Lecture 4 3 d stress tensor and equilibrium equations

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Transcript of Lecture 4 3 d stress tensor and equilibrium equations

Page 1: Lecture 4 3 d stress tensor and equilibrium equations
Page 2: Lecture 4 3 d stress tensor and equilibrium equations

Unit 1- Stress and Strain

  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

Topics Covered

Page 3: Lecture 4 3 d stress tensor and equilibrium equations

3-D Stress and Strain stress vector 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.

σ = limΔs−>0

ΔFΔs

=dFds

σ

Page 4: Lecture 4 3 d stress tensor and equilibrium equations

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

Page 5: Lecture 4 3 d stress tensor and equilibrium equations

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 =

σxy,σyx,σxz,σzx ,σyz,σzy

Page 6: Lecture 4 3 d stress tensor and equilibrium equations

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:

σxx σxy σxz

σyx σyy σyz

σzx σzy σzz

⎢ ⎢ ⎢

⎥ ⎥ ⎥

where shear stresses across the diagonal are identical 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).

In this course we are also denoting shear stresses as

τ

Page 7: Lecture 4 3 d stress tensor and equilibrium equations

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).

σxx σxy σxz

σxy σyy σyz

σxz σyz σzz

⎢ ⎢ ⎢

⎥ ⎥ ⎥

The off diagonal elements are equal i.e

σxy =σyx

Page 8: Lecture 4 3 d stress tensor and equilibrium equations

Equilibrium equations

σyy€

σxx

σxx +∂σxx

∂xdx

σyy +∂σyy

∂ydy

σxy +∂σxy

∂xdx

σyx +∂σyx

∂ydy

σyx

σxy

dx

dy

X

Y

Fx = 0∑

σxx +∂σxx

∂xdx

⎝ ⎜

⎠ ⎟ (dy ×1) −σxx dy ×1( ) +

σyx +∂σyx

∂ydy

⎝ ⎜

⎠ ⎟ dx ×1( ) −σyx dx ×1( ) + X dxdy ×1( ) = 0

X

Y

X, Y – body force such as weight of the body

Page 9: Lecture 4 3 d stress tensor and equilibrium equations

Equilibrium equations

σyy€

σxx

σxx +∂σxx

∂xdx

σyy +∂σyy

∂ydy

σxy +∂σxy

∂xdx

σyx +∂σyx

∂ydy

σyx

σxy

dx

dy X

Y

∂σ xx

∂x+∂σ xy

∂y+ X = 0

∂σ yx

∂x+∂σ yy

∂y+Y = 0

For 2 dimension

x

y

X, Y – body force such as weight of the body

Page 10: Lecture 4 3 d stress tensor and equilibrium equations

Equilibrium equations

∂σ xx

∂x+∂σ xy

∂y+∂σ xz

∂z+ X = 0

∂σ yx

∂x+∂σ yy

∂y+∂σ yz

∂z+Y = 0

∂σ zx

∂x+∂σ zy

∂y+∂σ zz

∂z+ Z = 0

For 3 dimension

Page 11: Lecture 4 3 d stress tensor and equilibrium equations

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.

Proof _ resilienceVolume_of _ the_body

Modulus of resilience =

Page 12: Lecture 4 3 d stress tensor and equilibrium equations

Impact Load   Strain energy when load is applied gradually.

Extension

Load P

x €

σ2V2E

Energy stored in a body=

O N

M

=σ2AL2E

Page 13: Lecture 4 3 d stress tensor and equilibrium equations

Impact Load   Strain energy when load is applied suddenly.

Extension

Load P

x €

σ2AL2E

Energy stored in a body=

O N

M

derivation in book - R.K Bansal

σ2AL2E

= P × x = P × σE× L

σ = 2 × PA

Page 14: Lecture 4 3 d stress tensor and equilibrium equations

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

Page 15: Lecture 4 3 d stress tensor and equilibrium equations

Impact Load   Strain energy when load is applied with impact.

σ2AL2E

Energy of impact =

Energy of impact = Potential energy of the falling load

Potential energy of the falling load =

P h +δL( )

σ =PA1+ 1+

2AEhPL

⎝ ⎜

⎠ ⎟

Page 16: Lecture 4 3 d stress tensor and equilibrium equations

Impact Load   PROBLEM- A vertical compound tie

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

1 2

30 mm

20 mm 2.5 m

21 mm

P=10kN

3 mm

Page 17: Lecture 4 3 d stress tensor and equilibrium equations

Impact Load   Strain energy in shear loading.

τ 2AL2C

Strain energy stored =

φ

D

A

D1 C C1

B l

h

φ

P

Page 18: Lecture 4 3 d stress tensor and equilibrium equations

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