Question

Beam AB is supported as shown in the figure. Tie rod (1) has a diameter of 60 mm, and it is attached at B and C with 24 mm diameter double-shear pin connections. The pin connection at A consists of a 37 mm diameter single-shear pin. The pins at A, B, and C each have an ultimate shear strength of 500 MPa, and tie rod (1) has a yield strength of 280 MPa. A uniformly distributed load of w is applied to the beam as shown. Dimensions are a=3.80 m, b=0.47 m, and c=1.75 m. A minimum factor of safety of 2.5 is required for all components. What is the maximum loading w that may be applied to the structure?

Answer 1

Draw the free-body diagram of the rigid beam \(A B\).

Calculate the angle in tie rod (1) with respect to the horizontal axis. \(\tan \theta=\frac{c}{a}\)

\(\tan \theta=\frac{1.75}{3.8}\)

$$ \theta=24.727^{\circ} $$

Calculate the moment equation at joint \(A\). \(\Sigma M_{A}=0\)

\(\left(F_{1} \cos \theta\right) b+\left(F_{1} \sin \theta\right) a=w a\left(\frac{a}{2}\right)\)

\(\left(F_{1} \cos 24.727^{\circ}\right) b+\left(F_{1} \sin 24.727^{\circ}\right) a=w a\left(\frac{a}{2}\right)\)

\(0.9083 F_{1} b+0.4183 F_{1} a=w\left(\frac{a^{2}}{2}\right)\)

\(F_{1}(0.9083 b+0.4183 a)=w\left(\frac{a^{2}}{2}\right)\)

\(w=2 F_{1}\left(\frac{0.9083 b+0.4183 a}{a^{2}}\right)\)

\(w=2 F_{1}\left(\frac{0.9083 \times 0.47+0.4183 \times 3.8}{3.8^{2}}\right)\)

\(w=0.279 F_{1}\)

\(F_{1}=3.584 w\)

Calculate the equilibrium equations in vertical axis. \(\Sigma F_{y}=0\)

\(A_{y}+F_{1} \sin \theta-w a=0\)

\(A_{y}=w a-F_{1} \sin \theta\)

\(A_{y}=w a-F_{1} \sin 24.727^{\circ}\)

\(\quad=w(3.8)-w(3.584) \sin 24.727^{\circ}\)

\(=(2.3) w\)

Calculate the equilibrium equations in horizontal axis. \(\Sigma F_{x}=0\)

\(A_{x}-F_{1} \cos \theta=0\)

\(A_{x}=F_{1} \cos \theta=3.584 w \cos 24.727^{\circ}\)

\(A_{x}=3.255 w\)

Calculate the resultant force of the pin reaction at \(A\) \(|A|=\sqrt{A_{x}^{2}+A_{y}^{2}}=\sqrt{(3.255 w)^{2}+(2.3 w)^{2}}\)

\(=w \sqrt{(3.255)^{2}+(2.3)^{2}}\)

\(=(3.985) w\)

(a)

Calculate the allowable normal stress for tie rod (1).

$$ \begin{aligned} \sigma_{\text {allow }} &=\frac{\sigma_{\Gamma}}{F S} \\ \sigma_{\text {allow }} &=\frac{280}{2.5} \\ &=112 \mathrm{MPa} \end{aligned} $$

Calculate the cross-sectional area of the tie rod.

\(A_{1}=\frac{\pi}{4}(d)^{2}\)

\(\begin{aligned} A_{1} &=\frac{\pi}{4}(60)^{2} \\ &=2827.433 \mathrm{~mm}^{2} \end{aligned}\)

Calculate the allowable force in the tie rod. \(\sigma_{\text {allow }}=\frac{F_{1}}{A_{1}}\)

$$ \begin{aligned} & F_{1}=\sigma_{\text {allow }} A_{1} \\ F_{1} &=(112)(2827.433) \\ &=316672.496 \mathrm{~N} . \\ &=316.672 \mathrm{kN} \end{aligned} $$

Calculate the distributed load on the beam while considering the rod (1).

\(w \leq(0.279) F_{1}\)

\(w \leq(0.279) 316.672\)

\(=88.351 \mathrm{kN} / \mathrm{m}\)

Calculate the allowable shear stress of the pin shear at \(B\) and \(C\). \(\tau_{\text {allow }}=\frac{\tau_{U}}{F S}\)

\(\tau_{\text {allow }}=\frac{500}{2.5}=200 \mathrm{MPa}\)

Calculate the cross-sectional area of the pin. \(A_{p i n}=\frac{\pi}{4}\left(d_{p i n}\right)^{2}\)

\(A_{\text {pin }}=\frac{\pi}{4}(24)^{2}=452.3893 \mathrm{~mm}^{2}\)

Calculate the minimum area required for pin \(B\) and \(C\). \(A_{V}=2 A_{p^{m}}\)

\(A_{V}=2(452.3893)=904.7786 \mathrm{~mm}^{2}\)

Calculate the strength of the double pin shear pin connection. \(V=\tau_{\text {olluw }} A_{t}\)

\(\begin{aligned} V &=(200)(904.7786) \\ &=180955.72 \mathrm{~N} \\ &=180.955 \mathrm{kN} \end{aligned}\)

Calculate the distributed load on the beam while considering the double shear pins at \(B\) and

\(C\)

Calculate the cross-sectional area of the pin at \(A\).

$$ \begin{aligned} A_{V} &=A_{p i n} \\ &=\frac{\pi}{4}\left(d_{p i n, A}\right)^{2} \\ A_{V} &=\frac{\pi}{4}(37)^{2} \\ &=1075.21 \mathrm{~mm}^{2} \end{aligned} $$

Calculate the strength of the double pin shear pin connection. \(V=\tau_{\text {ollow }} A_{V}\)

\(\begin{aligned} V &=(200)(1075.21) \\ &=215042 \mathrm{~N} \\ &=215.042 \mathrm{kN} \end{aligned}\)

Calculate the distributed load on the beam while considering the single shear pins at \(A\). \(w \leq \frac{\left|F_{1}\right|}{|A|}\)

\(\begin{aligned} w \leq \frac{215.042}{3.985} &\left(\because F_{1}=V=215.042 \mathrm{kN}\right) \\=53.962 \mathrm{kN} / \mathrm{m} & \end{aligned}\)

Calculate the maximum loading applied to the structure by comparing with the loads at tie rod (1), double shear pins at \(B\) and \(C\) and single shear pin at \(A\). \(\begin{aligned} w_{\text {rod }(1)} & \leq 88.351 \mathrm{kN} / \mathrm{m} \\ w_{\text {pin } B, C} & \leq 50.486 \mathrm{kN} / \mathrm{m} \\ w_{\text {pind }} & \leq 53.962 \mathrm{kN} / \mathrm{m} \end{aligned}\)

Choose minimum of obtained values \(w \leq 50.486 \mathrm{kN} / \mathrm{m}\)