Question 23.7: Design a cranked lever to be operated by two persons. The ma......

Assumptions: As the lever is to be operated by two persons of sound health, assume one person can apply the 400 N load. Hence the load applied by the two persons is 800 N. To accommode two persons the length of the handle is assumed as 500 mm.

Allowable strength

\sigma_{\text {all }}=\frac{\sigma_{u l t}}{\mathrm{FOS}}=\frac{580}{5}=116 \mathrm{MPa}

\sigma_{\text {all }}=\frac{\sigma_y}{\text { FOS }}=\frac{380}{2.5}=152 \mathrm{MPa}

Hence, lowest value of l l 6 MPa is taken as allowable strength.

As two persons are applying load on the handle, the net load applied at a distance of 2/3 of its length from the fixed end or 113rd of length from the free end.

Let d_h be the diameter of the handle

l_h be the length of handle = 500 mm

Let P be the load applied by the two persons

= 800 N

Maximum bending moment is

M=\frac{2}{3}\left(l_h\right) \times P=\frac{2}{3} \times 500 \times 800

=266.67 \mathrm{~N} \cdot \mathrm{m} (23.18)

Section modulus of handle, assuming it as circular one

Z=\frac{\pi}{32} d_h^3 (23 .18)

From bending relation

\sigma_b=\frac{M}{Z}=\frac{266.67 \times 10^3}{(\pi / 32) d_h^3}

Taking \sigma_b=116 ~\mathrm{MPa}

116=\frac{266.67 \times 10^3 \times 32}{\pi \times d_h^3}

Solving, we get

d_h=28.61 \mathrm{~mm}=30.0 \mathrm{~mm}