* Shear stress, τ = 50 N/mm2 * Shear modulus, C = 8x104 N/mm2 * Strain energy per unit volume = τ2/2C = (50)2 / 2(8x104) = 0.3125 J/mm3 Therefore, the local strain energy per unit volume stored in the material due to shear stress is 0.3125 J/mm3.

2. Unit 1- Stress and Strain
Topics Covered
 Lecture -1 - Introduction, state of plane stress
 Lecture -2 - Principle Stresses and Strains
 Lecture -3 - Mohr's Stress Circle and Theory of
Failure
 Lecture -4- 3-D stress and strain, Equilibrium
equations and impact loading
 Lecture -5 - Generalized Hook's law and Castigliono's

3. 3-D Stress and Strain
stress vector σ that represents the force per
unit area acting at a given location on the
body's surface.
In other words, a stress vector cannot be fully
€
described unless both the force and the
surface where the force acts on has been
specified.
ΔF dF
σ = lim =
Δs−>0 Δs ds
€

4. 3-D Stress and Strain
Suppose an arbitrary slice is made across the
solid shown in the above figure, leading to
the free body diagram shown at left. Stress
would appear on the exposed surface, similar
in form to the external stress applied to the
body's exterior surface. The stress at point P
can be defined using the same above
equation

5. 3-D Stress and Strain
Stresses acting on an plane, are typically
decomposed into three mutually orthogonal
components. One component is normal to
the surface and represents direct stress. The
other two components are tangential to the
surface and represent shear stresses.
Normal component = σxx ,σyy ,σzz
Tangential component = xy ,σyx ,σxz ,σzx ,σyz ,σzy
σ
€
€

6. 3-D Stress and Strain
Since each point on the cube is under static
equilibrium (no net force in the absense of
any body forces), only nine stress
components from three planes are needed to
describe the stress state at a point P.
These nine components can be organized
into the matrix:
⎡σ σxy σxz ⎤
xx
⎢ ⎥
⎢σyx σyy σyz ⎥
⎢σ
⎣ zx σzy σzz ⎥
⎦
In this course we are also where shear stresses across the diagonal are identical
denoting shear stresses as τ as a result of static equilibrium (no net moment). This
grouping of the nine stress components is known as
the stress tensor (or stress matrix).
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7. 3-D Stress and Strain
Shear stresses across the diagonal are
identical as a result of static equilibrium (no
net moment). The six shear stresses reduces
to 3 shear stresses.
This grouping of the six stress components is
known as the stress tensor (or stress matrix).
The off diagonal elements are equal i.e σxy = σyx
⎡σ σxy σxz ⎤
xx
⎢ ⎥
⎢σxy σyy σyz ⎥
€
⎢σ
⎣ xz σyz σzz ⎥
⎦
€

8. Equilibrium equations
∂σyy
σyy + dy
∂y X, Y – body force such as weight of the body
Y ∂σyx
σyx + dy
∂y
σxy Y
€
∂σxx
dy σ σxx + dx
xx € X ∂x
€
σxy +
∂σxy
∂x
dx ∑F x =0
σyx σyy€
€ X
⎛ ∂σxx ⎞
dx ⎜σxx + dx ⎟(dy × 1) − σxx ( dy × 1) +
€ ⎝ ∂x ⎠
€ € ⎛ ∂σyx ⎞ €
⎜σyx + dy ⎟( dx × 1) − σyx ( dx × 1) + X ( dxdy × 1) = 0
⎝ ∂y ⎠
€

9. Equilibrium equations
∂σyy
σyy + dy
∂y
∂σyx For 2 dimension
y σyx + dy
∂y
σxy
€
Y
∂σ xx ∂σ xy
dy σ σxx +
∂σxx
dx + +X =0
xx € X ∂x ∂x ∂y
€ ∂σxy
σxy +
∂x
dx ∂σ yx ∂σ yy
σyx σyy€ + +Y = 0
€ x ∂x ∂y
dx
€
€ €
X, Y – body force such as weight of the body
€

10. Equilibrium equations
For 3 dimension
∂σ xx ∂σ xy ∂σ xz
+ + +X =0
∂x ∂y ∂z
∂σ yx ∂σ yy ∂σ yz
+ + +Y = 0
∂x ∂y ∂z
∂σ zx ∂σ zy ∂σ zz
+ + +Z =0
∂x ∂y ∂z

11. Impact Load
 Definitions
 Resilience – Total strain energy stored in the system.
 Proof resilience – Maximum strain energy stored in a
body is known as proof resilience. Strain energy in the
body will be maximum when the body is stressed upto
elastic limit
 Modulus of resilience- Proof resilience of a material
per unit volume.
Pr oof _ resilience
Modulus of resilience =
Volume _ of _ the _ body
€

12. Impact Load
 Strain energy when load is applied gradually.
M
2
Energy stored in a body=
σV
Load
2E P
σ 2 AL N
=
2E O Extension
€ x
€

13. Impact Load
 Strain energy when load is applied suddenly.
M
2
Energy stored in a body=
σ AL
Load
2E P
σ 2 AL σ
=P×x=P× ×L N
2E E O Extension
€ P x
σ =2×
A
derivation in book - R.K Bansal
€

14. Impact Load
 PROBLEM- A steel rod is 2m long and 50mm in
diameter. An axial pull of 100 kN is suddenly
applied to the rod. Calculate the instantaneous stress
induced and also the instantaneous elongation
produced in the rod. Take E=200GN/mm2

15. Impact Load
 Strain energy when load is applied with impact.
Energy of impact = Potential energy of the falling load
σ 2 AL
Energy of impact =
2E
Potential energy of the falling load = P ( h + δL )
€
P ⎛ €2AEh ⎞
σ = ⎜1+ 1+ ⎟
A ⎝ PL ⎠
€

16. Impact Load
 PROBLEM- A vertical compound tie
member fixed rigidly at its upper end
20 mm consists of a steel rod 2.5 m long and
2.5 m P=10kN 30mm external diameter. The rod and
1 2
the tube are fixed together at the ends.
3 mm The compound member is then
suddenly loaded in tension by a weight
21 mm of 10 kN falling through a height of 3
mm on to a flange fixed to its lower
30 mm
end. Calculate the maximum stresses
in steel and brass. Assume Es=2x105
N/mm2 and Eb=1.0x105 N/mm2

17. Impact Load
 Strain energy in shear loading.
τ 2 AL
Strain energy stored =
2C P
D D1 C C1
€
h
φ φ
A l B
€ €

18. Impact Load
 PROBLEM- The shear stress in a material at a price
is given as 50N/mm2. Determine the local strain
energy per unit volume stored in the material due to
shear stress. Take C=8x104 N/mm2