Ch 1 Stress Strain 2015 Ivle
description
Transcript of Ch 1 Stress Strain 2015 Ivle
i
Founded 1905
NATIONAL UNIVERSITY OF SINGAPORE
Department of Mechanical Engineering
ME2113
MECHANICS OF
MATERIALS I
Course Lecturer: A/P CJ TAY
ii
Founded 1905
SESSION 2015-16
Semester 1
ME2113 – Mechanics of Materials I Modular Credits: 3
Part I Lecture Notes
A/P CJ TAY
iii
Recommended Books
Basic Text:
A. C. Ugural, Mechanics of Materials, McGraw-Hill, 1993
(Chapter 4, 7, 9 & 10 for part I)
Supplementary Readings:
1. F. P. Beer and E. R. Johnston, Mechanics of Materials, McGraw-Hill,
3rd Ed., 2003.
2. R. C. Hibbeler, Mechanics of Materials, Prentice Hall, 4th Ed., 2000.
3. J. M. Gere and S. P. Timoshenko, Mechanics of Materials, PWS
Publishing Company, 4th ed., 1997.
4. R. R. Craig, Jr., Mechanics of Materials, McGraw-Hill, 2nd ed., 2000.
v
TABLE OF CONTENTS
Chapter 1
ANALYSIS OF STRESS AND STRAIN 1.1 STRESS
1.2 DEFORMATION AND STRAIN
Chapter 2
BENDING OF BEAMS
(SINGULARITY FUNCTIONS AND SIGN CONVENTION)
2.1 SINGULARITY FUNCTIONS
2.2 SIGN CONVENTION
Chapter 3
STRESSES IN LOADED BEAMS 3.1 PURE BENDING
3.2 SHEAR STRESSES IN BEAMS
3.3 RELATIVE MAGNITUDES OF BENDING
AND SHEAR STRESSES
Chapter 4 DEFLECTION OF BEAMS
4.1 STATICALLY DETERMINATE BEAMS
4.2 STATICALLY INDETERMINATE BEAMS
4.3 BEAM DEFLECTION BY MOMENT – AREA
1
Chapter 1
ANALYSIS OF STRESS AND STRAIN
1.1 STRESS
Apply general force dF on dA
Defn:
dA
dF
SheardA
dF
NormaldA
dF
S
dAS
S
dAS
n
dAn
2
02
1
01
0
lim
lim
lim
As dA 0, stress state is at the point P.
Note: Stress values depend on magnitude of dF
and also the direction of dF.
dA
P
S2
S1
n
dF
2
1.1.1 Stress state at a point
For a small isolated element with planes perpendicular to coordinate axes and surrounding a point P, there exist 9 stress components.
They are
3 normal stresses
6 shear stresses
As size of parallelepiped reduces, in the limit, these 9 stress
components will define completely, the state of stress at the
point P.
y
x
z
yy
yx
yz
zx
zy
zz
xx
xy
xz
Stresses shown are
all positive on a
cube of 1 unit
length
P
3
From equilibrium (i.e. taking moment about any
axis), we can show that:
zyyz
zxxz
yxxy
Number of “unknown” stresses reduced to 6.
Take Moment about Z-axis
xy
· 1x1 - yx
· 1x1 = 0
i.e. xy
= yx
The cube is stationary (in equilibrium)
y
x
z
yy
yx
yz
zx
zy
zz
xx
xy
xz
4
1.1.2 Stress Component on Arbitrary plane (2
dimensional case)
z
y
yx x’
x’y’
x
y
xy
x
x
y
For an arbitrary plane whose normal makes an
angle with the horizontal, what are the values
of x’, x’y’ in terms of x, y and xy ?
5
Consider a cut section on the element
Resolving forces in the x-direction
We have 111 BCTABAC xyxx
Divide above Eq. by BC
sincos yxxyxxxBC
AB
BC
ACT
Note: from the geometry of triangle ABC,
sin,cos BC
AB
BC
AC
Ty
B A
C
X’
x
yx
y
Tx
y
x xy
6
Similarly, we can show that cossin xyyyT
Consider the equilibrium of forces in the x’ direction on
plane BC, we have
sincos..
sin1cos11
'
'
yxx
yxx
TTei
BCTBCTBC
Consider the equilibrium of forces in the y’ direction on
plane BC, we have
sincos..
sin1cos11
''
''
xyyx
xyyx
TTei
BCTBCTBC
Substituting for Tx and Ty into the Eqs for 'x and '' yx
and note that yxxy , we have
cossin2sincos 22
' xyyxx
and
cossin)()sin(cos 22'' xyxyyx
y’ can be found by substituting + /2 for in the
expression for x’ , i.e.
cossin2cossin 22
' xyyxy
7
Rewriting,
2sin2cos2
)(
2
)(xy
yxyx
x
(i)
2sin2cos2
)(
2
)(xy
yxyx
y
(ii)
2cos2sin2
)(,, xy
yx
yx
(iii)
Element A
y
y
xy
x x
y’
y'
x’y’ x’ x'
x
Stresses on element A inclined at counter-clockwise to x-axis
8
Note that
When ,
)(
2tan
2
1
)(
22tan..
02cos2sin2
)(
1
yx
xy
yx
xy
xy
yx
or
ei
0,, yx
σx’ which is denoted as σ1 is known as the maximum
principal stress
σy’ which is denoted as σ2 is known as the minimum
principal stress and
θ , which is denoted as α is known as the principal angle and
Eq. a
(a')
x
y
θ which is denoted as α is known as the principal
angle
The angle is positive for counter-clockwise rotation.
(Usually denoted as α)
9
y
x
z
yy
yx
yz
zx
zy
zz
xx
xy
xz
Convention for denoting stress
1. Normal stress, ij
i – indicates the direction of a normal to the plane on
which the stress component acts;
j – indicates the direction of the stress.
Usually denoted by ij , e.g. xx
Sign convention
Stress is positive Tension (e.g. xx)
negative Compression (e.g. -xx)
For simplicity
xx is written as x
yy is written as y
10
2. Shear stress
i – indicates the direction of a normal to the plane on
which the stress component acts;
j – indicates the direction of the stress. y
x
z
yy
yx
yz
zx
zy
zz
xx
xy
xz
e.g. xy
Denoting stresses
ij
11
x
y
x
y
Positive Shear Negative Shear
+ xy - xy
Sign convention
12
2sin2cos2
)(
2
)('xy
yxyxx
(i)
22' 2sin2cos
2
)(
2
)(
xy
yxyxx (a)
2sin2cos2
)(
2
)('xy
yxyxy
(ii)
22' 2sin2cos
2
)(
2
)(
xy
yxyxy (b)
2cos2sin2
)(,, xy
yx
yx
(iii)
22 2cos2sin
2
)(,,
xy
yxyx
(c)
We haveMohr’s Circle for 2-D Stresses
)(2sin2cos2
)(
2
)(22
' bxyyxyx
y
x =
y =
)(2sin2cos2
)(
2
)(22
bxy
yxyx
y
)(2sin2cos2
)(
2
)(22
axy
yxyx
x
13
22' 2sin2cos
2
)(
2
)(
xy
yxyxx (a)
22' 2sin2cos
2
)(
2
)(
xy
yxyxy (b)
22 2cos2sin
2
)(,,
xy
yxyx
(c)
Eqs. a) + c) gives
222''
2' )2
()2
( xyyx
yxyx
x
Eqs. b) + c) gives
222''
2' )2
()2
( xyyx
yxyx
y
(iv)
(v)
22
' 2sin2cos2
)(
2
)(
xy
yxyxy (b)
)(2sin2cos2
)(
2
)(22
axy
yxyx
x
)(2sin2cos2
)(
2
)(22
bxy
yxyx
y
222''
2 )2
()2
( xy
yx
yx
yx
x
222''
2 )2
()2
( xy
yx
yx
yx
y
14
Eqns (iv) and (v) represent a circle in the - plane
with center )0,
2(
yx
2
2
2xy
yxR
and radius R given by:
National University of SingaporeNational University of SingaporeNational University of Singapore
15
)0,2
(yx
2
2
2xy
yxR
National University of SingaporeNational University of SingaporeNational University of Singapore
A
P
l
an
e
B
B
Hence the state of stress on Plane A or Plane B as shown
above can be represented by a point (point A and point
B) on the circumference of a Mohr’s circle
With center at
and radius R
)0,2
(yx
2
2
2xy
yxR
16
Hence the state of stress along the x- and y-plane as
shown above can be represented by a point on the
circumference of a Mohr’s circle
of center
and radius
)0,2
(yx
2
2
2xy
yxR
τxy
σy
Y
Xσxσx
τxy
σy
Plane A
Plane B
Hence the state of stress at any plane on the above element can be
represented by a point on the circumference of a Mohr’s circle:
I.e. the stresses on plane A is represented by a point on the
circumference and the stresses on plane B by another point on the
circumference.
P
la
n
e
B
P
la
n
e
B
17
on plane A or plane B as
The normal stresses at this plane are the maximum
(1) and minimum (
2) principal stresses. From
Mohr’s circle
2
2
122
)(xy
yxyx
2
2
222
)(xy
yxyx
2
2
2xy
yxR
)(
2tan
2
1 1
yx
xy
Eq. a’
Compare with earlier derivations the
angles are similar
2
yx
xy
yx
xy
yx
xy
yx
xy
2tan
2
1
2tan2
2
2
2tan
1
1
18
Y
X
Plane A
Plane B
• On the above element, the stresses on plane A and B
are represented by points A and B on the Mohr ’ s circle
• Point D and E represent the maximum ( 1 ) and
minimum ( 2 ) principal stresses.
• Line CA represents the direction of x - axis
• Line CD represents the direction of
2 • θ represents twice the maximum principal angle
(also the direction of ) of the element x
principal stress the maximum
19
x
''' ,, yxyx
maximum principal angle = α
Plane E
Plane D
20
Stress components at any arbitrary plane can be determined from Mohr’s
circle:
)22cos(' ROCx
2
yxOC
2
2
2xy
yxR
On substituting and simplifying,
(cf. eqn(i))
Similarly,
2sin2cos2
)(
2
)('
xyyxyx
x
2sin2cos2
)(
2
)('xy
yxyxy
and
2cos2sin2
)(,, xy
yx
yx
Note that sense of direction of rotation of axes is same for Mohr’s circle and
element, but rotation for element is half that of Mohr’s circle.
2sin2cos2
)(
2
)(xy
yxyx
x
2sin2cos2
)(
2
)(
,
xy
yxyx
y
Similarly
and
but rotation on Mohr’s circle is twice that of element.
21
Sign convention for stresses when constructing and analyzing
Mohr’s circle:
Shear stresses – if the shear stresses on opposite faces of the
element produce forces that result in a clockwise couple, these
stresses are taken as positive.
on Plane B the shear stress
direction is anticlockwise,
hence it is -ve shear stress
on Mohr’s circle
on Plane A the shear stress
direction is clockwise, hence
it is a +ve shear stress on
Mohr’s circle
22
x
y
x
y
Positive Shear Negative Shear
+xy -xy
IMPORTANT:
Do not confuse Mohr’s circle sign convention with that of an entire element
Sign convention for normal stresses on Mohr’s circle
– Positive is tensile and plotted along positive x-axis
- negative is compressive and plotted along negative x-axis
23
1.1.4 Applications of Mohr’s circle
i) Spherical Pressure Vessel
Consider a spherical pressure vessel with radius
r and wall thickness t subjected to an internal
gage pressure p.
The normal stresses can be related to the
pressure p by inspecting a free body diagram of
the pressure vessel. To simplify the analysis, we
cut the vessel in half as illustrated.
24
From equilibrium, the stress around the wall
must have a net resultant to balance the internal
pressure across the cross-section.
ii) Cylindrical Pressure Vessel
25
Consider a cylindrical pressure vessel with
radius r and wall thickness t subjected to an
internal gage pressure p.
To determine the longitudinal stress, we make a
cut across the cylinder similar to analyzing the
spherical pressure vessel.
26
From equilibrium,
To determine the hoop stress, we make a cut
along the longitudinal axis.
27
From equilibrium, the hoop stress yields,
NOTE:
• The above formulas are good for thin-walled
pressure vessels. ie radius r is larger than 5
times its wall thickness t (r > 5 · t).
• When a pressure vessel is subjected to
external pressure, the stresses are negative
since the wall is now in compression instead
of tension.
• The hoop stress is twice as much as the
longitudinal stress for the cylindrical
pressure vessel. This is why an overcooked
hotdog usually cracks along the longitudinal
direction first (i.e. its skin fails from hoop
stress, generated by internal steam pressure).
28
Example 1
The state of plane stress at a point is represented
by the figure shown. Determine the stresses on
an element oriented at 300 counterclockwise
from the position shown. Illustrate your answer
on a diagram.
6 MPa
y
x
12 MPa
8 MPa
29
From Eqns. (i) to (iii), we have
2sin2cos2
)(
2
)('
xyyxyx
x
(i)
2sin2cos2
)(
2
)('xy
yxyxy
(ii)
2cos2sin2
)(,, xy
yx
yx
(iii)
Substituting θ = 300 , we have
MPax 2.8'
MPay 2.12'
MPayx 66.5,,
To construct the Mohr’s circle,
MPayx
avg 22
128
2
66.116102
2222
xy
yxR
2sin2cos2
)(
2
)(xy
yxyx
x
2sin2cos2
)(
2
)(xy
yxyx
y
x
y
30
σ
Clockwise,
+ve shear
12.2 MPa 2
12.2 8
τ
6 R = 11.66
600
12 8.2
σx
σy
σy’
σx’
τx’y’ = 5.66
31
From the Mohr’s circle,
MPax 2.8'
MPay 2.12'
MPayx 66.5,,
5.66 MPa
y'
x’
12.2 MPa
8.2 MPa
300
300
32
1.2 DEFORMATION AND STRAIN
“Deformation” is a physical phenomenon – it can be
measured.
“Strain” is a mathematical concept.
Basic modes of deformation (displacement)
Rigid body and Non rigid body
Rigid body Translation, Rotation
Non rigid body Elongation, Angular Distortion
O Angular Distortion
A
A'
B B'
P P'
A
B'
A'
B
P
ORotation
OElongation
B
B’
PP’ A A'
O
B
AP
P' A'
B'
Translation
33
1.2.1 Definition of Strain
O
P
Q
Q'
P'
Line element (direct strain)
Engineering strain of line element PQ.
PQ
PQQP
,,
34
Rotation between two line elements (shearing
strain)
Shearing strain ,
When , is small
)tan( ,,
35
(i) Normal Strains
The normal strains are given by:
x
ux
y
vy
z
wz
(i)
where u,v and w represent the displacements in
the x, y and z directions respectively.
(ii) Shear Strains
The shear strains are given by:
y
u
x
vxy
y
w
z
vyz
(ii)
x
w
z
uxz
1.2.2 Strain – Displacement Relationships
36
1.2.3 Assumptions
Convention for Strains
Normal Strains
A
A' B B'
D
D'
C
C'
i – i indicates the direction in
which the elongation or
contraction is required. (the
sides are of an undeformed
element)
Positive Strain for elongation
(e.g. x) Negative Strain for
contraction (e.g. -x)
1. Deformations are infinitesimally small
2. Displacement of a point on the element is continuous, i.e.
no cracks, overlapping, slippage, etc. and also body from
which element is isolated, is continuous throughout.
3. Element is small, i.e. surroundings within close
neighbourhood of point P.
37
Shear Strains
AA' B
B'
C
C'
DD'
x
y
ij – i, j indicate directions of two mutually
perpendicular sides of an undeformed element
whose extent of angular deformation is required
(e.g. xy)
38
1.2.4 Strains at a point
The strains x, y, z, xy, yz and xz are those of a
cube element surrounding the point of interest,
P(x, y, z). If the size of the cube is allowed to
become infinitesimally small, the strains can be
regarded as being the strains at point P(x, y, z)
within a body.
Thus, 6 components of strains are required to
define completely the state of strain at point.
Transformation of Axes.
Often the strain components at a point referred
to a set of axes are different from the original
axes.
For 2- D analysis, if x, y and xy are strains in
x-y plane, what are the equivalent strains x’, y’
and x’y’ referred to x’ and y’ axes that make an
angle with the x-y axes
39
The three strain components referred to x-y axes which
are at an angle to x-y axes are:
2sin2
2cos22
'xyyxyx
x
(d)
2sin2
2cos22
'xyyxyx
y
(e)
2cos2
2sin22
'' xyyxyx
(f)
1.2.5 Principal Strains
There will be a plane in the element which does not
experience any shear strains,
i.e. x’y’=0 .
From eqn(f)
0
2cos2sin''
xyyxyx
40
i.e.
)(2tan
yx
xy
or
)(
tan2
1 1
yx
xy
(g)
For this value of ,
22
'222
)(
xyyxyxx
(h)
22
'222
)(
xyyxyx
y
(i)
'x and 'y are called the maximum and minimum
principal strains respectively.
(Usually denoted by 1 and 2)
The particular angle x’θ
or y’θ
denotes the direction of the principal
axes, and is denoted by . The rotation is defined as positive for
counter-clockwise rotation.
41
1.2.6 Mohr’s Circle of Strain
Re-examining eqns (d), (e) & (f) and compare with eqns
(i), (ii) & (iii) for x’ , y’ & x’y’
The equations are similar in form. Therefore similar to
Mohr’s circle of stress, a Mohr’s circle of strain can
similarly be constructed with center at
0,
2
yx in the
x’ , 2
xy plane. The radius of the circle is
22
22
xyyxR
However for Mohr’s strain circle,
x – axis normal strain
y – axis half shear stain 2
.
42
Mohr’s Circle of Strain
The convention for constructing and reading shear strain values from
Mohr’s strain circle is similar to that used for shear stress in Mohr’s
stress circle.
However for Mohr’s strain circle,
x – axis normal strain
y – axis half shear strain
Sign convention for Mohr’s strain circle
Normal strain: elongation +ve strain
contraction -ve strain
Shear strain: Positive shear stress results in positive shear strain
i.e. A positive shear strain corresponds to a clockwise shear stress
couple.
A negative shear strain corresponds to an anti-clockwise shear stress
couple.
43
Shear strain: Positive shear stress results in positive
shear strain
y
x
y
x
+ve shear stress +ve shear strain
44
45
1.2.7 Stress – Strain Relationships
Stress and strain are related through the
engineering properties of the material of the
body.
Assumptions
All the stresses / strains are within the elastic
range of the material
Material is homogeneous (i.e. properties
uniform throughout)
Material is isotropic (i.e. properties
independent of direction)
Hooke’s law
If only x is applied in the x-direction, the strain
in the x-direction is:
E
xx
(such as in an uniaxial tensile test)
E is the Young’s modulus
The lateral contraction in the y direction = E
x
is the Poisson’s ratio.
46
If only y is applied in the y-direction, the strain
in the y-direction is:
E
yy
The lateral contraction in the x- direction = E
y
If x and y are applied simultaneously, we have
EE
EE
xyy
yxx
For three –dimensions,
)(1
)(1
)(1
yxzz
zxyy
zyxx
E
E
E
(ii)
For shear strains,
GGG
xzxz
yzyz
xyxy
;; (iii)
G – shear modulus of elasticity.
Eqns (ii) and (iii) represent the generalized
Hooke’s law (for isotropic, homogeneous
materials).
(i) (a)
(b)
(a)
(b)
(c)
47
where the directions , r, and z are perpendicular to
each other
where the directions 1, 2, and 3 are perpendicular to
each other
Substituting the values of σy in Eq. (i)b into Eq.
(i)a and express the stress in terms of strains, we
have
Likewise we can obtain σy by substituting the value of σx
in Eq (i)a into Eq (i)b and obtain :
)(1
3211 E
)(1
zrE
yxx
E
21
The Hooke’s law is applicable to any orthogonal stress system
(stresses which are inclined at 900 to each other).
For example we can write the normal strains as
xyy
E
21
48
EXAMPLE 2
The strain components at a point in a machine member are given by
x = 900 μ,
y = -100 μ,
xy = 600 μ .
Using Mohr’s circle, determine the principal strain and the maximum shearing strains.
Centre of circle: (x+
y )/2 = (900 -100)/2 = 400 μ
Radius of circle
5832
600
2
100900
22
2222
xyyxR
49
2
C 1
x
y
2
400
900 , -300
-100 , 300
Max shearing strain
+600μ
Plane B
x = 900 μ
y = -100 μ
xy
=
Plane A
R=583
A
B
x
From the Mohr’s circle
1 = 983 x 10
-6 ,
2 = -183 x 10
-6
Maximum shearing strain γmax
= 1166 x 10
-6
On plane A, x = 900 μ, on plane A the shear
direction is anticlockwise, hence it is indicated as a -ve shear stress on the Mohr’s circle
-(xy
)/2= -600/2 = -300 μ
On plane B, y = -100 μ, on plane B the shear
direction is clockwise, hence it is indicated as a +ve shear stress on the Mohr’s circle
clockwise shear on plane B
(xy
)/2= 600/2 = 300 μ