1. torsion - limba.wil.pk.edu.pllimba.wil.pk.edu.pl/~az/cwicz/class_1.pdf · torsion definitions...
TRANSCRIPT
Adam Paweł Zaborski
1. Torsion
Definitions
torsion angle – the angle between the final and initial radius vectors of a point, Fig. 1.1; please note, that
the angle is defined in the cross-section, its unit is [rad] or []; this is the rotation angle of one cross-
section with respect to another
Fig. 1.1 Definition of the torsion angle
unit torsion angle – the torsion angle per meter
torsion moment, twisting couple, torque – the moment the vector of which is parallel to the bar axis; it may
be shown in different ways, Fig. 1.2
Fig. 1.2 Different symbols used to present a torque
The most common application of torsion is provided by transmission shafts, which can be either solid or
hollow.
Solution of BVP
de Saint-Venant’s assumptions
1. The projection of the cross-section onto its initial plane remains unchanged (as the rigid membrane),
Fig.1.3
2. The cross-section distortion consists in the movement “out-of-plane” only and is the same for all cross-
sections.
w
v
A
A'
Fig. 1.3 de S-V’s assumptions
The displacement components can be written:
),,(
,,
,cos,sin
,1cos,sin,'
zyu
xywxzv
wv
,
where:
– torsion angle,
Adam Paweł Zaborski
– unit torsion angle,
– distortion function; it does not depend on x,
We calculate the strain matrix:
,
0
00
021
21
yz
Tzy
and the stress matrix:
0
00
0 yGzG
Tzy
From the first Navier’s equation we get:
02 (1)
the remaining equations are identically fulfilled. We see that the distortion function is harmonic.
The static boundary conditions are:
on the side surface:
0
nymz
zy, (2)
the remaining conditions are identically fulfilled,
on the bottoms:
1,1
yGqzGq
zvzyvy , (3)
(1st equation is identically fulfilled).
From the kinematic conditions of fixing at the point O, we have:
.0,0,00,0,0
zy (4)
The Laplace equation (1), with zero boundary conditions with the partial derivatives (2), makes so-called
Neumann’s problem, which has a solution accurate up to a constant. For the cross-section with one axis of
symmetry we have:
0
zy.
Because we can do nothing with the remaining conditions on the bottoms (3), we define the loading in the
same form as for the shear stresses:
., yGzGzvzyvy
(5)
Calculating the torque, we have:
F F
syzvyvzs JGdFzzGyyGdFzqyqM (6)
where the torsion inertia moment is defined as:
dFzyzyJF
yzs
22 (7)
In the general case the torsion inertia depends on the distortion function and indirectly on the BVP’s
solution.
Eventually we get the relationship between the unit torsion angle, torsion moment and the torsion rigidity
GIs:
s
s
GJ
M (8)
The unit of the unit torsion angle is [1/m = rad/m].
Adam Paweł Zaborski
Torsion of bars with axisymmetric cross-section
Basic formulae
In the particular case of axial symmetry (a circle, a ring), the distortion function is identically equal to zero
and the torsion inertia moment reduces to the polar inertia moment:
00),( JJzy s
It means that the plane cross-section remains a plane after the deformation, it doesn’t deplane.
The stresses, (5), become: ., yGzG vzvy
z
y
xy
xz
Fig. 1.4. Shear stress distribution
The resultant shear stress is:
rJ
MrGyzG s
xzxy0
2222 , [MPa].
The shearing stress varies linearly with the distance from the axis of the shaft, Fig. 1.4.
The torsion angle is calculated from the unit torsion angle provided that the torsion moment is constant:
00 GJ
lM
GJ
M
dx
d ss
l
s
dx [1 = rad]
This is the angle through which one section rotates with respect to the other. Note, that the torsion angle is
dimensionless, so the angle unit in the above formula is a radian, not a degree.
When the interval is loaded by a continuous twist moment of intensity )(xm , the formulae become:
x
dmxM0
)()( , and
x
dGJ
Mx
0 0
)()(
Design conditions
There are two main requirements:
the requirement of the strength and
the requirement of usability.
The first requirement means that every structure has to sustain applied load. The second one consists in
several demands of durability, rigidity, resistance to severe weather conditions and so on. From the point of
view of the strength of materials course two principal design conditions should be listed:
the ultimate limit state
the serviceability limit state, usually the stiffness of the structure.
In the particular case of the torsion, the ultimate limit state is checked by:
tR)max(
where tR is the shear strength of a given material. The suitable value can be found in the standards.
Because the maximum value of the stress is attained at the side surface for Rr , we introduce so-called
cross-section twist factor or modulus 0W :
Adam Paweł Zaborski
R
JW
def0
0 ,
so:
0
0
0
0)max(W
MR
J
M
The stiffness condition means that the unit torsion angle or the torsion angle doesn’t exceed acceptable
(permissible) values:
acceptable , or acceptable
Transmission shafts
The principal specifications to be met in the design of a transmission shaft are the power to be transmitted
and the speed of rotation of the shaft. The role of the designer is to select the material and the dimensions
of the cross-section of the shaft, so that the maximum shearing stress allowable in the material will not be
exceeded when the shaft is transmitting the required power at the specified speed.
The power P associated with the rotation of a rigid body subjected to a torque M0 is:
0MP ,
where is the angular velocity of the body expressed in radians per second. But, f 2 , where f is the
frequency of the rotation, i.e., the number of revolutions per second. We write:
fMP 20 .
Hollow shaft
The linear repartition of the shearing stresses in the twisted circular bar signifies that the outer fibers are the
most useful and the central ones are almost not used. Therefore, the use of hollow shafts makes sense,
especially in the case where the weight is more important than the price of the element.
The torsion inertia moment for the hollow shaft with outer diameter D and inner diameter d is:
4444
0323232
dDdD
J
,
and the cross-section torsion factor is:
4400
162 dD
DD
JW
.
Examples
Example 1.1
The shaft with the diameter of 4 cm and length of 2 m is fixed at one end and loaded at another end by such
torque that the point A on the side surface is displaced to the point A’. The arc AA’ length is 1 mm.
Determine the torsion angle, the unit torsion angle, the torque, the maximum shear stress and the angular
strain on the side surface. G = 80 GPa.
Solution
From the definition of the twist angle we calculate:
05.04
2.02
1.0
d[rad].
For constant torsion moment we have:
025.02
05.0
l [rad/m].
Using the formula of the unit torsion angle we get:
500025.032
04.01080
49 ss JGM [Nm]
Adam Paweł Zaborski
(it’s quite significant value, the average value for the VW golf is about 150 Nm). The maximum stress
value is:
69
0
0
0
0 104002.0025.010802
d
GRJ
GJ
W
Mmax [Pa] 40 [MPa].
From the Hooke’s equation we determine
0005.002.0025.022
22 max
d
G.
Example 1.2
Determine the rotation angle between section A and section E of the shaft in Fig. 1.5, 41 d cm, 32 d cm,
G = 80 GPa.
A
B
C D E
30 Nm 40 70
d1 d2
0.3 0.5 m 0.4 0.4
Fig. 1.5 Twisted shaft
Solution
The formula of the twist angle:
0
0
GJ
lM
is valid only in the case of all parameters constant, so we have to cut the shaft into the intervals with
constant cross-section inertia moment and constant torque: AB, BC, CD and DE.
We have:
AB: 4.0ABl m, 60704030 ABM Nm, 741 1051.2
32
d
J AB m4, 00120.0
1051.21080
4.06079
AB rad
BC: 4.0BCl m, 104030 BCM Nm, 741 1051.2
32
d
J BC m4, 000199.0
1051.21080
4.01079
BC rad
CD: 3.0CDl m, 104030 CDM Nm, 842 1095.7
32
d
JCD m4, 000472.0
1095.71080
3.01089
CD rad
DE: 5.0DEl m, 30DEM Nm, 842 1095.7
32
d
J DE m4, 00236.0
1095.71080
5.03089
DE rad
and the total angle of the twist is:
00289.000236.0000472.0000199.000120.0 DECDBCABAE rad.
Fig. 1.6 presents the diagram of the twist angle.
Adam Paweł Zaborski
Fig. 1.6 Diagram of the twist angle
Example 1.3
Determine the diameter of the shaft in Fig. 1.7, if the acceptable values of (a) the shear stress is 150tR
MPa, (b) unit twist angle is 05.0acc [rad/m] and (c) the twist angle between the sections B and D is 0.02
[rad]. The Kirchhoff module value is G = 80 GPa, 12 2.1 dd .
A B
C
D
0.3 1.2 m 0.7 m
300 Nm 800 Nm
Fig. 1.7 Shaft under torsion
Solution
the 1st condition will be fulfilled when:
1500
0max tR
W
MMPa
hence the following equations:
30
3
0
0
0 1616
tt
R
MdR
d
M
W
M
and
0217.010150
300163
6
ABd m, 0181.0
10150
30016
2.1
13
6
BCd m, 0214.0
10150
50016
2.1
13
6
CDd m
so
0217.0,,max CDBCAB dddd m
the 2nd
condition will be fulfilled when
05.00
0 accGJ
M
hence, we have:
49
0
4
0
05.01080
3205.0
32
Md
dG
M
angle
0
0,0005
0,001
0,0015
0,002
0,0025
0,003
0 0,5 1 1,5
Adam Paweł Zaborski
so
0296.005.01080
300324
9
ABd m, 0246.0
05.01080
30032
2.1
14
9
BCd m, 0280.0
05.01080
50032
2.1
14
9
CDd m
so 0296.0,,max CDBCAB dddd
the 3rd
condition will be fulfilled when
02.0 accBD rad
hence, we have:
02.0
2.11080
322.1500
2.11080
323.03004949
00
ddGJ
lM
GJ
lM
CD
CDCD
BC
BCBCCDBCBD
so
0354.03.03002.150002.01080
32
2.1
14
9
d m
Taking into account the results obtained we finally assume:
cm6.3m036.0 d
Example 1.4
For the shaft in Fig. 1.8 determine the function of the twist moment and draw the unit torsion angle
diagram. Determine the values and draw the diagram of the torsion angle. 31 d cm, 1.32 d cm, 80G
GPa.
d1 d2
0.2 0.5 1.2 m 0.6
150 Nm/m
Fig. 1.8 Shaft with loading
Solution
In order to solve the redundant problem we need to go back to the BVP and the complete set of its
equations. As we know, there are:
the statics equations of the equilibrium of the forces (Navier’s equations with the SBC),
the geometric equations of compatibility (Cauchy’s equations with KBC), and
the constitutive equations describing behavior of the material.
For the particular problem of the shaft twist we have:
one statics equation 00 M (the resultant torque should be zero for static equilibrium)
one geometric equation of compatibility 0 EAAE (the twist angle between one end and another,
both fixed)
the constitutive equations 0
0
GJ
lM
We release the shaft end from constraints:
Adam Paweł Zaborski
B A
MA ME
0.2 0.5 1.2 m 0.6
150 Nm/m
D C E
Fig. 1.9 Shaft under loading
We have two unknowns EA MM , and two equations, the static one and the kinematic one:
07.0150 EA MM
0 DECDBCAB
The second equation can be written as:
02.17.01505.05.05.01502.01502.02.05.01506.0
02020101
GJ
M
GJ
M
GJ
M
GJ
M AAAA
Multiplying both sides by 01GJ , with 8771.00201 JJ we get:
05.11005.16.295.08771.032.06.0 AAAA MMMM ,
hence we obtain:
47.62AM Nm, 53.42EM Nm.
Fig. 1.10, 1.11 and 1.12 present the diagrams of the twist moment, unit twist angle and twist angle.
Fig. 1.10 Twist moment diagram
Moment
-60
-40
-20
0
20
40
60
80
0 0,5 1 1,5 2 2,5
Adam Paweł Zaborski
Fig. 1.11 Unit twist angle diagram
Fig. 1.12 Twist angle diagram
theta
-0,008
-0,006
-0,004
-0,002
0
0,002
0,004
0,006
0,008
0,01
0,012
0 0,5 1 1,5 2 2,5
alpha
0
0,001
0,002
0,003
0,004
0,005
0,006
0,007
0,008
0 0,5 1 1,5 2 2,5
Adam Paweł Zaborski
Torsion of noncircular members
Membrane analogy
It has been proved by de Saint-Venant that the distortion function vanishes for some particular cross-
section shapes only, e.g. axisymmetrical, triangular, elliptical etc. In general, if the distortion function is
non-zero, the BVP should be solved.
By use of some kind of parametrization (of so-called Prandtl’s functions), the BVP may be written in
another form of Poisson’s equation:
G22
where the function is the Prandtl’s function:
yz
xzxy
, .
The distribution of shearing stresses in a noncircular member may be visualized more easily by using the
membrane analogy. An elastic membrane subjected to a uniform pressure constitutes an analog of the bar
in torsion. The same BVP describes the deformation of the membrane and the shearing stresses in the
twisted bar. The shearing stress at some point will have the same direction as the horizontal tangent to the
membrane and its magnitude will be proportional to the maximum slope of the membrane. Moreover, the
torque will be proportional to the volume between the membrane and its initial plane.
Rectangular cross-section
In the case of rectangular cross-section the steepest slope occurs at the midpoint of the larger side of the
rectangle, Fig.1.13
Fig. 1.13 Prandtl’s function for rectangular section
Furthermore, the shearing stress is zero at the corners and at the center of the section, Fig. 1.14.
Adam Paweł Zaborski
=0
=0
max
ma
x
Fig. 1.14 Shearing stress in rectangular cross-section
The solution for a rectangular section may be obtained by expansion of the warping function in series:
0
).()(,n
nn zgyfyzzy .
The solution can be presented as formulae:
max3
2max ,,
sbh
s
bh
s hbJhb
M,
where s is the shearing stress at the middle of the shorter side and the coefficients , and are stated in
the table 1.1.
h/b 1 1.25 1.5 2 3 5 10 ∞
0.2082 0.2212 0.2310 0.2459 0.2672 0.2915 0.3123 0.3333
0.1406 0.1717 0.1958 0.2287 0.2633 0.2913 0.3123 0.3333
1.0 0.9159 0.8591 0.7958 0.7533 0.7429 0.7423 0.7423
Table 1.1 Coefficients for maximum shear stress, torsion inertia moment and additional shearing stress
Thin-walled cross-section
A thin-walled cross-section member is a bar the cross-section of which is made from profile, for example
from bent sheet of steel. The exact solution of such twisted profiles is difficult and usually not needed in
practice. There are simplified methods used instead.
From the point of view of practical calculation there are three types of these cross-sections, Fig. 1.15:
(a) the open thin-walled profiles folded (which can be uncurled and so-called middle-line doesn’t fork),
(b) the open profiles unfolded (which cannot be uncurled and the middle-line branches),
(c) the closed profiles (the middle-line is a closed line).
a) b) c)
Fig. 1.15 Thin-walled profiles with middle-line (a), (b) and (c)
Profiles with middle-line not branched
Some of the thin-walled cross-sections may be substituted by one rectangle with the same area and the
longer side length is equal to the middle-line of the profile.
For example, a C profile with the web 300x10 mm and the flanges 90x16 mm may be substituted by a
rectangle with middle-line length l = 2x90 + 300 = 480 mm and average width b = 12.25 mm.
Hydrodynamic analogy
There is an analogy between the distribution of the shearing stresses in the transverse section of a shaft and
the distribution of flow lines and velocities in water flowing through a closed channel of unit depth and
variable width, Fig. 1.16. In the last case, the water flow as well as the shear flow are constant, const .
Adam Paweł Zaborski
v = const
Fig. 1.16 Hydrodynamic analogy: flow lines and flow velocities in hollow shaft
Profiles with middle-line branched
The thin-walled cross-sections with the branched middle-line can be solved applying the hydrodynamic
analogy. The section is cut into rectangles and it is assumed that:
each rectangle works independently, transmitting a part of the applied torque, siM ,
the unit torsion angle of each rectangle is the same, idemi .
With the static equilibrium equation, we get a set of n equations for n component rectangles:
i is
is
si
sisis ni
GJ
M
GJ
MMM 1,,1,,
)1(
)1( ,
hence we find siM and the unit torsion angle of the whole cross-section..
The shorter cut lines are, the better the approximation is.
Hollow profiles
The shearing stress in closed thin-walled cross-sections may be determined by means of the hydrodynamic
analogy. The flow of the stress through every section is constant: const2211 . The shearing stress,
running around the cross-section and acting on a different lever, constitutes the torque as a sum of the
shearing stress actions. Assuming that the stress repartition is constant in each section of the closed
channel, the torque may be written as:
c c
s AdsssdsM 2)()( ,
where F is the area inside the middle-line of the profile, Fig. 1.17.
1
ds2
Fig. 1.17 Shearing stress circulation in closed profile
Hence, the extreme shearing stress will be attained in the section of the channel with the smallest width:
min
max
A
M s
2.
The above formula is known as Bredt’s first formula.
To determine the deformation caused by the torsion we use the formula:
GJ
M s ,
Adam Paweł Zaborski
where GJ is the torsional stiffness, and the inertia moment is (Bredt’s second formula):
c
AJ
ds
AJ
c
22 4,
4const .
Bredt’s theory has been developed on the basis of a simplifying assumption for the stress distribution
which gets closer to the actual distribution when the wall becomes thinner.
Examples
Example 1.5
Determine the maximum shearing stress and the unit twist angle of the cross-section in Fig. 1.18. Compare
the solution for the profile with not branched middle-line with the Bredt’s solution.
10 300
100
16
16
Fig. 1.18 Twisted cross-section
Solution
a) We consider the cross-section as a rectangle with the length equal to the length of the middle-line:
4745100216300 l mm
To obtain the same area we take:
41.12474
588058803230010210016 A mm
We have:
3
12.38
41.12
474
l
The maximum shearing stress is:
ss
s
s MM
W
M 5
2
31max 1011.4
47441.12
G
M
G
M
GJ
M ss
s
s 6
3
31
1031.347441.12
1
b) We divide the cross-section into 3 rectangles, two flanges 16100 , 25.6
l and one web 10268 ,
8.26
l. The parameters and for the web are
31 and for the flanges (from linear interpolation)
are 2976.0 and 2977.0 . Using the Bredt’s assumptions, we have:
– the static equation
sMMM wfl2
– the compatibility equation
flwwfl
w
w
fl
flwfl MM
MM
GJ
M
GJ
M7326.0
268103333.0100162977.0 33
and, finally:
Adam Paweł Zaborski
ss M
M 5
2
31
1000.326810
2681.0
w , ss M
M 5
21080.4
100162976.0
366.0
fl , sM51080.4max
G
M
G
M ss 6
31000.3
100162977.0
366.0
Conclusions:
Both methods give different answers, the relative errors are: for maximum shearing stress 17% and for the
unit twist angle about 10%.
Example 1.6
Determine the shearing stress in the T-beam (the web 30x5 cm and the flange 6x20 cm), twisted by the
torque 20 kNm,. while l/ = 6: = 0.299, = 0.299, h/b = 3.33: = 0.272, = 0.269.
MPaMPa
kNm,kNm
flw 0.52,8.43
18.10,82.91020
1021620269.01012530299.0
0036.02.0272.010180
0025.03.299.09820
213
21
8
2
8
121
MMMM
G
M
G
M
Example 1.7
To get an idea on applicability of Bredt’s formulae let’s analyze a particular case of twisted ring of the
outer diameter R and inner diameter r, where both exact and approximate solutions can be found. For the
sake of analysis we will use a parameter which is the ratio of the cross-section thickness to the middle-
line radius: .
a) Bredt’s solution
2max
2
1
2
ssB M
A
M
32 2
1
2
2
G
M
GA
M ssB
b) exact solution
22max4
2
s
s
s M
W
Mexact
224
2
G
M
GJ
M s
s
s exact
The ratio of Bredt’ solution to the exact solution is:
4
4,
24
4 22
exact
Bredt
exact
Bredt .
The above formulae as functions of parameter are presented in Fig. 1.19.
The maximum value of shearing stress is underestimated and quickly starts to be significant (from 1.0 )
and error associated to the unit twist angle remains small up to the larger values of the parameter (up to
0.5).
Adam Paweł Zaborski
Fig. 1.19 Comparison of Bredt’s solution with exact solution
Review problems
Problem 1.1
An engine is connected to a generator by a large hollow shaft with inner and outer diameters of 100 mm
and 150 mm, respectively. Knowing that the allowable shearing stress is 85 MPa, determine the maximum
torque that can be transmitted: (a) by the shaft as designed, (b) by a solid shaft of the same weight, (c) by a
hollow shaft of the same weight and of 200 mm outer diameter.
Ans.: a) 45.2 kNm, b) 23.3 kNm, c) 70.5 kNm
Problem 1.2
Knowing that each of the shafts AB, BC and CD consists of solid circular rods, Fig. 1.20, determine the
shaft in which the maximum shearing stress occurs and the magnitude of that stress.
201 d mm, 252 d mm, 303 d mm.
Ans.: 58.7 MPa in the 2nd
shaft.
D
C
B
A
270 Nm
110 Nm
90 Nm
Fig. 1.20 Shafts with loads
Adam Paweł Zaborski
Problem 1.3
A gear, Fig. 1.21, has two cogs: a greater one and a smaller one, with the diameters of 20 mm and 8 mm,
respectively. Knowing that the allowable shearing stress is 85 MPa, determine the diameters of the shafts if
the torque 2201 M Nm.
M1 M2
Fig. 1.21 Gear with two cogs and shafts
Ans.: 0174.0,0236.0 21 dd m.
Problem 1.4
Knowing that the allowable shearing stress of a brass shaft with both ends fixed, Fig. 1.22, is 55 MPa,
determine the diameter of the shaft. Using the calculated diameter and knowing that the Kirchhoff’s
modulus is 37 GPa, determine the rotation angle of the section where the torque is applied.
0.6 m 1 m
170 Nm
Fig. 1.22 Brass shaft with load
Ans.: 0214.0d m, 79.4rad0837.0
Problem 1.5
Determine the diameter of the steel shaft, Fig. 1.23, knowing that the allowable shear stress is 80 MPa, and
the maximum value of the rotation angle should not exceed the value of 2 degrees, G = 80 GPa.
100 Nm
0.6 m 0.7 m
170 Nm
Fig. 1.23 Shaft with loading
Ans.: 0198.0d m (cf. the twist angle diagram).
Adam Paweł Zaborski
Problem 1.6
Using an allowable shearing stress of 40 MPa, design a solid steel shaft of a VW engine to transmit 100
kW at the speed of 5000 rpm
Ans.: 0290.0d m
Problem 1.7
Determine the maximum shearing stress and the unit twist angle for the cross-sections stated in Fig. 1.24.
Fig. 1.24 Different profiles
Addendum
Glossary
torsion angle, twist angle – kąt skręcenia
unit torsion angle – jednostkowy kąt skręcenia
torsion moment – moment skręcający (jako siła przekrojowa)
torque – moment skręcający (jako obciążenie)
wheel nut torque - moment dokręcenia śruby
torsion inertia moment – moment bezwładności na skręcanie
torsion modulus, torsion/twist factor – wskaźnik na skręcanie
distortion, warping – spaczenie, deplanacja
torsion rigidity – sztywność skręcania
shaft – wał, wałek
gear – przekładnia
cog – koło zębate, ząb (koła zębatego)
brass – mosiądz
rpm (rotations per minute) – (ilość) obrotów na minutę
middle-line – linia środkowa
web – środnik
flange – półka
profile – profil