Bài giảng Strength of materials - Chapter 4: State of Stress and Strength Hypothese - Trần Minh Tú

STRENGTH OF MATERIALS
TRAN MINH TU - University of Civil Engineering,
Giai Phong Str. 55, Hai Ba Trung Dist. Hanoi, Vietnam
1/10/2013 1
4
CHAPTER
1/10/2013
State of Stress and Strength Hypothese
Contents
4.1. State of stress at a point
4.2. Plane Stress
4.3. Mohr’s Circle
4.4. Special cases of plane stress
4.5. Stress – Strain relations
4.6. Strength Hypotheses
1/10/2013 3
4.1. State of stress at a point
1/10/2013 4
K
x
y
z
n


• External loads applied to the body =>
The body is deformed =>The stress is
occurred
• At a point K on the arbitrary section, there
are 2 types of stress: normal stress s and
shearing stress t
• The state of stress at a point K is a set of
all stresses components acting on all
sections, which go through this point
• The most general state of stress at a point
may be represented by 6 components,
, ,
, ,
, , )
normal stresses
shearing stresses
(Note: 
x y z
xy yz zx
xy yx yz zy zx xz
s s s
t t t
t t t t t t 
1/10/2013 5
• Principal planes: no shear stress acts on
4.1. State of stress at a point
• Principal directions: the direction of the principal planes
• Principal stresses: the normal stress act on the principal plane
• There are three principal planes , which are perpendicular to each other
and go through a point
• Three principal stresses: s1, s2, s3 with: s1 ≥ s2 ≥ s3
• Types of state of stress:
- Simple state of stress: 2 of 3 principal
stresses equal to zeros
- Plane state of stress: 1 of 3 principal
stresses equal to zeros
- General state of stress: all 3 principal
stresses differ from zeros
6• Plane Stress – the state of stress in which two
faces of the cubic element are free of stress.
For the illustrated example, the state of stress
is defined by
.0,, and xy zyzxzyx ttstss
• State of plane stress occurs in a thin plate
subjected to the forces acting in the mid-plane
of the plate.
4.2. Plane Stress
sx
txy
sy
x
y
z
tyx x
y
sx
txy
sy
O
tyx
1/10/2013 7
Sign Convention：
• Normal Stress: positive: tension; negative: compression
• Shear Stress: positive: the direction associated with its subscripts are
plus-plus or minus-minus; negative: the directions are plus-minus or
minus-plus
4.2. Plane Stress
y
4.2.1. Complementary shear stresses：
• The shear stresses with the same subscripts
in two orthogonal planes (e.g. txy and tyx)
are equal
1/10/2013 8
0uF 
2
2
cos cos sin
sin sin cos 0
u x xy
y yx
A A A
A A
s s t 
s t 
 su >0 – pull out
 t uv - clockwise
2
uv xy x
2
yx y
τ A- τ Acos α - σ Acosαsinα
+τ Asin α+σ Asinαcosα=0
vF 0 
x
y
sx
txy
sy
O
u
sy
tyx
sx
v
u
A
Acos 
Asin 
su
tuv
txy
4.2. Plane Stress
4.2.2. Stresses on Inclined Planes：
Sign Convention：
 >0 - counterclockwise；
1/10/2013 9
4.2. Plane Stress
tyx
sy
su
tuv
sy
sx
txy
4.2.2. Stresses on Inclined Planes：
x
y
v
u
- > 0: counterclockwise from the x axis to u axis
x y x y
u xycos sin
s s s s
s t 
 2 2
2 2
x y
uv xysin2 cos 2
2
s s
t t 
1/10/2013 10
4.2. Plane Stress
4.2.3. Principal stresses are maximum and minimum stresses ：
By taking the derivative of su to and setting it equal to zero:
xyu
p
x y
2d
0 => tg2 =-
d
ts
 s s
2
2
1,2(3)
2 2 xy
x y x y
max, min
s s s s
s s t
01, 2 90
p
p p
p
1/10/2013 11
4.2. Plane Stress
4.2.4. Maximum in-plane shear stresses
x y 0
s s p
xy
d
0 => tg2 = => = 45
d 2
s st
 t
=> An element subjected to maximum shear stresses will be 450 from
the position of an element that is subjected to the principal stress
2
2
2
s s
t t
xy
x y
max,min
4.2.5. The first invariant of plane stress
The sum of the normal stresses has the same value in each coordinate
system
s s s s x y u v const
1/10/2013 12
4.3. Mohr’s Circle
 - cos2 sin 2
2 2
 sin 2 cos2
2
x y x y
u xy
x y
uv xy
s s s s
s t 
s s
t t 
( )2 ( )2
( )2 ( )2
2 2
2 2 - 
2 2
x y x y
u uv xy
s s s s 
s t t 
,0
2
x y
I
s s 
2
2
2
x y
xyR
s s 
 t 
Using the transformation relations:
Mohr’ Circle
Center: Radius:
1/10/2013 13
O
B AI
s1
sx
sy
s2
M
txy
su
tuv
u
s
x
s
y
s
y
t
xy
t
yx
t
yx
t
xy
s
x
u
su
tuv
tmin
tmax
u1
u2
R
K
`
 01
 02
Điểm cực
Plane Stress
 ,y xyM s t
,0
2
x y
I
s s 
2
2
2
x y
xyR
s s 
 t 
4.3. Mohr’s Circle
1/10/2013 14
4.4. Special Cases of Plane Stress
4.4.1. Uniaxial tension
4.4.2. Pure shear
1/10/2013 15
ss
t
t
tt
s
t
s
t
I
smaxsmin
max,min 3
2
2
1,
2 2
s s 
 t 
s s 
2
21
ax
3
m
2 2
s s s 
 t 
t 
4.4. Special Cases of Plane Stress
4.4.3. Special plane stress
1/10/2013 16
4.5. Stress – Strain relations
1. Uniaxial stress
E
x
x
s
 xy
E
s

 
xz
E
s

 
2. Pure shear
G
xy
xy
t
 0 zxyz 
x
y
z
sx
x
y
z
tx y
1/10/2013 17
4.5. Stress – Strain relations
  
zyx
zyx
x
E
EEE
sss
s

s

s

1
sy
x
y
z
sz
txy
sx
3. General state of stress
- The Principle of superposition
- Assumption: The normal strain
causes only the normal stress. The
Shear strain causes only the shear
stress
1/10/2013 18
a. Normal stress – normal strain relation
b. Shear stress – shear strain relation
with E, , G are Young modulus, Poisson ratio,
shear modulus, which the relation among them:
1
x x y z
E
 s  s s 
1
y y x z
E
 s  s s 
1
z z x y
E
 s  s s 
xy
xy
G
t
 xz
xz
G
t
 
yz
yz
G
t
 
 2 1
E
G

  
1322
1
sss 
E
  
1233
1
sss 
E
  
3211
1
sss 
E
4.5. Stress – Strain relations
Generalized Hooke’s law
1/10/2013 19
1
x x y
E
 s s 
1
y y x
E
 s s 
 1 1 2
1
E
 s s 
 2 2 1
1
E
 s s 
xy
xy
G
t
 
4.5. Stress – Strain relations
- Plane stress
321 aaaV 
)(a)(a)(aV 3322111 111  
321
1  
V
VV
x y z( ) ( )
E E
 
 s s s s s s
 1 2 3
1 2 1 2
4. Normal stress – unit volume change relation
1/10/2013 20
4.5. Stress – Strain relations
5. Strain energy
s2
s3
s1
u u us t s t 
1 1
2 2
Principal element: t = 0 =>
332211
2
1
2
1
2
1
sss u
  
312321
2
3
2
2
2
1 2
2
1
sssssssss 
E
1/10/2013 21
s2
s3
s 1
a
c
s3 -stb
s 1 -stb
s2 -stb
stb
b
stb
stb
=
Deformation=> Change
Shape
Volume
tb ( )s s s s 1 2 3
1
3
4.5. Stress – Strain relations
1/10/2013 22
shape volu u u =
volu
E

s s s
 21 2 3
1 2
( )
6
 shapeu
E

s s s s s s
2 2 2
1 2 2 3 3 1
1
6
Strain-energy density u
Strain-energy density for 
changes of shape ushape
Strain-energy density for 
changes of volume uvol
4.5. Stress – Strain relations
1/10/2013 23
4.6. Strength Hypothesis
- For a bar under tensile loading, one can conclude at which stress
failure will occur from the stress – strain diagram . To prevent such
failure, an allowable stress sallow is introduced and it is postulated that
the stresses in the bar must not exceed sallow, i.e: s ≤ sallow
- In an arbitrary structural member, a spatial stress state is present and
it is necessary to determine the circumstances under which the load
carrying capacity is lost and the material starts to fail.
- There exists no experimental setup which can provide a general
answer, hypothesis on the basic of specific experiments are used.
These so-called strength hypothesis allow us to calculate according to
a specific rule, an equivalent stress se from the normal and shear
stresses. It is assumed that the stress se , when applied to the uniaxial
case of a bar, has the same effect regarding failure through plastic flow
or fracture as the given spatial stress state in the body under
consideration.
1/10/2013 24
4.6. Strength Hypothesis
- Since the stress state in the body and in a tensile bar are then said to
be equivalent, the stress se is called equivalent stress. Therefore, if a
structural element shall not lose its load carrying capacity, the
equivalent stress must not exceed the allowable stress:
e allows s 
1. Maximum – normal – stress hypothesis: It is assumed that the
material starts to fail when the largest principal stress reaches a critical
value. Strength condition:
2. Maximum – normal – strain hypothesis: This hypothesis is based
on the assumption that the failure occurs when the maximum normal
strain reaches a critical value. Strength condition :
1 1e allow s s s
 2 3 1 2e allow s s  s s s
1/10/2013 25
3. Maximum – shear – stress hypothesis: This hypothesis is based
on the assumption that the failure occurs when the maximum shear
stress reaches a critical value. Strength condition :
4.6. Strength Hypothesis
3 1 3e allows s s s 
4. Maximum – distortion – energy hypothesis: Here it is assumed
that the material state becomes critical when the energy needed for the
“distortion” of a material element reaches a critical value.
Strength condition :
2 2 2
4 1 2 3 1 2 1 3 2 3e allows s s s s s s s s s s 
4. Mohr strength hypothesis: Mohr considers that the main factor
causing failure is the maximum shear stress, however, the normal
stress on the section on which the maximum shear stress is has
important effect on failure.
1/10/2013 26
[s y]
su
uv
o
t
[s L]O1O2
Strength condition：
 ss
s
s
ss 31
][
][
y
L
M
O3 s1s3
M
K
L
P
N
4.6. Strength Hypothese
1/10/2013 27
Sample Problem 4.1
From the established sign convention, it is seen that
1/10/2013 28
Sample Problem 4.1
1/10/2013 29
Sample Problem 4.2
The state of plane stress state at a
point is represented by the element in
Fig. Determine the principal stresses
and the principal directions of this state
of plane stress. With β =60o
Solution
210 / ;u kN cms 
u
6KN/cm2
4KN/cm2
10KN/cm2
β
6KN/cm2
4KN/cm2
β
x
y
With coordinates xy shown in fig., We have
 is the angle measured from the x axis
to the normal axis u of the inclined plane
(counterclockwise)
We have:
24 / ;y kN cms 
26 / ;xy kN cmt 
150o 
1/10/2013 30
Sample Problem 4.2
 t 
ssss
s 2sin2cos
22
xy
yxyx
u 
218,928 /x kN cms 
• The Principal directions:
2
2
xy
x y
tg
t
s s
From Equations (*)
u
6KN/cm2
4KN/cm2
β
x
y
1 2 119,4 ; 90 109,4
o o o 
2
2
1,2
2 2
x y x y
xy
s s s s
s t
• The Principal stresses:
2
1 21,041 /kN cms 
2
2 1,887 /kN cms 
1/10/2013 31
Homework
1/10/2013 32
Homework
THANK YOU FOR 
ATTENTION !
1/10/2013 33