22-Mec-A4 Design and Manufacture of Machine Elements · December 2018

Question 2 of 6: Stretch Forming of a Sheet-Metal Vee (25 marks)

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

Notes on this paper

Paper format. National Examinations, December 2018 — 16-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 processes) and Part B (Q4–Q6, machine-element analysis). The rubric asks for two questions from each part — four questions constitute a complete paper, each worth 25 %. All six are solved here, since the set is intended as a study resource.

Reference texts.

Kalpakjian, S. & Schmid, S., Manufacturing Engineering and Technology, 8th ed. — Ch. 11–12 (casting processes and design), Ch. 16 (sheet-metal forming), Ch. 30 (fusion welding).
Groover, M. P., Fundamentals of Modern Manufacturing, 7th ed. — Ch. 10–11 (casting), Sec. 20.5 (stretch forming).
Budynas, R. & Nisbett, K., Shigley's Mechanical Engineering Design, 11th ed. — Ch. 9 (welding, weld-as-a-line, Tables 9-2 and 9-3).
Hibbeler, R. C., Engineering Mechanics: Statics, 14th ed. — Ch. 6 (frames and machines), Ch. 8 (dry friction).
Hibbeler, R. C., Mechanics of Materials, 10th ed. — Sec. 4.6 (thermal stress in statically indeterminate axial members).

Check: every boxed result in this paper, which also carries the independent closure checks noted in each question (moment closure on Q4, volume conservation on Q2, gap compatibility on Q6).

Question 2: Stretch Forming of a Sheet-Metal Vee (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
Starting length between grips	$L_0$	20.0 in
Starting thickness	$t_0$	0.125 in
Width (constant)	$w$	10.0 in
Strength coefficient	$K$	70,000 lb/in²
Strain-hardening exponent	$n$	0.25
Formed vee: overall span	—	20.0 in (half-span 10.0 in)
Formed vee: rise at the die	—	5.0 in

Find. (a) the stretching force at first yield, (b) the true strain in the finished part, (c) the stretching force at the end of forming, and (d) the force the die must react at that instant.

The blank is gripped at both ends and pulled over a stationary die. The half-span is 10.0 in and the rise is 5.0 in, so each leg subtends the angle $A$ with the horizontal.

Approach. Treat the sheet as a membrane in pure uniaxial tension along its length: the force is the current cross-sectional area times the flow stress at the current strain, $F = w\,t\,Y_f$ with $Y_f = K\varepsilon^{\,n}$, and the die force follows from resolving the two leg tensions at the apex.

Fix the flow stress at first yield. Yielding is defined at the 0.2 % offset, so the strain at which the flow curve is evaluated is $\varepsilon = 0.002$: $$Y_f = K\,\varepsilon^{\,n} = 70{,}000\,(0.002)^{0.25} = 14{,}804\ \text{lb/in}^2$$
Convert flow stress to the stretching force at the start. At this instant the sheet is still flat and unthinned, so the cross-section carrying the load is simply $w t_0$: $$F = w\,t_0\,Y_f = (10.0)(0.125)(14{,}804)$$ $$\boxed{F_{\text{yield}} = 1.85\times 10^{4}\ \text{lb} \;(18{,}505\ \text{lb})}$$
Get the final length from the vee geometry. Each leg spans 10.0 in horizontally and rises 5.0 in, so its stretched length is $$L_{\text{leg}} = \sqrt{(10.0)^2+(5.0)^2} = 11.180\ \text{in} \quad\Rightarrow\quad L_f = 2L_{\text{leg}} = 22.361\ \text{in}$$
Compute the true strain. With the sheet gripped at its ends and drawn over the die, all of the elongation shows up as axial true strain: $$\varepsilon = \ln\frac{L_f}{L_0} = \ln\frac{22.361}{20.0}$$ $$\boxed{\varepsilon = 0.1116}$$
Re-enter the flow curve at the final strain. The metal has work-hardened substantially, so the flow stress is now $$Y_f = 70{,}000\,(0.1116)^{0.25} = 40{,}417\ \text{lb/in}^2$$
Thin the sheet by constant volume. The width is restrained by the grips and does not change, so all of the elongation is taken out of the thickness: $$w t_0 L_0 = w t_f L_f \quad\Rightarrow\quad t_f = t_0\frac{L_0}{L_f} = 0.125\left(\frac{20.0}{22.361}\right) = 0.1118\ \text{in}$$
Assemble the stretching force at the end. Multiplying the reduced area by the raised flow stress, $$F = w\,t_f\,Y_f = (10.0)(0.1118)(40{,}417)$$ $$\boxed{F_{\text{end}} = 4.52\times 10^{4}\ \text{lb} \;(45{,}188\ \text{lb})}$$ The force has risen by a factor of 2.4 even though the section has thinned by 11 %, which shows how strongly the $n = 0.25$ hardening dominates the area loss.
Resolve the leg tensions at the die. Each leg pulls along its own axis at $$A = \arctan\!\left(\frac{5.0}{10.0}\right) = 26.57^{\circ}$$ to the horizontal. The horizontal components of the two leg tensions cancel at the apex, and their vertical components both push down on the die, so $$F_{\text{die}} = 2F\sin A = 2(45{,}188)\sin 26.57^{\circ}$$ $$\boxed{F_{\text{die}} = 4.04\times 10^{4}\ \text{lb} \;(40{,}418\ \text{lb})}$$

The die force is smaller than the stretching force because the legs are shallower than 30°; only if the vee were steeper than $A = 30^{\circ}$ would the die see more load than each grip.

Result	Value
(a) Stretching force at first yield	18,505 lb (18.5 kip)
(b) True strain in the finished part	0.1116
(c) Stretching force at the end of forming	45,188 lb (45.2 kip)
(d) Die force at the end of forming	40,418 lb (40.4 kip)
Supporting values	$Y_f$ = 14,804 → 40,417 lb/in²; $t_f$ = 0.1118 in; $A$ = 26.57°

Topic: Stretch forming — force prediction from the power-law flow curve.

Key relations: $Y_f = K\varepsilon^{\,n}$; $F = w\,t\,Y_f$; constant volume $t_f = t_0 L_0/L_f$; $\varepsilon = \ln(L_f/L_0)$; $F_{\text{die}} = 2F\sin A$.

Why this works: Stretch forming is deliberately a pure-tension process. The sheet is pulled past its yield point everywhere before it is wrapped on the die, so the whole section is plastic and springback is almost eliminated — which is why the process is used for aircraft skins and architectural panels where shape fidelity matters more than cycle time. Because the state of stress is uniaxial tension, the flow curve can be used directly with no von Mises correction factor.

Common pitfall: Applying $F = 2wtY_f\sin A$ to part (a). At the beginning of the stroke the sheet has not yet been wrapped on the die, so $A = 0$ and that expression returns zero force — it is the die-force relation, not the grip-force relation. Equally common is forgetting to thin the sheet in part (c) and using $t_0$ with the final flow stress, which overstates the force by 12 %.

Source: Groover, Fundamentals of Modern Manufacturing, 7th ed., Sec. 20.5 (stretch forming) and Sec. 3.1.2 (flow curve); Kalpakjian & Schmid, 8th ed., Sec. 16.6.

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