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 dimensions 1.2 in. × 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.)
We will model this bar as a slender column that is fixed at end A and free at end B. Therefore, its critical load is given by the following equation (see Fig. 11-19b):
P_{cr}=\frac{π^2EI}{4L^2} (a)
The moment of inertia for the axis about which bending occurs is
I=\frac{(1.2\ in.)(0.6\ in.)^3}{12}=0.02160\ in.^4
Therefore, the expression for the critical load becomes
P_{cr}=\frac{π^2(16,000,000\ psi)(0.02160\ in.^4)}{4L^2}=\frac{852,700\ lb-in.^2}{L^2} (b)
in which P_{cr} has units of pounds and L has units of inches.
The deflection at the end of the brass bar is given by Eq. (11-54), which applies to a fixed-free column as well as a pinned-end column;
δ=e\left[sec\left(\frac{π}{2}\sqrt{\frac{P}{P_{cr}}}\right)\ -\ 1\right] (c)
in which P_{cr} is given by Eq.(a). To find the maximum permissible length of the bar, we substitute for 8 its limiting value of 0.12 in. Also, we substitute numerical values for the eccentricity e and the load P. and finally, we substitute, the expression for the critical load (Eq. b). Thus.
0.12\ in.=(0.45\ in.)\left[sec\left(\frac{π}{2}\sqrt{\frac{1500\ lb}{852,700/L^2}}\right)\ -\ 1\right]
This equation can be solved for the length L (inches). We begin by carrying out the multiplications and divisions: 0.2667 = sec(0.06588L) – 1. from which we get
sec (0.06588L) = 1.2667 and cos (0.06588L) = 0.7895
Therefore,
0.06588L = 0.6608 and 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.
