EGR 236 Properties and Mechanics of Materials
Lecture 05: Poisson's Ratio and Modulus of Rigidity
Spring 2014
Today:
-- Homework questions:
-- New Topics:
-- Poisson's Ratio
-- Modulus of Rigidity
-- Shear Stress Testing
-- Homework: Read Section 3.6 to 3.8
Work Problems from Chap 3: 31, 35, 39
Following today's class you should be able to:
-- explain what Poisson's ratio, ν, represents
-- write the equation that relates E, G, and ν
-- know how to read the modulus of rigidity off of a shear stress-strain
diagram
-- be able calculate axial and lateral strain
-- to solve problems that involve E, G, and ν
Poisson's Ratio:
When a deformable is subjected to axial stress, the body's length changes.
At the same time there is also a small change in the body's shape along a
direction lateral to the axial force.
Essentially, as a body elongates, it also get narrower. If a force acts to
compress the body and shorten its length, the body gets fatter.
The ratio that gives the relative change deformation of the lateral
deformation to the axial deformation is called Poisson's ratio.
P
Deformed Length:
L '  L   long
δlong
δlateral
Deformed width:
r '  r   lateral
Lateral Strain:
 lateral 
L
L
 lateral
r
r
P
Axial (Longitudinal) Strain:
 longitudinal 
 long
L
Poisson's Ratio (ratio of lateral to longitudinal strain):
 
 lateral
 longitudinal
The negative sign indicates that as the length increases as the lateral
dimension decreases. Values for Poisson's ratio generally fall between 1/4
and 1/2. You will find Poisson's ratio listed for a variety of materials on
common Material Properties tables ( see inside back cover of your text.)
Example 1:
A rod of steel is 10 in long and has a 0.50 in diameter. A 20 kip load pulls
on each end of the bar. Given a modulus of elasticity of 30 x 106 psi and
Poisson's ratio of 0.30
a) what is the elongation of the bar, δx?
b) what is the strain along the bar, εx?
c) what is the strain along the lateral direction, εdia?
d) what is change of diameter experienced by the bar, δdia?
-----------------------------------------------------------------------------------------------
20 kip
L = 10 in
D = 0.50 in
20 kip
Example 1:
A rod of steel is 10 in long and has a 0.50 in diameter. A 20 kip load pulls
on each end of the bar. Given a modulus of elasticity of 30 x 106 psi and
Poisson's ratio of 0.30
a) what is the elongation of the bar, δx?
b) what is the strain along the bar, εx?
c) what is the strain along the lateral direction, εdia?
d) what is change of diameter experienced by the bar, δdia?
----------------------------------------------------------------------------------------------Given:
L = 10 in
L=10 in
D = 0.50 in
D= 0.50 in
20 kip
20 kip
P= 20 kip
E = 30 x 106 psi
ν = 0.30
Area of the specimen:
 D2  (0.5 in)2
A

 0.1964 in2
4
4
Axial Stress in bar:
P
20 kip
 

 101.8 ksi
A 0.1964 in2
From Hooke's Law find Longitudinal or Axial Strain
  E x

x 


E
101.8 ksi 1000 psi
 0.00339 in / in
30(10)6 psi 1 ksi
Axial Elongation:
 x  L  x  10 in  0.00339 in / in  0.0339 in
Lateral Strain (from Poisson's ratio)
 
 lat
x

 lat   x
 0.30  0.0039 in / in  0.00102 in / in
Lateral Deformation:
 dia   lat  D  lat  0.50 in  0.00102 in / in  0.000509 in
The previous example was for a simple axially loaded system.
Sometimes the stress state in a body will be compounded by stresses acting
along several axes or directions at the same time. If you were to show an
infinitesimally small stress element you might find both shear and normal
stresses acting along each of the six sides. This element showing these stresses
is known as the generalized stress element.
y
σy
τyx
τyz
σx
τxy
τzy
τzx
σx
x
τxz
σz
σy
z
For the example presented, you can model axial stress as a stress element with
only a normal stress acting along one of the axis directions.
y
P
P
The directional strain would be
x 
x
σx
E
 x
 y   x 
E
 x
 z   x 
E
σx
x
z
The more general stress model shows normal stresses on each of the six faces
of the element. This type of load is called a multiaxial loading. To determine
the longitudinal and lateral strains along each direction you need to include the
effect of Poisson's ratio for each normal stress.
This can be found by superposition of the effect of each individual normal
y
stress.
x 
x

 y

 z
E
E
E



y  y  x  z
E
E
E



z  z  x  y
E
E
E
σy
σz
σx
σx
x
σz
σy
z
This set of equations for the elemental strain is the form of Hooke's law for
multiaxial loading.
80 MPa
80 MPa
Example 2:
A 30 x 30 mm x 4 mm aluminum plate is subject to an 80 MPa and 40 MPa
stress as shown on the figure below. Given an Elastic modulus of 70 GPa and
Poisson's ratio of 0.33 determine:
y
a) the strain along the x direction
b) the strain along the y direction
c) the strain along the z direction
40 MPa
d) deformation along line AB
B
e) deformation along line BC
A
f) deformation along line AC.
C
z
40 MPa
x
Example 2: continued
80 MPa
80 MPa
Example 2:
A 30 x 30 mm x 4 mm aluminum plate is subject to an 80 MPa and 40 MPa
stress as shown on the figure below. Given an Elastic modulus of 70 GPa and
Poisson's ratio of 0.33 determine:
y
a) the strain along the x direction
b) the strain along the y direction
40 MPa
c) the strain along the z direction
d) deformation along line AB
B
A
e) deformation along line BC
f) deformation along line AC.
Solution: E = 70 GPa
ν = 0.33
 x  80 MPa
x
C
 y  40 MPa
 z  0 MPa
z
40 MPa
Using the generalized Hooke's Law for multiaxial stress



80
0.33(40)
x  x  y  z 

 0.000954 mm / mm
E
E
E
70000
70000



40
0.33(80)
y  y  x  z 

 0.0001943 mm / mm
E
E
E
70000
70000



0.33(80) 0.33(40)
z  z  x  y  

 0.000566 mm / mm
E
E
E
70000
70000
Deformation: AB
 AB   x  LAB x  (30 mm)(0.000954 mm / mm)  0.0286 mm
Deformation: BC
 BC   y  LBC  y  (30 mm)(0.0001943 mm / mm)  0.00583mm
Deformation: AC
42.426 mm
deformed
30 mm
undeformed
42.4508
mm
30.00583 mm
30.0286 mm
30 mm
 AC  LAC ' LAC
 42.451  42.426
Hooke's Law for Shear:
You were shown that during the linear portion of the tensile test, stress and
strain were related by the modulus of elasticity, E, which was found as the
slope of the stress-strain plot in the proportional range. This relationship gave
the form of Hooke's Law for the axial stress case as
  E
(for axial stress)
Another common material test is to load a circular rod into a machine which
can apply a torque to the specimen while measuring both the torque magnitude
and the angular twist that the rod undergoes. This is called a torsion test.
From this test a plot of the shear stress vs. the shear strain can be plotted.
A typical plot of a shear stress-strain test is shown below.
Like the tensile test, this test may be used to find a modulus that relates the
shear stress to the shear strain within the elastic region.
This modulus is called the shear modulus or the modulus of rigidity, G. This
relationship is given by
  G
(for shear stress)
This modulus of rigidity has the same units as the elastic modulus such as psi
or MPa.
A proof will not be shown here, but there is a fundamental relationship
between the elastic modulus, the shear modulus and the Poisson's ration given
by:
G
E
2(1   )
This means that if you know or can find both the modulus of elasticity and
rigidity of a material, you can determine the value of Poisson's ratio.
Example 3:
The elastic portion of the tension stress-strain diagram for an aluminum alloy is shown in the figure
below. The specimen used for the test has a gauge length of 2 in. and a diameter of 0.5 in. When a
load of 9 kip is applied, the new diameter of the specimen is 0.49935 in. Compute the shear modulus
G for the aluminum.
From the Stress Strain curve, the modulus of elasticity may be read:
For an axial loading, the lateral strain may be found as
For a 9 kip load the stress in the material is
The axial (or longitudinal) strain is found as
From the two strains, Poisson's ratio may be found.
therefore, this allows us to find modulus of rigidity, G as
Example 3:
The elastic portion of the tension stress-strain diagram for an aluminum alloy is shown in the figure
below. The specimen used for the test has a gauge length of 2 in. and a diameter of 0.5 in. When a
load of 9 kip is applied, the new diameter of the specimen is 0.49935 in. Compute the shear modulus
G for the aluminum.
From the Stress Strain curve, the modulus of elasticity may be read:
E  slope 
70
 11400 ksi
0.00614
For an axial loading, the lateral strain may be found as

D
 Do 0.49935  0.50000
 dia  dia  new

 0.0013 in / in
D0
Do
0.50000
For a 9 kip load, the stress in the material is
 
F
F
9 kip


 458.4 ksi
1
A 1
2
2
 D0
 (0.5 in)
4
4
The axial (or longitudinal) strain is found as
 45.84 ksi
 axial  
 0.00402 in / in
E
11400ksi
From the two strains, Poisson's ratio may be found.

(0.0013)
  dia 
 0.323
 axial
0.00402
therefore, this allows us to find modulus of rigidity, G as
E
11400 ksi
G

 4308 ksi
2(1   ) 2(1  0.323)
EGR 236 Mechanics of Materials
HW Set 05
Spring 2014
Problem 3:31
The plastic rod is made of Kevlar 49 and has a diameter of 10 mm. If an axial load of 80 kN is
applied to it, determine the change in length and the change in diameter.
____________________________________________________________________________
Solution:
EGR 236 Mechanics of Materials
Problem 3:35
HW Set 05
Spring 2014
The block is made of titanium Ti-6Al-4V and is subjected to a compression of 0.06 in along the y
axis, and its shape is given a tilt of 89.7 degrees. Determine εx εy γxy
__________________________________________________________________________
Solution:
EGR 236 Mechanics of Materials
HW Set 05
Spring 2014
Problem 3.39
The elastic portion of the stress strain diagram for a steel alloy is shown in the figure. The specimen
from which it was obtained had an original diameter of 13 mm and a gauge length of 50 mm. If a
load of P = 20 kN is applied to the specimen determine its diameter and gauge length. Take ν = 0.40.
-----------------------------------------------------------------------------------------------------------------Solution:
400
0.002