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 \(\sqrt{26.67 \mathrm{~N} / \mathrm{mm}^{2}}\).

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