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When a set of co-planer external forces and moments act on a body,

the stress developed at any point ‘P’ inside the body

can be completely defined by the two dimensional state of stress:

x = normal stress in X direction,

y = normal stress in Y direction, and

xy = shear stress which would be equal but opposite in

X (cw) and Y (ccw) directions, respectively.

The 2D stress at point P is described by a box drawn with its faces perpendicular to X & Y directions, and showing all normal and shear stress vectors (both magnitude and direction) on each face of the box. This is called the stress element of point P.

Two dimensional state of stress, and the stress element

X

Y

F1F2

F3

F4Fn

MM

P

Stress Element

xx X

xy

cw

Y

y

y

xy

xy

ccwxy

xy

xy

x

xy

y

y

P x

The stress formulae that we have learnt thus far, can determine the 2D stresses developed inside a part, ONLY ALONG A RECTANGULAR AXIS SYSTEM X -Y, that is defined by the shape of the part.

For example, X axis for a cantilever beam is parallel to its length,

and Y axis is perpendicular to X.For a combined bending and axial loading (F1, F2 etc.) of this cantilever beam:

the normal and shear stress at a point P, can be determined using the formulae, such as,

x= Mv/I+P/A,

xy=VQ/(Ib).

Note that, these formulae can only determine stresses parallel to X and Y axis, and the stress element is aligned with X-Y axis.

The question is, what would be the values of normal and shear stresses at the same point P, if the stresses are measured along another rectangular axis system U-V, rotated at an angle with the X-Y axis system ?

X

Y

F1

F2P

xy

xy

xy

x

xy

y

y

P x

X

YU

V

xy

X

Y

xy

xy

x x

y

y

y

x

u

v

uv

u

v

uv

uv

u

uv

v

v

XF

Knowing the 2D stresses at point P along XY coordinate system,

we want to determine the 2D stresses for the same point P, when measured along a new coordinate system UV,

which is rotated by an angle with respect to the XY system.

The Problem is: given x, y, xy and ,

can we determine u, v, uv ?

v

u

X

F1Y

F2PX

Y

u

F2P

1. We cut the stress element by an arbitrary

plane at an angle This plane is normal to u-axis

u

uv

xy

x

xy

y

x

xy

X

Y

xy

xy

x

xy

y

y

L

Lsin

Lcos

xy(LBsin

x(LBcos

xy(

LBco

s

y(LBsin

u (LB)

uv(L

B)

2. To maintain static equilibrium, let the internal normal and shear

stresses u & uv, respectively are developed on the cut plane

3. Let, L be the length of the cut side. Then

the other two sides are Lsin & Lcos

4. If the thickness of the element is B, then the force acting on each face of the

element will be equal to the stress multiplied by the area

of the face.

THIS IS HOW WE CAN ACHIEVE THAT

U

xy(LBsin

x(LBcos x

y(LB

cos

y(LBsin

u (LB)

uv(L

B)

xy LBsincos

xyLB

sin

2

xy LBsincos

xyLB

cos

2

u (LB)

uv(L

B)

yLB

sinc

os

y LBsin 2

x LBcos 2

xLB

cos

sin

Equating forces in u-direction:

uLB = xLBcos2 yLBsin2 + 2xyLBsincos

Or, u = xcos2 + ysin2 + 2xysincos ………..(1)

Equating forces in v-direction:

uvLB = xyLBcos2 - xyLBsin2 - xLBsincos+ yLBsincos

Or, uv = xy(cos2 - sin2) – (x-y) sincos ……. (2)

5. Forces acting on the

faces = force x area

6. Resolving each force in u & v directions

CONTINUING

Replacing the square terms of trigonometric

functions by double angle terms and rearranging :

Equations 3, 4 & 5 gives us the 2D stress values, if measured along U-V axis which is at an angle from X-Y axis.

Since both sets of stresses refer to the stress of the same point, the two sets of

stresses are also equivalent.

)5.......(2sin2cos22

,

,

)4(....................2sin2

2cos

cossin cossin)sin(cos

)3(..........2sin2cos22

2sin)2cos1(2

)2cos1(2

cossin2sincos

v

yx

yx22

u

22u

xyyxyx

xy

xyuv

xyyxyx

xyyx

xyyx

thatshownbecanitthenaxisvtheto

larperpendicuplaneabyelementstressthecutweifAlso

xy

X

Y

xy

xy

x x

y

y

y

u

u

uv

v

uv

uv

u

uv

v

v

X

2sin2cos22

2sin2

2cos

2sin2cos22

v

yx

u

xyyxyx

xyuv

xyyxyx

Mohr’s circle

implements these three equations

by a graphical aid, which simplifies computation and visualization of the

changes in stress values (u, v & uv) with the rotation angle of the

measurement axis.

Mohr circle is plotted on a rectangular coordinate system in which the positive horizontal axis represents positive (tensile) normal stress , and the positive vertical axis represents the positive (clockwise) shear stress ().

Thus the plane of the Mohr circle is denoted as plane.

In this plane, the stresses acting on two faces of the stress element are plotted.

xy

Y

x

y

x

yxy

X

xy

xx Xcw

xy

Y

y

yxy

ccw

For a stress element

Y faces have stress: (y,-xy)

x faces have stress:(x & xy)

u U

V

uv

v

u

vuv

X

1. Start by drawing the original stress element with its sides parallel to XY axis, and show the normal and the shear stress vectors on the element.

2. Draw the rectangular axis and label them.3. On the plane, plot X with normal and

shear stress values of x and xy, and Y with values y and –xy.

4. Join X and Y points by a straight line, which intersects the horizontal axis at C. C denotes the average normal stress avg=(x+y)/2 .

5. The line CX denotes X axis, and line CY denotes Y axis in Mohr circle. Name them.

6. Draw the Mohr circle using C as the center, and XY line as the diameter.

7. To find stress along the new UV axis system,

draw a line UV rotated at an angle 2from the XY line. CU line denotes U axis, and CV denotes V axis.

8. The normal and shear stress values of the points U and V on the plane denote the stresses in U and V directions, respectively.

9. This way we can find stresses for an element

rotated at any desired angle .

Y

x

xy

y

x

yxy

X

Normal stress axis (

Shear

stre

ss

axis

(

U

xy

2

Y(y,-xy)

x

y

avgx+y)/2

V

xy

C

u

uv

v

uv

X (x,xy

)

X a

xis

Y a

xis

DRAWING MOHR CIRCLE

Y

x

xy

y

x

yxy

X

2sin2cos22

2sin2

2cos

2sin2cos22

v

yx

u

xyyxyx

xyuv

xyyxyx

u U

V

uv

v

u

vuv

X

X (x,xy

)

2sin2cos2

)2sin2cos2

(

)2sin2sin2cos2(cos

)22cos(

xyyx

xyyx

aaa

a

a

xyX a

xis

2

Shear

stre

ssa

xis

(

Y(y,xy)

x

y

Normal Stress axis (

Y a

xis

avgx+y)/2

U (u,uv)

V (v,xy)

a

2sin2

2cos

)2sin2

2cos(

)2sin2cos2cos2(sin

)22sin(

yxxy

yxxy

aaa

a

a

PROOF

Y(y,-xy)

xy

avg

xy

xy

X axis

Y axis

X (x,xy)Similarly, if the XY axis line is rotated by an angle 2 ‘ to make it vertical, then the shear stress maximizes and the element will have normal stress = avg and Maximum shear stress = max

In the Mohr circle, for a rotation of 2angle, the XY axis line becomes horizontal. In the rotated axis , the shear stress vanishes.

The element will have only normal stresses 1 & 2, and 1 being the maximum normal stress. are called the

Principal normal stresses. max

Principal Normal Stresses and Max Shear Stress max

1

1

2

2

Y

Xavg

avgavg

avg

max

maxx

Y

’

min

max

2 1

2’

avg,max)

avg,-max)

2x y

avg

2

2

2x y

xyR

1 22 tan xy

x y

1

2

max

avg

avg

R

R

R

min

max

avg

max

avgavg

avg

maxx

Y

Y

x

xy

y

x

yxy X

1

2

1

2

Y

X

Formulea for Principal Normal Stresses & Max Shear Stress

X (x,xy)

Y(y,-xy)

x

2 1

y

avg

max

2’

xy

xy

X axis

Y axis

avg,max)

avg,-max)

Maximum shear stress element

Principal normal stress element

2902

Determining u, v & uv

Given x, y, xy &

u U

V

uv

v

u

vuv

X

Y

x

xy

y

x

yxy

X

U (u,uv)

xyX a

xis

2

Y(y,-xy)

x

y

Y a

xis

avgx+y)/2

V (v,xy) x

y

C

u

uv

vu

v

X (x,xy

)

2

2: yx

avgC

2

2

2 xyyxRRadius

yx

xy

2

tan2 1

)22sin( Ravgu

)22( SinRavgv

)22( CosRuv

Y

X

5,000 psi

20,000 psi20,000 psi

4,000 psi

4,000 psi

X (20k,5k)

Y(-4k,-5k)

20k

-5k 21k-4k 8k

max

5k

5k

X axis

Y axis

(8k,13k)

(8k,-13k)

R=13K

For a stress element with

1. Draw the stress element along XY axis.

2. Draw the axes for mohr circle

3. Plot point X for x=20K, xy=5k

4. Plot point Y for sy= -4K, txy=-5k

5. Draw line XY and show X & Y axes.

6. Draw the circle with XY as the diameter

20 48

2 2x y

avg

k kk psi

o

yx

xy6.22)417.0(tan

420

52tan

2tan2 111

KpsiR 13max

x=20,000 psi,

y= -4000 psi, and

xy= 5000 psi.Draw the Mohr Circle and, draw two stress elements properly oriented for (i) the principal normal stresses, and (ii) max shear stresses element.

oo

o

4.676.2290

2902

kpsiRavg 211381 kpsiRavg 51382

kpsiR xyyx 135

2

)4(20

22

22

2

This completes the Mohr circle. Next, the stress elements

Y

X

5,000 psi

20,000 psi20,000 psi

4,000 psi

4,000 psi

X (20k,5k)

Y(-4k,-5k)

20k

-5k 21k-4k 8k

max

5k

5k

X axis

Y axis

(8k,13k)

(8k,-13k)

R=13K

5k

5k

Y

X

11.3

min

max

8k

13k

8k8k

8k

13kx

Y

33.7

The principal normal stress axis will be rotated CW

Draw the principal stress axis 11.3o CW from XY axis.

Show the principal stresses.

o3.112

6.22

The max axis will be rotated CCW

Draw the max stress axis

33.7o CCW from XY axis.

Show the the stresses.

o7.332

4.67

21k

21k

PRINCIPAL NORMAL STRESS ELEMENT

STRESS ELEMENT FOR MAX

That completes the drawing of the two stress elements

This ends the presentation

and thanks for watching it