Question

The rigid structure ABD is supported at B by a 36-mm-diameter tie rod (1) and at A by a 29-mm-diameter pin used in a single shear connection. The tie rod is connected at B and C by 24-mm-diameter pins used in double shear connections. Tie rod (1) has a yield strength of 260 MPa, and each of the pins has an ultimate shear strength of 320 MPa. A concentrated load of P = 50 kN acts as shown at D. Determine:

(a) the normal stress in rod (1).
(b) the shearing stress in the pins at A andB.

Answer 1

Given that,

\(P=50 \mathrm{kN}\)

Yeild strength of the tie rod, \(\sigma_{y}=260 \mathrm{MPa}\) Ultimate shear strength of the pin, \(\tau_{w^{t}}=320 \mathrm{MPa}\)

Diameter of the rod, \(d_{A B}=36 \mathrm{~mm}=0.036 \mathrm{~m}\) Diameter of the pin at \(\mathrm{A}(\) single shear \(), d_{a}=29 \mathrm{~mm}=0.029 \mathrm{~m}\)

Diameter of the pins at \(\mathrm{B}\) and \(\mathrm{C}(\) double shear \(), d_{b}=24 \mathrm{~mm}=0.024 \mathrm{~m}\) \(\theta=\tan ^{-1}\left(\frac{5.2}{8.9}\right)=30.3^{\circ}\)

Consider free body diagram of \(\mathrm{ABD}\)

\(\sum M_{A}=0\)

\(\left(T \sin 30.3^{\circ}\right)(8.9)-\left(50 \sin 60^{\circ}\right)(7.8)-\left(50 \cos 60^{\circ}\right)(4.1)=0\)

\(4.49 T=440.25\)

\(T=98.05 \mathrm{kN}\)

\(\sum F_{x}=0\)

\(A_{x}+50 \cos 60^{\circ}-\)

\(98.05 \sin 30.3^{\circ}=0\)

\(A_{x}=24.47 \mathrm{kN}\)

\(\sum F_{y}=0\)

\(A_{y}-50 \sin 60^{\circ}-98.05 \cos 30.3^{\circ}=0\)

\(A_{y}=127.96 \mathrm{kN}\)

Reaction at \(A, R_{A}=\sqrt{(24.47)^{2}+(127.96)^{2}}=130.279 \mathrm{kN}\)

(a) Normal stress in the tie rod, \(\sigma=\frac{T}{\frac{\pi}{4}\left(d_{A B}\right)^{2}}\)

\(\sigma=\frac{98.05 \times 10^{3}}{\frac{\pi}{4}(0.036)^{2}} \mathrm{~N} / \mathrm{m}^{2}\)

\(\sigma=96.328 \times 10^{6} \mathrm{~Pa}\)

\(\sigma=96.328 \mathrm{MPa}<\sigma_{y}\)

Therefore, the normal stress in the rod is \(\sigma=96.328 \mathrm{MPa}\)

(b)Shearing stress at pin A, \(\tau_{A}=\frac{R_{A}}{\frac{\pi}{4}\left(d_{a}\right)^{2}}=\frac{130.279 \times 10^{3}}{\frac{\pi}{4}(0.029)^{2}}\)

\(\tau_{A}=197.237 \mathrm{MPa}<\tau_{u t}\)

Therefore, the shearing stress in the pin at A is \(\tau_{A}=197.237 \mathrm{MPa}\) Shearing stress at pins \(B\) and \(C, \tau_{B}=\frac{T}{2 \times \frac{\pi}{4}\left(d_{b}\right)^{2}}=\frac{98.05 \times 10^{3}}{2 \times \frac{\pi}{4}(0.024)^{2}}\)

\(\tau_{B}=108.369 \mathrm{MPa}<\tau_{w t}\)

Therefore, the shearing stress in the pins at \(\mathrm{B}\) and \(\mathrm{C}\)

is \(\tau_{B}=108.369 \mathrm{MPa}\)