25-Nav-A6 Advanced Strength of Materials (25-Mec-A6) · May 2016

Question 6 of 6: Bolt-Cutter Linkage — Pin Forces and Member Stresses

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

Notes on this paper

Reference texts: Kalpakjian & Schmid, Manufacturing Engineering and Technology, 8th ed.; Groover, Fundamentals of Modern Manufacturing, 7th ed.; Hibbeler, Mechanics of Materials, 10th ed.; Hibbeler, Engineering Mechanics: Statics, 14th ed.; Shigley's Mechanical Engineering Design, 11th ed.

Check: the exam is headed "98-Mar-A6, Design and Manufacture of Machine Elements" and Part A (Q1–Q3) is entirely machining/casting/forming content with zero naval-architecture material. Solved here as the paper actually printed; reference texts above are chosen for the real content. Part B (Q4–Q6) is genuine strength-of-materials/machine-design content.

Check: Q5 states the shaft modulus as "E = 30 ksi," which is off by three orders of magnitude for steel (actual E ≈ 29,000–30,000 ksi = 30×10⁶ psi); the shaft's computed self-weight, w = γA = 0.283 × (π/4)(3²) = 2.00 lb/in, comes out to a clean round number confirming the 3-in-diameter reading, and the printed "30 ksi" is treated as a dropped exponent (E = 30×10⁶ psi used throughout).

Question 6: Bolt-Cutter Linkage — Pin Forces and Member Stresses (equal value, Part B)

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. Symmetric two-handle bolt-cutter: each handle pivots on the shared central pin C and squeezes with input force $P$ at the grip; each handle connects through a two-force link (member 2 on top, member 3, its mirror, on the bottom) from pin B (on the handle) to pin A (on the jaw), which transmits the clamping force to the bolt.

Given data
Quantity	Value
Input squeeze force $P$	2 lb (per handle)
Dimensions $a,b,c,d,e$	1, 3, 2, 8, 1 in.
Yield strength $S_y$	60.9 ksi
Shear yield $S_{ys}=0.5S_y$	30.45 ksi
Modulus $E$	$30\times10^6$ psi

Find. The force exerted on the bolt, the pin forces at A, B, and C, and the maximum stress in members 2 and 3.

Free-body of the top handle (member 1), pivoting at C; the lower handle and member 3 mirror this exactly by symmetry.

Approach. Isolate one handle as a free body pivoting about the fixed pin C. Since member 2 (pin A to pin B) is a two-force member, the force it applies to the handle at B is purely axial (horizontal, along A–B); take moments about C to solve that axial force directly from the given lever arms, then use equilibrium of the handle to get the pin-C reaction, and equilibrium of the jaw (also fed only by the same axial link force) to carry the force straight through to the bolt.

Moment arms about the fixed pivot C. From the figure, $P$ acts vertically with horizontal lever arm $d=8$ in about C, and the horizontal link force at B acts with vertical lever arm $e=1$ in about C (the handle centreline sits $e$ above C). Taking moments of the handle about C: $$\sum M_C=0:\qquad P\,d=F_2\,e$$ $$\boxed{F_2=\frac{Pd}{e}=\frac{2(8)}{1}=16.0\text{ lb (tension in member 2)}}$$
Pin B and pin A. Only member 2 and the handle meet at B, and member 2 is a two-force member, so the pin force at B equals the member's axial force: $F_B=F_2=16.0$ lb. By Newton's third law along the same two-force member, the force delivered to the jaw at pin A is likewise $F_A=16.0$ lb, directed along A–B.
Pin C reaction (handle equilibrium). Summing forces on the handle, with $P=(0,-2)$ lb applied at the grip and the link force $(-16,0)$ lb applied at B (pulling the handle toward A, confirming tension): $$\sum F_x=0:\ C_x=+16.0\text{ lb}\qquad \sum F_y=0:\ C_y=+2.0\text{ lb}$$ $$\boxed{|F_C|=\sqrt{16.0^2+2.0^2}=16.1\text{ lb}}$$
Force on the bolt. The jaw half is loaded only by the link force at A and the bolt contact force, so (as a two-force body) the bolt contact force equals the link force transmitted straight through: $$\boxed{F_{bolt,\ per\ jaw}=16.0\text{ lb}\qquad F_{bolt,\ total\ clamping}=2(16.0)=32.0\text{ lb}}$$ (the top and bottom jaw halves squeeze the bolt with equal and opposite 16.0 lb contact forces; members 2 and 3 carry identical loads by the top–bottom symmetry of the mechanism.)
Maximum stress in members 2 and 3. The source figure does not give the link plate's width/thickness or the pin diameter (only $a,b,c,d,e$, which locate the pins, are legible) — a representative small flat-link section consistent with this scale of tool is assumed here (see check note) to complete the stress check: width $0.25$ in, thickness $0.0625$ in ($A=0.0156\text{ in}^2$), pin diameter $0.125$ in ($A_{shear}=0.0123\text{ in}^2$). $$\sigma_{2,3}=\frac{F_2}{A}=\frac{16.0}{0.0156}=1{,}024\text{ psi}\qquad \text{FS}_{yield}=\frac{S_y}{\sigma}=\frac{60{,}900}{1{,}024}\approx60$$ $$\tau_{pin}=\frac{F_2}{A_{shear}}=\frac{16.0}{0.0123}=1{,}304\text{ psi}\qquad \text{FS}_{shear}=\frac{S_{ys}}{\tau}=\frac{30{,}450}{1{,}304}\approx23$$ Both stresses sit far below yield — the very small 2 lb input load makes this a lightly stressed problem regardless of the exact assumed cross-section; the factor of safety would stay comfortably in the tens even for a plate half or double the assumed thickness.

Question 6 — final results
Quantity	Value
Link 2 (and, by symmetry, link 3) axial force	16.0 lb, tension
Pin B force	16.0 lb
Pin C (handle pivot) reaction	16.1 lb ($C_x=16.0$, $C_y=2.0$ lb)
Force on bolt, per jaw half	16.0 lb
Total bolt clamping force	32.0 lb
Max stress, members 2 & 3 (assumed section)	≈ 1,024 psi tensile (FS ≈ 60 on $S_y$)
Max pin shear stress (assumed pin dia.)	≈ 1,304 psi (FS ≈ 23 on $S_{ys}$)

Check: members 2/3's cross-section and the pin diameter are not legible in the source figure (only the pin-location dimensions $a$–$e$ survived); a small flat-link section (0.25 in × 0.0625 in) and a 0.125-in pin were assumed to complete the stress check. The pin/joint force results (16.0–16.1 lb) do not depend on this assumption; only the reported stress magnitudes and factors of safety do, and both remain comfortably safe under any reasonable re-scaling of the assumed section, given the very light 2 lb input load.

Topic: Force analysis of a pin-connected multi-force machine (frames-and-machines statics) and basic pin/member stress checks.

Key relations: two-force member carries only axial load along the line joining its two pins; moment equilibrium of a rigid body about any point, $\sum M=0$; direct stress $\sigma=F/A$; single-shear pin stress $\tau=F/A_{shear}$.

Why this works: Recognizing member 2 as a two-force member collapses what looks like a multi-unknown frame problem into a single moment equation about the fixed pivot C — the lever-arm ratio $d/e=8$ is exactly the tool's mechanical advantage, turning a 2 lb hand force into a 16 lb clamping force at the jaw, before any pin-force or stress bookkeeping is needed at all.

Common pitfall: trying to resolve the link force into components before recognizing it is a two-force member (which fixes its direction along A–B immediately), or assuming the bolt receives the full input force $P$ rather than the amplified force after the lever-and-link mechanical advantage.

Source: Hibbeler, Engineering Mechanics: Statics, Ch.6–7 (frames and machines, two-force members); Shigley's Mechanical Engineering Design, Ch.3–4 (pin/bearing stress, factor of safety).

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