Idea Transcript
UNIT 12 TORSION Structure 12.1 Introduction Objectives
12.2 Torsion of Circular Shafts 1221 TheoryofTmioll 12.22 Resisting Torque 122.3 1)eformations in a Circular Shaft 12.24 Stresses in a Circular Shaft
12.3 Power Transmission by Shafts 12.4 Hollow Circular Shafts 124.1 Strength ofa Hollow Shaft 12.4.2 Torsion of Thin Tubes of Circular Section
12.5 Torsion of Non-circular Sections 12.5.1 Solid Non-circular Section Shafts 12.5.2 Thin Walled Non-circular Section Shafts
12.6 Concept of Plastic Torsion 12.7 Summary 12.8 Answers to SAQs
12.1 INTRODUCTION The results obtained during the study of shear enable us to pass over to the study of strength under torsion. Members in torsion are encountered in many engineering applications. The most common application is provided by transmission shafts, which are used to transmit power from one point to another, as from a steam turbine to an electric generator, or from a motor to a machine tool, or from the engine to the rear axle of an automobile. These shafts may either be solid or they may be hollow. In practice, we come across torsion very often; a turning force is always applied to transmit energy by rotation. This turning force is applied either to rim of a pulley, keyed to the shaft, or to any other suitable point at some distance from the axis of the shaft. The product of this turning force, and the distance between the point of application of the force and the axis of the shaft is known as torque, tutning moment or twisting moment. A major part of this unit is devoted to the treatment of members with circular, or tubular, cross-sectional areas. Non-circular sections are discussed only briefly. In practice, members that transmit torque, such as shafts of motors, torque tubes of power equipment, etc. are predominantly circular or tubular in cross-section.
Objectives After studying this unit, you should be able to conceptualise the theory of torsion, calculate the strength of the solid and hollow circular shaft, deformations and stresses developed in the shafts, detennine the power transmitted by the shafts in MKS and SI units, design the shaft for the required torque as per the strength criteria and as per the stiffness criteria, determine the torque transmitted by the new shaft (should be equal to the torque transmitted by the replaced shaft) in replacing the shaft, draw the free body diagram for the aialysis of stepped shaft,' describe the behaviour of non-circular section shaft under torsion, compare the analogy between membrane problem and torsion problem, and calculate the fully plastic torque Tp for various structural sections.
St~ssesinShaPts Shells and Thermal Stresses
12.2 TORSION OF CIRCULAR SHAFTS A shaft of circular section is said to be in pure torsion when it is subjected to equal and opposite end couples whose axes coincide with the axis of the shaft. In other words, if the moment is applied in a vertical plane perpendicular to the longitudinal axis of the beam or a shaft, it will be subjected to a torque causing twist or torsion in the member. As the beam bends due to bending moment, the shaft twists due to twisting moment. Figure 12.1 shows a pulley of radius R subjected to a system of couple, i.e. equal and opposite force W.The couple attached to the shaft which cause the tuming effect on the pulley.
.
Twisting moinent or torque, T = Force x Lever arm
In order that the body should remain in static equilibrium, it must exert an equal amount of resisting moment. At any point in the section of the shaft, a shear stress is induced or more exactly, the state of stress at any point in the cross-section of the shaft is one of pure shear, the direction of which is tangential at any point in the shaft. By the principle of complementary sheaf stresses, we know that in a state of simple shear there are two planes carrying the shear stress of the same intensity. These planes must be perpendicular to each other. In the case of the shaft in torsion, the planes of shear at a point are (a) the cross-section itself, and (b) the plane containing the point and the axis of the shaft. To find internal torque or resisting moment, in statically determinate members, only one equations of statics Z M,= 0 is required, where Z axis is directed along the member. As in the case of determination of twisting moment at any point along the length of the member, pass a plane at the desired section perpendicular to the longitudinal axis of the member and remove everything to either side of the cut. The internal or resisting torque necessary to maintain equilibrium of the isolated part is determined.Considering,for example, the GENERATOR
/---->
ROTATION
system'consisting of the turbine A and the generator B connected by the transmission shaft AB (Figure 12.2), and breaking the system into its three component parts (Figure 1 2 3 , we note that the turbine exerts a twisting couple or torque Ton the shaft, and that the shaft exerts im equal torque on the generator.
The generator reacts by exerting the equal and opposite torque T' on the shaft, and the shaft by exerting the torque T' on the turbine. Following assumptions are made, while finding out shear stresses and &formations in a circular shaft subjected to torsion. (a) m e material of the shaft is homogeneous and isotropic. (b) The twist along the shaft is uniform throughout (i.e.) all normal cross-sections which are at the same axial distance suffer equal relative rotation. (c) Normal cross-sections of the shaft, which were plane and circular before twist, remain plane and circular after twist, i.e. no warping or distortion of parallel planes normal to the axis of the member takes place. (d) All diameters of the normal cross-section which were straight before twist, remain straight with their magnitude unchanged, after twist. (e) Stress is proportional to strain, i.e. all the stresses are within the elastic limit. (f) Intensity of stress varies uniformly from zero at the centre to a maximum at the outside surface and hence the stress is proportional to the distance of that point from the centre. A little consideration will show that the above assumptions are justified, if the torque applied is small and the angle of twist is also small.
12.2.1 Theory of Torsion For the purpose of developing the expressions for the torsional stress and strain, we shall assume that one end of the shaft is fixed and a moment is applied at the other end, the plane of application of moment being perpendicular to the longitudinal axis of the beam. This assumption is valid because whether it rotates at uniform speed to transmit the power or is at rest, the stress and strain due to equal and opposite couples at its ends will remain the same. Consider a shaft fixed at one end, and subjected to a torque (T)at the other end as shown in Figure 12.3.
Figure 123
Let T = Torque in kg cm 1 = Length of the shaft, and R = Radius of the shaft. A balancing torque of equal magnitude and opposite in direction will be induced at the fixed end. Let the line CA on the surface of the shaft be deformed to CA' and OA to OA' as shown in Figure 12.3. Let LAC$= (I and LAOA'= 8. As a result of the torque applied, every cross-section of the shaft will be subjected to shear stresses. Let f, = shear stress induced at the outermost surface, and C = modulus of the rigidity of the shaft material.
St~essesin~Lacts& Shells nrd Thermal Streaseo
We b o w that Shear strain = Deformation per unit length
- -AA'
- I
= tan$ = @ (@being very small) We also know that the length of the arc AA' = R 8 AA'
Re
/. @=7-- 1 Moreover, deformation =
Shear stress Modulus of Rigidity
Now from Eqs. (12.1) and (12.2). we find that
The shaft may be taken to consist of an infinite number of elemental hollow shafts, one surrounding the other. ' If the deformation of a line on the surface of any such interior cylinder, at a radius r be considered, the shear stress intensity 'q' at the radius 'r' is given by the relation,
Thus, it can be stated that the intensity of shear stress at any point in the cross-section of a shaft subjected to pure torsion is proportional to its distance from the centre. This means that the shear stress is maximum on the outside surface and variation of shear stress with radius is linear. C
12.2.2 Resisting Torque From conditiorls of equilibrium the external torque T must be balanced by resisting torque, i.e. by momerlts of tangential shearing stresses acting on any transverse section. Consider a solid circular shaft subjected to some torque. Let R = Radius of the shaft, and f, = Maximum shear stress developed in the outermost layer of the shaft material. Now, consider an elementary ring of thickness dx at a distance x from the centre as shown in Figure 12.4.
We know that the area of the ring, da = 2 luc dx
Shear stress at this section, (fJ would be as follows :
.-. Turning force = stress x area = f, x da
We b o w that turning moment of this element, or moment of resistance offered by the elemental ring, dT = Turning force x distance of the element from the axis of the shaft
:.
Total moment of resistance offered by the whole shaft is
R
where,
R
J=j2nr2dr=jda2 0
0
Here, J represents the moment of inertia of the shaft section about the axis of the shaft. The moment of inertia of a plane area, with respect to an axis perpendicular to the plane of the figure is called polar moment of inertia with respect to the point, where the axis intersects the plane. In a circular plane, the point is always the centre of the circle. Therefore, J is hown as polar moment of inertia, i.e. moment of inertia about ZZ axis. As per perpendicular axis theorem, I, = In + I, For a circular section,
lu =
x 1' = D4 64
where D is diameter of the circular shaft.
Torsion
S ~ w h S ~ & ~ r l h The term Pad Thennnl Strcrru
J is known as torsional section mddulus or polar modulus, denoted by 2, . R
It is similar to section modulus, 2 ,which is equal to
M j.
Thus, polar modulus for a solid sbaft
The resisting torque or torsional moment of resistance is given by,
This resisting torque is also known as strength of the shaft. Strength of the shaft is defined as the maximum torque or power the shaft can transmit from one pulley to another. Connecting the Eqs. (12.1), (12.2) and (12.3), we get
1
ar =Rf- ' J- -z -1 e
(12.4)
3 .
which is called the torsion equation.
i
E E-q. (12.4) cnn be compared with equation of bending I = y --R-' The expression corresponds to
r
II
M
C0
E
, --7correspondsto - and -corresponds to Y J I 1 R'
The expression CJ corresponds to expression EI. The term CJ is called torsional rigidity and the term EI is called flexural rigidity. It may be noted that : T (a) - ,i.e. torque required for unit twist, is called the torsional stiffness of the shaft. 0 T (b) -,i.e. torque divided by the angle of twist per unit length, is called the
an
torsional rigidity CJ.
12.2.3 Deformations in a Circular Shaft Consider a circular shaft which is attached to a fixed support at one end. If a torque T is applied to the other end, the shaft will twist with its free end rotating through an angle 0 called the angle bf twist. Observation shows that within a certain range of values of T, the angle of twist 0 is proportional to T. It also shows that 0 is proportional to the length of the shaft. In other words, the angle of twist for a shaft of the same material and same cross-section,but twice as long, will be twice as large under the same torque T. When a circular shaft is subjected to torsion, every cross-section remains plane and undistorted. In other words, while the various cross-sections along the shaft rotate through different amounts, each cross-section rotates as a solid rigid slab. When a bar of square cross-section is subjected to torsion, its various cross-sections are warped and do not remain plane. The fact that the cross-section of a circulac shaft remains plane and undistorted is due to the fact that a circular shaft is axisymmetric, i.e. its appearance remains the same when it is viewed from a fixed position and rotated about its axis through an arbitrary angle. Square bar, on the other hand, retains the same appearance only if they are rotated through 90" or 180". If all sections of the shaft, from one end to the other, are to remain plane and undistorted, we must make sure that the couples are applied in such a way that the ends of the shaft themselves remain plane and undistorted. This may be accompIished by applying the couples Tand T' to rigid plates, which are solidly attached to the ends of the shaft. We may then be sure that all sections will remain plane and undktorted when the loading is . applied, and that the resulting deformation will occur in a uniform fashion throughout the entire length of the shaft.
We shall now determine the distribution of shearing strains in a circular shaft of length L and radius R ,which has been twisted through an angle '8'. Detaching from the shaft a cylinder of radius 'h', we consider a small square element formed by two adjacent circles and two adjacent straight lines traced on the surface of the cylinder before any load is applied. As the shaft is subjected to a torsional load, the element deform into a rhombus as shown in Figure 12.5.
Figure 12.5
iI
The shearing strain 'g' in a given element is measured by the change in the angles formed by the sides of that element. Since the circles defining two of the sides of the element remain unchanged, the shearing strain 'g ' must be equal to the angle between line AB and A'B. As we discussed in the preceding section,
1
The above equation shows that the shearing strain 'g' at a given paint in a shaft subjected to torsion is proportional to the angle of twist 9. It also shows that 4 is proportional to the distance 'r ' from the axis of the shaft to the point under consideration. In other words, the shearing strain in a circular shaft varies linearly with the distance from the axis of the shaft. The shearing strain is maximum on the surface of the shaft, where r = R. Thus, we have, Eliminating 8 from above two equations, we may express the shearing strain + at a distance r from the axis of the shaft as
12.2.4 Stresses in a Circular Shaft
I
No particular stress-strain relationship has been assumed so far in the discussion of circular shafts in torsion. We shall now consider the case when the torque T is such that all shearing stresses in the shaft remain below the yield ~trength~f,.For all practical purposes, this means that the stresses in the shaft will remain below the proportional limit and below the elastic limit as well. Thus, Hooke's law will apply and there will be no permanent deformation.
Torsion
st~ssesin Sh* & Shela a d The& Stressrs
ASjxr Hooke's law for shearing stress and strain,
fr = where,
C$
.
C = modulus of rigidity or shear modulus of the material, and
fi
= shear stress at a radius r from the axis of the shaft.
We h o w that
Multiplying by Con both sides, we get,
where,& = shear stress at a radius R from the axis of the shaft. The equation obtained shows that as long as the yield strength (or proportional limit) is not exceeded in any part of a circular shaft, the shearing stress in the shaft varies linearly with the distance r from the axis of the shaft. Figure 12.6 shows the stress distribution in a solid circular shaft of radius R.
J From the discussion we had in the theory of torsion, we have, T = A R
or
/ . = ZJ R
and
fr = - r
T J
If F ! is expressed in N m,R or r in metres and J in m4, the resulting shearing stress will be
cxpreaaed In N/m2,that is, Pascal (Pa). Up to this pint, our analysis of stresses in a shaft has been limited to shearing stresses.
Thfs is due to the fact that the element we had selected was oriented in such a way that its faces were either parallel or perpendicular to the axis of ,theshaft.We know that the dotma1 stresses, shearing stresses or a combination of both may be found under the same loading conditimh, depending upon the orientation of the element which has been chosen. Consider the two elements a and b located on the surface of a circular shaft subjected to torsion, as shown in figure. Since the faces of element a are parallel and perpendicular to the axis of the shaft, the only stresses on the ele~nentwill be the shearing stresses given by
I, =
7T R, i.e. the element a is in pwe shear. On the other hand, the faces of elenlent 6,
which form arbitrary angles with the axis of the shaft, will be subjected to a combination of n o d and shearing stresses. TR \ We also note that all the stresses involved have the same magnitude, J'
I
Torsion
Ductile materials generally fail in shear. merefore, when subjected to torsion, a specimen made of a ductile material breaks along a plane perpendicular to its longitudinal axis. On the other hand, brittle materials are weaker in tension than in shear. Thus, when subjected to torsion, a specimen made of brittle material tends to break along surfaces which are perpendicular to the direction in which tension is maximum, i.e. along surfaces forming a 45"angle with the longitudinal axis of the specimen. Example 12.1 Find the torque which a shaft of 25 cm diameter can safely transmit, if the shear is not to exceed 460kg/cm2. Solutlon Diameter of shaft, D = 25 cm Maximum shear stress,f, = 460kg/cm2 Let Tis the torque transmitted by the shaft.
Example 12.2 A bar of magnesium alloy 28 mm in diameter was tested on a gauge length of 25 cm in tension and in torsion. A tensile load of 5 tonnes produced an extension of 0.4mrn and a torque of 1250kg cm produced a twist of 1.5 lo.
Determine (a) the Young's modulus, (b) the modulus of rigidity, (c) the bulk modulus, and (d) the Poisson's ratio for the material under test. Solution Diameter of bar, D = 28 mm = 2.8cm X Area of bar, A = (2.81~ = 6.1575 cm2 4
Length of bar, 1 = 25 cm Load on the bar, P = 5 t = 5000 kg Extension of the bar, 61= 0.4rnm = 0.04 cm Torque, T = 1250kg cm Angle of twist, 0 = 1.51°= 0.02635radian
Stresses in Shafts & Sk& and Thennal Stmses
Young's modulus for the alloy, E = 0.5075 x lo6kg/cm2.
C = 0.1965 x lo6kg/cm2 Modulus of rigidity for the alloy, C = 0.1965 x lo6 kg/cm2.
Poisson's ratio for the alloy, p = 0.29.
= 0.4058 x lo6 kg/cm2 Bulk modulus for the alloy, K = 0.4058 x lo6kg/cm2.
- -
12.3 POWER TRANSMISSION BY SHAFTS We have already discussed that tlie main purpose of a shaft is to transmit power from the shaft to another in factories and workshops. Power is the time rate of doing work.
Consider a shaft of radius R subjected to end couples which cause turning effect. Work done in one revolution = Work done by each force = Px21cR+Px21cR
:.
= 21c x 2PR
= 2nT (:. 2PR=T) If the shaft rotates at N-rpm, the work done per minute = 2 W ,
where, N = nunber of revolutions per minute (rpm), and
T = average torque in kg m.
Torsion
MKS Unit 1 hp = 75 kg d s e c = 4500 kg d m i n
Work done per minute = 2 I
M
where, N = number of revolutions per minute, and T = average torque in kg m. Since there are 4500 kg m per minute in one horsepower, Work done in kg mlmin Power, P = hP 4500
Angular displacement in radians = 2nN SI Unit In SI system, power (P) is measured in watts (W). 1W = 1 Joulefsec = 1 N mlsec = 60 N d m i n Work done per minute = 2 M where, N = number of revolutions per minute, and T = average torque in N m. Since there are 60 N d m i n in one watt, Work done in N d m i n watts Power, P =
60
=
To watts
r
where, o = angular displacement in radianslsec = -
):2 \
1
Design of Shafts The principal specifications to be met in the design of 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 specific speed.
We know that, Power, P = T o
where,f = frequency of the rotation, i.e. number of revolutions per second. The unit of frequency is thus 1 s-' and is called a Hertz (Hz).
:.
P = 2nf T
p =2m watts (in SI units) 60 p =2M hp (in MKS Units) 4500
After having determined the torque T from the above equations, and having selected the material to be used, the designer will carry the values of T and of the maximum allowable stress into the elastic torsion formula.
J The torsional section modulus -can be calculated from the above equation. R J Knowing 2 the diameter of the shaft can be easily calculated Thus, we get,
123.1 Stepped Shafts Sometimes a shaft, made up of differeht lengths having different cross-sectional areas, is required to transmit some torque (or horse power) from one pulley to andhet. For such a shaft, the torque transmitted by individual sections have to be calculated fust and minimum value of these torques will be the strength of such a shaft. Consider such a shaft as in Figure 12.8 which may also be considered as two shafts connected in series.
The total torque transmitted T in this case will be the same as the torque transmitted by each pottion. Let Tl = Torque transmitted by the shaft No. 1 L, = Length of the shaft No. 1
R, = li'adius of the shaft No. 1 8, = Angle of twist in the shaft No. 1
J1= Polar moment of inertia for the shaft No. 1 f,, = Maximum shear stress in the shaft No. 1
Similarly, T2 ,
,R2 ,q ,J2 and
fs2
are the corresponding values for the shafi No. 2.
The total torque, T = TI = T2
(12.5)
Also, the total twist 9 = twist in portion 1( 9 , )+ twist in portion 2 (02) i.e.,
9 = g1+e2
Now,
From Eq. (12.5).
T = & -J1= & - 52 lR1 zR2
Substituting values of J, and J2,
(12.6)
Torsion
Further, I
-T -C- 8 J -
I
8 = - TL
JC
From Eq. (12.6). we obtain, 8 =
!
I \
TLl TL, +CJ, CJ2
Therefore,
Jhample 123 Calculate the diameter of a solid shaft transmitting 150 kW at 25 rpm, if the maximum shear stress in the shaft is not to exceed 70 MPa. Compare this with the shaft delivering same power at 25000 rpm. Solution Power transmitted, P = 150 kW = 150 x lo3 watts Number of revolutions, N = 25 rpm
L
Since, 1 Pa = 1 N/m2and 1 Mega Pascal = lo6N/m2 = N / d Then, maximum shear stress,f, = 70 MPa = 70 ~ / m m ~
Let T be the torque transmitted in N m.
If N = 25000 rpm, then
D = 160.9 rnm p = -2 w 60
It is seen from this example that the size of the shaft is reduced very much if the power is transmitted at high speed. That is the reason for the modem tendency to use high speed machines, which results in considerable saving in the material cost. Example 12.4 A steel shaft transmits 105 kW at 160 rpm If the shaft is 100 mm diameter, find the torque on the shaft and the maximum shear stress induced. Find also the twist of the shaft in a length of 6 m Take C = 8 x lo4 N/mm2 Solution
P = 105 kW = 105 x lo3W N = 160rpm D = 100 mm 1 = 6 m = 6000 mrn
We know.
0 = 0.04786 radian = 2" 45' Example 195.5
Find the diameter of the shaft required to transmit 60 kW at 150 r.p.m., if the maximum torque is likely to exceed the mean torque by 25 % for a maximum permissible shear stress of 60 N1mm2. Find also the angle of twist for a length of 2.5 metres.
Take C = 8 x lo4 ~ l r n m ~ Solution
P = 60kW = 6 0 x 1 0 ~ ~ N = 150 rpm T,, = 1.25 T,,,,
Here,
Torsion
B '= 0.0507 radians
!
Example 12.6 Show that for a given maximum shear stress the minimum diameter required for a solid circular shaft to transmit P kW at N rpmcan be expressed as
What value of the maximum shear stress has been used if the constant equals 84.71, being in millimetres? Solutton We know,
where,
p = -2Rw
60
watts
Stmrm m Shafta & S h e l
a d Thermal St-
when K = 84.71, we obtain,
Example 12.7 The stepped steel shaft shown in Figure 12.9 is subjected to a torque (T) at the free end and a torque (2T) in the opposite direction at the junction of the two sizes. What is the total angle of twist at the Eree end, if maximum shear stress in the shaft is limited to 7 0 kg/cm2 ? Assume the modulus of rigidity to be 0.84x lo6 kg/cm2.
Toque at C = T (anticlockwise) Torque at B = 2T (clockwise) ~aximum shear stress,f, = 700 kg/cm2 Let 8 be the angle of twist at C
Free Body Diagram Let us first find out the value of torque Tat C. It may be noted that if the value of torque is obtained for the portion AB, it will induce more stress in the portion BC (because the portion BC is of less diameter). Therefore, we shall calculate the torque for the portion BC (because it will not induce more stress than the permissible in the portion AB). b
_
F l p r e 12.10 r
F m Body Magrsm
i
I
The angle of twist due to torque Tat C,
0 = 0.06248 radian The angle of twist due to torque of 2T at B ,
= 0.005radian The angle of twist due to torque at B, will continue to be the angle of twist at C also. Since the directions of the two twists are opposite to each other, therefore, the net angle of twist at C will be = 0.06248- 0.005 = 0.05748radian
xample 12.8 A solid shaft 6.5 m long is securely fixed at each end. A torque of 91 N m is applied to the shaft at a section 2.5 m from one end as shown in Figure 12.11.Find the fixing torques set up at the ends of the shaft.
Figure 1211
If the shaft is 35 mm diameter, find the maximum shear stresses the two portions. Find also the angle of twist for the section where the torque is applied. Take C = 8.4x lo4 ~ / r n r n ~ . Solutlon
Free Body Diagram Since the ends being fixed, the angle of twist in the length AC of the shaft must be equal to the angle of twist in the length BC.
Since J and C have the same value for the two portions. Thus, we get,
Torsion
S t l ~ r c e sin Sbaftq & Sheb nnd 1 h
e d Stresses
Rere 1212 r Free Body Magrata
Maximum shear stress in the porti011AC,
f, = 6.65 Nlmm2 Maximum shear stress in the portion CB,
9 = 0,0113 radiail = 0" 38'16"
Example 12.9 Toques are applied on the shaft as shown in Figure 12.13. Flnd out in which w o n of the shaft, maximum shear stress and angle of twist occur. Take c'=80 GPa.
Etpre 1213
SoIutim
Free Body Diggram
TR Maximum shear stress,.h = J
I
1-
Maximum shear stress in AB,fs =
Maximum shear stress in BC,fs =
TAB RM JAB
TBC
-RBC JBC
TCD Maximum shear stress inCD,fs = -RcD JCD
Maximum shear stress in DE,fs =
TDE RDE JDE
Maximum shear stress occurs in section CD, i.e. 6.036 N/mm2. C = 80GPa = 8 0 x 1 o 9 ~ a= 8 0 x 1 0 ~MPa = 8 0 x 1 0 ~N/mm2
-
CLOCK WISE
5 0 N.m C
B ANTICLOCKWISE
Y
750 N.m
CLOCKWISE
750N.m
ANTICLOCKWISE
M
500 N. m
500 N m
CLOCKWISE
Figure 1214 :Free Body Diagram
Angle of mist
50 lo' 32 =
~ B C = [ $ ] BC
1200
= 29.84 x lo4 rad (clockwise)
x ( 4 0 ) ~x 80 x 10'
750 lo' M X ) = 5.73 x lo4 rad (anticlockwise) -X(1~)4x8~x~~3 32
Torsion
Stresses in Shafts & Skelb and Thermal Stresses
l")
32 Thus,
lo'
= 0.72 x 10- rad (ciockwise)
x (75)4 x 80 x lo3
OB = 29.84 x
radian
Oc = 29.84 x
- 5.73 x
= 24.11 x
OD = 24.11 x
- 1.21 x
= 22.9 x
OE = 22.90 x 10-
+ 0.72 x loy4 = 23.62 x 10-
radian radian radian
Maximum twist occurs in portion AB and is equal to 29.84 x 10- radian, i.e. 0.171°. Example 12.10 A steel shaft of 200 rnrn diameter and 8 m long is fixed at its ends. Torques of 20 kN m (clockwise) and 30 kN m (anticlockwise)are applied at 3 m and 6 m respectively from one end as shown in Figure 12.15. Plot the variation of the torque, surface shear stress and angle of twist along the shaft length. Take C = 80 MPa. 20 KN.m
A
30KN.m
B
/
4
1
I
Figure 1215
Solution
Let TA and TBbe the reaction torques at A and B respectively (in kN m).
Free Body Diagram Here, TA = - 20 kN m (anticlockwise) TB =
+ 30 kN m
(clockwise)
Surface shear stress in AC, f, =
Surface shear stress in CD,f, = 44
-+A
= 0 (
:. TCD= 0)
J
2OKN.m (4NTl i l O T K W I S E l
2 0 KN.m (CLOCKWISE) C
D
3OKN,m (ANTELOCKWISE)
Figure 12.16 :Free Body Diagram
11
1 I
i
I
/
/
! a
30 KN.m (CLOCKWISE)
Surface shear stress is DB, f, =
['
-x R
Tor!iion
)m
Angle of twist, OAc = AC
!
Angle of twist, OcD = 0 Angle of twist ODB =
[']m
Note
If both the ends
fixed, the total angle of twist between the supports is zero.
Variation of torque, surface shear stress and angle of twist along the span of the shaft has been shown in Figure 12.17.
0 1
A
20
a
I
VARIATION OF TOROUE (711
I
I I
I
I
Lt,
!-
I
I I
11.7
130
I
I
11.7
I' I I
0.0¶
VARIATION OF SURFACE SHEAR 'STRESS
I
19.09
I
I
1 (fr
I I
VARIATION OF ANGLE OF TWIST ( 8 ) ( I N DEGREES)
Figure 12.17 :Variation of Torque, Surface Shear Stress and Angle of Twist
SAQ 2 A solid sl~aftmade of steel and of 2 m length is to transmit 50 kW at 150 rpm. If the shear stress in the shaft material is 11ot to excc2d 50 MPa and maximum allowable twist in the shaft is lo,calculate the shaft diameter.
Take C = 80 GPa.
Stresses in Shafts & She*
aod Thorn4 St1~occ8
SAQ 3 What must thc length of a 5 r r m dirtmeter alulninlum wire be so that it can be twisted through one complete revolution w~aioutexceedirlg a shearing stress of 42 ?d/rnrn2.Take C -- 2.7 x 1o4 ~llnrn*.
SAQ 4 Find the power that can he t-xnmitted by a shaft 60 nun dianleter at 180 rpn~,if riic permss~blsshear stress as X5 X;;!rlun-
.:$A
(b)
& , I
(c)
I---
a
---4
(dl tl
< t2
Fa+
-71 2
[p7
E2
k a - + Fa+ (e> tl > t2
+ ---I a
+'--I
Stress- in S b A & Shells
and Thermal Strevsea
Fully Plastic Torque (Tp)for various section codes (Table 12.3) are as follows : For (a)
T~ = fi.?(:+b-if)
[
\
1
TP = fsy 4 2[b-$]+$[a+~)-ilt:]
For (b)
For (d) t13 a t ~ + b t ~ -3 - - t ~ t l )
For (e)
12.7 SUMMARY (1) A shaft of circular section is said to be in pure torsion when it is subjected to equal and opposite end couples whose axes coincide with the axis of the shaft. (2) The product of the turning force and the distance between the point of application of the force and the axis of the shaft is known as torque, turning moment or twisting moment. (3) When a shaft is subjected to a torque, then
where, q = Intensity of shear stress on a layer, at a distance r from the centre of the shaft,
fs= 'Intensity of shear stress on the outermost layer of the shaft i.e. a distance R from the centre of Che shaft,
C = Modulus of rigidity of the shaft material, 0 = Angle (in radians) through which the cross-section of the shaft has been twisted as a.result of the torque, and
I = Length of the shaft. (4) Polar monlent of inertia denoted by J, is the moment of inertia of a plane area with respect to an a.xis perpendicular to the plane of the figure i.e. moment of inertia about Z axis. For shatt section, J is moment of inertia about the axis of the shaft. J =
It04 --
(for solid shaft)
32
)
k
where,
(for hollow shaft) J = : 32 (4'- D D = Diaiueter of the solid shaft, 11, = Intenlal diameter of the hollow shaft, and
D2 = Extenial diameter of the hollow shaft. ( 5 ) The ten11
J is known as torsional section modulus or polar modulus, Z,,,. R
Thus, polar modulus for n solid shaft.
7e03 5=1 6
Polar modulus for a hollow shaft. Zp =
1 6D2 (D; - D:
)
(6) Strength of the shaft is defined as the maximum torque or power the shaft can transmit from one pulley to another. It is also known as resisting torque or torsional moment of resistance.
For solid shaft. T =
Torsion
IC ig fs D'
For hollow shaft, T =
IC fs (D:- D;') 1602
where, fs = maximum shear stress at the outermost layer. The torsion equation is
(3
is called the torsional stiffness of the shaft.
The torque required for unit twist
.
?
T Torque divided by the angle of twist per unit length -is called the torsional 0/L rigidity. It is also equal to C x J, The shearing strain is maximum ($,,,) on the surface of the shaft where r = R.
The shearing strain @ at any distance r from the axis of the shaft as
9
=
, r
9max
Power transmitted by the shaft is as follows : p = - 2 M hp 4500
(in MKS Units)
p =2* watts (in SI units) 60 = T o watts
where,
o = -2rCN - - angular displacements in radiawsec.
60 N = Number of revolution per minute. When a thin tube of circular section of diameter D and thickness ' t ' subjected to a torque T, The shear stress,& =
~ Z R ~ T
and the angle of twist, 0 =
1
4 TL -
ICn3tc (14) For non circular cross-sections, the torque is equal to twice the volume between the stress function and the plane of the cross-section. T = 21 J @ d x d Y (15) For thin walled non-circular section shafts, T the shear stress.& = and
3
the angle of twist, 0 =
TL & T 4u C
t
where the integral is computed along the centre line of the wall section. n is the area bounded by the centre line of the wall cross-section and t is the thickness of the thin tube. (16) The Prandtl's membrane analogy is used for Inany torsion members, whose cross-sections are so complex that exact analytical solutions are difticult to obtain. (17) The Prandtl's niembrane analogy is based on the equivalence of following :
Str~ssesm S h d & SbeRe and Thennal Stresses
the torsion equation :
and, the membrane equation :
where, $ = the stress function,, Z = the lateral displacement due to lateral pressure p, and S = the surface tension. (18) When a torsion member made of an elastic-perfectly plastic material, subjected to a torque and if the entire cross-section becomes plastic at the limiting, then the member is said to be subjected to fully plastic torque.
12.8 ANSWERS TO SAQs SAQ 1
Torque based on shear stress,
Torque based on angle of twist,
The maximum torque then can be applied safely to the shaft is smaller of the above two values, i.e. 222.7 x lo6 N mm. SAQ 2 We know,
T = 3.183 x lo3N m = 3.183 x lo6N mm
Diameter of the shaft based on its strength, i.e. stress,
Diameter of the shaft based on its stiffness, i.e. angle of twist,
The required shaft diameter will be the larger of the above two values, i.e. 82.55 mm. SAQ 3
We have,
SAQ 4
We have,
T = -It f , ~ =~3605Nm 16
P = 67.95 kW SAQ 5
Power of the shaft, P = 45 kW = 45,000 W Let the torque between B and C of the shaft be TBC.
Also. Thus,
LBC= 25.94 ~ / m m ~
Similarly,
f,, = 58.36 ~ / m m ~
Hence, the greatest shear stress occurs in the 50 mm diameter shaft. Therefore, Maximum shear stress = 58.36 ~ / m r n ~ . Twist of the Shaft
Let OBc be the twist of the shaft portion BC,
Let OBA be the twist of the shaft BA.
:.
Angle of twist of A with respect to C = QBA+ QBc
0 = 0.10986 + 0.01626 = 0.12613 radian 0 = 7'14' SAQ 6
We have,
Maximum torque,
T = 3290 x lo3kg cm
Stwses m Shafb & Shells a d Thennal Stresses
SAQ 7
We have,
Therefore, T = Tm,
= 43.4 x lo6N mm
.
Tm, = 1.2Tm, = 52.08 x lo6 N mm We shall find out the diameter of the shaftboth for its strength and stiffness. For strength, We have,
D2 = 1 . 6 5 102mm ~ = 165mm
For stiffness,
T J
We have,
-
C8 1
Also,
Suitable diameter for the shaft is 298 mm, i.e. greater of the two values. SAQ 8 For strength,
For stiffness,
T - C8 J 1
We have,
Thus,
I
get,
16m2 .-~ f ,
32Tl -
RC8
We have,
p = - 2-
Torsion
60
Putting the value of T in following equation :
Thus, we get,
Dl = 120.9 mm
SAQ 9
We have,
T = 38.197 x 10' N m = 38.197 x lo6 N mm
Let D be the diameter of the solid shaft.
We have,
T --f s J* R*
D = 134.47 mm
Let D, and D, are the internal and external diameters of the hollow shaft.
.'.
Saving in weight =
(134.47)~- [(137.4212- (68.711~1 (134.47)~
= 21.7 % SAQ 10
Safe twisting moment, T =
7th 9 r --2
4TL Twist of the tube, 0 = nD3rc
= 0.016 radian = 0.917'
x 100
Stresses in Shafts & Shells and Thermal Stresses
96mm
SAQ 11
Tubing of Uniform Wall Thickness
-r(
The area bounded by the centre line. a = 96 mm x 56 mm = 5.376 x 1 0 - h 2 T shear stress in each wall,fr = 2ra
-
3x103~m
2x( 4x
m)x (5.376 x
m2
= 69.8 MPa
Tubing with Variable Wall Thickness Observing that the area a bounded by the centre line is the same as in the case of uniform wall thickness and substituting successively I = 3 mm and r = 5 mm. 3 x l d ~ m
fr*a = 'A'
= 2 x (3 x
m) (5.376 x
rn2)
= 93.0 MPa
and
Thus, we note that the stress in a given wall depends only upon its thickness.