mohr’s circle - principal strains

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RAJESH KUMAR.B

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Page 1: MOHR’S CIRCLE - Principal Strains

RAJESH KUMAR.B

Page 2: MOHR’S CIRCLE - Principal Strains

The Problem

An element in a stressed material has tensile strain of 0.00025 and a compressive strain of 0.000125 acting on two mutually perpendicular planes and equal shear strains of 0.0001 on these planes. Find principal strains and position of the principal planes.

Page 3: MOHR’S CIRCLE - Principal Strains

Step 1 Using a suitable scale, measure OL = = + 0.00025 xe

Y

L

xe = 0.00025

O

X

Page 4: MOHR’S CIRCLE - Principal Strains

ye = 0.000125

Step 2 Using a suitable scale, measure OM = = - 0.000125 ye

Y

L

X

M

Take in negative sideye

O

xe = 0.00025

Page 5: MOHR’S CIRCLE - Principal Strains

Step 3

Y

L

X

M O

ye = 0.000125 xe = 0.00025

T

2

es

2

es

At L , draw LT perpendicular to OX and equal to = 0.00005 in the downward direction

Page 6: MOHR’S CIRCLE - Principal Strains

Step 4

Y

L

XM

O

ye = 0.000125 xe = 0.00025

T

2

es

2

es

At M , draw MS perpendicular to OX and equal to = 0.00005 in the upward direction

S

2

es

Page 7: MOHR’S CIRCLE - Principal Strains

Step 5

Y

L

XM

O

ye = 0.000125 xe = 0.00025

T

2

es

S

2

es

Join ST to cut the line OX, at a point N .

N

Page 8: MOHR’S CIRCLE - Principal Strains

Step 6

Y

L

XM

O

ye = 0.000125 xe = 0.00025

T

2

es

S

2

es

N

With N as centre and NS or NT as radius, draw a circle

Mark U and V at the points where the circle meets OX

Mohr’s Circle

U

V

Page 9: MOHR’S CIRCLE - Principal Strains

Step 7

Y

L

XM

O

ye = 0.000125 xe = 0.00025

T

2

es

S

2

es

N

U

V

From N , draw a perpendicular to meet the circle at Z

Z

Page 10: MOHR’S CIRCLE - Principal Strains

Step 7

Y

L

XM

O

ye = 0.000125 xe = 0.00025

T

2

es

S

2

es

N

U

V

Z

Find the VNT = and UNT =

12θ

12θ 22θ

22θ

Page 11: MOHR’S CIRCLE - Principal Strains

Step 8

Y

L

XM

O

ye = 0.000125 xe = 0.00025

T

2

es

S

2

es

N

U

V

Z

12θ

22θ

NZemax

OV 1 e

OU 2 e

2e 1e