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Chapter 12

Q. 12.3

A solid circular shaft is subjected to an axial tensile force 27 kN and a bending moment 3165 N m. Based on an allowable tensile stress of 120 MPa, calculate the required shaft diameter.

Step-by-Step

Verified Solution

If d be the shaft diameter, we get tensile stress due to axial load assuming d is in mm as

\sigma_1=\frac{P}{A}=\frac{(27)\left(10^3\right)}{\frac{\pi}{4} d^2}  MPa

or        \sigma_1=\frac{34377.47}{d^2}  MPa              (1)

Axial stress due to bending (maximum) as

\sigma_2=\pm \frac{M}{z}=\pm \frac{(32) 3165\left(10^3\right)}{\pi d^3}  MPa =\pm \frac{32.24\left(10^6\right)}{d^3}            (2)

Maximum normal stress on the section =\left(\sigma_n\right)_{\max }=\sigma_1+\sigma_2 \text {. So, }

\left(\sigma_n\right)_{\max }=\frac{34.38\left(10^3\right)}{d^2}+\frac{32.24\left(10^6\right)}{d^3}  MPa

Now, putting \left(\sigma_n\right)_{\max }=120 , we get

\frac{34.38\left(10^3\right)}{d^2}+\frac{32.24\left(10^6\right)}{d^3}=120

\Rightarrow \quad 34.38\left(10^3\right) d+32.24\left(10^6\right)=120 d^3

\Rightarrow 120 d^3-34.38\left(10^3\right) d-32.24\left(10^6\right)=0

\Rightarrow \quad d^3-286.5 d-268.67\left(10^3\right)=0

Solving by trial and error, we get d = 66.0065 mm. Therefore, required shaft diameter is 66 mm.