MEMS 0040 Exam 1 Spring 2021 key (1)

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MEMS 0040 Materials and Manufacturing Exam 1 Spring 2021 Name: ____________________________________________________ • Closed book • You may use a calculator I have neither given nor received unauthorized assistance on this test: (Signature): ________________________________________________

Section 1: True/False (2 points each, 20 points total) __ T __ 1. The main chemical reducing agent in a blast furnace is carbon monoxide gas. __ F __ 2. The aluminum refining process requires aluminum metal to dissolve in molten cryolite. Aluminum oxide dissolves in the cryolite, not aluminum metal. __ F __ 3. For open die forging of a cylinder along its axis, the magnitude of the maximum in engineering stress is smaller than the magnitude of the maximum true stress. During open die forging, the area increases, so the true stress is smaller than the engineering stress. __ T __ 4. The work hardening exponent of a metal decreases with increasing temperature. __ T __ 5. The average flow stress is a measure of the work done per unit volume during permanent deformation of a metal. __ F __ 6. The area reduction for wire drawing is the ratio of the original cross-section area to the final cross-section area of the wire. Area reduction = (A o - A f )/A o . __ T __ 7. If 100 castings are to be made using the Expanded Polystyrene Process, then 100 patterns will be needed. __ F __ 8. Alloys are better than pure metals for making sand castings because they have better fluidity. Pure metals almost always have better fluidity than alloys. __ F __ 9. The most suitable risers for casting of metals have low volume-to-surface area ratios. Risers should have high V/A ratios. __ F __ 10. Hot chamber die casting is generally slower than cold chamber die casting because it takes more time for the casting to solidify. Hot chamber die casting is faster.

Section 2: Problems and Short Answer Questions (80 points total) 11. The Basic Oxygen Furnace (BOF) is used to make liquid steel from liquid pig iron and steel scrap. (a) List the equation for the main chemical reaction to reduce the amount of carbon in the liquid metal during BOF processing. (5 points) 2 C + O 2 ® 2 CO (b) List the chemical equation to reduce the amount of silicon in the liquid metal during BOF processing. (5 points) Si + O 2 ® SiO 2 (c) How are the carbon and silicon impurities in parts (a) and (b) removed from the BOF vessel? (5 points) The CO forms as gas bubbles, which enter the gas phase and exit the top of the vessel into a gas collection hood. The SiO 2 which forms is less dense than the liquid steel, so it floats to the top of the bath and is dissolved into the liquid slag phase, which is removed after the liquid steel is tapped by rotating the vessel nearly upside down to pour it out. 12. The refining of aluminum metal requires large amounts of electrical power because the voltage is relatively high for an electrochemical reaction and the current is also high (power = voltage x current). What is the reason for the applied current in aluminum refining being so high? (5 points) The production rate is proportional to the current used. In order to make a lot of aluminum in a short time, a high current is needed.

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13. A cylinder of low alloy steel is to be open die forged from its original dimensions of 100 mm in diameter and 200 mm high to a final height of 120 mm. The strength coefficient is 400 MPa, and the work hardening exponent is 0.2. Assume no friction. Show your calculations. (20 points total) (a) What is the engineering strain in this forging operation? (4 points) e = (h – h o )/h o = (120 mm – 200 mm)/(200 mm) = -0.40 (b) What is the true strain in this forging operation? (4 points) e = ln(h o /h) = ln(200/120) = 0.51 Since this is in compression, the value can also be expressed as a negative value, e = -0.51 (c) What is the maximum true stress required in this forging operation? (4 points) s max = Y f when e = e max , and this occurs at the end of the open forging process. Y f = K e n = (400 MPa)(0.51) 0.2 = 350 MPa (d) What is the average flow stress required in this forging operation? (4 points) Y fav = K e n /(1 + n) = (400 MPa)(0.51) 0.2 /(1 + 0.2) = 291 MPa (e) If the temperature used in the open die forging operation were increased, what would be the (qualitative) effects on the strength coefficient and the work hardening exponent? (2 points each, 4 points total) Circle one for each. Strength coefficient: Decreases / Increases Work hardening exponent: Decreases / Increases

14. (a) An extrusion process reduces the cross-section of a square metal rod from 50 mm by 50 mm to 40 mm by 40 mm. Calculate the required extrusion pressure for this operation if the strength coefficient of the metal is 160 MPa and the strain hardening exponent is 0.1. (14 points) P = [K e n /(1 + n)] e e = ln[(50) 2 /(40) 2 ] = 0.446 P = [(160 MPa) (0.446) 0.1 /(1 + 0.1)] (0.446) = 59.8 MPa (b) Explain the difference between direct and indirect extrusion. (6 points) For direct extrusion, the product moves in the same direction as the ram. For indirect extrusion, the product and the ram move in opposite directions. 15. A metal casting mold is being designed that includes a mold cavity with a rectangular section of dimensions 400 mm by 200 mm with a depth of 100 mm. (a) If the mold constant is 3 min/cm 2 and the mold exponent is 2, calculate how long will it take for the casting to solidify. (8 points) V = (400 mm)(200 mm)(100 mm) = (40 cm)(20 cm)(10 cm) = 8000 cm 3 A = 2 [(400 mm)(200 mm) + (400 mm)(100 mm) + (200 mm)(100 mm)] = 2 [(40 cm)(20 cm) + (40 cm)(10 cm) + (20 cm)(10 cm)] = 2800 cm 2 T TS = C (V/A) n = (3 min/cm 2 ) [(8000 cm 3 )/(2800 cm 2 )] 2 = 24.5 min (b) Determine the minimum size of a cube-shaped riser that will prevent a shrinkage cavity in the casting. (8 points) For a cube with side length L, V = L 3 and A = 6L 2 T TS = C (V/A) n = (3 min/cm 2 ) [L 3 /(6L 2 )] 2 = (1 min/cm 2 ) (L 2 /12) = 24.5 min L = [(24.5 min)(12 cm 2 /min)] 0.5 = 17.1 cm Note that this dimension is smaller than 400 mm or 200 mm and is greater than 100 mm, so it appears to be a reasonable value. (c) How could the size of the riser be further reduced without risking the formation of a shrinkage cavity in the casting? (no calculations needed) (4 points) The size of the riser could be further reduced by increasing the V/A ratio. This could be done by changing the shape to a cylinder (with aspect ratio near 1) or a sphere.

MEMS 0040 Exam 1 Equations Sheet F = Y fav A K f = 1 + (0.4 μ D/h) d = t o - t f r = d/t o P = 2 p NFL r = A o /A f P = Y fav e d = D o - D f r = (A o - A f )/A o Q = v A A A = v B A B mgh 1 + mv 1 2 /2 = mgh 2 + mv 2 2 /2 T TS = C m (V/A) n

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