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## Q. 5.2

A non-rotating shaft supporting a load of 2.5 kN is shown in Fig. 5.14. The shaft is made of brittle material, with an ultimate tensile strength of 300 N/mm². The factor of safety is 3. Determine the dimensions of the shaft.

## Verified Solution

$\text { Given } P=2.5 kN \quad S_{u t}=300 N / mm ^{2}(f s)=3$.

Step I Calculation of permissible stress

$\sigma_{\max .}=\frac{S_{u t}}{(f s)}=\frac{300}{3}=100 N / mm ^{2}$.

Step II Bending stress at fillet section
Due to symmetry, the reaction at each bearing is 1250 N. The stresses are critical at two sections—(i) at the centre of span, and (ii) at the fillet. At the fillet section,

$\sigma_{o}=\frac{32 M_{b}}{\pi d^{3}}=\frac{32(1250 \times 350)}{\pi d^{3}} N / mm ^{2}$.

$\frac{D}{d}=1.1 \text { and } \frac{r}{d}=0.1$.

$\text { From Fig. } 5.5, K_{t}=1.61$.

$\therefore \quad \sigma_{\max .}=K_{t} \sigma_{o}=1.61\left[\frac{32(1250 \times 350)}{\pi d^{3}}\right]$.

$=\left(\frac{7174704.8}{d^{3}}\right) N / mm ^{2}$           (i).

Step III Bending stress at centre of the span

$\sigma_{o}=\frac{32 M_{b}}{\pi d^{3}}=\frac{32(1250 \times 500)}{\pi(1.1 d)^{3}}$.

$=\left(\frac{4783018.6}{d^{3}}\right) N / mm ^{2}$          (ii).

Step IV Diameter of shaft
From (i) and (ii), it is seen that the stress is maximum at the fillet section. Equating it with permissible stress,

$\left(\frac{7174704.8}{d^{3}}\right)=100$.

or            d = 41.55 mm.