長
a
)
=
t
( 0 . 0282 :
b) 2
2
( 48 . 7
s
)
0
4
30 , 000
=
6 . 16 × 1o
. 16 x10
6
= 97 - 400 = ( 1
4
d
=
au
c) (48 . 7)(i 5x10- 4
.
ENGR 200: Materials of Engineering
Final Practice Exam (Fall 2025)
5x 10 im
. 028m
0
p 5x 10 imm
28mm
:
=
Chapter 8: Failure (25%) F= 30 , 000 N
a =
5 X 66
0 5 mm =
-
O )
979 ≈
San Francisco State University
フ
48 . 7
MP
4 . 87 × 10'Paz
School of Engineering
=
-
0
-
:
=
、
.
00157
=
.
93 MPa 「
11ㄱ
可
.
.
=
.
50 - 1.93
No , immediate catastrophic
Fracture is expected
Figure 1: A Tower Crane
(a) What is the stress (σo) experienced by the tie bar during operations?
(b) What is the maximum stress (σm) experienced by the crack tip during operations?
(c) Determine the Fracture toughness KC of the tie bar.
Q 1)
.
(d) Given that the tie bar is made of an alloy with a KIC of 50 MPa m and considering the
operating stress, should you immediately stop the crane from operating to prevent a potential
catastrophic failure?
AS
PART
←
Qu 2 Within any given day, the tie bar experiences a maximum load of 30,000 N 100 times. When it is
not operating, it is subjected to a minimum load of 2,000 N.
ㄱ
( i)
7 MPa
間 448 (i) What is the mean stress experienced by the tie bar?
omax = 3
.
Omin = 2 , 800
=
6 16 x10
-
-
3 25 MPa
.
4
EbMPa
.
. 25 .
48 .7 + 3
(ii) What is the stress amplitude?
(iii) Determine the maximum stress amplitude it operates in order to not fail prematurely if
his service life is 500,000 cycles (Figure 2).
2
. 25
(ii 48 .7- 3
_
EtMPa
2
)
N
=
5 x 105
=
200MPa
22 7 < 200
.
The die bar will not fail
prematurely
Maximum Stress Amplitude, S (MPa)
X 1o
Figure 2: Maximum Stress Amplitude vs Cycles to failure
.
.
Qu. 1 During a monthly inspection of the tie bar of a tower crane (Figure 1), surface crack of 0.5 mm
length was discovered on the tie bar. Diameter of the tie bar is equal 28 mm. Assume the tie bar is made
of 4340 steel and is subjected to a maximum load of 30,000 N during operation, the radius of the
curvature of the crack is 5 x 10-3 mm. d) bIVEN SOMPaTm
= 1 93 MPam
CALCULATED
Tie Bar
SAME
0 0396
48 7 ( 0 0396 )
=
um
0
San Francisco State University
School of Engineering
ENGR 200: Materials of Engineering
Final Practice Exam (Fall 2025)
Chapter 10: Phase Transformations (25%)
Qu. 3 Using the isothermal transformation diagram for an iron–carbon alloy of eutectoid composition
(Figure 3), please, identify and name the final microstructures resulting from the two cooling curves
labeled (a) and (b).
(a):
coarse
pearlite
(b): Bainite
(c): Draw and label on Figure 4 the cooling curve that yields the following microstructures: 50% of
microstructure A (Figure 4a) and 50% of microstructure B (Figure 4b).
(a)
(b)
Figure 3: Isothermal Transformation Diagram for an
Iron–Carbon Alloy of Eutectoid Composition
Figure 4a:
Microstructure A
Figure 4b:
Microstructure B
Chapter 14: Polymer Structures (25%)
Qu. 4 Epidermis is an Artificial skin for burned victims. It has a special keratinized layer responsible
for nutrition. To recreate something similar, scientists have to use different types of cells. When the
layer of the artificial skin is placed over the injured part of the body, it kick - starts the body’s own
regeneration process.
Figure 5: Epidermis
San Francisco State University
School of Engineering
ENGR 200: Materials of Engineering
Final Practice Exam (Fall 2025)
If the thickness of Epidermis is 1.5 mm and the pressure difference across the membrane is maintained
at 650Pa, what is the diffusion flux in [(cm3SPT)/cm2-s] at 298K? Assume conditions of steady state
and permeability constant of 4000x10-6 (cm3SPT) (cm)/cm2-s-Pa. 」
= 0 , 00 µ ( 6 SO
P 4000 X 10-6
0 004
4) 1 1 5 mm 0 15cm
0 15
2
:
=
=
=
.
.
AP
=
=
.
650 Pa
J
Chapter 16: Composite Materials (25%)
)
.
17 . 3 ( cm ㎥STP ) /
=
=
Ccms )
Qu. 5 Composite materials are becoming more widely used in aircraft industry due to their high
strength, low weight and excellent corrosion resistant properties. As an engineer who is given task to
design the I beam section of an aircraft (Figure 6) please, answer the following questions given the
material properties in Table 1.
미
176
carbon :
ARAMID
=
(a) Determine the Moduli of Elasticity of Carbon/Epoxy, Aramid/Epoxy, and Boron /Epoxy
composites in the longitudinal direction, given that the composites consist of 45 vol%
70 . a epoxy and 55 vol% fiber.
:
(b) What are the specific moduli of each of these composites?
= bl . a
1. 40
BORON
=
c
(c) What are the specific strengths (i.e. specific UTS) of each of these composites?
:
85 . 0
(d) What is the final cost of each of these composites?
) carbon
Aramid 20000
4
:
:
1
Boron :
.
=
=
9) Carbon/Epoxy
2222
0 55 [230
.
127 66 Pa
=
,
.
.
ARAMIDIEPOXY
130 ) +
0 55 [
2143
U
3600
) + 0 45 (2 4 )
0
45 (2 4 )
=
72 60 Pa
.
.
.
.
=
BORON/EpO XY
1385
0
.
55 ( 400 ) +
o
45 ( 2 4 )
=
.
_
221
d) Carbon
Avamid
Figure 6: I beam of the Aircraft to be designed
Boron :
:
0
.
16 Pa
0 55135 ( + 0 45
.
.
:
55(62) + 0 45 (5)
.
.
0 55(40) o . 45 (5)
.
+
=
Table 1: Properties of various materials used in composites
Materials
E[GPa]
UTS [MPa]
Specific Gravity
Unit Cost
($US/kg)
Epoxy (Matrix)
2.4
72
1.25
5
Carbon Fiber
230
4000
1.80
35
Aramid Fiber
130
3000
1.40
62
Boron Fiber
400
3600
2.60
40
=
(5)
36 . 3
24 3
.
=
2 1 5
.