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

Question 5 of 8: Force on the most heavily loaded rivet

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 5: Force on the most heavily loaded rivet (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
Rivets in the top row (measured from the left-hand rivet)	$x = 0,\;75,\;150,\;362.5,\;462.5$ mm, all at $y = 0$
Sixth (low) rivet	$x = 150$ mm, $y = -200$ mm
Number of rivets, $n$	6, all the same diameter
Applied load, $P$	50 000 N, horizontal, acting to the left
Line of action of $P$	through the centre of the low rivet, $y = -200$ mm

Find. The resultant shear force carried by the most heavily loaded rivet of the group.

Figure 5.1 — Rivet group, centroid and applied load. The eccentricity of the 50 kN load about the centroid G produces a torsional shear that adds to the direct shear.

Approach. Treat the group as an eccentrically loaded shear joint: locate the centroid of the rivet areas, resolve the applied load into a force through the centroid plus a couple, share the force equally among the rivets as direct shear, distribute the couple as torsional shear proportional to each rivet's distance from the centroid, and add the two vectorially rivet by rivet.

Locate the centroid of the rivet group. With all rivets of equal area the centroid is the simple average of the hole centres, $$\bar{x}=\frac{\sum x_i}{n},\qquad \bar{y}=\frac{\sum y_i}{n}.$$ Substituting the six positions, $$\bar{x}=\frac{0+75+150+362.5+462.5+150}{6}=\frac{1200}{6}=200\ \text{mm},$$ $$\bar{y}=\frac{0+0+0+0+0-200}{6}=-33.33\ \text{mm}.$$ So $G=(200,\,-33.33)$ mm, measured from the left-hand rivet of the top row.
Reduce the applied load to the centroid. Moving the 50 kN force to $G$ requires adding the couple it exerts about $G$. The load acts through $(150,-200)$, whose position relative to the centroid is $\mathbf{r}=(-50,\,-166.67)$ mm, and $\mathbf{P}=(-50\,000,\,0)$ N. Hence $$M = r_x P_y - r_y P_x = (-50)(0)-(-166.67)(-50\,000),$$ $$\boxed{M = -8.333\times10^{6}\ \text{N}\cdot\text{mm}}$$ the negative sign meaning a clockwise couple of magnitude 8.333 kN·m.
Compute the polar second moment of the rivet areas. For equal-area rivets it is convenient to work "per unit area", so that $$J = \sum \left(a_i^2 + b_i^2\right),\qquad a_i = x_i-\bar{x},\ \ b_i = y_i-\bar{y}.$$ Rivet by rivet: $(-200,33.33)$, $(-125,33.33)$, $(-50,33.33)$, $(162.5,33.33)$, $(262.5,33.33)$ and $(-50,-166.67)$ mm, giving $41\,111+16\,736+3611+27\,517+70\,017+30\,278$ and therefore $$J = 189\,271\ \text{mm}^{2}.$$
Direct shear. The centroidal force divides equally between the six rivets, $$\mathbf{F}_{d}=\frac{\mathbf{P}}{n}=\frac{-50\,000}{6}=-8333\ \text{N (i.e. 8.333 kN to the left on every rivet).}$$
Torsional shear. The couple produces on each rivet a force perpendicular to its radius from $G$, of magnitude $M r_i / J$; in components, $$F_{Mx} = -\frac{M\,b_i}{J},\qquad F_{My} = \frac{M\,a_i}{J}.$$ Because $b_i$ is the same ($+33.33$ mm) for the whole top row, every top-row rivet picks up the same horizontal component $F_{Mx}=+1468$ N, while their vertical components grow with distance from $G$. The low rivet, with $b_6=-166.67$ mm, instead picks up $F_{Mx}=-7338$ N — which acts in the same direction as the direct shear and is why that rivet governs.
Add the two contributions vectorially and pick the largest. Summing components and taking $F_i=\sqrt{F_{ix}^2+F_{iy}^2}$ gives the table below; the low rivet carries $$F_{6}=\sqrt{(-15\,671)^2+(2201)^2}\ \text{N},$$ $$\boxed{F_{\max}=F_{6}=15\,825\ \text{N}\approx 15.8\ \text{kN}}$$

The margin over the next-worst rivet is not large — rivet 5, the far right-hand one, carries 13.44 kN — so a design based on this joint would size all the rivets on 15.8 kN and would not rely on the distinction.

Rivet	Position rel. to G (mm)	$F_x$ (N)	$F_y$ (N)	Resultant (N)
1	(−200.0, +33.3)	−6866	+8806	11 166
2	(−125.0, +33.3)	−6866	+5504	8799
3	(−50.0, +33.3)	−6866	+2201	7210
4	(+162.5, +33.3)	−6866	−7155	9916
5	(+262.5, +33.3)	−6866	−11 558	13 443
6 (low rivet)	(−50.0, −166.7)	−15 671	+2201	15 825
Centroid of group				(200, −33.33) mm
Couple about the centroid, $M$				8.333 × 10⁶ N·mm (clockwise)
Polar second moment, $J$ (unit area)				189 271 mm²
Most heavily loaded rivet				No. 6, F = 15 825 N ≈ 15.8 kN

Topic. Eccentrically loaded riveted / bolted shear joints — the direct-plus-torsional shear superposition.

Key relations. Centroid $\bar{x}=\sum A_i x_i/\sum A_i$; reduction of an offset load to a centroidal force plus a couple $M = P e$; direct shear $F_d = P/n$ for equal fasteners; torsional shear $F_{M,i} = M r_i / J$ with $J=\sum(a_i^2+b_i^2)$ for equal areas; resultant $F_i = |\mathbf{F}_d + \mathbf{F}_{M,i}|$.

Why this works. The joint plate is assumed rigid relative to the fasteners, so under the couple it rotates about the centroid of the fastener group as a rigid body. Each fastener's shear deformation is then proportional to its distance from that centre of rotation, and, since the fasteners are elastic and identical, so is the force it carries — which is exactly the torsion formula $\tau = Tr/J$ applied to a discrete set of areas. The direct shear is uniform because a pure translation of a rigid plate strains every fastener equally. The two effects superpose because both are linear-elastic, but they must be added as vectors, not magnitudes: the governing rivet is the one where the two contributions happen to point the same way, which is not necessarily the one farthest from the centroid.

Common pitfall. Adding $F_d$ and $F_M$ arithmetically, or assuming the farthest rivet automatically governs. Here rivet 5 is farthest from $G$ ($r=264.6$ mm) yet carries only 13.4 kN, because its torsional component is nearly perpendicular to the direct shear, while rivet 6 at $r=174.0$ mm carries 15.8 kN because its torsional component is almost collinear with it. A second pitfall is forgetting to include the low rivet when locating the centroid, which throws $\bar{y}$ to zero and every subsequent number with it.

Source. Budynas & Nisbett, Shigley's Mechanical Engineering Design, 10th/11th ed., §8–11 "Shear joints with eccentric loading" (and the parallel treatment for welds in §9–5); Hibbeler, Mechanics of Materials, Ch. 5 for the underlying torsion analogy.

← Question 4 Next: Question 6 →

Quantity	Value
Rivets in the top row (measured from the left-hand rivet)	\(x = 0,\;75,\;150,\;362.5,\;462.5\) mm, all at \(y = 0\)
Sixth (low) rivet	\(x = 150\) mm, \(y = -200\) mm
Number of rivets, \(n\)	6, all the same diameter
Applied load, \(P\)	50 000 N, horizontal, acting to the left
Line of action of \(P\)	through the centre of the low rivet, \(y = -200\) mm