Chapter 6, Problem 6.8RP

A wooden beam has a square cross section as shown Determine which orientation of the beam provides the greatest strength at resisting the moment M. What is the difference in the resulting maximum stress in both cases?

R6–8

To determine

The orientation of the beam that provides the greatest strength at the resisting moment M and the difference in the resulting maximum stress in both cases.

Answer to Problem 6.8RP

Solution:

• The orientation of $\text{Case}\left(\text{a}\right)$ provides the greatest strength.
• The difference in the resulting maximum stress in both cases is $\Delta {\sigma }_{\max }=2.49\left(\frac{M}{a^3}\right)$.

Explanation of Solution

Given information:

• Case (a): The wooden beam is considered as a square.
• Case (b): the wooden beam is considered as a rhombus.

Maximum stress:

Determine the maximum stress in the wooden beam using the equation given below:

${\sigma }_{\max }=\frac{Mc}{I}$ (1)

Here, ${\sigma }_{\max }$ is maximum stress in the beam, M is the resisting moment, c is the distance between the neutral axis and the extreme fibre, and I is moment of inertia of cross-section of the beam about the neutral axis (NA).

Case (a): The wooden beam is considered as a square:

The free-body diagram of the square wooden beam is shown in Figure 1.

Distance between the neutral axis and the extreme fibre of the square beam is $c=a/2$.

Moment of inertia of the square section about the neutral axis is $I=\frac{a^4}{12}$.

Substitute $\frac{a}{2}$ for c and $\frac{a^4}{12}$ for I in Equation (1).

$\begin{array}{c}{\left({\sigma }_{\max }\right)}_{\text{square}}=\frac{M\times a\times 12}{2\times a^4}\\ =\frac{6M}{a^3}\end{array}$ (2)

Case (a): The wooden beam is considered as a rhombus:

The free-body diagram of the rhombus beam is shown in Figure 2.

Distance between the neutral axis and the extreme fibre of the rhombus beam is $c=a/\sqrt{2}$.

Moment of inertia of the rhombus section about the neutral axis is $I=\frac{a^4}{12}$.

Substitute $\frac{a}{\sqrt{2}}$ for c and $\frac{a^4}{12}$ for I in Equation (1).

$\begin{array}{c}{\left({\sigma }_{\max }\right)}_{\text{rhombus}}=\frac{M\times a\times 12}{\sqrt{2}\times a^4}\\ =\frac{8.4853M}{a^3}\end{array}$ (3)

From the given resisting moment M, the stress induced in the square section beam is less compared to the rhombus-shaped beam. Therefore, the square-shaped beam can take more load (resisting moment) compared to the rhombus-shaped beam before failure. Therefore, the square-shaped beam is stronger when compared to the rhombus-shaped beam.

Difference in stress: $\left(\Delta {\sigma }_{\max }\right)$:

Determine the difference in the resulting maximum stress using the relation given below:

$\Delta {\sigma }_{\max }={\left({\sigma }_{\max }\right)}_{\text{rhombus}}-{\left({\sigma }_{\max }\right)}_{\text{square}}$ (4)

Substitute $\frac{8.4853M}{a^3}$ for ${\left({\sigma }_{\max }\right)}_{\text{rhombus}}$ and $\frac{6M}{a^3}$ for ${\left({\sigma }_{\max }\right)}_{\text{square}}$ in Equation (4).

$\begin{array}{c}\Delta {\sigma }_{\max }=\frac{8.4853M}{a^3}-\frac{6M}{a^3}\\ =2.4853\left(\frac{M}{a^3}\right)\\ \approx 2.49\left(\frac{M}{a^3}\right)\end{array}$

Thus, the difference in the resulting maximum stress in both cases is $\Delta {\sigma }_{\max }=2.49\left(\frac{M}{a^3}\right)$.