Question

The beam is supported by two rods AB and CD that have cross-sectional areas of 18 mm² and 9 mm², respectively.

Determine the position d of the 6-kN load so that the average normal stress in each rod is the same.

Answer 1

Concepts and reason

Stress: When an alloy is loaded with a force, it produces a stress which makes the alloy to deform.

Strain: When the alloy is stressed in specific direction the response of the system for this stress is strain. The deformation of an alloy in specified direction divided by the original length gives the strain the alloy.

Free body diagram is the representation of all forces acting on the body.

When load is applied in a specific direction the body deforms in that direction.

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.

The change in position of a point under action of load is called as displacement.

Hooke’s law states that stress is directly proportional to strain.

Fundamentals

Calculate the stress using the following relation:

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

Here, the force applied on the body is $F$ and the cross sectional area is $A$ .

Write the equilibrium equations.

$\begin{array}{l}\\{F_R} = \sum {\bf{F}} = 0\\\\{\left( {{M_R}} \right)_O} = \sum {{{\bf{M}}_O}} = 0\\\end{array}$

Here, the resultant force is ${F_R}$ and the resultant moment about any arbitrary point is ${\left( {{M_R}} \right)_O}$ .

Draw the free body diagram as shown below:

Calculate the average normal stress acting in the rod AB.

${\sigma _{AB}} = \frac{{{F_{AB}}}}{{{A_{AB}}}}$

Here, ${\sigma _{AB}}$ is average normal stress in the rod AB, and ${A_{AB}}$ is the cross-sectional area of rod AB.

Substitute $\sigma$ for ${\sigma _{AB}}$ and $18 \times {10^{ - 6}}{\rm{ }}{{\rm{m}}^2}$ for ${A_{AB}}$ .

$\sigma = \frac{{{F_{AB}}}}{{18 \times {{10}^{ - 6}}}}$ ...... (1)

Calculate the average normal stress acting in the rod CD.

${\sigma _{CD}} = \frac{{{F_{CD}}}}{{{A_{CD}}}}$

Here, ${\sigma _{CD}}$ is average normal stress in the rod CD and ${A_{CD}}$ is the cross-sectional area of rod CD.

Substitute $\sigma$ for ${\sigma _{CD}}$ and $9 \times {10^{ - 6}}{\rm{ }}{{\rm{m}}^2}$ for ${A_{CD}}$ .

$\sigma = \frac{{{F_{CD}}}}{{9 \times {{10}^{ - 6}}}}$ ...... (2)

Equate the equations (1) and (2).

$\begin{array}{l}\\\frac{{{F_{AB}}}}{{18 \times {{10}^{ - 6}}}} = \frac{{{F_{CD}}}}{{9 \times {{10}^{ - 6}}}}\\\\{F_{AB}} = 2{F_{CD}}\\\end{array}$

Equate the sum of moments about O to zero.

$\begin{array}{l}\\\sum {{M_O}} = 0\\\\\left( {{F_{CD}} \times \left( {3 - d} \right)} \right) - \left( {{F_{AB}} \times d} \right) = 0\\\end{array}$

Substitute $2{F_{CD}}$ for ${F_{AB}}$ .

$\begin{array}{l}\\\left( {{F_{CD}} \times \left( {3 - d} \right)} \right) - \left( {2{F_{CD}} \times d} \right) = 0\\\\{F_{CD}}\left( {3 - d - 2d} \right) = 0\\\\3 - 3d = 0\\\\d = 1{\rm{ m}}\\\end{array}$

Ans:

Therefore, the position of the load is $1{\rm{m}}$ .