22-Mec-A4 Design and Manufacture of Machine Elements · May 2014

Question 6 of 8: Stress element at A, Mohr's circle, principal and maximum-shear elements

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 6: Stress element at A, Mohr's circle, principal and maximum-shear elements (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A cantilever beam of rectangular section, built in at the left-hand end:

Find. The stresses \(\sigma_x,\ \sigma_y,\ \tau_{xy}\) on a horizontal/vertical element at A; the Mohr circle; and the principal and maximum-shear elements with their correct orientations.

Approach. Compute the section properties, then superpose the three elementary stress states that act at A — uniform axial tension, bending from the transverse load, and transverse (flexural) shear — to obtain \(\sigma_x\), \(\sigma_y\) and \(\tau_{xy}\). Feed those into the stress-transformation equations, plot Mohr's circle, and read off the principal stresses, the maximum in-plane shear, and the two element orientations.

1. Section properties. For the rectangular section, \[A = b\,h = 75 \times 100 = 7500\ \text{mm}^2, \qquad I = \frac{b h^{3}}{12} = \frac{75 \times 100^{3}}{12} = 6.25\times10^{6}\ \text{mm}^4.\] The neutral axis is at mid-depth, so point A — 25 mm above the bottom — sits \(y_A = 25 - 50 = -25\ \text{mm}\), i.e. 25 mm below the neutral axis.
2. Internal actions on the section through A. Take the free body to the right of the section at \(x = 1300\ \text{mm}\). It carries the 67 500 N axial force and the 36 000 N transverse load, the latter at a lever arm of \(1500 - 1300 = 200\ \text{mm}\): \[N = 67\,500\ \text{N (tension)},\qquad V = 36\,000\ \text{N},\qquad M = 36\,000 \times 200 = 7.2\times10^{6}\ \text{N}\cdot\text{mm}.\] Because the load hangs off the free end of a cantilever, the moment is hogging: it puts the top fibres in tension and the bottom fibres in compression.
3. Axial stress. The horizontal load acts on the centroidal axis, so it produces uniform tension across the whole section: \[\sigma_{\text{axial}} = \frac{N}{A} = \frac{67\,500}{7500} = +9.0\ \text{MPa}.\]
4. Bending stress at A. The flexure formula gives the magnitude, and the hogging sense gives the sign; A lies below the neutral axis, so bending puts it in compression: \[\sigma_{\text{bend}} = -\frac{M\,|y_A|}{I} = -\frac{7.2\times10^{6} \times 25}{6.25\times10^{6}} = -28.8\ \text{MPa}.\] Superposing the two normal contributions, \[\sigma_x = 9.0 - 28.8 = \boxed{-19.8\ \text{MPa (compression)}}, \qquad \sigma_y = 0.\] The axial tension is not enough to overcome the bending compression at this point, so A ends up in net compression — a result worth noting, because a quick glance at the figure suggests "tension".
5. Transverse shear stress at A. Using \(\tau = VQ/(I\,b)\) with \(Q\) the first moment of the area below A about the neutral axis. That area is a 75 mm × 25 mm strip whose centroid is 37.5 mm below the neutral axis: \[Q = (75 \times 25)(37.5) = 70\,312.5\ \text{mm}^3,\] \[\tau_{xy} = \frac{V Q}{I b} = \frac{36\,000 \times 70\,312.5}{6.25\times10^{6}\times 75} = \boxed{5.4\ \text{MPa}}\] The shear on the vertical face of the element acts downward, in the same sense as the transverse load carried by the right-hand free body; with the usual convention (positive \(\tau_{xy}\) acting in \(+y\) on the \(+x\) face) this is \(\tau_{xy} = -5.4\ \text{MPa}\).
6. Draw the element at A with horizontal and vertical sides. The element carries 19.8 MPa of compression on its vertical faces, nothing on its horizontal faces, and 5.4 MPa of shear on all four faces, in the sense established in step 5. This is the first of the three sketches asked for, and it is shown at the left of Figure 6.3.
7. Construct Mohr's circle. The centre lies at the average normal stress and the radius is the hypotenuse of the half-difference and the shear: \[\sigma_{\text{avg}} = \frac{\sigma_x + \sigma_y}{2} = \frac{-19.8 + 0}{2} = -9.9\ \text{MPa},\] \[R = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^{2} + \tau_{xy}^{2}} = \sqrt{(-9.9)^{2} + (5.4)^{2}} = \sqrt{98.01 + 29.16} = 11.28\ \text{MPa}.\] The plotted points \(X(-19.8,\,-5.4)\) and \(Y(0,\,+5.4)\) are diametrically opposite, as they must be.
8. Principal stresses and maximum shear. Reading the ends of the horizontal diameter and the top of the circle, \[\sigma_1 = \sigma_{\text{avg}} + R = -9.9 + 11.28 = \boxed{+1.38\ \text{MPa}}\] \[\sigma_2 = \sigma_{\text{avg}} - R = -9.9 - 11.28 = \boxed{-21.18\ \text{MPa}}\] \[\tau_{\max}^{\text{in-plane}} = R = \boxed{11.28\ \text{MPa}}\] Since \(\sigma_1 > 0 > \sigma_2\) and the third principal stress is \(\sigma_3 = 0\) (a free surface state), the in-plane maximum shear is also the absolute maximum shear: \(\left(\sigma_1-\sigma_2\right)/2 = 11.28\ \text{MPa}\).
9. Orientation of the principal element. From the transformation equation, \[\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x - \sigma_y} = \frac{2(-5.4)}{-19.8} = 0.5455 \quad\Rightarrow\quad 2\theta_p = 28.61^{\circ},\ \theta_p = 14.30^{\circ}.\] Substituting \(\theta = 14.30^{\circ}\) back into \(\sigma(\theta) = \sigma_{\text{avg}} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta\) returns \(-21.18\ \text{MPa}\), which identifies this plane as the one carrying \(\sigma_2\). Therefore \[\boxed{\sigma_2 = -21.18\ \text{MPa acts on the face at }14.30^{\circ}\text{ CCW from }x;\ \ \sigma_1 = +1.38\ \text{MPa at }104.30^{\circ}}\]
10. Orientation of the maximum-shear element. Planes of maximum shear always lie 45° from the principal planes: \[\theta_s = \theta_p + 45^{\circ} = 14.30 + 45 = 59.30^{\circ}.\] On that element the shear is \(11.28\ \text{MPa}\) and both normal faces carry the average normal stress \(\sigma_{\text{avg}} = -9.9\ \text{MPa}\) — the maximum-shear element is never free of normal stress unless \(\sigma_x + \sigma_y = 0\).