Mohr's Circle, Principle Planes & Stresses

2. This graphical representation is extremely useful because it enables you
to visualize the relationships between the normal and shear stresses
acting on various inclined planes at a point in a stressed body.
Using Mohr’s Circle you can also calculate principal stresses, maximum
shear stresses and stresses on inclined planes.

3. NUMERICAL

5. Given Data:-
σ1 = 120 N/mm2 (Tensile),
σ2 = 60 N/mm2 (Tensile), θ = 30o
Scale:- 1cm = 10 N/mm2
Then σ1 = 120/10 = 12 cm
σ2 = 60/10 = 6 cm

6. σ1 = 120 N/mm2 (Tensile) = 12 cm, σ2 = 60 N/mm2 (Tensile) = 6cm, θ = 30o Scale:- 1cm = 10 N/mm2
1. Take any point A and
draw a horizontal line
through A.
A. B
σ1 (12 cm)
σ2 (6 cm)
O
E
D
2θ
C
θ
2. Take AB = σ1 = 12 cm
and AC = σ2 = 6 cm
3. With BC as diameter
(i.e. BC = 12-6 = 6cm)
describe a circle.
4. Let “O” is the centre of
the circle. Draw a circle
with radius 6 cm
5. Through O, draw a line
OE making an angle 2θ
(i.e 2 x 30 = 60o with
OB.
6. From E, draw ED
perpendicular to CB.
7. Join AE & EC
8. Measure
Length AD = 10.50 cm
Length ED = 2.60 cm
Length AE = 10.82 cm
Then,
1. Normal Stress = Length AD x Scale
= 10.50 x 10 = 105 N/mm2
Theoretical Results
σn = 105 N/mm2
σt = 25.98 N/mm2
σR = 108.16 N/mm2
3. Resultant Stress = Length AE x Scale
= 10.82 x 10 = 108.2 N/mm2
2. Tangential or Shear stress = Length ED x Scale
= 2.60 x 10 = 26 N/mm2
AD = 10.50 cm
DE=2.60cm
Check the
results with
the solution
ф
MOHR’S CIRCLE
σn
σt

7. MOHR’S CIRCLE PROOFMOHR’S CIRCLE PROOF

9. A
C
E
DO B
2θф
Max Normal Stress σn = σ1
σn
σt
σR
Min Normal Stress σn = σ2σ2
σ1
IMPORTANT POINTS

10. A
C
E
DO B
2θф
σn
σt
σR
σ2
σ1
σt max.
ф

11. NUMERICAL
NUMERICAL

14. σ1 = 200 N/mm2 (Tensile) = 10 cm, σ2 = -100 N/mm2 (Comp.) = 5cm, θ = 30o Scale:- 1cm = 20 N/mm2
Theoretical Results
σn = 125 N/mm2
σt = 129.9 N/mm2
σR = 180.27 N/mm2
(σt)max = 129.9 N/mm2
ф = 46o 6’
1. Take any point A and
draw a horizontal line
through A on both sides.
Check the
results with
the solution
MOHR’S CIRCLE
Compressive
(-) Tensile
(+)
y
x
2θ =60o
θ
ф
C
A
O
D
E
σ2= -5cm
(100)
σ1 = 10 cm
(200)
6.25 cm (σn) 6.5cm(σt)
2. Take AB = σ1 = 10 cm
and AC = σ2 = -5 cm
3. With BC as diameter
(i.e. OB = 7.5 cm)
describe a circle.
4. Let “O” is the centre of
the circle. Draw a circle
with OB or OC
5. Through O, draw a line
OE making an angle 2θ
(i.e 2 x 30 = 60o with
OB.
6. From E, draw ED
perpendicular to CB.
8. Measure
Length AD = 6.25 cm
Length ED = 6.5 cm
Length AE = 9.0 cm
B
7. Join AE & EC
σn = Length AD x Scale
= 6.25 x 20
= 125 N/mm2
σR = Length AE x Scale
= 9.0 x 20
= 180 N/mm2
σt = Length ED x Scale
= 6.5 x 20
= 130 N/mm2
ф = 46o

15. NUMERICAL

17. σ1 = 65 N/mm2 (Tensile) = 6.5 cm, σ2 = 35 N/mm2 (Tensile) = 3.5 cm, Scale:- 1cm = 10 N/mm2
Theoretical Results
σn = 75N/mm2
σt = 15 N/mm2
1. Take any point A and
draw a horizontal line
through A.
Check the
results with
the solution
MOHR’S CIRCLE
фA
L O
B
G
C
E
M
D
F
Ʈ = 25 N/mm2 (Shear Stress) = 2.5 cm , θ = 45o
σ2 = 3.5 cm
(35)
σ1 = 6.5cm
(65)
Ʈ = 2.5cm
(25)
Ʈ=2.5cm
(25)
2. Take AB = σ1 = 6.5 cm
and AC = σ2 = 3.5 cm
3. Draw perpendicular at B
& C and cut off BF &
CG = Ʈ = 2.5cm
4. Let “O” is the centre of
the circle at
intersection. Draw a
circle with OF or OG
5. Through O, draw a line
OE making an angle 2θ
(i.e 2 x 45 =90o with OF.
6. From E, draw ED
perpendicular to AB.
8. Measure
Length AD = 7.5 cm
Length ED = 1.5 cm
7. Join AE
σn = Length AD x Scale
= 7.5 x 10
= 75 N/mm2
σt = Length ED x Scale
= 1.5 x 10
= 15 N/mm2
7.5 cm (σn)
1.5cm(σt)

19. σ1 = 20 N/mm2 (Tensile) = 10 cm, σ2 = 10 N/mm2 (Tensile) = 5 cm, Scale:- 1cm = 2 N/mm2
1. Take any point A and
draw a horizontal line
through A.
MOHR’S CIRCLE
A L
O
B
G
C
M
F
Ʈ = 10 N/mm2 (Shear Stress) = 5 cm
σ2 = 5 cm
(10)
σ1 = 10 cm
(20)
Ʈ = 5cm
(10)
Ʈ=5cm
10)
2. Take AB = σ1 = 10 cm
and AC = σ2 = 5 cm
3. Draw perpendicular at B
& C and cut off BF &
CG = Ʈ = 5cm
4. Let “O” is the centre of
the circle at
intersection. Draw a
circle with OF or OG
7. Measure angle 2θ
angle FOB = 63.7o
6. Measure
Length AM = 13.17.5
cm
Length AL = 1.91 cm
Major Principal stresses = Length AM x Scale
= 13.1 x 2
= 26.2 N/mm2
13.1 cm
1.91 cm
Minor Principal stresses = Length AL x Scale
= 1.91 x 2
= 9.82 N/mm2
63.7o
5. Mark M & L at the
intersection of circle on
AB
Location of principal planes
2θ = 63.7o, θ = 31.85o
Second Principal Plane
θ + 90o = 121.85o

21. σ1 = 30 N/mm2 (Tensile) = 15 cm, σ2 = 10 N/mm2 (Tensile) = 5 cm, Scale:- 1cm = 2 N/mm2
1. Take any point A and
draw a horizontal line
through A.
MOHR’S CIRCLE
A L
O
B
G
C
M
F
Ʈ = 10 N/mm2 (Shear Stress) = 5 cm
σ2 = 5 cm
(10)
σ1 = 15 cm
(30)
Ʈ = 5cm
(10)
Ʈ=5cm
10)
2. Take AB = σ1 = 15 cm
and AC = σ2 = 5 cm
3. Draw perpendicular at B
& C and cut off BF &
CG = Ʈ = 5 cm
4. Let “O” is the centre of
the circle at
intersection. Draw a
circle with OF or OG
7. Measure angle 2θ
angle FOB = 45o
6. Measure
Length AM = 7.5 cm
Length AL = 1.5 cm
Major Principal stresses = Length AM x Scale
= 17.1 x 2
= 34.2 N/mm2
17.1 cm
2.93 cm
Minor Principal stresses = Length AL x Scale
= 2.93 x 2
= 5.86 N/mm2
45o
5. Mark M & L at the
intersection of circle on
AB
H
Max.
Shear
Stress
Location of principal planes
2θ = 45o, θ = 22.5o
Second Principal Plane
θ + 90o = 112.5o
8. Ʈmax = Length OH x
Scale
= 7.05 X 20 = 14.1 N/mm2