Question 23.6: Design a lever for a safety value. The lever loaded safety v......

Given l_2=200 \mathrm{~mm}, l_1=1000 \mathrm{~mm}

From moment equilibrium equation, taking moment about fulcrum, F (see Figure 23.12(a))

W \times l_1=F \times l_2

W=\frac{F \times l_2}{l_1}=\frac{3 \times 10^3 \times 200}{1000}=600 \mathrm{~N}

Considering the force equilibrium condition,

\Sigma F_V=0

-R+3 \times 10^3-W=0

3 \times 10^3-600=R \quad \therefore R=2400 \mathrm{~N}

The loads are shown in Figure 23. 12(b).

M_{\max }=M=480 \mathrm{~N} \cdot \mathrm{m}

\sigma_{\text {all }}=\frac{380}{4}=95 \mathrm{~N} / \mathrm{mm}^2

Assuming a rectangular cross-section arm,

Z=\frac{1}{6} t h^2

\sigma_b=\frac{M}{Z}

95=\frac{480 \times 10^3}{\frac{1}{6} t h^2}

Assuming t h^2=\frac{480 \times 10^3 \times 6}{95}=30315.79 \mathrm{~mm}^3

h=3 t

t \times(3 t)^2=30315.79

t \times 9 t^2=30315.79

Solving, we get t=14.99 \simeq 15.0 \mathrm{~mm}

\therefore \quad h=3 \times 15=45.0 \mathrm{~mm}

Design of Pin

Assuming l_p=d_p, form bearing consideration

l_p^2=d_p^2=\frac{R}{p_b}=\frac{2400}{20}

=120 \mathrm{~mm}^2

\therefore \quad l_p=d_p=10.95 \sim 12.0 \mathrm{~mm}

The lever may be weak due to the pin hole at the valve spindle axes. Hence, it should be checked for bending failure at this critical section.

I=\frac{1}{12}\left[15 \times 45^3-15 \times 12^3\right]

=111746.25 \mathrm{~mm}^4

y=\frac{45}{2}=22.5 \mathrm{~mm}

Z=\frac{111746.25}{22.5}=49665.5 \mathrm{~mm}^3

\sigma_b=\frac{M}{Z}=\frac{480 \times 10^3}{4966.5}

=96.65 \mathrm{MPa}

which is slightly more than the allowable stress.

Hence modifying the dimensions as

t=15 \times \frac{96.65}{95.00}=15.26

taking t=20 \mathrm{~mm} \text { for safe value }

h=20 \times 3=60 \mathrm{~mm}