Strength of Materials- Torsion of Circular Cross Section- Hani Aziz Ameen

Strength of Materials Handout No.5

Torsion of circular cross-section

Asst. Prof. Dr. Hani Aziz Ameen Technical College- Baghdad Dies and Tools Eng. Dept. E-mail:[email protected]

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Strength of materials- Handout No. 5 - Torsion of circular cross-section- Dr. Hani Aziz Ameen

5-1 Introduction Torsion is produced by a couple which consists of two forces equal in magnitude and opposite in direction and lies in a plane perpendicular to the longitudinal axis of the shaft as shown in fig (5-1)

Fig(5-1) Torque may by found in many applications like (1)In power station (all rotating shafts )

(2)

According to the action of twisting , shearing force will appear in each cross section of the shaft. The force are represented by stresses called torsion stress (shearing stress ) . The shearing stress is indicated on small element (complementary shear stress) as shown in Fig (5-2)

Fig(5-2)


5-2 Derivation of Torsion Formulae for Circular Cross-Section

In torsion derivation many assumptions are used :

1- Circular section remains circular after the torque application .

2- The material must be homogenous . 3- The plane cross-section remains as it is after the torque

application 4-

law G )

Gstrain shearstress shear ............ (5-1)

where G .....Modulus of rigidity Let us taken a bar subjected to torque, shown in Fig.(5-3)

Fig(5-3) From Fig(5-3) , it can be deduced that :

Arc ( AA )=R AOA ............. (5-2) Arc ( AA )=L ABA ............. (5-3)

From eqs (5-1), (5-2)&(5-3) , yields

LRG ........................ (5-4)

For a certain value of torque , = constant , L & G are constants , Hence , eq(5-4) , will be

KR ................................ (5-5)


where K=L

G

From eq (5-5)the distribution of shear stress through the cross section of the beam .

- a- -b- Fig(5-4)

0M 2o1o

0r dr r 2TR

01 ...... (5-6)

From similar triangles(Fig(5-4b)) , we have

Rr

1 .................... (5-7)

From eq(5-6)&eq(5-7) , we have

0rdr Rrr2T

R

0

R

0

3 0dr r R

2T

04

R R

2T4

...................(5-8)

TR.T

Form eq(5-8)and eq(5-4) ,gives

J G

L.T

where J is the polar moment of inertia


J=32D4

(for solid circular shaft)

J= shaftcircular hollowfor dD32

44

5-3 Representation of Torque on Shaft

There are many symbols to represent torque on the shaft as shown in Fig(5-5)

The start of the arrow

+ The end of the arrow Fig(5-5)


5-4 Torque Diagram To study the shaft which is subjected to the load shown in the Fig(5-6)

Fig(5-6) We must draw the torque diagram . Firstly identified the direction of the input torque i.e. ( ) as a clockwise direction and then the output torque ( ) as a counter clockwise direction as shown in Fig(5-6 b) . Secondly we draw a diagram with the specific scale as shown in Fig(5-6 c) . 5-5 Power Transmission by Shafts Rotating shaft are widely used for transmitting power . Power is defined as the work done per unit time. Work done by a torque acting on rotating shaft (Fig(5-7)) is equal to

W=T.

.Tt.T

tWP

where..... t

angular velocity


In shaft has N r.p.m

30N

60N2

hence, 30N*TP

Thus , T N.m r.p.mNwattP 30

TrpmNkWP1000*30m.N

Trpm NkW P9550.m.N .. (5-9)

The unit commonly used for the power is horsepower hp ,which is equivalent to

1hp = 745.7 Watt

.rpm Nhp P 7120N.mT . (5-10)

5-6 Calculation of Shaft Diameter

The shaft diameter can be found either from max .shear stress or from max . angle of twist. (a) maximum shear stress condition .

from eq(5-9) , N

P9550m.NT

max = w3D.T16

Jr.T

T16D 3w

D 3wN.

P 9550*16

D 3w

.NP5.36 m . (5-11)


b) Angle of twist condition

rad,J.GL.T

max , 32DJ

4

max4DGTL32 ,

4maxGL.T32D

180 ,

4max4 .GL.T92.4

180G

TL32D ,

We have from eq(5-9) N

P9550T

42 N.G..

L.P9550*180*32D

m GN

L.P09.49D 4 (5-12)

5 7 Composite Shafts ( a ) Series connection If two or more shafts of different materials diameters are connected together and each of them carries the same torque then the shafts are said

to be connected in series, see Fig.(5-8) Fig(5-8) In such case the composite shaft is treated by considering each component separately, applying the torsion theory to each in turn

2

222

1

111LJG

LJGT


2121 & TTT i.e.

(b) Parallel Connection If two or more materials are rigidly fixed together such that the applied torque is shared between them, the composite shaft so formed is said to be connected in parallel (see Fig.(5-9)).

Fig(5-9) BABA , TTT .e.i

In this case the angle of twist of each part is equal to each other and

2

2B

1

1AJ.GL.T

J.GL.T & T=TA+TB

5-8 Plane Bolt Coupling

A common connection for two shafts is the flange bolt coupling. Each flange is connected to the end of the shaft and both flanges connected to each other by means of bolts.

The transmitted torque occurs, in the bolts and will create a shearing force in each bolt. This connection can be clarified as shown in Fig.(5-10)


Fig(5-0) 0M0

F.R.n = T .................... (5-13) where n .. is the number of bolt Another equation can be obtained from the definition of stress

AF

F R n = T A R n = T (5-14)

In many cases, more than one bolt circles are used as shown in Fig(5-11) .

Fig(5-11) 0M0 T n F1R1 n F r = 0


From triangular geometry

RF

RF

1

1

& RR1

1

Or RARF 1

11 ................ (5-15)

Eqs (5-15) & (5-10) gives :

AAA if andA F putting ,RnARFRnT 211

21

RRRAnT

21 .. (5-16)

5 9 Springs The springs are used in many applications like

a- to absorb the energy b- to return the part of machine to its original position in the

case of reciprocating motion. There are several types of springs, according to the type of loading , springs are classified as :- 1 Tension spring 2 Compression spring 3 Torsion spring According to the nature of spring , the springs are classified as

1 Helical spring 2 Heal spring

Helical spring also can be classified :- a- Closed coil helical spring. b- Open coil helical spring.

5.9.1 Close Coiled Helical Springs For close coiled springs as in Fig(5-12) ... helix angle is small and it may be neglected


i.e. each turn may be considered to lie in a horizontal plane if the central axis of the spring is vertical.

Fig(5-12) Every cross section will be subjected to a torque F R tending to twist the section, so

2rR.r.F

Jr.T

4max

3max rFR2 .. (5-17)

5.9.2 Deflection of Spring Practically the deflection of spring is measured along longitudinal axis to determine the spring deflection, considering spring subject to either tensile or compression as shown in Fig(5-13) , for one cross section.

GJ

L.T

4

2

4 GrFR2

2rG

R.R.F

4

3

r GFR2R (for small )


Fig(5-13) Total deflection

nr G

FR4n2 4

3

where n ... is the number of coils 5 10 Stress Concentrations in Torsion

If the diameter of the shaft changes suddenly as in Fig(5-14) , a high stress concentration is developed at the transition point of the shaft. For circular cross section.

Jr.TK tmax

where Kt is the stress concentration factor .

Fig(5-14)


5-11 Longitudinal Shearing Stresses

If the shaft is subjected to external torque the shearing forces will be induced through any transverse cross section as shown in Fig(5-15)

Fig(5-15)

0Mo 0dx)rrd(rd )rdx(

It follows from the conditions of static equilibrium that the

shearing stresses acting on longitudinal sections are of the same distribution and intensity as the shearing stress acting on transverse sections .


5-12 Examples The following examples explain the differences ideas of the torsion of circular cross section problems. Example (5-1) Fig.(5-16) shows a solid steel shaft 40 mm in diameter is subjected to the torsional load , find the max. shear stress in the shaft

Fig(5-16) Solution

T = F * d T = 2* 0.5 = 1 kN.m

MPa 6.79)04.0(*

16*1Jr.T

3max

Example (5-2) Fig(5-17) shows a 50 mm diameter solid steel line shaft is used for power transmission purposes in a manufacturing plant. A motor inputs 100 hp to a pulley at A , which is transmitted by the shaft to pulleys at B and C and D . The output horsepower from pulleys located at B,C, & D is 45 hp , 25 hp and 30 hp respectively .

(a) Plot the torque diagram of the shaft (b) Find the max. shear stress in the shaft .

Fig(5-17)


Solution

The torque exerted on each gear is determined TA= 7120 * (100/300) = 2373.33 N.m TB = 7120 * (45/300) = 1068 N.m TC = 7120 * (25/300) = 593.33 N.m TD = 7120 * (30/300) = 712 N.m

(b) Since the shaft is of uniform cross section the maximum shear stress occurs in segment AC , thus ,

MPa18.53)05.0(

33.1305*16DT16

Jr.T

33max

Example (5-3) Fig(5-18) shows the hydraulic turbine, generates P = 30000 kW of electric power when rotating at N=250 rpm .Find the maximum shear stress in the tubular generator shaft with outside and inside diameter s indicated as shown .

Fig(5-18)


Solution

m.N114600250

300009550NP9550T

344

m0297.055.0

)3.0()55.0(16r

J

MPa5.380297.0

1146000Jr.T

max

Example (5-4) What must be the length of a 5 mm diameter aluminum wire so that it can be twisted through one complete revolution without exceeding a shearing stress of 42 MPa , take E=27 GPa Solution. Torque transmitted by the wire

m.N031.1)005.0(*16

*10*42D16

*T 363

41144 m10*136.6)005.0(32

D*32

J

We know that ,

m096.10LL

2*10*2710*136.6

031.1L

GJT 9

11

Example (5-5) A solid steel shaft has to transmit 75 kW at 200 rpm ,taking allowable shear stress as 70 MN/m2 . Find suitable diameter for the shaft, if the maximum torque transmitted on each revolution exceeds the mean by 30% . Solution

Tmax.=1.3Tmean

Nm3581TT1000*60

T*200*2751000*60NT2P mean

Tmax.= 1.3 * 3581 =4655.3 Nm Also,

DdD

16.T

44

max

4655.3 = 70 * 106/(16*D3) D3= ( 4655.3*16) /(70*106* ) = 338.7 *10 6 D = 0.069 m


Example(5-6) A solid circular shaft transmits 75 kW power at N=200 rpm . Find the shaft diameter, if the twist in the shaft is not to exceed 1° in 2 meters length of shaft and shear stress is limited to 50 MPa. Take G=100 GN/m2 Solution

m.N3581200*2

1000*60*75T1000*60

T*200*2751000*60NT2P

First case : considering allowable shear stress ( 50 MPa) ,

)D*16/(*10*503581D*16

*T 363

D= 0.0714 m = 71. 4 mm

Second case : considering angle of twist ( 1°) ,

2180/*1*10*100

)D*32/(3581

LG

TT 9

4P

D = 0.0804 m = 80.4 mm

From the two cases we find that suitable diameter for the shaft is 80.4 mm or say 80 mm ( i.e. greater of the two value) Example (5-7) A hollow shaft of diameter ratio 3/8 is required to transmit 600 kW at 110 rpm the maximum torque being 20% greater than the mean . The shear stress is not to exceed 63 MPa and the twist in a length of 3m not to exceed 1.4 degree , find the maximum external diameter satisfying these conditions . Take G = 84 GPa , d = 0.375 D ; Tmax=1.2 Tmean Solution

m.N52087T1000*60

T*110*26001000*60NT2P

T = Tmean = 52087 N.m Tmax = 1.27 Tmean = 1.2*52087=62504 N.m


First case : When shear stress is not to exceed 63 MPa

D2

RJT

DdD

16Tor

D2)dD(

32D2.JT

4444

62504 = 63*106* ( /16)*(((D4 (0.375D)4)/D)

D3 = (62504* 16) / (63*106* *0.9802)=5.155*10 3

D = 0.1727 m or 172.7 mm .................... (i) Second case : When angle of twist is not to exceed 1.4

180*3*4.1*10*84*)dD(

32LG*JT

LG

JT 9

44

62504 = ( /32) * ( D4 (0.375D)4 ) * ((84*104*1.4* )/(3*180)) D = 0.1755 m ............... (ii)

From eq(i) and eq(ii) we take a great of the two value D= 175.5 mm Example (5-8) A circular bar made of cast iron is to resist an occasional torque of 2.2 kN.m acting in transverse plane. If the allowable stresses in compression , tension and shear are 100 MPa , 35 MPa and 50 MPa respectively, find :

i) Diameter of the bar ii) Angle of twist under the applied torque permeter length of

bar Take G = 40 GPa ; T= 2.2 kN.m

MPa 35 , MPa 50 , MPa100 tc


Solution

i) Diameter of the bar Since cast iron is weakest in tension , it will fail due to tensile principle stress . Due to the tensile stress t , the maximum shear stress is also equal to t

allowabletallowablemax )()(

RJT max

i.e. )2/D(

10*35D*)32/(

1000*2.2 6

4

63

10*35*16*1000*2.2D

D = 0.0684 m ii) Angle of twist ,

46.1L

GRL

GR maxmax

Example(5-9) A hollow circular shaft 20mm thick transmits 294 kW at 200 rpm .Find the diameter of the shaft if shear strain due to torsion is not to exceed 8.6*10-4 . Take modulus of rigidity 80 GN/m2 Solution

DH dH = 2 t = 0.04m m0.04- D d or HH

P = Nm14037T1000*60

T*200*22941000*60NT2

64

H4H

H4

H4H

H

10*04.0DD

D8.71489

dD32

2D*14037

JTR or

RRT

2694 m/MN8.6810*10*80*10*6.8G*

68mm20*2108d

mm108m108.0D8.6810*04.0DD

D8.71489

H

H6

4H

4H

H


Example(5-10) A solid steel shaft is subjected to a torque of 45 kN.m . If the angle of twist is 0.5° per meter length of the shaft and the shear is not to be allowed to exceed 90MN/m2 Find :.

i) Suitable diameter for the shaft ii) Final maximum shear and angle of twist and iii) Maximum shear in the shaft.

Take G=80GN/m2 Solution

rad008727.0rad180

*5.05.0

i)Diameter of the shaft ,D :

59

310*445.6

008727.0*10*80L*10*45

GTLJ

LG

JT

m16.032*10*445.6D10*445.6D32

4/1554

3D16

TR

*JTRJ

T

mm5.136m1365.010*90*10*45*1616T D

3/1

6

33/1

Diameter of the shaft =160mm ii) Final maximum shear stress and angle of twist.

Since the diameter is given by the angle of twist, the final angle of twist is 0.5° per meter length, the maximum shear stress will be less than the given value of 90 MPa ,the final maximum shear stress is given by,

410*88.61

008727.0*2/16.0J

TRRJ

T MPa

Example(5-11) Two shafts of the same material and same length are subjected to the same torque. If the first shaft is of a solid circular section and the second shaft is of a hollow circular section ,whose internal diameter is 2/3 of the outside diameter and the maximum shear stress developed in each shaft is the same , compare the weights of the two shafts. Solution Ds = diameter of the solid shaft DH = external diameter of the hollow shaft dH = internal diameter of the hollow shaft = 2/3 DH


The torque transmitted by the solid shaft 3

ss D16

.T ..................( i)

The torque transmitted by the hollow shaft

3H

H

4

H4H

H

4H

4H

H D8165*

16.

D

D32D

16.

DdD

16.T ...(ii)

Since both the torques are equal , equate eq(i) and eq(ii) Ts = TH

SH3s

3H

3H

3s D08.1DD246.1DD

8165

16D

16.

H

s

HHH

sss

H

sAA

*L*A*L*A

WW

shaft hollow theofweigh shaft solid theof weigh

where :- . Weight density

Ls = LH s = H

543.1D08.1*

DD*3/2D

DdD

D2

s95

2s

2H

2H

2s

2H

2H4

2s4

Example (5-12) A solid cylindrical shaft is to transmit 30 kW at 100 rpm

i) If the shear stress is not to exceed 80 MPa , find its diameter.

ii) What percentage saving in weight would be obtained if this shaft is replaced by a hollow one whose internal diameter equals o.6 of the external diameter, the length material and maximum shear stress being the same ? Solution

i) Diameter of solid shaft, Ds :

Nm 28648T1000*60

T*100*23001000*60NT2P

3sD

16T ( Assuming T (Tmean) = Tmax )

28648 = 80*106 * 122mmor m122.0DD16 s

3s

ii) Percentage saving in weight


3s

H

4H

4H3

sH

4H

4H D

DD6.01Dor D

16.

DdD

16.

m 128.0D

10*824.11296.01D

H

33H

0768.0128.0*6.0dD6.0d HHH

1221.00768.0128.0

DdD

AA

L A L A

WW 22

2s4

2H

2H4

S

H

s ss

HHH

s

H

= 0.704

Percentage saving in weight = 6.29100*704.01100*1s

H

Example (5-13) A hollow steel shaft is made to replace a solid wrought iron shaft of the same external diameter, being 35 percent stronger than the iron , find what fraction of the outside diameter the internal diameter may be. Also, neglecting the coupling, find the percentage saving in weight by the substitution. Assuming that steel is 2 percent heavier than wrought iron. Solution

Ratio of internal diameter to external diameter of hollow shaft, H

HDd

Since the torque transmitted will be equal Ts = TH

H

4H

4H

163s D

dD.35.1D16

.

HsH

4H

4H3

H DDD

dD35.1D

4H

4H

4H

H

4H

4H4

H dD35.1DD

dD*35.1D

714.0Dd

35.135.0

DdD35.0d35.1

H

H

H

H4H

4H

41

Percentage saving in weight Area of cross section of the steel shaft

49.0*D4

714.01D4D

d1D4

dD4

2H

22H2

H

2H2

H2H

2H


Let be the weight of unit volume of wrought iron. Then 1.02 w is the weight of unit volume steel.

Weight of unit length of iron shaft = w*D4

4H

Weight of unit length of steel shaft = w02.1*49.0*D4

4H

w*D

w02.1*49.0*Dshaftiron of Weightshaft steel of Weight

2H4

4H4 = 0.5

Hence saving in weight = %50100*5.01 Example(5-14) A solid shaft of mild steel 200mm in diameter is to be replaced by hollow shaft of alloy steel for which the allowable shear stress is 22 percent greater. If the power to be transmitted is to be increased by 20 percent and the speeds of rotation increased by 6 percent , find the maximum internal diameter of the hollow shaft. The external diameter of the hollow shaft is to be 200mm. Solution We have : sHsHsH N06.1N , P2.1P , 22.1 Maximum internal diameter of hollow shaft, dH :

1000*60

TN2P&kW1000*60

TN2PkW1000*60NT2P HH

Hss

s

But ssHHssHH

sH TN2.1T*N*06.11000*60

TN2*2.11000*60

TN2P2.1P

TH = 1.132 Ts

ii... JT and

)i...( JT

RJT

22.0H

H

H

22.0s

s

s

Dividing eq(i) and eq(ii) we get

4

4H

4

s

s4s32s

4H

4H32s

H

s

sH

Hs

2.0132.1d2.08196.0

22.1D*T32.1

dD*TJ*T

J*T

mm7.103m1037.0d10*1554.12.0*132.1*8196.02.0d H4444

H


Example (5-15) Fig(5-19) shows the stepped steel shaft is subjected to a torque T at the free end and a torque of 2T in the opposite direction at the junction of two sizes . Find the total angle of twist at the free end, if the maximum shear stress in the shaft is limited to 70 MN/m2 , assume the modulus of rigidity to be 84 GN/m2 .

Fig(5-19) Solution The torque 2T at B is equivalent to two torques each of value T . Then BC is subject to a torque T at C and an opposite torque T at B while AB is also subjected to equal and opposite torque T at A and B For the length BC ; Torque T = T , L = 1.8 m , J = ( /32)*(0.05)4= 6.136*10 7 m4

1= angle of twist of C relative to B = (TL)/(GJ) = ( T*1.8)/(84*105*6.136*10 7) . (i) For the length AB Torque , T = T , L = 1.2 , J = ( /32)*(0.1)4= 9.817 * 10 6

2 = (T*1.2)/(84*109*9.817*10 6) . (ii) where 2 angle of twist of B relative to A

1 and 2 are in opposite direction. Hence c is the total angle of twist at C

c = 1 2 The maximum shear stress occurs in BC and its value is 70 MPa (given)

T/J = /R , T = J/R , T = (70*106*6.136*10 7)/0.025 = 1718.1 Nm

From eq(i) : 1 = (1718.1 * 1.8) / (84*109*6.136*10 7) = 0.06 rad From eq(ii) : 2 = (1718.1 * 1.2) / (84*109*9.817*10 6) = 0.0025 rad

c = 1 2 = 3.29 °


Example(5-16) Fig.(5-20) shows a steel shaft LMNP . If equal opposite torques are applied at the end of the shaft find the maximum permissible value of d1 for the maximum shearing stress in LM is not to exceed that in NP . If torque is 10 kN.m what is the total angles of twist ? Take : G = 80 GN/m2

Fig(5-20) Solution

For the shaft LM : 1.0

d1.016

*D

dD16

*T41

4

11

41

41

1

)d1.0(T*6.1

)d1.0(16*1.0*T

41

441

41

for the shaft NP : 33

333 )0875.0(

16*D16

*T

3)0875.0(*T*16

3

but 31

341

4 )0875.0(*T16

)d1.0(6.1

5544

1 10*3.310*7.61.0d d1 = 0.0758 m

Total angle of twist

JL

GT

GJTL

= (T/G) *[ (L1/J1)+(L2/J2)+(L3/J3)]

44449

3

0875.0*)32/(72.0

1.0*)32/(45.0

)0758.01.0(*)32/(36.0

10*8010*10

= (1/(8*106)) [ 54740.4 + 45836.6 + 125112.4]* (180/ ) degree = 1.616 ° or 1° 37


Example(5-17) Fig(5-21) shows a solid phosphor bronze shaft of 80 mm diameter is coupled to a hollow steel shaft of 80 mm outside diameter . The torque applied to the compound shaft develops a maximum shear stress of 40 MPa in the bronze shaft and a maximum shear stress of 72 MN/m2 in steel shaft . the length of steel shaft is 1 m and of bronze shaft is 1.2 m . Angle of twist for the steel shaft is not to exceed 1 ° , If Gsteel = 80 GN/m2 and Gbronze = 40 GN/m2 Find :

i) Internal diameter of the steel shaft ii) Total angle of twist for whole of the shaft

Fig(5-21) Solution

rad 01745.0180/*11bs i) Internal diameter of steel shaft , ds :

Torque transmitted by the phosphor bronze shaft

kNm02.4)10(*10*40*)08.0(*16

.D16T 363

b3bb

Torque transmitted , T = Tb = Ts = 4.02 kN m Internal diameter of the steel shaft on the basis of shear stress

ss

4s

4s

s *D

dD16

T

6

4s

43 10*72*

08.0)d(08.0

1610*02.4

or (0.08)4 (ds)4= (4.02*103*16*0.08)/( *72*106) = 2.2748*10 5 or (ds)4 = (0.08)4 2.2748*10 5 = 1.826*10 5 ds = 0.06537 m


Internal diameter of the steel shaft on the basis of angle of twist :

s = (Ts Ls)/(GsJs) 0.01745 = (4.02*103*1) / [80*109*( /32)* (0.084 (ds)4)] (0.084 (ds)4) = ( 4.02*103*1*32) / (0.01745*80*109* ) (ds)4=1.166*10 5 ds=58.43 mm

ii) Total angle of twist = s+ b = [(TsLs)/(GsJs)] + [(TbLb)/(GbJb)] = T [(Ls)/(GsJs)] + [(Lb)/(GbJb)] , since ( T = Ts = Tb )

494493

08.0*)32/(*10*402.1

}05843.008.0){32/(*10*80110*02.4

= 4.02*103[4.3449*10 6 + 7.46*10 6] = 0.0474 rad = 2.719° Example(5-18) A solid alloy shaft of 50 mm diameter is to be coupled in series with a hollow steel shaft of the same external diameter . If the angle of twist per unit length of the steel shaft is to be 70 percent of that of the alloy shaft, find the internal diameter of the steel shaft . Also find the speed at which the shaft should be driven to transmit 20 kW, if the allowable shearing stresses in alloy and steel are 56 MPa and 80MPa respectively . Take Gsteel = Galloy*2.25 Solution

i) Internal diameter of the hollow steel shaft , ds :

angle of twist per unit length of a shaft is given by

GJT

L

Ts / (GsJs) = 0.7 * [ Ta/(GaJa)]


(Ja/Js) = 0.7* (Gs/Ga)

[( /32) 0.054] / ( /32)(0.054 (ds)4)] = 0.7 * 2.25

0.054 (ds)4= (0.054)/(0.7*2.25) = 3.968*10 6

(ds)4=0.054 3.968*10 6=2.282*10 6 ds = 0.03887 m

Speed , N :

L

GR

for steel shaft : s

ss

s

sL

GR

.............(i)

and for alloy shaft : a

aa

a

aL

GR

............(ii)

Dividing eq(i) by eq(ii) , we get

a

a

s

s

a

s

a

s

a

s L*L

*RR*

GG = 2.25*1*0.7 = 1.575

as 575.1

2s MN/m 80 , MPa78.50575.1/80a

Torque : D)16/(*T 3

aaa = 50.78*( /16)*(0.05)3*106=1246 Nm Power transmitted , P=(2 NT)/(60*1000) 20 = (2 N * 1246) / (60*1000) N = (20*60*1000)/ (2* *1246) =153.3 rpm Example(5-19) Fig(5-22) shows a solid steel shaft 6 m long is securely fixed at each end. A torque of 1250 N.m , is applied to the shaft at a section 2.4 m from one end. find the fixing torques set up at the ends of the shaft . If the diameter of the shaft is 40mm what are the maximum shear stresses in the two portions? Find also the angle of twist for the section where the torque is applied . Modulus of rigidity = 84 GPa


Fig(5-22) Solution Angle of twist In this case 1= 2 (T1L1)/(GJ) = (T2L2)/(GJ) ............ (i) T1+T2 = 1250 ........... (ii) From eq(i) T2=(T1L1)/L2 = (T1*2.4)/3.6 T1 + (T1*2.4)/3.6 = 1250 T1 (1+0.667) = 1250 T1= 749.8 Nm T2 = 1250 749.8 = 500.2 Nm

= 1= 2= (749.8*2.4)/ (84*109*( /32)(0.04)4)=0.0852 rad = 4.88 degree Max. shear stress in the two portions ;

31

1 DT16 = ( 16*749.8)*10 6 / ( *0.043) = 59.66 MPa

32

2 DT16 = ( 16*500.2)*10 6 / ( *0.043) = 39.8 MPa

Example(5-20) Fig(5-23) shows a hollow shaft is 1 m long and has external diameter of 50 mm . It has 20 mm internal diameter for a part of length and 30mm internal diameter for the rest of the length . If the maximum shear stress is not to exceed 80 MPa , find the maximum power transmitted by it at a speed of 300 rpm if the twists produced in the two portions of the shaft are equal . Find the lengths of the two portions .


Fig(5-23) Solution To find the maximum power transmitted , P Torque for part ( 1 )

Nm170905.0

03.005.016

*10*805

03.005.016

T44

644

1

Torque for part ( 2 )

Nm191305.0

02.005.016

*10*80T44

62

The safe torque which the shaft can transmit = 1709 Nm thus, power transmitted P = ( 2 N T ) / ( 60* 1000 ) = ( 2* * 300 * 1709) / (60*1000) =53.7kW Length of two portion , L1, L2

T/J = G / L , = TL / GJ As per given condition is same for both parts T1L1/G1J1 = T2 L2 / G2 J2 L1 J2 = L2J1 Since T and G are same for both the part L1 * /32 [ (0.05)4 (0.02)4]=L2* /32 [ (0.05)4 (0.03)4] L1*6.09*10 9 = L2*5.44*10 6 L1+L2 = 1 L2 = 1 L1 6.09*10 6*L1=5.44*10 6(1 L1) L1 = 5.44 / 11.53 = 0.472 m L2= 1 0.472 = 0.528 m

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Strength of Materials- Torsion of Circular Cross Section- Hani Aziz Ameen