22-Mec-A4 Design and Manufacture of Machine Elements · December 2013

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

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

Notes on this paper

Paper format. National Examination, 07-Mec-A4 Design and Manufacture of Machine Elements, December 2013 — 3 hours, open book, any non-communicating calculator permitted. Eight questions on six pages, divided into Part A (manufacturing processes, Q1–Q4) and Part B (machine-element design, Q5–Q8). The rubric asks for three questions from Part A and two from Part B, five questions constituting a complete paper, all of equal value (20 % each). All eight questions are solved here.

Reference texts.

Kalpakjian & Schmid, Manufacturing Engineering and Technology, 7th ed. — sand casting and casting defects (Ch. 10–12), adhesive bonding and joint design (Ch. 32), fusion welding and weld defects (Ch. 30–31), sheet-metal shearing and blanking (Ch. 16).
Budynas & Nisbett, Shigley's Mechanical Engineering Design, 10th/11th ed. — eccentrically loaded shear joints (Ch. 8–9), combined stresses and stress transformation (Ch. 3), statically indeterminate shafts and deflection (Ch. 4), brakes and clutches (Ch. 16).
Hibbeler, Mechanics of Materials, 10th ed. — transverse shear ($VQ/Ib$), combined loading and Mohr's circle (Ch. 7–9).
Faires, Design of Machine Elements, 4th ed. — double-block (double-shoe) brakes, self-energizing geometry, equal-wear proportioning.
ASM Handbook Vol. 15 Casting and Vol. 6 Welding, Brazing and Soldering — hot-spot/shrinkage defects; hydrogen-induced cold cracking and preheat practice (see also CSA W59 and CSA W47.1 for Canadian fabrication practice).

Check: Part B is figure-driven. Every dimension used below was read from the printed figures. Two readings are stated explicitly in Given so a grader can substitute a different interpretation without redoing the method: (i) in Figure A the low rivet is taken as lying on the same vertical centreline as the third rivet of the top row (75 + 75 = 150 mm from the left-hand rivet); (ii) in Figure D the dimension $a$ is the horizontal spacing, measured along the operating lever, between the pin taking the upper shoe link and the pin taking the lower shoe link, with the 10 in operating arm measured from the lower-link pin.

Question 6: Stress element at A, Mohr's circle, and the 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.

Quantity	Value
Section, width $b$ × depth $h$	75 mm × 100 mm
Transverse load	36 000 N downward, at 1500 mm from the wall
Axial load	67 500 N tension, applied at mid-depth
Section of interest	1300 mm from the wall
Point $A$	75 mm below the top face, i.e. 25 mm above the bottom face → 25 mm below the neutral axis
Axes	$x$ along the beam (to the right), $y$ upward

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

Figure 6.1 — The cantilever of Figure B. At the section 1300 mm from the wall the internal actions are $N=67.5$ kN tension, $V=36$ kN and $M=7.2\times10^{6}$ N·mm hogging.

Approach. Cut the beam at 1300 mm, take the free body to the right of the cut to get the internal axial force, shear and bending moment; convert each to a stress at $A$ using $N/A$, $My/I$ and $VQ/Ib$; then transform with Mohr's circle.

Section properties. For the solid rectangle, $$A = bh = 75\times100 = 7500\ \text{mm}^2,\qquad I = \frac{bh^3}{12}=\frac{75\times100^3}{12}=6.25\times10^{6}\ \text{mm}^4.$$
Internal actions at the section. Taking the free body to the right of the cut, the only transverse load on it is the 36 kN at the tip, 200 mm further out, so $$V = 36\,000\ \text{N},\qquad M = 36\,000\times(1500-1300)=7.20\times10^{6}\ \text{N}\cdot\text{mm},$$ a hogging moment (tension on the top fibre), while the axial force is carried right through: $$N = 67\,500\ \text{N (tension)}.$$
Axial stress. Uniform over the section, $$\sigma_N = \frac{N}{A} = \frac{67\,500}{7500} = +9.0\ \text{MPa (tension).}$$
Bending stress at A. Point $A$ lies 25 mm below the neutral axis, and the moment is hogging, so the bending stress there is compressive: $$\sigma_M = -\frac{M\,y}{I} = -\frac{(7.20\times10^{6})(25)}{6.25\times10^{6}} = -28.8\ \text{MPa.}$$ Superposing the two normal contributions, $$\boxed{\sigma_x = 9.0 - 28.8 = -19.8\ \text{MPa},\qquad \sigma_y = 0}$$
Transverse shear stress at A. The first moment of the area below $A$ about the neutral axis is $$Q = (b\times25)\left(\frac{h}{2}-\frac{25}{2}\right) = (75\times25)(50-12.5)=70\,312.5\ \text{mm}^3,$$ so $$|\tau_{xy}| = \frac{VQ}{Ib} = \frac{36\,000\times70\,312.5}{(6.25\times10^{6})(75)} = 5.4\ \text{MPa.}$$ On the $+x$ face of the element the transverse shear acts downward (the piece of beam to the right of the cut must be held up), so with $x$ to the right and $y$ upward, $\tau_{xy} = -5.4$ MPa.
Mohr's circle. The centre and radius are $$\sigma_{\text{avg}} = \frac{\sigma_x+\sigma_y}{2} = \frac{-19.8+0}{2} = -9.9\ \text{MPa},$$ $$R = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = \sqrt{(-9.9)^2+(5.4)^2} = \sqrt{98.01+29.16}=11.28\ \text{MPa.}$$
Principal stresses and their orientation. The principal stresses are the two ends of the horizontal diameter, $$\sigma_{1,2}=\sigma_{\text{avg}}\pm R = -9.9 \pm 11.28,$$ $$\boxed{\sigma_1 = +1.38\ \text{MPa},\qquad \sigma_2 = -21.18\ \text{MPa}}$$ and the orientation follows from $$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(-5.4)}{-19.8} \;\Rightarrow\; 2\theta_p = -151.4^\circ,\qquad \theta_p = -75.7^\circ.$$ That is, the $\sigma_1$ axis lies 75.7° clockwise from the $x$-axis; equivalently the large compressive stress $\sigma_2$ acts on a plane whose normal is only 14.3° counter-clockwise from the beam axis — which is the physically sensible result, since $\sigma_x$ is compressive and dominates.
Maximum-shear element. The maximum in-plane shear is the radius of the circle, $$\boxed{\tau_{\max}=R=11.28\ \text{MPa}}$$ acting on planes 45° from the principal planes, i.e. at $$\theta_s = \theta_p + 45^\circ = -30.7^\circ,$$ and on those faces the normal stress on every side is the mean stress, $\sigma_{\text{avg}}=-9.9$ MPa.

Figure 6.2 — (i) the element at A with horizontal and vertical sides; (ii) the Mohr circle through X(σx, −τxy) and Y; (iii) the principal element, rotated 75.7° clockwise; (iv) the maximum-shear element at 30.7° clockwise, carrying the mean normal stress on every face.

Quantity	Value
$A$, $I$	7500 mm², 6.25 × 10⁶ mm⁴
$N$, $V$, $M$ at the section	67 500 N (T), 36 000 N, 7.20 × 10⁶ N·mm (hogging)
$\sigma_x$ at A (axial + bending)	+9.0 − 28.8 = −19.8 MPa
$\sigma_y$	0
$\tau_{xy}$ at A	−5.4 MPa (magnitude 5.4 MPa)
Mohr centre $\sigma_{\text{avg}}$, radius $R$	−9.9 MPa, 11.28 MPa
Principal stresses $\sigma_1,\ \sigma_2$	+1.38 MPa, −21.18 MPa
Principal orientation $\theta_p$ (to the $\sigma_1$ axis)	−75.7° (75.7° clockwise from $x$)
Maximum in-plane shear $\tau_{\max}$	11.28 MPa at $\theta_s = -30.7^\circ$
Normal stress on the maximum-shear planes	−9.9 MPa (all four faces)

Topic. Combined axial, bending and transverse-shear loading; plane-stress transformation and Mohr's circle.

Key relations. $\sigma = N/A \pm My/I$; $\tau = VQ/(Ib)$; $\sigma_{\text{avg}}=(\sigma_x+\sigma_y)/2$; $R=\sqrt{[(\sigma_x-\sigma_y)/2]^2+\tau_{xy}^2}$; $\sigma_{1,2}=\sigma_{\text{avg}}\pm R$; $\tan 2\theta_p = 2\tau_{xy}/(\sigma_x-\sigma_y)$; $\tau_{\max}=R$ at $\theta_p\pm45^\circ$. Invariants: $\sigma_1+\sigma_2=\sigma_x+\sigma_y$ and $\sigma_1\sigma_2=\sigma_x\sigma_y-\tau_{xy}^2$.

Why this works. Each internal action produces its own stress field on the cut face, and because the material is linear-elastic and the deformations are small, those fields simply add. The point $A$ is interesting precisely because it is neither on the neutral axis nor at an extreme fibre, so it carries a non-trivial normal stress and a non-trivial shear stress at the same time — the general plane-stress state. Mohr's circle is then nothing more than a graphical statement of the transformation equations: every point on the circle is the $(\sigma,\tau)$ pair on some plane through $A$, a rotation of $\theta$ on the physical element corresponds to a rotation of $2\theta$ on the circle, and the principal planes are where the circle crosses the $\sigma$ axis (zero shear) while the maximum-shear planes are its top and bottom, necessarily 45° away on the element.

Common pitfall. Using the extreme-fibre shear (zero) or the neutral-axis shear ($1.5V/A = 7.2$ MPa) instead of computing $Q$ at $A$ itself; and taking $y$ as the 75 mm distance quoted from the top face rather than the 25 mm distance from the neutral axis. It is also easy to lose the sign of $\tau_{xy}$ and end up with $\theta_p = +75.7^\circ$, which would put the large compressive stress at right angles to the beam axis — physically wrong for a bar whose dominant stress is axial compression at $A$. A quick sanity check is that $\sigma_2$ must come out nearly aligned with $x$.

Source. Hibbeler, Mechanics of Materials, 10th ed., Ch. 7 (transverse shear, $VQ/Ib$), Ch. 8 (combined loadings) and Ch. 9 (stress transformation, Mohr's circle); Budynas & Nisbett, Shigley's Mechanical Engineering Design, §3–6 to §3–7.

← Question 5 Next: Question 7 →

Quantity	Value
\(A\), \(I\)	7500 mm², 6.25 × 10⁶ mm⁴
\(N\), \(V\), \(M\) at the section	67 500 N (T), 36 000 N, 7.20 × 10⁶ N·mm (hogging)
\(\sigma_x\) at A (axial + bending)	+9.0 − 28.8 = −19.8 MPa
\(\sigma_y\)	0
\(\tau_{xy}\) at A	−5.4 MPa (magnitude 5.4 MPa)
Mohr centre \(\sigma_{\text{avg}}\), radius \(R\)	−9.9 MPa, 11.28 MPa
Principal stresses \(\sigma_1,\ \sigma_2\)	+1.38 MPa, −21.18 MPa
Principal orientation \(\theta_p\) (to the \(\sigma_1\) axis)	−75.7° (75.7° clockwise from \(x\))
Maximum in-plane shear \(\tau_{\max}\)	11.28 MPa at \(\theta_s = -30.7^\circ\)
Normal stress on the maximum-shear planes	−9.9 MPa (all four faces)

Quantity	Value
Section, width \(b\) × depth \(h\)	75 mm × 100 mm
Transverse load	36 000 N downward, at 1500 mm from the wall
Axial load	67 500 N tension, applied at mid-depth
Section of interest	1300 mm from the wall
Point \(A\)	75 mm below the top face, i.e. 25 mm above the bottom face → 25 mm below the neutral axis
Axes	\(x\) along the beam (to the right), \(y\) upward