Question

Determine the required thickness of member BC and the diameter of the pins at A and B if the allowable normal stress for member BC sallow = 29ksi is and the allowable shear stress for the pins is tallow = 10 ksi.

Answer 1

Concepts and reason

The concepts required to solve the question are free body diagram, uniformly distributed load, moment, equilibrium condition of forces, normal stress and shear stress.

Stress: It is defined as an internal force which is exerted by a material per unit area.

Normal stress:

Internal resistive force per unit area produced in a body is called as normal stress. It is due to the material being subjected to axial load.

Shear stress: It is defined as stress due to shearing of the material when force is applied parallel to the cross-section of the material.

Moment: The product of force and the perpendicular distance of the force from which moment is taken.

Uniformly distributed load (U.D.L.): It is the distributed load having a constant value over a span on the beam. It is denoted by load per unit length.

Equilibrium of a system:

A system is said to be in equilibrium when the sum of all the external forces and moments are zero.

A system to be in equilibrium in three dimensions, the sum of all external moments about any point should be zero.

A system to be in equilibrium in three dimensions, the sum of all the external forces acting along $x$ , $y$ and $z$ directions have to be zero.

Free body diagram: It is used to show the magnitude and direction of all the forces that act in the system. It is a graphical representation of all the forces due to contact or non-contact.

Contact force: It is the force that arises due to the contact of the surface. It acts as normal to that contact.

Non-contact force: The force that arises due to gravity, electric change etc. is known as non-contact force. Force due to gravity always acts in a downward direction.

First of all, draw the free body diagram of the system and then resolve all the forces into horizontal and vertical directions. After the resolution of forces, apply the equilibrium condition of forces in order to find the unknown reaction or force acting on the member.

Use the expression of normal stress in order to calculate the thickness of the member and also, use the expression of the shear stress in order to calculate the diameter of the pins.

Fundamentals

Write the expression of the equilibrium condition of force in the horizontal direction.

$\sum {{F_x}} = 0$

Here, $\sum {{F_x}}$ is the sum of all the forces acting in the horizontal direction.

Write the expression of the equilibrium condition of force in the vertical direction.

$\sum {{F_y}} = 0$

Here, $\sum {{F_y}}$ is the sum of all the forces acting in the vertical direction.

Write the expression of the equilibrium condition of the moment.

$\sum M = 0$

Here, $\sum M$ is the sum of all the moment about any point.

Write the expression of the moment.

$M = \left( {{\rm{Force}}} \right)\left( {{\rm{perpendicular distance}}} \right)$

Write the expression of the stress normal.

$\sigma = \frac{F}{A}$

Here, $\sigma$ is the stress produced in the material, $F$ is the applied force to the material and $A$ is the area of material at which the force exerts.

Write the expression of shear stress.

$\tau = \frac{F}{A}$

Here, $\tau$ is the shear stress, $F$ is the shearing force.

Write the expression of the cross-sectional area of the bar.

$A = \frac{{\pi {d^2}}}{4}$

Here, $A$ is the cross-section area of the bar and $d$ is the diameter of the bar.

Write the expression of the net force of the two forces, which are perpendicular to each other.

${F_R} = \sqrt {{F_x} + {F_y}}$

Here, ${F_x}$ is the force acting in the x-axis direction and ${F_y}$ is the force which is acting in the y-axis direction.

Uniform distributed load: In case of uniformly distributed load, the net force acts at mid of the uniform distributed load. This is due to the centroid of the uniformly distributed load being at mid of the uniform distributed load. Therefore, the uniform distributed load is replaced by a point load at the mid of the span of the uniform distributed load. The net force due to uniformly distributed load is calculated by multiplying of the intensity of the uniform distributed load to the length of the uniform distributed load.

Sign Conventions:

The clockwise moment is taken as the negative and counterclockwise moment is taken as positive.

Force pointing in vertically upward direction is taken as positive and force in vertically downward direction is taken as negative.

Similarly, force pointing in the right direction is taken as positive and force pointing in the left direction is taken as negative.

Draw the free body diagram of the member $AB$ .

From above free body diagram of the member $AB$ , take the moment about the point $A$ and apply the equilibrium condition of the moment. $\begin{array}{l}\\{\sum M _A} = 0\\\\\left[ {\left( {2{\rm{ kip/ft}}} \right)\left( {8{\rm{ ft}}} \right)} \right]\left( {4{\rm{ ft}}} \right) - \left( {{F_{BC}}\sin 60^\circ } \right)\left( {{\rm{8 ft}}} \right) = 0\\\\\left( {{F_{BC}}\sin 60^\circ } \right)\left( {{\rm{8 ft}}} \right) = \left[ {\left( {2{\rm{ kip/ft}}} \right)\left( {8{\rm{ ft}}} \right)} \right]\left( {4{\rm{ ft}}} \right)\\\\{F_{BC}} = 9.2376{\rm{ kip}}\\\end{array}$

Here, ${F_{BC}}$ is the force which is acting on the member $BC$ .

From above free body diagram of the member $AB$ , write the net force acting in member AB in the vertical direction and apply equilibrium condition of forces.

$\begin{array}{l}\\\sum {{F_y}} = 0\\\\{F_{BC}}\sin 60^\circ + {A_y} - \left( {2{\rm{ kip/ft}}} \right)\left( {8{\rm{ ft}}} \right) = 0\\\end{array}$

Here, ${A_y}$ is the vertical reaction at the point $A$ .

Substitute $9.2376{\rm{ kip}}$ for ${F_{BC}}$ .

$\begin{array}{l}\\\left( {9.2376{\rm{ kip}}} \right)\sin 60^\circ + {A_y} - \left( {2{\rm{ kip/ft}}} \right)\left( {8{\rm{ ft}}} \right) = 0\\\\{A_y} = 8{\rm{ kip}}\\\end{array}$

From above free body diagram of the member $AB$ , write the net force acting in member AB in the horizontal direction and apply the equilibrium condition of forces.

$\begin{array}{l}\\\sum {{F_x}} = 0\\\\{F_{BC}}\cos 60^\circ - {A_x} = 0\\\\{A_x} = {F_{BC}}\cos 60^\circ \\\end{array}$

Here, ${A_x}$ is the horizontal reaction at the point .

Substitute for ${F_{BC}}$ .

$\begin{array}{c}\\{A_x} = \left( {9.2376{\rm{ kip}}} \right)\cos 60^\circ \\\\ = 4.6188{\rm{ kip}}\\\end{array}$

Calculate the net force acting at a point $A$ using the Pythagoras theorem.

${A_R} = \sqrt {{{\left( {{A_x}} \right)}^2} + {{\left( {{A_x}} \right)}^2}}$

Substitute $4.6188{\rm{ kip}}$ for ${A_x}$ and $6.7624{\rm{ kip}}$ for ${A_y}$ .

$\begin{array}{c}\\{A_R} = \sqrt {{{\left( {4.6188{\rm{ kip}}} \right)}^2} + {{\left( {{\rm{8 kip}}} \right)}^2}} \\\\ = 9.237{\rm{ kip}}\\\end{array}$

Calculate the thickness of the member $BC$ .

$\begin{array}{c}\\{\sigma _{allow}} = \frac{{{F_{BC}}}}{A}\\\\ = \frac{{{F_{BC}}}}{{wt}}\\\end{array}$

Further, solve,

$t = \frac{{{F_{BC}}}}{{w{\sigma _{allow}}}}$ ……. (1)

Here, ${\sigma _{allow}}$ is the normal allowable stress of member BC, $w$ is the width of the member $BC$ and $t$ is the thickness of the member $BC$ .

Substitute $29{\rm{ ksi}}$ for ${\sigma _{allow}}$ , $1.5{\rm{ in}}$ for $w$ and $9.2376{\rm{ kip}}$ for ${F_{BC}}$ .

$\begin{array}{c}\\t = \frac{{\left( {9.2376{\rm{ kip}}} \right)}}{{\left( {1.5{\rm{ in}}} \right)\left( {29{\rm{ ksi}}} \right)}}\\\\ = 0.2123{\rm{ in}}\\\end{array}$

In case of designing, the thickness will be taken as $0.25{\rm{ in or }}\frac{{\rm{1}}}{4}{\rm{ in}}$ .

Calculate the diameter of the pin $A$ .

${\tau _{allow}} = \frac{{{A_R}}}{{\left( {\frac{{\pi {d^2}}}{4}} \right)}}$

Here, ${\tau _{allow}}$ is the allowable shear stress, and $d$ is the diameter of the pin $A$ .

Substitute ${\rm{10 ksi}}$ for ${\tau _{allow}}$ , and $9.237{\rm{ kip}}$ for ${A_R}$ .

$\begin{array}{l}\\{\rm{10 ksi}} = \frac{{9.237{\rm{ kip}}}}{{\left( {\frac{{\pi {d^2}}}{4}} \right)}}\\\\d = \sqrt {\frac{{4\left( {9.237{\rm{ kip}}} \right)}}{{\pi \left( {{\rm{10 ksi}}} \right)}}} \\\\ = 1.0845{\rm{ in}}\\\end{array}$

In case of designing, the diameter of pin A will be taken as $1.125{\rm{ in}}$ .

The shearing at the pin $B$ is double. Therefore, calculate the diameter of the pin $B$ . ${\tau _{allow}} = \frac{{{F_{BC}}}}{{2\left( {\frac{{\pi {d^2}}}{4}} \right)}}$

Here, ${\tau _{allow}}$ is the allowable shear stress, and $d$ is the diameter of the pin $B$ .

Substitute ${\rm{10 ksi}}$ for ${\tau _{allow}}$ , and $9.2376{\rm{ kip}}$ for ${F_{BC}}$ .

$\begin{array}{c}\\{\rm{10 ksi}} = \frac{{9.2376{\rm{ kip}}}}{{2\left( {\frac{{\pi {d^2}}}{4}} \right)}}\\\\d = \sqrt {\frac{{4\left( {9.2376{\rm{ kip}}} \right)}}{{\pi \left( {{\rm{20 ksi}}} \right)}}} \\\\ = 0.7568{\rm{ in}}\\\end{array}$

In case of designing, the diameter of pin B will be taken as $0.8125{\rm{ in}}$ .

Ans:

The thickness of the member $BC$ is $0.25{\rm{ in}}$ .