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## Q. 4.15

A lever-loaded safety valve is mounted on the boiler to blow off at a pressure of 1.5 MPa gauge. The effective diameter of the opening of the valve is 50 mm. The distance between the fulcrum and the dead weights on the lever is 1000 mm. The distance between the fulcrum and the pin connecting the valve spindle to the lever is 100 mm.
The lever and the pin are made of plain carbon steel $30 C 8\left(S_{y t}=400 N / mm ^{2}\right)$ and the factor of safety is 5. The permissible bearing pressure at the pins in the lever is 25 N/mm². The lever has a rectangular cross-section and the ratio of width to thickness is 3:1. Design a suitable lever for the safety valve.

## Verified Solution

$\text { Given } \quad S_{y t}=400 N / mm ^{2} \quad(f s)=5$,

$\text { For valve, } d=50 mm \quad p=1.5 MPa$.

$\text { For lever, } l_{1}=1000 mm \quad l_{2}=100 mm \quad d / b=3$,

$\text { For pin, } \quad p=25 N / mm ^{2}$

The construction of the lever-loaded safety valve is shown in Fig. 4.50. It is mounted on steam boilers to limit the maximum steam pressure. When the pressure inside the boiler exceeds this limiting value, the valve automatically opens due to excess of steam pressure and steam blows out through the valve. Consequently, the steam pressure inside the boiler is reduced.
Step I Calculation of permissible stresses for lever and pin

$\sigma_{t}=\frac{S_{y t}}{(f s)}=\frac{400}{5}=80 N / mm ^{2}$.

$\tau=\frac{S_{s y}}{(f s)}=\frac{0.5 S_{y t}}{(f s)}=\frac{0.5(400)}{5}=40 N / mm ^{2}$.

Step II Calculation of forces acting on lever
The valve is held tight on the valve seat against the upward steam force F by the dead weights P attached at the end of the lever. The distance $l_1$ and the dead weights P are adjusted in such a way, that when the steam pressure inside the boiler reaches the limiting value, the moment $\left(F \times l_{2}\right)$ overcomes the moment $\left(P \times l_{1}\right)$. As a result, the valve opens and steam blows out until the pressure falls to the required limiting value and then the valve is automatically closed.
The maximum steam load F, at which the valve blows off is given by,

$F=\frac{\pi}{4} d^{2} p=\frac{\pi}{4}(50)^{2}(1.5)=2945.24 N$         (a).

Taking moment of forces F and P about the fulcrum,

$F \times l_{2}=P \times l_{1} \quad \text { or } \quad 2945.24 \times 100=P \times 1000$.

$\therefore \quad P=294.52 N$              (b).

The forces acting on the lever are shown in Fig. 4.51(a). Considering equilibrium of vertical forces,

F = R + P
or          R = F – P = 2945.24 – 294.52 = 2650.72 N            (c).

Step III Diameter and length of pin
From (a), (b) and (c), the pin at the point of application of the force F is subjected to maximum force and as such, it is to be designed from bearing consideration.
Suppose, $d_{1} \text { and } l_{1}$ are the diameter and the length of the pin at F and assume,

$l_{1}=d_{1}$.

From Eq. (4.51),

$R=p\left(d_{1} \times l_{1}\right)$                (4.51).

$F=p\left(d_{1} \times l_{1}\right) \text { or } 2945.24=25\left(d_{1} \times d_{1}\right)$.

$\therefore \quad d_{1}=10.85 \text { or } 12 mm$.

$l_{1}=d_{1}=12 mm$             (i).

The pin is subjected to double shear stress, which is given by,

$\tau=\frac{F}{2\left[\frac{\pi}{4} d_{1}^{2}\right]}=\frac{2945.24}{2\left[\frac{\pi}{4}(12)^{2}\right]}=13.02 N / mm ^{2}$.

$\therefore \quad \tau<40 N / mm ^{2}$.

The force on the fulcrum pin (R) is comparatively less than the force acting on the spindle pin (F). Therefore, the dimensions $d_{1} \text { and } l_{1}$  of the pin at the fulcrum will be slightly less.
However, we will assume both pins of the same diameter and length to facilitate interchangeability of parts and variety reduction.

Step IV Width and thickness of lever
A gunmetal bush of 2-mm thickness is press fitted at both pin holes to reduce friction. Therefore, the inside diameter of the boss will be $\left(d_{1}+2 \times 2\right) \text { or }(12+2 \times 2)$ or 16 mm. The outside diameter of the boss is kept twice of the inside diameter, i.e., 32 mm. The bending moment diagram for the lever is shown in Fig. 4.51(b). The bending moment is maximum at the valve spindle axis. It is given by,

$M_{b}= P (1000-100)=294.52(1000-100)$.

= 265 068 N-mm
For a lever,

d = 3b

$\sigma_{b}=\frac{M_{b} y}{I} \quad 80=\frac{(265068)(1.5 b)}{\left[\frac{1}{12} b(3 b)^{3}\right]}$.

$\therefore \quad b=13.02 \text { or } 15 mm \quad d=3 b=45 mm$.

The lever becomes weak due to the pinhole at the valve spindle axis and it is necessary to check bending stresses at this critical section. The cross-section of the lever at the valve spindle axis is shown in Fig. 4.52. In this case, the length of the pin is increased from 12 mm to 20 mm to get practical proportions for the boss. For this cross-section,

$M_{b}=265068 N – mm \quad y=22.5 mm$.

$I=\frac{1}{12}\left[15(45)^{3}+5(32)^{3}-20(16)^{3}\right]$.

$=120732.92 mm ^{4}$.

Therefore,

$\sigma_{b}=\frac{M_{b} y}{I}=\frac{(265068)(22.5)}{(120732.92)}=49.40 N / mm ^{2}$.

$\text { Since, } \sigma_{b}<80 N / mm ^{2}$.

the design is safe.   