22-Mec-A4 Design and Manufacture of Machine Elements · Undated paper

Question 6 of 6: Stress element and Mohr circle for a cantilever I-beam

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

Question 6: Stress element and Mohr circle for a cantilever I-beam (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. A cantilever I-beam built into a wall, carrying a 6,000 lb transverse load 40 in from the wall and a 3,000 lb axial pull applied on the centroidal axis through a bolted end plate. Point A lies 20 in from the wall, 1 in below the top surface.

Find. The stresses on a horizontal/vertical element at A; the Mohr circle for that state; and the correctly oriented principal and maximum-shear elements.

Approach. Superpose three elementary stress fields at A — uniform axial tension, flexural stress $My/I$, and transverse shear $VQ/It$ — to build the plane-stress state $(\sigma_x,\ \sigma_y = 0,\ \tau_{xy})$, then transform it with the Mohr circle to obtain the principal stresses and their orientation.

1. Section properties. Treating the I-section as a $4 \times 8$ in rectangle less the two side voids of $(4 - 0.375)$ in width and 7 in height, $$I = \frac{(4)(8)^3 - (3.625)(7)^3}{12} = \frac{2048 - 1243.4}{12} = 67.05\ \text{in}^4$$ $$A = 2(4)(0.5) + (0.375)(7) = 4.000 + 2.625 = 6.625\ \text{in}^2$$
2. Internal actions at the section through A. Everything outboard of A must be carried across the cut. The whole 6,000 lb load is outboard, so the shear is $V = 6000\ \text{lb}$, and its moment arm to A is $40 - 20 = 20\ \text{in}$: $$M = (6000)(20) = 120{,}000\ \text{lb}\cdot\text{in}, \qquad N = 3000\ \text{lb\ (tension)}$$ The moment is hogging — a downward load on a cantilever puts the top fibres in tension — and A is above the neutral axis, so the bending stress there is tensile and adds to the axial stress rather than cancelling it.
3. Normal stress at A. With $y_A = 4 - 1 = 3\ \text{in}$, $$\sigma_{\text{bend}} = \frac{My_A}{I} = \frac{(120000)(3)}{67.05} = 5369\ \text{psi},\qquad \sigma_{\text{axial}} = \frac{N}{A} = \frac{3000}{6.625} = 453\ \text{psi}$$ $$\boxed{\sigma_x = 5369 + 453 = 5822\ \text{psi (tension)}},\qquad \sigma_y = 0$$
4. Shear stress at A. The first moment of the area above A comprises the whole top flange plus the sliver of web from $y = 3$ to $y = 3.5\ \text{in}$: $$Q = (4)(0.5)(3.75) + (0.375)(0.5)(3.25) = 7.500 + 0.609 = 8.109\ \text{in}^3$$ The width at A is the web thickness, not the flange width — the single most common slip in this calculation: $$\boxed{\tau_{xy} = \frac{VQ}{I\,t_w} = \frac{(6000)(8.109)}{(67.05)(0.375)} = 1935\ \text{psi}}$$ A sits usefully off both the neutral axis and the extreme fibre, which is exactly why the examiner chose it: all three stress fields are simultaneously alive there.
5. Build the Mohr circle. The centre lies at the average normal stress and the radius is the hypotenuse of the half-difference and the shear: $$\sigma_{\text{avg}} = \frac{\sigma_x+\sigma_y}{2} = 2911\ \text{psi},\qquad R = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2} = \sqrt{2911^2 + 1935^2} = 3495\ \text{psi}$$
   Q6 — Mohr circle for the stress state at A. Centre 2911 psi, radius 3495 psi; the X and Y face points are the ends of the red diameter.
6. Principal stresses and maximum shear. $$\sigma_1 = \sigma_{\text{avg}} + R = \boxed{6406\ \text{psi}},\qquad \sigma_2 = \sigma_{\text{avg}} - R = \boxed{-584\ \text{psi}},\qquad \tau_{\max} = R = \boxed{3495\ \text{psi}}$$ Two independent checks confirm the circle. The first stress invariant holds, $\sigma_1+\sigma_2 = 6406 - 584 = 5822 = \sigma_x + \sigma_y$; and the second invariant holds, $\sigma_1\sigma_2 = -3.74\times10^{6} = -\tau_{xy}^2$. Note that $\sigma_2$ is compressive even though the point is in net tension — pure shear always drags one principal stress across zero.
7. Orientation of the principal element. $$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(1935)}{5822} = 0.6647 \;\Longrightarrow\; 2\theta_p = 33.6^\circ,\qquad \boxed{\theta_p = 16.8^\circ}$$ measured counterclockwise from the $x$ face. Substituting back into the transformation equation returns $\sigma = 6406\ \text{psi}$, which confirms that this root lands on $\sigma_1$ and not on $\sigma_2$ — a check worth making every time, because the $\arctan$ root can belong to either.
8. Orientation of the maximum-shear element. Planes of maximum shear are always 45° from the principal planes: $$\theta_s = \theta_p - 45^\circ = \boxed{-28.2^\circ}$$ On that element the shear is $\tau_{\max} = 3495\ \text{psi}$ and, crucially, all four faces still carry the average normal stress $\sigma_{\text{avg}} = 2911\ \text{psi}$ — the maximum-shear element is not free of normal stress.
   Q6 — the three required elements: the x-y element as computed, the principal element rotated 16.8° counterclockwise, and the maximum-shear element at −28.2°.
9. Three-dimensional check. The out-of-plane principal stress is zero (a free surface region under plane stress), and because $\sigma_1$ and $\sigma_2$ straddle zero, the in-plane Mohr circle is already the largest of the three circles. The absolute maximum shear therefore equals the in-plane value, $\tau_{\text{abs,max}} = (\sigma_1-\sigma_2)/2 = 3495\ \text{psi}$. Had both principal stresses been of the same sign, the governing circle would have been one of the out-of-plane pair and the answer would be larger — a distinction worth stating explicitly.