Metal Forming Processes Questions and Answers

Answer: D

A cluster rolling mill is the one with two or more sets of rolls, in which small working rolls are supported by larger backing rolls, so the reduction in thickness can be large in a single pass. A planetary mill is a different arrangement in which a pair of backing rolls is surrounded by planet rolls. Two-high and three-high mills have only two or three rolls, so they do not have several sets of rolls. The key names the cluster mill.

In a cluster mill, such as the Sendzimir mill, each small work roll is backed by a group of larger rolls, which lets thin strip and hard metals be rolled.

Answer: B

A metalworker may prefer indirect extrusion because the billet remains stationary in it. In direct extrusion the billet slides in the container, and the friction along the container wall requires a larger force. In indirect extrusion the die moves toward the billet, so there is no relative motion with the container and the force is lower. The statement that indirect extrusion needs more force is the reverse of the truth, and the remaining options are not the reasons that favour the indirect process.

Answer: D

R is 0.632.

A perfectly plastic material does not strain-harden, so n = 0:

\[ R = 1 - e^{-(0 + 1)} = 1 - \frac{1}{e} = 1 - 0.368 = 0.632 \]

That is, at most about 63% reduction in area per pass. The larger values come from taking n greater than zero: n = 1 gives \(1 - e^{-2} = 0.865\), and n of about 0.75 and 0.5 give 0.826 and 0.777, reductions possible only for strain-hardening materials.

Answer: C

The correct match is P-4, Q-1, R-2, S-3.

• Toothpaste tubes are made by impact extrusion: a slug of soft metal such as aluminium is struck so that it flows up around the punch as a thin-walled tube (4).
• Metallic pipes are made by centrifugal casting, where the spinning mould throws the metal against its wall to form a hollow cylinder without a core (1).
• Plastic bottles are made by blow moulding, inflating a hot plastic parison inside a mould (2).
• Threaded bolts get their threads by thread rolling (3), which cold-forms the thread and improves its fatigue strength.

Option A gives bottles to centrifugal casting, option B gives toothpaste tubes to blow moulding, and option D gives pipes to rolling and bottles to extrusion.

Answer: A

The correct matching is 1-S, 2-P, 3-Q, 4-R.

• Coining squeezes the blank between closed dies, so the work is in compression (S).
• Wire drawing pulls the wire through the die, so the applied stress in the drawn wire is tensile (P).
• Blanking separates the blank by shearing between punch and die edges (Q).
• Deep drawing stretches the cup wall in tension while the flange is squeezed in the hoop direction as it is drawn in, so it has both tensile and compressive stress (R).

Options that pair coining with tension or blanking with combined stress get the basic mechanism of those processes wrong.

Answer: D

The diameter increases by about 41.4 %.

With frictionless dies the cylinder stays cylindrical, and volume is constant in plastic deformation:

\[ \begin{aligned} d_{1}^{2}h_{1} &= d_{2}^{2}h_{2} \\ d_{2} &= 100\sqrt{\frac{50}{25}} = 141.4\ \text{mm} \end{aligned} \]

The change is 41.4 mm on 100 mm, or 41.4 %. 20.7 % is half of that and 2.07 % a tenth of that again, and 0 would mean no lateral flow, which is impossible when the height is halved at constant volume.

Answer: C

The angle of bite is about 0.062 rad.

A 10 % reduction of 8 mm is a draft of 0.8 mm, and the roll radius is 205 mm. The contact angle satisfies:

\[ \begin{aligned} \cos\alpha &= 1-\frac{\Delta h}{2R} = 1-\frac{0.8}{410} = 0.99805 \\ \alpha &= 0.0625\ \text{rad} \end{aligned} \]

The small-angle form \(\alpha \approx \sqrt{\Delta h/R} = \sqrt{0.8/205} = 0.0625\) rad gives the same result.

0.031 is half the correct value and 0.006 and 0.600 are the right digits with the decimal point misplaced. The strip width plays no part in the bite angle.

Answer: B

The maximum reduction is about 63.2 percent.

For ideal drawing of a perfectly plastic material, the drawing stress is \(\sigma_{d} = Y\ln(A_{0}/A_{f})\). The drawn wire cannot carry more than its yield stress, so the limit is \(\sigma_{d} = Y\):

\[ \begin{aligned} \ln(A_{0}/A_{f}) &= 1 \\ A_{f}/A_{0} &= e^{-1} = 0.368 \\ r &= 1 - 0.368 = 0.632 \end{aligned} \]

That is 63.2 percent. 51.2, 75.0 and 93.2 percent correspond to true strains of 0.72, 1.39 and 2.69, which would require the wire to carry more than its yield stress or fail to use it fully.

Answer: A

The pass is feasible, with maximum draft 3.6 mm.

In flat rolling, the maximum draft that friction can drag into the roll gap is \(\Delta h_{max} = \mu^{2}R\), with roll radius \(R = 250\) mm:

\[ \begin{aligned} \Delta h_{max} &= 0.12^{2}\times 250 \\ &= 0.0144\times 250 = 3.6\ \text{mm} \end{aligned} \]

The required draft is \(30 - 27 = 3\) mm, which is less than 3.6 mm, so the pass is feasible. 3.0 mm is the required draft, not the maximum; 6.0 and 2.6 mm do not follow from \(\mu^{2}R\), and the pass is not infeasible.