Solution: MSEN 621 Mechanical Properties of Materials Midterm Exam

This document provides a detailed solution to the MSEN 621 Midterm Exam on Mechanical Properties of Materials. The solution covers a range of topics, starting with the derivation of the modulus of elasticity based on atomic bonding energy and plotting interaction energy for different materials. It then addresses point defects, dislocations, and deformation mechanisms like slip and twinning. Crystallographic planes and slip systems are analyzed, including planar density calculations for FCC and BCC structures. The document further explores stress-strain relationships, calculating engineering stress and strain, plotting stress-strain curves, and determining yield strength, tensile strength, modulus of elasticity, modulus of resilience, and ductility. The solution also delves into cold work calculations for both circular and rectangular specimens, followed by hardness calculations using Brinnel and Rockwell scales. Furthermore, it examines impact energy versus temperature curves, S-N plots for fatigue analysis, and phase diagrams, specifically the lever rule application for ferrite and cementite mass fraction calculations in iron-carbon alloys. The solution uses formulas, diagrams, and plots to illustrate the concepts and provide clear answers to the exam questions.

Question 1
To derive an expression for modulus of elasticity:
We first find the expression of FN by finding the derivative of EN with respect to r as [1].
FN = d EN
dr = d
dr (− A
r + B
rn )
¿ d
dr ( − A
r ) + d
dr ( B
rn )
FN = A
r2 − nB
rn+ 1
Equating FN = d EN
dr =0 and solving for r gives r0.
A
r2 − nB
rn+ 1 |r =r0
= A
r0
2 − nB
r 0
n+1 =0
A r0
n +1=nB r0
2
A r0
n+1
r 0
2 =nB
r0
n +1 −2= nB
A
r0
n − 1= nB
A
1
r0
n −1 = A
nB

r0 =( A
nB ) 1
1 − n
Finding the derivative of FN with respect to [1]:
d FN
dr = d
dr ( A
r2 − nB
rn+1 )= d
dr (− A
r2 )+ d
dr ( B
r n+1 )=− 2 A
r3 + n(n+1)
rn+ 2
d FN
dr |r =r0
= ( − 2 A
r3 + n ( n+1 )
rn+ 2 ) r=r0= ( A
nB ) 1
1 −n
¿ 2 A
( ( A
nB ) 1
1 −n
)3 + n ( n+1 ) B
( ( A
nB ) 1
1 −n
)n+ 2
The required expression is given by:
( d F N
dr )r =r0
=− 2 A
( A
nB ) 3
1 −n
+ n (n+1)B
( A
nB ) n+2
1 −n
Plotting EN vs FN for X, Y and Z using the derived formulas.
Excel plot:

0 0.5 1 1.5 2 2.5 3 3.5 4
0
0.5
1
1.5
2
2.5
3
3.5
4
E_ vs _𝑁 𝐹 𝑁
X
Y
Z
𝐹_𝑁
E_𝑁
Question 2
a) The major types of point defects in materials are
i) Interstitialcy
ii) Vacancy
iii) Impurity atoms
Vacancy defects occur when an atom is missing from its lattice site in the metal’s crystal
structure. The figure 2.1 illustrates this defect.
Interstitialcy defects occur when an atom of the material occupies an interstitial site (not a
normal site). The figure 2.2 illustrates this defect.
Impurity atoms present on the lattice may substitute the lattice atoms thus causing a point defect
in the material. The figure 2.3 illustrates this defect [2].

Figure 2.1: Vacancy defect
Figure 2.1: Interstitialcy defect

Figure 2.1: Large substitutional impurity atom defect
b) Edge dislocation is a line defect which occurs due to loss of half of a plane of the atoms
in the middle of the lattice or due to presence of an extra plane of atoms. Figure 2.4
illustrates edge dislocation [2].
Screw dislocation is a line defect in crystal lattice which occurs when the planes in the lattice
trace a helical path around the dislocation line. Figure 2.5 illustrates screw dislocation.
Figure 2.4: Edge dislocation.
Figure 2.5: Screw dislocation.

c) The changes in the shape and size of objects due to external forces or loads is called
plastic deformation. This is an irreversible process of deformation.
Plastic deformation occurs through twinning, slip or a combination of twinning and slip. The slip
mechanism involves the sliding of one plane over another plan and does not affect the metal’s
crystal structure and the arrangement of atoms in the crystal remains unchanged. Slip takes place
in one particular plane and direction called slip planes and slip directions respectively. The two
types of motion that causes slip are climb and slide motion [1], [2].
The twinning mechanism involves atomic arrangement where one part looks like an exact mirror
of another part. The twinning plane is the boundary of the plane acting like a mirror.
Deformation twinning is caused by deformation process while annealing twinning is caused by
heat treating. The figure 2.6 below shows the slip and twinning process [2].
Figure 2.6: slip and twinning mechanism.

Question 3
a) Planar densities [2]
For FCC structure:
P D100 ( FCC )= 1
4 R2 = 0.25
R2
P D110 ( FCC )= 1
4 R2 √2 =0.1768
R2
P D111 ( FCC )= 1
2 R2 √3 =0.2887
R2
For BCC structure:
P D100 ( BCC )= 3
16 R2 = 0.1875
R2
P D110 ( BCC ) = 3
8 R2 √ 2 = 0.2652
R2
P D111 ( BCC ) = √3
16 R2 = 0.1083
R2
b) The {111}<110> is a FCC unit cell having a slip system consisting of {111} plane and the
<110>.

From the table provided in part d) it is clear that {110}<111> is for metals with BCC
structures such as W, Mo and α − Fe [2].
c) The sketch below shows the atomic packing for a {0001}-type plane for HCP structure.
The arrows show the three different slip directions.
d) Considering the slip planes and the slip directions
Using Frank’s rule, the dislocation energy is proportional to the square of the burger vector.

Comparing the energy, ½ (110) dislocations have the energy a2
4 and the slip direction is 110
From the table, it can be concluded that the higher the number of slip systems, the more ductile
the material is. For example, Copper (no. of slip systems=12) is more ductile than Zinc (no. of
slip systems=3) [2].
Question 4
a) The engineering stress is calculated by the formula:
σ = F
A . Where F is the load applied and A is the area of the specimen.
The specimen is cylindrical hence:
A= π d2
4 . Where d is the diameter of the specimen.
Substituting A in the formula gives:
σ = F
( π d2
4 )= 4 F
π d2 eqn ( i )
The engineering strain is calculated as:
ε = li −lo
lo
eqn ( ii ) .
Where li is the specimen’s instantaneous length and l0 is the original specimen’s length [1].
Using the formula of equations (i) and (ii), the engineering stress and strain can be calculated and

tabulated as below.
Stress, σ (Mpa) Strain, ε
0 0
98.69471569 0.0005
197.3894314 0.001
296.0841471 0.0015
394.7788627 0.002
592.1682941 0.003
692.4172573 0.004
720.3937121 0.005
796.5518392 0.0075
837.7393977 0.01
927.8857522 0.015
997.0497656 0.0. 2
1163.354247 0.04
1235.626755 0.06
1246.506488 0.07
1239.512374 0.08
1177.342475 0.1
969.0733107 0.115
Excel plot of engineering stress vs engineering strain is given below [1].

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14
0
200
400
600
800
1000
1200
1400
Engineering stress,σ vs engineering strain,ε
strain,ε
stress,σ (Mpa)
b) The formula for modulus of elasticity is given by
E= ∆ σ
∆ ε
Using values from the second and third rows of the stress and strain table, E is calculated as:
E=197.39 − 98.69
0.001 − 0.0005 =197.40GPa ≈ 200 GPa
∴ E=200 GPa for stainless steel.
c) Using the graph of stress versus strain above, the yield strength of stainless steel at strain
offset of 0.002 is given by the y-intercept of a line parallel to the curve and with x-
intercept of 0.002 [1].

The yield strength is given as: 740 MPa
d) The maximum point/peak of the stress versus strain curve is the tensile strength of the
alloy
Tensile strengtℎ=1250 MPa
e) The modulus of resilience, Ur:
Ur = σ y
2
2 E
Where E is the modulus of elasticity and σ y is the yield strength of alloy.
From values calculated for E and σ y in part b) and part c) respectively, the modulus of
resilience is found as [1]:
Ur = (740 ×106 )2
2× 200× 109 =1.369 ×106 J /m3
f) Ductility
The percent elongation%EL, of the steel specimen is calculated as:
%EL=(ε f −ε e)× 100
Where ε eis the strain in elastic region and ε f is the strain at fracture point.
In this case, ε f =0.115 , εe=0.005;
%EL= ( 0.115− 0.005 ) × 100=11 %