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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:

sx = normal stress in X direction,

sy = normal stress in Y direction, and

txy = 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

F1 F2 

F3 F4 

Fn 

MM

P

Stress

Element

sxsx X

txycw

Y

sy

sy

txy

txy

ccw

txy 

txy   t  x  y

sx 

  t  x  y

sy 

sy 

P sx 

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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,

sx= Mv/I+P/A,

txy

=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 f  with the X-Y axis system ?

X

Y

F1

F2P

txy 

txy 

  t  x  y

sx 

  t  x

  y

sy 

sy 

P sx 

X

Y f

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f

txy X

Y

txy 

  t  x  y

sx 

  t  x  y

sy 

sy 

sx 

Xf

F

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 f with respect to the XY system.

The Problem is: given sx, sy, txy and f,

can we determine su

, sv

, tuv

 ?

X

F1 Y

F2PX

Y

f

F2P

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1. We cut the stress

element by an arbitraryplane at an angle f. This

plane is normal to u-axis

txy 

sx 

  t  x  y

sy 

f

sx 

txy X

Y

txy 

  t  x  y

sx 

  t  x  y

sy 

sy 

f

f

Lsinf 

   L  c  o  s   f 

txy(LBsinf) 

sx(LBcosf) 

  t  x  y   (

   L   B  c  o  s   f   )

sy(LBsinf) 

f

2. To maintain static equilibrium,

let the internal normal and shear

stresses su & tuv, respectively

are developed on the cut plane

3. Let, L be the length

of the cut side. Then

the other two sides are

Lsinf & Lcosf

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

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txy(LBsinf) 

sx(LBcosf) 

  t  x  y

   (   L   B  c  o

  s   f   )

sy(LBsinf) 

f

f

f

f

f

f

Equating forces in u-direction:

suLB = sxLBcos2f + syLBsin

2f + 2txyLBsinfcosf 

Or, su = sxcos2f + sysin

2f + 2txysinfcosf ………..(1) 

Equating forces in v-direction:

tuvLB = txyLBcos2f - txyLBsin

2f - sxLBsinfcosf+ syLBsinfcosf

Or,t

uv

 =

t

xy

 cos

2

f

 - sin

2

f

) – 

s

x

-

s

y

)

 

sin

f

cos

f

 ……. 2) 

5. Forces

acting on the

faces = force xarea

6. Resolving

each force in u

& v directions

CONTINUING

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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 isat an anglef

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

cossincossin)sin(cos

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

2sin)2cos1(2

)2cos1(2

cossin2sincos

v

yx

yx

22

u

22

u

f t f s s s s 

s 

f s s 

f t 

f f s f f s f f t t 

f t f 

s s s s 

f t f s 

f s 

s 

f f t f s f s s 

 xy

 y x y x

 xy

 xyuv

 xy

 y x y x

 xy

 y x

 xy y x

that  shownbecanit thenaxisvtheto

lar  perpendicu planeabyelement  stressthecut weif   Also

--

-+

-

--

+--

+

-

+

+

+-++

++

txy X

Y

txy 

  t  x  y

sx 

  t  x  y

sy 

sy 

f

X

f

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f t f s s s s 

s 

f s s 

f t t 

f t f s s s s 

s 

2sin2cos22

2sin2

2cos

2sin2cos22

v

yx

u

 xy

 y x y x

 xyuv

 xy

 y x y x

---+

--

+-

++

Mohr’s circle

implements these three equations

by a graphical aid, which simplifies

computation and visualization of thechanges in stress values (su, sv & tuv)

with the rotation angle f of the

measurement axis. 

Mohr circle is plotted on a rectangular coordinate system

in which the positive horizontal axis represents positive

(tensile) normal stress (s), and the positive vertical axis

represents the positive (clockwise) shear stress (t).

Thus the plane of the Mohr circle is denoted as s-t 

plane. 

In this s-t plane, the stresses acting on two faces of the

stress element are plotted.

txy

Y

sx

sy

sx

sytxy

X

txy

sxsx Xcw

txy

Y

sy

sytxy

ccw

s-s

t

-t

For a stress element

Y faces have stress:

(sy,-txy)

x faces have stress:(sx & txy)

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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 s-t rectangular axis and label them.3. On the s-t plane, plot X with normal and

shear stress values of sx and txy, and Y with

values sy and –txy.4. Join X and Y points by a straight line, which

intersects the horizontals

 axis at C. C

denotes the average normal stress

savg=(sx+sy)/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 2   from

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 s-t 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 f .

Y

sx

txy

sy

sx

sytxy

X

Normal stress

axis (s) 

   S   h  e  a  r  s   t  r

  e  s  s

  a  x   i  s   (  t   )

-s 

  t  x  y

Y(sy,-txy)

sxsy

savg(sx+sy)/2 

  t  x  y

C

su

  t  u  v

sv

  t  u  v

s 

t 

-t 

DRAWING MOHR CIRCLE

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Y

sxtxy

sy

sx

sy

txy

X

f t f s s s s 

s 

f s s 

f t t 

f t f s s s s s 

2sin2cos

22

2sin2

2cos

2sin2cos22

v

yx

u

 xy

 y x y x

 xyuv

 xy y x y x

--

-+

--

+-

++

X

f t f s s 

f t 

f s s 

f  f  

f  

2sin2cos2

)2sin2cos2

(

)2sin2sin2cos2(cos

)22cos(

 xy

 y x

 xy y x

aaa

a

a

+-

+-

+

-

-s 

  t  x  y

   S   h  e  a  r  s   t  r  e  s  s  a  x   i  s   (  t   )

Y(sy,txy)

sxsy

Normal Stress axis (s) 

savg(sx+sy)/2 

f s s 

f t 

f s s 

f t 

f  f  

f  

2sin2

2cos

)2sin2

2cos(

)2sin2cos2cos2(sin

)22sin(

 y x

 xy

 y x xy

aaa

a

a

--

--

-

-PROOF

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Y(sy,-txy)

sxsy

savg-s 

-t 

o 

txy

txy

t 

s 

X (sx,txy)

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 = savg andMaximum shear stress = tmax 

In the Mohr circle, for a rotation

of 2  angle, the XY axis line

becomes horizontal. In therotated axis s1-s2, the shear

stress vanishes.

The element will have only

normal stresses s1 & s2, and s1 

being the maximum normal

stress. s1 & s2 are called thePrincipal normal stresses. tmax

Principal Normal Stressess

1

  s

2

,

and Max Shear Stresstmax 

2 

x

Y

’ 

s2 s1

2’ 

(savg,tmax)

(savg,-tmax)

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2

 x y

avg 

s s s 

+

2

2

2

 x y

 xy Rs s 

t -

+

1  2

2 tan  xy

 x y

t  

s s 

- 

  -

1

2

max

avg 

avg 

 R

 R

 R

s s 

s s 

t 

+

-

x

Y

f

Y

sxtxy

sy

sx

sy

txy X

Formulea for Principal Normal Stresses & Max Shear Stress

X (sx,txy)

Y(sy,-txy)

sxs2 s1

sy

savg

tmax

2 -s 

-t 

o 

2’ txy

txy

(savg,tmax)

(savg,-tmax)

s 

t 

Maximum shear

stress element

Principal normal

stress element

     2902   -

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Determining su, sv & tuv 

Given sx, sy, txy & f 

X

Y

sx

txy

sy

sx

sytxy

X

-s 

  t  x  y

Y(sy,-txy)

sxsy

savg(sx+sy)/2 

  t

  x  y

C

su

  t  u  v

sv

  t  u  v

s 

t 

-t 

2:   y xavg C 

s s s 

+

2

2

2  xy

 y x R Radius   t 

s s +

 

  

    -

 

 

 

 -

  -

 y x

 xy

s s 

t  

2tan2   1

)22sin(   f  s s    -+   Ravg u

)22(   f  s s    --   Sin Ravg v

)22(   f  t    -   Cos Ruv

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Y

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Y

X

5,000 psi

20,000 psi20,000 psi

4,000 psi

4,000 psi

X (20k,5k)

Y(-4k,-5k)

20k

s2 -5k s121k-4k 8k

tmax

22.6 s -s 

t 

-t 

o 67.4 5k

5k

(8k,13k)

(8k,-13k)

For a stress element with

1. Draw the stress element along XY axis.

2. Draw the s-t axes for mohr circle

3. Plot point X for sx=20K, txy=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 2

 x y

avg 

k k k psi

s s s 

+   -

o

 y x

 xy6.22)417.0(tan

420

52tan

2tan2   111

 

  

 

+

 

 

 

 

-   ---

s s 

t  

 Kpsi R   13max   t 

  sx=20,000 psi,

sy= -4000 psi, and

txy= 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

-

-        kpsi Ravg    211381   ++ s s 

kpsi Ravg    51382   --- s s 

kpsi R  xy y x

135

2

)4(20

2

2

2

2

2

+

 

 

 

    --+

 

 

 

    -   t 

s s 

This completes the Mohr circle. Next, the stress elements

Y

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Y

X

5,000 psi

20,000 psi20,000 psi

4,000 psi

4,000 psi

X (20k,5k)

Y(-4k,-5k)

20k

s2 -5k s121k-4k 8k

tmax

22.6 s -s 

t 

-t 

o 67.4 5k

5k

(8k,13k)

(8k,-13k)

11.3

x

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 tmax axis will be

rotated CCW

Draw the tmax stress axis33.7o CCW from XY axis.

Show the the stresses.

o7.332

4.67 

PRINCIPAL NORMAL STRESS ELEMENT

STRESS ELEMENT FOR tMAX 

That completes the drawing of

the two stress elements

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This ends the

presentationand thanks for watching it