Theoretically when a homogeneous isotropic bar is subjected to axial loading, does it fail in shear?

Question

I m referrencing James M Gere, Timoshenko Mechanics of Materials Chapter 2 Page 106. The book talks about effect of uniaxial load on a small element that is inclined in one axis. So my question is if we look at an element that is inclined in all three axis wouldnt the element have shear in all 6 planes, and since normally shear modulus is lower would the bar fail in shear with shears occuring in all three axis of the element? or am i misunderstanding something.

Answer 1

Let's first clear up a misconception:

since normally shear modulus is lower would the bar fail in shear with shears occuring in all three axis of the element

The shear strength, not the shear modulus, is what determines shear failure. The shear modulus is only the slope of the stress stress–shear strain line within the elastic region (i.e., the constant of proportionality between these two parameters).

Let me rephrase the question, which I think is a pretty insightful one:

If we consider a cubic element exposed to uniaxial loading and rotate it around one of the other axes, we see shear arise on four faces (as graphically analyzed by Mohr's circle, for example):

(Images from my site.)

Ductile materials fail in shear, and a uniaxially stretched ductile rod will pull apart with angled failure surfaces because shear develops at an angle as described above:

Therefore, would rotating the element again—around the other axis—increase the shear even more and cause easier failure?

Is this an accurate rephrasing?

If so, one way to address this question is to look at the stress transformation from these rotations. We have $$\boldsymbol{\sigma^\prime=Q\sigma Q^T},$$ where $\boldsymbol{Q}$ is the direction cosine matrix, or a matrix $q_{i,i^\prime}$ of cosines between the original axis and the rotated axis.

Consider, as an example, a first rotation of 45° around the 2-axis for a unit uniaxial strain along the 1-axis:

$$\boldsymbol{\sigma^\prime=Q\sigma Q^T};$$

$$\boldsymbol{\sigma^\prime}=\left[\begin{array}{c}\frac{1}{\sqrt{2}}&0&-\frac{1}{\sqrt{2}}\\0&1&0\\ \frac{1}{\sqrt{2}}&0& \frac{1}{\sqrt{2}}\end{array}\right]\left[\begin{array}{c}1&0&0\\0&0&0\\ 0&0&0\end{array}\right]\left[\begin{array}{c}\frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}}\\0&1&0\\ -\frac{1}{\sqrt{2}}&0& \frac{1}{\sqrt{2}}\end{array}\right];$$

$$\boldsymbol{\sigma^\prime}=\left[\begin{array}{c}\frac{1}{2}&0&\frac{1}{2}\\0&0&0\\ \frac{1}{2}&0& \frac{1}{2}\end{array}\right].$$

The maximum shear stress is $\frac{1}{2}$.

Now apply a second rotation of 45° around the 3-axis:

$$\boldsymbol{\sigma^{\prime\prime}}=\left[\begin{array}{c}\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}&0\\\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0\\0&0&1\end{array}\right]\left[\begin{array}{c}\frac{1}{2}&0&\frac{1}{2}\\0&0&0\\ \frac{1}{2}&0& \frac{1}{2}\end{array}\right]\left[\begin{array}{c}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0\\-\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0\\0&0&1\end{array}\right];$$

$$\boldsymbol{\sigma^{\prime\prime}}=\left[\begin{array}{c}\frac{1}{4}&\frac{1}{4}&\frac{1}{2\sqrt{2}}\\\frac{1}{4}&\frac{1}{4}&\frac{1}{2\sqrt{2}}\\ \frac{1}{2\sqrt{2}}&\frac{1}{2\sqrt{2}}& \frac{1}{2}\end{array}\right].$$

The maximum shear stress is now $\frac{1}{2\sqrt{2}}$. Notably, the additional rotation has lessened the shear stress, not increased it; we obtained the maximum shear strain possible with a single rotation of 45°. This is presumably why Gere considers only single rotations.