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
y
A particular point
on the part
x
sy = -2 ksi
Some Part
sx = 6 ksi
x & y  orientation
txy = 3
ksi
Mohr’s Circle
t (CW)
sy = -2 ksi
x-axis
sx = 6 ksi
(6 ksi, 3 ksi)
2
6
3
txy = 3 ksi
s
Center of
Mohr’s Circle
3
(-2 ksi, -3 ksi)
y-axis
Mohr’s Circle
t (CW)
(savg, tmax)
sy = -2 ksi
x-face
sx = 6 ksi
(6 ksi, 3ksi)
txy = 3 ksi
s
s2
savg = 2 ksi
(-2 ksi, -3ksi)
y-face
s avg 
sx  sy
2
 2 ksi
(savg, tmin)
s1
Mohr’s Circle
(savg, tmax)
(2 ksi, 5 ksi)
t (CW)
sy = -2 ksi
x-face
sx = 6 ksi
(6 ksi, 3ksi)
R
txy = 3 ksi
s
s2
R 
( 3 ksi )  ( 4 ksi )
2
3 ksi
4 ksi
2
 5 ksi
R  t max
y-face
s1  savg + R  7 ksi
s2  savg – R  -3 ksi
(savg, tmin)
(2 ksi, -5 ksi)
s1
Mohr’s Circle
t (CW)
(savg, tmax)
(2 ksi, 5 ksi)
sy = -2 ksi
x-face
sx = 6 ksi
(6 ksi, 3ksi)
txy = 3 ksi
2q
s2
2 q  Tan
-1
 3 ksi 


 4 ksi 


2 q  36 . 869 
q  18 . 435 
3 ksi
4 ksi
y-face
(savg, tmin)
(2 ksi, -5 ksi)
s
s1
Principle Stress
t (CW)
(savg, tmax)
(2 ksi, 5 ksi)
s2 = -3 ksi
x-face
(6 ksi, 3ksi)
q = 18.435°
s1 = 7 ksi
Principle Stress
Element
s2
Rotation on element is
half of the rotation from
the circle in same
direction from x-axis
2q
3 ksi
4 ksi
(savg, tmin)
(2 ksi, -5 ksi)
s
s1
Shear Stress
(savg, tmax)
(2 ksi, 5 ksi)
t (CW)
savg = 2 ksi
 = 26.565°
tmax = 5 ksi
savg = 2 ksi
Maximum Shear
Stress Element
x-face
(6 ksi, 3ksi)
2
2q
s2
3 ksi
4 ksi
2   90  - 2 q
2   90 - 36 . 869 
2   53 . 130 
  26 . 565 
y-face
(savg, tmin)
(2 ksi, -5 ksi)
s
s1
Relationship Between Elements
savg = 2 ksi
tmax = 5 ksi
sy = -2 ksi
sx = 6 ksi  = 26.565°
q = 18.435°
txy = 3 ksi
q +  = 18.435 ° + 26.565 ° = 45 °
savg = 2 ksi
s2 = -3 ksi
s1 = 7 ksi
What’s the stress at angle of 15°
CCW from the x-axis?
y
A particular point
on the part
x
V
s = ? ksi
Some Part
s = ? ksi
15°
t = ? ksi
U & V  new axes @ 15° from x-axis
U
x
Rotation on
Mohr’s Circle
t (CW)
(savg, tmax)
(sU, tU)
x-face
30°
s
s2
savg = 2 ksi
y-face
15° on part and
element is 30° on
Mohr’s Circle
(sV, tV)
(savg, tmin)
s1
(savg, tmax)
t (CW)
Rotation on
Mohr’s Circle
(sU, tU)
x-face
sU = savg + R*cos(66.869°)
R
sU = 3.96 ksi
66.869°
s2
sV = savg – R*cos(66.869°)
sV = 0.036 ksi
tUV = R*sin(66.869°)
tUV = 4.60 ksi
s
savg = 2 ksi
y-face
(sV, tV)
(savg, tmin)
s1
What’s the stress at angle of 15°
CCW from the x-axis?
y
A particular point
on the part
x
Some Part
V
sV = .036 ksi
sU = 3.96 ksi
U
15°
x
t = 4.60 ksi
Questions?
Next: Special Cases
Special Case – Both Principle
Stresses Have the Same Sign
Y
X
sx 
Z
Cylindrical
Pressure Vessel
sy 
pD
4 t
pD
2 t
Mohr’s Circle
sy
t (CW)
t 
sx
2
sy
This isn’t the whole
story however…
sx - sy
sx
s
Mohr’s Circle for X-Y Planes
sx = s1 and sy = s2
Mohr’s Circle
sy
t (CW)
sz = 0 since it is
perpendicular to the free
face of the element.
sx
tmax  txz
sx
sz
sz = 0
s3
s1
s
sz = s3 and sx = s1
t max 
s1 - s 3
2
Mohr’s Circle for X-Z Planes
Mohr’s Circle
sy
sz
t (CW)
tmax  txz
sx
s3
sz = 0 since it is
perpendicular to the free
face of the element.
s2
s1 > s2 > s3
s1
s
Pure Uniaxial Tension
sy = 0
t max 
sx = P/A
s2 = 0
Ductile Materials Tend to Fail
in SHEAR
sx
2
s1= sx
Note when sx = Sy,
Sys = Sy/2
Pure Uniaxial Compression
sy = 0
t max 
sx
2
sx = P/A
s2= sx
s1 = 0
Pure Torsion
T
t max  t xy
T
t xy 
T c
J
s2 = -txy
s1
CHALK
s1 = txy
Brittle materials
tend to fail in
TENSION.
Uniaxial Tension & Torsional
Shear Stresses
• Rotating shaft with axial load.
• Basis for design of shafts.
t max  R
(sx, txy)
sx = P/A
s2 = sx/2-R
txy = Tc/J
(0, tyx)
2
R 
 sx 
2

  t xy  t max
 2 
sx/2
s1 = sx/2+R