# Solution to Problem 121 Shear Stress

**Problem 121**

Referring to Fig. P-121, compute the maximum force P that can be applied by the machine operator, if the shearing stress in the pin at B and the axial stress in the control rod at C are limited to 4000 psi and 5000 psi, respectively. The diameters are 0.25 inch for the pin, and 0.5 inch for the control rod. Assume single shear for the pin at B.

**Solution 121**

$6P = 2T \sin 10^\circ$

$3P = T \sin 10^\circ$ → Equation (1)

$\Sigma F_H = 0$

$B_H = T \cos 10^\circ$

$T = \dfrac{3P}{\sin 10^\circ}$

Thus,

$B_H = \left( \dfrac{3P}{\sin 10^\circ} \right) \cos 10^\circ$

$B_H = 3 \cot 10^\circ \, P$

$\Sigma F_V = 0$

$B_V = T \sin 10^\circ + P$

$T \sin 10^\circ = 3P$

Thus,

$B_V = 3P + P$

$B_V = 4P$

${R_B}^2 = {B_H}^2 + {B_V}^2$

${R_B}^2 = (3 \cot 10^\circ \, P)^2 + (4P)^2$

$R_B^2 = 305.47P^2$

$R_B = 17.48P$

$P = \dfrac{R_B}{17.48}$ → Equation (2)

Based on tension of rod (equation 1):

$P = \frac{1}{3} T \sin 10^\circ$

$P = \frac{1}{3} [ \, 5000 \times \frac{1}{4} \pi (0.5)^2 \, ] \sin 10^\circ$

$P = 56.83 \, \text{lb}$

Based on shear of rivet (equation 2):

$P = \dfrac{4000 [ \, \frac{1}{4} \pi (0.25)^2 \, ]}{17.48}$

$P = 11.23 \, \text{lb}$

Safe load P,

$P = 11.23 \, \text{ lb}$ *answer*