Question

A hollow steel shaft is to transmit 4200 n.m of torque and is to be sized so that the torsional stress does not exceed 120 mpa

Answer 1

A hollow steel shaft is to transmit 4200 n.m of torque and is to be sized so that the torsional stress does not exceed 120 Mpa

If the inside diameter is 70 percent of the outside diameter, what size shaft should be used?

Answer:

the preferred outer diameter  of the hollow steel shaft obtained a little bit higher than 61.67 mm is 80 mm

the preferred inner diameter of the hollow steel obtained a little bit lower than 56 mm is 50 mm

Explanation:

From above;

let the inside diameter be $d_i$

let the outside diameter be $d_o$

Then

$d_i$ = 70%  $d_o$

$\dfrac{d_i}{d_o} = \dfrac{70}{100}$    

$\dfrac{d_i}{d_o} =0.7$   ------ (1)

to determine the outer diameter of the hollow steel shaft by using the equation of the torque, we have:

$T = \dfrac{\pi}{16} \times \tau \times d_o^3 ( 1 - (\dfrac{d_i}{d_o})^4)$

where;

$\tau$ = torsoinal shear stress = 120 MPa

$4200= \dfrac{\pi}{16} \times 120\times 10^6 \times d_o^3 ( 1 - 0.7)^4)$

$4200= 0.19635 \times 120\times 10^6 \times d_o^3 (0.7599)$

$4200=17904763.8 \ d_o^3$

$d_o^3 = \dfrac{4200}{17904763.8}$

$d_o^3 = 2.3457 \times 10^{-4}$

$d_o =\sqrt[3]{2.3457 \times 10^{-4}}$

$d_o =0.06167 \ m$

$d_o =0.06167 \ m \times (\dfrac{1000 \ mm}{1 \ m})$

$d_o =61.67 \ mm$

From the tables of A-17 of sized and Renard Numbers,

the preferred outer diameter  of the hollow steel shaft obtained a little bit higher than 61.67 mm is 80 mm

However, from equation (1)

$\dfrac{d_i}{d_o} =0.7$   ------ (1)

replacing the value of $d_o$

$\dfrac{d_i}{80} =0.7$

$d_i = 80 \times 0.7$

$d_i = 56 \ mm$

From the tables of A-17 of sized and Renard Numbers,

the preferred inner diameter of the hollow steel obtained a little bit lower than 56 mm is 50 mm