Beam Bending

Abstract

A beam is a structural member whose length is greater than its other dimensions. It will bend when external loads are applied to it.

Appendix: Solutions

• 8.5.1 Plot the variation of bending moment for the simply supported  beam loaded as shown.

Solution

• 8.5.2 Plot the bending moment for the cantilevered beam loaded as shown.

Solution

• 8.5.3 What is the curvature in an aluminum, 2″ square bar subjected to a bending moment M = 150 ft-lbs?

Solution

Recall Eq. (8.7):

$$\begin{aligned} \rho (x) & = \frac{EI}{M(x)} \\ \rho & = \frac{{\left( {10*10^{6} } \right)\left( {\frac{{2*2^{3} }}{12}} \right)}}{150*12} = \frac{{\left( {10*10^{6} } \right)\left( {\frac{16}{12}} \right)}}{150*12} = \frac{{16*10^{6} }}{15*12*12} = 0.0074*10^{6} = 7,400\;{\text{in}}. = 616.67\;{\text{ft}}. \\ \end{aligned}$$

• 8.5.4 What is the maximum stress in the bar of Exercise 8.5.3?

Solution

$$\sigma_{xx} = - \frac{M}{I}y$$

For a positive moment, the maximum tensile stress is at the bottom of the beam:

$$\sigma_{xx}^{\hbox{max} } = - \frac{150*12}{{\frac{{\left( {2*2^{3} } \right)}}{12}}}\left( { - 1} \right) = \frac{150*144}{16} = 1,350\;{\text{psi}}$$

• 8.5.5 What is the minimum stress in the bar of Exercise 8.5.3?

Solution

The stress at the top of the bar is compressive (i.e., negative) and has the same magnitude as the stress at the bottom of the bar. Thus, $\sigma_{xx}^{Min} = - 1,350\;psi$