Answer:
The answer is explained below
Explanation:
Given that:
1 ft = 0.3048 m, 1 in = 0.0254 m, 1 pound = 4.44822 newton
[tex]b_1=3.6ft=1.1\ m[/tex], [tex]b_2=2.2 ft=0.67\ m[/tex], [tex]d_2=7.5 in=0.19\ m[/tex], [tex]d_1=6.8in=0.17\ m[/tex]. P = 2200 lb = 9786 N
The area (A) is given as:
[tex]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[/tex]
The moment of area is given as:
[tex]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[/tex]
The maximum tensile stress is given as:
[tex]\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[/tex]
The maximum compressive stress is given as:
[tex]\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[/tex]
The maximum shear stress is given as:
[tex]\tau_{max}=|\frac{\sigma_c}{2} |=\frac{47.13\ MPa}{2}=23.57\ MPa[/tex]
