torsion ingles
TRANSCRIPT
TorsionTorsion 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 RuleShafts 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 GeneratorGenerator: 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
1 1 1T P d= 2 2 2T P d=
Net Torque Due to Internal Stresses
( )T dF dAρ ρτ= =∫ ∫
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.
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.
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.
Axial Shear Components
TL
φφ
∝∝
From observation, the angle of twist of the shaft is proportional to the applied torque and to the shaft length.
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.
Torsional Deformations of Cylindrical Bars
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 ( )xφ
Torsional Deformations of a Circular Bar
ddxφθ =
maxrrLφγ θ= =
|
maxbb rd rab dx
φγ θ= = =
The rate of twist or angle of twist per unit length
Shear Strain at the outer surface of the bar
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
For Linear Elastic MaterialsFrom Hooke’s Law
G is shear modulus of elasticity and γ is shear strain
From Shear Strain equation :
Shear Stress at the outer surface of the bar :
Torsion Formula : To determine the relationship between shear stresses and torque, torsional formula is to be accomplished.
Shear Stresses in cylindrical bar with circular cross section.
max G rτ θ=
Gτ γ=
maxrrLφγ θ= =
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 area of all such elemental moments.
Polar moment of inertia of circle with radius r and diameter d
Maximum Shear Stress
Distribution of stresses acting on a cross section.
4 4
2 32pr dI π π
= = max 3
16
p
Tr TI d
τπ
= =
Ar
AM Max δτρρτδδ 2==
PolarMax I
Tr
ρτρτ ==
Generalized torsion formula
2max maxp
A A
T dM dA Ir r
τ τρ= = =∫ ∫∫= APolar AI δρ 2
PolarITρτ = Generalized
torsion formula
PP GITL
LGIT
G=⇒==⇒= φρφργτγ
Circular tube in torsion.
Total angle of twist
Lφθ =
Angle of twist per unit length
Equations derived above are only for bars of circular cross sections
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=11.5x106psia) what is the maximum shear stress in the bar b) what is the angle of twist between the ends?
a) From torsional formula
b) Angle of twist 4 4
4*1.5 .497032 32pdI inπ π
= = =
max 3 3
16 16*250*12 4530*1.5
T psid
τπ π
= = =
( )( )( )( ) rad
inpsiininftlb
GITL
P
02834.04970.0105.11
541225046 =
××−
==φ
Compare the solid and hollow shafts
468.027151273
14.18.581.67
2
2
===
===
mmmm
AAth unit lengWeight per
mmmm
ddSize
solid
hollow
solid
hollow
Hollow shafts vs Solid shafts
PMax I
TR=τ
PGIT
=θ
strength
Angle of twist per unit length
Both are proportional to PI
1
Weight is proportional to the cross-sectional area (A)
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
mNTmNTmNT
mmd
.175
.275.450
30
3
1
2
====
GPaGmmLmmL
CD
BC
80400500
===
Determine the maximum shear stress and the angle of twist between gears B and D.
mNmNmNTCD ⋅=⋅+⋅−= 175450275
( ) ( )( )
MPam
mmN
I
dT
p
CDCD 0.33
32030.0
2030.0.1752
4 ===π
τ
( )( )( ) rad
mPammN
IGLT
P
CDCDCD 0110.0
1095.710804.0.175
489 =××
=××
= −φ
mNTBC ⋅−= 275
( ) ( )MPa
m
mmN
I
dT
p
BCBC 9.51
1095.72
030.0.275248 −=
×
−== −τ
( )( )( ) rad
mPammN
IGLT
P
BCBCBC 0216.0
1095.710805.0.275
489 −=××
−=
××
= −φ
oCDBCBD rad 61.00106.0011.00216.0 −=−=+−=+= φφφ
A tapered bar in torsion
Change of diameter with length:
Evaluate the integral of this type of equation( ) ( )34 3
1bxabbxa
x+
−=+∫δ
Lddbda AB
A−
== ..........where
( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛−
−= 33
113
32BAAB ddddG
TLπ
φ
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).
Stress elements oriented at θ = 0 and θ= 45° for pure shear.Graph of normal stresses σθ and shear stresses
τθ versus angle θ of the inclined plane.
Torsion FailureIn 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: (a) shear distortion of an element oriented at θ = 0, and (b) distortion of an element oriented at θ = 45°.
Strains in Pure ShearThe 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:
( )ν
τγ
+=
=
12EG
GIf 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:
( ) ( ) 2121 0045
45
ooo
o GE===
==
+=+= θθθ
θ
γν
τν
σε
(a) Determine the maximum shear, tensile and compressive stresses in the tube.
( )( )( ) ( )[ ]
MPamm
mmNITr
PMax 2.58
060.0080.032
080.0.400044=
−== πτ
The maximum tensile stress occurs at an inclination of 45ocw: MPa
MPa
ncompressio
tension
2.582.58−=
=σσ
Aluminum tube with G=27GPa
(b) Determine the maximum strains
radrad
radMPa
MPaGMax
Max
MaxMax
0011.02
0022.02
0022.027000
2.58
===
===
γε
τγ
Relationship Between E and G.
Before shear the distance bd is equal to: 2hLbd =After shear the distance bd is equal to: ( )Maxbd hL ε+= 12
Using the Law of Cosines: ⎟⎠⎞
⎜⎝⎛ +−+= γπ
2..2 2222 CoshhhLbd
( ) ⎟⎠⎞
⎜⎝⎛ +−=+ γπε
211 2 CosMax
( )
γγπεεε
SinCos
MaxMaxMax
−=⎟⎠⎞
⎜⎝⎛ +
++=+
2
211 22
As the ε and γ are very small, the equation is reduced to:
2
211
γε
γε
=
+=+
Max
Max
Substituting for torsion
( ) ( )ντνσε
τγ
+=+=
=
11EE
G
Max( )ν+=12EG
Transmission of Power by Circular ShaftsThe 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).
ψTW = Power is the rate at which work is done (ω=rad/s): ωψ T
dtdT
dtdWP ===
)()...().....(...000,33
2
)()...()..(....60
2
hpHftlbTrpmnnTH
wattsPmNTrpmnnTP
−=
−=
π
π
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.
inlbrpm
hpn
HT −=×
×== 5042
500240000,33
2000,33
ππ
ind
inpsi
inlbTddT
allowMax
62.1
280.46000
5042161616 333
=
=×
−×==⇒=
ππτπτ
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.
( )
( )
( ) mNHz
wattsf
PT
mNHz
wattsf
PT
mNHz
wattsf
PTfTP
C
B
A
.239102
150002
.557102
350002
.796102
500002
2
===
===
===⇒=
ππ
ππ
πππ
( )
( ) ( )rad
mPa
mmNGI
LT
MPam
mNdT
Polar
ABABAB
ABAB
0162.0050.0
321080
0.1.796
4.32050.0
.7961616
49
33
=⎟⎠⎞
⎜⎝⎛×
×==
=×
==
πφ
ππτ
( )
( ) ( )rad
mPa
mmN
MPam
mN
BC
BC
0058.0050.0
3280000000
2.1.239
7.9050.0
.23916
4
3
=⎟⎠⎞
⎜⎝⎛×
×=
=×
=
πφ
πτ
Maximum shear stress = 32.4MPa.
oAC
BCABAC
rad 26.1022.00058.00162.0 ==+=
+=
φ
φφφ
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
21 TTT +=
Second Step : Formulate equations of compatibility. 21 φφ =
Third Step : Relate the angles of twist to the torques by torque-displacement relationships.
22
22
11
11
PP IGLT
IGLT
== φφ
Solving:
⎟⎟⎠
⎞⎜⎜⎝
⎛+
=⎟⎟⎠
⎞⎜⎜⎝
⎛+
=2211
222
2211
111
PP
P
PP
P
IGIGIGTT
IGIGIGTT
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: OBA TTT =+
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.
021 =+φφ
Torque-Displacement Equations:PB
BB
PA
AB
PA
AO
GILT
GILT
GILT
−−== 21 φφ
The minus sign appears because TB produces a rotation that is opposite in direction to the positive direction of φ2
Substitute in the equation: 021 =+φφ
0=−−PB
BB
PA
AB
PA
AO
GILT
GILT
GILT
Solution:
⎟⎟⎠
⎞⎜⎜⎝
⎛+
=⎟⎟⎠
⎞⎜⎜⎝
⎛+
=PBAPAB
PBAOB
PBAPAB
PABOA ILIL
ILTTILIL
ILTT
Maximum shear stresses
( ) ( )PBAPAB
BAOCB
PBAPAB
ABOAC
PB
BBCB
PA
AAAC
ILILdLT
ILILdLT
IdT
IdT
+=
+=
==
2
2
2
2
ττ
ττ
Angle of Rotation: The angle of rotation φC at section C is equal to the angle of twist of either segment of the bar.
( )PBAPAB
BAO
PB
BB
PA
AAC ILILG
LLTGI
LTGI
LT+
===φ
In the special case that the bar is prismatic LLTT
LLTT AO
BBO
A ==
( ) ( )ABP
AOCB
ABP
BOAC
P
BCB
P
AAC
LLIdLT
LLIdLT
IdT
IdT
+=
+=
==
2
2
2
2
ττ
ττ
( )ABP
BAO
P
BB
P
AAC LLGI
LLTGI
LTGI
LT+
===φ
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.
2φTWU ==
LGI
GILTU
GITL P
PP 22
22 φφ ===
Non-uniform torsion
( )∑∑==
==n
i iPi
iin
ii IG
LTUU1
2
1 2
The total energy of the bar is obtained by adding the strain energy of each segment
If either the cross section of the bar or the torque varies along the axis, then
( )[ ]
( )[ ]∫=
=
L
p
p
xGIdxxTU
xGIdxxTdU
0
2
2
)(2
)(2
Strain Energy Density in Pure ShearConsider an element of height h and thickness t. Then the forces V acting on each face is :
htV τ=The displacement ( δ )produced is hγδ =
2
22thU
VWU
τγ
δ
=
== Strain energy per unit volume:
2τγ
=u
2
22
2
γ
τ
Gu
Gu
=
=
Prismatic bar loaded with TA and TBsimultaneously
431052.7980
6.1.150.100
mmIGPaGmL
mNTmNT
P
B
A
×=
====
( )( ) ( ) ( )
P
B
P
BA
P
A
P
BA
P
An
i iPi
ii
GILT
GILTT
GILT
GI
LTTGI
LTIGLTU
22222
22
2
2222
1
2
++=+
+==∑=
When both loads act on the bar, the torque in segment CB is TA and in segment AC is (TA+TB)
( ) ( )( )( )
( )( )( )( )( )
( ) ( )( )( )43
2
4343
2
1052.798026.1.150
1052.798026.1.150.100
1052.798026.1.100
mmGPammN
mmGPammNmN
mmGPammNU
×+
×+
×=
JJJJU 56.441.189.126.1 =++=
Prismatic bar fixed at one end and loaded with a distributed torque of constant intensity t per unit distance.
4
6
18.17105.11
12/480
inIpsiG
ftLininlbt
P =
×=
=−=
txxT =)(
( )[ ] ( )
( ) ( )( )( ) lbin
inpsiinininlb
GILtU
GILtdxtx
GIGIdxxTU
P
P
L
P
L
p
−=×−
==
=== ∫∫
58018.17105.116
144/4806
621
2
46
3232
32
0
2
0
2
Thin-Walled Tubes under torsionStresses acting on the longitudinal faces ab and cd produce forces
dxtFdxtF cccbbb ττ ==
From equilibrium:ccbb tt ττ =
tconsftflowshear tan_ ===τ
m
L
fArdsfT
rfdsdTM
20
==
=
∫
Where LM denotes the length of the median line and AM is the area enclosed by the median line of the cross section.
MtAT
2=τ
The area enclosed by the median line is not the cross section area.
For a rectangular section:
bhtTbht
TbhA
horiz
vert
M
2
1
2
2
=
=
=
τ
τ
Area enclosed by the median line
trT
rAM
2
2
2πτ
π
=
=
Angle of Twist ( ) ( )trGTL
trGLTTUW 33
2
2222 πφ
πφ
=⇒===
Strain Energy of a Thin Walled TubeFind the strain energy of an element and then integrating throughout the volume of the bar.
dxt
dsGfdx
tds
Gttdsdx
GdU
222
2222
===ττ
Where f is the shear flow constant
∫∫∫∫∫ ==== MMM L
M
LLL
tds
GALT
tds
GLfdx
tds
GfdUU
02
2
0
2
00
2
822
Introducing a new property of cross section called the Torsion Constant
∫=
MLM
tds
AJ
0
24
GJLTU
2
2
=
For a constant thickness:M
M
LtAJ
24= 2
2
rA
rL
M
M
πFor a circular tube of thickness t
π
=
=
trJ 32π=
Comparison of circular and square tubes.
Circular Tube
rtAtrJ
rAM
ππ
π
22
1
31
21
==
=
Square Tube
rtbtA
trtbJ
rbAM
π
π
π
248
4
2
333
2
222
2
==
==
==
Both tubes have the same length, same wall thickness and same cross sectional area :
224 rbrtbt ππ =⇒=
Ratios of circular tube over square tube
79.04
42
22
1
2
2
1 ====π
π
π
ττ
r
r
AA
M
M
Stresses
Angles of twist 62.042
82
3
33
1
2
2
1 =⎟⎠⎞
⎜⎝⎛===π
π
π
φφ
tr
tr
JJ
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.
⎟⎟⎠
⎞⎜⎜⎝
⎛=== 3
1
16DTK
ITrKK
PNomMax π
ττ
stressshearalnoNom __min1 =
K = Stress Concentration Factor
=ττ
Stress-concentration factor K for a stepped shaft in torsion. (The dashed line is for a full quarter-circular fillet.)