22-Mec-A4 Design and Manufacture of Machine Elements · May 2015

Question 1 of 6: Stretch Forming of a Sheet-Metal Panel

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

Notes on this paper

Paper format. National Examinations, May 2015 — 07-Mec-A4, Design and Manufacture of Machine Elements. Three hours, open book, any non-communicating calculator permitted. Six questions in two parts: Part A (Q1–Q3, manufacturing processes) and Part B (Q4–Q6, machine-element design). The rubric asks for two questions from each part, and all questions carry equal value (25 % each). All six are solved here, because this set is a study resource rather than an examination script.

Reference texts.

S. Kalpakjian and S. R. Schmid, Manufacturing Engineering and Technology, 8th ed. — sheet-metal forming, deep drawing, abrasive machining.
M. P. Groover, Fundamentals of Modern Manufacturing, 7th ed. — stretch forming and the flow-curve treatment of forming loads.
R. G. Budynas and J. K. Nisbett, Shigley's Mechanical Engineering Design, 11th ed. — bolted and welded joints (Ch. 8, 9) and brakes (Ch. 16).
R. C. Hibbeler, Mechanics of Materials, 10th ed. — net-section bending and bearing stress.

Units. The paper mixes systems deliberately: Q1, Q5 and Q6 are in US customary units (inch, pound, psi) and Q4 is metric. Each question is solved in the units in which it is set, as the exam intends.

Question 1: Stretch Forming of a Sheet-Metal Panel (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 blank length (flat)	$L_0$	20.0 in
Starting stock thickness	$t_0$	0.125 in
Blank width	$w$	10 in
Strength coefficient	$K$	70,000 lb/in²
Strain-hardening exponent	$n$	0.25
Formed shape (Figure b)	—	symmetric vee, 20.0 in span, 5.0 in rise

Find. The stretching force at first yield, the true strain at the end of forming, the stretching force at that final condition, and the vertical force carried by the form die.

Figure 1 — stretch forming. (a) the flat blank gripped at both ends and pulled in pure tension at the moment of first yield; (b) the finished part wrapped over the form die, the sheet now inclined at angle A to the pull direction.

Approach. Stretch forming is a pure-tension operation: the grips supply a force equal to the current flow stress times the current cross-section, so the whole problem reduces to evaluating the flow curve $Y_f = K\varepsilon^{\,n}$ at the right strain and applying constant volume to get the current thickness, after which the die force follows from resolving the two inclined grip forces vertically.

Evaluate the flow stress at first yield. At the instant yielding begins the sheet has undergone only the conventional 0.2 % offset strain, so the flow curve is evaluated at $\varepsilon = 0.002$: $$Y_f = K\,\varepsilon^{\,n} = 70{,}000\,(0.002)^{0.25} = 70{,}000\,(0.2115) = 14{,}803\ \text{lb/in}^2$$
Convert the flow stress to a grip force. Near the beginning the blank is still flat, so the grip force is straight tension over the full original cross-section $A_0 = w\,t_0 = 10(0.125) = 1.25\ \text{in}^2$: $$F = w\,t_0\,Y_f = 10(0.125)(14{,}803)$$ $$\boxed{F \approx 18{,}500\ \text{lb}}$$
Find the stretched length from the finished geometry. The formed part is a symmetric vee of total span 20.0 in and rise 5.0 in, so each half is the hypotenuse of a 10-by-5 triangle: $$L_f = 2\sqrt{(10)^2 + (5)^2} = 2(11.180) = 22.361\ \text{in}$$
True strain experienced by the metal. Because the width is restrained by the grips, all of the elongation is taken up in length: $$\varepsilon = \ln\!\frac{L_f}{L_0} = \ln\!\frac{22.361}{20.0} = \ln(1.1180)$$ $$\boxed{\varepsilon = 0.1116}$$
Flow stress and thickness at the end of forming. Substituting the final strain into the flow curve, and using constant volume with the width held fixed, $$Y_f = 70{,}000\,(0.1116)^{0.25} = 70{,}000\,(0.5780) = 40{,}463\ \text{lb/in}^2$$ $$t_f = t_0\frac{L_0}{L_f} = 0.125\!\left(\frac{20.0}{22.361}\right) = 0.1118\ \text{in}$$ The sheet has thinned by about 10.6 %, which is exactly the thinning implied by a true strain of 0.1116.
Stretching force at the very end. The force is again the current flow stress acting on the current section: $$F = w\,t_f\,Y_f = 10(0.1118)(40{,}463)$$ $$\boxed{F \approx 45{,}240\ \text{lb}}$$ The load has risen by a factor of about 2.4 over the operation, entirely because of strain hardening.
Die force. At the end of the stroke each half of the sheet leaves its grip at $$A = \tan^{-1}\!\frac{5.0}{10.0} = 26.57^\circ$$ to the horizontal. The die must react the vertical components of the two grip forces: $$F_{die} = 2F\sin A = 2(45{,}240)\sin 26.57^\circ = 2(45{,}240)(0.4472)$$ $$\boxed{F_{die} \approx 40{,}460\ \text{lb}}$$

The die force is smaller than twice the grip force precisely because the sheet is still fairly shallow; a deeper form (larger A) would drive the die load toward $2F$.

Quantity	Result
(a) Stretching force at first yield	18,500 lb
(b) True strain at end of forming	0.1116
(c) Stretching force at end of forming	45,240 lb
(d) Die force at end of forming	40,460 lb
Final sheet thickness	0.1118 in
Sheet angle at the grips	26.57°

Topic. Stretch forming and the flow-curve description of plastic work in sheet-metal forming.

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

Why this works. In stretch forming the sheet is gripped at both ends and pulled over a form block, so every element of the sheet is in essentially uniaxial tension along its length. That means the stress state is the simplest one possible and the force is just stress times area — there is no redundant work, no friction term and no bending correction, unlike deep drawing or bending. The whole difficulty is bookkeeping: you must use the current strain to get the flow stress and the current thickness to get the area, and both change as the part is formed. Because the sheet is held in tension throughout, springback is very small, which is the reason aircraft skin panels and large automotive body panels are made this way rather than by pressing.

Common pitfall. Two errors dominate. The first is computing the final force on the original thickness of 0.125 in, which overstates the load by about 12 %; the sheet has thinned and the area has shrunk. The second is treating part (a) with the stretch-forming formula $2wtY_f\sin A$, which gives zero because the sheet is still flat — at first yield the operation is pure tension and $A = 0$.

Source. Groover, Fundamentals of Modern Manufacturing, 7th ed., §20.5 (stretch forming); Kalpakjian and Schmid, Manufacturing Engineering and Technology, 8th ed., Ch. 16.

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