\text { Given } \quad S_{y t}=400 N / mm ^{2} \quad(f s)=5 \quad F=5 kN .
For lever, long arm = 450 mm short arm = 100 mm d/b = 3.
\text { For pin } \quad p=10 N / mm ^{2} \quad l_{1} / d_{1}=1.25 .
Step I Calculation of permissible stresses for the pin and lever
\sigma_{t}=\frac{S_{y t}}{(f s)}=\frac{400}{5}=80 N / mm ^{2} .
\tau=\frac{S_{s y}}{(f s)}=\frac{0.5 S_{y t}}{(f s)}=\frac{0.5(400)}{5}=40 N / mm ^{2} .
Step II Calculation of forces acting on the lever
The forces acting on the lever are shown in Fig. 4.53. Taking moment of forces about the axis of the fulcrum,
\left(5 \times 10^{3}\right)(100)=P \times 450 \quad \therefore P=1111.11 N .
R=\sqrt{(5000)^{2}+(1111.11)^{2}}=5121.97 N .
Step III Diameter and length of fulcrum pin
Considering bearing pressure on the fulcrum pin,
R=p(\text { projected area of the pin })=p\left(d_{1} \times l_{1}\right)
or R=p\left(d_{1} \times l_{1}\right) (a).
where,
d_{1}=\text { diameter of the fulcrum pin }( mm ) .
l_{1}=\text { length of the fulcrum pin (mm) } .
p = permissible bearing pressure (N/mm²)
Substituting values in Eq. (a),
5121.97=10\left(d_{1} \times 1.25 d_{1}\right) .
∴ d_{1}=20.24 mm .
l_{1}=1.25 d_{1}=1.25(20.24)=25.30 mm (i).
Step IV Shear stress in pin
The pin is subjected to double shear. The shear stress in the pin is given by,
\tau=\frac{R}{2\left[\frac{\pi}{4} d_{1}^{2}\right]}=\frac{5121.97}{2\left[\frac{\pi}{4}(20.24)^{2}\right]}=7.96 N / mm ^{2} \text { (ii) } .
Step V Dimensions of the boss
The dimensions of the boss of the lever at the fulcrum are as follows:
inner diameter = 21 mm
outer diameter = 42 mm
length = 26 mm (iii)
Step VI Dimensions of cross-section of lever
For the lever
d=3 b \quad M_{b}=(5000 \times 100) N – mm .
Therefore,
\sigma_{b}=\frac{M_{b} y}{I} \quad \text { or } \quad 80=\frac{(5000 \times 100)(1.5 b)}{\left[\frac{1}{12}(b)(3 b)^{3}\right]} .
∴ b = 16.09 mm
d = 3b = 3 (16.09) = 48.27 mm (iv).
