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Chapter 12

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 } .

Step-by-Step

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

12.14