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Solid Mechanics
1. Shear force and bending moment diagrams
Internal Forces in solids
Sign conventions
• Shear forces are given a special symbol on yV 12
and zV
• The couple moment along the axis of the member is given
xM T= = Torque
y zM M= =bending moment.
Solid Mechanics
We need to follow a systematic sign convention for systematic development of equations and reproducibility of the equations
The sign convention is like this.
If a face (i.e. formed by the cutting plane) is +ve if its outward normal unit vector points towards any of the positive coordinate directions otherwise it is –ve face
• A force component on a +ve face is +ve if it is directed towards any of the +ve coordinate axis direction. A force component on a –ve face is +ve if it is directed towards any of the –ve coordinate axis direction. Otherwise it is –v.
Thus sign conventions depend on the choice of coordinate axes.
Shear force and bending moment diagrams of beams Beam is one of the most important structural components.
• Beams are usually long, straight, prismatic members and always subjected forces perpendicular to the axis of the beam
Two observations:
(1) Forces are coplanar
Solid Mechanics
(2) All forces are applied at the axis of the beam.
Application of method of sections
What are the necessary internal forces to keep the segment of the beam in equilibrium?
x
y
z
F PF V
F M
� = �
� = �
� = �
00
0
• The shear for a diagram (SFD) and bending moment diagram(BMD) of a beam shows the variation of shear
Solid Mechanics
force and bending moment along the length of the beam.
These diagrams are extremely useful while designing the beams for various applications.
Supports and various types of beams
(a) Roller Support – resists vertical forces only
(b) Hinge support or pin connection – resists horizontal and vertical forces
Hinge and roller supports are called as simple supports
(c) Fixed support or built-in end
Solid Mechanics
The distance between two supports is known as “span”.
Types of beams
Beams are classified based on the type of supports.
(1) Simply supported beam: A beam with two simple supports
(2) Cantilever beam: Beam fixed at one end and free at other
(3) Overhanging beam
(4) Continuous beam: More than two supports
Solid Mechanics
Differential equations of equilibrium
[ ]xFΣ = → +0
yFΣ� �= ↑ +� �0
V V V P xV P xV Px
∆ ∆∆ ∆∆∆
+ − + == −
= −
0
x
V dVP
x dxlim∆
∆∆→
= = −0
[ ]AP xM V x M M M ∆Σ ∆ ∆= − + + − =
20 0
2
P xV x M
M P xVx
∆∆ ∆
∆ ∆∆
+ − =
+ − =
20
2
02
Solid Mechanics
x
M dM Vx dxlim
∆
∆∆→
= = −0
From equation dV Pdx
= − we can write
D
C
X
D CX
V V Pdx− = − �
From equation dM Vdx
= −
D CM M Vdx− = −�
Special cases:
Solid Mechanics
Solid Mechanics
Solid Mechanics
Solid Mechanics
( ) ( )x≤ ≤ −0 2 1 1
A B
VVV ; V
− ==
= =
5 05
5 5
( ) ( )( )( )
( )B C
x
V . x
V . xV ; V
. xx .
≤ ≤ −− + − − == − + −= − =
− + − =� =
2 6 2 2
5 30 7 5 2 0
5 30 7 5 225 5
25 7 5 2 05 33
( ) ( )
C D
xVVV ; V
≤ ≤ −− + − − == += + = +
6 8 3 35 30 30 10 0
1515 15
( ) ( )
D E
xVVVV ; V
≤ ≤ −− + − − + =+ == −
= − = −
8 10 4 45 30 30 10 20 05 0
55 5
x ( ) ( )x ( ( )x ( ) ( )x ( ) ( )
≤ ≤ − −≤ ≤ − −≤ ≤ − −≤ ≤ − −
0 2 1 12 6 2 26 8 3 38 10 4 4
Solid Mechanics
Problems to show that jumps because of concentrated force and concentrated moment
( ) ( )
A B
xM xM xM ; M
≤ ≤ − −− + == − +
= + =
0 2 1 110 5 0
5 1010 0
( ) ( )
( ) ( )
( ) ( )
E x .
C x
x
. xM x x
. xM x x
M .
M=
=
≤ ≤ − −
−− + − − + =
−= − + − −
= +
=
2
2
5 33
6
2 6 2 2
7 5 210 5 30 2 0
2
7 5 210 5 30 2
241 66
40
( ) ( ) [ ]( ) ( ) ( )
C x
D x
x C D
M x x x x
M
M=
=
≤ ≤ − − −− + − − + − + − + =
= +
= −6
8
6 8 3 3
10 5 30 2 30 4 10 6 20 0
20
10
[ ] ( ) ( )( ) ( ) ( ) ( )
E x
x D E
M x x x x x
M =
≤ ≤ − −− + − − + − + − + − − =
=8
8 10 4 4
10 5 30 2 30 4 10 6 20 20 8 0
0
Solid Mechanics
We can also demonstrate internal forces at a given section using above examples. This should be carried first before drawing SFD and BMD.
[ ]x A B≤ ≤ −0 2
Solid Mechanics
A
B
VVVV
− ==
==
5 05
55
A B
M xM xM ; M
− + == −
= =
10 5 010 5
10 0
[ ]x B C≤ ≤ −2 6
( )( )
( )B C
V . x
V . xV ; V
. xx .
− + − − == − + −= − =
− + − ==
5 30 7 5 2 0
7 5 2 5 3025 5
25 7 5 2 05 33
( ) ( )
C
E
B
xM x x .
xM
M x . .xM
−− + − − + =
==
= ==
=
2210 5 30 2 7 5 0
26
40
5 33 41 662
0[ ]x C D≤ ≤ −6 8
C D
VVV , V
− + − − === =
5 30 10 30 01515 15
Solid Mechanics
[ ]x D E≤ ≤ −8 10
D E
VVV , V
− + − − + == −
= − = −
5 30 10 30 20 05
5 5
Solid Mechanics
[ ]
[ ]
x Ax
y Ay
Ay
F R
F R
R kN
M M .M k m
∆
� → + = � =
� �� ↑ + = � + − =� �
= ↑
� = � + − × == −
0 0
0 60 90 0
30
0 60 90 4 5 0285
( )( )
V x
V x
+ + − − == − −= × −= −=
30 60 30 3 0
30 3 9030 3 9090 900
( )B A
B A
M MM M
− = − −= + = −
= −
6060 60 285
225
Solid Mechanics
( )C B
C B
M MM M
− = − −= + = − +
= −
9090 225 90
135
( )D C
D C
M MM M
− = − −= + = − + =
135135 135 135 0
y
Ay Cy
Ay Cy
F
R R
R R ( )
� �� ↑ + =� �
+ − − =
+ =
0
200 240 0
440 1
[ ]A
Cy
Cy
Ay
MR
R kN
R kN
� =− × − × + × =
= ↑
= ↑
0
200 3 240 4 8 0
195
245
V xV xVV
+ − − == −= × − = −=
245 200 30 030 4530 8 45 240 45195
Solid Mechanics
*
M .M .M
− × + ×= × − ×=
245 3 90 1 5245 3 90 1 5600
[ ]Ay By
A By
By
By
Ay
R R
M R
R
R kN
R kN
+ =
� = − × + + + =
− + + =
=
=
32
0 32 2 18 8 4 0
64 16 4 0
12
20
Solid Mechanics
Problem:
[ ]
( )
x
Ax
y Ay Dy Ay Dy
FR
F R R R R
� → + ==
� �� = ↑ + + − − = � + =� �
0
0
0 60 50 0 110 1
( )C A
C A
M MM M
− = − −= + = − + =
5050 8 25 17
V xV x
xx / .
+ − == −− =
= =
20 8 08 20
8 20 020 8 2 5
[ ]A Dy
Dy
Ay
M . R
R kN
R kN
� = − × − × + × =
= = ↑
= ↑
0 60 1 5 50 4 5 0
29058
552
Solid Mechanics
( )y
B
F V x
V x x m
� �� = ↑ + + − =� �
� = − ≤ ≤ �
�
0 52 20 0
20 52 0 3
[ ]
( )
M
xM x
xM x x m
� =
+ − =
= − ≤ ≤
2
2
0
2052 0
220
52 0 32
y
B C
F
V
V kN x m
� �� = ↑ +� �
+ − =
� = ↑ ≤ ≤ ��
0
52 60 0
8 3 4
[ ] ( )
( )B C
M M x x .
M x x . x m
� = − + − =
� = − − ≤ ≤ ��
0 52 60 1 5 0
52 60 1 5 3 4
Solid Mechanics
B E
B
M M .M . .
− = −= − +
1 61 6 67 6
x / . m× − =
= =20 52 0
52 20 2 6
dM VdxdV Pdx
= −
= −
[ ] ( ) ( )( ) ( ) ( )
M M x x . x
M x x . x x
� = − + − + − == − − − − ≤ ≤
0 52 60 1 5 50 4 0
52 60 1 5 50 4 4 5
( )
yF
V
V kN x
� �� = ↑ +� �
+ − − == ≤ ≤
0
52 60 50 0
58 4 5
Solid Mechanics
D C
D C
M MM M
− = −= +
= − =
5858
58 58 0
C B
C B
M MM M
− = −= − +
= − + =
88
8 66 58
B E
B E
M M .M . M . .
− = −= − + = − +
=
1 61 6 1 6 67 6
66
x / .× − =
= =20 52 0
52 20 2 6
dM VdxdV Pdx
= −
= −
B AM M Vdx− = −�
Solid Mechanics
2. Concept of stress Traction vector or Stress vector
Now we define a quantity known as “stress vector” or “traction” as
∆
∆∆→
=�
� Rn
A
FTAlim
0 units aP N / m− 2
and we assume that the quantity
∆
∆∆→
→�
R
A
MAlim
00
(1) nT�
is a vector quantity having direction of RF∆�
(2) nT�
represent intensity point distributed force at the point "P" on a plane whose normal is n̂
(3) nT�
acts in the same direction as RF∆�
Solid Mechanics
(4) There are two reasons are available for justification of the
assumption that ∆
∆∆→
→�
R
A
MAlim
00
(a) experimental (b) as A∆ → 0, RF∆
� becomes resultant of a parallel
force distribution. Therefore RM∆ = 0�
for � force system.
(5) nT�
varies from point to point on a given plane
(6) nT�
at the same point is different for different planes.
(7) n nT T′ = −� �
will act at the point P
(8) In general
Components of nT�
R n t sˆˆ ˆF F n v t v s∆ ∆ ∆ ∆= + +�����
Solid Mechanics
∆ ∆ ∆ ∆
∆ ∆ ∆ ∆∆ ∆ ∆ ∆→ → → →
= = + +�
� R n t sn
A A A A
F F v vˆˆ ˆT n t sA A A Alim lim lim lim
0 0 0 0
n nn nt nsˆˆ ˆT n t sσ τ τ= + +�
where
∆
∆
∆
∆σ∆
∆τ∆∆τ∆
→
→
→
= = =
= = =
= = =
n nnn
A
t tnt
A
s sns
A
F dF Normal stresscomponentA dA
v dv Shear stresscomponentA dAv dv Another shear componetA dA
lim
lim
lim
0
0
0
στ
−−
NormalStressShear stress
n nndF dAσ=
t ntdV dAτ=
Notation of stress components
The magnitude and direction of nT�
clearly depends on the plane m-m. Therefore, stress components magnitude & direction depends on orientation of cut m-m.
(a) First subscript- plane on which σ is acting (b) Second subscript- direction
Solid Mechanics
Rectangular components of stress
Cuts ⊥ to the coordinate planes will give more valuable information than arbitrary cuts.
∆ ∆ ∆ ∆
∆∆ ∆ ∆∆ ∆ ∆ ∆→ → → →
= = + +�
� yR x zx
A A A A
vF F v ˆˆ ˆT i j kA A A Alim lim lim lim
0 0 0 0
x xx xy xzˆˆ ˆT i j kσ τ τ= + +
�
where
xxx
A
y zxy xz
A A
F NormalstressA
v vShear stress; Shear stressA A
lim
lim lim
∆
∆ ∆
∆σ∆
∆ ∆τ τ∆ ∆
→
→ →
= =
= = = =
0
0 0
Solid Mechanics
σ=x xxdF dA
y xydv dAτ=
z xzdv dAτ=
Similarly,
∆ ∆ ∆ ∆
∆∆ ∆ ∆∆ ∆ ∆ ∆→ → → →
= = + +� yR x zy
A A A A
FF v v ˆˆ ˆT i j kA A A Alim lim lim lim
0 0 0 0
τ σ τ= + +�y yx yy yz
ˆˆ ˆT i j k
τ τ σ= + +�z zx zy zz
ˆˆ ˆT i j k
xxσ and xyτ will act only on x-plane. We can see xσ and xyτ
only when we take section ⊥ to x-axis.
The stress tensor
σ τ τσ τ σ τ
τ τ σ
� �� �
� �= � �� �� �� �� �
xx xy xz
jj yx yy yz
zx zy zz
Rec tan gular stresscomponents
• This array of 9 components is called as stress tensor.
• It is a second rank of tensor � because of two indices
Components a point “P” on the x-plane in x,y,z directions
Solid Mechanics
• These 9 rectangular stress components are obtained by taking 3 mutually ⊥planes passing through the point “P”
• ∴ Stress tensor is an array consisting of stress components acting on three mutually perpendicular planes.
τ τ τ= + +
�n nx ny nz
ˆˆ ˆT i j k
What is the difference between distributed loading & stress?
RA
Fq limA∆
∆∆→
=0
yyq σ= can also be called.
No difference!
Except for their origin!
Solid Mechanics
Sign convention of stress components.
A positive components acts on a +ve face in a +ve coordinate direction
or
A positive component acts on a negative face in a negative coordinate direction.
Say x xy a;Pa Pσ τ= − = −20 10 and xz Paτ = 30 at a point P
means.
Solid Mechanics
State of stress at a point
The totality of all the stress vectors acting on every possible plane passing through the point is defined to be state of stress at a point.
• State of stress at a point is important for the designer in determining the critical planes and the respective critical stresses.
• If the stress vectors [and hence the component] acting on any three mutually perpendicular planes passing through the point are known, we can determine the stress vector nT
� acting on any plane “n” through that
point.
The stress tensor will specify the state stress at point.
x x x y x z
ij y x y y y z
z x z y z z
σ τ τσ τ σ τ
τ τ σ
′ ′ ′ ′ ′ ′
′ ′ ′ ′ ′ ′ ′
′ ′ ′ ′ ′ ′
� �� �
� �= � �� �� �� �� �
can also represent state of stress at a point.
Solid Mechanics
The stress element
Is there any convenient way to visualize or represent the state of stress at a point or stresses acting three mutually perpendicular planes say x- plane , y-plane and z-plane.
xx xy xz
ij yx yy yzP
zx zy zz
σ τ τσ τ σ τ
τ τ σ
� �+ + +� �
� � = + + +� �� �� �+ + +� �� �
( )( )
xx xx
yy yy
x,y ,zContinuous functionsof x,y ,z
x,y ,z
σ σσ σ
= ���= ��
Let us consider a stress tensor or state of stress at a point in a component as
Solid Mechanics
ijσ− −� �� �� �= −� � � �− − −� �� �
10 5 305 50 6030 60 100
Equilibrium of stress element
[ ]xF� = → +0
x yx zx x yx zxdydz dxdz dydx dydz dxdz dxdyσ τ τ σ τ τ+ + − − − = 0
Similarly, we can show that yF� = 0 and zF� = 0 is satisfied.
y
dz
dy
zdx
x
xyτ
xzτ xσ
Solid Mechanics
PzM
C.C.W ve
� =� �� �+� �
0
( ) ( )xy yxdydz dx dxdz dyτ τ− = 0
xy yxτ τ− = 0
xy yxτ τ=
Shearing stresses on any two mutually perpendicular planes are equal.
PxM� �� = �� �0 yz zyτ τ= and PyM� �� = �� �0 zx xzτ τ=
Cross-shears are equal- a very important result
Since xy yxτ τ= , if xy veτ = − yxτ is also –ve
Solid Mechanics
∴The stress tensor
xx xy xz
ij yx xy xy yz
zx xz zy yz yz
is sec ondrank symmetric tensor
σ τ τσ τ τ σ τ
τ τ τ τ σ
� �� �
� �= =� �� �� �= =� �� �
Differential equations of equilibrium
[ ]xF� → + = 0
yxx zxx yx zx
x xy zx x
x y z y x z z y xx y z
y z x z y x B x y z
τσ τσ τ τ
σ τ τ
∂� ∂ ∂� � + ∆ ∆ ∆ + + ∆ ∆ ∆ + + ∆ ∆ ∆ � � �∂ ∂ ∂� � �
− ∆ ∆ − ∆ ∆ − ∆ ∆ + ∆ ∆ ∆ = 0
yxx zxxx y z y x z x y z B x y z
x y z
τσ τ∂∂ ∆ ∆ ∆ + ∆ ∆ ∆ + � ∆ ∆ + ∆ ∆ ∆ =∂ ∂ ∂
20
Canceling x y∆ ∆ and z∆ terms and taking limit
yxx zxx
xyz
lim Bx y z
τσ τ∆ →∆ →∆ →
∂∂ ∂+ + + =∂ ∂ ∂0
00
0
Similarly we can easily show that
Solid Mechanics
[ ]yxx zxx xB F
x y z
τσ τ∂∂ ∂+ + + = � =∂ ∂ ∂
0 0
xy yy zyy yB F
x y z
τ σ τ∂ ∂ ∂� �+ + + = � =� �∂ ∂ ∂
0 0
[ ]yzxz zzz zB F
x y z
ττ σ∂∂ ∂+ + + = � =∂ ∂ ∂
0 0
• If a body is under equilibrium, then the stress components must satisfy the above equations and must vary as above.
For equilibrium, the moments of forces about x, y and z axis at any point must vanish.
pzM� �� =� �
0
xy yxxy xy yx
yx
yx xx y z y z y x zx y
yx z
τ ττ τ τ
τ
∂ ∂� � ∆∆+ ∆ ∆ ∆ + ∆ ∆ − + ∆ ∆ ∆ � �∂ ∂� �
∆− ∆ ∆ =
2 2 2
02
�
Solid Mechanics
xy xy yx yx
xy yxxy yx
y x z x y zx y z x y zx y
yxx y
τ τ τ τ
τ ττ τ
∆ ∆ ∆ ∂ ∆ ∆ ∆ ∂∆ ∆ ∆ ∆ ∆ ∆+ − − =∂ ∂
∂ ∂ ∆∆+ − − =∂ ∂
2 22 20
2 2 2 2
02 2
Taking limit
xy yxxy yx
xyz
yxlimx y
τ ττ τ
∆ →∆ →∆ →
∂ ∂ ∆∆+ − − =∂ ∂0
00
02 2
xy yxτ τ� − = �0 xy yxτ τ=
Relations between stress components and internal force resultants
Solid Mechanics
x xxA
F dAσ= � ; y xyA
V dAτ= � ; z xzA
V dAτ= �
xz xy xy dA dAz dMτ τ− =
( )x xz xyA
M y z dAτ τ= −�
y xzA
M dAσ= � ; z xyA
M dAσ= − �
Solid Mechanics
3. Plane stress and Plane strain Plane stress- 2D State of stress
If ( ) ( )( ) ( )
x xyij
xy yy
x,y x,yplane stress-is a --- state of stress
x,y x,y
σ τσ
τ σ� �
� �= −� �� �� �� �
All stress components are in the plane x y− i.e all stress
components can be viewed in x y− plane.
xy
x xyx xy
ij xy yyx y
D Stateof stress
Stresscomponents in plane xy
τ
σ τ σ τσ τ σ τ σ=
−
� �� �� �
� �= = � �� �� �� �� �� �
� �
2
0
0
0 0 0
x xy xz
ij yx yy yz
zx zy zz
D Stateof stress
components
σ τ τσ τ σ τ
τ τ σ
−
� �� �
� �= −� �� �� �� �� �
3
6
Solid Mechanics
This type of stress-state (i.e plane stress) exists in bodies whose z - direction dimension is very small w.r.t other dimensions.
Stress transformation laws for plane stress
The state of stress at a point P in 2D-plane stress problems are represented by
x xy nn ntij
xy y nt tt
σ τ σ τσ
τ σ τ σ� � � �
� �= =� � � �� �� �� �� �
Solid Mechanics
* We can determine the stress components on any plane “n” by knowing the stress components on any two mutually ⊥ planes.
Stress transformation laws for plane stress
In order to get useful information we take different cutting planes passing through a point. In contrast to 3D problem, all cutting planes in plane stress problems are parallel to x-
Solid Mechanics
axis. i.e we take different cutting plane by rotating about z- axis.
As in case of 3D, the state of stress at a point in a plane stress domain is the totality of all the stress. If we know the stress components on any two mutually ⊥ planes then stress components on any arbitrary plane m-m can be determined. Thus the stress tensor
x xyij
xy y
σ τσ
τ σ� �
� �= � �� �� �� �
is sufficient to tell about the state of stress
at a point in the plane stress problems.
dA Area of ABdACs Areaof BCdASin Area of AC
θθ
===
nF� + =� �� �0�
nn x xy xy
yy
dA dACos Cos dACos Sin dASin Cos
dASin Sin
σ σ θ θ τ θ θ τ θ θσ θ θ
− − − −
= 0
nn x xy yyCos Sin Cos Sinσ σ θ τ θ θ σ θ− − − =2 22 0
Solid Mechanics
nn x y xy
x y x ynn xy
Cos Sin Sin Cos
Cos Sin
σ σ θ σ θ τ θ θσ σ σ σ
σ θ τ θ
= + +
+ −= + +
2 2 2
2 22 2
nF� + =� �� �0�
nt x xy xy
y
dA dACos Sin dACos Cos dASin Sin
dASin Cos
σ σ θ θ τ θ θ τ θ θσ θ θ
− − + −
= 0
( )nt x y xyCos Sin Sin Cos Cos Sinτ σ θ θ σ θ θ τ θ θ= − + + −2 2
( ) ( )( )
nt x y xy
x ynt xy
Cos Sin Cos Sin
Sin Cos
τ θ θ σ σ τ θ θ
σ στ θ τ θ
= − − + −
−= − +
2 2
2 22
We shall now show that if you know the stress components on two mutually ⊥ planes then we can compute stresses on any inclined plane. Let us assume that we know that state of stress at a point P is given
x xyij
xy y
σ τσ
τ σ� �
� �= � �� �� �� �
This also means that
Solid Mechanics
Solid Mechanics
If θ θ= we can compute on AB
If πθ θ= +2
we can compute on BC
If θ θ π= + we can compute on CD
If πθ θ= + 32
we can compute on DA
• nnσ and ntτ equations are known as transformation laws for plane stress.
• They are not only useful in determination of stresses on any plane but also useful in transforming stresses from one coordinate system to another
• Transformation laws do not require an equilibrium state and thus are also valid at all points of the body under accelerations.
• These laws are true for any point P of a body.
Invariants of stress tensor
• Any quantity for which its 2D scalar components transform from one coordinate system to another according to nnσ and ntτ is called a two dimensional
Solid Mechanics
symmetric tensor of rank 2. Here in particular the tensor is a stress tensor.
• Moment of inertia if x xx y yy xy xyI , I ; Iσ σ τ= = = −
• By definition a tensor is a mathematical quantity that transforms according to certain laws, such that certain invariant properties are maintained for all coordinate systems.
• Tensors, as governed by their transformation laws, possess several properties. We now develop those properties for 2D second vent symmetric tensor.
x y x ynn xyCos Sin
σ σ σ σσ θ τ θ
+ −= + +2 2
2 2
x y x yt xyCos Sin
σ σ σ σσ θ τ θ
+ −= + −2 2
2 2
x ynt xySin Cos
σ στ θ τ θ
−= − +2 2
2
Solid Mechanics
n t x y x y Iσ σ σ σ σ σ′ ′+ = + = + = 1
I =1 First invariant of stress in 2D
n t nt x y xy x y x y Iσ σ τ σ σ τ σ σ τ′ ′ ′ ′− = − = − =2 22
I =2 Second invariant of stress in 2D
• I ,I1 2 are invariants of 2D symmetric stress tensor at a point.
• Invariants are extremely useful in checking the correctness of transformation
• Of I1 and I2 , I1 is the most important property : the sum of normal stresses on any two mutually ⊥ planes (⊥directions) is a constant at a given point.
• In 2D we have two stress invariants; in 3D we have three invariants of stresses.
Solid Mechanics
Solid Mechanics
Problem:
A plane-stress condition exists at a point on the surface of a loaded structure, where the stresses have the magnitudes and directions shown on the stress element. (a) Determine
the stresses acting on a plane that is oriented at a −15� w.r.t. the x-axis (b) Determine the stresses acting on an element
that is oriented at a clockwise angle of 15� w.r.t the original element.
Solution:
� it is in C.W.
x
y
xy
Q
σστ
= −=
= −
= −
4612
19
15�
Solid Mechanics
Substituting θ = −15� in ntτ equation
x y MPasσ σ+ − + −= = = −46 12 34
172 2 2
( ) ( )Sin Sin . ; Cos Cos .θ θ= − = − = − =2 2 15 0 5 2 2 15 0 866
x y x yn xyCos Sin
σ σ σ σσ θ τ θ
+ −� �= + +� �
� �2 2
2 2
n . .σ = − − × + ×17 29 0 866 19 0 5
n . MPasσ = −1
32 6
x ynt xySin Cos
σ στ θ τ θ
−� �= − +� �
� �2 2
2
n t MPaτ = −1 1
31
x y MPaσ σ− − − −= = = −46 12 58
292 2 2
n t . .τ = − × − ×1 1
29 0 5 19 0 866
Solid Mechanics
Now
As a check
t n nt θσ σ τ =� �= =� �2 75�
n Cos SinMPa
σ = − − × − ×= −
17 29 2 165 19 2 16532
nt
nt
. Sin CosMPa
ττ
= −= −
00 29 330 19 33031
n t x y . . MPa sσ σ σ σ+ = + = − − = − = − +32 6 1 4 34 46 12
θ = 145�
tn Sin CosMPa
τ = + × − ×=
29 150 19 15031
t cos sinσ∴ = − − −17 29 150 19 150
t . MPaσ = −1 4
tn n t nt θτ τ τ =� �= =� �2 2 75�
Solid Mechanics
4. Principal Stresses Principal Stresses
Now we are in position to compute the direction and magnitude of the stress components on any inclined plane at any point, provided if we know the state of stress (Plane stress) at that point. We also know that any engineering component fails when the internal forces or stresses reach a particular value of all the stress components on all of the infinite number of planes only stress components on some particular planes are important for solving our basic question i.e under the action of given loading whether the component will ail or not? Therefore our objective of this class is to determine these plane and their corresponding stresses.
(1) ( ) n y n yn n xyCos Sin
σ σ σ σσ σ θ θ τ θ
+ −= = + +2 2
2 2
(2) Of all the infinite number of normal stresses at a point, what is the maximum normal stress value, what is the minimum normal stress value and what are their
Solid Mechanics
corresponding planes i.e how the planes are oriented ? Thus mathematically we are looking for maxima and minima of
( )n Qσ function..
(3) n y n yn xyCos Sin
σ σ σ σσ θ τ θ
+ −= + +2 2
2 2
For maxima or minima, we know that
( )nx y xy
d Sin Cosdσ σ σ θ τ θθ
= = − − +0 2 2 2
xy
x ytan
τθ
σ σ=
−2
2
(4) The above equations has two roots, because tan repeats itself after π . Let us call the first root as Pθ
1
xyP
x ytan
τθ
σ σ=
−1
22
( ) xyP P
x ytan tan
τθ θ π
σ σ= + =
−2 1
22 2
Solid Mechanics
P P sπθ θ= +
2 1 2
(5) Let us verify now whether we have minima or minima at
Pθ1
and Pθ2
( )
( )P
nx y xy
nx y P xy P
d Cos Sind
d Cos Sind θ θ
σ σ σ θ τ θθσ σ σ θ τ θθ =
= − − −
∴ = − − −1 1
1
2
2
2
2
2 2 4 2
2 2 4 2
We can find PCos sθ1
2 and PSin sθ1
2 as
x yP
x yxy
Cosσ σ
θσ σ
τ
−=
−� + �
�
1 22
2
22
xy xyP
x y x yxy xy
Sinτ τ
θσ σ σ σ
τ τ
= =− −� �
+ + � �� �
1 2 22 2
22
22 2
Substituting PCos θ1
2 and PSin θ1
2
Solid Mechanics
( )( )
( )
P
x y x y xy xyn
x y x yxy xy
x y xy
x y x yxy xy
x yxy
x yxy
dd θ θ
σ σ σ σ τ τσθ σ σ σ σ
τ τ
σ σ τ
σ σ σ στ τ
σ στ
σ στ
=
− − −= −
− −� � + + � �
� �
− −= −
− −� � + + � �
� �
� �−� − � �= + �� �� −� � �+ ��
1
2
2 2 22 2
2 2
2 22 2
22
22
2 4
22 2
4
2 2
42
2
x ynxy
dd
σ σσ τθ
−� ∴ = − + �
�
222
2 42
(-ve)
( ) ( ) ( )
( )
P P
nx y P xy P
x y P xy P
d Cos Sind
Cos Sin
πθ θ θ
σ σ σ θ π τ θ πθ
σ σ θ τ θ
= = +
= − + − +
= − +
1 1
2 1
1 1
2
2
2
2 2 4 2
2 2 4 2
Substituting P PCos &Sinθ θ1 1
2 2 m we can show that
P
x ynxy
d sd θ θ
σ σσ τθ =
−� ∴ = − + �
� 2
222
2 42
(+ve)
Solid Mechanics
Thus the angles P sθ1
and P sθ2
define planes of either
maximum normal stress or minimum normal stress.
(6) Now, we need to compute magnitudes of these stresses
We know that,
P
x y x yn xy
x y x yn P xy P
Cos Sin
Cos Sinθ θ
σ σ σ σσ θ τ θ
σ σ σ σσ σ θ τ θ=
+ −= + +
+ −= = + +
1 111
2 22 2
2 22 2
Substituting PCos sθ1
2 and PSin θ1
2
x y x yxy
Max.Normalstress becauseof sign
σ σ σ σσ τ
+ −� = + + �
�
+
22
1 2 2
Similarly,
( )( )P P
x y x yn P
xy P
x y x yP xy P
Cos
Sin
Cos Sin
πθ θ θσ σ σ σ
σ σ θ π
τ θ π
σ σ σ σθ τ θ
= = =+ −
= = + + +
+
+ −= − −
12 1
1
1 1
22
22 2
2
2 22 2
Substituting PCos θ1
2 and PSin θ1
2
Solid Mechanics
x y x yxy
Min.normalsressbecauseof vesign
σ σ σ σσ τ
+ −� = − + �
�
−
22
2 2
We can write
x y x yxyor
σ σ σ σσ σ τ
+ −� = ± + �
�
22
1 2 2 2
(7) Let us se the properties of above stress.
(1) P P sπθ θ= +2 1 2
- planes on which maximum normal stress
and minimum normal stress act are ⊥ to each other.
(2) Generally maximum normal stress is designated by σ1 and minimum stress by σ2 . Also P P;θ σ θ σ→ →
1 21 2
alg ebraically i.e.,σ σσσ
>−
− −
1 2
1
2
01000
Solid Mechanics
(4) maximum and minimum normal stresses are collectively called as principal stresses.
(5) Planes on which maximum and minimum normal stress act are known as principal planes.
(6) Pθ1
and Pθ2
that define the principal planes are known as
principal directions.
(8) Let us find the planes on which shearing stresses are zero.
( )nt x y xySin Cosτ σ σ θ τ θ= = − − +0 2 2
xy
x ytan
directionsof principal plans
τθ
σ σ=
=
=
22
Thus on the principal planes no shearing stresses act. Conversely, the planes on which no shearing stress acts are known as principal planes and the corresponding normal stresses are principal stresses. For example the state of stress at a point is as shown.
Then xσ and yσ are
principal stresses because no shearing stresses are acting on these planes.
Solid Mechanics
(9) Since, principal planes are ⊥ to each other at a point P, this also means that if an element whose sides are parallel to the principal planes is taken out at that point P, then it will be subjected to principal stresses. Observe that no shearing stresses are acting on the four faces, because shearing stresses must be zero on principal planes.
(10) Since 1σ and 2σ are in two ⊥ directions, we can easily say that
x y x y Iσ σ σ σ σ σ′ ′+ = + = + =1 2 1
Solid Mechanics
5. Maximum shear stress Maximum and minimum shearing stresses
So far we have seen some specials planes on which the shearing stresses are always zero and the corresponding normal stresses are principal stresses. Now we wish to find what are maximum shearing stress plane and minimum shearing stress plane. We approach in the similar way of maximum and minimum normal stresses
(1) x ynt xySin Cos
σ στ θ τ θ
−� = − + �
� 2 2
2
( )ntx y xy
d Cos Cosdτ σ σ θ τ θθ
= − − +2 2
For maximum or minimum
( )ntx y xy
d Cos Sindτ σ σ θ τ θθ
= = − − −0 2 2 2
( )x y
xytan
σ σθ
τ− −
� =22
This has two roots
( )x yS
xytan
s stands for shear stressp stands for principalstresses.
σ σθ
τ−
= −
−−
12
2
Solid Mechanics
( ) ( )x yS S
xytan tan
σ σθ θ π
τ− −
= + =2 1
2 22
S Sπθ θ∴ = +
2 1 2
Now we have to show that at these two angles we will have maximum and minimum shear stresses at that point.
Similar to the principal stresses we must calculate
( )
( )S
ntx y xy
ntx y S xy S
d Sin Cosd
d Sin Cosd θ θ
τ σ σ θ τ θθ
τ σ σ θ τ θθ =
= − −
= − −1 1
1
2
2
2
2
2 2 4 2
2 2 4 2
xyS
x yxy
Cosτ
θσ σ
τ
=−�
+ ��
1 22
22
22
( )x yS
x yxy
Sinσ σ
θσ σ
τ
− −=
−� + �
�
1 22
2
22
Substituting above values in the above equation we can show that
Solid Mechanics
S
ntdd θ θ
τθ =
=
1
2
2 - ve
Similarly we can show that
S S
ntdd πθ θ θ
τθ = = +
=
2 1
2
2
2
+ ve
Thus the angles Sθ1and Sθ
2define planes of either maximum
shear stress or minimum shear stress. Planes that define maximum shear stress & minimum shear stress are again ⊥ to each other.. Now we wish to find out these values.
( )
( )S
x ynt xy
x ynt S xy S
Sin Cos
Sin Cosθ θ
σ στ θ τ θ
σ στ θ τ θ=
−= − +
−= − +
1 11
2 22
2 22
Substituting SCos θ1
2 and SSin sθ1
2 , we can show that
x ymax xy
σ στ τ
−� = + + �
�
22
2
( ) ( ) ( )S S
x ynt S xy SSin Cosπθ θ θ
σ στ θ π τ θ π= = +
−= − + + +
1 12 1 22 2
2
Substituting SCos θ1
2 and SSin θ1
2
x ymin xy
σ στ τ
−� = − + �
�
22
2
Solid Mechanics
maxτ is algebraically minτ> , however their absolute magnitude is same. Thus we can write
x ymax min xyor
σ στ τ τ
−� = ± + �
�
22
2
Generally
max S
min S
τ θτ θ
−
−1
2
Q. Why maxτ and minτ are numerically same. Because Sθ1 &
Sθ2
are ⊥ planes.
(2) Unlike the principal stresses, the planes on which maximum and minimum shear stress act are not free from normal stresses.
Solid Mechanics
x y x yn xyCos Sin s
σ σ σ σσ θ τ θ
+ −= + +2 2
2 2
S
x y x yn S xy SCos Sinθ θ
σ σ σ σσ θ τ θ=
+ −= + +
1 112 2
2 2
Substituting SCos θ1
2 and SSin θ1
2
S
x yn θ θ
σ σσ σ =
+= =
1 2
( )( )
S S
x y x yn S
xy S
Cos
Sin
πθ θ θσ σ σ σ
σ θ π
τ θ π
= = ++ −
= + +
+ +
12 1
1
22
2 2
2
Simplifying this equation gives
S
x yn θ θ
σ σσ σ =
+= =
2 2
Therefore the normal stress on maximum and minimum shear stress planes is same.
(3) Both the principal planes are ⊥ to each other and also the planes of maxτ and minτ are also ⊥ to each other. Now let us see there exist any relation between them.
Solid Mechanics
6. Mohr’s circle Mohr’s circle for plane stress
So far we have seen two methods to find stresses acting on an inclined plane
(a) Wedge method (b) Use of transformation laws.
Another method which is purely graphical approaches is known as the Mohr’s circle for plane stress.
A major advantage of Mohr’s circle is that, the state of the stress at a point, i.e the stress components acting on all infinite number of planes can be viewed graphically.
Equations of Mohr’s circle
We know that, x y x yn xyCos Sin
σ σ σ σσ θ τ θ
+ −= + +2 2
2 2
This equation can also be written as
x y x yn xyCos Sin
σ σ σ σσ θ τ θ
+ −− = +2 2
2 2
x ynt xySin Cos
σ στ θ τ θ
−� = − + �
� 2 2
2
( )
x y x yn nt xy
x a y R
σ σ σ σσ τ τ
+ +� �� � − + = +� � � �� � � �
↓ ↓ ↓
− + =
2 22 2
2 2 2
2 2
Solid Mechanics
The above equation is clearly an equation of circle with center at ( ),0a on τ σ− plane it represents a circle with
center at x y ,σ σ+� ��
02
and
having radius
x yxyR
σ στ
−� = + �
�
2
2
This circle on σ τ− plane- Mohr’s circle.
From the above deviation it can be seen that any point P on the Mohr’s circle represents stress which are acting on a plane passing through the point.
In this way we can completely visualize the stresses acting on all infinite planes.
Solid Mechanics
(3) Construction of Mohr’s circle
Let us assume that the state of stress at a point is given
A typical problem using Mohr’s circle i.e given x y,σ σ′ ′ and
x yτ ′ ′ on an inclined element. For the sake of clarity we
assume that, x y, sσ σ′ ′ and x yτ ′ ′ all are positive and x yσ σ>
Solid Mechanics
• Since any point on the circle represents the stress components on a plane passing through the point. Therefore we can locate the point A on the circle.
• The coordinates of the plane ( )x xyA ,σ τ= + +
Therefore we can locate the point A on the circle with
coordinates ( )x xy, sσ τ+ +
• Therefore the line AC represents the x-axis. Moreover,
the normal of the A-plane makes 0�w.r.t the x-axis.
• In a similar way we can locate the point B corresponding to the plane B.
Solid Mechanics
The coordinates of ( )y xyB , sσ τ= + −
Since we assumed that for the sake of similarity y xsσ σ< .
Therefore the point B diametrically opposite to point A.
• The line BC represents y- axis. The point A corresponds
to Q = 0� , and pt. B corresponds to Q = 90�(+ve) of the stress element.
At this point of time we should be able to observe two important points.
• The end points of a diameter represents stress components on two ⊥ planes of the stress element.
• The angle between x- axis and the plane B is 90° (c.c.w) in the stress element. The line CA in Mohr’s circle represents x- axis and line CB represents y-axis or plane B. It can be seen that, the angle between x-axis and y-axis in the Mohr’s circle is 180° (c.c.w). Thus 2Q in Mohr’s circle corresponds to Q in the stress element diagram.
Stresses on an inclined element
• Point A corresponds to 0Q = on the stress element. Therefore the line CA i.e x-axis becomes reference line from which we measure angles.
• Now we locate the point “D” on the Mohr’s circle such that the line CD makes an angle of 2Q c.c.w from the x-axis or line CA. we choose c.c.w because in the stress element also Q is in c.c.w direction.
Solid Mechanics
• The coordinates or stresses corresponding to point D on the Mohr’s circle represents the stresses on the x′ - face or D on the stress element.
x avg
x y
y avg
RCos
RSin
RCos
SinceD& D are planes inthestress element ,thenthey becomediametrically opposite point sonthecircle, just likethe planes A& Bdid
σ σ βτ βσ σ β
′
′ ′
′
= +
=
= −
′ ⊥
Calculation of principal stress
The most important application of the Mohr’s circle is determination of principal stresses.
The intersection of the Mohr’s circle --- with normal stress axis gives two points P1 andP2 . Thus P1 and P2 represents points corresponding to principal stresses. In the current diagram the coordinates the of
P , sP ,
σσ
==
1 1
2 2
00
avg Rσ σ= +1
avg Rσ σ= −2
The principal direction corresponding to σ1 is now equal to
pθ1
2 , in c.c.w direction from the x-axis.
Solid Mechanics
p pπθ θ= ±
2 1 2
We can see that the points P1 andP2 are diametrically opposite, this indicate that principal planes are ⊥ to each other in the stress element. This fact can also be verified from the Mohr’s circle.
In- plane maximum shear stress
What are points on the circle at which the shearing stress are reaching maximum values numerically? Points S1 and S2 at the top and bottom of the Mohr’s circle.
• The points S1 and S2 are at angles θ =2 90� from pointsP1 P2 and, i.e the planes of maximum shear stress
are oriented at ±45� to the principal planes.
• Unlike the principal stresses, the planes of maximum shear stress are not free from the normal stresses. For example the coordinates of
max avg
max avg
S , s
S ,
τ στ σ
= +
= −1
2
max Rτ = ±
avgσ σ=
Mohr’s circle can be plotted in two different ways. Both the methods are mathematically correct.
Solid Mechanics
Finally
• Intersection of Mohr’s circle with the σ -axis gives principal stresses.
• The top and bottom points of Mohr’s circle gives maximum –ve shear stress and maximum +ve shear stress.
• Do not forget that all these inclined planes are obtained by rotation about z-axis.
Solid Mechanics
Mohr’ circle problem
Solution:
A - (15000,4000)
B - (5000,-4000)
(a)
x y MPaσ σ+ += =15000 5000
100002 2
R MPa= 6403
x yxyR
σ στ
−� −� = + = + � �� �
= +
2 22 2
2 2
15000 50004000
2 2
5000 4000
x yσ σ−= 5000
2
Solid Mechanics
Point D : x Cos . MPaσ ′ = + =10000 6403 41 34 14807
x y Sin . MPaτ ′ ′ = − = −6403 41 34 4229
Point D′ : n y Cos . MPaσ σ ′= = − =10000 6403 41 34 593
nt x y Sin .τ τ ′ ′= = =6403 41 34 4229
b) P.; .σ θ= = =
1138 66
16403 19 332
MPaσ =2 3597
c) max SMPa . .τ θ= − = = −1
6403 25 67 25 67�
Solid Mechanics
(2) θ = 45�
Principal stresses and principal shear stresses.
Solution:
( )
x y
x yxyR MPa
σ σ
σ στ
+ − += = −
−� − −� = + = + − = � �� �
2 222
50 1020
2 2
50 1040 50
2 2
( )( )
A ,
B ,
→ − −→
50 40
10 40
x y
x y
p R s
p R
σ σσ
σ σσ
+= = + = − + =
+= = − = − − = −
1 1
2 2
20 50 302
20 50 702
Solid Mechanics
p
p
p
Q .
Q .
Q .
=
=
=
1
1
2
2 233 13
116 6
206 6
�
�
s
s
s
Q .
Q .
Q .
=
=
=
1
1
2
2 143 13
71 6
161 6
Solid Mechanics
Q. x y xyMPa, MPa and MPaσ σ τ= = − =31 5 33
Stresses on inclined element θ = 45�
Principal stresses and maximum shear stress.
Solution:
x yavg MPa
σ σσ
+ −= = =31 513
2 2
x yxyR . MPa
σ στ
−� = + = �
�
22 37 6
2
( )( )
A ,
B ,− −31 33
5 33
x avgRCos s
. Cos . MPa
σ β σ′ = +
= + =37 6 28 64 13 46
x y RSin . . .τ β′ ′ = − = − = −37 6 28 64 18 02
y avgRCos
MPa
σ β σ′ = −
= −20
Solid Mechanics
. MPaσ∴ =1 50 6
. MPaσ = −2 24 6
p .θ =1
30 68
max s
min
avg
. MPa .
. MPaMPa
τ θτσ σ
= − = −
= −= =
137 6 14 32
37 613
Solid Mechanics
7. 3D-Stress Transformation 3D-stress components on an arbitrary plane
Basically we have done so far for this type of coordinate system
x x x y x z
x x x y x z
n n n D i r . c o s i n e s o f x
ˆˆ ˆ ˆi n i n j n k
′ ′ ′
′ ′ ′
′−
′ = + +
y x y y y z
y x y y y z
n n n
ˆˆ ˆ ˆj n i n j n k
′ ′ ′
′ ′ ′′ = + +
z x z y z z
z x z y z z
n n n
ˆ ˆˆ ˆk n i n j n k
′ ′ ′
′ ′ ′′ = + +
Solid Mechanics
n x x x y x z
n x x x y x z
ˆˆ ˆT T i T j T ks
ˆˆ ˆT i j kσ τ τ
′ ′ ′
′ ′ ′ ′ ′ ′
= + +
′ ′ ′= + +
�
�
x x
x x
x z
ABC dAPAB dAnPAC dAnPBC dAn
′
′
′
−−−−
[ ]xF� → + = 0
x x x x x yx x y zx x zT da dAn dAn dAnσ τ τ′ ′ ′ ′= + +
x x x x x yx x y zx x z
x y xy x x y x y zy x z
x z xz x x yz x y z x z
T n n n
T n n n
T n n n
σ τ ττ σ ττ τ σ
′ ′ ′ ′
′ ′ ′ ′
′ ′ ′ ′
= + +
= + +
= + +
x x y y z
x y y y z
z x y z z
σ τ ττ σ ττ τ σ
′ ′ ′ ′ ′
′ ′ ′ ′ ′
′ ′ ′ ′ ′
� �� �� �� �� �� �
x x y x z, ,σ τ τ′ ′ ′ ′ ′
( ) ( )x n x x x y x z x x x y x zˆ ˆˆ ˆ ˆ ˆ ˆT i T i T j T k . n i n j n kσ ′ ′ ′ ′ ′ ′ ′′= = + + + +
� (1)
( ) ( )x y n x x x y x z y x y y y zˆ ˆˆ ˆ ˆ ˆ ˆT j T i T j T k . n i n j n kτ ′ ′ ′ ′ ′ ′ ′ ′′= = + + + +
� (2)
( ) ( )x z n x x x y x z z x z y z zˆ ˆ ˆˆ ˆ ˆ ˆT k T i T j T k . n i n j n kτ ′ ′ ′ ′ ′ ′ ′ ′′= = + + + +�
(3)
y x x y x yx y y zx y z
y y xy y y y y y zy y z
y z xz y y yz y y z y z
T n n n
T n n n
T n n n
σ τ ττ σ ττ τ σ
′ ′ ′ ′
′ ′ ′ ′
′ ′ ′ ′
= + +
= + +
= + +
( )( )y y x y y y z y x y y y zˆ ˆˆ ˆ ˆ ˆT i T j T k n i n j n kσ ′ ′ ′ ′ ′ ′ ′= + + + + (4)
( )( )z z x z y z z z x z y z zˆ ˆˆ ˆ ˆ ˆT i T j T k n i n j n kσ ′ ′ ′ ′ ′ ′ ′= + + + + (5)
Solid Mechanics
( )( )y z y x y y y z z x z y z zˆ ˆˆ ˆ ˆ ˆT i T j T k n i n j n kτ ′ ′ ′ ′ ′ ′ ′ ′= + + + + (6)
x x
x y
x z
n Cosn Sin
n
θθ
′
′
′
==
= 0
y x
y y
y z
n Sin
n Cos
n
θθ
′
′
′
= −
=
= 0
z x
z y
z z
nn
n
′
′
′
==
=
00
1
z x z y z
z
: :σ τ τσ
′ ′ ′ ′ ′= = =
=
0 0 0
( ) ( )
x x y xy
y x y xy
x y x y xy
Cos Sin Sin Cos
Sin Cos Sin Cos
Sin Cos Cos Sin
σ σ θ σ θ τ θ θ
σ σ θ σ θ τ θ θ
τ σ σ θ θ τ θ θ
′
′
′ ′
= + +
= + −
= − − + −
2 2
2 2
2 2
2
2
x xy
xy y
σ ττ σ� �� �� �� �� �
0
0
0 0 0
Principal stresses
x y zn ,n ,n
( )n x y z
n nx ny nz
ˆˆ ˆˆT n n i n j n k
ˆˆ ˆT T i T j T k
σ σ= = + +
= + +
�
�
Where
nx x x yx y zx z
ny xy x y y zy z
nz xz x yz y z z
T n n n
T n n n
T n n n
σ τ ττ σ ττ τ σ
= + +
= + +
= + +
x x y y z zTn n Tn n Tn nσ σ σ= = =
Solid Mechanics
( )
( )( )
x x yx y zx z
yx x y y zy z
xz x yz y z z
n n n
n n n Syst.of linear homog.eqns.
n n n
σ σ τ τ
τ σ σ τ
τ τ σ σ
− + + = ���+ − + = ��
+ + − = ��
0
0
0
x y z x y zn n n : n n n= = = + + =2 2 20 1
( )x xy zx x
xy y zy y
zx yz z z
nn
n
σ σ τ ττ σ σ ττ τ σ σ
� �− � �� �� �− =� �� �� �� �− � �� �� �
0
For non trivial solution must be zero.
( ) ( )( )
x y z x y y z z x xy yz zx
x y z xy yz zx x yz y zx z xy
σ σ σ σ σ σ σ σ σ σ σ τ τ τ σ
σ σ σ τ τ τ σ τ σ τ σ τ
− + + + + + − − −
− + − − − =
3 2 2 2 2
2 2 22 0
This has 3- real roots , ,σ σ σ1 2 3
( )
( )x x yx y zx z
yx x y y zy z
x y z
n n n
n n n
and n n n
σ σ τ τ
τ σ σ τ
− + + =
+ − + =
+ + =
1
1
2 2 2
0
0
1
x y zn ,n ,n σσ σ σ
� →
> >1
1 2 3
Stress invariants
I I Iσ σ σ− + − =3 21 2 3 0 (1)
Solid Mechanics
x y z
x y y z x z xy yz zx
x y z xy yz zx x yz y zx z xy
I
I stress inv ar iants
I
σ σ σ
σ σ σ σ σ σ τ τ τ
σ σ σ τ τ τ σ τ σ τ σ τ
�= + +��= + + − − − ��
= + − − − ��
1
2 2 22
2 2 23 2
I Iσ σ′ ′− + =3 21 3 0
x y z x y x z y z x y y z x zI Iσ σ σ σ σ σ σ σ τ τ τ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′′ ′= + + = + + − − −2 2 21 2
I I ; I I ; I I′ ′ ′= = =1 1 2 2 3 3
3D 2D
III σ
σ σ σσ σ σ σ σ σσ σ
= + += + +=
3
1 1 2 3
2 1 2 2 3 3 1
3 1 2
III
σ σσ σ
= +==
1 1 2
2 1 2
3 0
Principal planes are orthogonal
n nˆ ˆT n T .n′′ =� �
x y z
x y z
n nx ny nz
n n x n y n z
ˆˆ ˆn̂ n i n j n k
ˆˆ ˆn̂ n i n j n k
ˆˆ ˆT T i T j T k
ˆˆ ˆT T i T j T k
′ ′ ′
′ ′ ′ ′
= + +
′ = + +
= + +
= + +
�
�
Solid Mechanics
yx
n n
xy
ˆ ˆT n T n
ττ′
=
′ =� �
( ) ( )n nˆ ˆT n T n
ˆ ˆ ˆ ˆn n n nσ σ′′ =
′ ′=1 2
� �
( ) ( )x x y y z z x x y y z zn n n n n n n n n n n nσ σ′ ′ ′ ′ ′ ′+ + = + +1 2
σ σ≠1 2
x x y y z zn n n n n n′ ′ ′+ + = 0
ˆ ˆn .n′ must be ⊥ to each other.
The state of stress in principal axis
σ
σσ
� �� �� �� �� �
1
2
3
0 00 00 0
x
y
z
n x
n y
n z
T n
T n
T n
σσ
σ
=
=
=
1
2
3
n x y zn n nσ σ σ σ= + +2 2 21 2 3
x y zn n n n
x y z
T T T T s
n n nσ σ σ
= + +
= + +
2 2 2 2
2 2 2 2 2 21 2 3
n nTτ σ= −22 2�
Solid Mechanics
8. 3D Mohr’s circle and Octahedral stress 3-D Mohr’s circle & principal shear stresses
x xy
ij xy y
z
σ τσ τ σ
σ
� �� �
� �= � �� �� �� �
0
0
0 0
Once if you know andσ σ1 2
τ
σ στ
σ σσ
−=
+=1
2 31
1 3
2
2
τ
σ στ
σ σσ
−=
+=2
1 32
1 2
2
2
τ
σ στ
σ σσ
−=
−=3
1 23
1 2
2
2
max max , ,σ σ σ σ σ στ − − −= 1 2 2 3 3 1
2 2 2
σ σ σ> >1 2 3
Solid Mechanics
• The maximum normal stress 1σ and maximum shear stress maxτ and their corresponding planes govern the failure of the engineering materials.
• It is evident now that in many two-dimensional cases the maximum shear stress value will be missed by not considering σ =3 0 and constructing the principal circle.
Solid Mechanics
Problem:
The state of stress at a point is given by
x y zMPa, MPa, MPa andσ σ σ= = − =100 40 80
xy yz zxτ τ τ= = = 0
Determine in plane max shear stresses and maximum shear stress at that point.
Solution:
MPa, MPa MPasσ σ σ= = = −1 2 3100 80 40
MPaσ στ − −= = =1 212
100 8010
2 2
MPaσ στ − += = =1 313
100 4070
2 2
MPaσ στ − += = =2 323
80 4060
2 2
MPaMPa
σ σσ
σσ
+= =
==
1 212
13
23
902
3020
max max , ,τ τ τ τ= 12 13 23
max MPaτ = 70 This occurs in the plane of 1-3
Solid Mechanics
, ,τ τ τ →1 2 3 Principal shear stress in 3D
( )max max , ,τ τ τ τ= 1 2 3
Solid Mechanics
Plane stress
z
σ σσ σ
>= =
1
3 0
x yxy
σ στ τ
−� = ± + �
�
22
2 ---- in plane principal shear stresses.
maxσ σ στ −= =1 3 1
2 2
Solid Mechanics
Problem
At appoint in a component, the state of stress is as shown. Determine maximum shear stress.
Solution:
ijσ � �� �= � �� �
� �
100 00 50
- plane stress problem
We can also write the matrix as ija� �� �� �=� � � �� �� �
100 0 00 50 00 0 0
σσσ σ
==− −= =
1
2
1 2
10050
100 5025
2 2
max MPaτ = 25
Solid Mechanics
Now with , ,σ σ σ= = =1 2 3100 50 0
max MPaσ στ −= =1 3 502
Occurs in the plane 1-3 instead of 1-2
Solid Mechanics
Some important states of stresses
(1) Uniaxial state of stress: Only one non-zero principal stress.
σσ� �� �� �= � �� � � �
� �� �
11
0 00
0 0 00 0
0 0 0- plane stress.
(2) Biaxial state of stress: two non-zero principal stresses.
σσ
σσ
� �� �� �= � �� � � �
� �� �
11
11
0 00
0 00
0 0 0- plane stress
(3) Triaxial state of stress: All three principal stresses are non zero.
σσ
σ
� �� �−� �� �� �
1
2
3
0 00 00 0
3D stress
(4) Spherical state of stress: σ σ σ= =1 2 3 (either +ve or – ve)
Dσ
σσ
� �� �−� �� �� �
0 00 0 30 0
stress-special case of triaxial stress.
Solid Mechanics
(5) Hydrostatic state of stress
PP
P
+� �� �+� �
+� �� �
0 00 00 0
hydrostatic tension
PP
P
−� �� �−� �
−� �� �
0 00 00 0
hydrostatic compression.
(6) The state of pure shear
zy
x xy xz
ij xy y yz
zx z
σ τ τσ τ σ τ
τ τ σ
� �� �
� � � �=� �� �� �� �
x y x z
ij x y y z
z x z y
τ τσ τ τ
τ τ
′ ′ ′ ′
′ ′ ′ ′
′ ′ ′ ′
� �� �
� �= � �� �� �� �� �
0
0
0
Then we say that the point P is in state of pure shear.
I =1 0 is necessary and sufficient condition for state of pure shear
Solid Mechanics
Octahedral planes and stresses
If x y zn n n= = w.r.t to the principal planes, then these planes
are known as octahedral planes. The corresponding stresses are known as octahedral stresses.
Eight number of such planes can be identified at a given point --- Octahedron
x y z
n x y z
n n n
T n n n
σ σ σ σ
σ σ σ
= + +
= + +
2 2 21 2 3
2 2 2 2 2 2 21 2 3
x y z
x y z
n n n
n n n .
+ + =
= = = ± =
2 2 2
0
1
154 73
3
octσ σ σ σ
σ σ σ
� � � = + + � � �� � �
+ +=
2 2 2
1 1 1
1 2 3
1 1 13 3 3
3
Solid Mechanics
1I = meanstress3
σ σ σ+ + =1 2 33
oct canbeint erpreted meannormalstress at a pt.σ = − −
oct n octTτ σ= −2 2
( ) ( ) ( )octτ σ σ σ σ σ σ= − + − + −2 221 2 2 3 3 1
13
Therefore, the state of stress at a point can be represented with reference to
(i) stress components of x,y,z coordinate system
(ii) stress components of x’,y’z’ coordinate system
(iii) using principal stresses
(iv) using octahedral shear and normal stresses
We can prove that:
octτ is smaller than maxτ (exist only on 4 planes) but can exist on 8 planes at a point.
Solid Mechanics
Decomposition into hydrostatic and pure shear stress
x xy xz
ij yx z yz
zx zy z
σ τ τσ τ σ τ
τ τ σ
� �� �
� �= � �� �� �� �� �
Mean stress x y z IPσ σ σ+ +
= = 13 3
x xy xz x xy xz
yx y yz yx y yz
zx zy z zx zy z
PPP P
P P
Hydrostatic State of pureshearstat of stress Deviatoric state of stress
Dilitationalstress Stressdeviator
σ τ τ σ τ ττ τ τ τ σ ττ τ σ τ τ σ
� � � �−� �� � � �� �= + −� � � �� �� � � �� � −� �� � � �� � � �
0 00 00 0
Thus the state of the stress at a point can alos be represented by sum of dilational stress and stress deviator
Solid Mechanics
IP σ σ σ+ += =1 2 3 13 3
P PP P
P P
σ σσ σ
σ σ
−� � � � � �� � � � � �= + −� � � � � �
−� � � � � �� � � � � �
1 1
2 2
3 3
0 0 0 0 0 00 0 0 0 0 00 0 0 0 0 0
σ =1 mean stress + deviation from the mean
The deviatoric and octahedral shear stresses are the answer for the yielding behavior of materials – which is a type of failure of materials.
Solid Mechanics
9. Deformation and strain analysis
Two types of deformation have been observed for an infinitesimal element.
Deformation of the whole body = Sum of deformations of
Deformation is described by measuring two quantities.
(1)Elongation or contraction of a line segment
(2)Rotation of any two ⊥ lines.
Measure of deformations of an infinitesimal element is known as strain.
• The strain component that measures elongation or construction – normal strain -ε
• The strain component that measures rotation of any two ⊥lines is – shearing strain- γ
( )( )( )
u u x,y ,z
v v x,y ,z (x,y ,z) is the point in the undeformed geometry
w w x,y ,z
= ��
= ��= �
( ) ( ) ( )= + +� ˆˆ ˆu u x,y ,z i v x,y ,z j w x,y ,z k
Solid Mechanics
Normal strain ε - Account for changes in length between two points.
( )* * *
ns s
P Q PQ s sP lim limPQ s∆ → ∆ →
− ∆ − ∆∈ = =∆0 0
We can also define the same point x y z, ,∈ ∈ ∈
(1) By definition x∈ is + if *s s∆ > ∆
x∈ is - if *s s∆ > ∆
(2) It is immaterial how * *P Q is oriented finally. However for
n∈ we must consider PQ in the direction of n̂ in the undeformed geometry
(3) In general ( )n n x,y ,z s∈ =∈
(4) No units.
(5) Meaning of nn∈
Shearing strain -
Accounts the change in angle
( )nY P+ Change in angle between
⊥ lines in ˆn̂& t direction.
( )nt ntx xy y
Y P lim limπ φ α β∆ → ∆ →∆ → ∆ →
− = +0 00 0
2
Mm/mm,0.5%=0.005;
,µ µ−= 610 1000
. mm / mm−= × =61000 10 0 001
( )( )
*n
*n
n n
s s
s s if ss s s s
∆ = + ∈ ∆
∆ +∈ ∆ ∆ →∈ ∆ =∈ ∆
1
1 0
lim as s∆ → 0
Solid Mechanics
(1)We must select two ⊥ lines in the undeformed geometry.
(2)Units of ntY →radius.
(3)By deflection nt tnY Y=
(4)Two subscripts are required for
Y - to show directions of initial
infinitesimal line segments.
(5) ntY is +ve if angle is decreased
ntY is -ve if angle is more.
By taking two ⊥ lines
We can define n t nt, &Y∈ ∈
Rectangular strain components
x y xy
z y yz
x z xz
, andY PQRS
, andY QABS
, andY RSCD
∈ ∈ −
∈ ∈ −
∈ ∈ −
x xy xz
ij xy y yz
xz yz z
Y Y
E Y Y
Y Y
� �∈� �
� �= ∈� �� �� �∈� �� �
They represent the state of strain at a point , since we can determine strain along any direction n̂
- Rectangular strain components . - We then say that we have strain
computer associated with x,y ,z coordinate system.
Solid Mechanics
Strain displacement relations: Strains are due to deformation as displacement so there must be some relation between deformational displacements and strains. So let us consider the side of the elementPQRS . We shall demonstrate that ‘w’ has no impact. So it can be neglected.
P u,vu vQ u x ; v xx x
→∂ ∂→ + ∆ + ∆∂ ∂
* * *
PQ x
P Q x
= ∆
= ∆
( )*xx x∆ + ∈ ∆1
( )*x
xlim x x
∆ →∆ = + ∈ ∆
01
* u v wx x x xx x x
u u v w xx x x x
∂ ∂ ∂� � � ∆ = + ∆ + ∆ + ∆ � � �∂ ∂ ∂� � �
� �∂ ∂ ∂ ∂� � � = + + + + ∆ � � �� �∂ ∂ ∂ ∂� � � � �
2 2 2
2 2 2
1
1 2
Solid Mechanics
*
xx
x
x
y
z
x xlimx
u u v wlimx x x x
u u v wx x x x
v u v wy y y y
w u vz z z
∆ →
∆ →
∆ − ∆∈ =∆
∂ ∂ ∂ ∂� � � = + + + + − � � �∂ ∂ ∂ ∂� � �
∂ ∂ ∂ ∂� � � ∈ = + + + + − � � �∂ ∂ ∂ ∂� � �
� � � ∂ ∂ ∂ ∂∈ = + + + + − � � �∂ ∂ ∂ ∂� � �
∂ ∂ ∂� � ∈ = + + + � �∂ ∂ ∂� �
0
2 2 2
0
2 2 2
2 2 2
2
1 2 1
1 2 1
1 2 1
1 2wz
∂� + − �∂�
2 21
So far no assumption has been made except for size of x, y& z∆ ∆ ∆
*xy * *
yu x uCosx yx y
φ ∆∂ ∆ ∂� �� = + �� �∂ ∂� ∆ ∆� �1
* *yv x v
x yx y
� �� ∆∂ ∆ ∂� + +� � � �∂ ∂� ∆ ∆� � �1
* *yw x w
x yx y
� �∆∂ ∆ ∂� + � � �∂ ∂� ∆ ∆� �
*xy xy
xyz
Y lim π φ∆ →∆ →∆ →
= −000
2
Solid Mechanics
*xy xy
xyz
SinY lim Cosφ∆ →∆ →∆ →
=000
( )( )
xy * *xyz
*x
*y
x yu u v v w wSinY limx y y x x y x y
x x
y y
∆ →∆ →∆ →
� �� ∆ ∆∂ ∂ ∂ ∂ ∂ ∂� = + + + + �� � �∂ ∂ ∂ ∂ ∂ ∂� ∆ ∆� � �
∆ = + ∈ ∆
∆ = + ∈ ∆
000
1 1
1
1
( )( )xyx x yyz
u v u u v v w wy x x y x y x y
SinY lim∆ →∆ →∆ →
� ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂+ + + + �∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂� =+ ∈ + ∈0
00
1 1
( )( )xyx y
u v u u v v w wSiny x x y x y x y
Y
− � ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂+ + + + �∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂� =+∈ +∈
1
1 1
( )( )yzx y
u v u u v v w wy x x y x y x y
Y sin−
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂� �+ + + +� �∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂= � �+ ∈ + ∈� �
� �� �
1
1 1
( )( )xzx z
w u w w u u v vx w x z x w x zY sin−
∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂� �+ + + +� �∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂= � �+∈ +∈� �� �
1
1 1
All bodies after the application of loads under go “small deformations”
Solid Mechanics
Small deformations :
(1) The deformational displacements ˆ ˆu ui vj wk= + +� are
infinitesimally small.
(2) The strains are small
(a) Changes in length of a infinitesimal line segment are infinitesimal.
(b) Rotations of line segment are also infinitesimal.
x y zu u u v u u v, , , ; ; ; ;x u w x x x y
∂ ∂ ∂ ∂ ∂ ∂ ∂� ∈ ∈ ≤< ∈ �∂ ∂ ∂ ∂ ∂ ∂ ∂�
21 1 1 1� � � are
negligible compare to u v,x x
∂ ∂∂ ∂
quantities.
xux
ux
∂∈ = + −∂
∂ −∂= +
1 2 1
2 11
2
x
y
z
uxvywz
∂∈ =∂∂∈ =∂∂∈ =∂
xy xySinY Y≈
Solid Mechanics
( )xyx y
u vy x v uY
x y
∂ ∂+∂ ∂ ∂ ∂= = +
∂ ∂+ ∈ +∈1
xz
yz
w uYx zv wYz y
∂ ∂= +∂ ∂∂ ∂= +∂ ∂
Another derivation : Let us take plane PQRS in xy plane.
Also assume that ( ) ( )u u x,y & v v x,y= = only.
Small deformation
Displacements are small
Strains are small
* * *
xx
P Q PQ x xlimPQ x∆ →
− ∆ − ∆∈ = =∆0
Strains<0.001
* * *
xx
yy
yx P Q x
xy
x xuxlim
x x
v y yy vlim
y y
′
∆ →
∆ →
∂� ∆ = + ∆ �∂�
∂� + ∆ − ∆ � ∂∂� ∈ = =∆ ∂
� ∂+ ∆ − ∆ �∂ ∂� ∈ = =∆ ∂
0
0
1
1
1
Y .< 0 06�
*s .
s mm−
∆ =
= × 4
0 2002
2 10
Solid Mechanics
*xy xy
x xy y
Y lim limπ φ α β∆ → ∆ →∆ → ∆ →
= − = +0 00 0
2
v vxx xtan yy
xxx
α
∂ ∂∆∂ ∂= = ∂∂� ++ ∆ � ∂∂�
11
tanα α≈
vxuy
α
β
∂=∂∂=∂
xyu vYy x
∂ ∂= +∂ ∂
u u v v, , ,x y y x
u u v, ,x y yx
∂ ∂ ∂ ∂∂ ∂ ∂ ∂
� � ∂ ∂ ∂� � � �∂ ∂ ∂� � �
2 22
1�
We can define the state of strain at point by six components of strains
State of strain
- Engineering strain matrix - We can find n∈ in any
direction we can find ntY for any two arbitrary directions.
x y , z, xy xz yz
yx zx zy
, Y , Y , Y
Y Y Y
∈ ∈ ∈
↓ ↓ ↓
x xy xz
ij xy y yz
xz yz z
Y Y
E Y Y
Y Y
� �∈� �
� �= ∈� �� �� �∈� �� �
Solid Mechanics
2D- strain transformation
Plain strain: In which
x xy
xy y
Y
Y
∈� �� �∈� �� �
( )( )
( )
x x
y y
xy xy
x,y
x,y
Y Y x,y
∈ =∈
∈ =∈
=
z
yz
zx
Y
Y
∈ ==
=
00
0
implication of these equation is that a point in a given plane does not leave that plane all deformations are in to plane of the body.
Solid Mechanics
Given x y xy, & Y∈ ∈ what are n t nt, & Y∈ ∈ .
We can always draw PQRS for given n̂
If x y xy, & Y∈ ∈
As in case of stress we call these formulae as transformations laws.
x
x
x
dxSinds
dxsinds
sin cos
θα
θ
θ θ
∈=
=∈
=∈
1
y ydy
cos cos sinds
α θ θ θ=∈ =∈2
xy
xy
dyY sin
dsY sin sin
α θ
θ θ
=
=
3
Solid Mechanics
x y xy
n x y xy
x y xy
dL dxcos dy sin Y dycos
dy dydL dx cos sin Y cosdS ds ds ds
cos cos sin Y sin cos
θ θ θ
θ θ θ
θ θ θ θ θ
=∈ + ∈ +
=∈ =∈ + ∈ +
=∈ + ∈ +2
- state of strain at a point
- stress tensor
- strain tensor
Replace
x x
y y
xyxy xy
Y
σσ
τ
→∈→∈
→∈ =2
( ) ( )x y xy
x y xy
x y xy
sin cos sin cos Y sin
cos sin cos sin Y cos
cos sin cos sin Y cos
α θ θ θ θ θ
β θ θ θ θ θ
θ θ θ θ θ
= −∈ + ∈ −
= −∈ − + ∈ − −
=∈ −∈ −
2
2
2
( )x y xynt YY sin cosθ θ∈ −∈
= − +2 22 2 2
x xy
xy y
Y
Y
∈� �� �∈� �� �
xyx
xyy
Y
Y
� �∈� �� �� �∈� �� �
2
2
x xy
xy y
∈ ∈� �� �∈ ∈� �� �
xyxy
Y∈ =
2
x xy
xy y
σ ττ σ� �� �� �� �
x y x y xyn
Ycos sinθ θ
∈ + ∈ ∈ −∈∈ = + +2 2
2 2 2
Solid Mechanics
Principal shears and maximum shear In plane- principal strains
xy xyp
x y
/tan Q
ϒ∈ →=
∈ −∈2 2
2
p pθ θ− − ⊥1 2
to each other
,∈ ∈ ∈ >∈1 2 1 2
( )x ys
xy
s p
tan
/
θ
θ θ π
∈ −∈= −
∈
= ±1
22
4
s sθ θ− − ⊥1 2
to each other
x y
x y I
x y xy
y xy
xyx y
I
J
I
J
YJ
σ σ
σ σ τ
+ =
∈ + ∈ =
− =
∈∈ −∈ =
� ∈ ∈ − = �
�
1
22
22
2
22
x ymax min xy
maxmax s
minmin s
or R
Y
Y
θ
θ
∈ −∈� ∈ ∈ = ± = ± + ∈ �
�
=∈ −
=∈ −
1
2
22
2
2
2
Solid Mechanics
Mohr’s Circle for strain
3D-strain transformation
xyx x y y z z xy xy
Y; ; ;σ σ σ τ→∈ →∈ →∈ =∈ =
2
( )
( )( )
x xy xz
xy y yz
xz yz z
∈ −∈ ∈ ∈
∈ ∈ −∈ ∈ =
∈ ∈ ∈ −∈
0
, ,∈ ∈ ∈1 2 3 - ∈ >∈ >∈1 2 3
* * * * *
s x y
s P Q P R
u vx y x yx x
′ ′
∆ = ∆ + ∆
∆ = +
∂ ∂� � � �� � = + ∆ + + ∆ − ∆ + ∆ � �� � � �∂ ∂� � � � � �
2 2 2
2 2 2
2 22 21 1
x x y y,Y ,′ ′ ′ ′∈ ∈
Solid Mechanics
ny
. xx
yu v x x yx x x
∆� ∈ = + ∆ �∆�
∆∂ ∂� �� � = + + + ∆ − ∆ − ∆ � �� �∂ ∂ ∆� � � �
2
222 2 2
1
1 1
u u v vx y x yx x y y
yx
x
u vx y x yx y
y xx
� �� � ∂ ∂ ∂ ∂� � �+ + ∆ + + + ∆ − ∆ − ∆ � � �∂ ∂ ∂ ∂� � �� � � �=∆� + ∆ �∆�
� ∂ ∂� + ∆ + + ∆ − ∆ + ∆ � �∂ ∂� � =∆� + ∆ �∆�
222 2 2
2
2
2 2 2 2
2
1 2 1 2
1
1 2 1 2
1
Transformation
x x x x y y y z z z xy x x x y
yz x y x z zx x z x x
n n n n n
n n n n
σ σ σ σ ττ τ
′ ′ ′ ′ ′ ′
′ ′ ′ ′
= + + +
+ +
2 2 2
x x x x y x y z x z xy x x x y
yz x y x z zx x z x x
n n n n n
n n n n′ ′ ′ ′ ′ ′
′ ′ ′ ′
∈ =∈ + ∈ +∈ +∈
+ ∈ +∈
2 2 2
x yx y x y
Yτ ′ ′
′ ′ ′ ′→∈ →2
xy xy
yz yz
zx zx
τττ
→∈
→∈
→∈
x x
y y
z zx
σσσ
→∈→∈
→∈
Solid Mechanics
Principal strains:
( )
( )( )
x x xy y xz z
xy x y y yz z
xz x yz y z z
n n n
n n n
n n n
∈ −∈ + ∈ +∈ =
∈ + ∈ −∈ + ∈ =
∈ +∈ + ∈ −∈ =
0
0
0
( )
( )( )
x xy xz
xy y yz
xz yz z
∈ −∈ ∈ ∈
∈ ∈ −∈ ∈ =
∈ ∈ ∈ −∈
0
J J J∈ − ∈ + ∈− =3 21 1 2 3 0
x y zJ =∈ +∈ + ∈1
x xyx y x z y z xy yz zx
xy y
y yz x xz
yz z xz z
J∈ ∈
=∈ ∈ +∈ ∈ + ∈ ∈ −∈ −∈ −∈ +∈ ∈
∈ ∈ ∈ ∈+
∈ ∈ ∈ ∈
2 2 22
x y z xy yz zx x yz y xz
x xy xz
z xy yx y yz
zx zy z
J =∈ ∈ ∈ + ∈ ∈ ∈ −∈ ∈ −∈ ∈
∈ ∈ ∈
−∈ ∈ ∈ ∈ ∈
∈ ∈ ∈
2 23
2
∈ >∈ >∈1 2 3
System of linear homogeneous equations
Solid Mechanics
( )
( )x x xy y zx z
xy x y y zy z
x y z
n n n
n n n
n n n
∈ −∈ +∈ + ∈ =
∈ + ∈ −∈ + ∈ =
+ + =
1
1
2 2 2
0
0
1
x y zn ,n & n� unique
Decomposition of a strain matrix into state of pure shear + hydrostatic strain
x xy xz x xy xz
ij yx y yz yx y yz
zx zy z zx zy z
Stateof pureshear Hydrostatic
� � � �∈ ∈ ∈ ∈ −∈ ∈ ∈ ∈� �� � � � � �� �∈ = ∈ ∈ ∈ = ∈ ∈ −∈ ∈ + ∈� � � �� � � �� � � � ∈� �∈ ∈ ∈ ∈ ∈ ∈ −∈ � �� � � �� � � �
0 00 00 0
where x y z∈ +∈ + ∈∈=
3
JJJ
=∈ + ∈ + ∈=∈ ∈ + ∈ ∈ + ∈ ∈=∈ ∈ ∈
1 1 2 3
2 1 2 2 3 3 1
3 1 2 3
Solid Mechanics
Plane strain as a special case of 3D
∈ =3 0 is also a principal strain
z → is a principal direction
if ;∈ >∈ ∈ =∈1 2 1 2 +ve
if ∈1 +ve, ∈2 -ve.
if +ve, -ve∈ ∈1 2
P & z′ ′1 will come closer
to the maximum extent,
so that the included angle
is maxπ −∈2
Solid Mechanics
Transformation equations for plane-strain
Given state of strain at a point P.
xx xy
ijxy yy
YE
Y
∈� �� �= � �� � ∈� �� �
This also means that
Now what are the strains associated with x ,y′ ′ i.e
x x x y
i jx y y y
YE
Y′ ′ ′ ′
′ ′′ ′ ′ ′
∈� �� �= � �� � ∈� �� �
This also means that
deformation
Solid Mechanics
Assume that xx yy,∈ ∈ and x yY ′ ′ are +ve
Applying the law of cosines to triangular P* Q* R*
( ) ( ) ( ) ( )( )
( ) ( ) ( ) ( )
( )
xy
x x y x
y xy
P* R* P* R* Q* R* P* R* Q* R*
cos Y
x x y x
y cos Y
π
π′
= + −
� + ��
� �′∆ +∈ = ∆ + ∈ + ∆ +∈ − ∆ + ∈� � � � � �� � � � � �� �
� � �∆ +∈ + �� � �
2 2 2
22 2
2
2
1 1 1 2 1
12
x x cosθ′∆ = ∆ and y x sinθ′∆ = ∆
( )xy xy xycos Y sinY Yπ + = − ≈ −2
( ) ( ) ( )( )( )( )
x x y
x y xy
x x cos x sin
x sin cos Y
θ θ
θ θ
′′ ′ ′∆ + ∈ = ∆ + ∈ + ∆ + ∈
′− ∆ + ∈ + ∈ −
22 22 2 2 2 2
2
1 1 1
2 1 1
Solid Mechanics
( ) ( ) ( )( )( )( )
( ) ( )( )
x x y
x y xy
x x x x y y
xy x y x y
cos sin
sin cos Y
cos sin
sin Y
θ θ
θ θ
θ θ
θ
′
′
+ ∈ = + ∈ + + ∈
− + ∈ + ∈ −
+ ∈ + ∈ = + ∈ + ∈ + + ∈ + ∈
+ + ∈ + ∈ + ∈ ∈
22 22 2
2 2 2 2 2
1 1 1
2 1 1
1 2 1 2 1 2
2 1
( ) ( )( )
( ) ( )
x x y
xy x y
x y
xy
cos sin
Y sin
cos sin
Y sin
θ θ
θ
θ θ
θ
+ ∈ = + ∈ + + ∈
+ + ∈ + ∈
= + ∈ + + ∈
+
2 2
2 2
1 2 1 2 1 2
2 1
1 2 1 2
2
x x y xy
xyx x y
cos sin Y sin
Ycos sin sin
θ θ θ
θ θ θ
′
′
+ ∈ = + ∈ + ∈ +
∈ =∈ + ∈ +
2 2
2 2
1 2 1 2 2 2
22
x y x y xyx
Ycos sinθ θ′
∈ + ∈ ∈ −∈∈ = + +2 2
2 2 2
If yQ πθ ′= + �∈2
x y x y xyx
Ycos sinθ θ′
∈ + ∈ ∈ −∈∈ = + +2 2
2 2 2
x y x y xyy
Ycos sinθ θ′
∈ + ∈ ∈ −∈∈ = + −2 2
2 2 2
x y x y′ ′∈ + ∈ =∈ + ∈ J= =1 first invariant of strain.
Solid Mechanics
( )
x y xyx OBQ
OB x y xy
xy OB x y
Y
Y
Y
π′ =∈ +∈
∈ =∈ = +
∈ =∈ +∈ +
= ∈ − ∈ +∈
4 2 22
2
( )( )
OB x y x y
x y OB x y
OB x y
Y
Y
( )
′ ′ ′ ′ ′
′ ′ ′ ′
′
∈ =∈ +∈ +
= ∈ − ∈ +∈
= ∈ − ∈ +∈
2
2
2 3
x y x y xyx OBQ Q
Ysin cosπ θ θ′ ′= +
∈ +∈ ∈ −∈∈ =∈ = − +
42 2
2 2 2- (4)
Substituting (4) in (3)
( ) ( ) ( )x y x y x y xy x yY sin Y cosθ θ′ ′ = ∈ +∈ − ∈ −∈ + − ∈ + ∈2 2
( )x y x y xyY sin Y cosθ θ′ ′ = − ∈ −∈ +2 2 (5)
tensorial normal strain xx∈ =engineering normal strain
xx yy z, ,=∈ ∈ ∈
tensorial shear strain ( ) xyxy
YEngineeringshear strain� ∈ = �
� 2 2
Solid Mechanics
( )
xzxx xy xz
ij xy yy yz
zx zy zz zz
Y� �� ∈ ∈ ∈ = �� �� � �
� �∈ = ∈ ∈ ∈� �� �� �∈ ∈ ∈ =∈� �� �
2
( )
x y x yx xy
x y x yy xy
x yx y xy
cos sin
cos sin
sin cos
θ θ
θ θ
θ θ
′
′
′ ′
∈ +∈ ∈ −∈∈ = + +∈
∈ + ∈ ∈ −∈∈ = − −∈
∈ −∈∈ = − + ∈
2 22 2
2 22 2
2 22
Components.
- Strain tensors
Solid Mechanics
Problem:
An element of material in plane strain undergoes the following strains
x y xyY− − −∈ = × ∈ = × = ×6 6 6340 10 110 10 180 10
Show them on sketches of properly oriented elements.
Solution:
x−
′∈ = − × 6340 10 ; y−
′∈ = × 6110 10 ; x yY −′ ′ = × 6180 10
x−∈ = × 6340 10
Solid Mechanics
Problem:
During a test of an airplane wing, the strain gage readings
from a 45� rosette are as follows gage A, −× 6520 10 ; gage B −× 6360 10 and gage C −− × 680 10
Determine the principal strains and maximum shear strains and show them on sketches of properly oriented elements.
Solution:
(1)
x
OB
y
−
−
−
∈ = ×
∈ = ×
∈ = − ×
6
6
6
520 10
360 10
80 10
( )( )
xy OB x yY
rad
− − −
−
= ∈ − ∈ +∈
= × × − × − ×
= ×
6 6 6
6
2
2 360 10 520 10 80 10
280 10
x y− −
−∈ + ∈ × − ×= = ×6 6
6520 10 80 10220 10
2 2
Solid Mechanics
x y− −
−∈ −∈ × + ×= = ×6 6
6520 10 80 10300 10
2 2
xyp
x y
xyxy
etan
Y
θ−
−
−−
∈ × ×= =∈ −∈ ×
×∈ = = = ×
6
6
66
2 140 102
300 10
280 10140 10
2 2
p
p p
.
. .
θ
θ θ
∴ =
= =
2 25 02
12 51 102 51
�
�
( ) ( )
x y x yxyor
.
− − −
− −
∈ + ∈ ∈ −∈� ∈ ∈ = ± + ∈ �
�
= × ± × + ×
= × ± ×
22
1 2
2 26 6 6
6 6
2 2
220 10 300 10 140 10
220 10 331 06 10
.
.
−
−
∴ ∈ = ×
∈ = − ×
61
62
551 06 10
111 06 10
( ) ( )
x .
x y x yxyCos Sin
cos . Sin .
.
θ
θ θ
′ =
− − −
−
∈
∈ +∈ ∈ −∈= + + ∈
= × + × × + × ×
= ×
12 51
6 6 6
6
2 22 2
220 10 300 10 2 12 51 140 10 2 12 51
551 06 10
�
Solid Mechanics
p .θ =1
12 51 and p .θ =2
102 51
(b) In- plane maximum shear strains are
x yxyxymax xyminor
. −
∈ −∈� ∈ ∈ = ± +∈ �
�
= ± ×
22
6
2
331 06 10
( )( )
xy max
xy min
.
.
−
−
∈ = ×
∈ = − ×
6
6
331 06 10
331 06 10
( )x ys
xytan Q
.
−
−
∈ −∈ − ×= − =∈ ×
6
6300 10
22 140 10
s
s s
Q .
Q . Q .
=
= − =
2 64 98
32 5 57 5� �
( )( ) ( )x y
x y xyQ .Sin . Cos .
. . .
′ ′ =− − −
∈ −∈∈ = − + ∈
= − × − × = ×
57 5
6 6 6
2 57 5 2 57 52
271 89 10 59 17 10 331 06 10
�
Solid Mechanics
and
s .θ = −1
32 5� s .θ = −2
32 5�
x y −∈ + ∈∈= = × 6220 10
2
min
max
Y .
Y .
−
−
= − ×
= ×
6
6
662 11 10
662 11 10
Solid Mechanics
10. Stress strain diagrams • Bar or rod – the longitudinal direction is considerably
greater than the other two, namely the dimensions of cross section.
• For the design of the m/c components we need to understand about “mechanical behavior” of the materials.
• We need to conduct experiments in laboratory to observe the mechanical behavior.
• The mathematical equations that describe the mechanical behavior is known as “constitutive equations or laws”
• Many tests to observe the mechanical behavior- tensile test is the most important and fundamental test- as we gain or get lot of information regarding mechanical behavior of metals
• Tensile test Tensile test machine, tensile test specimen, extensometer, gage length, static test-slowly varying loads, compression test.
Stress -strain diagrams
After performing a tension or compression tests and determining the stress and strain at various magnitudes of load, we can plot a diagram of stress Vs strain.
Solid Mechanics
Such is a characteristic of the particular material being tested and conveys important information regarding mechanical behavior of that metal.
We develop some ideas and basic definitions using σ −∈ curve of the mild steel.
Structural steel = mild steel = 0.2% carbon=low carbon steel
Region O-A
(1) σ and ∈ linearly proportional.
(2) A- Proportional limit
pσ - proportionality is maintained.
(3) Slope of AO = modulus of elasticity “E” – N/m2,Pa
(4) Strains are infinites ional.
f o
o
L L
L
−∈=
Solid Mechanics
Region A-B
(1) Strain increases more rapidly than σ
(2) Elastic in this range
Proportionality is lost.
Region B-C
(1) The slope at point B is horizontal.
(2) At this point B, ∈ increases without increase in further load. I.e no noticeable change in load.
(3) This phenomenon is known as yielding
(4) The point B is said to be yield points, the corresponding stress is yield stress ysσ of the steel.
(5) In region B-C material becomes “perfectly plastic i.e which means that it deforms without an increase in the applied load.
(6) Elongation of steel specimen or ∈ in the region BC is typically 10 to 20 times the elongation that occurs in region OA.
(7) s∈ below the point A are said to be small, and s∈ above A are said to be large.
(8) s A∈ <∈ are said to be elastic strains and A∈>∈ are said to be plastic strains = large strains = deformations are permanent.
Solid Mechanics
Region C-D
(1)The steel begins to “strain harden” at “C” . During strain hardening the material under goes changes in its crystalline structure, resulting in increased resistance to the deformation.
(2)Elongation of specimen in this region requires additional load,
∴ σ −∈ diagram has + ve slope C to D.
(3) The load reaches maximum value – ultimate stress.
(4)The yield stress and ultimate stress of any material is also known as yield strength and the ultimate strength .
(5) uσ is the highest stress the component can take up.
Region-DE
Further stretching of the bar is needed less force than ultimate force, and finally the component breaks into two parts at E.
Solid Mechanics
Look of actual stress strain diagrams
C toE BtoC Oto A∈ >∈ >∈
(1) Strains from O to A are
so small in comparison to the
strains from A to E that they
cannot be seen.
(2) The presence of well defined
yield point and subsequent large
plastic strains are characteristics of mild – steel.
(3) Metals such a structural steel that undergo large permanent strains before failure are classified as ductile metals.
Ex. Steel, aluminum, copper, nickel, brass, bronze, magnesium, lead etc.
Aluminum alloys – Offset method
(1) They do not have clear cut yield point.
(2) They have initial straight line portion with clear proportional limit.
(3) All does not have obvious
yield point, but undergoes
large permanent strains after
proportional limit.
(4) Arbitrary yield stress is
Solid Mechanics
determined by off- set method.
(5) Off-set yield stress is not material property
Elasticity & Plasticity
(1) The property of a material by which it (doesn’t) returns to its original dimensions during unloading is called (plasticity) elasticity and the material is said to be elastic (plastic).
(2) For most of the metals proportional limit = elastic limit.
(3) For practical purpose proportional limit = elastic limit = yield stress
(4 )All metals have some amount of straight line portion.
Solid Mechanics
Brittle material in tension
(1) Materials that fail in tension at relatively low values of strain are classified or brittle materials.
(2) Brittle materials fail with only little elongation (elastic) after the proportional limit.
(3)Fracture stress = Ultimate stress for brittle materials
(4)Up to B, i.e fracture strains are elastic.
(5)No plastic deformation in case of brittle materials.
Ex. Concrete, stone, cast iron, glass, ceramics
Ductile metals under compression
Solid Mechanics
(1) σ −∈ curves in compression differ from σ −∈ in tension.
(2)For ductile materials, the proportional limit and the initial portion of the σ −∈ curve is same in tension and compression.
(3)After yielding starts the behavior is different for tension and compression.
(4)In tension after yielding – specimen elongates – necking and fractures or rupture. In compression – specimen bulges out- with increasing load the specimen is flattened out and offers greatly increased resistance.
Brittle materials in compression
(1)Curves are similar both in tension and compression
(2)The proportional limit and ultimate stress i.e fracture stress are different.
(3)In case of compression both are greater than tension case
(4)Brittle material need not have linear portion always they can be non-linear also.
Solid Mechanics
11. Generalized Hooke’s Law
(1) A material behaves elastically and also exhibits a linear relationship between σ and ∈ is said to be linearly elastic.
(2) All most all engineering materials are linearly elastic up to their corresponding proportional limit.
(3) This type of behavior is extremely important in engineering – all structures designed to operate within this region.
(4) Within this region, we know that either in tension or compression
EStress in particular direction straininthat dir.X Eσ = ∈
=
E =Modulus of elasticity –Pa,N / m2
= Young’s modulus of elasticity.
(5) x xEσ = ∈ or y yEσ = ∈
(6) Eσ = ∈ is known as Hooke’s law.
(7) Hooke’s law is valid up to the proportional limit or within the linear elastic zone.
Solid Mechanics
Poisson’s ratio
When a prismatic bar is loaded in tension the axial elongation is accompanied by lateral contraction.
Lateral contraction or lateral strain
f o
o
d d
d
−′∈ = this comes out to be –ve
( ) lateral strainPoisson's ratio = nu
axial strainis perpendicular to
ν ′−∈=− =∈
′∈ ∈
If a bar is under tension ∈ +ve, ′∈ -ve and ν = +
If a bar is under compression ∈ -ve, ′∈ +ve and ν = +
ν =always +ve = material constant
For most metals . to . sν = 0 25 0 35
Concrete . to .ν = 0 1 0 2
Rubber .ν = 0 5
ν is same for tension and compression
ν is constant within the linearly elastic range.
Solid Mechanics
Hooks law in shear
(1)To plot ,Yτ the test is twisting of hollow circular tubes
(2) ,Yτ diagrams are (shape of them) similar in shape to tension test diagrams ( )Vsσ ∈ for the same material,
although they differ in magnitude.
(3)From Yτ − diagrams also we can obtain material properties proportional limit, modulus of elasticity, yield stress and ultimate stress.
(4)Properties are usually ½ of the tension properties.
(5)For many materials, the initial part o the shear stress diagram is a st. line through the origin just in case of tension.
GYτ = - Hooke’s law in shear
G =Shear modulus of elasticity or modulus of rigidity.
Pa or N / m s= 2
Proportional limit
Elastic limit
Yield stress
Ultimate stress
Material properties
τ
ϒ
Proportional limit
G1
Yield point
Solid Mechanics
E,v, and G → material properties – elastic constants - elastic
properties.
Basic assumptions solid mechanics
Fundamental assumptions of linear theory of elasticity are:
(a) The deformable body is a continuum
(b) The body is homogeneous
(c) The body is linearly elastic
(d) The body is isotropic
(e) The body undergoes small deformations.
Continuum
Completely filling up the region of space with matter it occupies with no empty space.
Because of this assumption quantities like
( )( )
( )
u u x,y ,z
x,y ,z
x,y ,z
σ σ=
=
∈=∈
Homogeneous
Elastic properties do not vary from point to point. For non-homogenous body
( )( )( )
E E x,y ,z
v v x,y ,z
G G x,y ,z
=
=
=
Solid Mechanics
Linearly elastic
Material follows Hooke’s law
EGY
v Constant
στ
= ∈==
Isotropic
Material properties are same in all directions at a point in the body
E C for all
C for allG C for all
θν θ
θ
===
1
2
3
The meaning is that
x x
y y
EE
σσ
= ∈= ∈
The material that is not isotropic is anisotropic
( )( )( )
E E
G G
θν ν θ
θ
===
The meaning is that
x x
y y
EE
E E
σσ
= ∈= ∈
≠
1
2
1 2
Solid Mechanics
Small deformations
(a) The displacements must be small
(b) The strains must also be small
Generalized Hooke’s law for isotropic material
We know the following quantities from the tension and shear testing.
E
Tensiletestv
σ = ∈��
′∈ �= − ��∈
GYτ = - Shear test or torsion test.
What are the stress –strain relation for an element subjected to 3D state of stress. i.e what is the generalized Hooke’s law.
Hooke’s law – when only one stress is acting
Generalized Hooke’s law – when more than one stress acting
We assume that
Material is linearly elastic, Homogeneous, Continuum, undergoing small deformations and isotropic.
For an isotropic material the following are true
(1)Normal stress can only generate normal strains.
- Normal stresses for reference xyz cannot produce Y of this
reference
Solid Mechanics
(2)A shear stress say xyτ can only produce the corresponding
shear strain xyY in the same coordinate system.
Principal of superposition:
This principle states that the effect of a given combined loading on a structure can be obtained by determining separately the effects of the various loads individually and combining the results obtained, provided the following conditions are satisfied.
(1)Each effect is linearly related to the load that produces it.
(2)The deformations must be small.
Solid Mechanics
Let us know consider only xσ is applied to the element. From Hooke’s we can write
xx
xy
xz
E
vE
vE
σ
σ
σ
∈ =
∈ = −
∈ = −
Solid Mechanics
Only yσ applied
yy
yx
yz
E
vE
vE
σ
σ
σ
∈ =
∈ = −
∈ = −
Similarly, zσ alone is applied
zz
zx
zy
E
vE
vE
σ
σ
σ
∈ =
∈ = −
∈ = −
Contribution to x∈ due to all three normal stresses is
yxx
v vE E E
σσ σ∈ = − − 3
( )( )
( )
x x y z
y y x z
x z x y
Therefore
vE
vE
vE
σ σ σ
σ σ σ
σ σ σ
� �∈ = − +� �
� �∈ = − +� �
� �∈ = − +� �
1
1
1
Normal strains are not affected by shear stresses
Solid Mechanics
Now let us apply only xyτ
xyxyY
G
τ=
Similarly because of yz xzandτ τ
yzyz
xzxz
YG
YG
τ
τ
=
=
Therefore, when all six components of stresses and strains are acting on an infinitesimal element or at a point then the relation between six components of stresses and strains is
( )( )
( )
x x y z
y y x z
x z x y
xyxy
yzyz
xzxz
vE
vE
vE
YG
YG
YG
σ σ σ
σ σ σ
σ σ σ
τ
τ
τ
� �∈ = − +� �
� �∈ = − +� �
� �∈ = − +� �
=
=
=
1
1
1
These six equations are known as generalized Hooke’s law for isotropic materials.
Solid Mechanics
Matrix representation of generalized Hooke’s law for isotropic materials is therefore,
x x
y y
z z
xy xy
yz yz
xz xz
v vE E Ev v
E E Ev v
E E EY
GY
YG
G
σσστττ
− −� �� �� �− −∈ � �� � � �� �� � � �∈ � �� � � �− −� �� � � �∈� � � �� �=� � � �� �� � � �� �� � � �� �� � � �� �� � � �� � � �� �� �� �� �
10 0 0
10 0 0
10 0 0
10 0 0 0 0
10 0 0 0 0
10 0 0 0 0
Stress components in terms of strains
( ) ( )( )
x y z x y z x y z
x y z
v sE E
veE
σ σ σ σ σ σ
σ σ σ
∈ + ∈ + ∈ = + + − + +
−� �= + + � �� �
1 2
1 2
x y z e∈ + ∈ +∈ =
( )x x x y zv vE
σ σ σ σ� �∈ = − − +� �1
( )x x y z xv vE
σ σ σ σ σ� �= − + + +� �1
( ) ( )x x y zv vE
σ σ σ σ� �= + − + +� �1
1
Solid Mechanics
( )( )
( )( )
xveEv
E v
v veE v
σ
σ
� �= + −� �−� �
× += −−
11
1 2
11 2
x xve E
v vσ � �∴ = ∈ +� �− +� �1 2 1
Ev
µ=+1
(mu) where ( )( )
Evv v
λ =+ −1 1 2
,λ µ are Lames constants
x x
y y
z z
xy xy xy
xy yz yz
xy zx zx
ee
eY G Y
Y G Y
Y G Y
σ λ µσ λ µσ λ µτ µτ µτ µ
= + ∈= + ∈
= + ∈= =
= =
= =
2
2
2
Lame’s constants have no physical meaning
Stress-strain relations for plane stress
Solid Mechanics
( )( )( )
x x
y y
xy xy
z yz zx
x,y
x,y
x,y
σ σσ
τ τσ τ τ
=
=∈
=
= = = 0
( )( )
( ) ( )
x x y
y y x
z x y x y
xyxy
yz xz
vE
vE
v vE v
YG
Y Y
σ σ
σ σ
σ σ
τ
∈ = −
∈ = −
−∈ = − + = ∈ +∈−
=
= =
1
1
1
0
Stress- strain relations for plane strain
( )( )
( )
x x
y y
xy xy
xz yz
x y
x,y
x,y
Y Y x,y
Y Y
e
∈ =∈
∈ =∈
=
∈ = = =
=∈ +∈3 0
( )( )
( ) ( )( )( )
( )( )
x x x
y y y
z x y z
x y
e x,y
e x,y
v x,y
v e e
ve
v
σ λ µ σσ λ µ σ
σ σ σ σ
λ µλ µ
λ µ
= + ∈ =
= + ∈ =
= − + =
= − += − +
= − + ∈ + ∈
2
2
2
xy xy
xz yz
GYττ τ
=
= = 0
Solid Mechanics
• Therefore, the stress transformation equations for plane stress can also be used for the stresses in plane strain.
• The transformation laws for plane strain can also be used for the strains in plane stress. z∈ does not effect geometrical relationships used in derivation.
Example of Generalized Hooke’s law
Principal stress and strain directions of isotropic materials
τ is zero along those planes, therefore Y is also zero along these planes i.e normal strains of the element are principal strains. For isotropic materials - the principal strains and principal stresses occurs in the same direction.
σ λ µ� �∈ = − − ∈� �x x yv e vE1
σ σ=x y2
σ σ
σ λ µ
σ σ
σ λ µ
� �∈ = −� �
= + ∈
� �∈ = −� �
= + ∈
x x y
x x
y y x
y y
vE
e
vE
e
1
1
σ σ= −x y
σ σ
σ
� �∈ = +� �
+� = ��
x x y
x
vE
vE
1
1
Solid Mechanics
12. Volumetric strain and Bulk modulus Relation between E, andGν
( )
( )
xy
xy
vE
vE
σ τ σ σ
σ τ σ σ
= ∈ = −
= − ∈ = −
1 1 1 2
2 2 2 1
1
1
( ) ( )
( )
xyxy xy
xy
xy xyxy
xy
vv
E Ev
EY
G
G
ττ τ
τ
τ
τ
+∈ = + =
− +∈ =
∈ =∈ = =
−∈ =
1
2
1
2
11
1
2 2
2
( )xy xyv
E G
τ τ+= �
1
2
( )EG
v=
+2 1
Only two elastic constants are independent.
Solid Mechanics
Volumetric strain-dilatation
Consider a stress element size dx,dy ,dz
dv dxdydz=
After deformations
( )( )( )
*x
*y
*z
dx dx
dy dy
dz dz
= + ∈
= + ∈
= + ∈
1
1
1
In addition to the changes of length of the sides, the element also distorts so that right angles no longer remain sight angles. For simplicity consider only xyY .
The volume *dv of the deformed element is then given by
( )* * * * *dv Area OA B C dz= ×
( ) ( )* * * * *xy
* *xy
Area OA B C dx dy CosY
dx dy CosY
=
=
* * * *xydv dx dy dz CosY∴ =
For small xy xyY CosY ≈ 1
( )( )( )
* * * *
x y z
dv dx dy dz Volumechangedoesn't depend onY
dxdydz
∴ = −
= + ∈ +∈ + ∈1 1 1
dropping all second order infinitesimal terms
Solid Mechanics
( )*x y zdv dxdydz= + ∈ + ∈ + ∈1
Now, analogous to normal strain, we define the measure of volumetric strain as
final volume-initial volumeVolumetric strain
initialvolume=
*dv dvedv−=
x y ze =∈ + ∈ + ∈
• e =volumetric strain = dilatation. This expression is valid only for infinitesimal strains and rotations
• x y ze J first in varianceof strain.=∈ + ∈ +∈ = =1
• Volumetric strain is scalar quantity and does not depend on orientation of coordinate system.
• Dilatation is zero for state of pure shear.
Bulk modulus of elasticity
( ) ( )x y z x y zv
Eσ σ σ−∈ + ∈ + ∈ = + +1 2
Mean stress ( )x y zσ σ σ σ= = + +13
( )ve
Eσ−= 3 1 2
Keσ =
Solid Mechanics
Where ( )
EKv
=−3 1 2
bulk modulus of elasticity.
• Bulk modulus is widely used in fluid mechanics.
• From physical reasoning E ,G ,K> > ≥0 0 0
Steel : E = 200 Gpa
v = 0.3
Al : E = 70 Gpa
v = 0.33
Copper: E = 100 Gpa
v = 0.35
( )( )
EG SinG Eand Gv
v v
= >++ > → > −
02 1
1 0 1
Similarly SinG E & K> ≥0 0
( )EK v v .
v= → − ≥ → ≤
−1 2 0 0 5
3 1 2
∴ Theoretical bounds on v are
v .− < ≤1 0 5
asv . K α→ →0 5 and 0C → material is incompressible.
Solid Mechanics
13. Axially loaded members
Solid Mechanics
Geometry, locating and material properties
• A prismatic bar is subjected to axial loading
• A prismatic bar is a st. structural member having constant cross-section through out it length.
• Bar or rod →length of the member is � cross sectional dimensions.
Axial force is a load directed along the axis of the member – can create tension or compression in the member.
Typical cross sections of the members
- Solid Sections
- Hollow Sections
Solid Mechanics
Material properties: The member is homogenous linearly elastic and isotropic material.
Stresses, strains and deformations
Consider a prismatic bar of constant cross-sectional area A and length L, with material properties A & v. Let the rod be subjected to an axial force “p”, which acts along x-axis.
x y z
y z
F PM M M
V V
== = =
= =
0
0
The right of the section m-m exerts elementary forces or stresses on to the left of the section to maintain the equilibrium. Sum of all these elementary forces must be equal to the resultant F.
- Other sections
Solid Mechanics
xA
y x
z x
dA F
M zdA
M ydA
σ
σ
σ
=
= =
= − =
�
�
�
0
0
Above equation must be satisfied at every cross-section, however, it does not tell how xσ is distributed in the cross-section.
The distribution cannot determine by the methods of static or equations of equilibrium- statically indeterminate
To know about the distribution of xσ in any given section, it is necessary to consider the deformations resulting from the application of loads.
Since the body needs to develop only xσ component in order to maintain equilibrium, therefore the state of stress at any point of prismatic rod is
x
ij
σσ
� �� �� �=� � � �� �� �
0 00 0 00 0 0
Solid Mechanics
We make the following assumptions on deformation based on experimental evidence
(1)The axis of the bar remains straight after deformation
(2)All plane cross-sections remain plane and perpendicular to the axis of the bar
Key kinematical assumptions
• As a result of the above kinematic assumptions all points in a given y-z plane have the same displacements in the x-direction.
• Any line segment AB undergoes same strain x∈ therefore
x∈ cannot be a function of y or z, but at most is a function of x- only.
In the present case situation is same at all cross-sections of the prismatic bar, therefore
x Constant∈ =
at all points of the body i.e x∈ is also no a function of x.
Solid Mechanics
Since we are studying a homogenous, linearly elastic and isotropic prismatic bar
( )( )
( )
x x y z
y y x z
z z x y
vE
vE
vE
σ σ σ
σ σ σ
σ σ σ
� �∈ = − − →� �
� �∈ = − − →� �
� �∈ = − − →� �
1
1
1
In the present case, x∈ is independent of y and z coordinates, therefore xσ is also independent of y and z coordinates i.e
xσ is uniformly distributed in a cross-section
Moreover throughout the bar.
We know that internal resultant force
xA
F dAσ= �
Since xσ is a independent of y & z
xx
y x
z x
EVEVE
σ
σ
σ
∈ =
∈ = −
∈ = −
x xE Constantσ = ∈ =
Solid Mechanics
A
F da Aσ σ= =�
∴ F PA A
σ = =
y xA A
z xA A
M .zdA zdA
M .ydA ydA
σ
σ
= = � =
= − = � =
� �
� �
0 0
0 0 (1)
Eq. (1) indicates that moment are taken about the centroid of the cross-section.
Elongation or Contraction
xx
PE AE
σ∈ = =
Total elongation of the rod
( ) ( )L L
xP PLu L u da dx
AE AEδ− = = ∈ = =� �
0 0
0
Solid Mechanics
xPA
PLAE
AE Axialrigidity
σ
δ
=
=
=
If A,E and P are functions of x then
( )( ) ( )
L P xdx
A x E xδ = �
0
Stiffness and flexibility
These are useful in computer analysis of structural members.
kf
= 1
AEk
L= L
fAE
=
P kSS fP
==
Solid Mechanics
Extension of results: Non-uniform bars (non-prismatic)
For a prismatic bar
This is exact solution for prismatic bar.
( )( )
( )( )
( )( ) ( )
x
L
P x F xA x A x
P xS dx
A x E x
Approximate expression
σ = =
= �0
The above formula becomes a good approximation for uniformly varying cross-sectional area ( )A x member.
Above formula is quite satisfactory if the angle of taper is small
Plane sections remain plane and perpendicular to the x- axis is no longer valid for the case of non-prismatic rods.
xP PL
&A AE
σ δ= =
Solid Mechanics
( ) ( )x x yxF b y b xΣ σ τ= � ∆ − ∆ =0 0
( )xy yx xy
x . sx
τ τ σ ∆= =∆
Taking x∆ → 0 , we note that yxτ → 0 only if yx
∆ →∆
0 i.e at the
slope of the upper surface of the rod tends to zero.
Solid Mechanics
Case2
( )A BBC
AAB
P P LPLAE A EPL P LAE A E
δ
δ
− += =
−= =
2
2 2
1
1 1
( )A BBC
AB A
P PA
P / A
σ
σ
+= −
= −2
1
CA BC ABS Sδ = +
This method can be used when a bar consists of several prismatic segments each having different material, each having different axial forces, different dimensions and different materials. The change in length may be obtained from the equation
ni i i
ii i ii
PL PandA E A
δ σ=
= =�1
Solid Mechanics
Statically indeterminate problems
Equilibrium
y
a a s
F
F F F P
Σ� �=� �
+ + − =1 2
0
0
[ ]C
a a
MbF bF
Σ =− =1 2
0
0
(1)
For statically indeterminate problems we must consider the deformation of the entire system to obtain “compatibility equation”
The rigid plate must be horizontal after deformation
s A geometric compatibility equationδ δ= �
s s A As A
s s A A
F L F LandA E E A
δ δ= =
Then using the geometry compatibility
(2)
a aF F=1 2
a sF F P+ =2
s Aδ δ= � A A s As
A A s s
F L F LE A E A
=
Solid Mechanics
By solving (1) & (2) we can obtain internal forces sF & AF
Stresses in axially loaded members
Uniaxial state stress is a special case of plane stress
xij
σσ � �� �= � �� �
� �
00 0
x
xmax
σ σσ στ
=
= =
1
12 2
Occurs at 45�to x y− or x z− planes.
Solid Mechanics
A −Principal stress elements
B,C −maximum shear stress elements.
Ductile material are weak in shear. They fail along maxτ planes.
Brittle materials weak in normal tensile stresses. They fail along σ1 planes.
Limitations of analysis
xP PL
& SA AE
σ = =
(1)They are exact for long prismatic bars of any cross-section, when axial force is applied at the centroid of the end cross-sections.
Solid Mechanics
(2)They should not be employed (especially xPA
σ = ) at
concentrated loads and in the regions of geometric discontinuity.
(3)They provide good approximation if the taper is small.
(4)Above equations should not be applied for the case of relatively short rods.
(5)They are exact for relatively short members under compressive loading.
Solid Mechanics
Stress concentrations
• High stresses are known as stress concentrations
• Sources of stress concentrations- stress raisers
• Stress concentrations are due to :
(1)Concentrated loads
(2)Geometric discontinuities
Stress concentration due to concentrated loads
max
ave
nom
Stress concentration factor=K
Pbt
σσ
σ
=
=
Solid Mechanics
Stress concentration due to hole
Discontinuities of cross section may result in high localized or concentrated stresses.
maxnom
nom
PKdt
K Stressconcentration factor
σ σσ
= =
=
Solid Mechanics
Stress Concentration due to fillet
maxave
ave
PKdt
σ σσ
= =
Solid Mechanics
14. Torsion of circular bars Geometry, loading and Material properties
A prismatic bar of circular cross- section subjected to equal and opposite torques acting at the ends.
Whenever torques act on a member, then it will be twisted.
Torsion refers to the twisting of a straight bar when it is loaded by torques.
Material: Homogeneous, linearly elastic, and isotropic undergoing small deformations.
Presently theory is valid only for
Stresses and strains in polar coordinates
Stresses, strains and displacements in polar coordinates.
Since we are dealing with a circular member it is preferable to use polar coordinates
Solid Mechanics
r r rx
ij r x
xr x x
θ
θ θ θ
θ
σ τ τσ τ σ τ
τ τ σ
� �� �� �=� � � �� �� �
( )
( )
( )
x x r
r r x
x r x
rQ x rxr x x xr rx
vE
vE
vE
Y ; Y Y ; Y YG G G
θ
θ
θθ θ θ
σ σ σ
σ σ σ
σθ σ σ
τ τ τ
∈ = − +� �� �
∈ = − +� �� �
∈ = − +� �� �
= = = = =
1
1
1
Equilibrium and elementary forces
Since every cross-section of the bar is identical and since every cross-section is subjected to the same internal torque “T”, then the bar is said to be under “pure torsion”
To keep the body under equilibrium, elementary forces
xdF dAθτ= are only forces that are required to be exerted by the other section, so that
x y z y z
x
F V V M M
M T T
= = = = =
= = 0
0
Solid Mechanics
(1)
Direction of zθτ can be obtained from the direction of internal torque T at that section.
The state of stress in pure torsion is therefore
While the relation in (1) express an important condition that must be satisfied by the shearing stresses xQτ in any given
cross-section of the bar it does not tell how these stresses are distributed in the cross-section.
The actual distribution of stresses under a given load is statically indeterminate. So we must know about the deformation of the bar.
Presence of xθτ in polar coordinates means, presence of
xy xQ
xz xQ
Cos
Sin
τ τ θτ τ θ
=
=
x
xA
dT dF r rdA
T rdA
T T
θ
θ
σ
τ
= × =
=
=
�
0
x
x
θ
θ
ττ
� �� �� �� �� �
0 0 00 00 0
Solid Mechanics
Therefore the state of stress in case pure torsion in terms of rectangular stress components is then
xy xz
yx
zx
τ τττ
� �� �� �� �� �
0
0 0
0 0
- state of pure shear.
We must then ensure that
Deformation in pure torsion
Following observations can be made from the deformation of a circular bar subjected to equal and opposite end torques.
(1)The cross-sections of the bar do not change in shape i.e they remain circular.
(2)A line parallel to the x- axis or longitudinal line become a helical curve.
(3)All cross-sections remain plane.
(4)All cross-sections rotate about the axis of the bar as a solid rigid slab.
y xy
z xz
V dA
V dA
τ
τ
= =
= =�
�
0
0
Solid Mechanics
(5)However, various cross-sections along the bar rotate through different amount.
(6)The radial lines remain radial lines after deformation
(7)Neither the length of the bar nor the length of radius will change.
These are especially of circular bars only. Not true for non-circular bars.
Assumptions on deformation for pure torsion
(1)All cross –sections rotate with respect to the axis of the circular bar i.e x-axis.
(2)All cross-sections remain plane and remain perpendicular to the axis of the bar.
(3)Radial lines remain straight after the deformation.
(4)Neither the length of the bar nor its radius will change during the deformation.
These assumptions are correct only if the circular bar undergoes “small deformations” only.
Variation of shear strain ( xY θ )
Because of T0 , the right end will rotate through an infinitesimal angle
φ - angle of twist.
Solid Mechanics
*φ - varies along the axis of the bar.
angle of twist per unit length.
xQY dx Ydx rdφ= =
ddxφ = − rate of twist
xQY is independent of x and
dY rdxφ=
Solid Mechanics
In case of pure torsion the shear strain Y varies linearly with “r”
Maximum shear strain Y occurs at the outer surface of the circular bar i.e., r R=
Shear strain is zero at the center of the bar.
The equation dY rdxφ= is strictly valid to circular bars having
small deformations.
If the material is linearly elastic
Therefore, variation of shear stress xQτ in pure torsion is
given by
Shear stress τ is only function of “r” and varies linearly with radius r of the circular bar.
maxdY Rdxφ=
GYτ =
xQ xQdGY GYdxφτ τ= = =
maxmax xQdRGdxφτ τ= =
Solid Mechanics
The torsion formula
Relation between internal torque T and shear stressτ
A
T rdA
dT Gr rdAdx
τ
φ
=
=
�
�
Since G & ddxφ
are independent of area A then
A
dT G r dAdxφ= �
2
For solid circular bar,
PdT GIdxφ∴ =
∴
But dGrdxφτ =
P
TGr GIτ =
PA
I r dA
Polar moment of inertiaof across sec tion
=
−
�2
PI Dπ= 4
32 PI Rπ= 4
2
P
d Tdx GIφ = =
P
TrI
Torsion formula
τ =
Solid Mechanics
This is the relation between shear stresses xQτ and torque T
existing at the section.
Torsion formula is independent of material property.
Angles of twist
We now determine the relative rotation of any two cross-sections
P
d Tdx GIφ= =
maxmax xQP
TRI
τ τ= =
maxT
Dfor solidcircular bars
τπ
= 316
B
A
x
B / A B APx
Tdx
GIφ φ φ= − = �
Solid Mechanics
In case of prismatic circular bar subjected to equal opposite torques at the ends
Direction of φ at a section is same as that of T
Since P
d Tdx GIφ= = then, in case of pure torsion.
Thus in case of pure torsion ( )xφ varies linearly with x
In case of torsion
The product
Load
displacement
PGI −Torsional rigidity
B / A B AP
B A
TL nGI
if x x Lpuretorsion
φ φ φ= − =
− =
P P
TL T LGI GI
φ = = 0
d constantdx Lφ φ= = =
P
TLGI
φ =
P
P
GI Lk ; fL GI
= =
Solid Mechanics
xy xQ
xz xQ
Cos
Sin
τ τ θτ τ θ
=
=
We should ensure that distribution of xQτ should also gives
y zV V= = 0
y xy xA A
R
yP
R
P
V dA Cos dA
TrV Cos drdI
T rCos drdI
θ
π
π
τ τ θ
θ θ
θ θ
= =
=
= =
� �
� �
� �
2
0 02
0 0
0
R
zP
TV rSin drdI
πθ θ= =� �
2
0 0
0
Hollow circular bars: The deformation of hollow circular bars and solid circular bars are same. The key kinematic assumptions are valid for any circular bar, either solid or hollow. Therefore all equations of solid circular bars can be employed for hollow circular bars, instead of using
yV = 0∴
zV = 0∴
Solid Mechanics
Hollow bars are move efficient than solid bars of same “A”.
• Most of the material in soild shaft is stressed below the maximum stress and also have smaller moment arm “r”.
• In hollow tube most of the material is near the outer boundary, where τ is maximum values and has large moment arms “r”.
( )
P
P
o i
TrI
I D solid
D D hollow
τ
π
π
=
= −
= − −
4
4 4
32
32
( )P
P o i
I D Soild
I D D hollow
π
π
= −
= − −
4
4 4
32
32
omax
P
imin
P
TRI
TRI
τ
τ
=
=
Solid Mechanics
omax
P P
imin
P
TR TR;I I
TRI
τ
τ
=
=
( )
YG
,Y f r
τ
τ
=
−
P
d Tdx GIφ= =
B AB / AP
B A
TLGI
L x xconstantlinearly with x
φ φ φ
φ
= − =
= −==
(4) If weight reduction and savings of materials are important, it is advisable to use a circular tube.
(5) Ex large drive shafts, propeller shafts, and generator shafts usually have hollow circular cross sections.
Extension of results
Case-I Bar with continuously varying cross-sections and continuously varying torque
• Pure torsion refers to torsion of prismatic bar subjected to torques acting only at the ends.
Solid Mechanics
• All expressions are developed based on the key kinematic assumptions, these are therefore, strictly valid only for prismatic circular bars.
The above equations yield good approximations to the exact solution, provide if ( )R x doesn’t vary sharply with x.
( ) ( )( )
( ) ( )( )
( )( )
B
A
P
Px
B A B / APx
T x rx
I x
T xdxdx GI x
T xdx
GI x
τ
φ
φ φ φ
=
= =
− = = �
Solid Mechanics
Some special cases
( )
( )
( )( )
P
P
TrxI x
TxGI x
τ =
=
( ) ( )
( ) ( )P
P
T x rx
IT x
xGI
τ =
=
Case II
i
i ii
P
T rI
τ =
i
ni i
B / Ai Pi
T LG I
φ=
=�1
Solid Mechanics
Statically indeterminate problems
(1)
We note that within AAB, T T= and
within CBC T T=
• To solve the problem we must consider geometry of deformation to formulate the compatibility equation.
• Clearly the rotation of section B with respect to A must be same as that with respect to C i.e
AB BC
A AB C BCB / A B / C
AB P BC P
T L T L;G I G I
φ φ= =
(2)
A CT T T+ + = 0[ ]xMΣ = 0
B / A B / C
Compatibility equation
φ φ=
AB BC
A AB C BC
AB P BC P
T L T LG I G I
=
Solid Mechanics
Stresses in pure torsion
If a torsion bar is made up of brittle material, which is generally weak in tension, failure will occur in tension along
a helix inclined at 45�to the axis.
Ductile materials generally fail in shear. When subjected to torsion, a ductile circular bar breaks along a plane perpendicular to its longitudinal axis or x – axis.
Solid Mechanics
xPA
σ =
Torsion testing m/c
Solid Mechanics
Combined loading or combined stress
Principal of superposition
maxP
TRI
τ =x
PA
σ =
Solid Mechanics
Stress concentrations in torsion
Stress concentration effect is greatest at section B-B
avg nomT
K KD
τ τ τπ�
= = = ��
1 31
16
max avg nomK Kτ τ τ= =
Solid Mechanics
Limitations of torsion formulae
(1)The above solutions are exact for pure torsion of circular members (solid or hollow section)
(2)Above equations can be applied to bars (solid or hollow) with varying cross-sections only when changes in ( )R x are small and gradual.
(3)Stresses determined from the torsion formula are valid in regions of the bar away from stress concentrations, which are high localized stresses that occur whenever diameter changes abruptly and whenever concentrated torque are applied.
(4)It is important to recognize that, the above equation should not be used for bars of other shapes. Noncircular bars under torsion are entirely different from circular bars.
P P P
Tr T TL d, ; ;Y r
I GI GI dxφτ φ= = = =
Solid Mechanics
15. Symmetrical bending of beams Some basics
• Transverse loads or lateral loads: Forces or moments having their vectors perpendicular to the axis of the bar.
• Classification of structural members.
• Axially loaded bars :- Supports forces having their vectors directed along the axis of the bar.
• Bar in tension:- Supports torques having their moment vectors directed along the axis.
• Beams :- Subjected to lateral loads.
• Beams undergo bending (flexure) because of lateral loads.
Solid Mechanics
Roughly speaking, “bending” refers to a change in shape from a straight configuration to a non straight configuration.
Bending moments i.e zM and yM are responsible for
bending of beams.
The loads acting on a beam cause the beam to bend or flex, thereby deforming its axis into a curve-known as “ deflection curve” of the beam.
If all points inx y− plane remain in the xy − plane after deformation i.e after bending then xy − plane is known as “plane of bending”.
If a beam bend in a particular plane, then the deflection curve is a plane curve lying in the plane of bending.
Solid Mechanics
The y −direction displacement [i.e. v −component] of any point along its axis is known as the “deflection of the beam”.
Pure bending and non-uniform bending
If the internal bending moment is constant at all sections then beam is said to be under “pure bending”.
dM Vdx
= −
Pure bending (i.e., M=constant) occurs only in regions of a beam where the shear force is zero.
If ( )M M x= it is non- uniform bending
Solid Mechanics
Curvature of a beam
When loads are applied to the beam, if it bends in a plane say xy −plane, then its longitudinal axis is deformed into a curve.
O − Center of curvature
R − Radius of curvature
kR
= =1 Curvature
in general ( )R R x= and ( )k k x= .
RdQ dS=
dQk
R dS= =1
for any amount of R
The deflections of beams are very small under small deformation condition. small deflections means that the deflection curve is nearly flat.
under small deformations.
dQkR dX
= =1
Solid Mechanics
It is given that deflections at A and B should be zero.
Symmetrical bending of beams in a state of pure bending
Geometry, loading and material properties
A long prismatic member possess a plane of symmetry subjected to equal and opposite couples M0 (or bending moments) acting in the same plane of symmetry.
Solid Mechanics
Initially we choose origin of the coordinate system “O ” is not at the centroid of the cross-section.
The y −axis passing through the cross-section is an axis of symmetry. The XY plane is the plane of symmetry.
Material is homogeneous, linearly elastic and isotropic undergoing small deformations.
Stresses in symmetric member in pure bending
x y z
x y
z
F V V
M M
M M M
= = =
= =
= = 0
0
0
Solid Mechanics
Therefore, xdAσ are the only elementary forces that are required to be developed by right of the section on to the left of the section.
The distribution of Xσ any section should satisfy
x x
y x
z x
F dA
M z dA
M M y dA M
σ
σ
σ
= � =
= � =
= � − =
�
�
�
0 0
0 0
Actual distribution of stresses - cannot by statics - statically indeterminate - deformations should be considered.
Thus, the state of stress at any point within a prismatic beam (cross-section having an axis of symmetry) subjected to pure bending is a uniaxial state of stress.
xM y dAσ= −�
x
ij
σσ
� �� �� �=� � � �� �� �
0 00 0 00 0 0
Solid Mechanics
Deformations in a symmetric member in pure bending
Since the member is subjected to bending moments, it will bend under the action of these couples.
Since, the prismatic member possessing a plane of symmetry (i.e xy- plane) and subjected to equal and opposite couples M0 acting in the plane of symmetry, the member will bend in the plane of symmetry (i.e xy plane).
The curvature k at a particular point on the axis of the beam depends on the bending moment at that point. Therefore a prismatic beam in pure bending will have constant curvature.
The line AB, which was originally a straight line, will be transformed in to a circle of center O and so the line A B′ ′.
Solid Mechanics
Decrease in length of AB and increase in length of A B′ ′ in positive bending.
Cross-sections which are plane and ⊥ to the axis of the undeformed beam, remain plane and remain⊥ to the axis of the deformed beam i.e to the deflection curve.
Kinematic assumption
Variation of strain and M κ− relation
Elementary theory of bending or Euler-Bernoulli theory
Under the action of M0 , the beam deflects in the xy – plane (plane of symmetry) and any longitudinal fibers such as SS bent into a circular curve. The beam is bent concave upward (due to +ve bending) upon which is a +ve curvature.
Solid Mechanics
Cross-sections mn and pq remain plane and normal to the longitudinal axis of the beam. Cross-sections mn and pq rotate with respect to each other about z-axis.
Lower part of the beam is intension and upper part is in compression.
The x- axis lies along the neutral surface of undeformed beam
Variation of strain and M-k relations (contd.)
Initial length of fiber ef dx=
Final length of ( )* *ef e f R y dQ= = −
The distance dx between two planes is unchanged at the neutral surface,
dQRdQ dx k
R dx= � = =1
Solid Mechanics
Therefore, the longitudinal strain i.e x∈ at a distance “y” from the neutral axis is
( )* *
xR y dQ dxe f ef y
ef dx R− −− −∈ = = =
In case of pure bending ( ) ( )x x x xx and z , y∈ ≠∈ ∈ =∈
The preceding equation shows that the longitudinal strains ( )x∈ in the beam (in pure bending) are proportional to the
curvature and vary linearly with the distance y from the neutral axis or neutral surface.
x∈ = 0 at the neutral surface
Maximum compressive xyR
−∈ = 1
Maximum tensile xyR
+∈ = 2
However, we still do not know the location of neutral axis or neutral surface.
xyR
∴ ∈ = − � x ky∈ = −
Solid Mechanics
Stresses in beams in pure bending :- For linearly elastic and isotropic beam material
( ) xyx x y z xyv Y
E G
τσ σ σ� �∈ = − + =� �
1
( ) yzy y x z yzv Y
E G
τσ σ σ� �∈ = − + =� �
1
( ) zxz z x y xzv Y
E Gτσ σ σ� �∈ = − + =� �
1
The state of the stress at any point within a prismatic beam in pure bending is
x
ij
σσ
� �� �� �=� � � �� �� �
0 00 0 00 0 0
x xEy
E EkyR
σ −∴ = ∈ = = −
y x x
z x x
V VEV VE
σ
σ
∈ = − = − ∈
∈ = − = − ∈
From the above equation
( )( ) ( )
x
x x x
x
x
x,z
y ylinear f (y )linear f (y )
i.e.,var y linearly with the distance y from the neutral surface
σ σσ σ
σ
≠= ∈ =∈
∈ =∴ =
�
�
Solid Mechanics
xσ at y = 0 i.e on the neutral surface = 0
Maximum compressive xEC
Rσ = − 1
Maximum tensile xEC
Rσ = 2
Maximum normal stress xσ occurs at the points farthest from the neutral axis.
In order to compute the stresses and strain we must locate the neutral axis of the cross-section.
Solid Mechanics
Location of neutral axis
We must satisfy the following equations at any given section m-m
Considering first equation
The above equation shows that the distance y between neutral axis and centroid “C” of a cross-section is zero.
In other words, the neutral axis i.e z-axis pass through the centroid of the cross-section, provided if the material follows Hooke’s law.
x
x z
x y
dA
ydA M M M
zdA M
σ
σ
σ
=
− = = =
= =
�
�
�
0
0
0
xA A
A
EydA
R
ydA
σ = − =
=
� �
�
0
0
Solid Mechanics
The origin ‘O’ of coordinates is located at the centroid of the cross-sectional area.
Thus, when a prismatic beam of linearly elastic material is subjected to pure bending, the y and z (neutral axis) axes are principal centroidal axes.
Moment – Curvature relationship
Moment of inertia of cross-sectional area about neutral axis
Moment-Curvature relation
xA
M ydAσ= − �
A
EyM ydA
R= + �
A
EM y dA
R= �
2
zzA
y dA I= =�2
EIMR
∴ =
Mk
R EI= =1
Mk
R EI= = 01
Solid Mechanics
Curvature k is directly proportional to M- internal bending moment and inversely proportional to EI- flexural rigidity of the beam.
Flexural rigidity is a measure of the resistance of a beam to bending.
Relation between xσ and M - Flexure formula
x Ekyσ = −
and MkEI
=
- flexure formula.
Stresses evaluated from flexure formula are called bending stresses or flexural stresses.
xMy
Iσ∴ = −
Solid Mechanics
The maximum tensile and compressive bending stresses occur at points located farthest from the neutral axis.
The maximum normal stresses are
Cross- sectional properties of some common shapes
-Section moduli
S =Section modulus
z − axis – neutral axis
MC MI S
σ −= = −11
1
MC MI S
σ = =22
2
I IS and SC C
= =1 21 2
Solid Mechanics
zzbh bhI S= =
3 2
12 6
zzdI d Sπ π= =
34
64 32
zzbhI
h b / for eqilateral triangle
=
=
3
363 2
zzI . r= 40 1098
Solid Mechanics
Distribution xσ on various cross-sections
maxMS
σ =
max
ISy
=
alllowM Sσ=
square
circle
S.
S= 1 18
Solid Mechanics
• This result shows that a beam of square cross-section is more efficient in resisting bending then circular beam of same area.
• A circle has a relatively larger amount of material located near the neutral axis. This material is less highly stresses.
• I - Section is more efficient then a rectangular cross-section of the same area and height, because I - section has most of the material in the flanges at the greatest available distance from the neutral axis.
Extension of results
Long prismatic beam under pure bending, and symmetrical bending.
Elementary theory of bending
( )M M xM Constant
≠=
( )x
zz
Myy
II I
MkR EI
σ = −
=
= =1
xx
y x
z z
Ev
v
σ∈ =
∈ = − ∈
∈ = − ∈
Solid Mechanics
Bending of beams due to applied lateral loads
Consider now a beam subjected to typical arbitrary transverse loads acting. In this case the interval bending moment ( )M M x= and ( )V x ≠ 0, and thus we have non-uniform bending.
Non-uniform bending is a result of presence of transverse shear force ( )V y . If ( )V y = 0 then M = constant.
It can be shown that the above results can also be used for non-uniform bending problems.
dM Vdx
= −
( ) ( )
( )( )
xM x y
x,yI
M xk
R x EI
σ −=
= =1
( ) ( )xx
y x
z x
x,yx,y
Eσ
νν
∈ =
∈ = − ∈
∈ = − ∈
Solid Mechanics
The above results can also be used for non-uniform bending problems provided if they satisfy the following conditions.
• The cross-sections should have y-axis of symmetry
• All applied transverse or lateral loads should lie in the x-y plane of symmetry and all applied couples act about z-axis only.
• L h longslender beams− −�
• Bending that conforms to conditions (i) and (ii) is called symmetrical bending.
If these three conditions are satisfied then one can employ the following expressions for non-uniform bending as-well
Solid Mechanics
Application of above equations to the non-uniform bending problems is equivalent to the following two assumptions.
(a)That even under such loading conditions, plane sections still remain plane after deformation and they remain ⊥ to the deformed longitudinal axis or neutral surface.
Bending stresses in a non-prismatic beam
The above equation can also be applied to the case of non-prismatic beam subjected to either pure or non-uniform bending, provided cross-sectional properties do not vary sharply.
( ) ( )
( )( )
( )
x
zz
M x yx,y
II I
M xk x
R x EI
σ = −
=
= =1
( )
( )( )
xx
y x
z z
x,yE
x,y v
x,y v
σ∈ =
∈ = − ∈
∈ = − ∈
( )( )
( )( )
( )( )
xM x y
I xM x
k xR x EI x
σ = −
= =1
Solid Mechanics
Problem
Determine the maximum tensile and compressive stresses in the beam due to the uniform load.
Solution
Centroid :-
2A mm y 3yA mm
1 × =20 90 1800 50 × 390 10
2 × =40 30 1200 20 × 324 10
A AΣ= = 3000 yAΣ = × 3114 10
Ay yAΣ= y = × �33000 114 10� y mm= 38
( )zzI I I Ad sΣ= = + 2
bh Ad�
= + ��
= × + × + × × + ×
32
3 2 2 2
12
1 190 20 1800 12 30 40 1200 18
12 12
4 4zzI I mm m−= = × = ×3 9868 10 868 10
Solid Mechanics
C mm=1 22 and C mm=2 38
x
maxmax
MyI
M I: SS y
σ
σ
= −
= =
At maximum +ve bending moment i.e at (D)
at D:
At maximum -ve moment i.e at (B)
IS .C
IS .C
−−
−
−−
−
×= = = ××
×= = = ××
96
1 31
96
2 32
868 1039 45 10
22 10
868 1022 84 10
38 10
maxtM .s .
σ −= =× 6
2
1 89822 84 10
maxt. MPaσ = 83 1
maxCM .s .
σ −= =× 6
1
1 89839 45 10
maxC. MPaσ = 48 11
maxtM . . MPas .
σ −= = =× 6
1
3 37585 55
39 45 10
maxCM . . MPas .
σ −= = =× 6
2
3 375147 8
22 84 10
max maxt C. and . MPaσ σ= =85 55 147 8
Solid Mechanics
Problem
a wooden member of length L = 3m having a rectangular cross-section 3 cm × 6 cm is to be used as a cantilever with a load P = 240N acting at the free end. Can the member carry this load if the allowable flexural stress both in tension and in compression is allowσ = 50 Mpa ?
Solution
maxM N-m= 720
A. .S m
.−×= = ×
36 31 0 06 0 03
9 1012 0 015
max maxt CA A
M PLS S
σ σ= = =
max maxt C allowσ σ σ= =
∴The beam can carry P N= 240 only when oriented as in (B)
allow Aalow
SP N
Lσ ×= = 150
B. .S . m
.−×= = ×
35 31 0 03 0 06
1 8 1012 0 03
allow Balow
SP NL
σ ×= = 300
Solid Mechanics
Limitations
(1)The flexure formula is exact for a prismatic beam in pure bending.
(2)It provides very good approximation of xσ for long slender beams (L h)>> under symmetrical bending.
(3)The flexure formula can be employed for any shape of the cross-section, provided the cross-section has y-axis of symmetry.
(4)It should not be employed in regions close to geometric discontinuities and concentrated loads.
Solid Mechanics
16. Shear Stresses in Beams
( )y xyA
V x dAτ= �
It is reasonable to assume that
(1)The shear stresses acting on the cross-section are parallel to the shear force ( )yV x i.e ⊥ to the line PQ
(2)It is also reasonable to assume that the shear stresses xyτ
are uniformly distributed across the width of the beam, so that xM T= = 0 for symmetrical bending
( )( ) ( )
xy xy
y xyA
x,y such thats
V x x,y dA
τ τ
τ
∴ =
= �
Solid Mechanics
• Thus, there are horizontal shear stresses (or longitudinal shear stresses) acting between horizontal layers of the beam as well as vertical shear stresses acting on the cross-sections.
• At any point of the beam xy yxτ τ=
• Pattern of distribution of xyτ =pattern of distribution of
yxτ
• Since xy yxτ τ= , it follows that the vertical shear stresses
xyτ must vanish athy = ±2
, if the beam is subjected only
lateral loads.
Solid Mechanics
Derivation of shear stress formula
Beam under non-uniform bending i.e ( )M M x=
t = width or thickness of the beam at y y= 1
t = width or thickness of the beam at y y= 1
Solid Mechanics
We now wish to satisfy equilibrium in the x- direction.
Taking [ ]xFΣ → + = 0 we have then
( ) ( )
( ) ( )
( ) ( )
x x yxA A
yx x xA A
x
x x,y dA x,y dA t x
t x x,y dA x,y dAx
M x yx,y
I
σ σ τ
τ σ σ
σ
− + ∆ + + ∆ =
� �= + ∆ −� �
∆ � �� �
−=
� �
� �
0
1
( ) ( )
( ) ( )
( ) ( )
yxA A
yxA
yxA
t M x x ydA M x ydAx I I
t M x x M x ydAxI
M x x M xydA
It x
τ
τ
τ
� �= − + ∆ +� �
∆ � �� �
� �= − + ∆ −� �
∆ � �� �
+ ∆ −− � �= � �∆� �
� �
�
�
1 1 1
1
1
taking limit as x∆ → 0
( ) ( )yx
x A
yxA
M x x M xlim ydA
It x
dM ydAIt dx
τ
τ
∆ →
+ ∆ −−=∆
−=
�
�
0
1
1
( )ydM V xdx
= −
( )yyx
A
V xydA
Itτ∴ = �
Solid Mechanics
The above integral is by definition the first moment of are A about the z-axis, we denote it by symbol Q.
A
Q ydA= �
y
yx xyV Q
Itshear formula
τ τ τ∴ = = = (1)
in the above equation zzI I= stands for the moment of inertia of the entire cross sectional area around the neutral axis.
From (1)
yyx
V Q VQt fI I
τ = = =
The quantity “f” is known as the “shear flow”.
Shear flow is the horizontal shear force per unit distance along the longitudinal axis of the beam.
Solid Mechanics
Distribution of shear stresses in a Rectangular beam
An example of application of equations
A
h / yhQ udA b y y s
b hQ y
I bh
−� �� = = − + �� �� � �
� = − �
�
=
�
22
3
22 2
2 4
112
at xy yxhy τ τ= ± = = 02
The shear stresses in a rectangular beam vary quadratically with the distance y from the neutral axis.
Maximum value of shear stress occurs at the neutral axis where Q is maximum.
max maxxy yxVh V
I Aτ τ= = =
2 38 2
xy yxVQ V h yIt I
τ τ�
= = = − ��
22
2 4
Solid Mechanics
Thus maxτ in a beam of rectangular cross-section is 50%
larger than the average shear stress VA
It is always possible to express the maximum shear stress xyτ
as
maxxyV
KA
τ =
for most of the cross-sectional areas
K Rec tan gular= 32
K Circular= 43
K Triangular= 32
For most of the cross-section maxτ occurs at the neutral axis. This is not always true.
Solid Mechanics
Stress elements in non-uniform bending
Solid Mechanics
Problem
A wood beam AB is loaded as shown in the figure. It has a rectangular cross –section (see figure). Determine the maximum permissible value maxp of the loads if the
allowable stress is bending is allow MPaσ = 11 (for both tension and compression) and allowable stress in horizontal shear is
allow . MPaτ = 1 2
Solution
maxV occurs at supports and maximum BM occurs in between the loads.
Therefore, the maximum permissible values of the load P in dending and shear respectively are
maxV P= maxM . P Pa= =0 5
bhS =2
6A bh=
maxmax
M PaS bh
σ∴ = = 26
max maxmax
xy yx maxV P P
A A bhτ τ τ= = = = =3 3 3
2 2 2
allow allowallow allowb s
bh bhP Pa
σ τ= =2 2
6 3
Solid Mechanics
Substituting numerical values into these formulas,
Thus bending governs the design and the maximum allowable load is
Problem
An I –beam is loaded as in figure. If it has the cross-section as shown in figure, determine the shearing stresses at the levels indicated. Neglect the weight of the beam.
Solution
Vertical shear is same at all sections
allow b
allow s
P . kN
P . kN
=
=
8 25
8 25
maxP . kN= 8 25
Solid Mechanics
( )( ) ( )( )zzI I . mm s= = − = ×
3 36 4150 300 138 276
95 7 1012 12
The ratio V . N / mm sI .
−×= = ××
33 4
6
250 10 2 61 1095 7 10
Level ( )2A mm y
mm 3
Q Ay
mm
=
× 310
t mm xy
VQ MPaIt
τ =
1-1 0 150 0 150 0
2-2 ×=12 150
1800 144 259.2 150
12
4.5
56.4
3-3 ×=
×=
12 1501800
12 12144
144
132
259.2
19.0
12
60.5
4-4 ×=
×=
12 1501800
12 1381656
144
69
259.2
114.3
12
81.3
278.2
373.5
max . MPaτ = 81 3
Solid Mechanics
Warping of the cross sections due to shear stress
Plane sections will not remain plane and perpendicular to the axis of the beam in the deformed configuration due to the presence of shear force.
The cross-sections are wrapped with highest distortion at the axis.
It can be shown that if L h>> then distortion of cross-sections due to shear negligible.
Use all formulae developed so far only when L h>> - such beams are called slender beams.
Do not apply them if L h<< -- short beams.
Solid Mechanics
17. Theories of failure or yield criteria (1) Maximum shearing stress theory
(2) Octahedral shearing stress theory
(3) Maximum normal stress theory – for brittle materials.
Maximum shearing stress theory or Tresca Criterion
This theory says that:
Yielding occurs when the maximum shear stress in the material reaches the value of the shear stress at yielding in a uniaxial tension (or compression) test.
Maximum shearing stress under general state of stress is
( )max max , ,τ τ τ τ= 1 2 3
where ; ;σ σ σ σ σ στ τ τ− − −= = =2 3 1 3 1 21 2 32 2 2
The maximum shearing stress in uniaxial tension test at the moment of yielding is
yst
στ =
2
Tresca criterion is ysmax
στ ≥
2
Octahedral shearing stress theory or Hencky-Von-Mises failure criterion
This theory also known as “The maximum distortion strain-energy theory”
For ductile materials
Solid Mechanics
This theory states that
Yielding occurs when the octahedral shear stress in the material is equal to the value of the octahedral shear stress at yielding in a uniaxial tensile test.
( ) ( ) ( )octτ σ σ σ σ σ σ= − + − + −2 221 2 2 3 1 3
13
Octahedral shear stress in the uniaxial tension test at the moment of yielding i.e. y ysσ σ σ= = 1
( ) ( ) ( )t ys ys
t ys
τ σ σ
τ σ
= − + − + −
=
2 2210 0 0 0
32
3
Von Mises theory says that oct ysτ σ≥ 23
von octσ τ= 32
Von Mises theory says that von ysτ σ≥
Maximum Normal stress criterion or Rankine Theory:
This theory is generally used for design of components made up of brittle materials.
* Excellent experimental evidence is available for supporting maximum shearing stress and Von Mises criterion
Solid Mechanics
According to this theory, a given structural component fails when the maximum normal stress (tensile) in that component reaches the ultimate strength or ultimate stress ultσ obtained from the tensile test of a specimen of the same material.
Thus the structural component will fail when
Simple application of theories
ultσ σ≥1
Solid Mechanics
18. Combined loading Torsion + Direct shear
AMrI
σ =p
TrI
τ =1
VA
τ =243
Solid Mechanics
Bending + axial loading
Neutral surface is now shifted due to the application of axial load.
xPA
σ =
xMyI
σ −=
zzx
zz
M yPA I
σ � −= + ��
Solid Mechanics
19. Elastic strain energy Consider an infinitesimal stress element at point in a linearly elastic body, subjected to a normal stress xσ
The work done by this force
�
int
x xdistanceforce
dW dF dS
dydz dxσ
= ×
= × ∈
12
12�����
int x xdW dVσ= ∈12
This internal work is stored in the volume of the element as the internal elastic energy or the elastic strain energy.
x xdU dVσ∴ = ∈12
dV =volume of the element.
The strain energy density U0 is defined as the internal elastic energy stored in an elastic body per unit volume of the material.
x xdUStrainenergy density UdV
σ ∈∴ = = =0 2
Solid Mechanics
U0 can be interpreted as an area under the inclined line on the stress-strain diagram. Similar expressions can developed for yσ and zσ corresponding to strains y∈ and z∈ .
Elastic strain energy for shearing stresses:
Analogous expressions apply for the shearing stresses
xz zx,τ τ with the corresponding shear strains yzY and xzY
Strain energy for multiaxial states of stress
The strain energy expressions for a 3D state of stress follow directly by addition of the energies of each stress component.
x x y y z z xy xy yz yz zx zx
dU
Y Y Y dVσ σ σ τ τ τ
=
� �∈ + ∈ + ∈ + + +� �� �
1 1 1 1 1 12 2 2 2 2 2
The strain energy density for the most general case is
�shear xy xy
distanceaverage force
dU dxdz Y dyτ= ×12�����
shear xy xydU Y dvτ= 12
Solid Mechanics
x x y y z z xy xy
yz yz zx zx
dUU Ydv
Y Y
σ σ σ τ
τ τ
= = ∈ + ∈ + ∈ +
+ +
01 1 1 12 2 2 21 12 2
Substituting the values of strain components from generalized Hooke’s law, we can show that
It is the expression for elastic strain energy per unit volume for linearly plastic, homogeneous, isotropic materials.
In general, for a stressed body the total strain energy is obtained by integration of 0U over its volume.
Internal strain energy in axially loaded bars
x z xy xz yzσ σ τ τ τ= = = = = 0
xx x x xU
E Eσσ σ σ∴ = ∈ = = 2
01 1 12 2 2
∴The total internal energy xV V
U U dv dVE
σ= = =� �2
01
2
( ) ( )( )x y z x y y z z x
xy yz zx
vUE E
G
σ σ σ σ σ σ σ σ σ
τ τ τ
= + + − + +
+ + +
2 2 20
2 2 2
12
12
( )V
U elastic energy stored U dV= = � 0
Solid Mechanics
x P P LU AL .ALE EAEA
σ= == =2 2 2
22 22
P LUEA
=2
2
Strain energy in torsion of circular shafts
U .Y .G Gττ τ τ= = = 2
01 1 12 2 2
v v
U U dv dvG
τ= =� �2
01
2
p
TrI
τ = where pI Rπ= 4
2
R
p
TU . .r . r.dr.LG I
π= �2
22
0
12
2
Strain energy in bending
x
v v
M MU dv y dv y dA.LE EI EI
σ∴ = = =� � �2 2 2
2 22 22 2 2
p
T LUGI
=2
2
P
TYI
τ =
Solid Mechanics
Conclusion
Axially loaded bars P LUAE
=2
2
Torsion of shafts P
T LUGI
=2
2
Bending (pure) of beams M LU
EI=
2
2
We can use the following equations in case of non-uniform cases
L L L
P
P T MU dx ; U dx ; U dxAE GI EI
= = =� � �2 2 2
0 0 02 2 2
M LUEI
=2
2
Solid Mechanics
Problem:
( ) ( )P x Y.A L x= −
( )
L
L
L
PU dxAE
Y A L xdx
AE
Y A Y A LL x Lx.dx L L LAE E
Y A L Y ALL LAE E
=
−=
� �= + − = + −� �
� �
� �= + − =� �
� �
�
�
�
2
022 2
02 2 2 3
2 2 23
02 2 3 2 3
3 3
2
2
22 2 3
2 3 6
P LUAE
=2
2
( ) ( )P x Y.A L x P= − +
( ) ( )L Y A L x P YA L x .PU dx
AE− + + −= �
22 2 2
0
22
Y AL P L YAP LU LE AE AE
Y AL P L YPE AE E
� �= + + −� �
� �
= + +
2 3 2 2 22
2 3 2 2 2
26 2 2 2
6 2 2
Since U P or U δ∞ ∞2 2 principle of superposition should
not be used.
Solid Mechanics
20. Deflection of beams When a beam with a straight longitudinal axis is loaded by lateral loads, the axis is deformed into a curve, called the “deflection curve” or “elastic-curve”
Deflections: means u ,v displacement of any particle. In case of beams deflection means v displacement of particles located on the axis of the beam.
Deflection calculation is an important part of component design
Deflections -- useful in vibration, analysis of various engineering components ex. Earthquake loading.
Undesirable vibrations are due to excessive deflections.
Solid Mechanics
Approximate sketches of deflection curves
Approximate sketches of the deflection curve can be drawn if BM diagram is available for a given loading.
We know that +BM means
- BM means
Examples
(1)
Solid Mechanics
The objective is to find the shape of the elastic curve or deflection curve for given loads i.e., what is the function v(x).
There are two approaches
(1) Differential equations of the deflection curve
(2) Moment-area method
Differential equations of the deflection curve
Consider a cantilever beam: The axis of the beam deforms into a curve as shown due to load P.
Here we assume only symmetrical bending case. The xy plane is the plane of bending.
v↓ − deflection of the beam.
v ve↑ + and. v↓ −
To obtain deflection curve we must express v as a function of x.
Solid Mechanics
When the beam is bent, there is not only a deflection at each point along the axis but also a rotation.
The angle of rotation θ of the axis of the beam is the angle between x – axis and the tangent to the deflection curve at a point.
For given x-y coordinate system
ve anticlockwiseθ → + →
O Center of curvature′ =
Radiusof curvatureρ =
From geometry d dsρ θ =
dkds
curvature of the deflectioncurve
θρ
= =1
k - curvature - +ve when angle of rotation increases as we move along the beam in the +ve x – direction.
dvSlopeof thedeflectioncurve tandx
θ= =
Slope dvdx
is positive when the tangent to the curve slopes
upward to the right.
The deflection curves of most beams have very small angles of rotations, very small deflection and very small curvatures. That is they undergo small deformations.
When the angle of rotation θ is extremely small, the deflection curve is nearly horizontal
Solid Mechanics
ds dx≈
This follows from the fact that
( )ds dx dv v dx′= + = + 22 2 1
for small θ ( )v′ 2 can be neglected compared to 1
ds dx∴ ≈
Therefore, in small deflection theory no difference in length is said to exist between the initial length of the axis and the arc of the elastic curve.
dkdxθ
ρ= =1
Since θ is small tanθ θ≈
d d vkdx dxθ
ρ∴ = = =
2
21
dkdx only insmalldeformationtheorydu udx
ν ν
θ
�′′= = ����′= =��
2
2
If the material of the beam is linearly elastic and follows Hooke’s law, the curvature is
MkEIρ
= =1
dvdx
θ∴ =
Solid Mechanics
M+ → leads to +K and so on
d v MEIdx
∴ =2
2 or
d vEI Mdx
=2
2
The basic differential equations of the deflection curve.
Sign conventions used in the above equation:
(a) The (b) dvdx
and θ are
(c) k is + (d) M is +ve if beam bends
Another useful equations can be obtained by noting that
Non-prismatic beams
( ) ( )
( )( ) ( )
( )( ) ( )
d vEI x M xdx
EI x v v x
EI x v P x
=
′′′ = −
′′′′ = +
2
2
dM VdxdV pdx
= −
= −
Solid Mechanics
For prismatic beams.
( )( )( )
nd
rd
th
EIv M x BM equation( order )
EIv V x Shear force equation( order )
EIv P x Load equation( order )
′′ =
′′′ = −
′′′′ = +
2
3
4
Integrating the equations and then evaluating constants of integration from boundary conditions of the beam.
Assumptions involved in the above equations
(a) Material obeys Hooke’s law
(b) Slope of deflection curve small – small deformations
(c) Deformations due to bending only – shear neglected
When sketching deflection curve we greatly exaggerate the deflection for clarity. Otherwise they actually are very small quantities.
Solid Mechanics
Approximate sketching
(3) (4)
(5) (6)
Solid Mechanics
Boundary conditions
(1)Boundary conditions
(2)Continuity conditions
(3)Symmetry conditions
Boundary conditions
Pertain to the deflections and slopes at the supports of a beam:
(i)Fixed support or clamped support
(ii)
( )( ) ( ) ( )
v a
M a EIv a v a
=′′ ′′= = � =
0
0 0
(iii) ( ) ( )
( ) ( )M a EIv a
V a EIv a
′′= =′′= − =
0
0
( )( ) ( )
v a
a v aθ=
′= =0
0
Solid Mechanics
Continuity conditions
All deflection curves are physically
continuous. Therefore
Similarly at “C”
( ) ( )from side AC from side BCv c v c′ ′=
Symmetry conditions
Lv � ′ = ��
02
because of loading
and beam. This we should load
in advance.
The method for finding deflection using differential equations is known as “ method of successive integration”.
Application of principle of superposition: Numerous problems with different loadings have been solved and readily available. Therefore in practice the deflection of beam subjected to several or complicated loading conditions are solved using principle of superposition.
+ +
( ) ( )from side AC from side BCv c v c=
Solid Mechanics
Problem 1
Determine the equation of the deflection curve for a simple beam AB supporting a uniform load of intensity of acting through out the span of the beam. Also determine maximum deflection maxδ at the mid point of the beam and the angles of rotation AQ and BQ at the supports. Beam has constant EI.
Solution
qL
V qx= −2
(1)
qL qxM x− + =
20
2 2
qLx qx
M = −2
2 2 (2)
Differential equation of deflection curve.
( )EIv M x
qLx qxEIv
′′ =
′′ = −2
2 2
Slope of the beam
qLV qx+ − = 0
2
Solid Mechanics
qLx qxEIv C′ = − +
2 3
14 6
BC →Symmetry conditions
Lv x
qLL qLC
qL qLC
� ′ = = ��
= − +
= − +
2 3
1
3 3
1
02
016 48
016 48
qLC = −
3
1 24
Slope equation is
( )
qLx qx qLEIv s
qv L L x
EI
′ = − −
−′ = − +
2 3 3
3 2 3
4 6 24
624
Deflection of the beam
qLx qx qLEIv x C= − − +
3 4 3
212 24 24
B.C.
( )v xC
= == − − + �2
0 00 0 0 0
qLx qx qLEIv x= − −
3 4 3
12 24 24
C =2 0
Solid Mechanics
( )( )
qv L x Lx x
EIq
v x L x LxEI
−∴ = − +
−= + −
3 3 4
4 3 3
224
224
you can check v = 0 at x = 0 and L = 0
(b) From symmetry maximum deflection occurs at the
midpoint Lx =2
qLLv xEI
−� = = ��
452 384
-ve sign means that deflection is downward as expected.
maxqLLv x s
EIδ � = = = �
�
452 384
( )AqL
Q vEI
−′= =3
024
-ve sign indicates clock wise rotation as expected.
( )BqL qL qL
Q v x LEI EI EI
′= = = − −3 3 3
4 6 24
( ) qLv L
EI′ =
3
24 + ve sign means anticlockwise direction.
since the problem is symmetric, ( ) ( )v v L′ ′=0
Solid Mechanics
Problem: 2
Above problem using third order equation
( )EIv V x′′′ = −
qL qLEIv qx qx� ′′′ = − − = − �
� 2 2
Moment equation
qLx qxEIv C′′ = − +
2
12 2
B.C.
( ) ( )M x EIv xC
qLx qxEIv
′′= = � = =� =
′′ = −
12
0 0 0 00
2 2
Problem 3
Above problem using fourth order differential equation
P qEIv q
=′′′′ = −
Shear for a equation
EIv qx C′′′ = − + 1
From symmetry conditions
Solid Mechanics
L LV x EIv x
qLLq C C
qLEIv qx
� � ′′′= = � = = � �� �
= − + � = +
′′′∴ = − +
1 1
0 02 2
02 2
2
Problem 4
Determine the equation of the deflection curve for a cantilever beam AB subjected to a uniform load of intensify q. Also determine the angle of rotation and deflection at the free end. Beam has constant EI.
Solution:
qL qxM qLx+ − + �
2 2
2 2
Differential equation of deflection curve
( )EIv M x
qL qxEIv qLx
′′ =
′′ = − + −2 2
2 2
V qL qx+ − = 0V qx qL= −
qL qxM qLx= − −
2 2
2 2
Solid Mechanics
Slope equation:qL x qLx qx
EIv C′ = − + − +2 2 3
12 2 6
BC: ( )v x′ = = �0 0
qL x qLx qxEIv′ = − + −
2 2 3
2 2 6
Deflection equation
qL x qLx qxEIv C= − + − +
2 2 3 4
24 6 24
( )v xC
= == + − + �2
0 00 0 0 0
qL x qLx qxEIv∴ = − + −
2 2 3 4
4 6 24
( )v x L
qL qL qL qLEIv
′ = �
− −′ = + − =3 3 3 3
2 2 6 6
BqL
v QEI
′∴ = = −3
6
( )v x L
q qLv L L L
EI EI
= �
− −� �= − + =� �
44 4 4 3
6 424 24
-maximum deflection also.
C =1 0
C =2 0
qv L x Lx x
EI
− +− � �= +� �� �
2 2 3 46 424
qLv
EI=
4
8�( ) qL
v x LEI
−= =43
24
Solid Mechanics
Problem 5
Above problem using third order equation
( )EIv V xEIv qL qx
′′ = −′′′ = −
Moment equation
qxEIv qLx C′′ = − +
2
12
B.C. ( ) ( )M x L EIv x L′′= = � = =0 0
qL qL qLqL
qx qLEI v qLx
� = − = � = −
′ ′′ = − +
2 2 22
2 2
0 42 2 2
2 2
qx qLEIv qLx′′ = − +
2 2
2 2
Problem 6
Above problem with fourth order equation
( )EIv P x
EIv q−
′′′′ =
′′′′∴ = ⊕
Shear force equation
EIv qx C′′′ = − + 1
( ) ( )B.C V x L EIv x LqL C C qL
′′′= = � = == − + � = +1 1
0 00
Solid Mechanics
EIv qx qL′′′∴ = − +
Problem 7
A simple beam AB supports a concentrated load P acting at distances a and b from the left-hand and right-hand supports respectively. Determine the equations of the deflection curve, the angles of rotation and at the supports, the maximum deflection and the deflection at the midpoint C of the beam. Constant EI
Solution
PbM xL
− = �0
AQ BQ
maxδ Lδ
EI =
PbVL
+ = 0
PbVL
= −
PbxHL
=
PbV PL
+ − = 0
PbV PL
= −
Pb Pbx P x PL L
+ = � = −
Solid Mechanics
( )
( )
PbxM P x aL
PbxM P x aL
Pbx PxaM Px Pa PaL L
+ − −
= − −
= − + = − +
Differential equation of deflection curve
PbxEIv x aLPxaEIv Pa a x LL
′′ = ≤ ≤
′′ = − + ≤ ≤
0
Slope equations:
PbxEIv C o x aL
′ = + ≤ ≤2
12
Px aEIv Pax C a x LL
−′ = + + ≤ ≤2
22
B.C. ( ) ( )AP PBv x a v x a′ ′= = =
( )P L a a PaC Pa CL L
PLa Pa PaC Pa CL L L
PaC C
− −+ = + +
/ / // /− + = − + +/ / / / /
� = +
2 32
1 2
2 3 32
1 2
2
1 2
2 2
2 2 2
2
Solid Mechanics
Deflection curve equations:
PbxEIv C x C x aL
Px a PaxEIv C x C a x LL
′ = + + ≤ ≤
−= + + + ≤ ≤
3
1 3
3 2
2 4
06
6 2
B.C: ( )v x = =0 0 and ( )v x L= = 0
C= + + �30 0 0
PL a PaL C L CL
PL a PaL C L C
PaL C L C
= − + + +
= − + + +
= + +
3 2
2 4
2 2
2 4
2
2 4
06 2
06 2
3
( ) ( )
( )AP PBv x a v x a
P L a a Pa PaC a C a CL L
PLa Pa Pa PaC a C a CL L L
Pa PaC a C a C
Pa PaLC a C a C L
= = =
− −+ = + + +
/ /−/ /+ + = + + +/ //
+ = + +
= + − −
3 4 3
1 2 4
3 4 4 3
1 2 4
3 3
1 2 4
3 2
1 2 2
6 6 2
6 6 6 2
6 2
3 3
C =3 0
PaLC C L= − −2
4 23
Solid Mechanics
Pa Pa PaLC a C a C L
Pa PaL PaL PaC L C L
PaL PaC
/ /+ = + − −
= − − � = − −
= − −
3 3 2
2 2 2
3 2 2 3
2 2
3
2
2 3 3
6 3 3 6
3 6
Some important formulae to remember
(1)
(2)
(3)
(4)
(5)
Problem 8
A simple beam AB supports a concentrated load P acting at the center as shown. Determine the equations of the deflection curve, the angles of rotation AQ and BQ at the supports, the maximum deflection maxδ of the beam.
B BqL qL
,QEI EI
δ = =4 3
8 6
B BPL PL, Q
EI EIδ = =
3 2
3 2
B BM L M L, Q
EI EIδ = =
20 0
2
c max A BqL qL
; Q QEI EI
δ δ= = = =4 35
384 24
c max A BPL PL;Q Q
EI EIδ δ= = = =
3 2
48 16
Solid Mechanics
Solution
PV = −
2
PxM =
2
PxM =
2
PM x− = 0
2
PxM =
2
V P /= 2
PV P+ − = 0
2
Px LM P x
Px L Px PL PL PxM P x Px
� − + − = ��
� = − − = − + = − ��
02 2
2 2 2 2 2 2
PL PxM = −2 2
Solid Mechanics
Differential equation deflection curve
PxEIv x L /
PL Px LEIv x L
′′ = ≤ ≤
′′ = − ≤ ≤
0 22
2 2 2
Slope equations
PxEIv C x L /
PLx Px LEIv C x L
′ = + ≤ ≤
′ = − + ≤ ≤
2
1
2
2
0 24
2 4 2
AP PB
L Lv x v x� � ′ ′= = = � �� � 2 2
PL PL PLC C+ = − +2 2 2
1 216 4 16
PL PL PLC C C= + − = +2 2 2
1 2 24 8 8
Deflection equations:
PxEIv C x C x L /
PLx PxEIv C x C L / x L
= + + ≤ ≤
= − + + ≤ ≤
3
1 3
2 3
2 4
0 212
24 12
B.C: ( )v x = =0 0 and ( )v x L= = 0
PLC C= +2
1 2 8
Solid Mechanics
C= + + �30 0 0
PL PL C L C
PL C L C
= − + +
= + +
3 3
2 4
3
2 4
04 12
6
AP PB
L Lv x v x
PL C L PL PL LC C
L PL PL LC C C
� � ′ ′= = = � �� �
+ = − + +
= − + +
3 3 31
2 4
3 3
1 2 4
2 2
96 2 16 96 2
2 16 48 2
L PL PL L PLC C C L/ // /+ = + − −/ /
3 3 3
2 2 22 16 24 2 6
( )PLPL PL PL C L C− −− − = � =
23 3 3
2 22 8 3
24 6 16 48
C =3 0
PLC C L= − −3
4 26
L PL LC C C= + +3
1 2 42 24 2
PL PLC −= − =2 2
29 3
48 16
PLC = −2
23
16
Solid Mechanics
PL PL PLC∴ = − + = −2 2 2
13
16 8 16
( )
PL PLC L
PLPL PL
� −∴ = − − ��
− +−= + =
3 2
4
33 3
36 16
8 936 16 48
Deflection curves
Px PL LEIv x C x
PLx Px PL PL LEIv x x L
= − + ≤ ≤
= − + − + ≤ ≤
3 2
3
2 3 2 3
012 16 2
34 12 16 48 2
LxPL PL PLEIv =
−= − =3 3 3
2 96 32 48
( )Lx
PLPL PL PL PLEIv
PL
=− − += − − + =
= −
33 3 3 3
23
6 1 9 2316 96 32 48 96
48
PLC = −2
1 16
PLC = −3
4 48
LxPLv
EI=∴ = −3
2 48
Solid Mechanics
Slope equations:
Px PL LEIv x
PLx Px PL LEIv x L
′ = − ≤ ≤
′ = − − ≤ ≤
2 2
2 2
04 16 2
32 4 16 2
( )
( ) ( )A
PL PLEIv x
PLv x Q Clock wiseEI
′ = = − = −
′∴ = = = − −
2 2
2
0 016 16
016
( ) ( )
( ) ( )B
PLPL PL PL PLEIv x L
PLv x L Q +ve, CCW from x-axisEI
− −′ = = − − = =
′∴ = = =
22 2 2 2
2
8 4 332 4 16 16 16
16
Problem 9
A cantilever beam AB supports load of intensity of acting over part of the span and a concentrated load P acting at the free end. Determine the deflections Bδ and angle of rotation
BQ at end B of the beam. Beam has constant EI. Use principle of superposition.
Solution
( )B Bqa qL
L a , QEI EI
δ = − =1 1
3 34
24 6
B BPL PL, Q
EI EIδ = =
2 2
3 2
3 2
v PL / EI= − 3 48
Solid Mechanics
( )B B B
B B B
qa PLL aEI EI
qa PLQ Q QEI EI
δ δ δ= + = − +
= + = +
1 2
1 1
3 3
3 2
424 3
6 2
Solid Mechanics
21. Moment- Area Method This method is based upon two theorems related to the area of the bending moment diagram it is called moment-area method.
First moment area theorem
Consider segment AB of the deflection curve of a beam in region of + ve curvature.
The equation
d MEIdx
θ =2
2 can be written as
d d Mdx EIdx
θ θ= =2
2
Md dx
EIθ =
The quantity M dxEI
corresponds to an infinitesimal area of
the MEI
diagram. According to the above equation the area is
equal to the arrange in angle between two adjacent point m1 and m2 . Integrating the above equation between any two points A & B gives.
B B
B A BAA A
Md dx
EIθ θ θ θ= − = ∆ =� �
Solid Mechanics
This states that the arrange in angle measured in radius between the two tangents at any two points A and B on the
elastic curve is equal to the area of MEI
diagram between A &
B , If Aθ is known then
B A BAθ θ θ= + ∆
In performing above integration, areas corresponding to the M+ are taken + ve, area corresponding to the – ve M are
taken –ve
If B
A
M dxEI� is +ve- tangent B rotates c.c.w from A or Bθ is
algebraically larger than A.
If – ve – tangent B rotates c.w from A.
Second moment-area theorem
This is related to the deflection curve between A and B.
Solid Mechanics
We see that dt is a small contribution to BAt . Since the angles between the tangents and x-axis are very small we can take
The expression Mx dxEI
=1 first moment of infinitesimal area
M dxEI
w.r.t. a vertical line through B.
Integrating between the point A & B
B B
BAA A
Mt dt x dxEI
′= =� � 1 = First moment of the area of the
MEI
diagram between points A & B, evaluated w.r.t. B.
if M is +ve φ =+ve
if M is -ve φ = -ve
x and x1 are always taken +ve quantities.
∴Sign of tangential deviation depends on sign of M.
Mdt x d x dxEI
θ= =1 1
BA
ABB
A
t xt x
Mwhere dxEI
φφ
φ
==
= �
1
Solid Mechanics
A positive value of tangential deviation- point B is above A and vice versa – ve value means point B is below the point A.
In applying the moment area method a carefully prepared sketch of the elastic curve is always necessary.
Problem:1
Consider an aluminum cantilever beam 1600 mm long with a 10 –kN for a applied 400 mm from the free end for a distance of 600 mm from the fixed end, the beam is of greater depth
than it is beyond, having 4I mm= × 61 50 10 . For the
remaining 1000 mm of the beam 4I mm= × 62 10 10 . Find the
deflection and angular rotation of the free end. Neglect weight of the beam and E GPa= 70
Solution:
2
2
N/mm
N/mm
−× ×
= ×
9 6
3
70 10 10
70 10
EI .= × 243 5 10
Solid Mechanics
.A bhE E
.A bhE
. .A bhE E. .A bhE E
−� = = × × = − ��
= = −
−� = = × × = − ��
−� = = × × = − ��
1
2
3
4
1 1 0 12 36600
2 2129 6
1 1 0 48 115 2480
2 21 1 0 12 7 2
1202 2
B
BA B AA
MQ Q Q dx A A A AEI
∆ = − = = + + +� 1 2 3 4
B. . .Q
E E E E E= − − − − = −36 129 6 115 2 7 2 288
Solid Mechanics
BQ . radE
−= − = − = − ××
33
288 288 4 14 1070 10
from tangent at A.
BA Bt δ=
x mm;x ;x mm;x mm= = = =2 1 3 41060 1400 840 480
BA Bt A x A x A x A x. . .
E E E E
. mmE
δ= = + + +− − − −� � � � = + + + � � � �� � � �
−= = −
1 1 2 2 3 3 4 4
36 129 6 115 2 7 21400 1060 840 480
2880004 11
below the tangent at point A.
Problem 2
Find the deflection due to the concentrated force P applied as soon as figure, at the center of a simply supported beam EI constant.
Solution:
BQ . rad−= × 34 14 10
B . mmδ = −4 11
Solid Mechanics
c CB
AB
v c c t
c c t
′′ ′= −
′′ ′ = 12
Pa PaA bh a sEI EI
Pa PaA bh aEI EI
= = × × =
= = × × =
2
1
2
2
1 1 3 32 2 4 81 1 3 9
32 2 4 8
x a ; x a= =1 22
23
( )
ABPa Pat A x A x a aEI EI
Pa Pa Pa Pa veEI EI EI EI
= + = +
= + = = +
2 2
1 1 2 2
3 3 3 3
3 2 92
8 3 89 10 5
4 4 4 2
Since EI is constant MEI
diagram is same as M diagram.
Solid Mechanics
CBPa a Pat a sEI EI
� = × × × = ��
31 22
2 2 3 3
AB /Pac c tEI
′′ ′ = =3
254
( )c
PaPa Pa PavEI EI EI EI
−∴ = − = =33 3 315 45 11
4 3 4 12
The +ve sign of ABt & CAt indicate points A & C above the tangent through B.
(a) The slope of the elastic curve at C can be found from the slope of one of the ends as:
BC B C C B BCQ Q Q Q Q Q∆ = − � = − ∆
B
BC B CC
M Pa PaQ Q Q dx a sEI EI EI
∆ = − = × × =�21
22 2 2
B ABPa Pa Pa PaQ t / LEI a EI EI EI
≈ = − = −3 2 2 25 1 5
2 4 2 8 2
(b) If the deflection curve equations is wanted then by selecting an ordinary point E at a distance x
Ev E E EE′′ ′ ′′= −
cPavEI
=311
12
cPaQ
EI=
2
8
Solid Mechanics
E AB EBL xv t t
L−� = − �
�
In this way one
can obtain equation
of the deflection curve.
(c) To simplify the calculations some care in selecting the tangent at a support must be considered.
In this approach to find
CAt we need to consider
unhatched region which
is more difficult.
(d) The deflection at C can also be calculated as follows.
AC BCc
t tv +=2
∴C is at the center of the beam. However, this is also move complicated approach compared to first, as to find CAt we again need to consider unhatched region.
Solid Mechanics
Problem 3
Find the deflection of the end A of the beams shown in figure caused by the applied forces. The EI is constant.
Solution
Solid Mechanics
Pa PaA bh aEI EI
a Pa PaAEI EI
Pa PaA and AEI EI
−� = = × × = − ��
� = × × − = − ��
= =
2
1
2
2
2 2
3 4
1 12 2 2
12 2 4
4 2
a a a a ax a ; x a
a ax a a / ; x
= + = = + + =
= + = =
1 2
3 4
7 2 112
3 3 3 3 2 61 2
7 63 2 3
( )
CBt A x A x A x
Pa a Pa a Pa aEI EI EI
PaPa Pa PaEI EI EI EI
= + +
= − × + × + ×
− + += − + + =
2 2 3 3 4 42 2 2
33 33
11 7 24 6 4 6 2 3
11 7 811 724 24 3 24
CBPa Pat
EI EI= =
3 3424 6
The + sign of CBt indicates that the point C is above the tangent through B. Hence corrected sketch of the elastic curve is made.
Solid Mechanics
ABPa Pat a
EI EI= − × = −
2 322 3 3
A ABv t A A
Pa Pa PaEI EI EI
′′ ′∴ = −
= − =3 3 3
3 12 4
Note: Another method to find Av is shown. This may be simpler method than the present one.
APav
EI=
3
4
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