Lecture4 3-d Stress Tensor and Equilibrium Equations

7/30/2019 Lecture4 3-d Stress Tensor and Equilibrium Equations

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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, Equilibriumequations and impact loading

Lecture -5 - Generalized Hook's law and Castigliono's

Topics Covered


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3-D Stress and Strainstress vector that represents the force perunit area acting at a given location on thebody's surface.

In other words, a stress vector cannot be fullydescribed unless both the force and the

surface where the force acts on has been

specified.

= lims>0

F

s=

dF

ds


4/18

3-D Stress and StrainSuppose an arbitrary slice is made across the

solid shown in the above figure, leading to

the free body diagram shown at left. Stresswould appear on the exposed surface, similarin form to the external stress applied to the

body's exterior surface. The stress at pointP

can be defined using the same aboveequation


5/18

3-D Stress and StrainStresses acting on an plane, are typicallydecomposed into three mutually orthogonalcomponents. One component is normal to

the surface and represents direct stress. Theother two components are tangential to the

surface and represent shear stresses.

Normal component = xx,yy,zz

Tangential component =xy,yx,xz,zx,yz,zy


6/18

3-D Stress and StrainSince each point on the cube is under staticequilibrium (no net force in the absense ofany body forces), only nine stress

components from three planes are needed todescribe the stress state at a pointP.

These nine components can be organized

into the matrix:

xx

xy

xz

yx

yy

yz

zx

zy

zz

where shear stresses across the diagonal are identical

as a result of static equilibrium (no net moment). Thisgrouping of the nine stress components is known as

the stress tensor(or stress matrix).

In this course we are also

denoting shear stresses as


7/18

3-D Stress and StrainShear stresses across the diagonal areidentical as a result of static equilibrium (nonet 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).

xx

xy

xz

xy

yy

yz

xz

yz

zz

The off diagonal elements are equal i.e xy =yx


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Equilibrium equations

yy

xx

xx

+

xx

xdx

yy +

yy

ydy

xy +xy

xdx

yx +

yx

ydy

yx

xy

dx

dy X

Yxx

x+

xy

y+ X = 0

yx

x+

yy

y+Y = 0

For 2 dimension

x

y

X, Y body force such as weight of the body


10/18

Equilibrium equations

xx

x+

xy

y+

xz

z+

X= 0

yx

x+

yy

y+

yz

z+Y = 0

zx

x+

zy

y+

zz

z+Z= 0

For 3 dimension


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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 thebody will be maximum when the body is stressed upto

elastic limit

Modulus of resilience- Proof resilience of a materialper unit volume.

Proof _ resilience

Volume_of _ the_bodyModulus of resilience =


12/18

Impact Load Strain energy when load is applied gradually.

Extension

LoadP

x

2

V

2E

Energy stored in a body=

O

N

M

=

2AL

2E


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Impact Load

Strain energy when load is applied suddenly.

Extension

LoadP

x

2

AL

2E

Energy stored in a body=

O

N

M

derivation in book - R.K Bansal

2AL

2E= P x = P

E L

= 2 P

A


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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 stressinduced and also the instantaneous elongation

produced in the rod. Take E=200GN/mm2


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Impact Load

Strain energy when load is applied with impact.

2AL

2EEnergy of impact =

Energy of impact = Potential energy of the falling load

Potential energy of the falling load = P h +L( )

=

P

A1+ 1+

2AEh

PL


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Impact Load PROBLEM- A vertical compound tie

member fixed rigidly at its upper end

consists of a steel rod 2.5 m long and30mm 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 lowerend. Calculate the maximum stresses

in steel and brass. Assume Es=2x105

N/mm2 and Eb=1.0x105 N/mm2

1 2

30 mm

20 mm2.5 m

21 mm

P=10kN

3 mm


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Impact Load

Strain energy in shear loading.2AL

2CStrain energy stored =

D

A

D1 C C1

Bl

h

P


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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 toshear stress. Take C=8x104 N/mm2

Lecture4 3-d Stress Tensor and Equilibrium Equations

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