SOLUTIONS
The stress for option (a) is more and hence
One Mark Solutions critical.
0 1 . Ans: (a) 02. Ans: (d) 06. Ans: (d)
03. Ans: Increases , Decreases 07. Ans: (a)
04. Ans: (c)
08. Ans: (c)
05. Ans: (a)
Sol: Soderberg's equation 09. Ans: 100 MPa
1 Sol: O'min = 200 MPa
-
'ta
+-
't m
=-
s se S sy Fs. ' O'max = 400 MPa
1 Stress amplitude 'cra '
-
'ta
+ --
't m
=
S se 2x S se FS 400-200
= = lOOMPa
s se 2
+-
't m
=- .
FS
't a
2
10. Ans. 2
Sol: O'max = 100 MPa
I6Tm Bending stress is a completely reversible
't m = 3 � 'tm OC
Tm. stress and for bending stress mean stress,
1t d
Tm O'rn = 0
Option Ta Tm Ta + -
2 � Stress amplitude 'cra ' = 100 MPa
a T 0 T S yt = 300 MPa, Sut = 500 MPa
3T T S e = 200 MPa, F.S = ?
b - 2T
4 4 8
From Soderberg's equation
T T
IT cr a cr m 1
2 2 4 - +-=-
S e S yt FS
3T
d
T - �T 100 = _
� F.S = 2
1
-
4 4 8
200 FS
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: 212 : Machine Design
11. Ans: (a) (J' a + - 1
(J' m = -
-
Sol: Soderberg line: The most conservative, it S e S yt F.S
connects Se on aa axis with Sy on am axis. 49 98 1
=> --- + - = -
1 48.48 294 FS
12. Ans: (c) => FS = l . 50
Sol: crmin =
50 MPa
CJ'max = 2 50 MPa 02. Ans: (c)
(J'max - (J'min 2 50 - 50 Sol: Basquin's equation , A Sr LB.
100 MPa
=
CJ'a
2 2
= = =
A = 490 x 10 38 ----( 1 )
Stress ratio, A = 1 0 x 10 68 ----(2)
crm = 50 Solving 1 & 2, A = 3397, B = 0 .281
R= in � _!_ = 0.20
3397 = 100 x Lo.2si => L = 28 1026 cycles
(J'max 250 5
13. Ans: 2
03. Ans: (b)
Sol: crmin - 50MPa, CJ'max 50MPa,
Sol: D = 200mm, t 1 mm, Pmin 4 MPa
= =
Syt = 300MPa
= =
Se l OOMPa,
=
Pmax 8 MPa.
The stress is completely reversed stress and
=
Syt 600 MPa , Sut = 800 MPa ,
=
mean stress, crm = 0 and stress amplitude.
Se 400 MPa.
=
cra 50MPa.
= (YI max = - = --- = 400.MPa,
PD 8 x 200
=
(J'
se
=-
(Yh
4t 4x 1
a FS
· =
- 1 = --- = 200.MPa.
PD 4 x 200
50 = l => F.S = 2
OO (YI mm
4 4x 1
FS
cr 1 = cr 2 max = 400 MPa,
cr2 min = 200MPa.
Two Marks Solutions
(J' l m =
CJ' I max + CJ' I min
01. Ans: (b) 2
147 42 400 + 200
Sol: cr m = + = 98 MPa = = 300MPa
2 2
147 - 49
aa 49 MPa
2
= =
1 96 400-200
S = = 1 48. 48 MPa = = 1 OOMPa.
e 1 .32 2
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: 213 : Fatigue
CF2 m = 300.MPa, CF2 0 = 100.MPa 1 00 X 1 0 3 N
(jmax
According to distortion energy theory, A mm 2
20 x 1 0 3 N
(jmin
A mm 2
= 300MPa
(jmax - (j min 80 x l0
3
40 x l 0 3
=
(j a ---=:;;.;;__2---==- N / mm 2
2A A
Goodman's equation 60 x 10 3
(jmax + (jmin
(j m = N / mm 2
(j aes -
(jmes 1 2 A
- + -=-
Se S u1 FS 1
Soderberg's equation � + crm =-
Se S y1 FS
1 00 300 _1
� + = � F.S = 1 .6.
400 800 FS
04. Ans: (a)
160 - 40 p
Sol: pm = = 60kN � cr = m
2 m A
60 x 1 0 3 N
= = 84.88N / mm2
7t (30) 2 mm2
4
06. Ans: (d)
Pa = 1 60 - (- 40) = l OO kN. Sol: crmax = 150 MPa, crmin = 50 MPa
2
� cra = 1 4 1 .47N / mm2 S e = 200 MPa, Syt = 300 MPa,
Sut = 400 MPa
Soderberg's Equation: - 1 50 - 50
cr - (jmax (jmin - - 50 MPa
2 2
3 -
(j a -
- 1
(jm = -
+ cr cr 1 50 + 50
S e S yt FS cr m = max + min = = l OO MPa
2 2
1 4 1 .47 84.88 -
1
�
240
+ = � F.S = 1 _26_
420 FS Goodman's equation :
(Fa (Fm 1
- + -= -
05. Ans: (d) S e S ut FS
Sol: Pmax = 1 00 kN, Pmin = 20 kN, 50 100 _1_
+ = � FS = 2
Syt = 240 MPa, Se = 1 60 MPa, 200 400 FS
A =?, F.S = 2
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. A.CE . . : 214 : Machine Design
07. Ans: (385.42) 1 = 0.520x1038
Sol: Sut = 600MPa, Se = 250MPa, B = 0.094
L = 10000 cycles , sr = ? :. A = 250x106x0.094 = 916.09
0 094
Basquin's equation, : .916.09 = srL ·
A = srL8 916.09 = srxl 04x0.094
A = (0.8x600)x1038 . . . . . . . . . ( 1 ) sr = 385.42 MPa
A = 250x1068 . . • . . • • • (2)
Dividing equation (2) with equation (1)
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