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## Q. 12.6

A circular hollow shaft of outer diameter 200 mm and inner diameter 160 mm is subjected simultaneously to a torque T = 11.1 kN m and an axial compressive load, P = 362 kN. Calculate: (a) $\left(\sigma_T\right)_{\max },(b)\left(\sigma_C\right)_{\max }$ and (c) $\tau_{\max }$ .

## Verified Solution

Let us first compute the area properties as follows:
Cross-sectional area,

$A=\frac{\pi}{4}\left(d_{ o }^2-d_{ i }^2\right)=\frac{\pi}{4}\left(200^2-160^2\right) mm ^2=11309.73 mm ^2$

Polar moment of inertia,

$J=\frac{\pi}{32}\left(d_{ o }^4-d_{ i }^4\right)=\left(\frac{\pi}{32}\right)\left(200^4-160^4\right) mm ^4=92.74 \times 10^6 mm ^4$

Axial stress $\left(\sigma_{x x}\right)$ due to P is

$\sigma_{x x}=\frac{P}{A}=\frac{362\left(10^3\right)}{11309.73} N / mm ^2=32 MPa \text { (compressive) }$

or        $\sigma_{x x}=-32 MPa$

and shear stress $\left(\tau_{x y}\right)$ due to T is

$\tau_{x y}=\frac{T\left(d_0 / 2\right)}{J}=\frac{11.1\left(10^6\right)(200 / 2)}{92.74\left(10^6\right)} N / mm ^2=11.97 MPa$

The stress states are shown in Figure 12.14 on a differential element of the shaft:

Thus, the principal stresses are

$\sigma_{1,2}=\frac{\sigma_{x x}}{2} \pm \sqrt{\left\lgroup \frac{\sigma_{x x}}{2} \right\rgroup^2+\tau_{x y}^2}$

$=-16 \pm \sqrt{(-16)^2+11.97^2} MPa$

or        $\sigma_1=-36 MPa , \quad \sigma_2=4 MPa$

and the maximum shear stress is

$\tau_{\max }=\sqrt{(-16)^2+11.97^2}=20 MPa$

Thus,

$\left(\sigma_{ T }\right)_{\max }=4 MPa , \quad\left(\sigma_{ C }\right)_{\max }=36 MPa \text { and } \tau_{\max }=20 MPa$