Solid Mechanics
Internal Forces in solids
Sign conventions
• Shear forces are given a special symbol on given
V y
1
2
and V z
• The couple moment along the axis of the member is
M x
= = Torque
M y
= M z
= 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?
F x
F y
F z
=
=
0
0
= 0
P
V
M
• 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
[ Σ F x
= → + ]
Σ F y
= ↑ +
V + ∆ ∆ = 0
∆ V = − ∆
∆ V
∆ x
= − P lim
∆ x → 0
∆
∆
V dV x dx
= − P
[ Σ M
A
= 0
] ∆ + ∆ M −
∆ + ∆ M −
∆ M
∆ x 2
2
2
= 0
= 0
2
2
= 0

lim
∆ x → 0
∆
∆
M x
= dM dx
= − V
From equation dV dx
= − P we can write
V
D
− V
C
= −
X
D
X
C
Pdx
From equation dM dx
= − V
M
D
− M
C
= − Vdx
Special cases:
Solid Mechanics

Solid Mechanics

Solid Mechanics

Solid Mechanics

Solid Mechanics
0
2
6
8 x 2 −
6 ( 2 ( ) x 8 − x 10 − −
0
V
V
V
A
= 5
= 5 ; V
B
= 5
2
V
V
− + −
= − +
V
B
= −
− +
25
(
; V
C
− 2
=
)
5
= 0 x = 5 33
( ) 0
( − 2 )
6
V
V
− + − − =
= + 15
V
C
8
V
= + 15
10
; V
D
4
= +
− 4
15
( ) ( )
− + − − + =
V
V
V
D
= − 5
= − 5 ; V
E
= − 5

Solid Mechanics
2
M
M
M
−
0
M
M
M
A x 2 − ( ) ( )
− 10 5 x = 0
5 x 10
= + 10 ; M
B
= 0 x 6 − ( ) ( )
+ x
= − x + 30 ( x )
=
− 30 ( x
+
)
( − 2 ) 2
2
( − 2 ) 2
2
= 0
M
C x = 6
= 40
6
M x 8 ( ) ( ) [ ]
− + x − 30 ( x ) 30 ( x ) 10 ( x ) =
M
C x = 6
M
D x = 8
= 20 +
= − 10
8
M
M
E x = 8
10
= 0
[ ] ( ) ( )
− + x − 30 ( x ) 30 ( x ) 10 ( x ) − ( x ) 0
Problems to show that jumps because of concentrated force and concentrated moment

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.
0 2 [ ]

Solid Mechanics
2 6 [ ]
V
V = 5
V
A
V
B
=
= 5
5
M
M
− 10 5 x = 0
M
A
= − x
= 10 ; M
B
= 0
6
V − + −
V = ( )
(
V
B
= − 25
− + (
; V
C
=
)
5
0 x = 5 33
) 0 M − + x − 30 ( x ) .
x = 6
M
C
= 40
M x = .
x
M
= 2
B
= 0
=
( x − 2 ) 2
2
= 0
8 [ ]
V − + −
V
V
C
=
=
15
15 , V
D
=
−
15
=

8 10 [ ]
Solid Mechanics
V
V
V
D
− + − − + =
= − 5
= − 5 , V
E
= − 5

[
[
Solid Mechanics
F x
→ + = 0
]
F y
↑ + = 0
M
∆
= 0 ]
R
Ax
= 0
R
Ay
+ − =
R
Ay
= 30 kN ↑
M
M + 60 90 4 5 0
= 285
30
V =
V − ( x ) 0
30 ( x ) 90
= × −
= −
= 0
M
M
B
B
− M
60
A
= +
( 60 )
M
A
= −
=
225
−

M
M
C
C
− M
B
( )
=
B
+ 90 = − 225 90
= −
M
135
= − − 90
M
D
− M
C
M
D
= M
C
( 135 )
+ 135 = − + =
Solid Mechanics
F y
↑ + = 0
R
Ay
R
Ay
+ R
Cy
+ R
Cy
− − =
= 440
[
M
A
= 0
]
− × − × + R
Cy
R
Cy
= 195 kN ↑
R
Ay
= 245 kN ↑
V
V
V
V
+ 245 200 30 x = 0
= 30 x − 45
= × − = −
= 195

Solid Mechanics
*
[
R
Ay
+
M
A
R
By
= 32
] × + + +
− + + R
By
= 0
R
By
= 12 kN
R
Ay
= 20 kN
M
M
M
− × + ×
= × − ×
= 600
R
By
= 0

Solid Mechanics
Problem :
V
V = 8 x − 20
8 x − = x
+ 20 8 x = 0
= = .
M
C
− M
M
C
= M
A
A
( 50 )
+ 50 = − + =
[
[ F x
R
Ax
=
→ + =
0
F y
M
A
0
]
R
Ay
+ R
Dy
− − =
= 0 ] 60 1 5 50 4 R
Dy
R
Dy
R
Ay
R
Ay
+ R
Dy
=
=
290
5
=
52 kN
58 kN
↑
↑
= 110

Solid Mechanics
F y
V + 52 20 x = 0
V = 20 x − 52 0 x 3 m
[ M = 0 ]
M +
20 x 2
2
M
− 52 x = 0
= 52 x −
20 x
2
2
( 0 x 3 m )
F y
V + − =
V = 8 kN ↑
B
3 ≤ ≤ 4
C x m
[
M
M = 0 ] M − 52 x + 60 ( x − .
) = 0
= x − ( x − .
)
B
52 60 1 5 3 ≤ ≤ 4
C x m

Solid Mechanics
F y
V + − − =
V = 58 kN ( 4 5 )
[ M = 0 ] M − 52 x + 60 ( x − .
) + 50 ( x ) 0
M = 52 x − 60 ( x − .
) − 50 ( x − 4 ) ( 4 5 ) dM dx dV dx
= − V
= − P
20 × − = x = = . m
M
B
− M
E
= −
M
B
= − .
+

Solid Mechanics
M
B
− M
A dM dx dV dx
= −
= − V
= − P
Vdx
20 × − = x = = .
M
B
− M
E
= −
M
B
= − 1 6 + M
E
=
= −
66
.
+
M
C
− M
B
= − 8
M
C
M
B
= − + =
M
D
− M
C
= − 58
M
D
= M
C
+ 58
= − =

Solid Mechanics
Traction vector or Stress vector
Now we define a quantity known as “ stress vector ” or
“ traction ” as
T n
=
∆ lim
A → 0
∆
∆
F
R
A
units P N / m
2 and we assume that the quantity
∆ lim
A → 0
∆
∆
M
A
R → 0
(1) T n
is a vector quantity having direction of ∆ F
R
(2) T n
represent intensity point distributed force at the point
"P" on a plane whose normal is ˆn
(3) T n
acts in the same direction as ∆ F
R

Solid Mechanics
(4) There are two reasons are available for justification of the assumption that
∆ lim
A → 0
∆
∆
M
A
R → 0
(a) experimental
(b) as ∆ A → 0, ∆ F
R
becomes resultant of a parallel force distribution. Therefore ∆ M
R
= 0 for force system.
(5) T n varies from point to point on a given plane
(6) T n
at the same point is different for different planes.
(7) T n ′
= − T n
will act at the point P
(8) In general
Components of T n
∆ F
R
= ∆ ˆ + ∆ v t ˆ + ∆ ˆ

Solid Mechanics
T n
=
∆ lim
A → 0
∆
∆
F
R
A
=
∆ lim
A → 0
∆
∆
F n
A
+
∆ lim
A → 0
∆
∆ v t
A t ˆ + lim
∆ A → 0
∆
∆ v s
A s ˆ
T n
= σ nn
ˆ + τ nt
ˆ t + τ ns
ˆ s where
τ
τ
σ nn nt ns
=
=
=
∆
∆
∆ lim
A → 0 lim
A → 0 lim
A → 0
∆
∆
∆
∆
F n = dF n
A dA v t = dv
A dA t
∆
∆ v s = dv
A dA s
=
=
=
Normal stresscomponent
Shear stresscomponent
Another shear componet
τ
σ
−
− NormalStress
Shear stress dF n
= σ nn dA dV t
= τ nt dA
Notation of stress components
The magnitude and direction of T n
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
(b) Second subscript- direction
σ is acting

Solid Mechanics
Rectangular components of stress
Cuts ⊥ to the coordinate planes will give more valuable information than arbitrary cuts.
T x
=
∆ lim
A → 0
∆
∆
F
R
A
=
∆ lim
A → 0
∆
∆
F x
A i
ˆ +
∆ lim
A → 0
∆ v y
∆ A j
ˆ +
∆ lim
A → 0
∆
∆ A k
T x
= σ xx
ˆ i + τ xy
ˆ j + τ xz where
σ xx
τ xy
=
=
∆ lim
A → 0
∆
∆ lim
A → 0
∆
∆
F x
A
∆ v
A y
= Normalstress
= Shear stress; τ xz
=
∆ lim
A → 0
∆
∆ v
A z = Shear stress

dF x
= σ xx dA dv y
= τ xy dA
Solid Mechanics dv z
= τ xz dA
Similarly,
T y
=
∆ lim
A → 0
∆
∆
F
R
A
=
∆ lim
A → 0
∆
∆ v x
A i ˆ +
∆ lim
A → 0
∆ F y
∆ A j ˆ + lim
∆ A → 0
∆
∆ A k
T y
= τ yx
ˆ i + σ yy
ˆ j + τ yz
ˆ k
T z
= τ zx
ˆ i + τ zy
ˆ j + σ zz
ˆ
σ xx
and τ xy
will act only on x -plane. We can see σ x
and τ xy only when we take section ⊥ to x -axis.
The stress tensor
Components directions a point “ P ” on the x-plane in x , y , z
σ jj
=
σ xx
τ xy
τ xz
τ yx
σ yy
τ yz
τ zx
τ zy
σ zz
Rectan gular stresscomponents
• This array of 9 components is called as stress tensor.
• It is a second rank of tensor because of two indices

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.
T n
= τ nx i
ˆ + τ ny j
ˆ + τ nz
ˆ
What is the difference between distributed loading & stress? q =
∆ lim
A → 0
∆
∆
F
R
A q = σ yy
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 σ x means.
= − 20 ;Pa τ xy
= − 10 P a
and τ xz
= 30 Pa at a point P

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 T n
acting on any plane “ n ” through that point.
The stress tensor will specify the state stress at point.
σ ij ′
=
σ
τ
τ
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.
σ ij
P
+ σ xx
= + τ yx
+ τ zx
+ τ xy
+ σ yy
+ τ zy
+ τ xz
+ τ yz
+ σ zz
σ xx
= σ xx
σ yy
= σ yy
(
( x,y,z x,y,z
)
)
Continuous functions of x,y,z
Let us consider a stress tensor or state of stress at a point in a component as

σ ij
=
− 10 5 − 30
5 50 − 60
− 30 − 60 − 100
Solid Mechanics
[
Equilibrium of stress element dy dz y
τ xz
τ xy
σ x x dx z
F x
0
]
σ x dydz + τ yx dxdz + τ zx dydx − σ x dydz − τ yx dxdz − τ zx dxdy = 0
F y
= 0 and F z
= 0 is satisfied. Similarly, we can show that

Solid Mechanics
M z
P
= 0
(
τ xy
) (
τ yx
)
τ xy
− τ yx
= 0
τ xy
= τ yx
= 0
Shearing stresses on any two mutually perpendicular planes are equal.
M x
P
= 0 τ yz
= τ zy
and M y
P
= 0 τ zx
= τ xz
Cross-shears are equal- a very important result
Since τ xy
= τ yx
, if τ xy
= − ve τ yx
is also –ve

Solid Mechanics
∴ The stress tensor
σ ij
=
σ xx
τ yx
= τ xy
τ xy
σ xy
τ xz
τ yz
τ zx
= τ xz
τ zy
= τ yz
σ yz issec ondrank symmetrictensor
Differential equations of equilibrium
[ F x
→ + = ]
σ x
+
∂
∂
σ x x τ yx
+
∂ τ yx
∂ y y x z τ zx
+
∂ τ
∂ zx z
σ
∂
∂
σ x x y z τ xy x z
∂ τ
∂ yx y
τ zx
∆ ∆ + ∆ ∆ ∆ =
2 τ
∂ zx z
0
∆ ∆ + ∆ ∆ ∆ = 0
Canceling x y
and
∆ z
terms and taking limit lim x y
∆ →
0
0
0
∂
∂
σ x x +
∂ τ yx
∂ y
+
∂ τ
∂ z zx + B x
= 0
Similarly we can easily show that

Solid Mechanics
∂
∂
∂
σ x x
τ xy
∂ x
∂ τ
∂ xz x
+
+
+
∂ τ yx
∂
∂
σ y yy
∂ y
∂ τ yz
∂ y
+
+
∂ τ
∂ zx z
∂ τ zy
∂ z
+ B x
=
+ B y
=
0
0
+
∂ σ
∂ z zz + B z
= 0
[
[ F x
=
F y
= 0
F z
=
• 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.
τ
M z p
= 0 xy
+
∂ τ xy
∂ x x y z
2 x
+ τ xy y z
∆
2 x
−
− τ yx x z
∆
2 y
= 0
τ yx
+
∂ τ
∂ y yx y x z
∆ y
2

2 τ xy
Solid Mechanics
2 y x z
+
∂ τ xy
∂ x
τ xy
2
+
2
∂ τ xy
∂ x
∆ x
2
−
2 τ yx
− τ yx
−
2 x y z
−
∂ τ yx
∂ y
∂ τ yx
∂ y
∆ y
2
= 0
2
2
= 0
Taking limit lim x y z
0
0
0
τ xy
+
∂ τ xy
∂ x
τ xy
− τ yx
= 0 τ xy
= τ yx
∆ x
2
− τ yx
−
∂ τ yx
∆ y
∂ y 2
= 0
Relations between stress components and internal force resultants

Solid Mechanics
F x
=
A
σ xx dA ; V y
τ
=
− τ xy
A
τ xy dA ; V z x
M x
=
A
( y τ xz
− τ xy
=
A
τ xz dA
M y
=
A
σ xz dA ; M z
= −
A
σ xy dA

Solid Mechanics
Plane stress- 2D State of stress
3
σ ij
=
σ x
τ xy
τ xz
τ yx
σ yy
τ yz
τ zx
τ zy
σ zz
− 6 components
σ ij
=
2
σ x
τ xy
τ xy
σ y
0
0
0 0 0
=
τ
σ x yx = τ xy
Stresscomponentsin plane xy
τ xy
σ y
If
σ ij
=
τ
σ x xy
( )
( )
τ
σ xy yy
( )
( )
− plane stress-is a --- state of stress
All stress components are in the plane x y i.e all stress components can be viewed in x y plane.

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 2 D -plane stress problems are represented by
σ ij
=
σ x
τ xy
τ xy
σ y
=
σ nn
τ nt
τ nt
σ tt

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 3 D 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
σ ij
=
σ x
τ xy
τ xy
σ y
is sufficient to tell about the state of stress at a point in the plane stress problems. dACs θ = Areaof BC dASin θ = Area of AC
F n
+ = 0
σ nn dA − σ x
θ θ τ xy
σ yy
= 0
σ nn
− σ x
Cos 2 θ − 2 τ xy
θ θ σ
− xy yy
Sin 2 θ
θ θ −
= 0

Solid Mechanics
σ
σ nn nn
=
=
σ x
Cos
2 y
Sin
2 θ + 2 τ xy
σ x
+
2
σ y
+
σ x
−
2
σ y
Cos 2 θ τ
θ θ xy
Sin 2 θ
F n
+ = 0
σ nt dA − σ x
θ θ τ xy
σ y
θ θ = 0
θ θ τ xy
θ θ −
τ nt
= − σ x
θ θ σ y
θ θ τ xy
τ nt
τ nt
(
Cos
2 θ − Sin
2 θ
= −
= −
(
θ
σ x
− σ
2 y
θ σ
) x
Sin
− σ y
2 θ τ
)
+ τ xy
(
Cos
2 θ − Sin
2 θ xy
Cos 2 θ
)
)
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
σ ij
=
σ x
τ xy
τ xy
σ y
This also means that

Solid Mechanics

Solid Mechanics
If θ θ we can compute on AB
If θ θ
π
2
we can compute on BC
If θ θ π we can compute on CD
If θ θ
3 π
2
we can compute on DA
• σ nn
and τ nt
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 σ nn
and τ nt
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 σ x
= I , σ y
= I ; τ xy
= − I xy
• 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.
σ nn
=
σ x
+
2
σ y
+
σ x
−
2
σ y
Cos 2 θ τ xy
Sin 2 θ
σ t
=
σ x
+
2
σ y
+
σ x
−
2
σ y
Cos 2 θ τ xy
Sin 2 θ
τ nt
= −
σ x
− σ
2 y
Sin 2 xy
Cos 2 θ

σ n
+ σ σ x
+ σ y
= σ x ′
+ σ y ′
= I
1
Solid Mechanics
I
1
= First invariant of stress in 2 D
σ σ τ 2 nt
= σ σ − τ 2 xy
= σ σ
′
− τ = I
2
I
2
= Second invariant of stress in 2 D
• I ,I are invariants of 2 D symmetric stress tensor at a point.
• Invariants are extremely useful in checking the correctness of transformation
• Of I
1
and I
2
, I
1
is the most important property : the sum of normal stresses on any two mutually
⊥ directions) is a constant at a given point.
⊥ planes
(
• In 2 D we have two stress invariants; in 3 D 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 :
σ x
σ y
τ xy
= − 46
= 12
= − 19
Q = − 15
it is in C.W.

Sin 2 θ
σ x
+ σ y
2
σ x
− σ y
2
=
− +
2
=
− 34
2
= − 17 MPas
=
− −
2
=
− 58
2
= − 29 MPa
= Sin ( − ) = − . ; Cos 2 θ
σ n
= − −
σ n
1
= −
×
= Cos
+ × .
( −
Substituting θ = − 15 in τ nt
equation
τ nt
= −
σ x
− σ
2 y
Sin 2 θ τ xy
Cos 2 θ
τ = − × − × .
τ = − 31 MPa
Solid Mechanics
)
σ n
=
σ x
+
2
σ y
+
σ x
−
2
σ y
Cos 2 θ τ xy
Sin 2 θ
= .

Solid Mechanics
σ σ n
2
= τ nt θ = 75
∴ σ t
= − − cos − sin 150
σ t
= − 1 4
τ tn
= τ n t
= τ nt θ = 75
τ tn
=
= + ×
31
Sin − ×
MPa
Cos 150
Now θ = 145
σ n
= − −
= − 32 MPa
Cos × − Sin ×
τ nt
τ nt
= . Sin
= − 31 MPa
− Cos 330
0
As a check
σ n
+ σ σ x
+ σ y
= − .
− .
= − 34 MPa s

Solid Mechanics
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) σ n
= ( ) =
σ n
+
2
σ y
+
σ n
−
2
σ y
Cos 2 θ τ xy
Sin 2 θ
(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
σ n function..
(3) σ n
=
σ n
+
2
σ y
+
σ n
−
2
σ y
Cos 2 θ τ xy
Sin 2 θ
For maxima or minima, we know that d d
σ
θ n 0
(
σ x
− σ y
)
Sin 2 θ + 2 τ xy
Cos tan 2 θ =
2 τ xy
σ x
− σ y
(4) The above equations has two roots, because itself after π . Let us call the first root as θ
P
1 tan repeats tan 2 θ
P
1
=
2 τ xy
σ x
− σ y tan 2 θ
P
2
= tan
(
2 θ
P
1
+ π
)
=
2 τ xy
σ x
− σ y

Solid Mechanics
θ
P
2
= θ
P
1
+
π
2 s
(5) Let us verify now whether we have minima or minima at
θ
P
1
and θ
P
2
∴ d d
2 σ
θ 2 n d d
2 σ
θ 2 n
= − 2
(
σ x
− σ y
)
Cos 2 θ − 4 τ xy
Sin 2 θ
= − 2
(
σ x
− σ y
)
Cos 2 θ
P
1
− 4 τ xy
Sin 2 θ
P
1
P
1
We can find Cos 2 θ
P
1 s and Sin 2 θ
P
1 s as
Cos 2 θ
P
1
=
2
σ x
− σ y
2 σ x
− σ y
2
+ τ 2 xy
Sin 2 θ
P
1
=
2
2 τ xy
σ x
− σ
2 y
2
+ τ 2 xy
=
Substituting Cos 2 θ
P
1
and Sin 2 θ
P
1
τ xy
σ x
− σ
2 y
2
+ τ 2 xy

d d
2 σ
θ 2 n
=
=
P 1
Solid Mechanics
=
− 2
(
σ x
− σ y
)(
σ x
− σ y
)
2
2
σ x
− σ y
2
+ τ 2 xy
−
−
(
σ x
− σ y
) 2
σ x
− σ y
2
2
+ τ 2 xy
−
4 τ 2 xy
σ x
− σ
2 y
4
σ x
− σ
2 y
2
+ τ 2 xy
2
+ τ 2 xy
− 4
σ x
− σ y
2
2
+ τ 2 xy
σ x
− σ y
2
2
+ τ 2 xy
∴ d d
2 σ
θ 2 n = − 4
σ x
− σ y
2
2
+ τ 2 xy
(-ve) d 2 d θ
σ
2 n
θ θ θ
π
2
= 2
(
σ σ y
)
Cos
(
2 θ
P
1
π
)
4 τ xy
Sin
(
2 θ
P
1
+ π
)
= 2
(
σ σ y
)
Cos 2 θ
P
1
+ 4 τ xy
Sin 2 θ
P
1
Substituting Cos 2 θ
P
1
&Sin 2 θ
P
1 m we can show that
∴ d d
2 σ
θ 2 n
P 2
= − 4
σ x
− σ
2 y
2
+ τ 2 xy s (+ve)

Solid Mechanics
Thus the angles θ
P
1 s and θ
P
2 s define planes of either maximum normal stress or minimum normal stress.
(6) Now, we need to compute magnitudes of these stresses
We know that,
σ n
=
σ x
+
2
σ y
+
σ x
−
2
σ y
Cos 2 xy
Sin 2 θ
σ n
P 1
= σ
1
=
σ x
+
2
σ y
+
σ x
−
2
σ y
Cos 2 θ
P
1
+ τ xy
Sin 2 θ
P
1
Substituting Cos 2 θ
P
1 s and Sin 2 θ
P
1
σ
1
=
σ x
+ σ
2 y
+
σ x
− σ
2 y
2
+ τ 2 xy
Similarly,
τ
σ n xy
θ θ
P 2
= θ
P 1
=
π
2
Sin
(
2 θ
P
1
+ π
)
= σ
2
=
σ x
+
2
σ y
+
σ x
−
2
σ y
Cos
(
2 θ
P
1
+ π
)
+
=
σ x
+
2
σ y
−
σ x
−
2
σ y
Cos 2 θ
P
1
− τ xy
Sin 2 θ
P
1
Substituting Cos 2 θ
P
1
and Sin 2 θ
P
1

σ =
σ x
+ σ
2 y
−
Solid Mechanics
σ x
− σ
2 y
2
+ τ 2 xy
We can write
σ
1 or σ
2
=
σ x
+ σ y
2
±
σ x
− σ
2 y
2
+ τ 2 xy
(7) Let us se the properties of above stress.
(1) θ
P
2
= θ
P
1
+
π
2 s - planes on which maximum normal stress and minimum normal stress act are ⊥ to each other.
(2) Generally maximum normal stress is designated by and minimum stress by σ
2
. Also θ
P
1
→
P
2
→ σ
2
σ
1
σ
1
> σ
0 −
2
− 1000 − alg ebraically i.e.,
σ
σ
1
2

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) θ
P
1
and θ
P
2
that define the principal planes are known as principal directions.
(8) Let us find the planes on which shearing stresses are zero.
τ nt
0
(
σ x
− σ y
)
Sin 2 xy
Cos θ tan 2 θ =
2 τ xy
σ x
= σ y
= directionsof principal plans
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 σ x
and σ y
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 σ
1
and say that
σ
2
are in two
⊥
directions, we can easily
σ x
+ σ y
= σ
1
+ σ
2
= σ x ′
+ σ y ′
= I
1

Solid Mechanics
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) τ nt
= −
σ x
− σ
2 y
Sin 2 xy
Cos 2 θ d τ d θ nt = −
(
σ x
− σ y
)
Cos 2 θ τ xy
Cos
For maximum or minimum d τ d θ nt 0
(
σ x
− σ y
)
Cos 2 θ − 2 τ xy
Sin tan 2 θ =
−
(
σ x
2 τ
− xy
σ y
)
This has two roots tan 2 θ
S
1
= −
(
σ x
− σ
2 τ xy y
)

tan 2 θ
S
2
= tan
(
2 θ
S
1
+ π
)
=
Solid Mechanics
−
(
σ x
2 τ
− xy
σ y
)
∴ θ
S
2
= θ
S
1
+
π
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 d d
2 τ
θ nt
2
= 2
(
σ x
− σ y
)
Sin 2 θ − 4 τ xy
Cos 2 θ
Cos 2 θ
S
1 d d
2 τ
θ nt
2
=
2
= 2
(
σ x
− σ y
)
Sin 2 θ
S
1
S
1
2 τ xy
σ x
− σ
2 y
2
+ τ 2 xy
− 4 τ xy
Cos 2 θ
S
1
Sin 2 θ
S
1
=
2
−
(
σ x
− σ y
)
σ x
− σ y
2
2
+ τ 2 xy
Substituting above values in the above equation we can show that

Solid Mechanics d
2 τ d θ nt
2
= - ve
S 1
Similarly we can show that d d
2 τ
θ nt
2
θ θ
S 2
= θ
S 1
+
π
2
= + ve
Thus the angles θ
S
1 and θ
S
2 define 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.
τ nt
τ nt
= −
θ θ
S
1
(
σ x
− σ y
)
2
= −
Sin 2
(
σ x
− σ y
)
2 xy
Cos 2 θ
Sin 2 θ
S
1
+ τ xy
Cos 2 θ
S
1
Substituting Cos 2 θ
S
1
and Sin 2 θ
S
1 s , we can show that
τ nt θ θ
S 2
= θ
S 1
+
π
2
τ max
= +
= −
(
σ x
− σ y
)
2
σ x
− σ y
2
Sin
(
Substituting Cos 2 θ
S
1
and Sin 2 θ
S
1
2
+ τ 2 xy
2 θ
S
1
+
) xy
Cos
(
2 θ
S
1
+ π
)
τ min
= −
σ x
− σ
2 y
2
+ τ 2 xy

Solid Mechanics
τ max
is algebraically > τ min
, however their absolute magnitude is same. Thus we can write
τ max or τ min
= ±
σ x
− σ y
2
2
+ τ 2 xy
Generally
τ
τ max
− θ
θ
S min
−
S
2
1
Q. Why τ max
and τ min
are numerically same. Because
θ
S
2
are ⊥ planes.
θ
S
1
&
(2) Unlike the principal stresses, the planes on which maximum and minimum shear stress act are not free from normal stresses.

Solid Mechanics
σ n
=
σ x
+
2
σ y
+
σ x
−
2
σ y
Cos 2 θ τ xy
θ
σ n θ θ
S
1
=
σ x
+
2
σ y
+
σ x
−
2
σ y
Cos 2 θ
S
1
+ τ xy
Sin 2 θ
S
1
Substituting Cos 2 θ
S
1
and Sin 2 θ
S
1 n
S 1
=
σ x
+ σ y
2
σ
+ τ n θ θ
S
2
= θ
S
1
+
π
2 xy
Sin
(
=
2 θ
S
1
+ π
)
σ x
+
2
σ y
+
σ x
−
2
σ y
Cos
(
2 θ
S
1
+ π
)
Simplifying this equation gives n
S
2
=
σ x
+ σ
2 y
Therefore the normal stress on maximum and minimum shear stress planes is same.
(3) Both the principal planes are planes of τ max
and min
⊥ to each other and also the
τ are also ⊥ to each other. Now let us see there exist any relation between them.

Solid Mechanics
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, σ n
=
σ x
+
2
σ y
+
σ x
−
2
σ y
Cos 2 θ τ xy
Sin 2 θ
This equation can also be written as
σ n
τ
−
σ x
+
2
σ y
=
σ x
−
2
σ y nt
= −
σ x
− σ
2 y
Cos 2
Sin 2 xy xy
Cos
Sin 2 θ
2 θ
σ n
−
( x
σ x
+ σ y
2
↓
− a ) 2
+
2
+ τ 2 nt
=
↓ y 2 =
σ x
+ σ y
2
↓
R 2
2
+ τ 2 xy

The above equation is clearly an equation of circle with center at a on τ σ plane it represents a circle with center at
σ x
+
2
σ y
, 0 and having radius
R =
σ x
− σ
2 y
+ τ 2 xy
This circle on
Mohr’s circle.
σ τ plane-
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

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 σ σ y ′
and
τ on an inclined element. For the sake of clarity we assume that, σ σ y ′ s and τ all are positive and σ x
> σ y

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 A
(
σ x
, + τ xy
)
Therefore we can locate the point A on the circle with coordinates
(
+ σ x
, + τ xy
) s
• 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 B
(
σ y
, − τ xy
) s
Since we assumed that for the sake of similarity σ y
< σ x s .
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 2 Q in
Mohr’s circle corresponds to Q in the stress element diagram.
Stresses on an inclined element
• Point A corresponds to
Q = 0
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 2 Q 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.
σ x ′
= σ avg
τ =
+
RSin β
RCos β
σ y ′
= σ avg
− RCos β stress element,thenthey become diametrically opposite point son thecircle, just likethe planes A& Bdid
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 P
1
and P
2
. Thus P
1
and P
2
represents points corresponding to principal stresses. In the current diagram the coordinates the of
P
1
P
2
=
=
σ
σ
1
2
σ
1
= σ avg
, 0
+ R
σ
2
= σ avg
− R
The principal direction corresponding to σ
2 θ p
1
, in c.c.w direction from the x-axis.
1
is now equal to

Solid Mechanics
θ p
2
= θ p
1
±
π
2
We can see that the points P
1
and P
2
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 the top and bottom of the Mohr’s circle.
S
1
and S
2
at
• The points S
1 points P
1
P
2
and S
2
are at angles 2 θ = 90 from
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
S
1
S
2
= + τ max
, σ avg s
= − τ max
, σ avg
τ max
= ± R avg
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 principal stresses.
σ -axis gives
• 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.

Mohr’ circle problem
Solution:
σ x
+
2
σ y
=
2
A - (15000,4000)
B - (5000,-4000)
R =
σ x
− σ y
2
2
+ τ 2 xy
=
(a)
= 5000
2 + 4000
2
R
σ x
= 6403
− σ y
2
MPa
= 5000
= 10000 MPa
2
Solid Mechanics
2
+ 4000
2

Point D : σ x ′
τ
= + Cos .
= − 6403 Sin .
Solid Mechanics
= MPa
= − 4229 MPa
Point D ′ : σ n
= σ y ′
=
τ nt
= τ
− Cos .
= 6403 Sin .
=
= 4229
MPa b) σ
1
= 16403 ; θ
P
1
=
σ
2
= 3597 MPa
2
= c) τ max
= 6403 MPa − θ
S
1
= .
= − .

Solid Mechanics
(2) θ = 45
Principal stresses and principal shear stresses.
Solution :
σ x
+
2
σ y
=
R =
σ x
− σ y
2
2
= − 20
2
+ τ 2 xy
=
A
B → (
( 50 40
)
)
2 p
1 p
2
= σ
1
=
σ x
+ σ y
= σ
2
=
σ x
2
+ σ y
2
2
( 40 ) 2 = 50 MPa
R = s
R 20 50 = − 70

Solid Mechanics
2 Q p
1
Q p
1
=
=
Q p
2
=
2 Q
Q s
1 s
1
=
=
Q s
2
=

Solid Mechanics
Q. σ x
= 31 MPa, σ y
= − 5 MPa and τ xy
= 33 MPa
Stresses on inclined element θ = 45
Principal stresses and maximum shear stress.
Solution :
σ avg
=
σ x
+ σ y
2
=
2
= 13 MPa
R =
σ x
− σ
2 y
2
+ τ 2 xy
=
A ( 31 33 )
B ( 5 33 )
τ
σ x ′
=
= RCos β σ avg s
+ = MPa
= − RSin β = − .
.
= − .
σ y ′
= RCos β σ avg
= − 20 MPa

∴ σ
1
=
σ
2
= −
θ p
1
=
τ max
=
τ min
σ σ
= − avg
= 13
−
MPa
θ s
1
= −
Solid Mechanics

Solid Mechanics
3D-stress components on an arbitrary plane
Basically we have done so far for this type of coordinate system
ˆ i n
′ = n n
ˆ i + n n ˆ
− j
D i r . c o s i n e s o f x ′
+ n
ˆ k n y x n
′ n
′
ˆ ˆ ˆ n z x n
′ n
′
ˆ ˆ ˆ

T n
= ˆ + ˆ + ˆ
T n
= σ i
ˆ ′ + τ ˆ j ′ + τ ˆ k ′
Solid Mechanics
[
F x
→ + = ]
T da = σ x dAn x x
+ τ yx dAn x y
+ τ zx dAn
T
T
T
= σ
= τ
= τ n + τ n
′
+ τ n
′ n + σ n + τ n
′ n + τ n
′
+ σ n
τ
σ x ′
τ τ
σ y ′
τ
τ τ σ z ′
σ τ x y
, τ x z
τ
τ
σ x ′
= T i
ˆ ′ =
(
ˆ + ˆ + ˆ
) (
=
=
ˆ ′ =
(
ˆ + ˆ + ˆ
ˆ ′ =
(
ˆ + ˆ + ˆ
) (
) (
σ y ′
σ z ′
=
(
=
(
ˆ
ˆ
ˆ
ˆ
T
T
T
ˆ +
ˆ +
ˆ +
ˆ +
ˆ +
ˆ +
ˆ
)
(1)
ˆ
ˆ
)
(2)
)
(3)
= σ
= τ
= τ
)(
)( n + τ n
′
+ τ n
′ n + σ n + τ n
′ n + τ n
′
+ σ n
ˆ
ˆ
ˆ
ˆ ˆ
)
(4)
ˆ
)
(5)

τ =
(
ˆ n = Cos θ n = Sin θ n n n = 0 n
ˆ
)(
ˆ
= − Sin θ
= Cos θ n n
= 0 n
ˆ
= 0
= 0
= 1
ˆ
Solid Mechanics
)
(6)
σ z ′
= 0
= σ z
: τ = 0 : τ = 0
σ x ′
= σ x
Cos
2 y
Sin
2 θ + 2 τ xy
θ θ
τ
σ y ′
= σ
= − x
(
Sin
σ x
2
− σ y
) y
Cos
2 θ − 2 τ
θ θ τ xy xy
θ θ
(
Cos
2 θ − Sin
2 θ
)
Principal stresses n ,n ,n z
T n
= σ = σ
( x
ˆ + ˆ +
T n
= nx
ˆ + ˆ + ˆ z
ˆ
)
σ x
τ xy
τ xy
σ y
0
0
0 0 0
Where
T nx
T ny
T nz
=
=
= τ
τ
σ n + τ n + τ n
+ σ n + τ n n + τ n + σ
Tn x
= σ n Tn y
= σ n Tn z
= σ n z

Solid Mechanics
τ
τ
( σ x
− σ )
( n x
+ τ
+ σ y
− σ
) n + τ n n + τ n n y
+ τ
+ ( σ z
− σ ) n z
= 0
= 0
= 0
Syst.of linear homog.eqns.
n x
= n y
= n z
= 0 : n
2 x
+ n
2 y
+ n
2 z
= 1
σ x
− σ τ xy
τ xy
σ y
− σ τ
τ zx zy
τ zx
τ yz n x n y n z
=
For non trivial solution must be zero.
−
σ 3 −
(
σ x
+ σ y
+
)
2 +
(
σ σ + σ σ + σ σ − τ 2 xy
− τ 2 yz
−
)
(
σ σ σ + 2 τ τ τ − σ τ 2 − σ τ 2 − σ τ 2 z xy
)
= 0
This has 3- real roots σ σ σ
3
τ
( σ x
− σ
+
1
(
σ
) n x
+ τ n + τ n y
− σ
1
) n y
+ τ n
= 0
= 0 and n + n + n = 1 n ,n ,n z
→
σ
1
> σ
2
> σ
3
σ
1
Stress invariants
σ 3 − I
1
σ 2 + I
2
σ I
3
0 (1)

Solid Mechanics
I
1
= σ x
+ σ y
+ σ z
I
2
= σ σ + σ σ + σ σ τ 2 − τ 2 − τ 2 x z xy yz zx
I
3
= σ σ σ + 2 τ τ τ − σ τ 2 − σ τ 2 − σ τ 2 z xy
σ 3 − I ′
1
σ 2 ′
3
0 stress inv ar iants
I ′
1
= σ x ′
+ σ y ′
+ σ z ′
I ′
2
= σ σ
′
+ σ σ + σ − τ 2
′ ′
− τ 2 − τ 2
I
1
I
2
I
3
=
=
=
σ
1
σ σ
σ σ
+ σ
σ
+
3
3D
2
+ σ
σ σ
3
I
1
=
1 2
= ′
2 3
= I ′
3
+ σ σ
Principal planes are orthogonal
2D
I
1
I
2
I
3
= σ
1
+ σ
=
2
= 0
ˆ ˆ
= ˆ + y
ˆ + z
ˆ
′ =
′
ˆ + y ′
ˆ + z ′
ˆ
T n
= ˆ + ˆ + ˆ
T n ′
= ˆ + ˆ + ˆ

Solid Mechanics
τ xy
ˆ
= n ′
ˆ
τ yx
ˆ n n ′ n ′
ˆ
( σ
1
) = ( σ
2
′ )
σ
1
( n n
′
+ n n
′
+ n n
′ n n
′
+ n n
′
+ n n
′
)
σ
1
≠ σ
2 n n
′
+ n n
′
+ n n
′
= 0
′ must be ⊥ to each other.
The state of stress in principal axis
σ
0
1
0 0
σ
0 0
2
σ
0
3
T n x
T n y
T n z
= σ
1 n x
= σ
2 n y
= σ
3 n z
σ n
= σ
1 n 2 x
+ σ
2 n 2 y
+ σ
3 n 2 z
T n
2
= T
2 n x
+ T
2 n y
+ T s
= σ 2 2
1 x
+ σ 2 2
2 y
+ σ 2 2
3 n z
τ 2 = T n
2
− σ 2 n

Solid Mechanics
3-D Mohr’s circle & principal shear stresses
σ ij
=
σ x
τ xy
τ xy
σ y
0 0
0
0
σ z
Once if you know σ
1 and σ
2
τ
τ
τ
1
σ
τ
2
σ
τ
3
σ
τ
2
1
=
σ
σ
2
−
2
σ
=
=
σ
1
1
+
−
2
2
σ
σ
3
=
σ
1
+ σ
2
3
3
2
3
=
=
σ
1
− σ
2
σ
1
− σ
2
2
2
σ
1
> σ
2
> σ
3
τ max
= max
σ σ σ
2
2
−
2
σ σ
3
−
2
σ
1

Solid Mechanics
• The maximum normal stress stress
σ
1
and maximum shear
τ max
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
τ
σ x xy
=
=
100
τ yz
MPa,
= τ zx
=
σ
0 y
= − 40 MPa, σ z
= 80 MPa and
Determine in plane max shear stresses and maximum shear stress at that point.
Solution:
σ
1
= 100 MPa, σ
2
= 80 MPa
τ
τ
12
13
=
σ
1
− σ
2
2
=
σ
1
− σ
2
3
=
=
τ
σ
23
12
σ
σ
13
23
=
=
σ
2
− σ
2
3
=
=
σ
30
1
+ σ
2
2
MPa
20 MPa
=
= 90
2
2
2
σ
3
= − 40 MPas
= 10 MPa
= 70 MPa
= 60 MPa
τ max
= max τ τ τ
τ max
= 70 MPa This occurs in the plane of 1-3

τ τ τ → Principal shear stress in 3 D
τ max
= max
( τ τ τ )
Solid Mechanics

Solid Mechanics
Plane stress
σ
σ
1
3
>
=
σ
σ z
= 0
τ = ±
σ x
− σ
2 y
2
+ τ 2 xy
---- in plane principal shear stresses.
τ max
=
σ σ
2
3 =
σ
2
1

Solid Mechanics
Problem
At appoint in a component, the state of stress is as shown.
Determine maximum shear stress.
Solution:
σ ij
=
100 0
0 50
- plane stress problem
We can also write the matrix as a ij
=
100 0 0
0 50 0
0 0 0
σ
σ
1
2
=
=
100
50
σ
1
− σ
2
2 =
2
τ max
= 25 MPa
= 25

Now with σ
1
= 100 , σ
2
= 50 , σ
3
= 0
τ max
=
2
3 = 50 MPa
Occurs in the plane 1-3 instead of 1-2
Solid Mechanics

Solid Mechanics
Some important states of stresses
(1) Uniaxial state of stress : Only one non-zero principal stress.
σ
1
0 0
0 0 0
0 0 0
=
σ
1
0
0 0
- plane stress.
(2) Biaxial state of stress : two non-zero principal stresses.
σ
0
1
0 0
σ
1
0
0 0 0
=
σ
0
1
0
σ
1
- plane stress
(3) Triaxial state of stress : All three principal stresses are non zero.
σ
0
1
0 0
σ
0 0
2
σ
0
3
− 3 D stress
(4) Spherical state of stress : σ
1
= σ
2
= σ
3
(either +ve or – ve)
σ
0
0 0
σ
0 0
0 − 3 D stress-special case of triaxial stress.
σ

Solid Mechanics
(5) Hydrostatic state of stress
+
0
0
P 0 0
+ P 0
0 + P
hydrostatic tension
−
0
0
P 0 0
− P 0
0 − P
hydrostatic compression.
(6) The state of pure shear
σ ij
=
σ x
τ xy
τ xz
τ xy
σ y
τ yz
τ zx
τ zy
σ z
σ ij
=
0 τ
τ
τ τ
0 τ
τ
0
Then we say that the point P is in state of pure shear .
I
1
= 0 shear
is necessary and sufficient condition for state of pure

Solid Mechanics
Octahedral planes and stresses
If n x
= n y
= n z
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
T n
2
=
1 n
2 x
σ
+
2 2
1 n x
σ
2 n
2 y
+ σ
+
2 2
2 n y
σ
3 n
2 z
+ σ 2 2
3 n z n
2 x
+ n
2 y
+ n
2 z
= 1 n x
= n y
= n z
= ±
1
3
=
σ oct
=
=
σ
σ
1
1
+ σ
1
3
2
3
+
2
σ
+
3
σ
1
1
3
2
+ σ
1
0
1
3
2

Solid Mechanics
σ
1
+ σ
3
2
+ σ
3 =
3
1
σ oct
= canbeint erpreted − − meannormalstress at a pt.
τ oct
= T n
2
− σ 2 oct
τ oct
=
1
3
(
2
) 2 + ( σ
2
− σ
3
+ σ
3
− σ
1
) 2
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:
τ oct is smaller than planes at a point.
τ max
(exist only on 4 planes) but can exist on 8

Solid Mechanics
Decomposition into hydrostatic and pure shear stress
σ ij
=
σ x
τ xy
τ xz
τ yx
σ z
τ yz
τ zx
τ zy
σ z
Mean stress P =
σ x
+ σ y
+ σ z
3
=
I
3
1
σ x
τ xy
τ xz
τ yx
τ y
τ yz
τ zx
τ zy
σ z
=
P 0 0
0 P
0 0
0
P
+
Hydrostatic stat of stress
Dilitationalstress
σ x
−
τ yx
τ zx
P τ xy
σ y
−
τ zy
P
σ z
τ xz
τ yz
− P
State of pureshear
Deviatoric state of stress
Stress deviator
Thus the state of the stress at a point can alos be represented by sum of dilational stress and stress deviator

σ
0
1
0 0
σ
0 0
2
σ
0
3
=
Solid Mechanics
P =
σ
1
+ σ
3
2
+ σ
3 =
I
3
1
P
0
0 0
P 0
0 0 P
+
σ
1
0
− P
0
0
σ
2
0
− P
0
σ
3
0
− P
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
( ) ˆ + ( ) ˆ + ( ) ˆ
(
( )
)
( )
(x,y,z) is the point in the undeformed geometry
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 -
• The strain component that measures rotation of any two
⊥ lines is – shearing strain-

Solid Mechanics
Normal strain points.
ε - Account for changes in length between two
∈ n
( ) = lim
∆ → 0
P Q − PQ
PQ
= lim
0
∆ − ∆ s
∆ s
We can also define the same point x
, , z
(1) By definition ∈ x
is + if
∈ x
is - if s
* s
* s s s lim as
* ( 1
∆ → n
)
0 s
∆ s
* ( 1 ) s if s s s s
0
(2) It is immaterial how P Q is oriented finally. However for
∈ n
we must consider PQ in the direction of ˆn in the undeformed geometry
(3) In general ∈ =∈ n
( )
(4) No units.
Mm/mm,0.5%=0.005;
µ = 10
− 6 , µ
(5) Meaning of ∈ nn
Shearing strain -
= × − 6 = .
mm / mm
Accounts the change in angle
Y n +
( ) Change in angle between
⊥ lines in ˆn& t ˆ direction.
( ) x y
0
0
π φ nt
= lim x y
0
0

Solid Mechanics
(1)We must select two ⊥ lines in the undeformed geometry.
(2)Units of Y nt
→ radius.
(3)By deflection Y nt
= Y tn
(4)Two subscripts are required for
Y - to show directions of initial infinitesimal line segments.
(5) Y nt
is +ve if angle is decreased
Y nt
is -ve if angle is more.
By taking two ⊥ lines
We can define ∈ ∈ nt
Rectangular strain components
∈ ∈ y andY xy
− PQRS
∈ ∈ y andY yz
− QABS
∈ ∈ xz
− RSCD
E ij
=
∈
Y
Y x xy xz
Y xy
Y xz
∈ y
Y yz
Y yz
∈ z
Rectangular strain components .
We then say that we have strain computer associated with x,y,z coordinate system.
They represent the state of strain at a point , since we can determine strain along any direction ˆn

lim x
*
∆ → 0 x
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 element PQRS . We shall demonstrate that ‘ w ’ has no impact. So it can be neglected.
P → u,v
Q u
∂
∂ u x
∆ x ; v +
∂
∂ v x
∆ x
PQ = ∆ x
= ∆ x
*
∆ x * ( 1 ) x
( ) x
=
1
∂ u
∂ x
2 ∂ v
∂ x
2
+ ∆ x + ∆ x +
+
∂
∂ u x
+
∂
∂ u x
2
+
∂
∂ v x
2
+
∂ w
∂ x
∆ x
2 ∂ w
∂ x
2
∆ x

Solid Mechanics lim
∆ → 0
∆ − ∆ x
∆ x
= lim
0
∈ = 1 2
∂
∂ u x
+
∂
∂ u x
+
∂
∂ u x
∂
∂ u x
2
+
∈ = 1 2
∂
∂ v y
+
∂
∂ u y
2
+
2
+
∂
∂ v x
∂ v
∂ x
2
+
∂
∂ v y
2
+
2
+
∂
∂ w x
∂ w
∂ x
2
− 1
2
− 1
∂
∂ w y
2
− 1
∈ = 1 2
∂
∂ w z
+
∂
∂ u z
2
+
∂ v
∂ z
2
+
∂
∂ w z
2
− 1
So far no assumption has been made except for size of x, y& z
Cos φ * xy
= 1 +
∂
∂ u x
∆
∆ x x
*
∂
∂ u y
∆
∆ y y
*
+
∂ ∆
∂ x ∆ x *
1 +
∂ v
∂ y ∆
∆ y y
*
+
∂ ∆
∂ x ∆ x *
∂
∂ w y ∆
∆ y y
*
Y xy
= lim x y
∆ →
0
0
0
π φ
* xy

Solid Mechanics
SinY xy
= lim Cos x y
∆ →
0
0
0
φ * xy
SinY xy x * y *
= lim x y
∆ →
0
0
0
( 1 )
(
1
) x y
1 + + 1 + v v
∂ ∂
+ w w
∂ ∂
∆ ∆
∆ ∆ *
SinY xy
= lim x y
∆ →
0
0
0
∂
∂ u y
+
∂
∂ v x
+ u u x y
+ v v x y
+ w w x y
( 1 + ∈ x
) (
1 + ∈ y
)
Y xy
=
Sin
− 1
Y yz
= sin
− 1
∂
∂ u y
+
∂
∂ v x
+ u u x y
+ v v x y
+ w w x y
( 1 +∈ x
) (
1 +∈ y
)
∂
∂ u y
+
∂
∂ v x
+ u u x y
+ v v x y
+ w w x y
( 1 + ∈ x
) (
1 + ∈ y
)
Y xz
= sin
− 1
∂
∂ w x
+
∂
∂ u w
(
+
1 w w
∂ ∂
+∈ x
)(
+
1 u u
∂ ∂
+∈ z
)
+ v v
∂ ∂
All bodies after the application of loads under go “small deformations”

Solid Mechanics
Small deformations :
(1) The deformational displacements infinitesimally small .
= ˆ + + are
(2) The strains are small
(a) Changes in length of a infinitesimal line segment are infinitesimal.
(b) Rotations of line segment are also infinitesimal.
∂ ∂ ∂ ∂
∂ ∂ ∂ ∂
1 ∈ x
1 ; 1 ; z
1 ;
∂
∂ u x
2
; are negligible compare to quantities.
∈ = 1 2
2 ∂ u
∂
∂ u x
∂ x
2
− 1
− 1
∂ u
∂
∂ x v
∂ y
∂ w
∂ z
SinY xy
≈ Y xy

Solid Mechanics
Y xy
=
∂
∂ u y
+
∂
∂ v x
(
1 + ∈ +∈ y
) =
∂
∂ v x
+
∂
∂ u y
Y xz
Y yz
=
=
∂
∂
∂
∂ w v z x
+
+
∂
∂
∂
∂ u w y z
Another derivation : Let us take plane PQRS in xy plane.
Also assume that
( ) ( )
only.
Small deformation
Displacements are small
Strains are small lim
∆ → 0
P Q −
PQ
PQ
= x
∆ x x
∆ x
*
P Q
′ lim
∆ → 0
1
= 1 +
∂
∂ y x
∆ x
+
∂ y
∂ x
∆ x x x
=
∂ u
∂ x lim
0
1 +
∂
∂ v y y y
∆ y
=
∂ v
∂ y
Strains<0.001
Y < .
∆ = 0 2002 s
− 4 mm

Solid Mechanics
Y xy
= x y lim
0
0
− * xy
= x y lim
0
0 tan α =
1
∂
∂ x
+ v
∂ y
∂
∆ x x
∆ x
=
1
∂ v
∂ x
+
∂ y
∂ x
α
β
=
=
∂ v
∂
∂ x u
∂ y
Y xy tan α α
=
∂
∂ u y
+
∂
∂ v x
∂ ∂ ∂ ∂
∂ ∂ ∂ ∂
2
2
∂ u
∂ x
,
∂ u
∂ y
,
1
∂ v
∂ yx
2
We can define the state of strain at point by six components of strains
∈ ∈ ∈ Y , Y , Y yz
↓ ↓ ↓
Y yx
Y zx
Y zy
State of strain
∈ x
Y xy
E ij
= Y xy
∈ y
Y xz
Y yz
Y
Y
∈ xz yz z
-
-
Engineering strain matrix
We can find ∈ n
in any direction we can find Y nt for any two arbitrary directions.

∈ =
Y yz
=
0
0
Y zx
= 0
2D- strain transformation
Plain strain : In which
∈ x
Y xy
Y xy
∈ y
∈ =∈ x
∈ =∈ y
( )
( )
Y xy
= Y xy
( ) 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

Solid Mechanics
Given ∈ ∈ xy what are ∈ ∈ nt
.
We can always draw PQRS for given ˆn
If ∈ ∈ xy
As in case of stress we call these formulae as transformations laws.
α
1
=
∈ x
=∈ x
=∈ x dxSin θ ds sin θ dx ds
θ θ
α
2
=∈ y cos θ
α
3
=
= dy ds
=∈ y
θ dy ds
θ θ
θ θ

x
x x
θ
y
θ
θ θ
θ
Solid Mechanics
θ
y
y
θ
xy
θ
2 θ
θ θ
∈ +∈
2 y
+
∈ −∈
2 y cos 2 θ +
Y xy
2 sin 2 θ
α = −∈ x
θ θ + ∈ y
θ θ − Y sin
2 θ
β = −∈ x cos θ ( − sin θ ) + ∈ y
=∈ x
θ θ −∈ y cos θ ( − sin θ ) − Y cos
2 θ
θ θ − Y cos 2 θ
Y nt
2
= −
(
∈ −∈
2 y
) sin 2 θ +
Y xy
2 cos 2 θ
∈ x
Y xy
Y xy
∈
Y
2 x
∈ y
Y
∈ xy
2 y
- state of strain at a point
∈ =
Y
2
- strain tensor
∈
∈ x xy
∈
∈ xy y xy xy
τ
σ x xy
τ
σ
Replace
τ
σ
σ x y xy
→∈ x
→∈ y xy y
Y xy
2
- stress tensor

Principal shears and maximum shear
In plane- principal strains
2 p
=
2 y
ϒ xy
/ 2
θ p
1
− θ p
2
− ⊥ to each other
1
,
2
∈ >∈
2 tan 2 θ s
= −
θ θ p
1
± π
( y
)
2 ∈ xy
/ 4
θ s
1
− θ s
2
− ⊥ to each other
σ x
+ σ y
= y
J
I
I
1
σ σ − τ 2 xy
= I
2
∈∈ −∈ = J
2
2
Y xy
2
= J
2
∈ max or ∈ min
R
Y max
2
Y min
2
=∈ max
− θ s
=∈ min
− θ s
2
1
Solid Mechanics
2 y
2
+ ∈ 2 xy

Mohr’s Circle for strain
∈ x ′
,Y , ∈ y ′
3D -strain transformation
σ x
→∈ x
; σ y
→∈ y
; σ z
→∈ τ
(
∈ xy
∈ xz
) ∈ xy
( )
∈ yz
(
∈ xz
∈ yz
)
= 0
Y xy
2
1
, ,
3
- ∈ >∈ >∈
3
2 x 2 y 2
∆ s *
2
= P Q
′ 2
+ P R
′ 2
= 1
∂ u
∂ x
2
1
∂ v
∂ x
2
+ ∆ x + + ∆ y x 2
Solid Mechanics y 2

Solid Mechanics
∈ = 1 +
= 1 +
∂ u
∂ x
∆ y
∆ x
2
2
+ 1 +
∂ v ∆ y x x
2
2 x 2 y 2
=
=
1 +
∂ u
∂ x
2
+ 2
∂ u
∂ x
∂
∂ u x
2
1
∂
∂ v y
2
+ 2
∂ v
∂ y
1 +
∆ y
∆ x x
2
1 2
∂ v
∂ y
1 +
∆
∆ y x
2
∆ x
2
2
∆ x x
2 y
2
Transformation
σ x ′
= σ
+ τ n
2 + σ n
2 + σ n
2 n n + τ n n
′
+ τ n n
′
σ x
σ y
σ z
→∈ x
→∈ y
→∈ zx
2 x
2
τ xy
τ yz
τ zx
→∈ xy
→∈ yz
→∈ zx y
2
τ
′
+ ∈ n 2 + ∈ n 2
′
+∈ n 2 n n +∈ n n
′
+∈ n n
′
→∈ →
Y
2

Solid Mechanics
Principal strains:
( ∈ −∈ ) n x
+ ∈ n +∈ n
∈
∈
( ) n y
+ ∈ n n +∈ n
=
( ) n z
=
=
0
0
0
( )
∈ xy
∈ xz
∈ xy
( )
∈ yz
J
2
System of linear homogeneous equations
(
∈ xz
∈ yz
)
= 0 x z y z
2
1 1
J
2
J
J
1
=∈ +∈ +∈ z
0
2 xy
2 yz
2 zx
∈ x
∈ xy
∈ xy
∈ y
+
∈
∈ y yz
∈
∈ yz z
+
∈
∈ x xz
∈
∈ xz z
J
3
=∈ ∈ ∈ + ∈ ∈ ∈ −∈ ∈ −∈ ∈
∈ x
∈ xy
∈ xz
−∈ ∈ ∈ ∈ y
∈ yz
∈ zx
∈ zy
∈ z
∈ >∈ >∈
3

∈
( ∈ −∈
1
) n x
+∈ n + ∈
(
1
) n y
+ ∈ n
2 x
+ n
2 y
+ n
2 z
= 1
= 0
Solid Mechanics
= 0 n ,n & n z unique
J
J
2
3
=∈ + ∈ + ∈
1 2
=∈ ∈ ∈
3
=∈ ∈ +∈ ∈ +∈ ∈
Decomposition of a strain matrix into state of pure shear + hydrostatic strain
∈ x
∈ xy
∈ xz xy
∈ xz
∈ 0 0
∈ = ∈ yx
∈ y
∈
∈ zx
∈ zy
∈ yz z
= ∈
∈ yx zx
∈ −∈ ∈ yz
+ 0 ∈
∈ zy
0 0
0
∈
State of pureshear Hydrostatic where ∈=
3 y z

Solid Mechanics
Plane strain as a special case of 3 D
∈ = 0 is also a principal strain z → is a principal direction
if
1 2
;
1 2
+ve if ∈
1
+ve, ∈
2
-ve. if ∈
1
+ve, ∈
2
-ve
P & z ′ will come closer to the maximum extent,
so that the included angle is
π
2
−∈ max

Solid Mechanics
Transformation equations for plane-strain
Given state of strain at a point P.
E ij
=
∈ xx
Y xy
Y
∈ xy yy
This also means that deformation
Now what are the strains associated with x ,y i.e
E =
∈
Y
Y
∈
This also means that

Solid Mechanics
Assume that xx
, yy
and Y are +ve
Applying the law of cosines to triangular P* Q* R*
( P* R* cos
π
2
2
+ Y xy
P* R*
2
Q* R* ) 2 − 2 ( )
∆ x ′ ( 1 +∈ x ′
)
∆ y
(
1 +∈ y
)
2
= ∆ x ( 1 + ∈ x
) 2
+ ∆ y
(
1 +∈ y
) 2
2 x ( 1 + ∈ x
) cos
π
2
+ Y xy
∆ = ∆ x cos θ and ∆ = ∆ x sin θ cos
(
π
2
+ Y xy
)
= − sinY xy
≈ − Y xy
∆ x ′ 2 ( 1 + ∈ x ′
) 2
= ∆ 2 2 θ ( 1 +∈ x
) 2
+ ∆
2 θ θ ( 1 + ∈ x
) (
1
2 2
+ ∈ y
θ
(
−
1
Y
+ ∈ y
)( )
) 2

Solid Mechanics
( 1 + ∈ x ′
) 2
=
1
2 x ′
2 cos 2
−
+
2 cos
θ
2
( 1 + ∈ x
) 2
+
θ θ (
1 sin
+∈ x
2
) (
θ
1
(
1 + ∈ y
+∈ y
)
−
2
)( )
θ
(
1
2 θ xy
(
1
2 x
2
) y sin
2 θ x y
(
1
)
2 y
1 2
+ cos
2 θ ( 1 2
(
= cos 2 θ ( 1 2 )
) sin sin 2 θ
2 θ
(
1 2 y
) y
)
(
1 2 y
)
+ 2 θ
2 y
)
1 2
∈ =∈ x
′
1 2 cos
2 θ x
+ ∈ y cos
2 θ sin
2 θ
2 y
+
Y xy
2 sin
2 sin 2 θ
θ +
If θ
′
∈ +∈
2 y
+
∈ −∈
2 y
π
2
∈ y ′
∈ + ∈
2 y
+
∈ −∈
2 y cos 2 θ cos 2 θ
+
Y xy
2
+
Y xy
2 sin 2 θ sin 2 θ
2 θ
′
∈ + ∈
2 y
+
∈ −∈
2 y cos 2 θ −
Y xy
2 sin 2 θ
∈ +∈ =∈ +∈ y
= J
1
= first invariant of strain.

Solid Mechanics
∈ x Q
2
π
4 y
2 y
Y
+ xy
Y xy
2
Y xy
2
( x y
)
2
Y
OB ′ x ′ ′
Y
2
OB ′
− ∈ +∈ y
(
(
2
OB ′
− ∈ +∈ y
)
′
)
∈ x ′
π
4
∈ +∈
2 y
−
∈ −∈
2 y sin 2 θ +
Y xy
2 cos 2 θ - (4)
Substituting (4) in (3)
Y
Y
( y
) ( x y
) sin 2 θ
( y
) sin 2 θ +
+ tensorial normal strain
=∈ ∈ ∈ z
2 θ
( y
)
∈ = engineering normal strain tensorial shear strain
( )
=
Engineeringshear strain
2
Y xy
2

∈ xx
∈ xy
∈ = ∈ xy
∈ yy
∈ zx
∈ zy
Y xz
2
∈ yz
( ∈ =∈ zz
)
Components.
- Strain tensors
Solid Mechanics
∈ +∈
2
∈ +∈
2 y y
+
−
∈ −∈
2
∈ −∈
2 y y
∈ = −
( y
) sin
2
2 θ cos cos
2 θ
2 θ
+∈ xy
+∈ xy
−∈ xy cos 2 θ sin sin
2 θ
2 θ

Solid Mechanics
Problem:
An element of material in plane strain undergoes the following strains
∈ = 340 10
− 6 ∈ = 110 10
− 6 Y xy
= × − 6
Show them on sketches of properly oriented elements.
Solution:
∈ = 340 10
− 6
∈ = − × − 6 ∈ = ×
− 6
; Y
′ =
180 10

Solid Mechanics
Problem :
During a test of an airplane wing, the strain gage readings from a
45
rosette are as follows gage A,
− 6
− 6 − 6
Determine the principal strains and maximum shear strains and show them on sketches of properly oriented elements.
Solution :
∈ = 520 10
− 6
∈ = 360 10
− 6
80 10
− 6
Y xy
= × ×
= × − 6
( x y
)
− 6 − rad
(
× − 6 − × − 6
)
(1)
2 y
=
× − 6 − ×
2
− 6
= × − 6

2 y
=
×
− 6 + ×
2
Solid Mechanics
− 6
= × − 6 tan 2 θ p
=
Y xy
2
=
2 ∈ xy y
= e × ×
− 6
− 6
2
− 6
= × − 6
∴ 2 θ p
=
θ p
= θ p
=
∈
1 or y
±
2
= × − 6 ±
(
= × − 6 ± .
2
× y
2
+∈ 2 xy
− 6
− 6
∴ ∈ = ×
×
− 6
− 6
∈ x
=
=
∈ +∈
2
= × y
×
+
∈ −∈
− 6
2 y
220 10
− 6 + ×
Cos
− 6
2 θ + ∈ xy cos ( ×
Sin 2 θ
.
) + ×
− 6
− 6
Sin ( ×
) 2
)

Solid Mechanics
θ p
1
= .
and θ p
2
= .
(b) In- plane maximum shear strains are
∈ xy max or y
2
+∈ 2 xy
2
= ± × − 6
( )
( )
=
= −
×
×
− 6
− 6
2 s
= −
( y
)
2 .
∈ xy
=
− × − 6
− 6
∈
2 Q s
=
Q s
= − Q s
=
Q .
= −
= −
(
.
×
2 y
− 6
)
Sin
− .
(
×
.
) +∈
− 6 xy
=
Cos
.
( )
×
− 6

θ = − θ s
2
= −
Y min
= −
Y max
=
Solid Mechanics
×
×
− 6
− 6
∈=
2 y
= ×
− 6

Solid Mechanics
• 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 curve of the mild steel.
σ −∈
Structural steel = mild steel = 0.2% carbon=low carbon steel
∈=
L f
L
− o
L o
Region O-A
(1) σ and ∈ linearly proportional.
(2) A - Proportional limit
σ p
- proportionality is maintained.
(3) Slope of AO = modulus of elasticity “ E ” – N/m 2 ,Pa
(4) Strains are infinites ional.

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 σ ys
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 are said to be large.
∈ s
above A
(8) ∈ <∈
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) σ u
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
(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
σ = ∈
E = Modulus of elasticity – Pa,N / m
2
= Young’s modulus of elasticity.
(5) σ x
= ∈ x
or σ y
= ∈ y
(6) σ = ∈ 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
′ d f
− d o d o
this comes out to be –ve
( )
∈ ′ is perpendicular to ∈
=− lateralstrain axialstrain
=
−∈ ′
∈
If a bar is under tension ∈ +ve, ∈ ′ -ve and ν = +
If a bar is under compression ∈ -ve, ∈ ′ +ve and ν = +
= always +ve = material constant
For most metals ν = 0 25 0 35
Concrete 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 τ
(1)To plot τ ,Y the test is twisting of hollow circular tubes
Yield point
Proportional limit
G
1
ϒ
(2) τ ,Y diagrams are (shape of them) similar in shape to tension test diagrams
( σ Vs ∈ )
for the same material, although they differ in magnitude.
(3)From τ − Y 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.
τ = GY - Hooke’s law in shear
G = Shear modulus of elasticity or modulus of rigidity.
= Pa or N / m s
Proportional limit
Elastic limit
Yield stress
Ultimate stress
Material properties

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
∈=∈
( )
( x,y,z
)
( x,y,z
)
Homogeneous
Elastic properties do not vary from point to point. For nonhomogenous body
( )
( )
( )

Solid Mechanics
Linearly elastic
Material follows Hooke’s law
τ
σ
= GY
Isotropic
Material properties are same in all directions at a point in the body
ν = C
2
1
3 for all θ for all θ for all θ
The meaning is that
σ
σ x y
E x
E y
The material that is not isotropic is anisotropic
( )
The meaning is that
σ
σ x y
= E
1
∈ x
= E
2
∈ y
E
1
≠ E
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.
σ = ∈ v = −
∈ ′
∈
Tensiletest
τ = GY - Shear test or torsion test.
What are the stress –strain relation for an element subjected to 3 D 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 τ xy
can only produce the corresponding shear strain Y xy
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
From Hooke’s we can write
σ x
is applied to the element.
σ
E x
∈ = − v v
σ
σ
E x
E x

Solid Mechanics
Only σ y applied
σ y
E
∈ = − v v
σ
σ
E y
E y
Similarly, σ z
alone is applied
σ
E z
∈ = − v v
σ
σ
E z
E z
Contribution to ∈ x
due to all three normal stresses is
∈ =
σ x − v σ
E E y
− v σ
E
3
Therefore
1
E
1
E
1
E
σ
σ
σ x y z
− v
(
σ y
+ σ z
)
− v ( σ x
+ σ z
)
− v
(
σ x
+ σ y
)
Normal strains are not affected by shear stresses

Solid Mechanics
Now let us apply only τ xy
Y xy
=
τ xy
G
Similarly because of τ yz and τ xz
Y yz
Y xz
=
=
τ yz
τ
G xz
G
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
Y xy
Y yz
Y xz
=
1
E
1
E
1
σ
σ
=
E
τ
=
σ xy
τ
G yz
τ
G xz
G x z y
− v
(
σ y
+ σ z
)
− v ( σ x
+ σ z
)
− v
(
σ x
+ σ y
)
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,
∈
∈
∈
Y x y z xy
Y yz
Y xz
=
1 − v − v
−
E E E v 1 − v
−
E E E v − v 1
E E E
0
0
0
0
0
0
0
0
0
0
0
0
1
G
0
0
0
0
0
0
1
G
0
0
0
0
0
0
1
G
τ
σ
σ
σ x y z xy
τ yz
τ xz
Stress components in terms of strains y e z
1
E
(
σ x
+ σ y
+ σ z
−
=
(
σ x
+ σ y
+ σ z
)
E v
∈ +∈ +∈ = e
σ x
+ σ y
+ σ z
) s
1
E
σ x
− v σ x
− v
(
σ y
+ σ z
)
=
1
E
=
1
E
σ x
− v
(
σ x
+ σ y
+ σ z
)
+ v σ x
σ x
( 1 v ) v
(
σ x
+ σ y
+ σ z
)

Solid Mechanics
=
=
σ
1
E
σ x
( 1 )
(
E v )
−
(
( veE v ) ve v )
∴ σ x
= ∈ + ve
− v 1
E
+ v
1
E
+ v
= µ (mu) where λ =
1 + v
Ev
( )(
λ µ are Lames constants
τ
σ z xy
= e λ z
=
µ
2 µ Y xy
τ
σ
σ x y e λ
λ x e y
µ
µ xy
= = 2 µ Y yz
τ xy
= = 2 µ Y zx v )
Lame’s constants have no physical meaning
Stress-strain relations for plane stress

σ x
= σ x
σ y
=∈ y
( )
( )
τ xy
= τ xy
σ z
= τ yz
= τ zx
= 0
Solid Mechanics
Y
Y xy yz
=
=
1
E
1
E
τ v
E
Y
(
(
σ
σ xy
G xz
( x y
σ
= x
−
−
0 v v
+
σ
σ
σ y x y
)
)
)
=
1
−
− v v
( y
)
Stress- strain relations for plane strain
∈ =∈ x
∈ =∈ y
( )
( )
Y xy
= Y xy
( )
∈ = Y xz
= Y yz
= 0 e y
σ x
= e λ µ σ x
σ y
= e λ µ σ
σ z
= −
= −
= − v v
= − ve
(
σ x
+ σ y
) y
=
( )
( )
σ z
( )
(
( e )
(
2
2
λ µ )
)
(
∈ +∈ y
)
τ xy
= GY xy
τ xz
= τ yz
= 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
σ x
=
1
E
λ e
σ x
µ
− v
+ ∈ x
σ
1
E
σ y
− v σ y x
σ y
= λ e µ y
σ
σ
1
E
σ x
− λ µ ∈ y
= σ
1
E
σ x
+ v σ y x
1 +
E v
Principal stress and strain directions of isotropic materials x x
= 2
= −
σ
σ y y
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.

Solid Mechanics
Relation between
σ
1
= τ xy
σ
2
= − τ xy
1
E
− τ
(
τ xy xy
+ v τ xy
( 1 + v )
E
Y xy
2
=
τ
2 xy
G
− τ xy
)
=
τ xy
1
E
1
E
( σ
1
− v σ
( σ
2
− v σ
1
2
)
)
( 1 + v )
E
τ xy
2 G
( 1 + v )
E
=
τ xy
2 G
G =
(
E
+ v )
Only two elastic constants are independent.

Solid Mechanics
Volumetric strain-dilatation
Consider a stress element size dx,dy,dz
After deformations dx * dy * dz *
( 1 x
)
(
1 y
) dx dy
( 1 z
) dz
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 Y xy
.
The volume dv *
of the deformed element is then given by dv
* =
( )
× dz
*
( )
=
( xy
)
= xy
∴ dv * = dx dy dz CosY xy
For small Y CosY xy
≈ 1
∴ dv * = −
( 1 x
) (
1 +∈ y
) ( 1 +∈ z
) dxdydz dropping all second order infinitesimal terms

dv
*
( y z
) dxdydz
Solid Mechanics
Now, analogous to normal strain, we define the measure of volumetric strain as
Volumetric strain = final volume-initial volume initialvolume e = dv
* − dv dv e =∈ +∈ +∈ z
• e = volumetric strain = dilatation. This expression is valid only for infinitesimal strains and rotations
• e =∈ +∈ +∈ = J
1
= first in varianceof strain.
• 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
( y z
Mean stress e =
(
E v ) (
σ x
+ σ y
+ σ z
E
1
3
(
σ x
+ σ y
+ σ z
) v )
σ
)
σ = Ke

Solid Mechanics
Where K =
(
E v )
bulk modulus of elasticity.
• Bulk modulus is widely used in fluid mechanics.
• From physical reasoning E > 0 ,G > 0 ,K ≥ 0
Steel : E = 200 Gpa
v = 0.3
Al : E = 70 Gpa
v = 0.33
Copper: E = 100 Gpa
v = 0.35
G =
(
E
+ v )
SinG Eand G
( 1 + v ) 0 v > − 1
> 0
Similarly SinG E > 0 & K ≥ 0
K =
(
E v )
1 2 v 0
∴
Theoretical bounds on v are v .
as v → 0 5 → α
1 v 0 5 and
C → 0
material is incompressible.

Solid Mechanics

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
- Other sections
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.
M x
= M y
= M z
=
V y
= V z
= 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.

Solid Mechanics
A
σ x
M y
=
M z
= −
σ x zdA = 0
σ x ydA = 0
Above equation must be satisfied at every cross-section, however, it does not tell how
σ x
is distributed in the crosssection.
The distribution cannot determine by the methods of static or equations of equilibrium- statically indeterminate
To know about the distribution of σ x
in any given section, it is necessary to consider the deformations resulting from the application of loads.
Since the body needs to develop only σ x
component in order to maintain equilibrium, therefore the state of stress at any point of prismatic rod is
σ ij
=
σ x
0 0
0 0 0
0 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
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
∈ =
E
1
E
1
E
1
σ
σ
σ x z y
−
−
− v
(
σ y
− σ z
) v ( σ x
− σ z v
(
σ x
− σ y
)
)
→
→
→
σ
E x
V
E
V
E
σ
σ x x
In the present case, ∈ x
is independent of y and z coordinates, therefore σ x
is also independent of y and z coordinates i.e
σ x
is uniformly distributed in a cross-section
σ = ∈ =
We know that internal resultant force
F =
A
σ x dA
Since σ x
is a independent of y & z

Solid Mechanics
F = σ
A da = σ A
∴
σ =
F
=
P
M
M y z
=
A
σ x
.zdA
= −
A
σ x
.ydA
= 0
= 0
A zdA = 0 ydA = 0
(1)
A
Eq. (1) indicates that moment are taken about the centroid of the cross-section.
Elongation or Contraction
∈ =
σ
E x =
P
AE
Total elongation of the rod
( ) − u ( ) δ
L
= = ∈ x da =
0
L
0
P
AE dx =
PL
AE

Solid Mechanics
σ
δ x
=
P
A
PL
=
AE
=
If A,E and P are functions of x then
δ =
L
0
( )
( ) ( ) dx
Stiffness and flexibility
= k = f
1 k =
AE
L f =
L
AE
These are useful in computer analysis of structural members.

Solid Mechanics
Extension of results: Non-uniform bars (non-prismatic)
For a prismatic bar
σ x
=
P
A
& δ =
PL
AE
This is exact solution for prismatic bar.
σ x
=
( )
( )
=
( )
( )
S =
L ( )
( ) ( ) dx
0
Approximate expression
The above formula becomes a good approximation for uniformly varying cross-sectional area ( ) 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.

Solid Mechanics
Σ F x
= 0 σ x
( ) τ yx
( )
τ xy
= τ yx
= σ x
( )
∆
∆ y x s
0
Taking x 0, we note that τ yx
→ 0 only if
∆ x slope of the upper surface of the rod tends to zero.
0 i.e at the

Solid Mechanics
Case2
δ
BC
δ
AB
=
PL
AE
=
− ( P
A
=
PL − P L
1
AE
=
A E
+ )
2
σ
BC
σ
AB
= −
= −
( P
A
+ P
B
A
2
1
δ
CA
= S
BC
+ S
AB
)
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
δ n
= i = 1 and σ i
=
P i
A i

Statically indeterminate problems
Equilibrium
Σ F y
= 0
F a 1
+ F a 2
+ − = 0
[ Σ M
C
= 0
] bF a 1
− bF a 2
= 0
F a 1
= F a 2
Solid Mechanics
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
A geometric compatibility equation
δ s
= s s
A E and δ
Then using the geometry compatibility
A
=
δ s
= δ
A
F L
E A
=
F L
E A

Solid Mechanics
By solving (1) & (2) we can obtain internal forces F s
& F
A
Stresses in axially loaded members
σ ij
Uniaxial state stress is a special case of plane stress
=
σ x
0
0 0
τ
σ
1
= σ max
= x
σ
2
1 =
σ
2 x
Occurs at 45 to x y or − planes.

Solid Mechanics
A − Principal stress elements
B,C − maximum shear stress elements.
Ductile material are weak in shear. They fail along planes.
τ max
Brittle materials weak in normal tensile stresses. They fail along σ
1
planes.
Limitations of analysis
σ x
=
P
A
& S =
PL
AE
(1)They are exact for long prismatic bars of any cross-section, when axial force is applied at the centroid of the end crosssections.

Solid Mechanics
(2)They should not be employed (especially σ x
=
P
A
) 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
Stress concentration factor=K =
σ
σ max ave
σ nom
=
P bt

Solid Mechanics
Stress concentration due to hole
Discontinuities of cross section may result in high localized or concentrated stresses.
K =
σ
σ max nom
σ nom
=
P dt

Stress Concentration due to fillet
K =
σ
σ max ave
σ ave
=
P dt
Solid Mechanics

Solid Mechanics
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
σ ij
= τ
τ
σ r
τ r θ
τ
θ r xr
τ
σ x
θ
θ
τ
σ
θ rx x x
Y r θ
1
E
1
E
1
=
E
τ
σ
σ rQ
G r x
σθ
−
−
− v v v
(
(
( σ
σ
σ x r r
+
+
+
σ
σ
σ
θ
θ x
; Y x θ
= Y
θ x
=
)
)
)
τ x θ
G
; Y xr
= Y rx
=
τ rx
G
Equilibrium and elementary forces
F x
= V y
= V z
= M y
= M z
= 0
M x
= =
0
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 dF = τ x θ dA are only forces that are required to be exerted by the other section, so that

Solid Mechanics
σ x θ rdA
T =
A
T T
0
τ rdA
Direction of τ z θ
can be obtained from the direction of internal torque T at that section.
The state of stress in pure torsion is therefore
0 0 0
0 0
0 τ x θ
τ
θ
0 x
While the relation in (1) express an important condition that must be satisfied by the shearing stresses τ xQ
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 τ x θ
in polar coordinates means, presence of
τ xy
= τ xQ
Cos θ
τ xz
= τ xQ
Sin θ

Solid Mechanics
Therefore the state of stress in case pure torsion in terms of rectangular stress components is then
0
τ yx
τ zx
τ xy
τ xz
0 0
0 0
- state of pure shear.
We must then ensure that
V y
=
V z
= τ
τ xy dA xz dA
= 0
= 0
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.

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 noncircular 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 ( Y x θ
)
Because of T
0
, the right end will rotate through an infinitesimal angle
φ - angle of twist.

* φ - varies along the axis of the bar.
φ
Y xQ is independent of x and
Y dx Ydx rd φ d φ dx
Solid Mechanics

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
Y max
= R d φ dx
Shear strain is zero at the center of the bar.
The equation d φ dx small deformations.
is strictly valid to circular bars having
If the material is linearly elastic
τ = GY
Therefore, variation of shear stress τ xQ
in pure torsion is given by
τ τ xQ
= GY xQ
= GY d φ dx
Shear stress
τ
is only function of “ r ” and varies linearly with radius r of the circular bar.
τ max
= τ xQ max
= RG d φ dx

Solid Mechanics
The torsion formula
Relation between internal torque T and shear stress τ
T = τ rdA
A
T = d φ dx
Since G & d φ dx
are independent of area A then d φ dx
A
I
P
=
A
For solid circular bar, I
P
=
π
32
D 4
∴ T GI
P d φ dx
∴ d φ dx
= =
T
GI
P
But τ = Gr d φ dx
τ T
I
P
=
π
2
R
4
P
τ =
Tr
I
P
Torsion formula

Solid Mechanics
This is the relation between shear stresses τ xQ
and torque T existing at the section.
Torsion formula is independent of material property.
τ max
= τ xQ max
=
TR
I
P
τ max
=
16
π D
T
3 for solidcircular bars
Angles of twist
We now determine the relative rotation of any two crosssections
= d φ T dx GI
P
φ
B / A
= φ φ
A
= x
B x
A
T
GI
P dx

Solid Mechanics
In case of prismatic circular bar subjected to equal opposite torques at the ends
φ
B / A
= φ φ
A
=
GI
P if x
B
− x
A
= L puretorsion
Direction of φ at a section is same as that of T
φ =
TL
=
T L
GI
P
GI
P
Since = d φ T dx GI
P
then, in case of pure torsion.
= d φ φ dx L constant
Thus in case of pure torsion φ ( ) varies linearly with x
In case of torsion
φ = displacement
TL
GI
P
Load k =
GI
L
P ; f =
L
GI
P
The product GI
P
− Torsional rigidity

Solid Mechanics
τ xy
= τ xQ
Cos θ
τ xz
= τ xQ
Sin θ
We should ensure that distribution of τ xQ
should also gives
V y
= V z
= 0
V y
V y
= τ xy dA
=
=
I
A
2 π
T
0 0
2 π
I
P
R
P
R
Tr
0 0
=
A
τ x θ
θ
θ
θ
θ
∴ V y
= 0
V z
=
T
I
P
2 π R
0 0
∴ V z
= 0
θ θ
= 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

Solid Mechanics
I
P
I
P
=
=
π
32
π
32
(
D
4 − Soild
D o
4 − D i
4
)
− hollow
τ
τ max min
=
=
TR o
I
P
TR i
I
P
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 ”.
τ
I
P
Tr
=
=
=
I
P
π
D
4 − solid
32
π
32
(
D
4 o
− D i
4
)
− hollow

Solid Mechanics
τ max
τ min
=
=
I
P
TR i
I
P
Y
τ
=
τ
G
( )
I
P o
= d φ T dx GI
P
φ
φ
B / A
= φ φ
A
=
=
B
− x
=
A constant
= linearly with x
TL
GI
P
(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.
τ =
( )
( )
= d φ
=
φ φ
A
= φ
B / A
( )
( ) x
B
= x
A
( )
( ) dx
The above equations yield good approximations to the exact solution, provide if ( )

Solid Mechanics
Some special cases
τ ( ) =
=
Tr
( )
T
( )
τ ( ) =
=
I
P
( )
GI
P
Case II
τ i
=
I i i
P i
φ
B / A
= n i = 1 i i
G I i

Solid Mechanics
Statically indeterminate problems
[ Σ M x
= 0
]
We note that within
T
A
+ T
C
+ = 0 (1)
=
A
and
within =
C
• 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
φ
B / A
= φ
B / C
Compatibility equation
φ
B / A
= A AB
G I
AB
; φ
B / C
=
G I
BC
T L
G I
AB
= (2)
G I
BC

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.

σ x
=
P
A
Torsion testing m/c
Solid Mechanics

Solid Mechanics
Combined loading or combined stress
τ max
=
TR
I
P
Principal of superposition
σ x
=
P
A

Solid Mechanics
Stress concentrations in torsion
Stress concentration effect is greatest at section B-B
τ max
= K τ avg
= K τ nom
τ avg
= τ nom
= K τ
1
= K
π
16 T
D
3
1

d φ dx
Solid Mechanics
τ =
Tr
I
P
, =
T
GI
P
; φ =
TL
GI
P
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 ( ) 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.

Solid Mechanics
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 bending of beams.
M z
and M y
are responsible for
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 in x 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 dx
= − V
Pure bending (i.e., M=constant) occurs only in regions of a beam where the shear force is zero .
If ( ) 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 k Curvature in general
RdQ dS
( ) and k
( ) . dQ
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.
R dX

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 M
0 moments) acting in the same plane of symmetry.
(or bending

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
F x
= V y
=
M z
= =
V
M x
= M y
= z
0
=
0
0

Solid Mechanics
M = − σ
Therefore, σ x dA 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 σ
X
any section should satisfy
F x
= 0
M y
= 0
M z
= M
σ x dA = 0
σ = 0
− σ
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.
σ ij
=
σ x
0 0
0 0 0
0 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
M
0
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 M
0
, 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 = ( )
The distance dx between two planes is unchanged at the neutral surface,
RdQ dx k dQ

Solid Mechanics
Therefore, the longitudinal strain i.e ∈ x
at a distance “y” from the neutral axis is e f − ef ef
=
( ) dx
=
−
R y y
R
In case of pure bending ∈ ≠∈ x
( ky
) ∈ =∈ x
The preceding equation shows that the longitudinal strains
( ) 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.
∈ = 0 at the neutral surface
Maximum compressive ∈ =
− y
R
1
Maximum tensile ∈ =
+ y
R
2
However, we still do not know the location of neutral axis or neutral surface.

Solid Mechanics
Stresses in beams in pure bending :- For linearly elastic and isotropic beam material
1
E
1
E
1
E
σ x
σ y
σ z
− v
(
σ y
+ σ z
)
− v ( σ x
+ σ z
)
− v
(
σ x
+ σ y
)
Y xy
Y yz
Y xz
=
τ xy
G
=
τ yz
G
=
τ zx
G
The state of the stress at any point within a prismatic beam in pure bending is
σ ij
=
σ x
0 0
0 0 0
0 0 0
∴ σ x
E
− Ey
R
= − Eky
V
E
V
E
σ
σ x x
V x
V x
From the above equation
σ x
≠ σ
σ x
= σ
( x,z
( )
)
∈ =∈ x linear f (y)
∴ σ x
= linear f (y) i.e.,var y linearly with the distance y from the neutral surface

Solid Mechanics
σ x
at y = 0 i.e on the neutral surface = 0
Maximum compressive σ x
= −
EC
1
R
Maximum tensile σ x
=
EC
R
2
Maximum normal stress from the neutral axis.
σ x
occurs at the points farthest
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
−
σ
σ
σ x dA = 0 x ydA M M
0
= M z x zdA M y
= 0
Considering first equation
A
σ x dA = −
A
Ey
R
= 0
A ydA = 0
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.

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
∴
M
M = −
A
σ x ydA
= +
A
Ey
R ydA
M =
E
R
A
= zz
=
A
Moment of inertia of cross-sectional area about neutral axis
M =
EI
R k
M k
1 M
R EI
0
Moment-Curvature relation

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 σ x
and M - Flexure formula
σ x
= − Eky and k =
M
EI
σ
Stresses evaluated from flexure formula are called bending stresses or flexural stresses.

Solid Mechanics
The maximum tensile and compressive bending stresses occur at points located farthest from the neutral axis.
The maximum normal stresses are
σ
1
=
− MC
1
I
= −
M
S
1
σ
2
=
MC
I
2 =
M
S
2
S
1
=
C
I
1 and S
2
=
I
C
2
-Section moduli
S = Section modulus
Cross- sectional properties of some common shapes z − axis – neutral axis

I zz
= bh 3
12
S = bh 2
6
Solid Mechanics
I zz
=
π
64 d
4
S =
π d 3
32
I zz
= h = bh
3
36
3 b / 2 for eqilateral triangle
I zz
= r
4

Solid Mechanics
Distribution σ x on various cross-sections
σ
M max
S =
=
M
S
I y max
= σ alllow
S
S square
S circle
= .

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 crosssection 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.
( )
σ x k zz
= −
My
I
M
σ
E x v v z x
Elementary theory of bending

Bending of beams due to applied lateral loads
Solid Mechanics dM dx
= − V
Consider now a beam subjected to typical arbitrary transverse loads acting. In this case the interval bending moment ( ) and ( ) ≠ 0, and thus we have nonuniform bending.
Non-uniform bending is a result of presence of transverse shear force
( ) ( ) = 0 then M = constant .
It can be shown that the above results can also be used for non-uniform bending problems.
σ x k =
1
( )
=
=
−
I
( )
EI
∈ x
ν
ν x x
=
σ x
E

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
σ x zz
=
= −
1
( )
=
I
EI
∈ x
∈ y
∈ z
=
σ
E x v x v z
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 nonprismatic beam subjected to either pure or non-uniform bending, provided cross-sectional properties do not vary sharply.
σ x
= −
=
1
( )
( )
( )
=
( )
( )

Solid Mechanics
Problem
Determine the maximum tensile and compressive stresses in the beam due to the uniform load.
Solution
Centroid :-
A mm 2 y yA mm 3
1 × = ×
2 × = ×
A = Σ A = 3000
Ay = Σ yA y
Σ yA = × 3
= × 3 y = 38 mm
I zz
= = Σ
( )
= Σ bh
3
12
+ Ad 2
=
1
12
90 20
3 + 1800 12
2 +
1
12
30 40
2 + ×
I zz
I × 3 mm 4 = × − 9 m 4
2

C
1
= 22 mm and C
2
= 38 mm
σ x
σ
= − max
=
My
I
M
S
: S =
I y max
At maximum +ve bending moment i.e at (D)
S
1
S
2
=
I
C
1
=
− 9
− 3
=
C
I
2
=
× −
− 3
9
=
=
×
×
− 6
− 6 at D:
σ t max
=
M
= s
2
σ t max
= 83 1
σ
σ
C
C max max
=
M
= s
1
= 48 11
×
×
− 6
− 6
Solid Mechanics
At maximum -ve moment i.e at (B)
σ t max
=
M
= s
1
σ
σ
C max t max
=
M
= s
2
= .
and
×
×
− 6
− 6
σ
C max
=
=
=
.
MPa

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 σ allow
= 50 Mpa ?
Solution
M max
= 720 N-m
S
A
=
12 .
× .
3
− m
σ t max
σ t max
= σ
C max
= σ
C max
=
M
S
A
=
PL
S
A
= σ allow
P alow
=
σ allow
L
× S
A
S
B
=
P alow
×
12 .
=
σ allow
L
×
.
S
B
3
= 150 N
= 1 8 10
− 5 m
3
= 300 N
∴ The beam can carry P = N

Solid Mechanics
Limitations
(1)The flexure formula is exact for a prismatic beam in pure bending.
(2)It provides very good approximation of slender beams (L h)
σ x
for long
>> 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
( ) =
A
τ xy dA
It is reasonable to assume that
(1)The shear stresses acting on the cross-section are parallel to the shear force ( ) i.e ⊥ to the line PQ
(2)It is also reasonable to assume that the shear stresses τ xy are uniformly distributed across the width of the beam, so that M x
= = 0 for symmetrical bending
∴ τ xy
= τ xy
( ) =
( ) such thats
A
τ xy

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 τ xy
= τ yx
• Pattern of distribution of τ xy
= pattern of distribution of
τ yx
• Since τ xy
= τ yx
, it follows that the vertical shear stresses
τ xy
must vanish at lateral loads. y = ± h
2
, if the beam is subjected only

Solid Mechanics
Derivation of shear stress formula
Beam under non-uniform bending i.e t = width or thickness of the beam at
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
[ Σ F x
→ + = ]
−
A
σ x
( x + ∆ ) +
A
σ x
+ τ yx t x 0
τ yx t =
σ x
1
∆ x
=
A
−
σ x
I
( x + ∆ ) −
A
σ x
τ yx t =
∆
1 x
−
1
I
A
( + ∆ ) +
1
I
A
τ yx t = −
1
∆ xI
τ yx
=
− 1
It
( x ) ( ) ydA
A
( x
∆ x
) M x
A ydA taking limit as x 0
( x ) ( )
∆ x
A ydA
∴
τ yx
=
− 1
It lim
∆ → 0
τ yx
=
− 1
A dM dx
= − ( )
τ yx
=
It
( )
A ydA

Solid Mechanics
The above integral is by definition the first moment of are
A about the z-axis, we denote it by symbol Q.
Q = ydA
A
∴ τ yx
= τ xy
τ shear formula
It (1) in the above equation I I zz
stands for the moment of inertia of the entire cross sectional area around the neutral axis.
From (1)
τ yx
= =
I
=
I
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
Q = =
Q =
A b h
2
2 4
− y 2 h
2
− y y + h / 2 − y
2
I =
1
12 bh
3
τ xy
= τ yx
= s
VQ V h
It 2 I 4
2
− y 2 at y = ± h
2
τ xy
= τ yx
= 0
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.
τ xy max
= τ yx max
=
Vh
8 I
2
=
3
2
V
A

Solid Mechanics
Thus τ max
in a beam of rectangular cross-section is 50% larger than the average shear stress
V
A
It is always possible to express the maximum shear stress τ xy as
τ xy max
= K
V
A for most of the cross-sectional areas
K =
3
2
K =
4
3
Rectan gular
Circular
K =
3
2
Triangular
For most of the cross-section τ max
occurs at the neutral axis.
This is not always true.

Stress elements in non-uniform bending
Solid Mechanics

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 allowable stress is bending is p max
of the loads if the
σ allow
= 11 MPa (for both tension and compression) and allowable stress in horizontal shear is
τ allow
= 1 2
Solution
V max
occurs at supports and maximum BM occurs in between the loads.
V max
∴
τ
σ
S = max
τ bh
=
=
6
2
P
M max
S
τ
=
M
6 max
Pa bh
2
= xy max
= yx max
= max
=
3
2
V max
A
=
3
2
P
A
=
3
2
P bh
Therefore, the maximum permissible values of the load P in dending and shear respectively are
P allow b
=
σ allow
6 a bh
2
P allow s
=
2 τ allow
3 bh

Solid Mechanics
Substituting numerical values into these formulas,
P allow b
=
P allow s
=
Thus bending governs the design and the maximum allowable load is
P max
= . kN
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

Solid Mechanics
I zz
= =
( )( ) ( )( ) 3
12 12
=
The ratio
V
I
Level
=
( )
× mm
3
6 y
= .
× − 3
× 10
3 mm
3
150 0 1-1 0
2-2
= 1800
144 259.2
3-3
= 1800
144
132
259.2
19.0
× t mm
6
τ xy
=
150 0
150 4.5
12 56.4
278.2
12 60.5
4-4
= 144
= 1800
144
69
259.2
114.3
373.5
12
= 1656
τ max
= 81 3
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
(1) Maximum shearing stress theory
(2) Octahedral shearing stress theory
For ductile materials
(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
τ max
= max ( τ τ τ ) where τ
1
=
σ
2
− σ
2
3 ; τ
2
=
2
3 ; τ
3
=
2
2
The maximum shearing stress in uniaxial tension test at the moment of yielding is
τ t
=
σ ys
2
Tresca criterion is τ max
≥
σ ys
2
Octahedral shearing stress theory or Hencky-Von-Mises failure criterion
This theory also known as “The maximum distortion strainenergy theory”

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.
τ oct
=
1
3
(
2
) 2 + ( σ
2
− σ
3
+ σ σ
3
) 2
Octahedral shear stress in the uniaxial tension test at the moment of yielding i.e. σ y
= σ ys
= σ
1
τ t
τ t
=
=
1
3
(
σ ys
3
2
σ ys
− 0
) 2
( 0 0 ) 2 +
(
σ ys
− 0
) 2
Von Mises theory says that τ oct
≥
3
2
σ ys
σ von
=
3
2
τ oct
Von Mises theory says that τ von
≥ σ ys
* Excellent experimental evidence is available for supporting maximum shearing stress and Von Mises criterion
Maximum Normal stress criterion or Rankine Theory :
This theory is generally used for design of components made up of brittle materials.

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 σ ult
obtained from the tensile test of a specimen of the same material.
Thus the structural component will fail when
σ
1
≥ σ ult
Simple application of theories

Solid Mechanics
Torsion + Direct shear
σ
A
=
Mr
I
τ
2
=
4 V
3 A
τ
1
=
Tr
I p

Bending + axial loading
Solid Mechanics
σ x
σ x
=
− My
I
σ x
=
P
A
−
A I zz
Neutral surface is now shifted due to the application of axial load.

Solid Mechanics
Consider an infinitesimal stress element at point in a linearly elastic body, subjected to a normal stress σ x
The work done by this force dW int
=
=
1
2
σ
1
2 x dydz force
× ∈ x dx distance dW int
=
1
2
σ dV
This internal work is stored in the volume of the element as the internal elastic energy or the elastic strain energy.
∴ dU =
1
2
σ dV = volume of the element. dV
The strain energy density U
0
is defined as the internal elastic energy stored in an elastic body per unit volume of the material.
∴ Strainenergy density U
0
= dU dV
=
σ x x
2
∈

Solid Mechanics
U
0
can be interpreted as an area under the inclined line on the stress-strain diagram. Similar expressions can developed for σ y
and σ z
corresponding to strains ∈ y
and ∈ z
.
Elastic strain energy for shearing stresses: dU shear
=
1
2
τ xy dxdz × Y dy average force distance dU shear
=
1
2
τ
Analogous expressions apply for the shearing stresses
τ τ with the corresponding shear strains Y yz
and Y xz
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. dU =
1
2
σ
1
2
σ
1
2
σ
1
2
τ +
1
2
τ +
1
2
τ Y dV
The strain energy density for the most general case is

U
0
= dU dv
+
1
2
1
2
τ
σ
+
1
2
1
2
τ y
σ y
Solid Mechanics
1
2
σ +
1
2
τ
Substituting the values of strain components from generalized Hooke’s law, we can show that
U
0
=
2
+
1
E
1
2
(
G
σ
(
2 x
τ 2
+ σ
+
2 y
τ
+
2
σ
+ z
2
τ 2
− xy yz zx
) v (
σ σ + σ σ + σ σ
)
E
)
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
U
0
over its volume.
( ) =
V
Internal strain energy in axially loaded bars
σ x
= σ z
= τ xy
= τ xz
= τ yz
= 0
∴ U
0
=
1
2
σ
∴
The total internal energy
1
2
σ x
σ
E x =
2
1
E
σ 2 x
U U dv
V
=
V
2
1
E
σ 2 x dV

U =
σ
2
2 x
E
AL ==
2
P 2
EA
2
.AL
=
2 EA
Solid Mechanics
U =
2 EA
Strain energy in torsion of circular shafts
U
0
=
1
2
τ .Y
=
1
2
τ .
τ
G
=
2
1
G
τ 2
τ
U = U dv v
= v
1
2 G
τ 2 dv
=
Tr
I p
where I p
=
π
2
R
4
U =
R
0
2
1 T
G I 2 p
2
2 π
Strain energy in bending
U =
2 GI p
∴ U = v
σ
2
2 x
E dv = v
2
M
2
EI 2
=
2
M
2
EI 2
τ =
TY
I
P

Solid Mechanics
Conclusion
U =
2 EI
Axially loaded bars U =
2 AE
Torsion of shafts U =
2 GI
P
Bending (pure) of beams U =
2 EI
We can use the following equations in case of non-uniform cases
U =
0
L
P 2
2 AE dx ; U =
L
0
T 2
2 GI
P dx ; U =
0
L
M 2
2 EI dx

Solid Mechanics
Problem:
( ) = ( − )
U =
L
P 2
2 AE dx
=
0
L
0
2 2
2
( ) 2
AE
=
2 AE
0 dx
L
L 2 + x 2 − 2 Lx.dx
=
2 AE
L 3
=
2 E
+
L
3
3
− L 3 =
Y AL
3
6 E
+
L
3
3
− L
3
U =
2 AE
( ) = ( ) P
U =
L
0
( − ) 2 +
2
P 2
AE
+ 2 ( − ) dx
U =
Y AL
6 E
3
+
P L
2 AE
+
2
2
YAP
AE
L
2 −
L 2
2
=
Y AL
6 E
3
+
P L
2 AE
+
YP
2 E
2
Since U P
2 or U ∞ δ 2
principle of superposition should not be used.

Solid Mechanics
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.

Examples
(1)
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

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
θ → + → anticlockwise
ρ = Radiusof curvature
From geometry ρ θ = ds k
ρ d θ ds curvature of the deflectioncurve k - curvature - +ve when angle of rotation increases as we move along the beam in the +ve x – direction.
Slopeof thedeflectioncurve = dv dx
= tan θ
Slope dv dx
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 for small θ ds = dx
2 + dv
2 = 1 + ( ) 2
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. k
ρ d θ dx
Since θ is small tan θ θ
∴ dv dx
= θ k
1
ρ d θ d v dx dx 2
θ k =
= d
2 dx
ν ν
=
= dx du u
′
′′ only insmalldeformationtheory
If the material of the beam is linearly elastic and follows
Hooke’s law, the curvature is k
ρ
M
EI

Solid Mechanics
M leads to +K and so on
∴ d v M dx 2 EI
or
EI d v dx
2
= M
The basic differential equations of the deflection curve.
Sign conventions used in the above equation:
(a) The (b) dv dx
and θ are
(c) k is + (d) M is +ve if beam bends
Another useful equations can be obtained by noting that dM dx dV dx
= − V
= − p
Non-prismatic beams
(
( d v dx
2
=
( ) ) ′ ( )
( ) ) ′′

Solid Mechanics
For prismatic beams.
EIv
EIv
EIv
′′ =
′′′ = −
′′′′ = +
( )
( )
BM equation( 2 nd order )
Shear force equation( 3 rd order )
Load equation( 4 th order )
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.

Approximate sketching
(3) (4)
(5) (6)
Solid Mechanics

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
θ
= 0
( ) = ( ) = 0
(ii)
= 0
( ) = ( ) = 0 ( ) = 0
(iii)
( )
( )
=
= − ′′
( )
( )
=
=
0
0

Continuity conditions
All deflection curves are physically continuous. Therefore
( ) from side AC
= from side BC
Similarly at “C”
( ) from side AC
= ( ) from side BC
Solid Mechanics
Symmetry conditions v ′
L
2
= 0 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.
+ +

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 δ max
at the mid point of the beam and the angles of rotation Q
A
and Q
B at the supports. Beam has constant EI.
Solution
V + qL
2
− qx = 0
= − qL
2
(1)
M − qL
2 x + qx
2
2
= 0
M =
2 2
2
(2)
Differential equation of deflection curve.
EIv ′′ = ( )
EIv ′′ = qLx qx
2 2
2
Slope of the beam

EIv ′ = qLx
4
2
− qx
6
3
+ C
1
BC → Symmetry conditions
Solid Mechanics v x =
L
2
=
0
0
0
= qLL
16
2
− qL
48
3
+
= qL
16
3
− qL
48
3
+ C
C
1
1
C
1
= − qL
3
24
Slope equation is
EIv ′ = v ′ = qLx
4
2
− qx
6
3
− qL
24
3
− q
24 EI
(
L 3 − 6 L 2 + x 3
) s
Deflection of the beam
EIv = qLx
12
3
− qx
24
4
− qL
24
3
2
B.C.
( = 0 ) = 0
C
2
C
2
= 0
EIv = qLx
12
3
− qx
24
4
− qL
24
3 x

Solid Mechanics v =
− q
−
24 q
24 EI
EI
(
(
L x − 2 Lx
3 + x x
4 + L x − 2 Lx
3
)
4
) you can check v = 0 at x = 0 and L = 0
(b) From symmetry maximum deflection occurs at the midpoint x =
L
2 v x =
L
2
=
− 5 qL 4
384 EI
-ve sign means that deflection is downward as expected.
δ max
= v x =
L
2
=
5 qL
384
4
EI s
Q
A
= v ′ =
− qL 3
24 EI
-ve sign indicates clock wise rotation as expected.
Q
B
= ′ ( = ) = qL
4
3
EI
− qL
6
3
EI
− qL
24
3
EI
( ) = qL 3
24 EI
+ ve sign means anticlockwise direction. since the problem is symmetric, v ′ ( ) = ( )

Solid Mechanics
Problem: 2
Above problem using third order equation
EIv ′′′ = − ( )
EIv ′′′ = − qx − qL
2
= qL
2
− qx
Moment equation
EIv ′′ = qLx qx
2 2
2
+ C
1
B.C.
( = 0 ) = 0
C
1
= 0
( = 0 ) = 0
2
EIv ′′ =
2 2
Problem 3
Above problem using fourth order differential equation
EIv ′′′′ = − q
Shear for a equation
EIv qx C
1
From symmetry conditions

V x =
L
2
= 0
0
∴
= − q
L
2
+ C
1
C
EIv ′′′ qL
2
1
= + qL
2
=
L
2
= 0
Solid Mechanics
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:
V qL qx = 0
M + qL
2
2
− qLx + qx 2
2
M qLx − qL
2
2
− qx
2
2
Differential equation of deflection curve
EIv ′′ = ( )
EIv ′′ = − qL
2
2
+ qLx − qx
2
2

Solid Mechanics
Slope equation: EIv ′ = − qL x qLx 2 qx
2 2 6
3
− + C
1
BC: ( ) C
1
= 0
EIv ′ = − qL x qLx
2 2
2
− qx
6
3
Deflection equation
EIv = − qL x
4
+ qLx
6
3 qx
24
4
− + C
2
( = 0 ) = 0
C
2
∴ EIv = − qL x
4
+ qLx
6
3
− qx
24
4 v =
−
24 q
EI
C
2
6
2 2
−
+ 4
3
+
L x Lx x
4
= 0
( )
EIv ′ =
− qL
2
3
+ qL
2
3
− qL
6
3
=
− qL
6
3 v Q
B
= − qL
3
6 EI
( ) v =
− q
24 EI
6 L 4 − 4 L 4 + L 4 =
( )
− 3 qL
4 v x L =
24 EI v =
4
8 EI
− 3 qL 4
24 EI

Solid Mechanics
Problem 5
Above problem using third order equation
EIv ′′ = − ( )
EIv ′′′ = qL qx
Moment equation
B.C. ( ) = 0
EIv ′′ = qLx − qx
2
2
+ C
1
( = ) = 0
0 = qL 2 − qL
2
2
= qL
2
2
EI v qLx − qx
2
2
+ qL
2
2
4 = −
EIv ′′ = qLx − qx
2
2
+ qL
2
2 qL 2
2
Problem 6
Above problem with fourth order equation
∴
EIv ′′′′ = ( )
EIv ′′′′
−
= ⊕ q
Shear force equation
EIv qx C
1
0
( ) = 0 qL C
1
C
1
( = ) =
= + qL
0

Solid Mechanics
∴ EIv ′′′ qx qL
Problem 7
EI = distances a and b from the left-hand and right-hand supports respectively. Determine the equations of the deflection
δ δ maximum deflection and the deflection at the midpoint C of the beam. Constant EI
Solution
Pb
L x P = −
M −
Pb
L
V
V
Pb
L
+
Pb
L
= 0
= −
Pb
L x = 0 H =
Pbx
L
V +
Pb
L
P 0
Pb
L

Solid Mechanics
M =
M =
(
Pbx
L
Pbx
L
−
)
(
Pbx
L
)
− Px Pa = −
Pxa
L
+ Pa
Differential equation of deflection curve
Slope equations:
EIv
EIv
′′ =
Pbx
′′ = −
L
Pxa
L
+
0
Pa a x L
B.C.
EIv ′ =
Pbx 2
2 L
+ C
1 o x a
( )
EIv ′ =
AP
=
−
2 L
+ Pax C
2
≤ ≤
( )
PB
2
(
PLa
L
C
2
2
1
L
−
=
)
/
2
/
/ /
Pa
2
2
3
2
+
+
+
C
C
C
1
1
2
=
−
= −
Pa
2 L
Pa
2
/
3
3
+
+
Pa
Pa
2
2
+
+
C
C
2
2

Solid Mechanics
Deflection curve equations:
B.C:
EIv ′ =
EIv =
−
Pbx 3
6 L
3
6 L
( = 0 ) = 0 and
+ C x C
3
(
2
2
) = 0
0
+ C x C
4
≤ ≤
C
3
C
3
= 0
0
0
=
= −
= −
PaL
2
6
3
6
3
L
2
2
2
2
2
+ C L C
+
+
4
C L C
2
4
4
C
4
= −
PaL
2
3
−
( )
AP
= ( )
PB
( ) 3
6 L
PLa 3
+
Pa /
6
Pa
L
3
6
+ C a
4
=
+ C a =
+ C a
Pa
2
3
+
=
−
−
Pa
6
6
L
Pa
/ L
4
4
C a C
+
+
4
Pa
2
Pa
2
3
3
=
Pa
3
3
+ C a −
PaL
2
3
−
+ C a C
4
+ C a C
4

(2)
(3)
Solid Mechanics
Pa 3
2
Pa
3
C
6
2
+ C a
= −
=
PaL
2
3
Pa 3
3
−
+
2
−
PaL
C L C L
3
2
= −
−
PaL
3
2
−
Pa
6
3
3
= −
3 6
Some important formulae to remember
(1)
δ
B
= qL
4
8 EI
,Q
B
= qL
3
6 EI
δ
B
=
δ
B
=
PL
3
3 EI
M L
2 EI
2
, Q
B
=
PL
2
2 EI
, Q
B
=
EI
(4)
δ c
= δ max
=
5 qL 4
384 EI
; Q
A
= Q
B
= qL 3
24 EI
(5)
Problem 8
δ c
= δ max
=
PL
3
48 EI
;Q
A
= Q
B
=
PL
2
16 EI
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 Q
A
and Q
B
at the supports, the maximum deflection δ max
of the beam.

Solid Mechanics
Solution
M =
Px
2
M =
Px
2
M
M
V = −
P
2
M −
P
2 x = 0
V
M
P
2
=
Px
2
P 0
V P / 2
−
Px
2
+
=
Px
2
−
P x
P x
−
L
2
= 0
−
L
2
=
Px
2
− Px +
PL PL Px
2 2 2
M =
PL Px
2 2

Differential equation deflection curve
EIv
EIv
′′ =
′′ =
Px
2
0
PL Px L
2 2 2
Slope equations
2
EIv ′ =
Px 2
4
+ C
1
EIv ′ =
2
=
L
2
4
2
+ C
2
AP
=
0
=
L
2
L
2
2
PB
PL 2
16
+ C
1
=
PL
4
2
−
PL
16
2
+ C
2
C
1
= C
2
+
PL
4
2
−
PL
8
2
= C
2
+
PL
2
8
C
1
= C
2
+
PL 2
8
Solid Mechanics
Deflection equations:
B.C:
EIv =
EIv =
Px
3
12
+ C x C
PLx
4
2
−
Px
12
3
+
3
0
C x C
4
L / 2
2
≤ ≤
( = 0 ) = 0 and ( ) = 0

C
3
Solid Mechanics
0
=
=
PL
4
3
−
PL
12
3
+ +
PL 3
6
+ C L C
4
C
4
= −
PL 3
6
−
C
3
= 0
4
=
L
2
AP
= =
L
2
PB
PL
96
3
+
C L PL
2 16
3
−
PL
96
3
C
1
L PL
2 16
3
−
PL
48
3
+ C
2
L
2
+
+
C
2
C
4
L
2
+ C
4
C
1
L PL
3
2 24
+ C
2
L
2
+ C
4
2
L PL
2 16
3
=
PL
24
3
+
2
/
/
L PL
2 6
3
−
PL
24
3
−
PL
6
3
−
PL
16
3
= C L C
2
=
(
48
) PL 2
C
2
= −
9 PL 2
48
=
− 3 PL 2
16
C
2
= −
3 PL 2
16

Solid Mechanics
∴ C
1
= −
3 PL
16
2
+
PL
8
2
= −
PL 2
16
C
1
= −
PL 2
16
∴ C
4
= −
PL
6
3
− L
Deflection curves
− 3 PL 2
16
=
− PL
6
3
+
3 PL
16
3
=
C
4
= −
PL
3
48
(
48
) PL 3
EIv =
EIv =
Px
12
3
−
PL
16
2
PLx
4
2
− +
3
Px
12
3
−
3 PL
16
2
0 x +
L
2
PL
3
L
48 2
EIv x =
L
2
=
PL
96
3
−
PL
32
3
=
− PL
3
48
∴ v x =
L
2
= −
PL 3
48 EI
EIv x =
L
2
=
PL
16
3
= −
PL
3
48
−
PL
96
3
−
3 PL
32
3
+
PL
48
3
=
(
96
) PL
3

Solid Mechanics v = − PL / 48 EI
Slope equations:
EIv ′ =
EIv ′ =
Px
4
2
−
PL
16
2
2
2 4
0
−
3 PL
2
L
16 2
L
2
∴
( 0 ) 0
PL 2
16
( = 0 ) = Q
A
= −
= −
PL 2
PL 2
16
16 EI
Clock wise
∴ (
( )
)
=
=
PL
2
2
−
PL
4
2
−
3 PL
16
2
Q
B
=
PL
2
16 EI
(
=
(
16
) PL
2
+ve, CCW from x-axis
=
)
PL 2
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 δ
B
and angle of rotation
Q
B
at end B of the beam. Beam has constant EI. Use principle of superposition.
Solution
δ
B
1
= qa
3
24 EI
( 4
δ
B
2
=
PL
3
3 EI
,
) Q
B
1
= qL
3
6 EI
Q
B
2
=
PL
2
2 EI

δ
B
= δ
B
1
+ δ
B
2
= qa
3
24 EI
( 4
Q
B
= Q
B
1
+ Q
B
1
= qa
6
3
EI
+
PL
2 EI
2
)
Solid Mechanics
PL 3
3 EI

Solid Mechanics
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
2 θ dx 2
=
M
EI
can be written as d
2 θ θ dx
2
= d M dx EI d θ =
M
EI dx
The quantity corresponds to an infinitesimal area of the
M
EI
diagram. According to the above equation the area is equal to the arrange in angle between two adjacent point m
1 and m
2
. Integrating the above equation between any two points A & B gives.
B
A d θ θ
B
− θ
A
= ∆ θ
BA
=
B
A
M
EI dx

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
M
EI
diagram between A &
B , If θ
A
is known then
θ
B
= θ
A
+ ∆ θ
BA
In performing above integration, areas corresponding to the
+ M are taken + ve, area corresponding to the – ve M are taken –ve
B
If is +ve- tangent B rotates c.c.w from A or θ
B
is
A 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 t
BA
. Since the angles between the tangents and x-axis are very small we can take
=
1
θ = x
1
M
EI dx
The expression x
1
M
EI dx = first moment of infinitesimal area
w.r.t. a vertical line through B.
Integrating between the point A & B t
BA
=
B
A dt =
B ′
A x
1
M
EI dx = First moment of the area of the diagram between points A & B, evaluated w.r.t. B.
M
EI t
BA t
AB
=
=
φ
φ x
1 x where φ =
B
A
M
EI dx if M is +ve φ = +ve if M is -ve φ = -ve x and x
1
are always taken +ve quantities.
∴
Sign of tangential deviation depends on sign of M.

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 I
1
= × 6 mm 4
. For the remaining 1000 mm of the beam I
2
= × 6 mm 4
. Find the deflection and angular rotation of the free end. Neglect weight of the beam and E = 70 GPa
Solution:
EI = 3 5 10
24
70 10
9 × 10
− 6 N/mm 2
3
N/mm
2

A
1
=
1
2 bh
2
600 ×
−
E
= −
36
E
A
2
= bh = −
A
3
A
4
=
=
1
2 bh
1
2 bh
E
2
480 ×
2
120 ×
−
−
E
E
= −
= −
E
E
∆ Q
BA
= Q
B
− Q
A
=
B
A
M
EI dx A
1
+ A
2
+ A
3
+ A
4
Q
B
= −
E
−
E
.
−
E
.
−
.
E
= −
288
E
Solid Mechanics

Solid Mechanics
Q
B
= −
288
E
= −
288
3
= − 4 14 10
− 3 rad
Q = .
− 3 rad from tangent at
A. x
2
= 1060 mm;x
1
= 1400 ;x
3 t
BA
= δ
B
= 840 mm;x
4
= 480 mm t
BA
= δ
B
=
=
=
A x + A x + A x + A x
− 36
E
− 288000
E
1400 +
= −
−
E
δ = − 4 11
1060 +
−
E
840 +
−
E
480
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:

Solid Mechanics
Since EI is constant
M
EI diagram is same as M diagram. v c
= ′′ ′ −
CB c c
1
2 t
AB
A
1
A
2
=
1
2 bh
=
1
2 bh x
1
1
2
1
2 a
3
4
Pa
EI
=
3
8
Pa
EI
2
3
3
4
Pa
EI
=
9
8
Pa
EI
2 s
=
2
3 a ; x
2
= 2 a t
AB
=
= A x + A x
Pa
4
3
EI
+
9
4
Pa
EI
3
=
=
10
4
3
8
Pa
EI
2
2
3 a +
9
8
Pa
EI
2
Pa
EI
3
=
5
2
Pa
3
EI
( + ve
2
) a

t
CB
1
2
2 a
2
Pa
EI
× v c
=
11 Pa
3
12 EI
2
3 a
=
Pa
3
3
EI s
Solid Mechanics
5
4
Pa
3
−
Pa
3
EI 3 EI
AB / 2
=
5 Pa 3
4 EI
=
(
4 EI
) Pa 3
=
11 Pa
3
12 EI
The +ve sign of t
AB
& tangent through B. t
CA
indicate points A & C above the
(a) The slope of the elastic curve at C can be found from the slope of one of the ends as:
∆ Q
BC
= Q
B
− Q
C
Q
C
= Q
B
− ∆ Q
BC
B
∆ Q
BC
= Q
B
− Q
C
C
M
EI dx
1
2
2 a
2
Pa
EI
=
Pa
2
2
EI s
Q
B
≈ t
AB
/ L =
5
2
Pa
3
4
1
−
Pa
2
2
EI
=
5
8
Pa
EI
2
−
Pa
2
2
EI
Q c
=
Pa 2
8 EI
(b) If the deflection curve equations is wanted then by selecting an ordinary point E at a distance x v
E
= ′′ ′ − ′′

Solid Mechanics v
E
=
L t
AB
− t
EB
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 t
CA we need to consider unhatched region which is more difficult.
(d) The deflection at C can also be calculated as follows. v c
= t
AC
+ t
BC
2
∴
C is at the center of the beam. However, this is also move complicated approach compared to first, as to find t
CA
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 x
1 x
3
A
1
A
2
A
3
=
=
1
2 bh
1
2
− Pa
EI
= −
1 a
2 2
Pa
EI
= −
Pa 2
4 EI and A
4
=
Pa 2
4
2
EI
Pa 2
EI
Pa
2
2 EI
= a
3
1 a
3 2
2 a a
=
7
3
7 a
; x
2 a / ; x
=
2 a a
4
3 3 2
=
2
3 a a
11 a
6 t
CB
= A x + A x + A x
= −
Pa
4
2
EI
×
11
6 a Pa
4 EI
2
×
7
6 a Pa
2
2
EI
×
2
3 a
= −
11
24
Pa
EI
3
+
7 Pa
24 EI
3
+
Pa
3 EI
3 =
(
24 EI
) Pa 3 t
CB
=
4 Pa
24
3
EI
=
Pa
6
3
EI
The + sign of t
CB
indicates that the point C is above the tangent through B. Hence corrected sketch of the elastic curve is made.

Solid Mechanics t
AB
= −
Pa
2
2
EI
×
2
3 a = −
Pa 3
3 EI
∴ v
A
= t
AB
=
−
Pa
3
3
EI
−
Pa
12
3
EI
=
Pa
4 EI
3 v
A
=
Pa 3
4 EI
Note: Another method to find v
A
is shown. This may be simpler method than the present one.