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

Question 5 of 8: Force on the most heavily loaded rivet in an eccentrically loaded group

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, May 2014 — 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 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, because this document is a study resource rather than an examination script.

Reference texts.

Kalpakjian & Schmid, Manufacturing Engineering and Technology, 7th ed. — sand casting and casting defects (Ch. 10–12), sheet-metal shearing and fine blanking (Ch. 16), bulk deformation and hydrostatic extrusion (Ch. 15), fusion welding and weld defects (Ch. 30–31).
Budynas & Nisbett, Shigley's Mechanical Engineering Design, 10th/11th ed. — eccentrically loaded shear joints (Ch. 8), statically indeterminate beams and deflection superposition (Ch. 4), welded joints in torsion (Ch. 9).
Hibbeler, Mechanics of Materials, 10th ed. — combined loading, transverse shear and stress transformation / Mohr's circle (Ch. 7–9).
AWS D1.1 Structural Welding Code — Steel and CSA W59 — preheat, hydrogen control and minimum fillet-weld sizes (Canadian practice).
ASM Handbook Vol. 15 Casting and Vol. 6 Welding, Brazing and Soldering — hot tearing, solidification cracking and riser/chill practice.

Check: Part B is entirely figure-driven. Every number below was read from the printed figures (Figures A, B, C and S7). Two readings are worth stating explicitly so a grader can substitute a different interpretation without redoing the method: (i) in Figure A the rivet group is five rivets in the top row plus one rivet 200 mm below, the lower rivet lying on the same vertical line as the third top rivet; (ii) in Figure B the 67 500 N horizontal force acts on the centroidal axis of the section, so it produces pure tension and no additional bending.

PART A — Manufacturing Processes

Question 5: Force on the most heavily loaded rivet in an eccentrically loaded group (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. The bracket of Figure A is attached by six rivets: five in a horizontal top row and one 200 mm below that row, on the same vertical line as the third top rivet. Reading the figure with the leftmost top rivet as the origin, \(x\) positive to the right and \(y\) positive upward:

Rivet	\(x\) (mm)	\(y\) (mm)	Comment
R1	0	0	top row — pitches 75, 75, 212.5, 100 mm
R2	75	0
R3	150	0	lies above R6
R4	362.5	0
R5	462.5	0
R6	150	−200	lower rivet, on the load line

The applied load is \(F = 50\,000\ \text{N}\), horizontal, directed to the left (\(-x\)), with its line of action passing through the level of R6, i.e. \(y = -200\ \text{mm}\). All rivets are of the same size, so each carries load in proportion to its distance from the group centroid.

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

Figure 5.1 — Rivet group of Figure A, with the group centroid G and the 50 kN eccentric load. Coordinates in mm, \(y\) positive upward.

Approach. Locate the centroid of the rivet group, resolve the applied load into a force through the centroid plus a couple about it, distribute the force equally among the rivets (primary shear) and the couple in proportion to each rivet's radius from the centroid (secondary shear), then add the two contributions vectorially at every rivet and take the largest resultant.

Locate the centroid of the rivet group. With all rivets the same size the centroid is the simple average of the hole centres, \[\bar{x} = \frac{\sum x_i}{n} = \frac{0+75+150+362.5+462.5+150}{6} = \frac{1200}{6} = 200\ \text{mm},\] \[\bar{y} = \frac{\sum y_i}{n} = \frac{0+0+0+0+0-200}{6} = -33.33\ \text{mm}.\] The single lower rivet pulls the centroid 33.3 mm below the top row, which is what makes R6 the critical fastener rather than the far-right rivet R5.
Transfer the applied load to the centroid. Sliding \(F\) to the centroid requires the addition of a couple equal to the force times its perpendicular offset. The load is horizontal and acts 200 mm below the top row, i.e. \(200 - 33.33 = 166.67\ \text{mm}\) below the centroid, so \[M = F\,e = 50\,000 \times 166.67 = 8.333 \times 10^{6}\ \text{N}\cdot\text{mm}.\] Because the force points to the left and passes below G, the couple is clockwise. The joint is therefore loaded by a 50 kN force through G plus a clockwise moment of 8.333 kN·m, giving \[\boxed{V = 50\,000\ \text{N} \quad\text{and}\quad M = 8.333\times10^{6}\ \text{N}\cdot\text{mm (CW)}}\]
Primary (direct) shear. The centroidal force is shared equally by all six rivets, \[F' = \frac{V}{n} = \frac{50\,000}{6} = 8333.3\ \text{N},\] acting on every rivet in the direction of the applied load, i.e. horizontally to the left.

Compute \(\sum r_i^{2}\) for the secondary shear. The moment is resisted by forces proportional to the radius from the centroid, so the distribution constant needs the polar sum. Taking \(r_x = x_i - 200\) and \(r_y = y_i + 33.33\):

Rivet	\(r_x\) (mm)	\(r_y\) (mm)	\(r^2\) (mm²)
R1	−200.0	33.33	41 111
R2	−125.0	33.33	16 736
R3	−50.0	33.33	3 611
R4	162.5	33.33	27 517
R5	262.5	33.33	70 017
R6	−50.0	−166.67	30 278
\(\sum r_i^{2}\)			189 271

Secondary (moment) shear. The moment-induced force on rivet \(i\) is \[F''_i = \frac{M\,r_i}{\sum r_j^{2}},\] directed perpendicular to \(r_i\), in the sense of the moment. It is convenient to work with the constant \[k = \frac{M}{\sum r_j^{2}} = \frac{8.333\times10^{6}}{189\,271} = 44.03\ \text{N/mm},\] so that the components of the secondary force are \(\left(+k\,r_y,\ -k\,r_x\right)\) for a clockwise moment. For the far rivet R5, for example, \(F''_5 = 44.03 \times \sqrt{262.5^2+33.33^2} = 44.03 \times 264.6 = 11\,650\ \text{N}\).
Add the two contributions vectorially at each rivet. The primary shear is \((-8333.3,\ 0)\) N at every rivet; the secondary components follow from step 5:

Rivet \(F_x\) (N) \(F_y\) (N) Resultant (N)

R1 −6866 8806 11 166

R2 −6866 5504 8 799

R3 −6866 2201 7 210

R4 −6866 −7155 9 916

R5 −6866 −11 558 13 443

R6 −15 671 2201 15 825

Notice why R6 governs even though R5 is further from the centroid. At R5 the secondary force is almost vertical while the primary force is horizontal, so the two are nearly perpendicular and combine as a hypotenuse. At R6 the radius is nearly vertical, so its secondary force is nearly horizontal — and it points in the same direction as the primary shear, so the two add almost arithmetically.
Report the governing rivet. \[F_{\max} = \sqrt{15\,671^{2} + 2201^{2}} = \boxed{15\,825\ \text{N} \approx 15.8\ \text{kN at rivet R6}}\]

Rivet	\(F_x\) (N)	\(F_y\) (N)	Resultant (N)
R1	−6866	8806	11 166
R2	−6866	5504	8 799
R3	−6866	2201	7 210
R4	−6866	−7155	9 916
R5	−6866	−11 558	13 443
R6	−15 671	2201	15 825

Quantity	Value
Group centroid G	(200, −33.33) mm from R1
Eccentricity of the load from G	166.67 mm
Moment about G	8.333 × 10⁶ N·mm (clockwise)
\(\sum r_i^{2}\)	189 271 mm²
Primary shear per rivet	8 333 N
Most heavily loaded rivet	R6 (the lower rivet)
Force on that rivet	15 825 N ≈ 15.8 kN
Next most loaded	R5, 13 443 N

Topic. Eccentrically loaded shear joints (riveted or bolted) — the primary/secondary shear decomposition.

Key relations. Centroid \(\bar{x} = \sum A_i x_i / \sum A_i\) (reduces to the mean for equal fasteners); primary shear \(F' = V/n\); secondary shear \(F''_i = M r_i / \sum r_j^{2}\), perpendicular to \(r_i\); resultant by vector addition.

Why this works. The connected member is treated as rigid, so the plate's motion relative to the bracket is a rotation about some instantaneous centre plus a translation. Decomposing that motion into a pure translation of the centroid plus a rotation about the centroid makes the fastener forces linear in the rivet coordinates: translation gives every rivet the same displacement and therefore the same force, while rotation gives each rivet a displacement proportional to its radius, hence a force proportional to \(r_i\) and perpendicular to it. Equating the moment of those forces to the applied moment produces the \(M r_i / \sum r_j^2\) distribution — formally identical to the torsion formula \(\tau = Tr/J\), with \(\sum r_j^2\) playing the part of the polar second moment.

Common pitfall. Adding the primary and secondary magnitudes arithmetically, or assuming the rivet furthest from the centroid always governs. Both errors appear here: R5 is furthest but R6 governs, because at R6 the two vectors are nearly collinear. Always add as vectors and always check every fastener — the check is trivial once \(k\) is known.

Source. Budynas & Nisbett, Shigley's Mechanical Engineering Design, 10th ed., §8-11 (Shear joints with eccentric loading); Hibbeler, Mechanics of Materials, §8.4.

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