Answer to Question #128732 in Mechanics | Relativity for Ariyo Emmanuel

Question

) A steel shaft ABCD having a total length of 3.5 m consists of three lengths having different sections as follows: AB is hollow having outside and inside diameters of 100 mm and 62.5 mm respectively, and BC and CD are solid. BC has a diameter of 100 mm and CD has a diameter of 87.5 mm. If the angle of twist is the same for each section, determine the length of each section. Find the value of the applied torque and the total angle of twist, if the maximum shear stress in the hollow portion is 47.5 MPa and shear modulus, C = 82.5 GPa. (Assuming the torque T and shear modulus are the same for all the sections).

Answer 1

Diagrammatically the question is as shown below

First find the polar moment of inertia of the hollow shaft AB

$j_1=\frac{\pi}{32}[(d_o)\space ^4-(d_i)^4]=\frac{\pi}{32}[(100)^4-(62.5)^4]=8.32\times10^6mm$

polar moment of inertia of solid shaft BC

$J_2=\frac{\pi}{32}(d_2)^4=\frac{\pi}{32}(100)^4=9.82\times10^6mm$

polar moment of inertia of solid shaft CD

$J_3=\frac{\pi}{32}(d_3)^4=\frac{\pi}{32}(87.50)^4=5.75\times10^6mm$

Now obtain the angle of twist of hollow shaft AB , solid shaft BC and CD

Angle of twist $\theta_1=\frac{T\times l_1}{C\times J_1}$

The angle of twist is the same for each section i.e

$\theta_1=\theta_2=\frac{T\times l_1}{C\times J_1}=\frac{8.32\times10^6}{9.82\times10^6}=0.847$

Now to find the length of each section

$l_1+l_2+l_3=L=3500mm$

factoring out $l_1$ the equation becomes

$l_1(1+\frac{l_2}{l_1}+\frac{l_3}{l_1})=3500$

$l_1(1+\frac{1}{0.847}+\frac{1}{1.447})=3500$

$l_1$ becomes $\frac{3500}{2.8717}=1218.8mm$

$l_2=\frac{l_1}{0.847}=\frac{1218.8}{0.847}=1439mm$

$l_3=\frac{l_1}{1.447}=\frac{1218.8}{1.447}=842.2mm$

length = $l_1=1218.8mm\space l_2=1439mm\space l_3=842.2mm$

Now use the given maximum shear stress in the hollow portion to get the torque of the hollow shaft

$T=\frac{\pi}{16}\times \tau[\frac{(d_o)^4-(d_i)^4}{d_o}]=\frac{\pi}{16}\times47.5[\frac{100^4-62.5^4}{100}]=7.9\times10^6 N/m$

Applied torque=$7.9\times10^6N/m$

Total angle of twist is equal to the sum of the angle of twists of the individual shafts

angle of twist

$\theta=\frac{T\times l}{C\times J}$

$\theta=\frac{T\times l_1}{C\times J_1}+\frac{T\times l_2}{C\times J_2}+\frac{T\times l_3}{C\times J_3}$

$=\frac{T}{C}[\frac{l_1}{j_1}+\frac{l_2}{j_2}+\frac{l_3}{j_3}]$

$=\frac{7.9\times10^6}{82.5\times10^3}[\frac{1218.8}{8.32\times10^6}+\frac{1439}{9.82\times10^6}+\frac{842.2}{5.75\times10^6}]=0.042rad$

$\theta=0.042\times\frac{180}{\pi}=2.406\degree$

Total angle of twist =$2.406\degree$