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## Q. 4.7

The frame of a hacksaw is shown in Fig. 4.26(a). The initial tension P in the blade should be 300 N. The frame is made of plain carbon steel 30C8 with a tensile yield strength of 400 N/mm$^2$ and the factor of safety is 2.5. The cross-section of the frame is rectangular with a ratio of depth to width as 3, as shown in Fig. 4.26(b).
Determine the dimensions of the cross-section.

## Verified Solution

Given    $P=300 N \quad S_{y t}=400 N / mm ^{2} \quad\left(f_{s}\right)=2.5$.

(depth/width) = 3.

Step I Calculation of permissible tensile stress

$\sigma_{t}=\frac{S_{y t}}{(f s)}=\frac{400}{2.5}=160 N / mm ^{2}$.

Step II Calculation of direct and bending stresses
The stresses at section XX consist of direct compressive stress and bending stresses. The tensile stress is maximum at the lower fibre. At the lower fibre,

$\sigma_{c}=\frac{P}{A}=\frac{300}{(t)(3 t)}=\frac{100}{t^{2}} N / mm ^{2}$            (i).

$\sigma_{b}=\frac{M_{b} y}{I}$  .

$=\frac{(300 \times 200)(1.5 t)}{\left[\frac{1}{12}(t)(3 t)^{3}\right]}=\frac{40000}{t^{3}} N / mm ^{2}$              (ii).

Step III Calculation of dimensions of cross-section
Superimposing the two stresses and equating it to permissible stress

$\frac{40000}{t^{3}}-\frac{100}{t^{2}}=160$.

$\text { or } 160 t^{3}+100 t-40000=0$.

Solving the above equation by trail and error method,

$t \cong 6.3 mm$.