ME 310
WASHINGTON STATE UNIVERSITY FALL SEMESTER, 2012 Campus: Pullman MANUFACTURING PROCESSES Exam 1 (Time allowed: allowed: FIFTY minutes) NOTE: Answer
questio tions ns.. Circl Circlee only only one answ answer. er. The The marks marks for each each quest question ion are ALL ques
as shown. shown. Multip Multiple le answers answers carry zero points. points. If you make a mistake, mistake, mark a cross cross through your wrong choice and circle your next alternative.
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Student’s Name:
ME 310
ID No:
SECTION A TRUE–FALSE QUESTIONS 1. Circle
the preferred choice.
T
F
Yield stress is where plastic flow begins.
T
F
During elastic deformation the volume of the specimen is constant.
T
F
The plastic deformation is caused by the massive motion of dislocations.
T
F
An alloy is a metal composed of two or more elements.
T
F
Hardness tests involve the indentation of the surface of a test specimen.
T
F
The theoretical strength of a material is smaller than the experimental. (6 marks)
SECTION B MULTIPLE CHOICE QUESTIONS 2. Which
of the following is not point defect in a crystal lattice structure
(a) Dislocation. (b) Interstitial. (c) Vacancy. (d) Shottky defect. 3. Dyring
plastic deformation, plastic work is
(a) Stored. (b) Converted to heat. (c) Converted to kinetic energy. (d) Non existant.
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4. In
ME 310
strain–rate sensitive materials
(a) As the strain rate increases, the yield stress remains constant. (b) As the strain rate increases, the yield stress increases. (c) As the strain rate increases, the yield stress decreases. (d) As the strain rate increases, the ultimate tensile stress decreases. 5. An
FCC unit cell has
(a) One atom at the center of the cell. (b) Atoms centered on all faces of its cell. (c) No atoms on its faces. (d) A triangular shape. 6. In
steels, as the carbon content increases
(a) The ductility increases. (b) The steel becomes more brittle. (c) The steel softens. (d) The mechanical properties of the steels remain unchanged. (10 marks) 7. What
is the di ff erence between true and engineering strain? (2 marks)
Answer: Engineering
8. 2.7
strain: e =
−0 0
, True strain: ε = ln
0
Show schematically the di ff erence between open-die and impresion-die forging. (2 marks)
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ME 310
SECTION C PROBLEMS 9.
A cylindrical bar specimen is tested in tension. The initial diameter of the bar is D0 = 12mm, and its initial length is 0 = 50mm. The following results were obtained: Yield load P yield = 35kN, final length at fracture frac = 69 mm, maximum load P max = 45 kN and occured at a length max = 60 mm. The modulus of elasticity E = 1011 Pa. Find: (a) The engineering yield stress, (b) The final true stain, (c) The bar diameter at the maximum load (neglect elasticity), (d) The true Ultimate Tensile Stress, (e) The true elastic strain at yield. (15 marks)
Answer:
(a) The yield stress is Y =
35 kN
2
π (0.12/2)
= 309 MPa
(3 marks)
69 (b) The ε = ln 50 = 0.322
(3 marks)
2 (c) Neglecting elasticity, the diameter at the maximum load is: V 0 = V f ⇒ π r02 L0 = π rf Lf
or r f = r0
L0 Lf
⇒ 2rf = 2r0
L0 Lf
⇒ ⇒ Df = D0
(d)The true ultimate tensile stress is
σU
=
(e) The true elastic strain at yield is
εel =
P max A
=
L0 Lf
π
309×106 Pa 1011 Pa
⇒ Df = 12
√ 45 kN 6
50 × 60
10−3
= 0.00309
50 60
= 10.95 mm
= 477 MPa
(3 marks)
(3 marks)
(3 marks)
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ME 310
strip of material obeys the power law relation σ = 130ε0.15 MPa. It has a thickness h0 = 25 mm and a width W = 500 mm. Its thickness is to be reduced to hf = 13 mm by a two-high rolling process with N = 30 rpm. If the radius of the rolls is R = 350 mm, determine the following: (a) The minimum coe fficient of friction, (b) The total power of the rolls, (c) The plastic work per unit volume.
10. A
(15 marks)
Answer:
cient of friction is: µ = h − h R =
(a) The minimum coe ffi
0
(b) The total power of the rolls is P = 2T ω, T =
FL 2
and F = Y¯ LW , thus T =
2
L 2
ω =
2πN 60
25−13 350
f
=
2π(30) 60
= 0.185
(4 marks)
= π rad/s.
¯
W Y . Also P = 2(π ) W ¯ Y = π L W ¯ Y . L = R (h − h ) = 350(25 − 13) = 64 mm. = 0.654. ε = ln
0
L2 2
2
f
25 13
¯ = Y
130ε0.15 1.15
0
= 130(0.654) 1.15
Finally, P = π (64 × 10
.15
3
−
MPa = 106.1 MPa.
)(0.5)(106.1 × 106 ) = 682 kW.
(c) The plastic work per unit volume is U =
ε
ε
(7 marks)
0.15
σdε = 130ε 0
0
dε =
130ε1.15 1.15
= 69.37MJ/m3
(4 marks)
ATTACHMENT
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ME 310
Equation Sheet h0 − hf = µ 2 R Draft F = Y¯ LW P = F v = T ω ω =
Roll Force (per roll) Roll Power (per roll)
2πN/60
Angular Velocity
T = F L/2
Torque (per roll)
L =
R (h − h ) σdε U =
Contact Length
ε =
True Strain
0
f
ε
0
ln (f /0 )
¯ = K εn /(n + 1) Y
F f = Y f 1 +
2µrf 3hf
Af
Deformation Energy
Mean Yield Stress Forging Force
