Introduction to Materials
1. State the standard (SI) units for the following:
Force, Area, Stress, Strain, Stiffness, Density
Answer . Force
Newtons (N) Area
m 2 Stress
Pa = N/m 2 Strain
no units Stiffness
Pa Density Kg/m 3 g/cm 3
2. State the equations for the following
Stress, Strain, Stiffness, Density
Answer .
Stress
Strain
Stiffness
σ = E ε
Density
=mass / volume =volume mass =ρ 3. Explain the difference between elastic strain and plastic strain
Answer .
Elastic strain the materials can return to its original form and dimensions once the force is removed
Plastic strain the material is deformed and can’t return to its original form once the force is removed.
4. Comment on whether elastic strain or plastic strain is important in terms of the
flexibility of golf club shafts.
Answer .
1)..(A F a initialare force e i load n ==σ1
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2l l l ngth Originalle ngth Changeinle n -==εl A Fl l l A F ain tensilestr ess tensilestr E ∆=∆===1111εσ
英语b级词汇
Yes if plastic once ud past the point of elastic the club is ruined as it will be bent out of shape for ever.
If elastic the club can be ud over and over and still retain the shape of the club.
5. On the same t of axes, sketch stress-strain graphs for (i) a plastic material,
(ii) a ceramic (brittle) material and (iii) a strong, ductile metal.
Answer.
A: brittle material
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C: strong and ductile
D: Plastic (weak ductile)
6. Define the following terms: (i) toughness (ii) yield strength (iii) stiffness
(iv)ductility
Answer.
Toughness: A measure of the amount of energy absorbed by a material as it fractures.
LOAD
LOAD Or ability of a material to absorb a sudden load. Toughness is indicated by the total area under the material’s tensile stress – strain curve.
Yield strength : The stress required to produce a very slight yet specified
amount of plastic strain.
Stiffness: This is stress per unit elastic strain
< a measure of how difficult it is to stretch the material while the material is behaving like a spring.
Ductility: The ability to be moulded without breaking.alevel国际学校
7. You have a material with a cross-ctional area of 30cm 2 and a force of 500N is exerted on it what stress is prent on the material.
Answer .
F/A = 500N/0.3m = 1667Pa
8. You have a material with a volume of 300cm 3 and a mass of 7kg, what is the density of the material?
Answer .
Mass/Vol = 7/3 = 2.3Kg/m 3
9. Describe how ionic bonding works?
Answer . In ionic bonding, atoms of one element give up their outer electron(s) to the atoms of another element to form complete outer shells.
10. What is the difference between crystalline and noncrystalline structures in materials?
Answer . The atoms in a crystalline structure are located at regular and repeating lattice positions in three dimensions; thus, the crystal structure posss a long-range order which allows a high packing density. The atoms in a noncrystalline structure are randomly positioned in the material, not posssing any repeating, regular pattern.
11. What is the difference between engineering stress and true stress in a tensile test?
Answer . Engineering stress divides the load (force) on the test specimen by the original area; while in true stress divides the load by the instantaneous area which decreas as the specimen stretches.
12. Define tensile strength of a material.
Answer . The tensile strength is the maximum load experienced during the tensile test divided by the original area.
13. Define yield strength of a material.
Answer. The yield strength is the stress at which the material begins to plastically deform. It is usually measured as the .2% offt value - the point at which the stress-strain for the material intercts a line which is offt from the elastic region of the stress-strain curve by 0.2%.
14. Why cannot a direct conversion be made between the ductility measures of
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elongation and reduction in area using the assumption of constant volume? Answer. Becau of necking that occurs in the test specimen.
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15. In what ca does the strength coefficient have the same value as the yield
strength?
Answer. When the material does not strain harden.
16. What is a polymer?
Answer. A polymer is a compound that consists of long-chain molecules. The molecules consist of repeating units, called mers, connected end to end.
17. What are the three basic categories of polymers?
Answer. The categories are: (1) thermoplastics, (2) thermotting polymers, and (3) elastomers.
18. How do the properties of polymers compare with tho of metals? Answer. In general, polymers have lower strength, hardness, stiffness, density, and temperature resistance compared to metals. In addition, polymers are low in electrical and thermal conductivity.
19. What are the two methods by which polymerization occurs? Briefly describe
the two methods.
Answer. The two types of polymerization are: (1) addition or chain polymerization and (2) step polymerization, also known as condensation polymerization.
20. Describe
a) Three plastic fabrication techniques
b)Describe and draw 3 types of chain structure in plastics
Answer.
Describe any below
∙Extrusion
∙Injection moulding
∙Extrusion blow moulding
∙Injection blow moulding
∙Thermoforming
∙Blown film extrusion
∙Rotational moulding
∙Welding
∙Others can be added
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21. A tensile test us a test specimen that has a gage length of 50 mm and an
area = 200 mm2. During the test the specimen yields under a load of 98,000 N.one love mp3
The corresponding gage length = 50.23 mm. This is the 0.2 percent yield point.
The maximum load of 168,000 N is reached at a gage length = 64.2 mm.
Determine (a) yield strength, (b) modulus of elasticity, and (c) tensile strength.
(d) If fracture occurs at a gage length of 67.3 mm, determine the percent
elongation. (e) If the specimen necked to an area = 92 mm2, determine the
percent reduction in area.
cm1Answer: (a) Y = 98,000/200 = 490 MPa.
(b) σ = E ε
Subtracting the 0.2% offt, ε= (50.23 - 50.0)/50.0 - 0.002 = 0.0026
E = σ/ε = 490/0.0026 = 188.5 x 103 MPa.
(c) TS = 168,000/200 = 840 MPa.
(d) EL = (67.3 - 50)/50 = 17.3/50 = 0.346 = 34.6%
(e) AR = (200 - 92)/200 = 0.54 = 54%
22. A test specimen in a tensile test has a gage length of 2.0 in and an area = 0.5
in2. During the test the specimen yields under a load of 32,000 lb. The
corresponding gage length = 2.0083 in. This is the 0.2 percent yield point. The maximum load = 60,000 lb is reached at a gage length = 2.60 in.
Determine: (a) yield strength Y, (b) modulus of elasticity E, and (c) tensile
strength TS.
Answer: (a) Y = 32,000/0.5 = 64,000 lb/in2
(b) σ = E e
Subtracting the 0.2% offt, e = (2.0083 - 2.0)/2.0 - 0.002 = 0.00215
E = σ/e = 64,000/0.00215 = 29.77 x 106 lb/in2
(c) TS = 60,000/0.5 = 120,000 lb/in2
sunshine的意思23. In Problem 22 (a) determine the percent elongation. (b) If the specimen
necked to an area = 0.25in2, determine the percent reduction in area. Answer: (a) % elongation = (2.60 - 2.0)/2.0 = 0.6/2.0 = 0.3 = 30%
(b) % area reduction = (0.5 - 0.25)/0.5 = 0.50 = 50%
24. S tate FIVE good and FIVE bad properties of ceramic materials.
