Torsion

Torsion
Torsion is the twisting of a straight bar when it is loaded
by twisting moments or torques that tend to produce
rotation about the longitudinal axes of the bar.
For instance, when we turn a screw driver to produce
torsion our hand applies torque ‘T’ to the handle and
twists the shank of the screw driver.
Sign convention – Right Hand Rule
Shafts are structural members with length
significantly greater than the largest
cross-sectional dimension used in
transmitting torque from one plane to
another.
Turbine: Exerts a Torque on the Shaft.
Shaft: Transmit the Torque to the Generator
Generator: Creates an equal and opposite
Torque.
Torsion: twisting of a straight bar when
loaded by moments (or torques)
Moments or torques tend to produce
rotation about the longitudinal axis of the
bar (twisting).
Couple: the pair of forces that tends to
twist the bar about its longitudinal axis.
Moment of a couple:
T (Nm)= Force (N) x Arm (m) or
T (lb in.)= Force (lb) x Arm (in.)
Representation of the moment of a couple:
a vector with a double-headed arrow
Then we can write the moments as
T1 = P1 d1
T2 = P2 d 2
Net Torque Due to Internal Stresses
Net of the internal shearing stresses is an internal torque,
equal and opposite to the applied torque,
Although the net torque due to the shearing
stresses is known, the distribution of the stresses is not.
( )
T =∫ρ dF =∫ ρ τ dA
Distribution of shearing stresses is statically indeterminate –
must consider shaft deformations.
Unlike the normal stress due to axial loads, the distribution
of shearing stresses due to torsional loads can not be
assumed to be uniform.
Axial Shear
Components
Torque applied to shaft produces shearing stresses on the faces
perpendicular to the axis.
Conditions of equilibrium require the existence of equal
stresses on the faces of the two planes containing the axis of
the shaft.
The existence of the axial shear components is demonstrated
by considering a shaft made up of axial slats.
The slats slide with respect to each other when equal and
opposite torques are applied to the ends of the shaft.
From observation, the
angle of twist of the
shaft is proportional to
the applied torque and
to the shaft length.
φ ∝ T
φ ∝ L
Torsional Deformations of Cylindrical Bars
When subjected to torsion, every cross-section of a
circular shaft remains plane and undistorted then the
bar is said to be under pure torsion.
Cross-sections for hollow and solid circular shafts
remain plain and undistorted because a circular shaft
is axisymmetric.
Cross-sections of noncircular (non-axisymmetric)
shafts are distorted when subjected to torsion.
Consider a cylindrical bar of circular cross
section twisted by the torques T at both the
ends.
Since every cross section of the bar is
symmetrical, we say that the bar is in pure
torsion.
Under action of torque T the right end of the
bar will rotate through small angle φ known
as the angle of twist.
The angle of twist varies along the axis of
the bar at intermediate cross section denoted
by
φ ( x)
Torsional Deformations of a Circular Bar
The rate of twist or angle of twist per unit
length
dφ
θ=
dx
Shear Strain at the outer surface of the bar
bb | rdφ
γ max =
=
= rθ
ab dx
For pure torsion the rate of twist is
constant and equal to the total angle of
twist φ divided by the length L of the bar
rφ
γ max = rθ =
L
For Linear Elastic Materials
From Hooke’s Law
τ = Gγ
G is shear modulus of elasticity and γ is
shear strain
From Shear Strain equation :
γ max = rθ =
rφ
L
Shear Stress at the outer surface of the
bar :
τ max = G r θ
Shear Stresses in cylindrical bar with circular cross
section.
Torsion Formula : To determine the
relationship between shear stresses and
torque, torsional formula is to be
accomplished.
Longitudinal and transverse shear stresses in a circular bar subjected to torsion.
Torsional Formula
Since the stresses act continously they have
a resultant in the form of moment.
The Moment of a small element dA located
at radial distance ρ and is given by
The resultant moment ( torque T ) is the
summation over the entire cross sectional Distribution of stresses acting on a cross section.
area of all such elemental moments.
T = ∫ dM =
A
τ max
r
∫ρ
2
dA =
A
τ max
r
Ip
δM = ρτδ A = ρ 2
τ Max
r
I Polar = ∫ ρ 2δA
Polar moment of inertia of circle with
radius r and diameter d
A
Maximum Shear Stress
Ip =
π r4
2
=
πd4
32
τ max
Tr 16T
=
=
Ip πd3
δA
τ=
ρ
r
τ Max
Tρ
=
I Polar
Generalized torsion formula
Tρ
τ=
I Polar
Generalized
torsion formula
θ=
φ
L
Tρ
TL
ρφ
γ = ⇒γ =
=
⇒φ =
G
GI P
L
GI P
Angle of twist
per unit length
τ
Total angle of twist
Equations derived above
are only for bars of
circular cross sections
Circular tube in torsion.
A solid steel bar of circular cross section has diameter d=1.5in and a length of
l=54in.The bar is subjected to torques T acting at the ends if the torques have
magnitude T=250 lb-ft . G=11.5x106psi
a) what is the maximum shear stress in the bar
b) what is the angle of twist between the ends?
a) From torsional formula
τ max
16T 16* 250*12
=
=
= 4530 psi
3
3
πd
π *1.5
b) Angle of twist
φ=
Ip =
πd4
32
=
π *1.54
32
= .4970in 4
(250lb − ft × 12in )(54in ) = 0.02834rad
TL
=
GI P 11.5 × 106 psi 0.4970in 4
(
)(
)
Compare the solid and hollow shafts
d hollow 67.1mm
Size =
=
= 1.14
d solid
58.8mm
Ahollow 1273mm 2
Weight per unit length =
=
= 0.468
2
Asolid
2715mm
Hollow shafts vs Solid shafts
strength
τ Max
Angle of twist per
unit length
Both are proportional to
Weight is proportional to the cross-sectional area (A)
TR
=
IP
T
θ=
GI P
1
IP
Non-Uniform Torsion
Bars consisting of prismatic segments with
constant torque throughout each segment.
Non-Uniform Torsion
Bars with continuously varying cross sections
and constant torque.
Bars with continuously varying cross sections
and continuously varying torque.
A shaft/gear assembly
Shaft is driven by a gear at C.
Gears at B and D are driven by the
shaft. It turns freely at A and E
d = 30mm
LBC = 500mm
T2 = 450 N .m L = 400mm
CD
T1 = 275 N .m G = 80GPa
T3 = 175 N .m
Determine the maximum shear stress and the angle of
twist between gears B and D.
TCD = −275 N ⋅ m + 450 N ⋅ m = 175 N ⋅ m
τ CD =
( 2 ) = 175N .m(0.030m 2 ) = 33.0MPa
TCD d
Ip
π (0.030m )4
32
φCD =
TCD × LCD
175 N .m(0.4m )
=
= 0.0110rad
9
−8 4
G × IP
80 × 10 Pa 7.95 × 10 m
(
)(
)
TBC = −275 N ⋅ m
τ BC =
φBC
( 2 ) = − 275N .m(0.030m 2 ) = −51.9MPa
TBC d
7.95 × 10−8 m 4
Ip
TBC × LBC
− 275 N .m(0.5m )
=
=
= −0.0216rad
9
−8 4
G × IP
80 × 10 Pa 7.95 × 10 m
(
)(
)
φBD = φBC + φCD = −0.0216 + 0.011 = −0.0106rad = −0.61o
A tapered bar in torsion
Change of diameter with length:
Evaluate the integral of this type of equation
where
a = d A ..........b =
dB − d A
L
δx
∫ (a + bx )
4
⎛ 1
32TL
1 ⎞
⎜⎜ 3 − 3 ⎟⎟
φ=
3πG (d B − d A ) ⎝ d A d B ⎠
1
=−
3
3b(a + bx )
Stresses and Strains in Pure Shear
Analysis of stresses on inclined planes: (a) element in pure shear, (b) stresses
acting on a triangular stress element, and (c) forces acting on the triangular stress
element (Free-body diagram).
Graph of normal stresses σθ and shear stresses
τθ versus angle θ of the inclined plane.
Stress elements oriented at θ = 0 and θ
= 45° for pure shear.
Torsion Failure
In torsion, a ductile material will break along
the plane of maximum shear, that is, a plane
perpendicular to the shaft axis.
A brittle material, will break along planes
whose normal direction coincide with maximum
tension, that is, along surfaces at 45o to the
shaft.
Note: Brittle materials are weaker in tension than in
shear. Ductile materials generally fail in shear.
Torsion failure of a brittle material by tension
cracking along a 45° helical surface.
Stress elements oriented at θ = 0 and θ = 45° for pure shear.
Strains in Pure Shear
The shear strain γ is the change
in angle between two lines that
were originally perpendicular to
each other. The element changes
its shape while the volume is
maintained constant (shear
distortion). For a elastic
material:
γ=
G=
τ
G
E
2(1 + ν )
Strains in pure shear: (a) shear distortion of an element
oriented at θ = 0, and (b) distortion of an element oriented
at θ = 45°.
If we rotate the element 45o, there will be no shear stresses and
the normal stresses will be maximum. When inclined 45o, the
elongation is given by:
ε θ = 45 =
o
σ θ = 45
E
o
(1 + ν ) =
τθ =0
o
2G (1 + ν )
=
γ θ =0
2
o
Aluminum tube with G=27GPa
(a) Determine the maximum shear, tensile and
compressive stresses in the tube.
(4000 N .m )(0.080m ) = 58.2MPa
Tr
=
π
IP
(0.080m )4 − (0.060m )4
32
σ
= 58.2 MPa
The maximum tensile stress occurs at an tension
σ compression = −58.2MPa
inclination of 45ocw:
τ Max =
[
(b) Determine the maximum strains
ε Max
τ Max
58.2 MPa
= 0.0022rad
G
27000MPa
0.0022rad
γ
= Max =
= 0.0011rad
2
2
γ Max =
=
]
Relationship Between E and G.
Before shear the distance bd is equal to:
Lbd = h 2
After shear the distance bd is equal to:
Using the Law of Cosines:
(1 + ε Max )2 = 1 − Cos⎛⎜ π + γ ⎞⎟
⎝2
⎠
Lbd = h 2 (1 + ε Max )
⎛π
⎞
L2bd = h 2 + h 2 − 2.h 2 .Cos⎜ + γ ⎟
⎝2
⎠
2
2
(1 + ε Max ) = 1 + 2ε Max + ε Max
As the ε and γ are very small,
⎞
⎛π
Cos⎜ + γ ⎟ = − Sinγ
the equation is reduced to:
⎠
⎝2
Substituting for torsion
1 + ε Max = 1 +
ε Max =
γ
2
γ
2
γ=
τ
G
ε Max =
σ
E
(1 + ν ) = τ (1 + ν )
E
E
G=
2(1 + ν )
Transmission of Power by Circular Shafts
The most important use of circular shafts is to transmit mechanical power from
one device or machine to another. The work done by a torque of constant
magnitude is equal to the product of the torque (T) and the angle through which
it rotates (ψ).
W = Tψ
Power is the rate at which work is done
(ω=rad/s):
2πnT
P=
....n(rpm)..T ( N − m)...P ( watts )
60
2πnT
...n(rpm).....T (lb − ft )...H (hp)
H=
33,000
dW
dψ
P=
=T
= Tω
dt
dt
Steel shaft in torsion that
transmit 40hp to gear B. The
allowable stress in the steel
is 6000psi.
(a) Find the diameter (d) of the shaft
if it operates at 500rpm.
33,000 H 33,000 × 40hp
T=
=
= 5042lb − in
2πn
2π × 500rpm
τ Max
16T
16T
16 × 5042lb − in
3
= 3 ⇒d =
=
= 4.280in 3
πd
πτ allow
π × 6000 psi
d = 1.62in
The diameter of the shaft must be larger than 1.62in if the allowable shear
stress is not to be exceeded.
Solid steel shaft in torsion. Motor transmits
50kW to the shaft ABC of 50mm diameter
at 10Hz. The gears at B and C extract
35kW and 15kW respectively.
Calculate the maximum shear stress in the
shaft and the angle of twist (φAC) between the
motor and the gear C. Use G=80GPa.
P
50000 watts
=
= 796 N .m
2πf
2π (10 Hz )
16 × 239 N .m
= 9.7 MPa
τ BC =
35000 watts
3
=
= 557 N .m
π (0.050m )
2π (10 Hz )
239 N .m × 1.2m
=
= 0.0058rad
φ
BC
15000 watts
π
⎛
⎞
=
= 239 N .m
(80000000 Pa ) × ⎜ (0.050m )4 ⎟
2π (10 Hz )
⎝ 32
⎠
P = 2πfT ⇒ TA =
P
2πf
P
TC =
2πf
TB =
16TAB 16 × 796 N .m
=
= 32.4 MPa
Maximum shear stress = 32.4MPa.
πd 3
π (0.050m )3
φ AC = φ AB + φBC
TAB LAB
796 N .m × 1.0m
= 0.0162rad
=
=
π
GI Polar
⎛
⎞
φ AC = 0.0162 + 0.0058 = 0.022rad = 1.26o
4
9
80 × 10 Pa ⎜ (0.050m ) ⎟
⎝ 32
⎠
τ AB =
φ AB
(
)
Statically Indeterminate
Torsional Members
The equilibrium equations are not enough for
determining the torque. The equilibrium equations
need to be supplemented with compatibility equations
pertaining to the rotational displacements.
First step : Write the equilibrium equations
T = T1 + T2
Second Step : Formulate equations of compatibility.
Third Step : Relate the angles of twist to the
torques by torque-displacement relationships.
Solving:
⎛
⎞
G1I P1
⎟⎟
T1 = T ⎜⎜
⎝ G1I P1 + G2 I P 2 ⎠
T1L
φ1 =
G1I P1
φ1 = φ2
T2 L
φ2 =
G2 I P 2
⎛
⎞
G2 I P 2
⎟⎟
T2 = T ⎜⎜
⎝ G1I P1 + G2 I P 2 ⎠
Bar ABC is fixed at both ends and loaded at C by a
torque TO.
Find the reactive torques TA and TB., the maximum
shear stresses and the angle of rotation.
Equilibrium equations:
TA + TB = TO
Compatibility equations
Separate the bar from the
support at the end B. The
torque TO produces an
angle of twist φ1. Then
apply the reactive torque
TB. It alone produces an
angle of twist φ2.
φ1 + φ2 = 0
Torque-Displacement Equations:
φ1 =
TO LA
GI PA
φ2 = −
TB LA TB LB
−
GI PA GI PB
The minus sign appears because TB produces a rotation that is opposite in direction
to the positive direction of φ2
Substitute in the equation:
φ1 + φ2 = 0
TO LA TB LA TB LB
−
−
=0
GI PA GI PA GI PB
Solution:
⎛
⎞
LB I PA
⎟⎟
TA = TO ⎜⎜
⎝ LB I PA + LA I PB ⎠
⎛
⎞
LA I PB
⎟⎟
TB = TO ⎜⎜
⎝ LB I PA + LA I PB ⎠
Maximum shear stresses
τ AC
TAd A
=
2 I PA
τ AC
TO LB d A
=
2(LB I PA + LA I PB )
τ CB
TB d B
=
2 I PB
τ CB
TO LAd B
=
2(LB I PA + LA I PB )
Angle of Rotation:
The angle of rotation φC at section C is equal to the
angle of twist of either segment of the bar.
TA LA TB LB
TO LA LB
φC =
=
=
GI PA GI PB G (LB I PA + LA I PB )
In the special case that the bar is prismatic TA =
τ AC
TAd
=
2I P
τ AC
TO LB d
=
2 I P (LB + LA )
φC =
τ CB
TO LB
L
TB =
TO LA
L
TB d
=
2I P
τ CB
TO LAd
=
2 I P (LB + LA )
TA LA TB LB
TO LA LB
=
=
GI P
GI P GI P (LB + LA )
Strain Energy in Torsion
and Pure Shear
The work W done by the torque as it rotates
through the angle φ is equal to the area below
the torque-rotation diagram.
U =W =
TL
φ=
GI P
Tφ
2
T 2 L GI Pφ 2
U=
=
2GI P
2L
Non-uniform torsion
The total energy of the bar is obtained by adding the
strain energy of each segment
Ti 2 Li
U = ∑U i = ∑
i =1
i =1 2Gi (I P )i
n
n
If either the cross section of the bar or the torque
varies along the axis, then
2
[
T ( x )] dx
dU =
2GI p ( x)
U =∫
L
0
[T (x )]2 dx
2GI p ( x)
Strain Energy Density in Pure Shear
Consider an element of height h and thickness t.
Then the forces V acting on each face is :
V = τht
The displacement ( δ )produced is
Vδ
2
τγh 2t
U=
2
U =W =
u=
τ2
2G
Gγ 2
u=
2
δ = γh
Strain energy per unit volume:
u=
τγ
2
Prismatic bar loaded with TA and TB
simultaneously
TA = 100 N .m
TB = 150 N .m
L = 1.6m
G = 80GPa
I P = 79.52 × 10 mm
3
n
2
4
When both loads act on the bar, the torque in segment CB is
TA and in segment AC is (TA+TB)
( 2 ) + (T
TA2 L
T L
U =∑ i i =
2GI P
i =1 2Gi (I P )i
( 2) = T L + T T L + T L
2 L
)
T
+
A
B
2
A
A B
2
B
2GI P
2GI P
2GI P
2GI P
2
2
(
(
(
100 N .m ) (1.6m )
100 N .m )(150 N .m )(1.6m )
150 N .m ) (1.6m )
U=
+
+
3
4
3
4
2(80GPa )(79.52 × 10 mm ) 2(80GPa )(79.52 × 10 mm ) 2(80GPa )(79.52 × 103 mm 4 )
U = 1.26 J + 1.89 J + 1.41J = 4.56 J
Prismatic bar fixed at one end and loaded
with a distributed torque of constant
intensity t per unit distance.
t = 480lb − in / in
L = 12 ft
T ( x) = tx
G = 11.5 × 106 psi
I P = 17.18in 4
U =∫
L
0
[T (x )]2 dx =
2GI p
1
2GI P
∫
L
0
2 3
t
L
2
(tx ) dx =
6GI P
(
t 2 L3
480lb − in / in ) (144in )
U=
=
= 580in − lb
6
4
6GI P 6 11.5 × 10 psi 17.18in
2
(
)(
3
)
Thin-Walled Tubes
under torsion
Stresses acting on the longitudinal
faces ab and cd produce forces
Fb = τ btb dx
Fc = τ ctc dx
From equilibrium:
τ b tb = τ c t c
shear _ flow = τt = f = cons tan t
dT = rfds
T= f
LM
∫ rds = 2 fA
m
0
Where LM denotes the length of the
median line and AM is the area enclosed
by the median line of the cross section.
T
τ=
2tAM
The area enclosed by the median line is not the cross
section area.
For a rectangular section:
AM = bh
τ vert
τ horiz
Area enclosed by the median line
T
=
2t1bh
T
=
2t2bh
AM = πr 2
T
τ=
2πr 2t
Strain Energy of a Thin Walled Tube
Find the strain energy of an element
and then integrating throughout the
volume of the bar.
τ2
τ 2t 2 ds
f 2 ds
dU =
tdsdx =
dx =
dx
2G
2G t
2G t
Where f is the shear flow constant
f 2 LM ds L
f 2 L LM ds
T 2L
U = ∫ dU =
dx =
=
∫
∫
∫
0
0
0
t
2G
2G
t
8GAM2
4 AM2
J= L
M ds
∫0 t
Introducing a new property of cross
section called the Torsion Constant
For a constant thickness:
Angle of Twist
T 2L
U=
2GJ
2
M
4tA
J=
LM
For a circular tube of
thickness t
Tφ
T 2L
TL
=
⇒
=
W =U =
φ
2
2 G 2π r 3 t
G 2π r 3 t
(
)
(
)
∫
LM
0
ds
t
LM = 2πr
AM = πr 2
J = 2πr 3t
Comparison of circular and square tubes.
Both tubes have the same length, same
wall thickness and same cross sectional
area :
πr
4bt = 2πrt ⇒ b =
Circular Tube
AM 1 = πr 2
J1 = 2πr t
3
A1 = 2πrt
2
Square Tube
AM 2 = b 2 =
π 2r 2
Stresses
4
π 3r 3t
3
J2 = b t =
8
A2 = 4bt = 2πrt
π 2r 2
τ 1 AM 2
4 = π = 0.79
=
=
4
τ 2 AM 1
πr 2
Ratios of circular tube over square tube
π 3r 3t
2
8 = ⎛⎜ π ⎞⎟ = 0.62
Angles of twist φ1 = J 2 =
2πr 3t
φ2 J 1
⎝4⎠
The circular tube not only
has 21% lower shear
stress than does the square
tube but it also has a
greater stiffness against
rotation.
Stress Concentration
in Torsion
Stepped shaft in torsion.
The effects of stress concentration are
confined to a small region around the
discontinuity.
τ Max = Kτ Nom
⎛ 16T ⎞
Tr
=K
= K ⎜⎜ 3 ⎟⎟
IP
⎝ πD1 ⎠
K = Stress Concentration Factor
τ Nom = τ 1 = no min al _ shear _ stress
Stress-concentration factor K for a stepped shaft in
torsion. (The dashed line is for a full quartercircular fillet.)