Question

If the beam is subjected to a bending moment of M - 20 kN . m, determine the maximum bending stress in the beam.

Answer 1

Concepts and reason

Stress:

When a force is applied on the body, a resistance is offered by the body to the deformation which is called as stress.

Moment of Inertia:

It is the tendency of the body to resist angular acceleration.

Bending moment:

The algebraic sum of the moments of all the forces acting to the left or right of the section is known as Bending moment.

Fundamentals

Moment of inertia for a rectangle is given as:

$I = \frac{{b{d^3}}}{{12}}$

Here, width is $b$ and the height is $d.$

Bending stress:

The uniform loading of a beam supported at two ends is shown in Figure (1).

When a beam experiences a load like the one shown in Figure (1), a normal compressive stress is seen in the top fibers, a normal tensile stress is seen in the bottom fibers, and there is no stress in the horizontal plane of the neutral axis. The bending stress can be written as,

${\sigma _b} = \frac{{My}}{I}$

Here, bending stress is ${\sigma _b}$ , bending moment is M, vertical distance away from the neutral axis is y, and moment of inertia around the neutral axis is I.

Moment of inertia of the section is calculated as follows,

$\begin{array}{c}\\I = 2\left[ {\frac{1}{{12}} \times \left( {0.02} \right) \times {{\left( {0.2} \right)}^3}} \right] + \left[ {\frac{1}{{12}} \times \left( {0.26} \right) \times {{\left( {0.02} \right)}^3}} \right]\\\\ = 2\left( {1.333 \times {{10}^{ - 5}}} \right) + 1.733 \times {10^{ - 7}}\\\\ = 2.67 \times {10^{ - 5}} + 1.73 \times {10^{ - 7}}\\\\I = 2.6873 \times {10^{ - 5}}{\rm{ }}{{\rm{m}}^4}\\\end{array}$

Vertical distance away from the neutral axis is written as follows,

$\begin{array}{c}\\c = \frac{{0.2}}{2}\\\\ = 0.1{\rm{ m}}\\\end{array}$

The maximum bending stress equation is written as follows,

${\sigma _{\max }} = \frac{{Mc}}{I}$

Substitute $20 \times {10^3}{\rm{ N}} \cdot {\rm{m}}$ for M, 0.1 m for c, and $2.6873 \times {10^{ - 5}}{\rm{ }}{{\rm{m}}^4}$ for I.

$\begin{array}{c}\\{\sigma _{\max }} = \frac{{20 \times {{10}^3} \times 0.1}}{{2.6873 \times {{10}^{ - 5}}}}\\\\ = 74.42 \times {10^6}{\rm{ Pa}}\\\end{array}$

Ans:

The maximum bending stress $\left( {{\sigma _{\max }}} \right)$ is $74.42 \times {10^6}{\rm{ Pa}}$