ASEN 3112 - Structures
Torsion Of
Circular
Sections
ASEN 3112 Lecture 7 – Slide 1
ASEN 3112 - Structures
Notation and Terminology
We consider a bar-like, straight, prismatic structural member of constant
cross section subjected to applied torque T:
y
y
T(+)
z
(a)
T(+)
(b)
z
x
T(+)
x
T(+)
This kind of member, designed primarily to transmit torque, is called a shaft.
Axis x is directed along the longitudinal shaft dimension, as shown. The applied
torque is a moment about that axis. It is positive when it acts as shown,
complying with the right-hand screw rule. Axes y and z lie on a cross
section conventionally taken as origin.
The right figure shows another symbol (double-headed arrow) for torque.
Both symbols (harpoon and double arrow) are used in this course.
ASEN 3112 Lecture 7 – Slide 2
ASEN 3112 - Structures
Shafts Are Used for Power Transmission
An important use of shafts is to transmit power between parallel planes, as in cars,
aircraft engines or helicopters. See figures above (taken from Google Images)
ASEN 3112 Lecture 7 – Slide 3
ASEN 3112 - Structures
Circular Cross Sections
2R
2Ri 2Re
In this lecture we restrict consideration to shafts of circular cross section,
which may be either solid or hollow (also called annular & tubular)
ASEN 3112 Lecture 7 – Slide 4
ASEN 3112 - Structures
Cylindrical Coordinate System
y
P(x,ρ,θ)
z
ρ
θ
x
Point P referred to cylindrical
coordinate system (looking at
cross section from +x down)
In addition to the {x,y,z} Rectangular Cartesian Coordinate (RCC) system
shown in the previous slide we shall also use frequently a cylindrical coordinate
system {x,ρ,θ} as depicted above for a typical cross section
The position coordinates of an arbitrary shaft point P
are {x,y,z} in RCC
and {x,ρ,θ} in cylindrical coordinates. Here ρ is the radial coordinate
(an unfortunate notational choice because of potential confusion with density,
but it is that used in most textbooks) and θ the angular coordinate,
taken positive CCW from y as shown. Note that y = ρ cos θ and z = ρ sin θ.
ASEN 3112 Lecture 7 – Slide 5
ASEN 3112 - Structures
Twisting A Shaft
To visualize what happens when a circular shaft is torqued, consider
first the undeformed state under zero torque. A rectangular-like mesh
is marked on the surface
The rectangles have sides parallel to the longitudinal axis x
and the circumferential direction θ. Then apply torques to the end.
The rectangles shear into parallelograms
This figure displays the main effect: shear stresses and associated shear strains
develop in the "cylinders" ρ = constant. Next slide goes into details
ASEN 3112 Lecture 7 – Slide 6
ASEN 3112 - Structures
Stress & Strain Components
y
z
T
Material element aligned with
cylindrical coordinate axes
θ
ρ
All other stress
components
are zero
radial
direction ρ
x
T
τxθ
circumferential
direction θ
τθx
longitudinal
direction x
we will often call
τ θx = τ xθ = τ
Cut out a material element aligned with the cylindrical coordinate axes as
shown above. The only nonzero stress component is the shear stress τ xθ = τ θx = τ.
ASEN 3112 Lecture 7 – Slide 7
ASEN 3112 - Structures
Stress and Strain Distribution
Consequently the stress and strain matrices have the form
0 τ xθ 0
0
γ
⇒ Hooke’s law ⇒
τθx 0 0
θx
0
0 0
0
γxθ
0
0
0
0
0
Stresses and strains are connected, point by point, by Hooke's law:
τ = Gγ or γ = τ/G
Because of circular symmetry τ and γ do not depend on θ. They may depend on x
if either the torque or the cross section redius change along the shaft length, but the
dependence will not be made explicit. Both τ and γ depend linearly on ρ, and we
scale that dependence as follows:
ρ
ρ
γ = γmax
τ = τmax ,
R
R
Here t max and gmax are the maximum values, which occur at ρ = R, the radius of the
shaft. (For an annular shaft, R = R e , the exterior or outer radius.)
ASEN 3112 Lecture 7 – Slide 8
ASEN 3112 - Structures
Stress Formula (1)
T
dA = ρ dρ dθ
θ
ρ
x
θ
T
τθx
τxθ
τ=τxθ ρ
90
x
τ=τxθ
dρ
ρ
θ
dθ
x
As shown above, cut out a material element aligned with the cylindrical coordinate
axes. The only nonzero component on the faces of the material element is the shear
stress τxθ = τ θx. It follows that the stress and strain matrices have the form
0 τxθ 0
0 γxθ 0
γθx
⇐ Hooke’s law ⇒
τ θx 0 0
0 0
0
0 0
0
0 0
in which the axes are reordered as (x,q,r) for convenience in identifying with plane
stress and strain later. In the sequel we often call τ = τxθ = τ θx and γ = γ xθ = γ θx
to reduce subscript clutter. Stresses and strains are connected, point by point, by
Hooke's law
τ
τ = G γ,
γ =
G
ASEN 3112 Lecture 7 – Slide 9
Stress Formula (2)
ASEN 3112 - Structures
Call T the internal torque acting on a cross section x = constant. The shear
stress τ = τ xθ acting on an elementary cross section area dA produces
an elementary force dF = τ dA and an elementary moment dM = dF ρ =
(τ dA) ρ about the x axis. Integration of the elementary moment over the
cross section
must balance the internal torque:
T =
(shear stress × elemental area ) × lever-arm
τmax
ρ
τmax d A =
ρ2 d A
= (τ d A) ρ =
R A
A
A R
τmax J
τmax
τmax
,
=
ρ 2 (ρ dρ dθ ) =
ρ 3 dρ d θ =
R
R A
R A
in which J =
ρ 3 dρ d θ is the polar moment of inertia of the cross section.
A
A
Solving for the maximum shear stress gives the stress formula
τmax =
TR
,
J
For an solid circular cross section
For an annular cross section
τ=
ρ
Tρ
τmax =
R
J
J=
J=
π
2
π
2
R4
(Re4 − Ri4 )
ASEN 3112 Lecture 7 – Slide 10
Twist Angle and Twist Rate (1)
T
ASEN 3112 - Structures
x
A
φ at x
B
γmax =(γxθ)max =(γθx)max
L
B'
φBA
T
BB' ≈ R φ ≈ γmax x or φ = (x/R) γ max . The derivative dφ/dx is
called the twist rate.
γmax
γ
dφ
dφ
=
=
so γ = ρ
dx
R
ρ
dx
But
Tρ
dφ
τ
=
=ρ
γ =
G
GJ
dx
T
dφ
=
dx
GJ
This is the twist rate formula.
ASEN 3112 Lecture 7 – Slide 11
ASEN 3112 - Structures
Twist Angle and Twist Rate (2)
To find the twist angle φ from end A to end B, integrate along
ΒΑ
the length of the shaft,
L
φB A =
L
dφ =
0
0
dφ
dx =
dx
0
L
T
dx
GJ
If T, G and J are constant along the shaft
φB A
T
=
GJ
0
L
dx =
TL
GJ
This is the twist angle between ends.
ASEN 3112 Lecture 7 – Slide 12
ASEN 3112 - Structures
Example
A
Material: steel, G = 12 x 10 6 psi
Solid circular x section, R = 1 in
L = 50 in
B
T = 5000 lbs-in
Find: maximum shear stress τ max , maximum shear strain γ max and twist angle φ BA
τmax =
γmax =
φ
BA
=
5000 lbs-in x 1 in
TR
TR
=
=
= 3183 psi
4 4
1
4
J
1.57
x
1
in
π
R
2
τmax
G
=
3183 psi
= 265 µ
12 x 10 6 psi
5000 lbs-in x 50 in
TL
◦
=
= 0.0133 rad = 0.76
6
4
GJ
12 x 10 psi x 1.57 1 in
ASEN 3112 Lecture 7 – Slide 13
ASEN 3112 - Structures
Torqued Shaft Failure Modes
Ductile material (mild steel). Brittle material (cast iron).
Failure by shear stress.
Failure by tensile normal
Solid cross section
stress. Solid cross section
ASEN 3112 Lecture 7 – Slide 14
Wall buckling failure
Tubular cross section
ASEN 3112 - Structures
Material Failure: Ductile vs. Brittle
Mohr's circle for (x,θ) plane
at ρ=R (outer shaft surface)
radius
of circle
is τmax
τ
Material failure surfaces for two
extreme cases: ductile vs. brittle
τ max
σ2 = −τmax
45
σ1 = τmax
T
ο
T
σ
−τ max
Ductile material failure
surface (plane of cross section)
ASEN 3112 Lecture 7 – Slide 15
Brittle material failure
surface (=helicoid)
ASEN 3112 - Structures
Photos of Specimen That Failed in Torsion
(a) Ductile failure by shear
(mild steel)
(b) Brittle failure by tensile
normal stress (cast iron)
ASEN 3112 Lecture 7 – Slide 16
(c) Brittle failure by tensile
normal stress (chalk)