Machine Component Design (INME 4011)
2nd Exam-Part B-060
Student Number :___________________ Name: _________________Date: November 27th, 2024
Notes: No means of communication between students are allowed. Please ensure that you have a complete copy of the
exam and use this format to submit your answers Only clear and concise answers will be considered.
------------------------------------------------------------------------------------------------------(30 points) A surface crack of length ai (mm) was detected in a C-frame of rectangular section ( depth of
section AB (h), measured from point A to B and thickness of the section AB (b)) made of a steel alloy (yield
strength=860 in MPa and fracture toughness = Kic in MPa.m-0.5 . The C-frame is subjected to the loads a
variable load P that acts in the directions as shown and it fluctuates from Pmin to Pmax (in N) .
Data (based on the student number last digit):
Student
#
ai (mm)
Kic
h (mm)
b (mm)
L (mm)
Pmax
Pmin
Use table
Odd
0.1
90
38
30
130
18000
5000
Table 1
Even
0.05
70
45
25
160
12000
4000
Table 2
In the region of crack propagation, the values measured for the relationship between da/dN and DKi in
this alloy were: (Hint: use it to determine the Paris coefficient C and exponent m)
Table 2
e
Table 1.
Delta Ki
10
15
20
40
60
da/dN
3.1773E-08
1.5446E-07
4.7433E-07
7.0810E-06
3.4423E-05
Delta Ki
da/dN
10
15
2.1453E-08
1.0014E-07
20
40
60
90
2.9881E-07
4.1620E-06
1.9429E-05
9.0699E-05
Hint: as the geometric factor Y depends on the crack length… then use the initial crack ai to calculate the Y1
for the different loads. Use the Y1’s to determine the final crack size af1.
With af1 calculate again the geometric factors Y2’s.. Take the average (Y1+Y2)/2 as the final value of the
geometric factor Y. calculate the final crack length af
Determine the life of the component
p
9 MPa m, sY = 1500 MPa.
Figure 2.19 C-frame.
t load P and bending moment M can be calculated using the
s, respectively.
p
paYM ,
Pp
K IP =
paYP
Bh
a = a/h = 5/40 = 0.125
#50
Y
1 12 +
.
=
(0
00111)(2 91(0 00111)
-
0 93(0
.
p=
on5)
.
1 12045
=
.
10 66 MPa
.
=
3 55
.
MP
Y af
k, =
12043))
F(yi) = (1
=
1 86551
.
.
af
=
06 41 455
=
2f
.
5 229
+
Yang =
2
Kic
.
00111)
=
==
000045)
Tmin
0 64)
.
1
Tmax =
-
.
.
Tmax
=
.
266
Yang #fnew
:
at new = (Yag)
At
.
=
=
-3 5185
.
(358
.
YAT
=
Y
=
+ x(52
-
588
23
5
[
=
+ Cd -588
23
=
Youg At
:
Bending
==
Thin
d (08
=
25.
.
*
Kic
=
Tmax-Tmin
=
Tmax
Yi
=
af
TYiaf
= (v) =
5
= 01
-
1
.
11859)
=
0 08315
.
Yi
2(3432 -1a)
+
=
=
11o
=12.