The beam is supported by a pin at A and a short link BC. If P = 10kN , determine the average shear stress developed...

Draw the free body diagram of the beam BA.

Take moment about A.

$\begin{array}{l}\\\sum {{M_A} = 0} \\\\ - \left( {{F_B}\sin 30^\circ \times 5} \right) + \left( {10 \times 4.5} \right) + \left( {40 \times 3.5} \right) + \left( {40 \times 2} \right) + \left( {20 \times 0.5} \right) = 0\\\\{F_B} = 110{\rm{ kN}}\\\end{array}$

Apply equilibrium conditions along x axis.

$\begin{array}{l}\\\sum {{F_x} = 0} \\\\{F_{Ax}} - 110\cos 30^\circ = 0\\\\{F_{Ax}} = 95.26{\rm{ kN}}\\\end{array}$

Apply equilibrium conditions along y axis.

$\begin{array}{l}\\\sum {{F_y} = 0} \\\\{F_{Ay}} + 110\sin 30^\circ - 10 - 40 - 40 - 20 = 0\\\\{F_{Ay}} = 55{\rm{ kN}}\\\end{array}$

Determine the total reaction force at pin A.

${F_A} = \sqrt {{{\left( {{F_{Ax}}} \right)}^2} + {{\left( {{F_{Ay}}} \right)}^2}}$

Substitute $55{\rm{ kN}}$ for ${F_{Ay}}$ , and $95.26{\rm{ kN}}$ for ${F_{Ax}}$ .

$\begin{array}{c}\\{F_A} = \sqrt {{{\left( {{F_{Ax}}} \right)}^2} + {{\left( {{F_{Ay}}} \right)}^2}} \\\\ = \sqrt {{{\left( {55} \right)}^2} + {{\left( {95.26} \right)}^2}} \\\\ = 110\;{\rm{kN}}\\\end{array}$

Consider the free body diagram of the link BC and pins at A, B and C.

Consider the free body diagram of link B and write equilibrium of forces along the link.

${F_B} = 110{\rm{ kN}}$

Calculate the shear stress in the pins due to double shear.

$\begin{array}{c}\\{\left( {{\tau _B}} \right)_{avg}} = \frac{{{F_B}}}{{2 \times {A_C}}}\\\\ = \frac{{{F_B}}}{{2 \times \frac{\pi }{4}{D^2}}}\\\end{array}$

Here, diameter of the pin is D.

Substitute 17mm for D and $110{\rm{ kN}}$ for ${F_B}$ .

$\begin{array}{c}\\{\left( {{\tau _B}} \right)_{avg}} = \frac{{110 \times {{10}^3}}}{{2 \times \frac{\pi }{4} \times {{\left( {17} \right)}^2}}}\\\\ = 242.312\,{\rm{N/m}}{{\rm{m}}^2}\\\end{array}$

Consider force equilibrium in link BC.

$\begin{array}{l}\\{F_B} = 110{\rm{ kN}}\\\\{F_C} = 110{\rm{ kN}}\\\end{array}$

Calculate the shear stress in the pins due to double shear.

$\begin{array}{c}\\{\left( {{\tau _C}} \right)_{avg}} = \frac{{{F_C}}}{{2 \times {A_C}}}\\\\ = \frac{{{F_C}}}{{2 \times \frac{\pi }{4}{D^2}}}\\\end{array}$

Substitute 17mm for D and $110{\rm{ kN}}$ for ${F_C}$ .

$\begin{array}{c}\\{\left( {{\tau _C}} \right)_{avg}} = \frac{{110 \times {{10}^3}}}{{2 \times \frac{\pi }{4} \times {{17}^2}}}\\\\ = 242.312\,{\rm{N/m}}{{\rm{m}}^2}\\\end{array}$

Calculate the average shear stress in pin at A.

${\left( {{\tau _A}} \right)_{avg}} = \frac{{{F_A}}}{{2 \times {A_A}}}$

Substitute 17mm for D and $110{\rm{ kN}}$ for ${F_A}$ .

$\begin{array}{c}\\{\left( {{\tau _A}} \right)_{avg}} = \frac{{110 \times {{10}^3}}}{{2 \times \frac{\pi }{4} \times {{17}^2}}}\\\\ = 242.312\,{\rm{N/m}}{{\rm{m}}^2}\\\end{array}$

The average shear stress in the pin at point B is $242.312\,{\rm{N/m}}{{\rm{m}}^2}$ .