Question 11.4: 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 = 7 kN acting with an eccentricity e = 11 mm (Fig. 11-26). The bar has a rectangular cross section with height h = 30 mm and width b = 15 mm .
What is the longest permissible length L_{\max }of the bar if the deflection at the end is limited to 3 mm? (For the brass, use E = 110 GPa.)
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. 11-20b) is
P_{\mathrm{cr}}=\frac{\pi^{2} E I}{4 L^{2}} (a)
The moment of inertia for the axis about which bending occurs is
I=\frac{h b^{3}}{12}=\frac{(30 \mathrm{~mm})(15 \mathrm{~mm})^{3}}{12}=0.844 \mathrm{~cm}^{4}Therefore, the expression for the critical load becomes
P_{\mathrm{cr}}=\frac{\pi^{2}(110 \mathrm{~GPa})\left(0.844 \mathrm{~cm}^{4}\right)}{4 L^{2}}=\frac{2.29 \mathrm{~kN} \cdot \mathrm{m}^{2}}{L^{2}} (b)
in which P_{\mathrm{cr}} has units of KN and L has units of meters.
Deflection. The deflection at the end of the bar is given by Eq. (11-62), 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_{c r}}}\right)-1\right] (c)
In this equation , P_{\mathrm{cr}} is given by Eq. (a).
Length. To find the maximum permissible length of the bar, we substi-tute e = 11 mm for δ its limiting value of 3 mm. Also, we substitute and P = 7 kN , and we substitute for P_{\mathrm{cr}} from Eq. (b). Thus,
The only unknown in this equation is the length L (meters). To solve for L, we perform the various arithmetic operations in the equation and then rearrange the terms. The result is
0.2727 = sec (2.746L) – 1
Using radians and solving this equation, we get L = 0.243 m . Thus, the max-imum permissible length of the bar is
L_{\max }=0.243 \mathrm{~m}If a longer bar is used, the deflection will exceed the allowable value of 3 mm.
