Principal Stresses and Mohrs Circle Questions and Answers

Answer: B

The maximum shear stress is 100 MPa.

The element carries principal stresses of +100 MPa (tension) and -100 MPa (compression). The maximum in-plane shear stress is half their difference:

\[ \begin{aligned} \tau_{max} &= \frac{\sigma_{1} - \sigma_{2}}{2} = \frac{100 - (-100)}{2} \\ &= 100\ \text{MPa} \end{aligned} \]

With the out-of-plane principal stress zero, the other two candidate values are 100/2 = 50 MPa, so 100 MPa is also the absolute maximum. 200 MPa forgets the factor of one half, 50 MPa treats both stresses as tensile, and 0 would apply only if the two principal stresses were equal.

Answer: A

The normal stresses are \(\tau_{o}\) and \(-\tau_{o}\).

The left element is in pure shear: \(\sigma_{x} = \sigma_{y} = 0\) and a positive \(\tau_{xy} = \tau_{o}\). The principal stresses are

\[ \begin{aligned} \sigma_{1,2} &= \frac{\sigma_{x} + \sigma_{y}}{2} \\ &\quad \pm \sqrt{\left(\frac{\sigma_{x} - \sigma_{y}}{2}\right)^{2} + \tau_{xy}^{2}} \\ &= \pm\tau_{o} \end{aligned} \]

and they act at 45° to the axes. With positive shear, the tensile principal stress lies along the 45° diagonal running into the first quadrant, which is where the figure places \(\sigma_{1}\); \(\sigma_{2}\), at right angles, is compressive. So \(\sigma_{1} = \tau_{o}\) and \(\sigma_{2} = -\tau_{o}\).

The reversed pair puts tension on the wrong diagonal. The values \(\pm\tau_{o}/\sqrt{2}\) come from resolving the shear force onto the diagonal without also accounting for the diagonal face's larger area, which restores the factor \(\sqrt{2}\).

Answer: B

The volumetric strain is \(3 \times 10^{-4}\).

For small strains, the volumetric strain is the sum of the three normal strains (the trace of the strain tensor). Shear strains change shape, not volume, so \(\varepsilon_{xy}\), \(\varepsilon_{yz}\) and \(\varepsilon_{zx}\) do not enter:

\[ \begin{aligned} \varepsilon_{v} &= \varepsilon_{xx} + \varepsilon_{yy} + \varepsilon_{zz} \\ &= (2.5 + 2.0 - 1.5) \times 10^{-4} \\ &= 3 \times 10^{-4} \end{aligned} \]

\(4 \times 10^{-4}\) adds the shear strains in as well. The negative options reverse the sign of the sum, which is positive because the two tensile strains outweigh the one compressive strain.

Answer: C

The shear strain is \(300 \times 10^{-6}\).

For a 0-45-90 rosette, gauges p and r read the normal strains directly, \(\varepsilon_x = \varepsilon_p\) and \(\varepsilon_y = \varepsilon_r\). The 45-degree gauge reads \(\varepsilon_q=\tfrac{1}{2}(\varepsilon_x+\varepsilon_y)+\tfrac{1}{2}\gamma_{xy}\), so

\[ \begin{aligned} \gamma_{xy} &= 2\varepsilon_q - \varepsilon_p - \varepsilon_r \\ &= (360 - 150 + 90) \times 10^{-6} \\ &= 300 \times 10^{-6} \end{aligned} \]

\(120 \times 10^{-6}\) results from getting the sign of \(\varepsilon_r\) wrong (\(360 - 150 - 90\)), and \(180 \times 10^{-6}\) is just the reading of gauge q.

Answer: C

The plane PQ is at 45 degrees to the x-axis.

The Mohr's circle is centred at the origin and cuts the normal stress axis at \(+\sigma_{0}\) and \(-\sigma_{0}\). So the principal stresses are \(\sigma_{0}\) and \(-\sigma_{0}\), and with the element drawn on x and y axes they act on the x and y faces. A plane in pure shear carries no normal stress, which on this circle means the top or bottom point, where the shear is \(\pm\sigma_{0}\).

Those points are 90 degrees round the circle from the principal points. Angles on Mohr's circle are twice the angles in the element, so the plane is \(90/2 = 45\) degrees from the x face.

At 90 degrees the plane is the other principal plane, with normal stress \(-\sigma_{0}\) and no shear. At 60 and 120 degrees the circle angle is 120 or 240 degrees, where the normal stress is \(-\sigma_{0}/2\), not zero.

Answer: A

The maximum shear stress is 500 MPa.

For plane stress the in-plane maximum shear stress is the radius of Mohr's circle:

\[ \begin{aligned} \tau_{max} &= \sqrt{\left(\frac{\sigma_{xx}-\sigma_{yy}}{2}\right)^{2} + \sigma_{xy}^{2}} \\ &= \sqrt{400^{2} + 300^{2}} = 500\ \text{MPa} \end{aligned} \]

The principal stresses are \(400 \pm 500\), that is 900 MPa and -100 MPa. Since they have opposite signs, the third principal stress (zero, out of the plane) lies between them and the absolute maximum shear stress is also (900 + 100)/2 = 500 MPa.

400 MPa is only the half-difference of the normal stresses, ignoring the shear. 300 MPa is just the given shear stress, and 800 MPa is the normal stress \(\sigma_{xx}\), not a shear stress.

Answer: B

The equivalent state is (B): tension \(\sigma\) along x and compression \(\sigma\) along y.

Pure shear of magnitude \(\sigma\) on planes at 45° to the axes is equivalent to equal and opposite principal stresses on the x and y planes: Mohr's circle for pure shear is centred at the origin with radius \(\sigma\), so the principal stresses are \(+\sigma\) and \(-\sigma\), 45° away from the planes of maximum shear.

Checking the direction: with \(\sigma_{x} = \sigma\) and \(\sigma_{y} = -\sigma\), the traction on the upper-right face (normal (1, 1)) is \((\sigma, -\sigma)/\sqrt{2}\), pointing down and to the right along the face, which is the shear shown in the figure.

(A) and (C) have both stresses of the same sign, which produces no shear on 45° planes, and (D) has unequal tensions, which would add a normal stress to the 45° faces.

Answer: D

The shear stress on a principal plane is zero.

A principal plane is, by definition, a plane on which the shear stress is zero and only a normal stress, the principal stress, acts. Maximum shear stress occurs on planes at 45 degrees to the principal planes.

The expressions in the other options combine shear stresses in a way that has no standard meaning; the familiar formula of this shape, \((\sigma_{1} - \sigma_{2})/2\), gives the maximum shear stress on the 45-degree planes, not the shear on a principal plane.

Answer: B

The centre of Mohr's circle lies at \(\frac{\sigma_x + \sigma_y}{2}\) from the y-axis.

Mohr's circle plots normal stress along the horizontal axis and shear stress along the vertical (y) axis. The two perpendicular normal stresses \(\sigma_x\) and \(\sigma_y\) are marked on the horizontal axis, and the circle drawn through them has its centre midway between them:

\[ \begin{aligned} \text{centre} &= \frac{\sigma_x + \sigma_y}{2} \\ \text{radius} &= \frac{\sigma_x - \sigma_y}{2} \end{aligned} \]

So \(\frac{\sigma_x - \sigma_y}{2}\) is the radius, not the position of the centre. The options adding or subtracting \(\sigma_z\) bring in a third stress that a two-dimensional Mohr's circle for \(\sigma_x\) and \(\sigma_y\) does not use.

Answer: C

The centre of Mohr's circle is at (0, 0).

The centre lies on the normal stress axis at the average normal stress, with zero shear.

\[ \begin{aligned} \sigma_{c} &= \frac{\sigma_{x} + \sigma_{y}}{2} \\ &= \frac{40 + (-40)}{2} = 0 \end{aligned} \]

So the centre is (0, 0). The shear stress of 10 MPa affects the radius, \(\sqrt{40^{2} + 10^{2}} \approx 41.2\) MPa, not the centre, which is why (0, 10) is wrong; (40, 40) and (40, -40) are points of the stress state, not the centre.