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## Q. 11.8

A solid circular shaft of steel $\left(\sigma_{y p}=400 MPa \right)$ has a 20 mm diameter and is subjected to a combination of static loading
Axial load, P = 25.0 kN
Bending moment, M = 50.0 Nm
Torque, T = 120 Nm
Calculate the factor of safety for design based on the
(a) maximum octahedral shearing stress criterion and
(b) maximum shearing stress criterion.

## Verified Solution

We first calculate resultant normal stress on the shaft element as

$\sigma_n=\frac{4 P}{\pi d^2}+\frac{32 M}{\pi d^3}=\frac{4 P}{\pi d^2}\left[P+\frac{8 M}{d}\right]$

$=\frac{4}{\pi(20)^2}\left[25\left(10^3\right)+\frac{8}{20}(50)\left(10^3\right)\right]$

=143.24 MPa

and the shearing stress developed on the shaft is

$\tau=\frac{16 T}{\pi d^3}=\frac{(16)(120)\left(10^3\right)}{\pi(20)^3}=76.39 MPa$

(a) Hence, by octahedral shear stress theory

$\sqrt{\sigma_n^2+3 \tau^2}=\frac{\sigma_{y p }}{\text { factor of safety }}$

or        $\text { Factor of safety }=\frac{\sigma_{ yp }}{\sqrt{\sigma_n^2+3 \tau^2}}=\frac{400}{\sqrt{143.24^2+3(76.39)^2}}=2.05$

So, the required factor of safety is 2.05.

(b) According to maximum shear stress criterion, we get

$\tau_{\max }=\sqrt{\left(\frac{\sigma_n}{2}\right)^2+\tau^2}=\frac{\tau_{ yp }}{\text { factor of safety }}=\frac{\sigma_{ yp }}{2(\text { factor of safety })}$

Therefore,

$\text { Factor of safety }=\frac{\sigma_{ yp }}{2 \sqrt{\left(\sigma_n / 2\right)^2+\tau^2}}$

$=\frac{\sigma_{ yp }}{\sqrt{\sigma_n^2+4 \tau^2}}=\frac{400}{\sqrt{143.24^2+4(76.39)^2}}=1.91$

So the required factor of safety is 1.91.