Question

Determine the largest load P that can be applied to the frame without causing either the

average normal stress or the average shear stress at section a-a to exceed σ = 150 Mpa and τ =

60 Mpa, respectively. Member CB has a square cross section of 25 mm on each side.

Answer 1

Concepts and reason

The concepts required to solve the question are equilibrium condition of forces, free body diagram, normal stress, shear stress, geometry and trigonometrical relations.

Free body diagram: It is used to show the relative magnitude and direction of all the forces that act on the system. It is a graphical representation of all the forces due to contact or non-contact. To draw the free body diagram, replace all supports with the forces that prevent the translation of body in a given direction and also place a force in a downward direction due to the weight of an object.

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: It is the force that arises due to gravity, electrical charge etc. The non-contact force due to gravity always acts in a downward direction.

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.

Normal stress:

It is internal resistive force per unit area produced in a body due to externally applied force is called as normal stress. It is simply called stress.

Shear stress:

It is internal resistive force per unit area produced in a body due to externally applied shear force, which causes deformation of angle when shear force is applied parallel to the body.

First of all, draw free body diagram of point C of the frame and then resolve all the forces into horizontal and vertical directions.

Apply the equilibrium condition of force at point C on the frame along the vertical direction in order to calculate the force acting in the member BC after substituting the value of required parameters.

Take a section a-a on the member BC and draw the free body diagram of member BC after taking section. And then calculate the normal force and shear force that are acting in the member BC at section a-a by applying the equilibrium condition of forces on the member BC in both directions.

Calculate the magnitude of the maximum applied a load by using the expressions of stress due to normal and shear stresses. And then Choose the maximum value of applied load among calculated values of applied load in order to find the maximum applied load.

Fundamentals

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

$\sum {{F_x}} = 0$

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

Write the expression of the normal stress.

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

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

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

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

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

Write the expression of the shear stress:

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

Here, $V$ is the shearing force.

Sign Conventions:

Take a positive sign of the force when a force acts in vertically upward direction and take negative of the force when a force acts in vertically downward direction.

Take a positive sign of the force when a force acts in the right direction and take negative of the force when a force acts in the left direction.

Draw free body diagram of point C of the frame.

In the above free body diagram, ${F_{BC}}$ is the force acting in the member BC, ${F_{AC}}$ is the force acting on the member AC and $\theta$ is the angle of the force ${F_{BC}}$ from negative x-axis in a clockwise direction.

From above diagram,

$\tan \theta = \frac{{AB}}{{AC}}$

Substitute $2{\rm{ m}}$ for $AB$ and $1.5{\rm{ m}}$ for $AC$ .

$\begin{array}{l}\\\tan \theta = \frac{{2{\rm{ m}}}}{{1.5{\rm{ m}}}}\\\\\theta = {\tan ^{ - 1}}\left( {1.33} \right)\\\\ = 53^\circ \\\end{array}$
‎

From above free body diagram, apply the equilibrium condition of force at point C on the frame along the y-direction:

$\begin{array}{l}\\\Sigma {F_y} = 0\\\\{F_{BC}}\sin \theta - P = 0\\\\{F_{BC}} = \frac{P}{{\sin \theta }}\\\end{array}$

Substitute $53^\circ$ for $\theta$ .

$\begin{array}{c}\\{F_{BC}} = \frac{P}{{\sin 53^\circ }}\\\\ = 1.25P\\\end{array}$

Draw the free body diagram of the member BC of the frame.

In the above diagram, ${N_a}$ is the normal force acting in the member BC during taking the section a-a and ${V_a}$ is the shear force acting in the member BC during taking the section a-a.

From above diagram, apply equilibrium condition of force on the member BC of the frame along the horizontal direction:

$\begin{array}{l}\\\Sigma {F_x} = 0\\\\{N_a} - {F_{BC}}\cos 53^\circ = 0\\\\{N_a} = 0.60{F_{BC}}\\\end{array}$

Substitute $1.25P$ for ${F_{BC}}$ .

$\begin{array}{c}\\{N_a} = 0.60\left( {1.25P} \right)\\\\ = 0.75P\\\end{array}$

From above diagram, apply equilibrium condition of force on the member BC of the frame along the vertical direction:

$\begin{array}{c}\\\Sigma {F_y} = 0\\\\{F_{BC}}{\rm{sin}}53^\circ - {V_a} = 0\\\\{V_a} = 0.80{F_{BC}}\\\end{array}$

Substitute $1.25P$ for ${F_{BC}}$ .

$\begin{array}{c}\\{V_a} = 0.80\left( {1.25P} \right)\\\\ = P\\\end{array}$

Draw the diagram of the member BC.

From above diagram, calculate the depth of the member BC at section a-a.

$\begin{array}{l}\\\sin 37^\circ = \frac{b}{d}\\\\d = \frac{b}{{\sin 37^\circ }}\\\end{array}$

Substitute ${\rm{25 mm}}$ for $b$ .

$\begin{array}{c}\\d = \frac{{25{\rm{ mm}}}}{{\sin 37^\circ }}\\\\ = 41.55{\rm{ mm}}\\\end{array}$

Calculate the cross-sectional area of the member BC.

${A_a} = bd$

Here, $b$ is the width of the member BC at section a-a and $d$ is the depth of the member BC at section a-a.

Substitute ${\rm{25 mm}}$ for $b$ and $41.55{\rm{ mm}}$ for $d$ .

$\begin{array}{c}\\{A_a} = \left( {{\rm{25 mm}}} \right)\left( {41.55{\rm{ mm}}} \right)\\\\ = 1038.75{\rm{ m}}{{\rm{m}}^{\rm{2}}}\left( {\frac{{1{\rm{ }}{{\rm{m}}^{\rm{2}}}}}{{{{10}^6}{\rm{ m}}{{\rm{m}}^{\rm{2}}}}}} \right)\\\\ \approx 1.04 \times {10^{ - 3}}{\rm{ }}{{\rm{m}}^{\rm{2}}}\\\end{array}$

Calculate the applied load by using the expression of the normal stress.

$\sigma = \frac{{{N_a}}}{{{A_a}}}$

Substitute $0.75P$ for ${N_a}$ , $1.04 \times {10^{ - 3}}{\rm{ }}{{\rm{m}}^{\rm{2}}}$ for ${A_a}$ and $150{\rm{ MPa}}$ for $\sigma$ .

$\begin{array}{l}\\150{\rm{ MPa}} = \frac{{0.75P}}{{1.04 \times {{10}^{ - 3}}{\rm{ }}{{\rm{m}}^{\rm{2}}}}}\\\\150{\rm{ MPa}}\left( {\frac{{{{10}^6}{\rm{ N/}}{{\rm{m}}^{\rm{2}}}}}{{1{\rm{ MPa}}}}} \right) = \frac{{0.75P}}{{1.04 \times {{10}^{ - 3}}{\rm{ }}{{\rm{m}}^{\rm{2}}}}}\\\\P = 208000{\rm{ N}}\left( {\frac{{1{\rm{ kN}}}}{{{{10}^3}{\rm{ N}}}}} \right)\\\\P = 208{\rm{ kN}}\\\end{array}$

Calculate the applied load by using the expression of the normal stress.

$\tau = \frac{{{V_a}}}{{{A_a}}}$

Substitute $P$ for ${V_a}$ , $1.04 \times {10^{ - 3}}{\rm{ }}{{\rm{m}}^{\rm{2}}}$ for ${A_a}$ and $60{\rm{ MPa}}$ for $\sigma$ .

$\begin{array}{l}\\60{\rm{ MPa}} = \frac{P}{{1.04 \times {{10}^{ - 3}}{\rm{ }}{{\rm{m}}^{\rm{2}}}}}\\\\60{\rm{ MPa}}\left( {\frac{{{{10}^6}{\rm{ N/}}{{\rm{m}}^{\rm{2}}}}}{{1{\rm{ MPa}}}}} \right) = \frac{P}{{1.04 \times {{10}^{ - 3}}{\rm{ }}{{\rm{m}}^{\rm{2}}}}}\\\\P = 62.40 \times {10^3}{\rm{ N}}\left( {\frac{{1{\rm{ kN}}}}{{{{10}^3}{\rm{ N}}}}} \right)\\\\P = 62.40{\rm{ kN}}\\\end{array}$

Choose the minimum value among the two calculated values of applied load in order to find the maximum applied load.

Therefore, $P = 62.4{\rm{ kN}}$

Ans:

The largest value of the applied load is $62.40{\rm{ kN}}$ .