Question

The beam is supported by a pin at A and a short link BC.

If P = 10kN, determine the average shear stress developed in the pins atA, B, and C . All pins are in double shear as shown, and each has a diameter of 17mm

Answer 1

Concepts and reason

A body is in equilibrium if vector sum of all the forces is equal to zero or moment of all force vectors about any point is equal to zero.

Magnitude of moment can be calculated by multiplying the magnitude of force with the perpendicular distance between the point where the force is applied and the point where the moment to be found.

Shear stress:

It is the ratio of a force and area tangential to the action of force.

Double shear stress:

Consider a pin on which two supports are connected.

Here, two shear stress acts on the pin at location 1 and 2. Shear 1 on pin due to left support and top right support, shear 2 due to left support and bottom right support.

So double shear is given as follows:

${\left( \tau \right)_{avg}} = \frac{F}{{2A}}$

Fundamentals

Write the equilibrium equation of force vector.

$\begin{array}{l}\\\sum {{\bf{F}} = 0} \\\\{{\bf{F}}_1} + {{\bf{F}}_2} + .... + {{\bf{F}}_n} = 0\\\end{array}$

Moment of can be calculated as follows:

It can be calculated by multiplying the magnitude of force with the distance between the point of application of force and point about which moment is to be calculated.

General sign convection for moment: The moment is considered positive in counter-clockwise direction and negative in clockwise direction.

General sign convection for axis: Distance along the axis is positive and opposite to the axis is negative.

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 ${F_A}$ .

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

Ans:

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