2}\sqrt{\frac{P}{P_{cr}}}\right)-1\right] (c)
In this equation, P_{cr} is given by Eq. (a).
Length. To find the maximum permissible length of the bar, we substitute for δ its limiting value of 0.12 in. Also, we substitute e = 0.45 in. and P = 1500 lb, and we substitute for P_{cr} from Eq. (b). Thus,
The only unknown in this equation is the length L (inches). To solve for L, we perform the various arithmetic operations in the equation and then rearrange the terms. The result is
0.2667 = sec (0.06588 L) – 1
Using radians and solving this equation, we get L = 10.03 in. Thus, the maximum permissible length of the bar is
L_{\max}=10.0 in.If a longer bar is used, the deflection will exceed the allowable value of 0.12 in.
