Purpose of Mohr’s Circle

• Visual tool used to determine the stresses that exist at a given point in relation to the angle of orientation of the stress element.

• There are 4 possible variations in Mohr’s Circle depending on the positive directions are defined.

Sample Problem

x = 6 ksi

y = -2 ksi

xy = 3 ksi

Some Part

A particular point on the part

x

y

x & y orientation

Mohr’s Circle (CW)

x-axis

y-axis

x = 6 ksi

y = -2 ksi

xy = 3 ksi

(6 ksi, 3 ksi)

63

(-2 ksi, -3 ksi)

2

3 Center of of Mohr’s CircleMohr’s Circle

Mohr’s Circle (CW)

avg=

2ksi

x-face

y-face

(6 ksi, 3ksi)

(-2 ksi, -3ksi)

(avg, max)

x = 6 ksi

y = -2 ksi

xy = 3 ksi

(avg, min)ksiyxavg 2

2

Mohr’s Circle (CW)

x-face

y-face

(6 ksi, 3ksi)

x = 6 ksi

y = -2 ksi

xy = 3 ksi

4 ksi

avg + R7 ksiavg – R ksi

(avg, max)(2 ksi, 5 ksi)

(avg, min)(2 ksi, -5 ksi)

3 ksi

maxR

ksi

)ksi()ksi(R

5

43 22

R

Mohr’s Circle (CW)

x-face

y-face

(6 ksi, 3ksi)

x = 6 ksi

y = -2 ksi

xy = 3 ksi 2

4 ksi

(avg, max)(2 ksi, 5 ksi)

(avg, min)(2 ksi, -5 ksi)

3 ksi

43518

869362

4

32 1

.

.

ksi

ksiTan

Principle Stress (CW)

x-face

(6 ksi, 3ksi)

1 = 7 ksi

2 = -3 ksi

2

4 ksi

(avg, max)(2 ksi, 5 ksi)

(avg, min)(2 ksi, -5 ksi)

3 ksi = 18.435°

Principle Stress Element

Rotation on element is half of the rotation from the circle in same direction from x-axis

Shear Stress (CW)

x-face

y-face

(6 ksi, 3ksi)

avg = 2 ksi

avg = 2 ksi

max = 5 ksi

2

4 ksi

(avg, max)(2 ksi, 5 ksi)

(avg, min)(2 ksi, -5 ksi)

3 ksi

56526

130532

86936902

2902

.

.

.

2

Maximum Shear Stress Element

= 26.565°

Relationship Between Elements

avg = 2 ksi

avg = 2 ksi

max = 5 ksi

= 26.565°

1 = 7 ksi

2 = -3 ksi

x = 6 ksi

y = -2 ksi

xy = 3 ksi = 18.435°

+ = 18.435 ° + 26.565 ° = 45 °

What’s the stress at angle of 15° CCW from the x-axis?

= ? ksi

= ? ksi

= ? ksi

Some Part

A particular point on the part

x

y

U & V new axes @ 15° from x-axis

15°

U

x

V

Rotation on Mohr’s Circle

(CW)

avg=

2ksi

x-face

y-face

(avg, max)

(avg, min)

30°

15° on part and element is 30° on Mohr’s Circle

(U, U)

(V, V)

U = avg + R*cos(66.869°)

U = 3.96 ksi

V = avg – R*cos(66.869°)

V = 0.036 ksi

UV = R*sin(66.869°)

UV = 4.60 ksi

Rotation on Mohr’s Circle

(CW)

avg=

2ksi

x-face

y-face

(avg, max)

(avg, min)

66.869°

R

(U, U)

(V, V)

What’s the stress at angle of 15° CCW from the x-axis?

U = 3.96 ksiV= .036 ksi

= 4.60 ksi

Some Part

A particular point on the part

x

y

15°

U

x

V

Questions?

Next: Special Cases

Special Case – Both Principle Stresses Have the Same Sign

Cylindrical Pressure Vessel

X

Y

Z t

Dpx 2

t

Dpy 4

Mohr’s Circle

x

(CW)

y

x

y

2yx

Mohr’s Circle for X-Y PlanesThis isn’t the whole story however…

x = 1 and y = 2

Mohr’s Circle

1

(CW)

y

x

z

z = 0 since it is perpendicular to the free face of the element.

z = 3 and x = 1

3

Mohr’s Circle for X-Z Planes

x

z = 0

maxxz

231

max

Mohr’s Circle

1

(CW)

2

y

x

z

z = 0 since it is perpendicular to the free face of the element.

3

maxxz

1 > 2 > 3

Pure Uniaxial Tension

y = 0

x = P/A

Ductile Materials Tend to Fail in SHEAR

1= x

2x

max

2 = 0

Note when x = Sy, Sys = Sy/2

Pure Uniaxial Compression

y = 0

x = P/A

2x

max

1 = 02= x

Pure TorsionT

T

J

cTxy

xymax

1 = xy2 = -xy

CHALK1

Brittle materials tend to fail in TENSION.

Uniaxial Tension & Torsional Shear Stresses

• Rotating shaft with axial load.

• Basis for design of shafts.

x = P/A

xy = Tc/J

Rmax

1 = x/2+Rx/2

x, xy)

2 = x/2-R

0, yx)

maxxyxR

2

2

2