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

Question 4 of 6: Stress element at A and Mohr's circle

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

Question 4: Stress element at A and Mohr's circle (25 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.

Find. The state of stress on a horizontal/vertical element at A; the corresponding Mohr's circle; and the principal and maximum-shear elements with their correct orientations.

Approach. Take a section through A, resolve the internal actions there (axial force, shear force and bending moment), superpose the three elementary stress distributions at the exact coordinate of A, then transform that plane stress state using Mohr's circle.

1. Section properties of the I-beam. Treating the section as a $4 \times 8$ in rectangle less the two side notches $(b_f - t_w)$ wide and $(d - 2t_f)$ deep, $$I = \frac{b_f d^{3} - (b_f - t_w)(d - 2t_f)^{3}}{12} = \frac{4(8)^{3} - 3.625(7)^{3}}{12} = 67.05\ \text{in}^4 ,$$ $$A = 2 b_f t_f + (d - 2t_f)t_w = 2(4)(0.5) + 7(0.375) = 6.625\ \text{in}^2 .$$
2. Internal actions on the section through A. Cutting at $x = 20$ in and taking the free body to the right, the 6,000 lb load lies 20 in beyond the cut while the 3,000 lb load acts along the axis, so $$V = 6{,}000\ \text{lb}, \qquad M = 6{,}000 \times 20 = 120{,}000\ \text{lb}\cdot\text{in}, \qquad N = 3{,}000\ \text{lb (tension).}$$ Because the axial load is applied on the neutral axis it adds no moment.
3. Normal stress at A. Point A lies 1 in below the top face, i.e. $y_A = 4 - 1 = 3$ in above the neutral axis, which is inside the web (the flange ends at 3.5 in). The cantilever carries tension in the top fibres, so the bending contribution at A is tensile: $$\sigma_{b} = \frac{M y_A}{I} = \frac{120{,}000(3)}{67.05} = 5{,}370\ \text{psi}, \qquad \sigma_{a} = \frac{N}{A} = \frac{3{,}000}{6.625} = 453\ \text{psi}.$$ Superposing, $$\sigma_x = 5{,}370 + 453 = \boxed{5{,}823\ \text{psi (tension)}}, \qquad \sigma_y = 0 .$$
4. Transverse shear stress at A. The first moment of the area above A is the flange plus the 0.5 in of web lying between $y = 3$ and $y = 3.5$ in: $$Q = b_f t_f\left(\frac{d}{2} - \frac{t_f}{2}\right) + t_w (0.5)\left(\frac{3.5 + 3.0}{2}\right) = 7.500 + 0.609 = 8.109\ \text{in}^3 .$$ The width at A is the web thickness, so $$\tau = \frac{VQ}{I t_w} = \frac{6{,}000 (8.109)}{67.05 (0.375)} = \boxed{1{,}935\ \text{psi}} .$$ With the sign convention that a positive $\tau_{xy}$ acts upward on the $+x$ face, the transverse shear here gives $\tau_{xy} = -1{,}935$ psi.
5. Mohr's circle. The centre and radius are $$\sigma_{\text{avg}} = \frac{\sigma_x + \sigma_y}{2} = 2{,}911\ \text{psi}, \qquad R = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^{2} + \tau_{xy}^{2}} = \sqrt{2{,}911^{2} + 1{,}935^{2}} = 3{,}496\ \text{psi}.$$ The circle is plotted through $X(5{,}823,\ -1{,}935)$ and $Y(0,\ +1{,}935)$, which are the ends of a diameter.

6. Principal stresses and their orientation. Reading the circle, $$\sigma_1 = \sigma_{\text{avg}} + R = \boxed{6{,}407\ \text{psi}}, \qquad \sigma_2 = \sigma_{\text{avg}} - R = \boxed{-584\ \text{psi}} ,$$ so the point is in tension along one principal direction and in modest compression along the other. The orientation follows from $$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x - \sigma_y} = \frac{2(-1{,}935)}{5{,}823} = -0.6646 \;\Longrightarrow\; \theta_p = -16.8^\circ ,$$ that is, the $\sigma_1$ face is reached by rotating the original element $16.8^\circ$ clockwise. Substituting $\theta_p$ back into the transformation equations returns $\sigma = 6{,}407$ psi with $\tau = 0$, which confirms the root chosen is $\sigma_1$ and not $\sigma_2$.
7. Maximum shear stress and its orientation. The top of the circle gives $$\tau_{\max} = R = \boxed{3{,}496\ \text{psi}} ,$$ accompanied on those faces by the mean normal stress $\sigma_{\text{avg}} = 2{,}911$ psi on all four sides. The maximum-shear planes lie at $45^\circ$ to the principal planes, hence at $\theta_s = \theta_p + 45^\circ = 28.2^\circ$ counter-clockwise from the original element.

The three elements requested by the question are drawn below. Note the check that the sum of the normal stresses is invariant under rotation: $\sigma_1 + \sigma_2 = 6{,}407 - 584 = 5{,}823 = \sigma_x + \sigma_y$.