Stresses in a Plane
• Some parts experience normal stresses in
two directions.
directions These types of problems are
called
– Plane Stress or
– Biaxial Stress
Mohr’s Circle
MET 210W
E. Evans
Cross Section
thru Body
Tangent and normal
to the surface
z
• Force ΔFy → τzy
• Force ΔFx → τzx
• Force ΔFz → σz
Convenient
coordinate
system
x
y
MET 210W – Mohr’s Circle
• Make two more cuts in the original body, each
perpendicular to the others.
1
General State of Stress
General State of Stress & Strain
• Cubic volume
element
• Characterized by
three components
on each face
• Changes
Ch
with
ith a
different coord.
system
σy
τyz
τyx
τzy
σx
τzx
τxz
σx
τxy
σz
σy
σ x νσ y νσ z
−
−
E
E
E
νσ x σ y νσ z
εy = −
+
−
E
E
E
νσ x νσ y σ z
εz = −
−
+
E
E
E
εx = +
Positive
orientations
shown
ε = strain
ν = Poisson’s Ratio
Rotating the element in any direction changes the state of stress.
Finding Maximum Stress
• There are three maximum stresses to be
found:
– The maximum normal stress, σ = P/A ± Mc/I
– The orientation of maximum stress – Mohr’s
Circle
– Time
Ti dependent
d
d t maximum
i
stress
t
– endurance
d
strength
MET 210W – Mohr’s Circle
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.
2
Mohr’s Circle
Sample Problem
τ (CW)
σy = -2 ksi
y
A particular point
on the part
x-axis
σx = 6 ksi
(6 ksi,
k i 3 ksi)
k i)
2
6
3
τxy = 3 ksi
x
σ
σy = -2 ksi
Some Part
σx = 6 ksi
Mohr’s Circle
(-2 ksi, -3 ksi)
y-axis
τxy = 3
ksi
x & y Æ orientation
τ (CW)
(σavg, τmax)
σy = -2 ksi
Mohr’s Circle
(σavg, τmax)
(2 ksi, 5 ksi)
τ (CW)
σy = -2 ksi
x-face
σx = 6 ksi
Center of
Mohr’s Circle
3
x-face
σx = 6 ksi
(6 ksi, 3ksi)
(6 ksi, 3ksi)
R
τxy = 3 ksi
τxy = 3 ksi
σ
σ2
σavg = 2 ksi
σ
σ2
σ1
3 ksi
4 ksi
σ1
R = (3 ksi
k i) 2 + ( 4 ksi
k i) 2
(-2 ksi, -3ksi)
= 5 ksi
R = τmax
y-face
σ avg =
σx + σy
2
= 2 ksi
MET 210W – Mohr’s Circle
(σavg, τmin)
y-face
σ1 = σavg + R = 7 ksi
σ2 = σavg – R = −2 ksi
(σavg, τmin)
(2 ksi, -5 ksi)
3
Mohr’s Circle
(σavg, τmax)
(2 ksi, 5 ksi)
τ (CW)
σy = -2 ksi
Principle Stress
2θ
σ2
3 ksi
σ
σ1
4 ksi
⎛ 3 ksi
s ⎞
⎜⎜
⎟⎟
2θ = Tan
T
⎝ 4 ksi ⎠
2θ = 36.869°
y-face
θ = 18.435°
Shear Stress
(σavg, τmax)
(2 ksi, 5 ksi)
σavg = 2 ksi
φ = 26.565°
τmax = 5 ksi
2φ = 90 − 36.869°
2φ = 53.130°
σ
τmax = 5 ksi
σy = -2 ksi
σx = 6 ksi φ = 26.565°
σ1
θ = 18.435°
y-face
φ = 26.565°
MET 210W – Mohr’s Circle
(σavg, τmin)
(2 ksi, -5 ksi)
σ1
(σavg, τmin)
(2 ksi, -5 ksi)
τxy = 3 kksii
2φ = 90° − 2θ
σ
σavg = 2 ksi
3 ksi
4 ksi
3 ksi
Relationship Between Elements
(6 ksi, 3ksi)
2φ
2θ
4 ksi
x-face
2θ
σ2
σ2
Rotation on element is
half of the rotation from
the circle in same
direction from x-axis
(σavg, τmin)
(2 ksi, -5 ksi)
τ (CW)
(6 ksi, 3ksi)
θ = 18.435°
σ1 = 7 ksi
Principle Stress
Element
−1
σavg = 2 ksi
x-face
(6 ksi, 3ksi)
τxy = 3 ksi
Maximum Shear
Stress Element
(σavg, τmax)
(2 ksi, 5 ksi)
σ2 = -3 ksi
x-face
σx = 6 ksi
τ (CW)
θ + φ = 18.435 ° + 26.565 ° = 45 °
σavg = 2 ksi
σ2 = -3 ksi
σ1 = 7 ksi
4
What’s the stress at angle of 15°
CCW from the x-axis?
τ (CW)
Rotation on
Mohr’s Circle
y
(σU, τU)
A particular
i l point
i
on the part
f
x-face
30°
V
x
σ
σ2
σ = ? ksi
Some Part
σ = ? ksi
15°
y-face
15° on part and
element is 30° on
Mohr’s Circle
x
U & V Æ new axes @ 15° from x-axis
(σavg, τmax)
τ (CW)
σavg = 2 ksi
(σV, τV)
(σU, τU)
y
σU = 3.96 ksi
66.869°
σ2
σ
σavg = 2 ksi
(
)
σV = σavg – R*cos(66.869°)
τUV = 4.60 ksi
A particular point
on the part
R
σU = σavg + R*cos(66.869°)
τUV = R*sin(66.869°)
(σavg, τmin)
What’s the stress at angle of 15°
CCW from the x-axis?
x-face
f
σV = 0.036 ksi
σ1
U
τ = ? ksi
Rotation on
Mohr’s Circle
(σavg, τmax)
x
σ1
Some Part
y-face
(σV, τV)
MET 210W – Mohr’s Circle
(σavg, τmin)
V
σV = .036 ksi
σU = 3.96 ksi
U
15°
x
τ = 4.60 ksi
5
Special Case – Both Principle
Stresses Have the Same Sign
Y
Questions?
X
σx =
Z
σy =
Cylindrical
Pressure Vessel
Mohr’s Circle
σy
τ (CW)
τ=
σx
σy
This isn’t the whole
story however…
σx − σy
σx
2
σ
Mohr’s Circle for X-Y Planes
σx = σ1 and σy = σ2
MET 210W – Mohr’s Circle
σz = 0
τ (CW)
σz = 0 since it is
perpendicular to the free
face of the element.
σx
τmax = τxz
σx
σz
pD
4t
Mohr’s Circle
σy
pD
2t
σ3
σ1
σ
σz = σ3 and σx = σ1
τmax =
σ1 − σ 3
2
Mohr’s Circle for X-Z Planes
6
σz
Pure Uniaxial Tension
Mohr’s Circle
σy
τ (CW)
τmax = τxz
σx
σy = 0
σ2
σ3
σz = 0 since it is
perpendicular to the free
face of the element.
σ1
τmax =
σx = P/A
σ
σ1= σx
σ2 = 0
σ1 > σ2 > σ3
Note when σx = Sy,
Sys = Sy/2
Ductile Materials Tend to Fail
in SHEAR
Pure Uniaxial Compression
σx
2
Pure Torsion
T
σy = 0
τmax =
σx
2
σx = P/A
σ2= σx
τmax = τ xy
T
τ xy =
σ1 = 0
Tc
J
σ2 = -τxy
σ1
CHALK
MET 210W – Mohr’s Circle
σ1 = τxy
Brittle materials
tend to fail in
TENSION.
7
Uniaxial Tension & Torsional
Shear Stresses
• Rotating shaft with axial load.
load
• Basis for design of shafts.
Equivalent Torque
• Special case of body subjected to
BENDING and TORSION.
TORSION
• Point on top of bar is in shear and in tension.
τmax = R
(σx, τxy)
σx = P/A
σ2 = σx/2-R
τxy = Tc/J
σx/2
σx = M/S
τxy = T/Zp
σ1 = σx/2+R
T
(0, τyx)
Fixed
2
⎛σ ⎞
R = ⎜ x ⎟ + τ 2xy = τmax
⎝ 2 ⎠
F
Example
Equivalent Torque
2
R=
τmax =
⎛ M ⎞ ⎛⎜ T ⎞⎟
⎟ +
⎜
⎝ 2S ⎠ ⎜⎝ Z p ⎟⎠
τmax =
T
M2 + T 2
= e
Zp
Zp
2
Strain Gage
Torque on nut is
10000 in-lbs
τmax = R
( x, τxy)
(σ
⎛ σx ⎞
⎜ ⎟ + τ 2xy
⎝ 2 ⎠
2
σ2 = σx/2-R
σx/2
M = Bending Moment
S = Section Modulus
T = Torque
Zp = Polar Section Modulus
σ1 = σx/2+R
(0, τyx)
1”- 8 Bolt
We want to apply a strain gage to the
bolt to measure the maximum strain.
Determine the proper orientation and
predict the strain to be measured.
Te = M2 + T 2
For round bars, Zp = 2S, Te = Equivalent Torque
MET 210W – Mohr’s Circle
Strain gages only measure normal strain, NOT shear strain!
8
Axial Force in Bolt
The axial force in the bolt can be found using Eqn. 18-3:
T=K
K*D*P
D P
(Pg 719)
(Pg.
where T = torque in in-lbs
D = nominal outside diameter of threads, in.
P = Clamping load, lbs
p
g on the lubrication present
p
K = constant depending
K = .15 for average commercial applications with
any lubrication present at all.
K = .20 for well cleaned and dried threads.
Values of K are approximate & should be verified
MET 210W – Mohr’s Circle
9