How to use Mohr's circle to find stresses on a slanted plane?

Question

Here is my attempt to the problem:

Does this make sense? rotation clockwise 60* from horizontal to normal of the slanted plane , therefore $2 \times 60$ clockwise as well on Mohr's circle.

I am unsure why shear stress is positive downwards (as given in my textbook), also does the angle make sense? Should this angle drawn between the right triangle form or my angle on the diagram? Is the shear stress ($\tau$) negative or positive? Its positive on my diagram because positive is downwards. On my teacher's solution, shear stress ($\tau$) is drawn negative downwards, its shear stress is therefore negative.

Answer 1

Your confusion is caused by the "sign conventions" as shown below. Note, in the usual case, the angle ($\theta$) between the x-axis to the normal stress on the inclined plane is measured "counterclockwise" from the x-axis in the positive direction. Following the usual convention, the shear stress should be positive on the inclined plane.

Note: The direction of shear stress is positive when it occurs on the "positive face" of the element, on which (positive face) the normal stress is projected towards the positive axis ($+x_1$ or $+y_1$ axis).

ADD:

Given $\sigma_x = 3.5, \sigma_y = 2.5, \tau_{xy} = 0$

The inclined plane is $120^o$, measured CCW, from the $x-axis$, the stresses on the plane are:

$\sigma_{x_1} = \dfrac{\sigma_x + \sigma_y}{2} +\dfrac{sigma_x - \sigma_y}{2}cos2\theta + \tau_{xy}sin2\theta$

$\sigma_{x_1} = \dfrac{3.5+2.5}{2} + \dfrac{3.5-2.5}{2}sin(240^o) + 0$

$\sigma_{x_1} = 3 - 0.25 = 2.75$

$\tau_{x_1y_1} = -\dfrac{\sigma_x - \sigma_y}{2}sin 2\theta + \tau_{xy}cos 2\theta$

$\tau_{x_1y_1} = -\dfrac{3.5-2.5}{2}sin(240^o) + 0 = **+0.433**$

Using your convention, $\theta = 60^o$, then

$\tau_{x_1y_1} = -\dfrac{3.5-2.5}{2}sin(120^o) + 0 = **-0.433**$

The resulting stresses are as shown below:

Answer 2

While deriving the stress transformation equations, the angle $\theta$ (the angle that the outward normal of a plane makes with the positive x axis) is usually taken as positive when measured anticlockwise from the +ve x axis.

If we plot the shear stress as positive downwards in the Mohr's Circle, an angle $\theta$ in the stress element measured anticlockwise, when corresponded with an angle 2$\theta$ swept anticlockwise in the Mohr's Circle will give results which conform with the stress transformation equations.

If you consider the shear stress as negative in the downward direction, an angle $\theta$ measured antickw in the stress element will need to be corresponded with an angle 2$\theta$ swept in the Mohr's Circle in clockwise direction for the results to conform with those obtained from transformation equations.

So there is nothing wrong with taking the shear stress as -ve downwards, it's just that now anticlockwise angles in the stress element would correspond to clockwise angles in the Mohr's circle. In order to maintain a consistency, we take the shear stress as postive downwards, because now an antickw angle in the element would correspond to an antickw angle in the Mohr's Circle.

Your teacher if draws shear stress as positive upwards will take the antickw angles on stress element as clockwise angles on Mohr's Circle and vice versa, and when he/she does that his/her results will be the same as yours - a positive shear stress on the inclined plane.

If shear stress is taken +ve downwards:

If shear stress is taken +ve upwards:

Note: A positive shear stress in my answer means that a shear stress which tries to rotate the material in the anticlockwise direction and a negative shear stress means a shear stress which tries to rotate the material in clockwise direction. Make sure that what a +ve shear stress means to you is the same as what a +ve shear stress means to your teacher.

Answer 3

Your original stress state is the principal stress state because the shear is zero.

$$\sigma X_{max}=3.5kPa \quad \ and \quad \sigma Y_{max}=2.5kPa$$

The plane of the weld is at 210 degrees from the X-axis so your $$\theta= 210*2=420\circ \quad \theta =420-360$$ $$\theta = 60\circ$$