An axial load is applied to a solid circular bar that contains an offset in order to fit in a tight space in a machine. Compute the maximum tensile and compressive normal stresses at section a.

Given: Magnitude of axial force and bent bar geometry.

Find: Maximum normal stresses at specified section.

Assume: Hooke’s law applies.

Step-by-Step

Learn more on how do we answer questions.

Although we have learned about bars subject to axial loading, rods subject to torsion, and beams subject to bending in separate chapters, in practice elements are often acted upon by multiple types of loads. In this example, the center section is subject to both axial loading and bending, as we can see by constructing an FBD of the section to the left of a.

Solve for the internal forces and moment at a:

\begin{aligned} & \sum F_x=0=N_a-5000 \mathrm{~N} \rightarrow N_a=5000 \mathrm{~N}, \\ & \sum F_z=0=V_a, \\ \curvearrowleft & \sum M_{\mathrm{about} a}=0=-M_a+(5000 \mathrm{~N})(0.01 \mathrm{~m}) \rightarrow M_a=50 \mathrm{Nm} . \\ & \end{aligned}The normal stress due to N_a is

\sigma_{x x}=\frac{N_a}{A}=\frac{5000 \mathrm{~N}}{\pi(0.015 \mathrm{~m})^2}=7.1 \mathrm{MPa} .

The normal stress due to M_a varies, with the maximum tensile stress at the top surface (z = +0.015 m) and the maximum compressive at the bottom (z = −0.015 m). For the circular cross section, Appendix A tells us that I = πr^4/4.

\sigma_{x x}=\frac{M_a\left(\pm z_{\max }\right)}{I}=\frac{(50 \mathrm{Nm})(\pm 0.015 \mathrm{~m})}{\frac{\pi}{4}(0.015 \mathrm{~m})^4}=\pm 18.9 \mathrm{MPa} .

Because both of these stresses contribute to the same component of the stress tensor, in our linear setting we use the principle of superposition to add them. The maximum tensile and compressive stresses are

\begin{aligned} & \sigma_{x x, \text { top }}=\frac{N_a}{A}+\frac{M_a\left(+z_{\max }\right)}{I}=7.1 \mathrm{MPa}+18.9 \mathrm{MPa}=26 \mathrm{MPa} \text { (maximum tension), } \\ & \sigma_{x x, \text { bottom }}=\frac{N_a}{A}+\frac{M_a\left(-z_{\mathrm{max}}\right)}{I}=7.1 \mathrm{MPa}-18.9 \mathrm{MPa}=-11.8 \mathrm{MPa} \text { (maximum compression). } \\ & \end{aligned}In several of the problems below you are asked to consider superposition of multiple sources of stress. In some cases there will be multiple contributions to a single element of the stress tensor, which may be added as in this example. In others, loads cause contributions to different elements of the stress tensor.

Question: 7.8

The first of the equilibrium equations represents ...

Question: 7.7

We begin with a general statement of the fact that...

Question: 7.6

First we need to consult our FBD and find the reac...

Question: 7.5

We obtained a formula for shear stress at a given ...

Question: 7.4

The gage at A, in the upper portion of the cross s...

Question: 7.3

The symmetry of the cross section shown suggests t...

Question: 7.2

In each case, we will first find the external reac...

Question: 7.1

Our strategy is to find the reactions at the suppo...