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

Question 6 of 6: Bolt Cutter — Force on the Bolt, Pin Loads, and Lever Stresses

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

Notes on this paper

Paper format. National Examinations, May 2016 — 07-Mec-A4, Design and Manufacture of Machine Elements. Three hours, open book, any non-communicating calculator. Six questions in two parts: Part A (Q1–Q3, manufacturing-process theory) and Part B (Q4–Q6, machine-element analysis). Candidates answer two from Part A and two from Part B; four questions constitute a complete paper and all questions carry equal value (25 %). All six are solved here.

Reference texts. Kalpakjian & Schmid, Manufacturing Engineering and Technology (Part A); Groover, Fundamentals of Modern Manufacturing (Part A); Budynas & Nisbett, Shigley's Mechanical Engineering Design, 11th ed. (Part B); Hibbeler, Mechanics of Materials, 10th ed. (Q4, Q5); Norton, Machine Design: An Integrated Approach, 6th ed. (Q6).

Check: two corrections carried through Part B. (1) Q5 prints the shaft modulus as “E = 30 ksi”; a 30 ksi modulus is physically impossible for steel and would make the shaft a rubber band, so it is read as the standard E = 30 × 10⁶ psi (30 Mpsi). (2) Q6 supplies the linkage dimensions but not the cross-section of the levers, so the stress in members 2 and 3 is worked for an explicitly stated assumed section and the required section is also reported. Both readings are flagged where they are used, in the spirit of Note 1 on the cover page.

Question 6: Bolt Cutter — Force on the Bolt, Pin Loads, and Lever Stresses (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.

Quantity	Symbol	Value
Hand load applied to each handle	$P$	2 lbf
Bolt contact to fulcrum pin A	$a$	1 in
Fulcrum pin A to handle pin B	$b$	3 in
Handle pin B to the line of pin C	$c$	2 in
Pin C to the point of application of P	$d$	8 in
Offset of pin C below the A–B pin line	$e$	1 in
Material, AISI 1080 HR — yield strengths	$S_y,\ S_{ys}$	60.9 ksi, 30.45 ksi
Elastic modulus	$E$	$30\times10^{6}$ psi

Find. The force the tool exerts on the bolt, the resultant force carried by each of the pins at A, B and C, and the maximum stress in members 2 and 3 (the two mirror-image levers that carry the jaws).

Q6 schematic of the upper and lower halves of the tool. The two handles (member 1 and its mirror) pin to each other at C; each handle pins at B to the lever on the opposite side; each lever pivots on the fulcrum pin at A and grips the bolt a distance a beyond it.

Check: interpretation of the linkage and of the section. Two readings are stated explicitly. (1) The tool is taken to be mirror-symmetric about the horizontal axis through pin C, with the two handles pinned to each other at C, each handle pinned at B to a lever, and each lever pivoting at A and reaching past it to the bolt. This is the only topology consistent with all five printed dimensions and with a functioning cutter. A consequence of the symmetry is that the pin force at C is purely vertical, so $e$ locates C but drops out of the force result. (2) The figure does not dimension the cross-section of members 2 and 3. The stress is therefore worked for a stated assumed section (a rectangular lever 0.50 in deep by 0.25 in wide with 0.25 in diameter pin holes), and the required section modulus is also reported so the answer stands whatever the true section turns out to be.

Approach. Take the upper half only, exploit mirror symmetry to kill the horizontal force at C, take moments about C on the handle to find the force it delivers at B, then take moments about the fulcrum A on the lever to find the force on the bolt; finish with pin resultants from force equilibrium and with $\sigma = M/S$ on the lever.

Set coordinates and exploit symmetry. Put the origin at the bolt contact with $x$ along the tool axis and the A–B pin row on $y = 0$. Then the bolt acts at $(0,0)$, pin A is at $(1,0)$, pin B at $(a+b,0) = (4,0)$, pin C at $(a+b+c,\,-e) = (6,-1)$ and the hand load acts at $x = 6 + 8 = 14$ in. Because the tool and its loading are mirror images about the horizontal axis through C, the force the lower handle exerts on the upper handle at C can have no horizontal component: $C_x = 0$. Horizontal equilibrium of the handle then gives $B_x = -C_x = 0$, so the force at B is purely vertical.
Take moments about C for the handle (member 1). The handle carries only the pin force at B, the pin force at C and the hand load $P$ acting downward at $x = 14$. Summing moments about C, with the position of B relative to C being $(-c,\,+e)$ and that of the load being $(+d,\,+e)$: $$\sum M_C = -c\,B_y - e\,B_x + d\,(-P) = 0 \quad \Longrightarrow \quad B_y = -\frac{P d}{c} = -\frac{(2)(8)}{2} = -8\ \text{lbf}$$ The negative sign means the lever pushes down on the handle at B with 8 lbf. Note that $B_x = 0$ has removed $e$ from the equation entirely. $$\boxed{\text{Force carried by pin B} = 8\ \text{lbf}}$$
Complete the handle equilibrium to size pin C. Summing vertical forces on the handle, $$C_y = P - B_y = 2 - (-8) = 10\ \text{lbf}$$ and since $C_x = 0$ the resultant is vertical: $$\boxed{\text{Force carried by pin C} = 10\ \text{lbf}}$$ The check is intuitive: the handle is a lever pivoted at C with the hand load 8 in out on one side and the link reaction 2 in out on the other, so the pin must carry the sum of the two, 10 lbf.
Take moments about the fulcrum A for the lever (member 2). The lever carries the bolt reaction $Q$ acting upward at $x = 0$, the reaction to the handle force at B (equal and opposite to $B_y$, hence $+8$ lbf upward at $x = 4$), and the fulcrum pin reaction at A. Moments about A, with the bolt at $-a$ and B at $+b$ relative to A: $$\sum M_A = -a\,Q + b\,(-B_y) = 0 \quad \Longrightarrow \quad Q = \frac{b}{a}\left(-B_y\right) = \frac{3}{1}(8) = 24\ \text{lbf}$$ $$\boxed{\text{Force exerted on the bolt} = 24\ \text{lbf}}$$
Find the fulcrum pin load and check the mechanical advantage. Vertical equilibrium of the lever gives $$A_y = -\left(Q + (-B_y)\right) = -(24 + 8) = -32\ \text{lbf}, \qquad A_x = 0$$ $$\boxed{\text{Force carried by pin A} = 32\ \text{lbf}}$$ so the fulcrum pin is the most heavily loaded joint in the tool — it carries more than the bolt force itself, because it must react both the bolt force and the handle force acting on opposite sides of it. Combining the two lever ratios gives the closed-form mechanical advantage $$\frac{Q}{P} = \frac{d}{c}\cdot\frac{b}{a} = \frac{8}{2}\cdot\frac{3}{1} = 12$$ which is the compact design statement: the handle multiplies by 4 and the jaw lever by a further 3.
Find the bending moment distribution in the levers. Member 2 is a simple lever loaded by $Q = 24$ lbf at its tip, reacted by $A_y = 32$ lbf at the fulcrum and by 8 lbf at B. Measuring $x$ from the bolt contact, $M(x) = Qx$ on $0 \le x \le 1$ and $M(x) = Qx + A_y(x-1)$ beyond it, which falls linearly to zero at $x = 4$ as it must. The peak is therefore at the fulcrum: $$M_{\max} = Q\,a = (24)(1) = 24\ \text{lbf}\cdot\text{in}$$ Member 3 is the mirror image of member 2 and carries exactly the same moment. $$\boxed{M_{\max} = 24\ \text{lbf}\cdot\text{in at the fulcrum pin A, in both members 2 and 3}}$$
Convert the moment to a stress. Adopting the stated assumed section — a rectangular lever $h = 0.50$ in deep by $t = 0.25$ in wide — the gross section modulus is $$S = \frac{t h^{2}}{6} = \frac{(0.25)(0.50)^{2}}{6} = 1.0417\times10^{-2}\ \text{in}^{3}$$ $$\sigma_{\max} = \frac{M_{\max}}{S} = \frac{24}{1.0417\times10^{-2}} = 2304\ \text{psi} = 2.30\ \text{ksi}$$ Through the 0.25 in diameter pin hole at A the net section modulus falls to $t(h^{3}-d_h^{3})/(6h) = 9.115\times10^{-3}\ \text{in}^{3}$ and the stress rises to 2633 psi, so the hole rather than the gross section governs: $$\boxed{\sigma_{\max} \approx 2.6\ \text{ksi in members 2 and 3, at the net section through pin A}}$$
Assess the margins. Against $S_y = 60.9$ ksi the factor of safety on the levers is $60\,900/2633 = 23$, and even the fulcrum pin is grossly over-strength: a pin in double shear at A needs only $$d_{\text{pin}} = \sqrt{\frac{4\,(F_A/2)}{\pi S_{ys}}} = \sqrt{\frac{4(16)}{\pi (30\,450)}} = 0.026\ \text{in}$$ to reach yield in shear. The general conclusion — which holds for any plausible section, since the required section modulus for yielding is only $M_{\max}/S_y = 3.94\times10^{-4}\ \text{in}^{3}$ — is that at a 2 lbf hand load the tool is nowhere near its structural limit. Such tools are sized by stiffness, by the jaw and edge geometry needed to bite into the bolt, and by the hand force a user can actually apply at full grip (which is closer to 50–100 lbf), not by the 2 lbf case analysed here. At 100 lbf of hand load the bolt force would be 1200 lbf and the lever stress 132 ksi — well past yield — which shows how quickly the margin evaporates and why the design case matters more than the arithmetic.

Quantity	Symbol	Result
Force exerted on the bolt	$Q$	24 lbf
Force carried by pin A (fulcrum)	$F_A$	32 lbf
Force carried by pin B	$F_B$	8 lbf
Force carried by pin C	$F_C$	10 lbf
Mechanical advantage	$Q/P$	12 : 1
Maximum bending moment, members 2 and 3	$M_{\max}$	24 lbf·in at pin A
Maximum stress, members 2 and 3 (assumed section)	$\sigma_{\max}$	2.63 ksi (net section at A)
Required section modulus for yielding	$S_{req}$	$3.94\times10^{-4}$ in³

Topic. Free-body analysis of a compound pin-connected linkage, and bending stress in a lever.

Key relations. Moment equilibrium $\sum M = 0$ taken about the pin whose force is unknown; two-lever mechanical advantage $Q/P = (d/c)(b/a)$; bending stress $\sigma = M/S$ with $S = th^{2}/6$ for a rectangle and $S_{net} = t(h^{3}-d_h^{3})/(6h)$ through a central hole; pin double shear $\tau = 2F/(\pi d^{2})$.

Why this works. The whole analysis is a chain of single-unknown moment equations, and each one is set up by taking moments about the pin whose reaction you do not want to appear. Symmetry supplies the extra piece of information that would otherwise be missing: because the loading and the geometry are mirror images about the axis through C, the internal force at C on that axis can have no component along the axis of reflection's normal — it must be purely vertical — which immediately makes the force at B vertical too and reduces each free body to three forces. The mechanical advantage then separates cleanly into two independent lever ratios, which is exactly how such a tool is designed: pick $d/c$ from the handle length available to the hand, and pick $b/a$ from how much jaw force the application demands.

Common pitfall. Using $e$ as the moment arm for a horizontal force at B. That is the natural first move, but symmetry has already forced $B_x = 0$, so the horizontal-force term vanishes and $e$ never enters the answer — a good example of checking which equilibrium equations are actually independent before grinding through arithmetic. A second pitfall is expecting the fulcrum pin at A to carry less than the bolt force; it carries the sum of the bolt force and the handle force, 32 lbf, because both act on the same side of the pivot in the sense of vertical equilibrium.

Source. Norton, Machine Design: An Integrated Approach, 6th ed., Ch. 3 (static load analysis of linkages and hand tools) and Ch. 4 (stress: bending of levers, net section through a hole); Budynas & Nisbett, Shigley's Mechanical Engineering Design, 11th ed., §3.9–3.10 and §4.4.

← Question 5 Back to the paper →