Due Date: July 24 2013 Name: - Course Notescoursenotes.mcmaster.ca/1M03_Summer_2013/Materials... ·...
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Due Date: July 24 th 2013 Name: ____________________ Student #: _________________ Materials 1M03 – Assignment 4 Please show your work for all the calculation questions and keep track of your units. 1. (3 mark) A stress-strain graph for a plain carbon steel is shown below. A cylindrical specimen of this alloy 565 mm long is pulled in tension. What is the radius of the specimen (in mm) if an elongation of 1.13 mm occurs when a load of 28.2 kN is applied? Answer: ! = ∆! ! ! ! = 1.13 !! 565 !! ! = 2×10 !! From the graph the corresponding stress is about 400 MPa. ! ! = !! ! ! = ! ! ! !! ! = 28,200 ! 400×10 ! !/! ! r = 4.74 mm
Transcript of Due Date: July 24 2013 Name: - Course Notescoursenotes.mcmaster.ca/1M03_Summer_2013/Materials... ·...
Due Date: July 24th 2013 Name: ____________________ Student #: _________________
Materials 1M03 – Assignment 4
Please show your work for all the calculation questions and keep track of your units.
1. (3 mark) A stress-strain graph for a plain carbon steel is shown below.
A cylindrical specimen of this alloy 565 mm long is pulled in tension. What is the radius of the specimen (in mm) if an elongation of 1.13 mm occurs when a load of 28.2 kN is applied?
Answer: ! = ∆!!!
! = 1.13 !!565 !!
! = 2×10!!
From the graph the corresponding stress is about 400 MPa.
!! = !!!
! = !!!
!!! = 28,200 !
400×10! !/!!
r = 4.74 mm
Due Date: July 24th 2013 Name: ____________________ Student #: _________________
2. (4 marks) A cylindrical specimen of a metal alloy 12 mm in diameter and 1.2 m
long is to be pulled in tension. Calculate the force (in N) necessary to cause a 0.04 mm reduction in diameter. Young's modulus and Poisson's ratio of the metal are 110 GPa and 0.32, respectively. Use 3 significant figures in your answer.
Answer:
! = ! × !! ! = !" × !!! ! = −
!!!!
0.32 = −
∆!!!!!
!! = −(−0.04!!12!! )
0.32 !! = 0.01042
! = 110×10! !" × 0.01042 × !(0.012!
2 )!
F = 129,632.166 N = 1.30*105 N
3. (1 mark) Which of the following statements is incorrect?
a) For a tensile test, true stress is always less than engineering strain. b) Foam used to absorb impact to protect high jumpers, must have a low
elastic modulus c) Yield stress is the stress at which dislocations become mobile d) Youngs modulus is primarily dictated by bonding
Answer: a
4. (1 mark) The atoms surrounding a screw dislocation experience what kinds of strains?
a) Shear strain b) Tensile strains c) Compressive strains d) All of the above e) Both A and C
Answer: a
Due Date: July 24th 2013 Name: ____________________ Student #: _________________
5. (1 Mark) True or False?
The process by which plastic deformation is produced by dislocation motion is called Slip.
Answer: True
6. (2 marks) A single crystal of a metal undergoes a tensile test. It is oriented so that the normal to its slip plane makes an angle of 22.7° with the tensile axis. Three possible slip directions make angles of 58.3°, 76.2° and 87.4° with the same tensile axis. One of these directions will be favoured. If plastic deformation begins at a tensile stress of 26 MPa, determine the critical resolved shear stress (in MPa) for the metal. Use 4 significant figures in your answer.
Answer: !!"## = !! × cos ! cos(!)
!!"## = (26 !"#) × cos 58.3 cos(22.7)
!!"## = 12.60 MPa
7. (3 marks) The yield strength for an alloy that has an average grain diameter of 7.25 x 10-2 mm is 245 MPa.At a grain diameter 3.42 x 10-2 mm, the yield strength increases to 320 MPa. At what grain diameter (in mm) will the yield strength be 400 MPa? Use 3 significant figures in your answer.
Answer: !! = !! + !!!!
245!"# = !! + !!1
7.25 × 10!! !!
320!"# = !! + !!1
3.42 × 10!! !!
Setting the two equations equal to !!, ky can be solved
245!"# − !!1
7.25 × 10!! !! = 320!"# − !!1
3.42 × 10!! !!
Due Date: July 24th 2013 Name: ____________________ Student #: _________________
75!"# = !!(1
3.42 × 10!! !! −1
7.25 × 10!! !!)
!! = 44.28766
Now !! !"# !" !"#$%& !"#
245!"# = !! + 44.287661
7.25 × 10!! !!
!! = 80.51976
400!"# = 80.51976 + 44.287661!
! = 0.0192 !!
8. (1 mark) For a cold worked sample of copper, to maximize both the tensile strength and ductility at the same time, the sample should be recrystallized for
a) 100 minutes at 113°C b) 20 minutes at 113°C c) 40 minutes at 102°C d) 9 minutes at 135°C e) 104 minutes at 43°C
Answer: d
9. (1 mark) True or False?
Due Date: July 24th 2013 Name: ____________________ Student #: _________________
Picture B has a higher surface energy than Picture A Answer: True, a larger grain boundary area causes a larger surface energy
10. (3 marks) Plot the stress-strain curve for a ductile metal. On your curve show the elastic modulus, the 0.2% off-set yield stress, the tensile strength, the maximum uniform elongation and the fracture strain. Note: This was an old midterm question!
Answer:
Mark: / 20
A B
