Question

A bracket ABCD having a hollow circular cross section consists of a vertical arm AB (L 5 6 ft), a horizontal arm BC parallel to the x0 axis, and a horizontal arm CD parallel to the z0 axis (see gure). The arms BC and CD have lengths b1 5 3.6 ft and b2 5 2.2 ft, respectively. The outer and inner diameters of the bracket are d2 5 7.5 in. and d1 5 6.8 in. Aninclined load P 5 2200 lb acts at point D along line DH. Determine the maximum tensile, compressive, and shear stresses in the vertical arm.

Answer 1

Answer:

The answer is explained below

Explanation:

Given that:

1 ft = 0.3048 m, 1 in = 0.0254 m, 1 pound = 4.44822 newton

$b_1=3.6ft=1.1\ m$, $b_2=2.2 ft=0.67\ m$, $d_2=7.5 in=0.19\ m$, $d_1=6.8in=0.17\ m$. P = 2200 lb = 9786 N

The area (A) is given as:

$A=\frac{\pi}{4} (d_2^2-d_1^2)=\frac{\pi}{4}(0.19^2-0.17^2)=5.65*10^{-3}m^2$

The moment of area is given as:

$l=\frac{\pi}{64} (d_2^4-d_1^4)=\frac{\pi}{64}(0.19^4-0.17^4)=2.3*10^{-5}m^4$

The maximum tensile stress is given as:

$\sigma_1=-\frac{P}{A}+\frac{M(\frac{d_2}{2} )}{l} =-\frac{9786\ N }{5.65*10^{-3}m^2}+\frac{11kNm(0.19 \ m)/2}{2.3*10^{-5}m^4} =-1.73\ MPa+45.4\ MPa=43.67\ MPa\\\sigma_1=43.67\ MPa$

The maximum compressive stress is given as:

$\sigma_c=-\frac{P}{A}-\frac{M(\frac{d_2}{2} )}{l} =-\frac{9786\ N }{5.65*10^{-3}m^2}-\frac{11kNm(0.19 \ m)/2}{2.3*10^{-5}m^4} =-1.73\ MPa-45.4\ MPa=47.13\ MPa\\\sigma_c=47.13\ MPa$

The maximum shear stress is given as:

$\tau_{max}=|\frac{\sigma_c}{2} |=\frac{47.13\ MPa}{2}=23.57\ MPa$