58
CHAPTER 5. TWODIMENSIONAL ELASTICITY
where the last condition is concluded from the fact 33 = 0. This is inserted to Hookes
law (4.19) and taken to express 33 . The tensors look
11 12 0
= 12 22 0
78
CHAPTER 6. ENERGY PRINCIPLES
1D:
Strain energy:
1
1
E
E
E
uk,l Cklij ui,j =
xx xx = (u,x )2 = (z (x)2 = z 2 (w (x)2
2
2
2
2
2
with dV = A dx = b dz dx:
h
2
1
uk,l Cklij ui,j dV
2
V
=
z = h
2
Eb
2
A.1. CHAPTER 1
83
(b)
2
f = f,ii = (f,i ),i =
f (r) xi
r
,i
(f (r) xi ),i r f (r) xi r,i
r2
f (r),i xi r + f (r) xi,i r f (r) xi r,i
=
r2
r
r
f (r) xi xi r + f (r) xi,i r f (r) xi xi
=
r2
xi
f (r) r x
ME2112 (Part 1)
Chap 2 - 1
1) Which of the following statement is incorrect?
a)
The principal stresses represent the maximum and minimum normal
stress at the point.
b)
When the state of stress is repr
ME2112 (Part 1)
Chap 1 - 1
ME2112 (Part 1)
Chap 1 - 2
Zero-Force Members
If a joint has only two non-collinear
members and there is no external load
or support reaction at that joint, then
those two m
ME2112 (Part 1)
Chap 1 - 1
Topics covered in previous lectures
ME2112 (Part 1)
Chap 1 - 2
Proof of forces being equal, opposite in direction and acting along
the line joining two points of force appli
ME2112 (Part 1)
Chap 2 - 1
y
x y
2
x y
2
xy
x y
2
x y
2
x y
2
Chap 2 - 2
In-Plane Principal Stress
2D Stress Transformation:
x
ME2112 (Part 1)
cos 2 xy sin 2
(2.1)
cos 2 xy sin 2
(2.2)
sin 2 xy
ME2112 (Part 1)
Chap 1 - 1
ME2112 (Part 1)
Chap 1 - 2
External and Internal Forces
2.2 Free Body Diagram
A free-body diagram (FBD) is a sketch of an object or a connected
group of objects, modeled as
ME2112 (Part 1)
Chap 2 - 1
ME2112 (Part 1)
Chap 2 - 2
Applications
Chapter 2 Transformation of Stress and Strain
Learning Objectives:
Study the states of stress and strain at points located
on obliqu
ME2112 (Part 1)
Chap 2 - 1
Applications (Stress transformation)
ME2112 (Part 1)
Chap 2 - 2
A cube of 1 unit length is stationary (in moment equilibrium)
Take moment about z-axis
y
xy (1 1) 1 yx (1 1)
ME2112 (Part 1)
Chap 1 - 1
Faculty of Engineering
Department of Mechanical Engineering
ME2112 Strength of Materials (Part I)
Course Lecturer: A/Prof C. Quan
Office: EA 05-28
Tel: 65168089
Email: mpeqc
ME2112 (Part 1)
Chap 2 - 1
ME2112 (Part 1)
Chap 2 - 2
Applications
Chapter 2 Transformation of Stress and Strain
Learning Objectives:
Study the states of stress and strain at points located
on obliqu
6.3. APPROXIMATIVE SOLUTIONS
73
with the element-length le . So, the approximation in e is
e
w(e) (x) = w1 N1
x xe
+w1e N2
le
x xe
e
+w2 N3
le
x xe
+w2e N4
le
x xe
le
(6.72)
The test functions Ni are
68
CHAPTER 6. ENERGY PRINCIPLES
also the surface A of the elastic body will not change due to the virtual displacement,
and, also, that the prescribed boundary traction ti as well as the body forces f
6 Energy principles
Energy principles are another representation of the equilibrium and boundary conditions
of a continuum. They are mostly used for developing numerical methods as, e.g., the
FEM.
6.1
108
A.6
APPENDIX A. SOLUTIONS
Chapter 6
1-D-Beam:
xx = E xx = E u,x
u = z tan z (x)
z
xx = Ez (x)
u
xx = Ezw (x)
tan
z 2 dA
zxx dA = Ew (x)
My =
A
A
:=Iy (moment of inertia)
My = EIy w (x)
Principl
1 Introduction and mathematical
preliminaries
1.1
Vectors and matrices
A vector is a directed line segment. In a cartesian coordinate system it looks like
depicted in gure 1.1,
x3
z
az
a3
P
P
a
a
y
e
8
CHAPTER 1. INTRODUCTION AND MATHEMATICAL PRELIMINARIES
1.4
Coordinate transformation
Assumption:
2 coordinate systems in one origin rotated against each other (g. 1.3).
x3
x3
x2
x2
x1
x1
Figure 1.3:
18
CHAPTER 2. TRACTION, STRESS AND EQUILIBRIUM
At any element An in or on the body (n indicates the orientation of this area) a resultant
force Fn and/or moment Mn produces stress.
Fn
dFn
=
= tn
An 0
2.2. EQUILIBRIUM
23
or
ui dV
(fi + ji,j )dV =
(2.25)
V
V
using the divergence theorem (1.56). The above equation must be valid for every element
in V , i.e., the dynamic equilibrium is fullled. Conse
28
CHAPTER 2. TRACTION, STRESS AND EQUILIBRIUM
In indical notation (I = ij : itendity-matrix (3x3):
kk
sij = ij ij
3
(2.60)
where kk /3 are the components of the hydrostatic stress tensor and sij the
3 Deformation
3.1
Position vector and displacement vector
Consider the undeformed and the deformed conguration of an elastic body at time t = 0
and t = t, respectively (g. 3.1).
deformed
undeformed
x3
38
CHAPTER 3. DEFORMATION
are equal to onehalf of the familiar engineering shear strains ij . However, only with
the denitions above the strain tensor
11 12 13
= 12 22 23
(3.33)
13 23 33
has tensor
ME2112 (Part 1)
Chap 1 - 1
Faculty of Engineering
Department of Mechanical Engineering
ME2112 Strength of Materials (Part I)
Course Lecturer: A/Prof C. Quan
Office: EA 05-28
Tel: 65168089
Email: mpeqc