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Q. 5.1

A flat plate subjected to a tensile force of 5 kN is shown in Fig. 5.13. The plate material is grey cast iron FG 200 and the factor of safety is 2.5. Determine the thickness of the plate.

Verified Solution

$\text { Given } \quad P=5 kN \quad S_{ ut }=200 N / mm ^{2}(f s)=2.5$.

Step I Calculation of permissible tensile stress

$\sigma_{\max .}=\frac{S_{u t}}{(f s)}=\frac{200}{2.5}=80 N / mm ^{2}$.

Step II Tensile stress at fillet section
The stresses are critical at two sections—the fillet section and hole section. At the fillet section,

$\sigma_{o}=\frac{P}{d t}=\left(\frac{5000}{30 t}\right)$

$\frac{D}{d}=\frac{45}{30}=1.5 \text { and } \frac{r}{d}=\frac{5}{30}=0.167$.

$\text { From Fig. 5.3, } K_{t}=1.8$.

$\therefore \sigma_{\max .}=K_{t} \sigma_{o}=1.8\left(\frac{5000}{30 t}\right)=\left(\frac{300}{t}\right) N / mm ^{2} \text { (i) }$.

Step III Tensile stress at hole section

$\sigma_{o}=\frac{P}{(w-d) t}=\frac{5000}{(30-15) t} N / mm ^{2}$.

$\frac{d}{w}=\frac{15}{30}=0.5$.

From Fig. 5.2,

$K_{t}=2.16$.

$\sigma_{\max }=K_{t} \sigma_{o}=2.16\left[\frac{5000}{(30-15) t}\right]=\left(\frac{720}{t}\right) N / mm ^{2}$             (ii).

Step IV Thickness of plate
From (i) and (ii), it is seen that the maximum stress is induced at the hole section.
Equating it with permissible stress, we get

$\left(\frac{720}{t}\right)=80$.

or         t = 9 mm.