Question

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

Answer 1

Step-by-step 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 = (P)/(A)+(Mc)/(I)`

where;

`A = \pi c^2`

`I = (\pi)/(4)c^4`

`\sigma_A = (P)/(\pi c^2)+(Mc)/( (\pi)/(4)c^4)`

`\sigma_A = (2500)/(\pi c^2)+(1500*12c)/( (\pi)/(4)c^4)`

`\sigma_A = (2500)/(\pi c^2)+(72000c)/(\pi c^3)}`

`\tau_A = (T_c)/(\tau)`

where;

`\tau = (\pi c^4)/(2)`

`\tau_A = (T_c)/((\pi c^4)/(2))`

`\tau_A = (2300*12 c)/((\pi c^4)/(2))`

`\tau_A = (55200 )/(\pi c^3)}`

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

`\sigma_(1,2) = (2500+72000)/(2 \pi c ^3) \pm \sqrt{((2500 +72000)^2)/(2 \pi c^3)+ (55200)/(\pi c^3)} \ \ \ \ \ ------(1)`

Let say :

`|\sigma_1 - \sigma_2| = \sigma_y`

Then :

`2\sqrt{( (2500c + 72000)/(2 \pi c^3))^2+ ( (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) = (2500+72000)/(2 \pi (0.75057) ^3) \pm \sqrt{((2500 +72000)^2)/(2 \pi (0.75057)^3)+ (55200)/(\pi (0.75057)^3)}`

`\sigma_(1,2) = (2500+72000)/(2 \pi (0.75057) ^3) + \sqrt{((2500 +72000)^2)/(2 \pi (0.75057)^3)+ (55200)/(\pi (0.75057)^3)} \ \ \ OR \\ \\ \\ \sigma_(1,2) = (2500+72000)/(2 \pi (0.75057) ^3) - \sqrt{((2500 +72000)^2)/(2 \pi (0.75057)^3)+ (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