- #1
jojosg
- 37
- 5
- Homework Statement
- The illustrated rocker arm is subjected to the action of loads F1 and F2. The given loads are F1=50kN and F2,=35.4kN. The allowable shear stress τ=100 MPa ,and the allowable bearing stress σ=240 MPa. Determine the diameter d of the pin B.
- Relevant Equations
- τ= V/A, σ = F/t*d
Hi, just need somebody to check my homework solution. Not entirely sure of my answer. My teacher used Mechanics of Materials by RC Hibbler, but I could not find an example question to solve this homework question.
Equilibrium at pin B
Take moments about B:
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F_1 \cdot 40 = F_2 \cdot (80\sin45^\circ)
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50 \cdot 40 = 35.4 \cdot 56.57 = 2000\ \text{kN·mm}
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Force equilibrium:
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\sum F_x = 0:\quad R_{Bx} + 35.4\cos45^\circ - 50 = 0 \quad\Rightarrow\quad R_{Bx} = 25\ \text{kN}
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\sum F_y = 0:\quad R_{By} - 35.4\sin45^\circ = 0 \quad\Rightarrow\quad R_{By} = 25\ \text{kN}
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R_B = \sqrt{25^2+25^2}=25\sqrt{2}=35.36\ \text{kN}
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Pin design (double shear) – two outer plates t=6 mm, center plate t=10 mm
Shear:
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\tau = \frac{R_B/2}{\pi d^2/4} = \frac{2R_B}{\pi d^2} \le 100\ \text{MPa} \quad\Rightarrow\quad d \ge \sqrt{\frac{2\times35360}{\pi\times100}} = 15.0\ \text{mm}
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Bearing – center plate:
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\sigma_b = \frac{R_B}{d\cdot10} \le 240\ \text{MPa} \quad\Rightarrow\quad d \ge \frac{35360}{2400} = 14.73\ \text{mm}
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Bearing – outer plates:
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\sigma_b = \frac{R_B/2}{d\cdot6} = \frac{R_B}{12d} \le 240\ \text{MPa} \quad\Rightarrow\quad d \ge \frac{35360}{2880} = 12.28\ \text{mm}
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Result shear governs
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\boxed{d = 15\ \text{mm}}
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Equilibrium at pin B
Take moments about B:
##
F_1 \cdot 40 = F_2 \cdot (80\sin45^\circ)
##
##
50 \cdot 40 = 35.4 \cdot 56.57 = 2000\ \text{kN·mm}
##
Force equilibrium:
##
\sum F_x = 0:\quad R_{Bx} + 35.4\cos45^\circ - 50 = 0 \quad\Rightarrow\quad R_{Bx} = 25\ \text{kN}
##
##
\sum F_y = 0:\quad R_{By} - 35.4\sin45^\circ = 0 \quad\Rightarrow\quad R_{By} = 25\ \text{kN}
##
##
R_B = \sqrt{25^2+25^2}=25\sqrt{2}=35.36\ \text{kN}
##
Pin design (double shear) – two outer plates t=6 mm, center plate t=10 mm
Shear:
##
\tau = \frac{R_B/2}{\pi d^2/4} = \frac{2R_B}{\pi d^2} \le 100\ \text{MPa} \quad\Rightarrow\quad d \ge \sqrt{\frac{2\times35360}{\pi\times100}} = 15.0\ \text{mm}
##
Bearing – center plate:
##
\sigma_b = \frac{R_B}{d\cdot10} \le 240\ \text{MPa} \quad\Rightarrow\quad d \ge \frac{35360}{2400} = 14.73\ \text{mm}
##
Bearing – outer plates:
##
\sigma_b = \frac{R_B/2}{d\cdot6} = \frac{R_B}{12d} \le 240\ \text{MPa} \quad\Rightarrow\quad d \ge \frac{35360}{2880} = 12.28\ \text{mm}
##
Result shear governs
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\boxed{d = 15\ \text{mm}}
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