##### Question

In: Mechanical Engineering

# Rigid bar ABC shown in Figure P1.30 is supported by a pin at bracket A and by tie rod (1).

Rigid bar ABC shown in Figure P1.30 is supported by a pin at bracket A and by tie rod (1). Tie rod (1) has a diameter of 5 mm, and it is supported by double-shear pin connections at B and D. The pin at bracket A is a single-shear connection. All pins are 7 mm in diameter. Assume a = 600 mm, b = 300 mm, h = 450 mm, P = 900 N, and θ = 55°. Determine the following:

(a) the normal stress in rod (1)

(b) the shear stress in pin B

(c) the shear stress in pin A

FIGURE P1.30

## Solutions

##### Expert Solution

Draw the free body diagram of the bar:

Calculate the angle $$\phi,$$ from triangle $$A B D$$.

$$\begin{array}{l} \tan \phi=\frac{A D}{A B} \\ \phi=\tan ^{-1}\left(\frac{450}{600}\right) \\ =\tan ^{-1}(0.75) \\ =36.87^{\circ} \end{array}$$

Take moments about $$A$$

$$F \sin \phi \times a=P \sin \theta \times(a+b)$$

Here, $$F$$ is the tension in tie rod, $$\phi$$ and $$\theta$$ are the angles made by the $$F$$ and $$P$$ with the

horizontal, and $$P$$ is the force applied.

Substitute $$900 \mathrm{~N}$$ for $$P, 36.87^{\circ}$$ for $$\phi, 55^{\circ}$$ for $$\theta, 600 \mathrm{~mm}$$ for $$a,$$ and $$300 \mathrm{~mm}$$ for $$b$$.

$$\begin{array}{l} F \sin \phi \times a=P \sin \theta \times(a+b) \\ F=\frac{900 \sin 55^{\circ} \times(600+300)}{\sin 36.87^{\circ} \times 600} \\ =1843.08 \mathrm{~N} \end{array}$$

Calculate the normal stress $$\sigma$$ in rod:

\begin{aligned} \sigma &=\frac{F}{\mathrm{~A}} \\ &=\frac{F}{\frac{\pi}{4} \times d_{1}^{2}} \end{aligned}

Here, $$d_{1}$$ is the diameter of a tie rod (1).

Substitute $$1843.08 \mathrm{~N}$$ for $$F,$$ and $$5 \mathrm{~mm}$$ for $$d_{1}$$.

\begin{aligned} \sigma &=\frac{1843.08}{\frac{\pi}{4} \times 5^{2}} \\ &=93.867 \mathrm{~N} / \mathrm{mm}^{2} \end{aligned}

Therefore, the normal stress in rod (1) is $$93.867 \mathrm{~N} / \mathrm{mm}^{2}$$

Calculate the shear stress $$\tau_{B}$$ on pin $$\mathrm{B}$$.

Pin $$B$$ is in double shear connection.

\begin{aligned} \tau_{B} &=\frac{F}{2 \times A_{p}} \\ &=\frac{F}{2 \times \frac{\pi}{4} \times d^{2}} \end{aligned}

Here, $$A_{p}$$ is area of the pin, and $$d_{2}$$ is the diameter of a pin.

Substitute $$1843.08 \mathrm{~N}$$ for $$F$$, and $$7 \mathrm{~mm}$$ for $$d_{2}$$.

\begin{aligned} \tau_{B} &=\frac{1843.243}{2 \times \frac{\pi}{4} \times 7^{2}} \\ &=23.94 \mathrm{~N} / \mathrm{mm}^{2} \end{aligned}

Therefore, the shear stress in pin at $$B$$ is $$23.94 \mathrm{~N} / \mathrm{mm}^{2}$$

Apply equation of equilibrium in $$x$$ -direction.

$$A_{x}+P \cos \theta-F \cos \phi=0$$

Substitute $$900 \mathrm{~N}$$ for $$P, 36.87^{\circ}$$ for $$\phi, 55^{\circ}$$ for $$\theta$$, and $$1843.08 \mathrm{~N}$$ for $$F$$.

\begin{aligned} A_{x} &+P \cos \theta-F \cos \phi=0 \\ A_{x} &=\left(1843.08 \times \cos 36.87^{\circ}\right)-\left(900 \times \cos 55^{\circ}\right) \\ &=958.243 \mathrm{~N} \end{aligned}

Apply equation of equilibrium in $$y$$ -direction.

$$A_{y}-P \sin \theta+F \sin \phi=0$$

Substitute $$900 \mathrm{~N}$$ for $$P, 36.87^{\circ}$$ for $$\phi, 55^{\circ}$$ for $$\theta,$$ and $$1843.08 \mathrm{~N}$$ for $$F$$.

$$\begin{array}{l} A_{y}-P \sin \theta+F \sin \phi=0 \\ A_{y}=\left(900 \times \sin 55^{\circ}\right)-\left(1843.08 \times \sin 36.87^{\circ}\right) \\ \quad=-368.61 \mathrm{~N} \end{array}$$

Calculate the reaction force at support $$A$$.

$$R_{A}=\sqrt{A_{x}^{2}+A_{y}^{2}}$$

$$\begin{array}{l} =\sqrt{(958.24)^{2}+(-368.61)^{2}} \\ =1026.69 \mathrm{~N} \end{array}$$

Calculate the shear stress $$\tau_{A}$$ on pin $$A$$.

Pin $$A$$ is in single shear connection.

\begin{aligned} \tau_{B} &=\frac{R_{A}}{A_{p}} \\ &=\frac{R_{A}}{\frac{\pi}{4} \times d^{2}} \end{aligned}

Substitute $$1026.69 \mathrm{~N}$$ for $$R_{A},$$ and $$7 \mathrm{~mm}$$ for $$d_{2}$$

$$\tau_{A}=\frac{R_{A}}{\frac{\pi}{4} \times d^{2}}$$

$$=\frac{1026.69}{\frac{\pi}{4} \times 7^{2}}$$

$$=26.67 \mathrm{~N} / \mathrm{mm}^{2}$$

Therefore, the shear stress in pin $$\mathrm{A}$$ is $$\sqrt{26.67 \mathrm{~N} / \mathrm{mm}^{2}}$$

The normal stress in rod (1) is $$93.867 \mathrm{~N} / \mathrm{mm}^{2}$$.

The shear stress in pin at $$B$$ is $$23.94 \mathrm{~N} / \mathrm{mm}^{2}$$.

The shear stress in pin $$\mathrm{A}$$ is $$\sqrt{26.67 \mathrm{~N} / \mathrm{mm}^{2}}$$.