Question

The beam is subjected to a moment of M = 4.8kip·ft . (Figure 1) 

Part A

Determine the maximum tensile bending stress in the beam. 

Part B 

Determine the maximum compressive bending stress in the beam. 

Answer 1

Concepts and Reason

When loads are applied on a beam, it develops an internal shear force and bending moment which vary from point to point along the axis of the beam. The shear and bending moment functions can be plotted on a graph paper, taking the beam as reference, called shear force and bending moment diagrams.

Bending on the beam also causes tensile and compressive stresses on the beam and these stresses depend on the beam’s geometry.

The bending stress distribution in the cross-section of a beam subjected to bending moment varies linearly from zero at neutral axis to a maximum at a distance farthest from the neutral axis when the beam material is assumed to behave in a linear elastic manner.

Centroid is the mean position of all points in a geometric figure. The summation of the displacements of every point in the geometric figure from the centroid is zero. Centroid of an area is nothing but the centroid of mass. It is the point whose coordinates are the mean of all points in the geometric area.

Fundamentals

The formula for calculating the centroid of the cross section is,

$\bar y = \frac{{\sum {{A_i}{y_i}} }}{{\sum {{A_i}} }}$

Here, the area of each member is ${A_i}$ and the location of the member’s centroid is ${y_i}$ .

The formula for calculating the moment of inertia is given by,

$I = \sum {\left[ {\frac{{{b_i}h_i^3}}{{12}} + {A_i}{{\left( {{y_i} - \bar y} \right)}^2}} \right]}$

Here the base length and height of each member are ${b_i}$ and ${h_i}$ respectively.

The formula for bending stress is given by,

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

Here, the bending moment is M, the moment of inertia is I, and the location of the beam’s outermost fiber from the neutral axis is y.

Draw the diagram of the cross-section.

Find the y-coordinate of the neutral axis using the formula,

$\begin{array}{c}\\\bar y = \frac{{{A_1}{y_1} + {A_2}{y_2} + {A_3}{y_3}}}{{{A_1} + {A_2} + {A_3}}}\\\\ = \frac{{\left( {10 \times 0.5} \right)\left( {\frac{{10}}{2}} \right) + 2\left( {3 \times 0.5} \right)\left( {10 - \frac{3}{2}} \right) + \left( {4 \times 0.5} \right)\left( {10 + \frac{{0.5}}{2}} \right)}}{{\left( {10 \times 0.5} \right) + 2\left( {0.5 \times 3} \right) + \left( {4 \times 0.5} \right)}}\\\\ = \frac{{71}}{{10}}\\\\ = 7.1{\rm{ in}}{\rm{.}}\\\end{array}$

Find the moment of inertia of section – 1 about the neutral axis.

$\begin{array}{c}\\{I_1} = \frac{{{b_1}h_1^3}}{{12}} + \left( {{b_1}{h_1}} \right){\left( {\bar y - {y_1}} \right)^2}\\\\ = \frac{{0.5 \times {{10}^3}}}{{12}} + \left( {10 \times 0.5} \right){\left( {7.1 - 5} \right)^2}\\\\ = 63.717{\rm{ in}}{{\rm{.}}^4}\\\end{array}$

Find the moment of inertia of section – 2 about the neutral axis.

$\begin{array}{c}\\{I_2} = \frac{{{b_2}h_2^3}}{{12}} + \left( {{b_2}{h_2}} \right){\left( {\bar y - {y_2}} \right)^2}\\\\ = \frac{{0.5 \times {3^3}}}{{12}} + \left( {3 \times 0.5} \right){\left( {7.1 - 8.5} \right)^2}\\\\ = 4.065{\rm{ in}}{{\rm{.}}^4}\\\end{array}$

Find the moment of inertia of section – 3 about the neutral axis.

$\begin{array}{c}\\{I_3} = \frac{{{b_3}h_3^3}}{{12}} + \left( {{b_3}{h_3}} \right){\left( {\bar y - {y_3}} \right)^2}\\\\ = \frac{{4 \times {{\left( {0.5} \right)}^3}}}{{12}} + \left( {4 \times 0.5} \right){\left( {7.1 - 10.25} \right)^2}\\\\ = 19.887{\rm{ in}}{{\rm{.}}^4}\\\end{array}$

Calculate the total moment of inertia of the beam.

$\begin{array}{c}\\I = {I_1} + 2{I_2} + {I_3}\\\\ = 63.717 + 2 \times 4.065 + 19.887\\\\ = 91.734{\rm{ in}}{{\rm{.}}^4}\\\end{array}$

Find the distance of the bottommost fiber from the neutral axis.

$\begin{array}{c}\\y = \bar y\\\\ = 7.1{\rm{ in}}{\rm{.}}\\\end{array}$

Calculate the bending stress at the bottom most fiber.

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

Here, $M$ is the bending moment.

Substitute $4.8{\rm{ kip}} \cdot {\rm{ft}}\left| {\frac{{12{\rm{ in}}{\rm{.}}}}{{1{\rm{ ft}}}}} \right|$ for $M$ , $7.1{\rm{ in}}{\rm{.}}$ for $y$ , and $91.734{\rm{ in}}{{\rm{.}}^4}$ for $I$ .

$\begin{array}{c}\\\sigma = \frac{{\left( {4.8 \times 12} \right) \times 7.1}}{{91.734}}\\\\ = \frac{{408.96}}{{91.734}}\\\\ = 4.458{\rm{ ksi}}\\\end{array}$

Find the distance of the uppermost fiber from the neutral axis.

$\begin{array}{c}\\y = \left( {10 + 0.5} \right) - \bar y\\\\ = 10.5 - 7.1\\\\ = 3.4{\rm{ in}}{\rm{.}}\\\end{array}$

Calculate the bending stress at the bottom most fiber.

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

Here, $M$ is the bending moment.

Substitute $4.8{\rm{ kip}} \cdot {\rm{ft}}\left| {\frac{{12{\rm{ in}}{\rm{.}}}}{{1{\rm{ ft}}}}} \right|$ for $M$ , $3.4{\rm{ in}}{\rm{.}}$ for $y$ , and $91.734{\rm{ in}}{{\rm{.}}^4}$ for $I$ .

$\begin{array}{c}\\\sigma = \frac{{\left( {4.8 \times 12} \right) \times 3.4}}{{91.734}}\\\\ = \frac{{195.84}}{{91.734}}\\\\ = 2.13487{\rm{ ksi}}\\\end{array}$

Ans: Part A

Therefore, the maximum tensile bending stress in the beam is ${\rm{4}}{\rm{.458 ksi}}$ .

Part B

Therefore, the maximum compressive bending stress in the beam is ${\rm{2}}{\rm{.13487 ksi}}$ .