The Plot So Far
Up to this point, we have looked at
elements that have been loaded axially only
This has produced an axial/normal strain
and stress
We looked at shear stress as produced on
an axis that was inclined at an angle to the
axis
Pure Shear and Torsion
“Every exit is an entry somewhere else.”
Tom Stoppard
23 September 2002
Meeting Twelve - Pure Stress and Torsion
Shear Stress Only
Shear Stress Only
Now we are going to begin looking at other
conditions and see how the types of
stresses and strains that they develop
If we start by looking at some element that
has a shear stress applied at two parallel
faces
23 September 2002
Meeting Twelve - Pure Stress and Torsion
3
23 September 2002
Meeting Twelve - Pure Stress and Torsion
2
4
1
Shear Stress Only
Shear Stress Only
Remember, we are looking at the face of a
cube here
The force on each of the faces would be
the shear stress, τ, times the area of the
face, which we will call A
23 September 2002
Meeting Twelve - Pure Stress and Torsion
5
23 September 2002
Meeting Twelve - Pure Stress and Torsion
Shear Stress Only
Shear Stress Only
If we were to draw the system with the
forces shown, rather than the stresses, we
would have
This would be two forces equal in
magnitude and opposite in direction
This is a couple
τA
23 September 2002
τA
Meeting Twelve - Pure Stress and Torsion
τA
7
23 September 2002
6
τA
Meeting Twelve - Pure Stress and Torsion
8
2
Shear Stress Only
Shear Stress Only
Since the width of the cube in not 0, this
couple produces a moment
If the moment is not balanced, the element
is not in equilibrium
To offset this couple moment, we will need
to have something else producing a
moment of equal and opposite magnitude
τA
23 September 2002
τA
τA
τA
Meeting Twelve - Pure Stress and Torsion
9
23 September 2002
Meeting Twelve - Pure Stress and Torsion
Shear Stress Only
Shear Stress Only
A set of forces acting on the top and
bottom of the cube with the same
magnitude as the forces acting on the right
and left faces will produce this moment
So for equilibrium we would have
τA
23 September 2002
Meeting Twelve - Pure Stress and Torsion
τA
τA
τA
τA
11
10
23 September 2002
τA
Meeting Twelve - Pure Stress and Torsion
12
3
Shear Stress Only
Shear Stress Only
Notice the orientation of the forces on each
plane
The two couples must offset one another
τA
If we convert these forces back to shear
stress representations we have
τA
τA
23 September 2002
Meeting Twelve - Pure Stress and Torsion
τA
13
23 September 2002
Meeting Twelve - Pure Stress and Torsion
14
Shear Stress Only
Shear Stress Only
Again, notice the direction of the shear
stresses and how they generate the two
offsetting couples
Now we started by assuming a cube, but
this shear stress diagram will also hold true
if we take any rectangular shape
23 September 2002
Meeting Twelve - Pure Stress and Torsion
15
23 September 2002
Meeting Twelve - Pure Stress and Torsion
16
4
Shear Strain
Shear Stress Only
If we subject the element to shear stress, we
will develop a shear strain in the element
Remember that the strain here is also
dimensionless because it is the change in the
angle (in radians) of that corner angle
If you want to prove this to yourself, look
at the couples (remember to change both
the moment arm and the forces)
23 September 2002
Meeting Twelve - Pure Stress and Torsion
17
Shear Strain
Meeting Twelve - Pure Stress and Torsion
Meeting Twelve - Pure Stress and Torsion
18
Shear Strain
In a manner similar to the stress/strain
relationship for axial stress we have a
relationship in a limited region between shear
stress and shear strain
23 September 2002
23 September 2002
19
G is the shear modulus or modulus of rigidity
τ is the shear stress
γ is the shear strain
τ = Gγ
23 September 2002
Meeting Twelve - Pure Stress and Torsion
20
5
Shear Stress on an Inclined Plane
Shear Stress on an Inclined Plane
If we return to our drawing of the cube with
shear stress on the four faces
We now pass a plane through the cube at an
angle of θ
τ
τ
τ
θ
τ
23 September 2002
Meeting Twelve - Pure Stress and Torsion
21
23 September 2002
Meeting Twelve - Pure Stress and Torsion
22
Shear Stress on an Inclined Plane
Shear Stress on an Inclined Plane
First we can generate the areas for each of the
faces in terms of the inclined face
Remember the depth of the cube into the page
is constant
The vertical face will have an area equal to
Aθcosθ
The horizontal face will have an area equal to
Aθsinθ
τ
23 September 2002
θ
τ
Aθcosθ
Aθ
τ
Meeting Twelve - Pure Stress and Torsion
23
23 September 2002
θ
Aθ
τ
Aθsinθ
Meeting Twelve - Pure Stress and Torsion
24
6
Shear Stress on an Inclined Plane
Shear Stress on an Inclined Plane
The forces on the planes can then be
developed as
Now if we add the shear and normal forces on
the inclined plane in terms of the shear and
normal stresses on that plane we have a FBD
τAθcosθ
θ
τθAθ
Aθ
τAθcosθ
σθAθ
θ
Aθ
τAθsinθ
23 September 2002
τAθsinθ
Meeting Twelve - Pure Stress and Torsion
25
23 September 2002
Meeting Twelve - Pure Stress and Torsion
26
Shear Stress on an Inclined Plane
Shear Stress on an Inclined Plane
If we now set our axis along the inclined plane
we have
Summing forces in the x-direction (normal to
the inclined plane) we have
0 = σ θ Aθ − τ Aθ cosθ sin θ − τ Aθ sin θ cosθ
τθAθ
τAθcosθ
θ
τθAθ
σθAθ
τAθcosθ
Aθ
Meeting Twelve - Pure Stress and Torsion
σθAθ
Aθ
τAθsinθ
23 September 2002
θ
27
23 September 2002
τAθsinθ
Meeting Twelve - Pure Stress and Torsion
28
7
Shear Stress on an Inclined Plane
Shear Stress on an Inclined Plane
Solving for σθ we have
Using the same logic, we sum forces along the
inclined plane and we have
σ θ = 2τ cos θ sin θ
0 = τ θ Aθ − τ Aθ cosθ cos θ + τ Aθ sin θ sin θ
τθAθ
τθAθ
σθAθ
θ
τAθcosθ
τAθcosθ
σθAθ
θ
Aθ
Aθ
τAθsinθ
23 September 2002
Meeting Twelve - Pure Stress and Torsion
29
τAθsinθ
23 September 2002
σ θ = 2τ cos θ sin θ
Meeting Twelve - Pure Stress and Torsion
30
Shear Stress on an Inclined Plane
Shear Stress on an Inclined Plane
And solving for the shear stress on the inclined
plane we have
So for any angle θ we can solve for the shear
and normal stress if we know the stress on the
normal face
(
τ θ = τ ( sin θ ) − ( cos θ )
2
2
)
τθAθ
τAθcosθ
θ
σθAθ
Aθ
23 September 2002
τAθsinθ
(
τ θ = τ ( sin θ ) − ( cos θ )
τθAθ
τAθcosθ
θ
σθAθ
2
2
)
σ θ = 2τ cosθ sin θ
Aθ
τAθsinθ
σ θ = 2τ cos θ sin θ
Meeting Twelve - Pure Stress and Torsion
31
23 September 2002
Meeting Twelve - Pure Stress and Torsion
32
8
Shear Stress on an Inclined Plane
Shear Stress on an Inclined Plane
If we plot the shear and normal stresses on the
inclined plane with respect to the angle of
inclination and the shear stress on the normal
face we have
This shows that the normal stress on the
inclined plane are at a maximum when the
plane is inclined at 45 degrees
σ θ = 2τ cosθ sin θ
(
τ θ = τ ( sin θ ) − ( cos θ )
23 September 2002
2
2
σ θ = 2τ cosθ sin θ
)
Meeting Twelve - Pure Stress and Torsion
(
τ θ = τ ( sin θ ) − ( cos θ )
33
23 September 2002
2
2
)
Meeting Twelve - Pure Stress and Torsion
34
Shear Stress on an Inclined Plane
Shear Stress on an Inclined Plane
It also shows that the normal stress on the
inclined face is equal to the original shear
stress on the normal face when the angle is 45
degrees
We can also plot the relationship between the
shear stress on the inclined face and the
original shear stress on the normal face
(
τ θ = τ ( sin θ ) − ( cos θ )
σ θ = 2τ cosθ sin θ
(
τ θ = τ ( sin θ ) − ( cos θ )
23 September 2002
2
2
2
)
σ θ = 2τ cosθ sin θ
)
Meeting Twelve - Pure Stress and Torsion
2
35
23 September 2002
Meeting Twelve - Pure Stress and Torsion
36
9
Shear Stress on an Inclined Plane
Shear Stress on an Inclined Plane
The shear stress on the inclined face is at a
maximum when the angle of inclination is
either 0 degrees (normal) or some integer
multiple of 90 degrees
We developed these relationships in the same
manner that we developed those for an
element in pure axial stress
The basis is still equilibrium and the methods
those developed in Statics
(
τ θ = τ ( sin θ ) − ( cos θ )
2
2
)
σ θ = 2τ cosθ sin θ
σ θ = 2τ cos θ sin θ
2
2
τ θ = τ ( sin θ ) − ( cos θ )
(
23 September 2002
Meeting Twelve - Pure Stress and Torsion
37
23 September 2002
)
Meeting Twelve - Pure Stress and Torsion
38
Torsion
Torsion
We are going to limit our analysis of torsional
effects in this class to bars with circular cross
sections
Other types of cross-sections can be analyzed
but it is beyond the scope of this class
We will start with a bar with a fixed end
support subjected to some moment (torsion) T
The bar has a radius of R and a length of L
23 September 2002
Meeting Twelve - Pure Stress and Torsion
39
23 September 2002
Meeting Twelve - Pure Stress and Torsion
40
10
Torsion
Torsion
We will assume that the bar cannot deform in
any way at the support.
Remember, however, that there is a resistive
moment provided by the support
It is still in equilibrium
Now if we look at the unsupported end of the
beam end on
And look at a point on the edge of the circle
23 September 2002
Meeting Twelve - Pure Stress and Torsion
41
r
23 September 2002
Meeting Twelve - Pure Stress and Torsion
Torsion
Torsion
Now we apply the moment/torsion to the beam
and the point moves to some new location
The angle through which the point moves is
known as the angle of twist, φ
r
T
T
φ
T
T
r
42
23 September 2002
Meeting Twelve - Pure Stress and Torsion
43
23 September 2002
Meeting Twelve - Pure Stress and Torsion
44
11
Torsion
Torsion
The circumferential distance, c, is the distance
that the point moves along the edge of the
circle
It is equal to the arc length of rφ
If we looked along the length of the cylinder,
we would have a approximate relationship
c = rφ
T
L
φ
c
T
r
c = rφ
23 September 2002
Meeting Twelve - Pure Stress and Torsion
45
23 September 2002
Meeting Twelve - Pure Stress and Torsion
46
Torsion
Torsion
The tan of γ is equal to c/L
For very small angles, the tangent of the angle
is approximately equal to the angle in radians
So for small values of γ, γ is equal to c/L
Error of Approximation on θ to tan(θ)
0.05
0.045
0.04
0.035
L
tan( )-
0.03
γ
0.025
0.02
0.015
c
0.01
0.005
23 September 2002
c = rφ
Meeting Twelve - Pure Stress and Torsion
0
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
θ
47
23 September 2002
Meeting Twelve - Pure Stress and Torsion
48
12
Torsion
Torsion
Making this assumption, we can use our
previous value for angle of twist and develop
If we assume that we are still in the linear
relationship region, the relationship between
shear stress and shear strain is given by
c = rφ
c rφ
γ= =
L L
L
τ = Gγ
L
γ
γ=
γ
c
23 September 2002
c = rφ
c rφ
=
L L
c
Meeting Twelve - Pure Stress and Torsion
49
Torsion
23 September 2002
Meeting Twelve - Pure Stress and Torsion
50
Meeting Twelve - Pure Stress and Torsion
52
Homework
So the shear stress can be related to the angle
of twist and the dimensions of the beam by
τ = Gγ =
L
Grφ
L
c = rφ
γ=
γ
4-1.2
4-1.4
4-1.7
c rφ
=
L L
c
23 September 2002
Meeting Twelve - Pure Stress and Torsion
51
23 September 2002
13