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Naddika [18.5K]
3 years ago
8

The internal loadings at a critical section along the steel drive shaft of a ship are calculated to be a torque of 2300 lb⋅ft, a

bending moment of 1500 lb⋅ft, and an axial thrust of 2500 lb. If the yield points for tension and shear are σY= 100 ksi and τY = 50 ksi, respectively, determine the required diameter of the shaft using the maximum-shear-stress theory
Engineering
1 answer:
Arturiano [62]3 years ago
6 0

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

\sigma_A = \dfrac{P}{A}+\dfrac{Mc}{I}

where;

A = \pi c^2

I = \dfrac{\pi}{4}c^4

\sigma_A = \dfrac{P}{\pi c^2}+\dfrac{Mc}{ \dfrac{\pi}{4}c^4}

\sigma_A = \dfrac{2500}{\pi c^2}+\dfrac{1500*12c}{ \dfrac{\pi}{4}c^4}

\sigma_A = \dfrac{2500}{\pi c^2}+\dfrac{72000c}{\pi c^3}}

\tau_A = \dfrac{T_c}{\tau}

where;

\tau = \dfrac{\pi c^4}{2}

\tau_A = \dfrac{T_c}{\dfrac{\pi c^4}{2}}

\tau_A = \dfrac{2300*12 c}{\dfrac{\pi c^4}{2}}

\tau_A = \dfrac{55200 }{\pi c^3}}

\sigma_{1,2} = \dfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\dfrac{(\sigma_x - \sigma_y)^2}{2}+ \tau_y^2}

\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)

Let say :

|\sigma_1 - \sigma_2|  = \sigma_y

Then :

2\sqrt{(   \dfrac{2500c + 72000}{2 \pi c^3})^2+ ( \dfrac{55200}{\pi c^3})^2 } = 100(10^3)

(2500 c + 72000)^2 +(110400)^2 = 10000*10^6 \pi^2 c^6

6.25c^2 + 360c+ 17372.16-10,000\ \pi^2 c^6 =0

According to trial and error;

c = 0.75057 in

Replacing  c into equation (1)

\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}}

\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}}

\sigma _1 = 22193 \ Psi

\sigma_2 = -77807 \ Psi

The required diameter d  = 2c

d = 1.50 in   or   0.125 ft

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