22-Mec-A4 Design and Manufacture of Machine Elements · December 2016

Question 6 of 6

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

Question 6 (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. (a) the force F transmitted per pitch, set by rivet shear; (b) the values of t, t′ and P that make every other failure mode reach its capacity at that same F; (c) the efficiency of the resulting joint.

Approach. Analyse one repeating strip of joint width equal to the pitch P. Within that strip there is exactly one rivet on each side of the butt, and because the joint is double-strapped that rivet is in double shear. Compute the shear capacity first, then force each of the remaining modes — bearing on the main plate, bearing on the straps, tension across the net section of the main plate, tension across the net section of the straps — to reach exactly the same capacity. That is what "balanced design" means.

1. (a) Rivet shear capacity per pitch. One rivet per pitch, two shear planes: $$A_{r}=\frac{\pi d^{2}}{4}=\frac{\pi(10)^{2}}{4}=78.54\ \text{mm}^{2}$$ $$F=2A_{r}S_{ys}=2(78.54)(170)=26\,703\ \text{N}$$ $$\boxed{F=26.7\ \text{kN per pitch of joint width}}$$ The factor of two is the whole reason double-strap butt joints are used: the same rivet does twice the work it would in a lap joint.
2. (b) Bearing on the main plate fixes t. The full force F is transferred from the main plate to the rivet shank over the projected bearing area dt: $$F=d\,t\,S_{yc}\quad\Rightarrow\quad t=\frac{F}{dS_{yc}}=\frac{26\,703}{10(300)}=8.90\ \text{mm}$$ $$\boxed{t=8.90\ \text{mm}}$$
3. Bearing on the straps fixes t′. Each of the two straps carries only half of F, so its projected bearing area need only be half as large: $$\frac{F}{2}=d\,t'\,S_{yc}\quad\Rightarrow\quad t'=\frac{F}{2dS_{yc}}=\frac{26\,703}{2(10)(300)}=4.45\ \text{mm}$$ $$\boxed{t'=4.45\ \text{mm}=\tfrac{1}{2}t}$$ The straps are exactly half the thickness of the main plate, which is the standard proportion for a balanced double-strap joint.
4. Tension across the net section of the main plate fixes P. Over one pitch, the material left to carry tension is a width \(P-d\): $$F=(P-d)\,t\,S_{yt}\quad\Rightarrow\quad P-d=\frac{26\,703}{8.90(300)}=10.0\ \text{mm}$$ $$\boxed{P=d+10.0=20.0\ \text{mm}=2d}$$ The result P = 2d is not a coincidence: equating the bearing capacity \(dtS_{yc}\) with the net-section capacity \((P-d)tS_{yt}\) gives \(P-d=d\,S_{yc}/S_{yt}\), and here the two yield strengths are equal.
5. Check the remaining mode — strap net-section tension. With the same pitch, each strap must carry F/2 across its own net section: $$(P-d)t'S_{yt}=(20.0-10.0)(4.45)(300)=13\,352\ \text{N}=\frac{F}{2}\ \checkmark$$ It is satisfied exactly by the t′ already found, so no criterion is left over-strength and the design is genuinely balanced: rivet shear, main-plate bearing, strap bearing, main-plate tension and strap tension all reach capacity at F = 26.7 kN together.
6. (c) Efficiency. A continuous plate of the same thickness and of width equal to one pitch would carry $$F_{\text{solid}}=P\,t\,S_{yt}=20.0(8.90)(300)=53\,407\ \text{N}$$ so $$\eta=\frac{F}{F_{\text{solid}}}=\frac{26\,703}{53\,407}=0.500$$ $$\boxed{\eta=50\ \%}$$ Equivalently \(\eta=(P-d)/P=10/20=0.50\): with a single row of rivets, the efficiency is simply the fraction of plate width remaining at the rivet line. That is the fundamental limitation of single-row riveting, and it is why pressure-vessel and boiler seams use double- or triple-riveted zig-zag patterns, which push the efficiency to 70–85 % by spacing the holes further apart and adding rows.