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ME2114
NATIONAL UNIVERSITY OF SINGAPORE
ME2114 MECHANICS OF MATERIALS II
(Semester II : AY2012/2013)
Time Allowed : 2 Hours
Matriculation Number
___________________________________________________________________________
INSTRUCTIONS TO CANDIDATES:
1.
Write your Matriculation Number in the box above.
2.
This examination paper contains FOUR (4) questions and comprises TWENTY-
THREE (23)printed pages.
3.
AnswerALLFOUR (4) questions.
4. Write your answers in the space provided in this question booklet.
5. This is a CLOSED-BOOK EXAMINATION.
Question
Number
Marks
Obtained
Maximum
Marks
1 25
2 25
3 25
4 25
Total 100
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PAGE 2 ME2114
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PAGE 3 ME2114
LIST OF EQUATIONS
1. Strain energy in slender rods due to axial loads
dxx
uEA2
1dxEA
N
2
1U
L
0
2L
0
2
2.
Strain energy in slender rods due to bending
dxx
vEI
2
1dx
EI
M
2
1U
L
0
2
2
2L
0
2
3. Strain energy in slender rods due to torsion
dxGJ
T
2
1U
L
0
2
4. Stiffness matrix of a truss member
L
EAksins,cosc
v
u
v
u
scsscs
csccsc
scsscs
csccsc
k
vuvu
22
22
22
22
andwhere
localk
5.
Stiffness matrix of a plane stress triangle element
BEBk ~~A4
t Tlocal
where
yxyxyx
x0x0x0
0y0y0y~B
and
)1(2
E00
01
E
1
E
01
E
1
E
22
22
E
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PAGE 4 ME2114
QUESTION 1
Figure 1 shows a solid beam ABC supported by two truss elements BD and BE. A force P is
applied at C. Using Castiglianos theorem, determine the reaction force from the simple
support at E and the vertical displacement of C. Take into consideration strain energy due to
axial forces only for truss elements BD and BE and strain energy due to bending momentsonly for beam ABC. The Youngs modulus, cross sectional area and second moment of area
for the truss elements and beam areE,AandI, respectively.
(25 marks)
Figure 1
2L
L
L 2L
A
E
P
DB
C
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PAGE 9 ME2114
QUESTION 2
The stiffness matrix of the plane stress triangle element in Figure 2a with a Youngs modulus
ofEand thickness tis
u v u v u v
[klocal]= 2L
Et
0.375 0 0.375 -0.1875 0 0.1875 u
0 1.125 -0.1875 -1.125 0.1875 0 v
0.375 -0.1875 0.6563 0.375 -0.2813 -0.1875 u
-0.1875 -1.125 0.375 1.2188 -0.1875 -0.0938 v
0 0.1875 -0.2813 -0.1875 0.2813 0 u
0.1875 0 -0.1875 -0.0938 0 0.0938 v
Show that the stiffness matrix of the element in Figure 2b is the same as that for the elementin Figure 2a. Hence, solve for the displacement of node C in Figure 2c which shows the finite
element mesh of a plate clamped along the top and left edges carrying a load P.
(Hint : Compare the matrix ]~
[B as defined in the given list of equations for both the elements
in Figures 2a and 2b)
(25 marks)
Figure 2a Figure 2b
Figure 2c
L
2L
a
b
c
d
P45
o
L
2L
L
2L
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PAGE 10 ME2114
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PAGE 13 ME2114
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PAGE 14 ME2114
QUESTION 3
A solid shaft which has a radius of 30 mm is made of an elastic perfectly plastic material. It is
loaded by a slowly increasing torque Tpuntil a plastic zone has occurred partially in the shaft.
Derive an expression for the torque Tp and hence the torque Tu required to cause full
plasticity.(10 marks)
Assuming the yield stress in shear =150 MPadetermine:
(a) the fully plastic torqueTu ,
(b)
the residual shearing stress at the outer surface after the torque is removed and plot
the residual stress distribution indicating the residual stress values at the outer surface
and center of the shaft cross-section.
(15 marks)
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PAGE 19 ME2114
QUESTION 4
A rigid horizontal bar pivoted at the left-hand end, supported by two columns P and Q, is
loaded at point A by a load of W (see Figure 3). The columns have a hollow circular cross-
section with inner and outer diameters of 80 mm and 100 mm respectively. Column Q is
pinned at both ends and column P is pinned at one end and fixed at the other. The columnsare fabricated from a material with a Youngs modulus of 180 GPa. Using a safety factor of
1.5 determine which column will buckle first and hence the maximum allowable value of W.
(25 marks)
The Eulers buckling loadPcris given by (usual notations apply):
(for a pinned-pinned column)
(for a pinned-fixed column)
2
2
L
EI
2.5 m
W
0.8 m
Figure 3
0.8 m 0.8 m
A
Column Q
Column P
3.5 m
crP
2
205.2
L
EIcr
P
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PAGE 23 ME2114
- END OF PAPER -
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Pg.
1
NATIONAL
UNIVERSITY
OF SINGAPORE
Department
of
Mechanical Engineering
Module Code: ME2114 Module Title: Mechanics
of
Materials II
Solution to Question No. _
c
Sh
em
- s.lreSS distribution
For elastic-plastic shaft, the torque in the shaft is given by:
T
=
S
r 2nr )dr
.
r +
LR y nr
2
dr
Elastic)
Plastic)
c
R
2
r
[r
r]
r
2
T
=
o nr dr + Jc y nr dr
2Jr I 3 fR 2
T
=
-ry
r dr
2Jrry
r dr
c 0 C
1 [ 4
21 [ 3 3
=
r y c
2c
-ry
R
3
c
=
R
3
c3 J
2Jrry 3 12
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-----------------------
Pg 2
a) For fully plastic shaft, C = 0 and the torque in the shaft is given by:
R3
_ =
nTyR3
=
2n
x 150 x 30
3
= 8 48x
10
6
Nmm
u = 2nT y
3
12
3 3
b) For elastic unloading we have:
TyR
3
[
2n
JR 4 _4x15 =2 MPa
R
3 =
-
3
~ =
3
y
TR 1[
R4
2
I.e. Residual stress at r
=
30
mm
is 150-200)
=
-50 MPa
Residual stress distribution:
150 MPa
-50 MPa
Set By :
_ J
Tay _
Date:
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NATIONAL UNIVERSITY OF SINGAPORE
Department of Mechanical Engineering
Module Code:
t \b-21
t / ~ . . 2 i l ' f M o d u l e
Title:.
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Solution to Question No.
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SetBy:
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Date: ______________
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----------------------------
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NATIONAL UNIVERSITY OF SINGAPORE
Department of Mechanical Engineering
Module Code:
Module Title:
Solution to Question No. _
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Ear.
t l
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Date: ___________________
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- - - - - - - - - - - - - - - - - - - - -
NATIONAL UNIVERSITY OF
SINGAPORE
Department of Mechanical Engineering
Module Code:
Module Title:
Solution to uestion
No.
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SetBy: ________________________
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Date: