Answer:
Explanation:
Given that:
Torque T = 2300 lb - ft
Bending moment M = 1500 lb - ft
axial thrust P = 2500 lb
yield points for tension σY= 100 ksi
yield points for shear τY = 50 ksi
Using maximum-shear-stress theory
[tex]\sigma_A = \dfrac{P}{A}+\dfrac{Mc}{I}[/tex]
where;
[tex]A = \pi c^2[/tex]
[tex]I = \dfrac{\pi}{4}c^4[/tex]
[tex]\sigma_A = \dfrac{P}{\pi c^2}+\dfrac{Mc}{ \dfrac{\pi}{4}c^4}[/tex]
[tex]\sigma_A = \dfrac{2500}{\pi c^2}+\dfrac{1500*12c}{ \dfrac{\pi}{4}c^4}[/tex]
[tex]\sigma_A = \dfrac{2500}{\pi c^2}+\dfrac{72000c}{\pi c^3}}[/tex]
[tex]\tau_A = \dfrac{T_c}{\tau}[/tex]
where;
[tex]\tau = \dfrac{\pi c^4}{2}[/tex]
[tex]\tau_A = \dfrac{T_c}{\dfrac{\pi c^4}{2}}[/tex]
[tex]\tau_A = \dfrac{2300*12 c}{\dfrac{\pi c^4}{2}}[/tex]
[tex]\tau_A = \dfrac{55200 }{\pi c^3}}[/tex]
[tex]\sigma_{1,2} = \dfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\dfrac{(\sigma_x - \sigma_y)^2}{2}+ \tau_y^2}[/tex]
[tex]\sigma_{1,2} = \dfrac{2500+72000}{2 \pi c ^3} \pm \sqrt{\dfrac{(2500 +72000)^2}{2 \pi c^3}+ \dfrac{55200}{\pi c^3}} \ \ \ \ \ ------(1)[/tex]
Let say :
[tex]|\sigma_1 - \sigma_2| = \sigma_y[/tex]
Then :
[tex]2\sqrt{( \dfrac{2500c + 72000}{2 \pi c^3})^2+ ( \dfrac{55200}{\pi c^3})^2 } = 100(10^3)[/tex]
[tex](2500 c + 72000)^2 +(110400)^2 = 10000*10^6 \pi^2 c^6[/tex]
[tex]6.25c^2 + 360c+ 17372.16-10,000\ \pi^2 c^6 =0[/tex]
According to trial and error;
c = 0.75057 in
Replacing c into equation (1)
[tex]\sigma_{1,2} = \dfrac{2500+72000}{2 \pi (0.75057) ^3} \pm \sqrt{\dfrac{(2500 +72000)^2}{2 \pi (0.75057)^3}+ \dfrac{55200}{\pi (0.75057)^3}}[/tex]
[tex]\sigma_{1,2} = \dfrac{2500+72000}{2 \pi (0.75057) ^3} + \sqrt{\dfrac{(2500 +72000)^2}{2 \pi (0.75057)^3}+ \dfrac{55200}{\pi (0.75057)^3}} \ \ \ OR \\ \\ \\ \sigma_{1,2} = \dfrac{2500+72000}{2 \pi (0.75057) ^3} - \sqrt{\dfrac{(2500 +72000)^2}{2 \pi (0.75057)^3}+ \dfrac{55200}{\pi (0.75057)^3}}[/tex]
[tex]\sigma _1 = 22193 \ Psi[/tex]
[tex]\sigma_2 = -77807 \ Psi[/tex]
The required diameter d = 2c
d = 1.50 in or 0.125 ft
