dr. c. w. ong
TRANSCRIPT
-
8/14/2019 Dr. c. w. Ong
1/34
Jump to first page
1
Mechanics
AP200Dr. C. W. Ong
-
8/14/2019 Dr. c. w. Ong
2/34
Jump to first page
2
Textbook :
Engineering Mechanics, Dynamics,by Meriam, J.L., Kraige, L.G.,John Wiley 1997 4th Ed.
Mechanics of EngineeringMaterials , by P.R. Benham andR.J. Crawford, ELBS, 1988
-
8/14/2019 Dr. c. w. Ong
3/34
Jump to first page
3
Assessment method :
s Course Work 30% (Exercises and tests)
s Examination 70%
-
8/14/2019 Dr. c. w. Ong
4/34
Jump to first page
4
VectorAA vector consists of two things :
1. Magnitude, represented by a
number A
2. Direction, represented by a
unit vectorAe
AeAA =
e
-
8/14/2019 Dr. c. w. Ong
5/34
Jump to first page
5
In Cartesian coordination system
yAxA
yA
eAA A
)sinx(cos
yx +=
+=
=
Y
X
A
YA
XA
-
8/14/2019 Dr. c. w. Ong
6/34
Jump to first page
6
jBAiBABA
jBiBB
jAiAA
YYXX
YX
YX
)()( +++=+
+=
+=
Y
X
YB
xBB
AYA
XA A
'B
CBA
=+
Addition of vectors
-
8/14/2019 Dr. c. w. Ong
7/34Jump to first page
7
kAjAiA
kAjAiA
eAA
ZYX
A
++=
++=
=
coscoscos
Direction cosine : cos , cos ,cos
1coscoscos
)coscos(cos
)coscoscos(
222
2/1222
2/1222222
222
=++
++=
++=
=++=
A
AAA
AAAAAZYX
3-D Cartesian coordination system
Y
Z
A
X
e
-
8/14/2019 Dr. c. w. Ong
8/34Jump to first page
8
Result of dot product is a scalar
Dot product :
CBA =
cosABBA
A
B
cosA
Definition is :
-
8/14/2019 Dr. c. w. Ong
9/34Jump to first page
9
Cross product
rule.handrightbydetermineddirectionwithvectors,twothetois
and,sin:Definition
vectornewaisResult
=
=
C
ABBA
CBA
AB
C
Fsin
d
F
O
-
8/14/2019 Dr. c. w. Ong
10/34Jump to first page
10
kBjBB
kCC
kAjAiAA
ZY
Z
ZYX
+=
=
++=A
B
C
CBA
)(Z
Exercise : Show that CBA
)(
is the volume of the parallelepiped
formed by the three vectors
-
8/14/2019 Dr. c. w. Ong
11/34Jump to first page
11
Presentation of vector product inrectangular coordination system
kAjAiAA zYX ++=kBjBiBB zYX
++=
ZZYYXX BABABABA ++= )(
jABkBAiBA ZXYXZY
++= iABjBAkAB ZYZXYX
ZYX
ZYX
BBB
AAA
kji
BA
=
Dot product :
Cross product :
-
8/14/2019 Dr. c. w. Ong
12/34Jump to first page
12
Fundamental of Mechanics ofMaterials
Stress and strain :
Assume F is uniformly distributed over thecross-sectional area A.
A
F
F
F
F
F
F
-
8/14/2019 Dr. c. w. Ong
13/34
Jump to first page
13
Normal stress
[ ] )(/ 2 PapascalmNA
F
dA
dF
A
F
A==
0lim
F
A
FP
Normal stress at a point :
-
8/14/2019 Dr. c. w. Ong
14/34
Jump to first page
14
Total force F acting on a cross-sectional
area is :
= A dAF
F F
F
F
Tensile stress (+)
Compressive stress (-)
-
8/14/2019 Dr. c. w. Ong
15/34
Jump to first page
15
Shear stress :
Define shear stress :
A
A
Fixed area
F
F
A
F
dA
dF=
Force F is tangential to the area.
At a local point :
-
8/14/2019 Dr. c. w. Ong
16/34
Jump to first page
16
Normal strain :A body deformed along the direction of stress. Normal strain is defined to describe the deformation :
x
x
x
F
F
l x
F
Ffixed
Shear strain :Deformation under shear stress is defined as
-
8/14/2019 Dr. c. w. Ong
17/34
Jump to first page
17
Elastic modulus
Stress-strain curve
In the elastic region, (Hooke's law)
Youngs modulus (modulus of elasticity) E = /
Shear modulus : (modulus of rigidity) G = /
Stress
Straino
Yield pt.
Workhardening
breakElastic
deformation
d
-
8/14/2019 Dr. c. w. Ong
18/34
Jump to first page
18
le
Find the totalextension ofthe bar. 3 The width of a cross-sectional element at x :
3 The stress in this element :
3 The strain of this element:
15mmW5mm
X
1.2m0.6mo
kN2
)(120
)105(6.0
3 mx
mm
xW ==
Paxmx
N2
7
22
31088.2
)120/(
102 =
=
dx
2
4
9
271092.1
10150
/1088.2
x
x
E
=
==
-
8/14/2019 Dr. c. w. Ong
19/34
Jump to first page
19
The extension of this element :
The total extension for the whole bar becomes :
= 2.13 x 10-4 m #
dxx
dxde2
41092.1 ==
==8.1
6.0 2
41092.1dx
xdee
-
8/14/2019 Dr. c. w. Ong
20/34
Jump to first page
20
Bulk modulus :)/( VV
pK
=
VV +
p
lim V 0 then
dV
dp
VK =At a local point :
-
8/14/2019 Dr. c. w. Ong
21/34
Jump to first page
21
Poisson's ratio :For homogeneous isotropic materials
s normal strain :
s lateral strain :
s
Poisson's ratio :s value of : 0.2 - 0.5 d
dL
= /L
x=
F F
dd +
x
d
-
8/14/2019 Dr. c. w. Ong
22/34
Jump to first page
22
Analysis of stress and strain inIsotropic Materials.
EEE
EEE
EEE
yxzz
zxy
y
zyxx
=
=
= ......(*)
z
zy
yz
yx
zx
xz
x
y
xy xy
z
-
8/14/2019 Dr. c. w. Ong
23/34
Jump to first page
23
Rewrite (*) by xxx, yyy , z zz :
Eyyxxzzzz
Exxzzyyyy
Ezzyyxxxx
/)(
/)(
/)(
==
=
-
8/14/2019 Dr. c. w. Ong
24/34
Jump to first page
24
)()1(
)(
)1(
)(1
zzyyxxzzzz
zzyyxxyyyy
zzyyxxxxxx
EE
EE
EE
+++
=
++
+
=
+++
=
)()21( zzyyxxzzyyxxE
++=++
Exercise : Verify)21(3
=
EK
Solution
From (*)
-
8/14/2019 Dr. c. w. Ong
25/34
Jump to first page
25
For hydrostatic pressure
pzzyyxx ===
=== zzyyxx
3
33
34
3
4)(
3
4
r
rrr
V
V
=
r
r
3)(3 =
r
r
)3()21( pE
=
)21(3/
=
E
VV
pK
(symmetry)
-
8/14/2019 Dr. c. w. Ong
26/34
Jump to first page
26
Similarly, take moment about x-axis, yz = zy
Take moment about y-axis, xz = zx
Shear stress and strain
Take moment about the z axis, total torque = 0,
so xy = yx
z
y
x yx
xy
x
y
-
8/14/2019 Dr. c. w. Ong
27/34
Jump to first page
27
Gxyxy / =Gyzyz / =
Gzxzx / =
Shear strain :
-
8/14/2019 Dr. c. w. Ong
28/34
Jump to first page
28
One better method is to select axes appropriately, such that
/2===
= yxxy
y
dx
x
dy
Shear deformation is usually defined by : .,..., etcy
dx
x
dy
However, pure rotation may cause non-zero
which cannot be used to represent any deformation
ydx
xdy
and
2/ x
y
2
x
y
xdy
x
y
xdy
Pure
rotation
-
8/14/2019 Dr. c. w. Ong
29/34
Jump to first page
29
(AC)2(1+n)2 = (AD)2 + (AD)2 -2(AD)2 cos(90o+ )
2(AD)2 (1+ n)2 = 2(AD)2+2(AD)2 sin
1+2 n= 1 + sin 1 +
2 n
since = 2 yx
n= yx
Example : Show that n = yx
2/ x
y
A
C
C
D
D2
-
8/14/2019 Dr. c. w. Ong
30/34
Jump to first page
30
yx (lW) sin 45o x2 =LW n (equilibrium along n-
direction)
= 2(lcos 45o) W n
Therefore yx = n
G
nn
2 =n2=== GG
nxyxy
From definition :
l
Ll
n
yx
xy
2/ x
y
A
2/
yx
xy
G
nn
2
=Example : Show that
-
8/14/2019 Dr. c. w. Ong
31/34
Jump to first page
31
GE
21
=+
(from p.24))(1 zzyyxxxxxxEE
+++=
n n
- n
- n
Set xx = n = - yy ,
zz = 0,
xx = n
n = (1+ ) n /E = n /2G (previous
example)
Example : Show GE
21
=+
xamp e 3 (i) Here = 0 = 0 and
-
8/14/2019 Dr. c. w. Ong
32/34
Jump to first page
32
xamp eThe cube of isotropicmaterial has side 20mm,E = 60 GPa, = 0.3.
(i) Find the force exerted bythe restraining walls upon the
block.
(ii)Find the strain in ydirection
3 (i) Here x = 0, y = 0 and
3 Since :
3 x = -9 x 106 Pa (compressive stress)
3 The force exerted by the restraining wall :
3 A x = (20 x 10-3 m)2 x (-9 x 106 Pa)
3 = -3.6 x 103 N (compressive force)
3
(ii) The strain in y direction :
z
y
12kN
x
Pam
Nz
7
23
3
103)1020(
1012=
=
)(1
zyxx
E =
)]103(3.00[1060
10 7
9
= x
#4
67
9
1095.1
)]109(3.0)103(3.00[1060
1
)(1
=
=
= xzyyE
El ti St i E
-
8/14/2019 Dr. c. w. Ong
33/34
Jump to first page
33
Elastic Strain Energy3 The energy store in the material :
3 The energy store :
3 Energy density within the material :
FdxdW=dx
F Fdx
xAEFdxdU )(
==
/
/
x
AFE=
VE
AeEAEe
dxxAE
Ue
=
==
=
2
2
2
0
2
1
)()(21
21
)(
EE
V
Uu
22
2
1
2
1 ==
e=extension
-
8/14/2019 Dr. c. w. Ong
34/34
34
Similarly for shear strain :
F
dx
= xdFU
= Fdx
==
/
/
x
AFG
G
Gu2
2
2
1
2
1 ==