Strength of Materials by R.K.Bansal_text

674 STRENGTH OF MATERIALS torsion of shafts and springs

T = Sx~C* 0 .-. Turning force on the elementary circular ring
= Shear stress acting on the ring x Area of ring
Now for a given shaft subjected to a given torque (D, the values of C, 9 and L are
= q x dA
constant. Hence shear stress produced is proportional to the radius R.

—t « R x “- x x x 2 jirdr q = xx

or R = constant ...(Hi) j

VIf q is the shear stress induced at a radius from the centre of the shaft then = x 2nr 2 dr

...(16.2) R
Now turning moment due to the turning force on the elementary ring,
— —But
x ce dT = Turning force on the ring x Distance of the ring from the axis

R = from equation (16.1) = R x 2nr2dr x r
Li

x CQ q ...(16.3)

RL r = jj;x 2nr>dr ...[16.3 (A)]

From equation {iii), it is clear that shear stress at any point in the shaft is proportional

to the distance of the point from the axis of the shaft. Hence the shear stress is maximum at The total turning moment (or total torque) is obtained by integrating the above

the outer surface and shear stress is zero at the axis of the shaft. equation between the limits O and R

16.2.1. Assumptions Made in the Derivation of Shear Stress Produced in a rR rR

dT =
—T = x x 2nr 3o dr

Circular Shaft Subjected to Torsion. The derivation of shear stress produced in a circular R\

Jo
Jo

shaft subjected to torsion, is based on the following assumptions :

1. The material of the shaft is uniform throughout. xr

2. The twist along the shaft is uniform. Tr x2k

3. The shaft is of uniform circular section throughout. —x R4 xx-xR3

4. Cross-sections of the shaft, which are plane before twist remain plain after twist. R x 2 rex 4

5. All radii which are straight before twist remain straight after twist. = TX-Tt X

16.3. MAXIMUM TORQUE TRANSMITTED BY A CIRCULAR SOLID SHAFT

The maximum torque transmitted by a circular solid shaft, is obtained from the maxi- -XX - x ...(16.4)

mum shear stress induced at the outer surface of the solid shaft. Consider a shaft subjected to 28

a torque T as shown in Fig. 16.3. A mmProblem 16.1. solid shaft of 150 diameter is used to transmit torque. Find the

Let x = Maximum shear stress induced at the outer surface maximum torque transmitted by the shaft if the maximum shear stress induced to the shaft is
mm45 N!
R = Radius of the shaft 2

.

Vq = Shear stress at a radius from the centre. Sol. Given :

VConsider an elementary circular ring of thickness 'dr' at a distance from the centre as mmDiameter of the shaft, D = 150

shown in Fig. 16.3. Then the area of the ring, Maximum shear stress, x = 45 N/mm!

cLA - 2jtrdr 's\ Let T = Maximum torque transmitted by the shaft.

From equation (16.2), we have / ^777777r^^y\ Using equation (16.4), ~T= xD3 = x45 x 1503
16 16
= 29820586 N-mm = 29820.586 N-m. Ans.

Shear stress at the radius r. N/mmProblem 16.2. The shearing stress is a solid shaft is not to exceed 40 2 when the

q= R r = X R torque transmitted is 20000 N-m. Determine the minimum diameter of the shaft.

Sol. Given :

Maximum shear stress, x = 40 N/mm2

Fig. 16.3 Torque transmitted, T = 20000 N-m = 20000 x 103 N-mm
Let D = Minimum diameter of the shaft in mm.

r TORSION OF SHAFTS AND SPRINGS

STRENGTH OF MATERIALS

Using equation (16.4), T \Ro -R?

16 x 20000 x 10 J 136.2 mm. Ans. * f Rl~Ri4

'=(fT

16.4. TORQUE TRANSMITTED BY A HOLLOW CIRCULAR SHAFTS Let D = Outer diameter of the shaft
0

D = Inner diameter of the shaft.
;

Torque transmitted by a hollow circular shaft is obtained in the same way as for a solid tfn=— and R.= ^L.

... shaft. Consider a hollow shaft. Let it is subjected to a torque T as shown in Fig. 16.4. Take an Then 02

elementary circular ring of thickness ‘dr’ at a distance r from the centre as shown in Fig. 16.4. 12

• R RSubstituting the values of
... and in equation (16.5),

RLet 0 = Outer radius of the shaft 0 t
R - Inner radius of the shaft
/ V Al
i \ It ” 16 16
\ T
r = Radius of elementary circular ring 2

f dr - Thickness of the ring Da
| Mx = Maximum shear stress induced at outer
/
» V J Msurface of the shaft
1 |

\ j

J ADo* - 4
\ /
q = Shear stress induced on the elementary ring 16

dA = Area of the elementary circular ring

= 2nr x dr .*
Fig. 16.4. Hollow shaft.
, . . trom
Shear stress at is obtained equa-
the elementary ring

tion (16.2) as

— -SL RR( v Here outer radius - 0) 16.5. POWER TRANSMITTED BY SHAFTS

Ro r

q ~ ~p~ x r —x Once the expression for torque (T) for a solid or a hollow shaft is obtained, power trans-
tig mitted by the shafts can be determined.
a=
.-. Turning force on the ring = Stress x Area = q x.dA * NLet = r.p.m. of the shaft

r x 2nrdr T = Mean torque transmitted in N-m

Then to = Angular speed of shaft. ...(16.7)

= 2it TT“ r2 dr ———P„ower = 2kNT* watts
60

Turning moment (dT) on the ring, = oixT I" -6o""“J
dT = Turning force x Distance of the ring from centre
= T x u>
...[16.7 (A)]

TX Problem 16.3. In a hollow circular shaft of outer and inner diameters of 20 cm and
r2dr x r = 2n -jr- r3 dr
= 2ji

Kq mm10 cm respectively, the shear stress is not to exceed 40 N/ 2 Find the maximum torque which

.

The total turning moment (or total torque T) is obtained by integrating the above equa- the shaft can safely transmit.

Rtion between the limits R and 0. Sol. Given :
t

—fRQ <*/?0 3 mmDOuter diameter, 0 = 20 cm = 200
dT = 22x r
Rj dr D mmInner diameter,

JR,
JRR. 0 = 10 cm = 100
;

i

—X r 3 dir Maximum shear stress, t = 40 N/mm2

-: 2ir Let T - Maximum torque transmitted by the shaft.

(-.• x and R„ are constant and can be taken outside the integral) The torque T obtained by this formula is the average (or mean) torque.

D

678 STRENGTH OF MATERIALS TORSION OF SHAFTS AND SPRINGS 679

Wrr r 3 i'rn\ ys
Using equation (16.6), — —D= d 3 D D
0 0 0
[81 J
A AJT 4 - = = 0.929
{8lJ
A
WNow weight of solid shaft, = Weight density x Volume of solid shaft

—= x 40 l6xl° ~ lx = w x Area of cross-section x Length

= 58904860 Nmm = w x ~ D2 x L
4
16 200 Weight of hollow shaft,

[ Wh = w x Area of cross-section of hollow shaft x Length

= 58904.86 Nm. Ans.

Problem 16.4. Two shafts of the same material and ofsame lengths are subjected to the wxj= w x D L[D 2 - (2/3
same torque, if the first shaft is of a solid circular section and the second shaft is of hollow - xL = 0
circular section, whose internal diameter is 2/3 of the outside diameter and the maximum shear j [D 2 Df\ 2 x
0) ]
stress developed in each shaft is the same, compare the weights of the shafts. 0

(AMIE, Summer 1989) *fna 31

Dividing equation (iv) by equation (i>), x LT — w x -7 x
D2 2*L
4f o
Sol. Given :
~Dw L, ir x 2 x
Two shafts of the same material and same lengths (one is solid and other is hollow)
transmit the same torque and develops the same maximum stress. Wr*" wx x a xL ”5A

Let T = Torque transmitted by each shaft J | j0o

x - Max. shear stress developed in each shaft 2 [v D D= 0.929 from equation (iii)]

D = Outer diameter of the solid shaft _ 9 .. (0.929£>0 )
D0 = Outer diameter of the hollow shaft
DD. = Inner diameter of the hollow shaft = -| 0 0

W = Weight of the solid shaft ^= - x 0.9292 x =
s
Weight of solid shaft = 1.55 Ans.
WA = Weight of the hollow shaft .

L = Length of each shaft Weight of hollow shaft 1

w = Weight density of the material of each shaft. AProblem 16.5. solid circular shaft and a hollow circular shaft whose inside diameter

Torque transmitted by the solid shaft is given by equation (16.4) |Co of the outside diameter, are of the same material, of equal lengths and are required to

—T= x 3 fc,
16
Torque transmitted by the hollow shaft is given by equation (16.6), transmit a given torque. Compare the weights of these two shafts if the maximum shear stress

4 -A 4 D Dn 4 0r developed in the two shafts are equal. (AMIE, Winter 1988)
K 0 - (2/3
A,
X Sol. Given :
16 Dn

A4 - ~ A n_ 65 Z?0 —g
16
81 x ZL Dia. of hollow shaft, D, = Dia. at outside
41

T* |A= = 0.75A

— A= 3 Let L = Length of both shaft (equal length),
it 65
xx T = Torque transmitted by each shaft (equal torque),
...(h) x - Maximum shear stress developed in each shaft (equal max. shear stress),

16 81 D = Dia. of solid shaft,

As torque transmitted by solid and hollow shafts are equal, hence equating equations W = Weight of solid shaft, and
s
(i) and (it),
Wh = Weight of hollow shaft.
— — — A™ xDn3s = Jt x x 65 n 3
Torque transmitted by a solid shaft is given by equation (16.4) as
—71
; T=™xtxD3 ...ft)

Cancelling x to both sides
lb

Aor = ||a3

A0

680 STRENGTH OF MATERIALS TORSION OF SHAFTS AND SPRINGS 681

Torque transmitted by a hollow shaft is given by equation (16.6) as Speed, N = 200 r.p.m.

— U—a a-Tm= * XTX r 4- 44 * Fa A)4 -(Q-75 4 Shear strain, <]> = 0.00086
1
d~1i6e
Modulus of rigidity, C = 0.8 x 10 a N/mm2

[ D
0
~ x t x -=f- [1 - (0.75) 4] = Let = External dia. of shaft and

DX X X 3 X [1 - 0.3164] Then A = Internal dia. of shaft
n

D D D0 =
+ 2t = + 2 x 20

i t

= x x x Dr? x 0.6836 AD, = - 40

16 Using equation (16.7),

But torque transmitted by solid shaft 2ji x 200 x T
T(or x ai) or 300000 =
= Torque transmitted by hollow shaft.
60 60
Hence equating equations <i) and (ii), we get
T = 300000 x 60 , 14323 9 Nm
D D DDx x x 2jt x 200
= 14323.9 x 1000 Nmm = 14323900 Nmm.
16
3= xxx 3 x 0.6836 or 3 = 0.6836 3
a
16

D D A=
(0.6836) 1/3 x J5 or = 0.8809

0

Now weight of solid shaft, — —G„ = Shear stress
Shear strain
W = p x g x Volume of solid shaft , we k, now,

Also

^pxfx D2 xl _ Shear stress
0.,8 x 10 s =
Weight of hollow shaft,
0.00086
Wh — P x S x Volume of hollow shaft
Shear stress (x) = 0.8 x 10 s x 0.00086 = 68.8 N/mm2

Now using equation (16.6),

= p X g X £jCD02 - D?)xL ID02 - 2 L (A 4 4
D0
= p xg x ^ (O.75D ) ] x -Ar= it )
0 XtX
16

= pxgxj D[ 2- D0.5625 2 xL = pxgx-~x D 2 - L (A 4 4
0] )
( Unq
1 0.5625) x — -A14323900 = it

lb x 68.8 x

= p x g x — x D 2 x 0.4375 x L A14323900 x 16 x _ D i p*

it x 68.6 0'

Dividing equation (iv) by equation (u), A A A A1060334.6

(A (A D= 4 - 4 = 2- 2
+2 2
pxgx—xDxL ) )
i
\y
2 DSubstituting the value of from equation (i) into the above equation, we get
i
j-j

ADp x g x - D D1060334.6 2 + (D - 40)2 ][Do2 - (D - 40)2 ]
°'4375 0=[ 0
2 x 0.4375 x L 0
0
[D 2 D 2 - 80D ID 2- D 2 - 1600 + 8OD0 ]
= + + 1600 0

AD= (°'8809 °l D D[ -• From (Hi), = 0.8809 0) = (2D 2 + 1600 - 80AX80A - 1600)
0.4375
= 2(D 2 + 800 - 40D 0 )80(A - 20)

D2 = 1, .„7„73„7„. = 2 - 40D + 800XA - 20)
0
= 0.776 0 . 160(A

D0.4375 2 Ans.
0
1060334.6 = {D 2 _ + gO0)(£) - 20)
40jDo
mm kWAProblem 16.6. hollow circular shaft 20 o
thick transmits 300 power at 200 r.p.m.
A A A A A A6627 = 3 - 20 2 ~ 40 2 + 800 + 800 - 16000
Determine the external diameter of the shaft if the shear strain due to torsion is not to exceed

mm0.00086. Take modulus of rigidity = 0.8 x 10s N/ 2 A A= 3- - 16000
60D 2 + 1600
. 0

(AMIE, Summer 1989 Converted to S.I. units) A3 - 60A2 + 1600A - 6627D - 16000 = 0
0

Sol. Given : t = 20 mm A3 -6°a2 - 16000 =
Thickness,
Power transmitted, WP = 300 kW = 300,000 The equation (ii) is solved by trial and error method.

682 STRENGTH OF MATERIALS TORSION OF SHAFTS AND SPRINGS 683

D(£) Let q = 100 mm. Problem 16.8. Find the maximum shear stress induced in a solid circular shaft of

DSubstituting this value of in the L.H.S. of equation (ii), we get diameter 15 cm when the shaft transmits 150 kW power at 180 r.p.m.

Q Sol. Given :
Diameter of shaft,
L.H.S. = 100 s - 60 x 1002 - 5027 x 100 - 16000 Power transmitted, D = 15 cm = 150 mm
Speed of shaft,
= 1000000 - 600000 - 502700 - 16000 = 1000000 - 1118700 = - 118700 WP = 150 kW = 150 x 103

mmD(ii) Let N = 180 r.p.m.
= 110

0

Substituting this value in the L.H.S. of equation (it), we get Let t = Maximum shear stress induced in the shaft

L.H.S. = 1103 - 60 x 1102 - 5027 x 110 - 16000 Power transmitted is given by equation (16.7) as

= 1331000 - 726000 - 552970 - 16000 = 1331000 - 1294970 = 36030 _ 2jiNT

DWhen = 100 mm, the L.H.S. of equation (ii), is negative but when D = 110 mm, the 60
Q
0 2ir x 180 x T
D DL.H.S. is positive. Hence the value of is more
lies between 100 and 110 mm. The value of 150 x 10 3 =
0 60
0
mmnearer to 110
as 36030 is less than 118700.

D(tit) Let 0 = 108 mm. 150 x 10 3 x 6( = 7957.7 Nm = 7957700 Nmm

Substituting this value in the L.H.S. of equation (ii), we get 2ji x 180

L.H.S. - 108 3 - 6 x 1082 - 5027 x 108 - 16000 Now using equation (16.4) as,

= 1259910 - 699840 - 542916 - 16000 = 1259910 - 1258716 = 1194

DThe value of 0 will be slightly less than 108 mm, which may be taken as 107.5 mm. Ans. —7957700 =

mmProblem A16.7. hollow shaft of external diameter 120 transmits 300 kW power at x t x 150 3

200 r.p.m. Determine the maximum internal diameter if the maximum stress in the shaft is not 16

mmto exceed 60 N! 2 (AMIE, Summer 1990) —16 x 7957700
N/mmr = s 2
.
it x 150 3 .
= 12 Ans.

Sol. Given : D = 120 mm A kWProblem 16.9. solid cylindrical shaft is to transmit 300 power at 100 r.p.m.
External dia.. 0
Power, N/mm(a) If the shear stress is not to exceed 80 2
Speed, WP = 300 kW = 300,000 find its diameter.
,

N - 200 r.p.m. (b ) What percent saving in weight would be obtained if this shaft is replaced by a hollow

one whose internal diameter equals to 0.6 of the external diameter, the length, the material and

Max. shear stress, t = 60 N/mm2 maximum shear stress being the same ? (AMIE, Winter 1983)

DLet ( - Internal dia. of shaft Sol. Given :
Power,
Using equation (16.7), Speed, WP = 300 kW = 300 x 103
N= 100
0I 3OD.0002n«x52^00|x ^T
Max. shear stress, t = 80 N/mm 2

mor= _qo_1 6Q ==114433233.99Nm (a) Let D - Dia. of solid shaft
2ji x 200
= 14323.9 x 1000 Nmm = 14323900 Nmm Using equation (16.7),

Now using equation (16.6), 60

^—r= * X -A4 4 2ji x 100 x T
l6
(On ) 300 x 10 3 =
60
dT~
300 x 10 3 x 60
~ a14323900 = 4 : 28647.8 Nm = 28647800 Nmm
k 4 - 2jt x 100
)
—(120 Now using equation (16.4),
x 60 x

14323900 x 16 x 120 = 1204 - D; 4 —xxx D3 or 28647800 = Dx 80 x s
16
jt x 60 1

145902000 = 207360000 - D, 4 16 x 2nn8r6.47800 \ 1/3

j
Dd = 207360000 - 145902000 = 61458000 mm, =_
| 121.8

D = (61458000) 1/4 = 88.5 mm. Ans. [ n x 80 J
{
= say 122.0 mm. Ans.

STRENGTH OF MATERIALS TORSION OF SHAFTS AND SPRINGS

(6) Percent saving in weight

Let Dq = External dia. of hollow shaft 14884 -(16364 -5898)

D - Internal dia. of hollow shaft 122 2 14884 * 100
l
D= 0.6 x 0. (given) = 14884 - 10486 x 100 = 29.55%. Ans.

The length, material and maximum shear stress in solid and hollow shafts are given the 1r4T8r8Q4.

same. Hence torque transmitted by solid shaft is equal to the torque transmitted by hollow kWProblem 16.10. A solid steel shaft has to transmit 75 at 200 r.p.m. Taking allowable

shaft. But the torque transmitted by hollow shaft is given by equation (16.6). shear stress as 70 N/mm2 find suitable diameter for the shaft, if the maximum torque trans-
,

mitted at each revolution exceeds the mean by 30%. (AMIE, Summer 1978)

.-. Using equation (16.6), Sol. Given :
Power transmitted,
m* (A 4 -A4 > R.P.M, of the shaft, WP =75 kW = 75 x 103
N = 200

— ^ A)* x 8-0n0n x [A4 -(0.6 4 Shear stress, t = 70 N/mm2

] A(v = °- 6 A)

[A= x x 50 x 4 -(0-6 fl0 ) I Let T - Mean torque transmitted

Tmax = Maximum torque transmitted - 1.3 T
D = Suitable diameter of the shaft

But torque transmitted by solid shaft Power is given by the relation,

= 28647800 Nmm 2nNT

.. Equating the two torques, we get D= n x 50 x 0.8704 r; ——60
X 200 X T
A4
or 75 x 10 3 ==
( 0.8704 60
28647800 = it x 50 x
10 3
— — Nm Nmm_75Sx x 60 =

T=
2ji x 72707707 3580.98 = 3580980

28647800 = 127.6 mm = say 128 mm Tmax = 1-3 T= 1.3 X 3580980 = 4655274 Nmm.

n x 50 x 0.8704 Maximum torque transmitted by a solid shaft is given by equation (16.4) as.

mmZ1= =
Internal dia. 0.6 xfl = 0.6 x 128 76.8
0
Now let W = Weight of solid shaft,
s
Similarly
Wh = Weight of hollow shaft. —or 4655274 = Dx 70 x s
W = Weight density x Area of solid shaft x Length 16
s

= w x — D2 x L (where w = weight density) — mmV/3
D_ f 16 x 4655274
= = 69.57 ~ 70 mm. Ans.
4 :

Wh = Weight density x Area of hollow shaft x Length ^ a x 70 )

kWProblem 16.11. A hollow shaft is to transmit 300 power at 80 r.p.m. If the shear

= w x j [D02 - Df] x L stress is not to exceed 60 N/mm 2 and the internal diameter is 0.6 of the external diameter, find

(v Both shafts are of same lengths and of same material) the external and internal diameters assuming that the maximum torque is 1.4 times the mean.

Now percent saving in weight (AMIE, Winter 1986)

W, - W,. Sol. Given : WP = 300 kW = 300 x 103
Power transmitted,
Speed of the shaft, N = 80 r.p.m.

Awx-D 2 x L - w x — [A2 " ^2 Maximum shear stress, x = 60 N/mm2
1x

44 Internal diameter, Di = 0.6 x External diameter = 0.6 D0

w x — D2 x L Maximum torque, ^' ~ L4 times the mean torque = 1.4 x T

max

D A2 - (D0 2 - 2 Cancelling wx — xL Power is given by the relation,

) 4

P„ ~ 2-kNT

60

—

STRENGTH OF MATERIALS TORSION OF SHAFTS AND SPRINGS 687
686

60xP 60x300x101 4

2ixN 2ji x 80 0.12967),>0
r = = =358og8Nm ji T_ D ~- 1 = n T X °l.8704D 3
16
n 16 0

Nmp = 1.4 T = 1.4 x 35809.8 Nra = 50133.7 = 50133700 Nmm. j

Now maximum torque transmitted by a hollow shaft is given by equation (16.6) as, Since torque transmitted is the same and hence equating equations (i) and (ii )

—Tmax = 16 x x x ^Tft 3 =^tx°. 8 7°4 Dl
0

D D= (0.8704)wa 0 = 0.9548D0 .

—50133700 = x 60 x -(Q. 6£>0 ) 4 Area of solid shaft = ^ D2 = D(0.9548D 0 )2 = 0.716 2
16 Do 0

Area of hollow shaft ^ D ~=
D3 [D 2 - 2= [D 2 - (0.6D0 ) 2 ]
^X 60 X .8704 0 ]
0
;

|= [D 2 - 0.36D Z x 0.64D 2 = 0.502£>02 .

0 J = ^ 0

16 x 50133700 169.2 ~ 170 mm. Ans. For the shafts of the same material, the weight of the shafts is proportional to the areas.
7i x 60 x .8704
Saving in material = Saving in area
Di x 170 = 102 mm. Ans.
an(i = 0.6 x J9 = 0.6
0
_ Area of solid shaft - Area of hollow shaft
Problem 16.12. A hollow shaft, having an inside diameter 60% of its outer diameter, is Area of solid shaft

to replace a solid shaft transmitting the same power at the same speed. Calculate the percent-

age saving in material, if the material to he used is also the same. = 0.716ZL 2 - 0.502DS2-

Sol. Given : 0.716I> 2 = 0.2988.

DLet 0 = Outer diameter of the hollow shaft 0
B D= Inside diameter of the hollow shaft = 60% of g
Percentage saving in material = 0.2988 x 100 = 29.88. Ans.
i

D = Diameter of the solid shaft 16.6. EXPRESSION FOR TORQUE IN TERMS OF POLAR MOMENT OF INERTIA
P = Power transmitted by hollow shaft or by solid shaft
Polar moment of inertia of a plane area is defined as the moment of inertia of the area
N - Speed of each shaft about an axis perpendicular to the plane of the figure and passing through the C.G. of the
area. It is denoted by symbol J.
t = Maximum shear stress induced in each shaft. Since material of both shafts
The torque in terms of polar moment of inertia (J) is obtained from equation 116 3 (All of
is same and hence shear stress will be same.
Power by solid shaft or hollow shaft is given by Art. 16.3.

—2nNT The moment (dT) on the circular ring is given by equation [16.3 (A)] as

p — ~ ~dT = ri 2nr3dr = K 2nr x r2dr = R r2 x 2nr x dr

60 = R^dA (v dA = 2jtr dr see Fig. 16.3)

60 xP Total torque, ~T = f dT = f 2 dA = f^r 2 dA
r
NT = , .. = constant Jo Jo
23t R R io
N(v P and are same for solid and hollow shafts)
r2 dA = Moment of inertia of the elementary ring about an axis perpendicu-
Torque transmitted by solid shaft is the same as the torque transmitted by hollow
lar to the plane of Fig. 16.3 and passing through the centre of the
Torque transmitted by solid shaft is given by equation (16.4) as

circle.

ffl dA = Moment of inertia of the circle about an axis perpendicular to the

Torque transmitted by hollow shaft is given by equation (16.6) as r
Jo
t-a
plane of the circle and passing through the centre of the circle

—= Polar moment of inertia (J) = Di
32 .

—

STRENGTH OF MATERIALS TORSION OF SHAFTS AND SPRINGS

Hence equation (i) becomes as Let a twisting moment T produces a twist of 6 radians in a shaft of length L.

Using equation (16.9), we have

— —J1 C«
= -r- or C- x Jr = Tr-L

J L 0

JR ...(16.8) But C x J - Torsional rigidity
...(16.9)
But from equation (16.1), we have TxL

ce .’. Torsional rigidity =

R~ L 0

If L = one metre and 0 = one radian

Z = 1.^99 Then torsional rigidity = Torque.

where JR L kWProblem 16.13. Determine the diameter ofa solid steel shaft which will transmit 90
C = Modulus of rigidity
at 160 r.p.m. Also determine the length of the shaft if the twist must not exceed 1° over the entire
0 = Angle of twist in radiation
length. The maximum shear stress is limited to 60 N/mm2. Take the value of modulus of rigid-
ity = 8 x 104 N/mm2.

L = Length of the shaft. Sol. Given :
Power,
16.7. POLAR MODULUS Speed, WP = 90 kW = 90 x 10s

Polar modulus is defined as the ratio of the polar moment of inertia to the radius of the Angle of twist, N = 160 r.p.m.
Max. shear stress,
Zshaft It is also called torsional section modulus. It is denoted by . Mathematically, —0=1° or 0 1° = radian
p • radian
loU
V* 180

t = 60 N/mm2

—J(a) For a solid shaft, Modulus of rigidity, C = 8 x 104 N/mm 2

- D* Let D - Diameter of the shaft and

L = Length of the shaft.

•tc tj (i) Diameter of the shaft

32 = 32 = JL ds ...(16.10) Using equation (16.7),
16 ...(16.11)
R “ D/2
(Here R is the outer radius)
— D(b) For a hollow shaft, J = 2nNT

( 4 - Df) 60
0
90 x i103a = ^ll60xr
”[^o 4 -A4 ]
60

90 x 10 3 x 60

r—r • = 5371L.48 NN--m = 5371.48 x 103 N-mm
=~h —mT
zjt x 160

2 Now using equation (16.4),

£lA)4 -A4 ] ...(16.12) '-S'*

x[D 4 -£M] ~5371.48 x 103 =
0

16 x 60 x D3

16.8. STRENGTH OF A SHAFT AND TORSIONAL RIGIDITY ™D = 5371.48 x 103 x 16 = 455945
jt x 60
The strength of a shaft means the maximum torque or maximum power the shaft can D = (455945) 173 = 76.8 mm. Ans.

transmit. ( ii ) Length of the shaft

Torsional rigidity or stiffness of the shaft is defined as the product of modulus of rigidity Using equation (16.7),

(O and polar moment of inertia of the shaft (J). Hence mathematically, the torsional rigidity is x ce

given as. =

Torsional rigidity = C x J. R~ L

Torsional rigidity is also defined as the torque required to produce a twist of one radian

per unit length of the shaft.

STRENGTH OF MATERIALS TORSION OF SHAFTS AND SPRINGS 691
690

8 x 10 4 x re —R„ = D = 76.8 mm, D4 = 32x2000x 11459100 = 13„3„7„7.81 x 10 4
22
' 7

( 76.8 ^ Lx 180 10 5 x re x 0.01745

l2 mmD = (13377.81 x 10 4) 1/4 = 107.5 ...(«)

mm.r = O8 Ax X1V0 4 Ax Jrle ax 7i 6u.8u = 8r»9n3o.f6* The suitable diameter of the shaft is the greater* of the two values given by equations

AAns. (i) and (ii).

or 60x180x2 mmDiameter of the shaft = 124.5 say 125 mm. Ans.

Problem 16.14. Determine the diameter of a solid shaft which will transmit 300 kW at
The maximum shear stress should not exceed 30 N/mm2 *(If diameter is taken smaller of the two values say 107.5 mm, then from equation
m250 r p and twist should not be
1 ° in a shaft length of 2 m. Take modulus of rigidity = 1 x
more than 10s N/mm2. D7Z

Sol. Given : T= x 3 the value of shear stress will be
Power transmitted,
Speed of the shaft, 16 ,

WP = 300 kW = 300 x 103 -tx11459100 = (107.5) 3

N = 250 r.p.m. 16

Maximum shear stress, x = 30 N/mm2 11459100 = 243920 x

Twist in shaft, 0=1°= = 0 01745 radian _11459100

Length of shaft, L = 2 m = 2000 mm °r T ~ 243920 ~ 46.' 978 N/mm2
Modulus of rigidity, which is more than the given value of 30 N/mm2).
Let C = 1 x 105 N/mm2
D = Diameter of the shaft. AProblem 16.15. hollow shaft of diameter ratio —(internal dia. to outer dia.) is to

O

Power is given by the relation, transmit 375 kW power at 100 r.p.m. The maximum torque being 20% greater than the mean.

„ 2aNT mThe shear stress is not to exceed 60 N/mm2 and twist in a length of 4 not to exceed 2°. Calcu-

60 late its external and internal diameters which would satisfy both the above conditions. Assume

2re x 250 x T modulus of rigidity, C = 0.85 x 10s N/mm2 . (AMIE, Summer 1988 Converted to S.I. units)

300 x 103 =

—300 x 103 * = 11459.1 N-m = 11459.1 x 103 N-mm Sol. Given : —=—
Diameter ratio,
T=
2re x 250

(i) Diameter of the shaft when maximum shear stress,

x = 30 N/mm2 D‘-6 D
o
Maximum torque transmitted by a solid shaft is given by equation (16.4) as
WP = 375 kW = 375000
Power,

Speed, IV = 100 r.p.m.

—11459100 = x 30 x D3 Max. torque, Tmax =1.2 Tmean
Length,
lb L = 4 m = 4000 mm

1/3 —0 = 2° = 2 x - radians = 0.0349 radians
loO
ie x 11459100 ^1
f
mmD
= _ 124 5 Max. twist.

^ re x 30 )

(H) Diameter of shaft when twist should not be more than 1°. Modulus of rigidity, C = 0.85 x 10 s N/mm2

Using equation (16.9), Power is given by, 2kNT
P = Here torque is T
r_ ce
bU

JL P X 60 375000 X 60 = 35810 Nm
where J = Polar moment of inertia of solid shaft 2jiN 2a x 100

= 35810 Nm

,

11459100 10 D x 0.01745 : = 1.2 X Tmean = 1.2 X 35810

= 42972 Nm = 42972 x 1000 Nmm.

STRENGTH OF MATERIALS TORSION OF SHAFTS AND SPRINGS

(£) Diameters of the shaft when shear stress is not to exceed GO N/mm2 . Speed of the shaft, N = 200
Twist in the shaft,
For the hollow shaft, the torque transmitted is given by
radians = 0.01745 rad.
(D0 *-Dl) 180

Length of the shaft. L = 2m = 2000 mm

Maximum shear stress, X = 50 N/mm 2

Modulus of rigidity, C = 1 x 105 N/mm2

42972 x 1000 = • x 60 x - =: —P„ = 2nNT 2it x 200 x T
16
D or 75 x 10 3 =
o

4 60

42972000x 16 DA"l0 (. 81 'l =Z^, , X Nm75 x 10 3 x 60 = 3580.98 = 3580980 Nmm.
jt x 60
1 0'> 4096

“4096j

2it x 200

3 _ 42972000 x 16 x 4096 (i) Diameter of the shaft when maximum shear stress is limited to 50 N/mm2
.
0 " itx 60x4015
Using equation (16.4),

D ( 42972000 x 16 x 4096 = 154.97 mm say 155 mm ~T = TT x D 3 or 3580980 = Dx 50 x 3
=
« 3tx60x4015 .

[ 16 16

D. = - D = - x 155 ~ 58.1 mm. D= 16x3580980 mm= 71 . 3
0
^ n x 50 J
(ii) Diameters of the shaft when the twist is not to exceed 2 degrees.
Using equation (16.9) in terms of torque and 0, we get (ii) Diameter of the shaft when the twist in the shaft is not to exceed 1°

T Cx0 Using equation (16.9),
J~ L
T CO

_

J~ L

42972000 (0.85 xl0 b )x 0.0349 3580980 10° x 0.01745
2000
4000 j = —D*

32

42972000 x 4000 x 32 _ D4 _ D 4 ' DD —. 4 _ - ^1 4 32 x 2000 x 3580980
itx 10 s x 0.01745
it X 0.85 X 10 6 X 0.0349 0 ‘ 4096 = 80.4 mm.

_~ n4 r, 81 1 4015 n „ The suitable diameter of the shaft is the greater value of the two diameters given by
l 4096 J
0 -s

L 4096 0 mmequations (i) and (ii) i.e., 80.4 say 81 mm. Ans.

4 _= 42972000 x 4000 x 32 x 4096 Problem 16.17. A hollow shaft, having an internal diameter 40% of its external

0 x 0.85 x s x 0.0349 x 4015 kWdiameter, transmits 562.5 power at 100 r.p.m. Determine the external diameter of the

it 10

D = 156.65 mm say 157 mm mshaft if the shear stress is not to exceed 60 N/mm 2 and the twist in a length of 2.5
0
should not exceed 1.3 degrees. Assume maximum torque - 1.25 mean torque and modulus

D = —g x 156.65 = 58.74 of rigidity - 9 x 104 N/mm2 .
mmand say 59 mm.

;

The diameters of the shaft, which would satisfy both the conditions are the greater of Sol. Given : D D D= 40% of external diameter, (
Internal diameter, ;
the two values. Power transmitted, WP - 562.5 kW = 562.5 x 103 0 ) = 0.40 0
Speed of the shaft,
External dia., D = 157 mm. Ans. N = 100 r.p.m.
0

Internal dia., D = 59 mm. Ans.
i Maximum shear stress, t = 60 N/mm2

kWProblem A16.16. solid circular shaft transmits 75 power at 200 r.p.m. Calculate

the shaft diameter, if the twist in the shaft is not to exceed 1° in 2 metres length of shaft, and Twist in the shaft, 0 = 1.3° = 1.3 x radians = 0.02269 rad
Length of shaft,
shear stress is limited to 50 N/mm 2 Take C= 1 x 10s N/mm2. 180

.

Sol. Given : 2500 mm
Power transmitted,
WP = 75 kW - 75 x 103

694 STRENGTH OF MATERIALS torsion of shafts and springs 695

mm mmThe external diameter of the shaft should be 179.6
Maximum torque, TmM = 1.26 x Tm€an say 180 (i.e., greater of the

C = 9 x 104 N/mm2 two values given by equations (i) and (ift)). Ans.

Modulus of rigidity,

The horse power transmitted is given by 16.9. FLANGED COUPLING
A flange coupling is used to connect two shafts. Fig. 16.5 shows such a coupling.
2nNT

562.5 x 10s = 60 T)T(••• Here =

T2n: x 100 x mav

60

h60U x 50602^.-50 x 1J-0u 3 = 53714 7
. 2ji x 100
Nm NmmTnor
mean _ 53714700

' Tmax = 1 25 x Tmean = 1.25 x 53714700 = 67143375 Nmm.
(i) Diameter of the hollow shaft when maximum shear stress is 60 N/mm-5 .

Using equation (16.6) for torque in case of hollow shaft

'

4 4

r Ho
-a«

D16 [ 0

'

where T = T:mx = 67143375 and D = 0.4 D t = 60 N/mm2
{ 0,

— —— "
67143375 = x 4 4 Fig. 16.5

lb [As -(0J.4Z)0 )

x 60 7

|_ The flanges of the two shafts are joined together by bolts and nuts (or rivets) and torque
is then transferred from one shaft to another through the bolts. Connection between each
a [H 4 - 0.0256D0 4 1 shaft and coupling is provided by the key. The bolts are arranged along a circle called the pitch
0 circle. The bolts are subjected to shear stress when torque is transmitted from one shaft to

x 60 x 0.9744D 3 = 11.479D 3 smother.
0 0
16 Let t = Shear stress in the shaft
q = Shear stress in the bolt
3 ‘ d = Diameter of bolt

D«-7-6{71-43^3r75)V' = 179 - 6ram —™ ,. D = Diameter of shaft
(Ii)
D* = Diameter of bolt pitch circle
U( ) Diameter of the shaft when the twist in the hollow shaft is not to exceed 1.3°. n - Number of bolts

Using equation (16.9), we have Maximum load that can be resisted by one bolt

1 = 5! ...(ft-

JL
3 T = Tmax = 67143375, 0 = 1.3° = 0.02269 rad., L = 2500 mm, C = 9 x 10 4 N/mm2
9*7*= Stress in bolt x Area of one bolt
J - Polar moment of inertia for hollow shaft

UV-H 4! (see equation 16.11) Torque resisted by one bolt = Load resisted by one bolt x Radius of pitch circle

[D 4 - (0.4 D„)4] (v D= 0.4 0) —= q x - ft- x D*

...(Hi) Total torque resisted by n bolts

— D= x 0.9744 4 = 0.09566 D*. D= n x q x —n c^r2 x * 2 xfl*
&2i 0
-nxqx Jtft
Substituting these values in equation (ft), we get

4 But the torque transmitted by the shaft,

67143375 9x 10 x 0.02269 —T - XT X D3
lb
0.09566D0 ...(ft)

„_ 2500 x 67143375 _ 85928.215 x 10 4 Since the torque resisted by the bolts should be equal to the torque transmitted by the

:

0 0.09566 x 9 x 104 x 0.02269 shafts, therefore equating (i) and (ft), we get

mmD0 = 85928.215 x 104 = 171.2

—

:

696 STRENGTH OF MATERIALS TORSION OF SHAFTS AND SPRINGS 697

2
red
—Dnx q x * re xTxfl1 1349900
V5L3 = 7.57 i
x =
7500 x re
8 16
|
From the above equation the unknown value of any parameter (say number of bolts or
Torque transmitted by the hollow shaft is given by equation (16.6).
diameter of bolt) can be calculated. .'. Using equation (16.6),

Problem 16.18. Two shafts are connected end to end by means of a flanged, coupling in

which there are 12 bolts, the pitch circle diameter being 25 cm. The maximum shear stress is Ddj«r ~ 4
limited to 55 N/mm2 in the shafts and 20 N/mm2 in the bolts. If one shaft is solid of 5 cm. :]

diameter and the other is hollow of 10 cm external diameter, calculate the internal diameter of —= -re-x55x f 100 4 - D; 4

' the hollow shaft and the bolt diameter so that both shafts and the coupling are all equally = 0.10799 x (108 - £t 4) D(v t = 55, 0 = 100)
Equating equations (i) and (Hi),
strong in torsion. (AMIE, Summer 1983) .. liii )
1349900 = 0.10799 (108 - D*)
Sol. Given :

Number of bolts, n = 12

mmPitch circle diameter = 25 cm = 250

Maximum shear stress in the shaft,

t = 55 N/mm2 or Z> 4 = 108 - = 10 8 - 1250 xX 10 4 = 8750 x 104

;

Maximum shear stress in the bolts, wD = (8750 x 104) = 96.72 mm. Ans.
i
q = 20 N/mm2
AProblem 16.19. shaft is to be fitted with a flanged coupling having 8 bolts on a circle
D = 5 cm = 50 mm
Dia. of solid shaft, of diameter 150 mm. The shaft may be subjected to either a direct tensile load of 400 kN or a

External dia. of hollow shaft, twisting moment of 18 kNm. If the maximum direct and shearing stresses permissible in the
bolt material are 125 N/mm2 and 55 N/mm2 respectively, find the minimum diameter of the
D = 10 cm = 100 mm
0

Let D, = Internal dia. of hollow shaft bolt required. Assume that each bolt takes an equal share of the load or torque.

d = Dia. of the bolt in mm. (AMIE, Summer 1985)

In case of a couping, the torque from one shaft to other shaft is transmitted through Sol. Given
bolts. As the shafts and coupling are all equally strong in torsion, the torque transmitted by
the solid shaft must be equal to the torque transmitted by the bolts which must be equal to the Number of bolts, n=8
torque transmitted by hollow shaft.
Pitch circle diameter, D* = 150 mm
The torque transmitted by solid shaft is given by equation (16.4).
.•. Radius of pitch circle = — = 75 mm
Using equation (16.4), Direct tensile load,
Twisting moment, W = 400 kN = 400 x 103 N

x 55 x 503 = 1349900 Nmm Maximum direct stress, T = 18 kNm = 18 x 103 Nm
Maximum shear stress,
Nmm Nmm= 18 x 103 x 103
Let = 18 x 106

Area of one bolt (v Dia. of bolt = d) p = 125 N/mm2

x = 55 N/mm2

Area of 12 bolts - dre „2 mmd - Diameter of bolt in

12 x
4
—71
Shear force in 12 bolts = Shear stress in bolts x Area of 12 bolts Area of one bolt
= d2
4

N= 20 x 12 x - d2 = 60 nd2 Area of 8 bolts mm—= 8 x d2 = 2 red 2 2

4 .
Torque transmitted by the bolts
4
Pitch circle dia.
First consider the strength of the bolt material from shearing effect.
= Shear force in bolts x
Max. shear force in 8 bolts = Max. shear stress in bolts x Area of 8 bolts

= 55 x 2 red2 N.

— Nmm Nmm= 60 red2 x — Torque transmitted by the bolts = Shear force in bolts x Radius of pitch circle
= 7500 red2
(v Radius = 75 mm)
= 55 x 2 red 2 x 75

Equating equations (i) and (ii), we get Nmm= 110 x 75 x red2
1349900 = 7500 red2

d: o

!

STRENGTH OF MATERIALS TORSION OF SHAFTS AND SPRINGS 699

NmmBut torque transmitted = Twisting moment T = 18 x 106

Equating the two values of the torque,
110 x 75 x ncP = 18 x 106

d2 = 18 x .--— = 694.4

110 x75xit

d = V6944 = 26.35 mm —(i)

Now consider the strength of the bolt material from direct tensile load.

Max. tensile load in 8 bolts = Max. tensile stress x Area of 8 bolts

N= 125 x 2 ltd 2 Fig. 16.6

W NBut direct tensile load, The torque transmitted by a hollow shaft is given by equation (16.6) as
= 400 x 10 3
—T= ,t. D±!°n *-D<
.-. Equating the two values of direct tensile load

125 x 2 it 2 = 400000

or d,2 = 400000 = 5_0n9Q.3„

-

125 x 2 x n Hence the torque transmitted by hollow shaft AB is given by,

mmd = J509.3 = 22.56 —Hi) . = - T r^-A4 !

Comparing the results of (i) and (ii), the minimum diameter of the bolt required is the 16 (v Here D = D,D = df)
0 t

maximum of the two values. "

Minimum diameter should be 26.35 mm. Ans. n [40“- 204 Nmm

= 16 X 80 X 40

16.10. STRENGTH OF A SHAFT OF VARYING SECTIONS = 942500 Nmm = 942.500 Nm.

When a shaft is made up of different lengths and of different diameters, the torque Similarly torque transmitted by hollow shaft BC is given by,

transmitted by individual sections should be calculated first. The strength of such a shaft is “ 16 *‘ ' D* -d.24 40 4 -30 4

the minimum value of these torques. 40

mm mmProblem 16.20. A shaft ABC of 500
length and 40 external diameter is bored,

mm mmfor a part of its length AB, to a 20 = 687200 Nmm = 687.2 Nm.
diameter and for the remaining length BC to a 30

diameter bore. If the shear stress is not to exceed 80 N/mm2 find the maximum power, the shaft According to Art. 16.10, the safe torque transmitted by the complete shaft is the mini-
,
mum of the above two torques.
can transmit at a speed of 200 r.p.m. Nm.'. Safe torque ( T) transmitted by the shaft is 687.2

mm mmIf the angle of twist in the length of 20 T = 687.2 Nm.
diameter bore is equal to that in the 30

mm mmdiameter bore, find the length of the shaft that has been bored to 20
and 30 diameter.

Sol. Given The power transmitted (P) is given by equation (16.7) as
Total length,
External dia., L - 500 mm — WP= 2nNT- TW„= 2nx200 x687.2 = ,14A 3„n90nm
Let D = 40 mm 60 60
- 14.39 kW. Ans.
L, = Length of shaft AB

d = Internal dia. of AB Now from equation (16.9), we have
j
= 20 mm T Cx0 T.L
J L °f C.J
L - Length of shaft BC
2

d = Internal dia. of BC The safe torque T and shear modulus (C) are same for the given shaft.
2

= 30 mm Hence angle of twist in shaft AB T.L,

t = Maximum shear stress

= 80 N/mm2 and angle of twist in shaft BC = T L<

N = Speed = 200 r.p.m. C * tJr-

T\ = Torque transmitted by shaft AB

T = Torque transmitted by shaft BC.
2

23 3

STRENGTH OF MATERIALS TADOIPIM nCOUACTC' AklA nnninrv>

But angle of twist in shaft AS = Angle of twist in shaft BC Shaft CD : Length = L 3
T.L T.L
mmDDiameter, s = 70
X2
Angle of twist is same for each section.
C.J ~C.J
X2

Hence 0j = 0 = 0

2 3

Max. shear stress in hollow portion,

where J is the polar moment of inertia. From equation (16.11), its value for a hollow shaft is Value of C r = 50 N/mm2
x
given by = 8.2 x 104 N/mm2

—J- [DJ-D- 4 Polar moment of inertia of each shaft is given as :
01
32 ]

J = Polar moment of inertia for the shaft AB
l

~= [404 - 204 ]

J = Polar moment of inertia for the shaft BC
2

~= [404 - 30 4 ]
Oi

Substituting these values in equation (i), we get

jn L2 Fig. 16.7

~[40 4 -204 ^[40 4 -304 ]
]

L = 4 - 20 4 )
x
(40 =L37

^01 ( 40 4 - 30 4 ) For shaft BC, ^ ^J = D = mmx 804 = 402.4 x 104 4
2 z

or L = 1.37 x L= 1.37 (500 -Lj)
x 2
^ = ^£> 4 = mm: 704 = 235.8 x 10 4
(v L + L 500 L = 500 - L x) For shaft CD, 4
x 2 2
=

= 1.37 x 500 - 1.37 L< Now using equation,

L + 1.37 L = 1.37 x 500 or 2.37 L ~ 1.37' x 500 —— XLT = Cx 0
xx x
— —Lr^ = 1.3-7 x 500 = 289 mm. Ans. JL
or 0=

J.C

T x L2
~dT2 ~x nC
1 T„ T x L,
and L = 500 - 289 = 211 mm. Ans. xC6 = „ = and, „ Txl,
2 1 0 =
77 > C0
2 3
J J2 X 7;
mProblem 16.21. A steel shaft ABCD having a total length of 2.4 consists of three x

lengths having different sections as follows : ®1 = 0 = 0

2 3

mm mmAB is hollow having outside and inside diameters of80 T.LX T.L2 T.L
and 50 respectively and 3
[Torque T and C are same for each portion]
mmBC and CD are solid, BC having a diameter of80 and CD a diameter of 70 mm. Ifthe angle J, .C J2 C ~ J3 C

of twist is the same for each section, determine the length of each section and the total angle

of twist if the maximum shear stress in the hollow portion is 50 N/mm2. Take C = 8.2 x Jy J2 J

104 N/mm2. Li
340.9 x 10 4
Sol. Given : L2 L3

Total length of shaft, L = 2.4 m = 2400 mm 402.4 x 4 235.8 x 10 4

10

Shaft AB : Length = L or
Shaft BC : x 340.9 402.4 235.8

mmDOuter dia., = 80
x

mmInner dia., rf = 50 r “ 340.9 L = 1-44L
L 235.8 3 3
Ll

Length = L
2

mmDiameter, 402.4
D = 80 Lr “ 235.8 L, = 1-71 L
2 *

=

702 STRENGTH OF MATERIALS torsion of shafts and springs

m mmLSilt same for both portions of the shaft, the shear stress will be maximum for the portion for which
LL+ 2.4 = 2400 polar moment of inertia (i.e., J) is less. Polar moment of inertia for a hollow shaft is given by

l 3' equation (16.11) as
2+

1.44 Lo + 1.71B. + L :2400 or 4.15 L = 2400
3 3

2400 = mm.

L, = T4.T1T5 578.35
J

Substituting the value of L in equations (i) and (ii), we get mm.. Polar moment of inertia of the portion having 20 internal diameter.
3
mmL, = 1.44 x 578.3 = 832.75
— mm= [504 - 204] 4
mmL = (v D =5Omm,Z) = 20mm)
2 o i
= 1.71 x 578.3 988.80

As the shear stress is given in shaft AS. The angle of twist of shaft AB can be obtained — mm= 4
[6250000 - 160000] = 597884
by using equation .

T cxe mmSimilarly polar moment of inertia of the portion having 30 internal diameter,

R L ~ mmJ =
Cx 0 2
For shaft AB, [504 — 304] 4
!
*1 ~ mm= 4
[6250000 - 810000] = 534070
.

(A) Li mmAs the polar moment of inertia of the portion having 30 internal diameter is less,

Tj x L 50 x 832.75 mmthe maximum shear stress will develop in the portion of 30 internal diameter. Also maxi-
x
mmmum torque transmitted will be given by the portion of 30
f¥W —f^i Ix 8 . 2 xl 0 4 internal diameter.

l2J Now using equation (16.9),

= 0.01269 radians = 0.7273° T__x_

Total angle of twist of the whole shaft J~R

= e + 0 + e = 0.7273 x 3 = 2.1819°. Ans. .'. Maximum torque,
t
2 3

mProblem 16.22. A hollow shaft is 1 long and has external diameter 50 mm. It has ^xj= x 534070 ('' R = ~~ = 25 mml

mm mm20 internal diameter for a part of the length and 30 internal diameter for the rest of the J/i yu J

length. If the maximum shear stress in it is not to exceed 80 M/mm2 determine the maximum = 1709024 N-mm - 1709.024 N-m (v 1000 mm = 1 m)
, .-. Maximum power is given by equation (16.7) as

power transmitted by it at a speed of 300 r.p.m. If the twists produced in the two portions of the

shafts are equal, find the lengths of the two portions. (AMIE, Winter 1984) 2kNT 2 it x 300 x 1709.024

Sol. Given : L = 1 m = 1000 mm 60 60
Total length,
External dia., mm W= 53688 = 53.688 kW. Ans.
Let
D = 50 Lengths of the two portions
0
Using equation (16.9), we have
Lj = Length of the portion of the shaft whose internal dia. is 20 mm.

L = Length of the portion of the shaft whose internal dia. is 30 mm. T ce
2
mm= (1000 - Lf)
J~ L - (l)

Maximum shear stress, For the two portions the shear modulus (C), the twist ( 6 ) and torque T transmitted are

Speed, t so N/mm2 T V!!!!!!!!!!!J1111111l11111lf[I] For the portion having internal diameter 20 mm,
jj.i////////J
N - 300 r.p.m.

Twists produced in the two portions of the mm ~20 T ™30 DIA mmPolar moment of inertia, J - 597884 4 L - Length.
shaft are given equal. y ,x

DIA Substituting these values in equation (j),

Using equation (16.9), we get mMhnnmnnmn 7,,, , // > T ce
,
t.t 1

JR X L,mm (1000 - L,)mm ) For the portion having internal diameter 30 mm.
H
—j, H 1000 mm Polar moment of inertia,

x = x R. ^ig- 16- 8 J2 = 534070 mm4

(J Length, L = (1000 - mm.
2
RFor a hollow shaft, is the outer radius of

the shaft. As the outer radius of the hollow shaft is Lft

STRENGTH OF MATERIALS

Substituting these values in equation (i ), we get 9 - ^1 x L\

JT _ C0 where ^1 = jc d4 =~ jt x 7S4
1i
L2 32^ 32

Dividing equations (ii) and {iii), 7\ x 2000 T\ x 2000 x 32

L2 =_ 2 x 75^ x 0.3 x 10s a x 75 4 x 0.3 x 10 5
£.

Ji by For the shaft BC, the value of 02 is given by,

534070 1000 - L L - mm] wh<W„ T„ X Ln
t 2
[v = (1000
.597884 ” Lfi
Lj ~J
2 2 ^C 2 32 2 " 32
2

0.8932 = 1 ~° tJ'.L or o0.i8932 L = 1000 - L ~= Tz x 1000 _ T X 1000 X 32
xl x 55 4 x 0.9 x 10 6 2
L\
71 x 554 x 0.9 * 10*
0.8932 L + L = 1000 or 1.8932 L = 1000
11 t

L 1000 = 528.2 mm. Ans.
x
or = 18939 t x 2000 x 32 T2 x 1000 x 32
i
_

and L = 1000 - L = 1000 - 528.2 a x 75 4 x 0.3 x 5 a x 55 4 x 0.9 x 10 5
2 1
10

= 471.8 mm. Ans.

Problem 16.23. Two solid shafts AB and BC of aluminium and steel respectively are 75 4 x 0.3 55 4 x 0.9
rigidly fastened together at B and attached to two Arigid supports at and C. Shaft AB is 7.5 cm
™ 75 4 x 0.3
m m Ain diameter and 2 in length. Shaft BC is 5.5 cm in diameter and 1 in length. torque of
T T1 ~ 55 4 x 0.9 x 2 =" 20 0.576 * 20 .'
20000 N-cm is applied at the junction B. Compute the maximum shearing stresses in each

material. What is the angle of twist at the junction ? Take modulus of rigidity of the materials Substituting this value in equation (i),

as C = 0.3 x 10s N/mm2 and C = 0.9 x 10 s N/mm2. (AMIE, Summer 1987) 0.576 T + T = 200000
Al St 2 2

Sol. Given : 1.576 T = 200000
2
Solid Shaft AB Solid Shaft BC
„ 200000
--^2 “ = 126900 Nmm
Material = Aluminium Material = Steel ^ gyg

m mmLength, L = 2 = 2000 Length, Li == l1 m = 1000 mm But T + T = 200000
2 i 2
t Dia.,

mmDia., mmd
2
d = 7.5 cm = 75 = 5.5 cm = 55 2\ = 200000 ~T = 200000 - 126900 = 73100 Nmm.
2
1

Modulus of rigidity, Modulus of rigidity, — —RFrom equation (16.9),

C = 0.3 x 105 N/mm2 C = 0.9 x 10 s N/mm2 =
x 2
J
T= 20000 N cm
Torque at junction B, ~T

= 200000 Nmm. For shaft AB J i£j
l
The torque is applied at junction B, hence angle of twist in shaft AB and in shaft BC

will be same (be., 0, = 0, = 6) T = 20000 N cm R71 x x 73100 x 37.5

wh, ere 9 = A,ngi,e oi twist in shaft AB, and / . S..A.I5,
t 1o
<*>

0 = Angle of twist in shaft BC. j

2

Let Tj = Torque transmitted to shaft AB, and ; BV 73100 x 37.5 x 32

T - Torque transmitted to shaft BC d, -= 75 mm — mmd 0.882 N/mm2. Ans.
2 2
= 75 ;
NmmT+Tj.-. = 20000
2 ...(i) — ——; _____ i

Using equation (16.9), we get J Ll>2 = 11 000 mmmm^ For shaft BC

T ~_ C X 0 % ^Xmm- L, = 2000 ->

~J ~L~ T “=_ 20000 N cm _ T2 xR2 126900 X 27.5 p. ^2 55
.
For the shaft AB, the above equation becomes as — x 55 4

T C; x e Fig. 16.8 (a) 32
rx
126900 x 27.5 x 32
Ji~ Li
= 3.884 N/mm2. Ans.

,

STRENGTH OF MATERIALS

16.11. COMPOSITE SHAFT

sd m .......N/mm N/mmC
ffor steel =8.4*10* 2 and C for brass = 4.2 x 104 .
each mater/a/
Find a/so tfte maximum shear stress and common ang/e of twist

length of 4 m.

Sol. Given : d = 60 mm
T = 1000 Nm = 1000 x 103 Nmm
Dia. of steel rod,
Torque,

Value of C for steel, C = 8.4 x 10* N/mm2
s

Value of C for brass, C = 4.2 x 104 N/mm2
b
m mmLength of composite shaft, L = 4 = 4000
mmLet O = Outside dia.' of brass tube in

r = Shear stress in steel
s

t. = Shear stress in brass
The inner dia. of brass tube will be equal to dia. of steel rod.

Inner dia. of brass tube,

d = 60 mm

Polar moment of inertia for steel rod is given by,

—J = d4 = mmx 604 4

* 32 32

Polar moment of inertia for brass tube is given by

D |R,
mm
M
r-
T

Fig. 16.9

: Torque transmitted by steel rod, and
: Torque transmitted by brass tube.

But total torque,

T = T + T = T + T
s b s s

y ^ looo x io3 Nmm = 500 x 103 Nmm
s
2
2
T. = T = 500 x 103 Nmm
and bs

:

708 STRENGTH OF MATERIALS TORSION OF SHAFTS AND SPRINGS 709
500 x 1 0'’ x 39.5 _ 77 6 f{/mm2. Ans.
T = Torque transmitted by copper rod

T = Torque transmitted by steel shaft.
s

Now for a composite shaft,

(ii) Common angle of twist Total torque,
Using equation (t), we get
T = T' + T,
C, x 9, x J\.
T + T = 1000 x 10 s ( v T - 1000 x 103 )
c s ...(j)

Using the relation,

e = = 500 ’'I0 x 4000
=Lor = 4000) T Cx0 T xL
C, (v L

—s
X tJ*. 8O.4A x 11 /0\4 x ^ x 60 J L °r 6 ~ U x C

32

= 0.01871 radians For copper rod,
For steel shaft.
= 0.01871 x 180 degrees = 1.072°. c T xC
cc
it
T.-L,
But angle of twist in each shaft will be equal. The common angle to twist will be equal to
Q>~Ts .Cs

the angle of twist in any shaft. (v Angle of twist are same)

Common angle of twist = 1.072°. Ans. L T L_c s . s

mmProblem 16.25. A composite shaft consists of copper rod of 30 diameter enclosed in Jc xC J .C
c SS
mm mma steel tube of external diameter 50 thick. The shaft is required to transmit a
and 10 But L = L = L
cs
torque of 1 000 Nm. Determine the shear stresses developed in copper and steel, if both the shafts

have equal lengths and welded to a plate at each end, so that their twists are equal. Take Hence the above equation becomes as

modulus of rigidity for steel as twice that of copper. T T± c ‘•s

Sol. Given Jr xC, " J„.C
S

Dia. of copper rod, d - 30 mm —rrp V.c X 30 4
32
External dia. of steel, D = 50 mm p= X (• C = 2C
0 s c
)
J* C« ’ ^-x 44x10 4 2C„
DInternal dia. (D ) of steel shaft = - 2 x Thickness
; 32
mm= 50 - 2 x 10 = 30

D — 30 mm 81 1 81
i
" 44 * 2 X 2 “ X Ts
T = 1000 Nm = 1000 x 103 Nmm
» 88

Total torque, Substituting the value of T in equation ii), we get
c

Let t = Shear stress in steel shaft

t = Shear stress in copper ;T, + T = 1000,000
c

C = Modulus of rigidity of steel
5

C = Modulus of rigidity of copper 8 IT, + 88 T,
c = 1000,000

L = Common length

0 = Common angle of twist = 0 = 0 169 T = 1000,000 x 88
CS s

J = Polar moment of inertia of copper rod 1000,000 x 88
c 1
NmmmT
= = 5„„2„0„7„10„

169

i x30W= xa!4 =-3! Substituting this value in (t), we get

J = Polar moment of inertia of steel shaft T + 520710 = 1000,000
s c

T = 1000,000 - 520710 = 479290 Nmm
C5

£ mm= |r[7V - D*] = [50'1 - 304] 4 Now using the relation,

— — mm[125 x 104 - 81 x 104 ] = 4 T T_
x 44 x 10 4
_

J~ R

— TORSION OF SHAFTS AMD SPRINGS 711

STRENGTH OF MATERIALS

From Art. 3.4.4, we know that the angle 9 made by the plane of maximum shear with

the normal cross-section is given by,

For a copper rod, we have —2t

tan 20 =
a

The bending stress and shear stress is maximum at a point on the surface of the shaft,

(- -I) where r = R - — —D D
and, y =

mmN479290 : JL _ 479290 x 15 x 32 = 90 407 / 2 Ans. Let a = Maximum bending stress i.e., on the surface of the shaft
b
_M_ D M D 32M
i X 2 ~ nD 3
— x 30 4 * " n_
2
DI
64

And for a steel shaft, we have t - Maximum shear stress i.e., on the surface of the shaft

?>*[%) 520710 4
V £ 7_

JL * 44 x 104

520710 x 25 x 32 = 301.358 N/mm2. Ans.

re x 44 x 10 4

16.12. COMBINED BENDING AND TORSION

When a shaft is transmitting torque or power, it is subjected to shear stresses. At the
same time the shaft is also subjected to bending moments due to gravity or inertia loads. Due

to bending moment, bending stresses are also set up in the shaft. Hence each particle in a shaft

is subjected to shear stress and bending stress. For design purposes it is necessary to find the

principal stresses, maximum shear stress and strain energy. The principal stresses and maxi-

mum shear stress when a shaft is subjected to bending and torsion, are obtained as :

Consider any point on the cross-section of a shaft.

Let T = Torque at the section

D = Diameter of the shaft

M = B.M. at the section.

The torque T will produce shear stress at the point whereas the B.M. will produce bend-

ing stress. q = Shear stress at the point produced by torque T and
Let
a = Bending stress at the point produced by B.M. (M)
.

—Major principal stress
= 16Do

j
D—re[D0 - i ] j- (M + ...(16.17)

The shear stress at any point due to torque (D is given by

Minor principal stress = ...(16.18)

r R JJ re[D0 4

9 = 7 xr —Maximum shear stress =
-4r-
[D 4
0 -Di l
re

The bending stress at any point due to bending moment (M) is given by mmProblem A16.26. solid shaft of diameter 80 is subjected to a twisting moment of

— —M M- —a or a = xy MN MNmm mm8 and a bending moment of 5 at a point. Determine :

-

Iy I (i) Principal stresses and

(it) Position of the plane on which they act.

712 STRENGTH OF MATERIALS torsion of shafts and springs

Sol. Given : 16J xl0 6 ['/9 + 16] 16 x 10 6 x 5
Diameter of shaft, 0
Twisting moment, D = SO mm
Bending moment, rt.ZV 1-
T = 8 MN mm = 8 x 106 Nmm
16 x 10 s x5x 16
M — 5 MN mm = 5 x 10s Nmm.
= 0.3395 X 10 6
The major principal stress is given by equation (16.14), as v k x 15 x 80

16 D mm.a = (0.3395 x 10 6 ) 1'3 = 0.3395 1/3 x 102 = 69.78

=--(
M+JJmM(M= Tf2
Major principal stress I + 2 Ans.

xD 3 + )

—= 7V((55xx110 6 ) 2 10 6 ) 2 and D= = 34.39 nrm. Ans.
5x1100 6 +( 8 x i
jcx80H +

16 x 10s 16.13. EXPRESSION FOR STRAIN ENERGY STORED IN A BODY DUE TO TORSION
TtxSO'1
v = 143.57 N/mm2. Ans. Consider a solid shaft which is in torsion. Take an elementary ring of width dr at a

Minor principal stress is given by equation (16.15). radius r as shown in Fig. 16.10.

™.’. Minor principal stress {M - 2 T2 Let D = Diameter of shaft /
-
-Jm l = Length of shaft

3 + )

jtD —R = Radius of shaft = ( uf

16 f_ ,„s VI(75Zx~l70 6 ) 2 + (8 x10 s ) 2 R\

) x = Shear stress on the surface of the shaft i.e., at

16 x 10 6 radius R \ J/j

— x x 8^0_o3 N/mm—= —-V ^1 1 1 \ ^ZQTrrT/Zr J
44.1TTrit . AX/LUIli 2

= 44.1 N/mm 2 (tensile). Ans. C = Shear modulus or modulus of rigidity

U = Total shear strain energy in the shaft.

Position of plane is given by equation (16.13), as Then shear stress due to torsion at a radius r from the
centre is given by
raT Fig. 16.10

ta 20 s 8 x 6

10

slJi V ^RTr - t
~R
20 = tan- 1 1.6 = 57° 59.68', or 237° 59.68' °r Xr r

XV

or 0 = 28° 59.84' or 118° 59.84’. Ans. Area of elementary ring, dA = 2nr.dr

Problem 16.27. The maximum allowable shear stress in a hollow shaft of external :. Volume of ring, V = dA.l = 2itr x dr xl

N mmdiameter equal to twice the internal diameter, is 80 / 2 Determine the diameter of the The shear strain energy is given by equation (16.9) as,

.

Nmmshaft if it. is subjected to a torque of 4 x 10e and a bending moment of 3 x 10° Nmm. (Shear stress) 2

Sol. Given : Shear strain energy x Volume

Maximum shear stress = 80 N/mm2 Shear strain energy in the ring of radius r

Torque, T = 4 x 10® Nmm

MBending moment, = 3 x 10G Nmm fr

fI
XX i 2
L= Xi?
DLet 0 = External diameter of shaft x 2 jtr . dr.l = -?-±- x r2 x 2nr dr.
D = Internal diameter of shaft 2C 2CR 2

i Total shear strain energy stored in the shaft is obtained by integrating the above equa-

DThen L>0 = 2 ; tion between limits 0 to R.

Using equation (16.19) for a follow shaft, we get

.'. Total strain energy stored in the shaft.

Maximum shear stress = R I*- 1 „2 --- f* ;-2 2xrdr
x(DQ4 - Df) .
f
A16
, 2CR- Jo

[\[(3 x 10 s ) 2 + (4 x 10 6 ) 2 J (v dA - 2ji.dr)

2 CR 2 Jo

torsion of shafts and springs 715

STRENGTH OF MATERIALS

Length of shaft, ml = 1.25 = 125 cm

-Ax mmV.-. Volume of shaft, 3
dA Jr R „2 1 1 = 78.54 x 125 = 9817.5 cm3 = 9817.5 x 10 s
=
[r .
Jo Maximum allowable shear stress,
x = 60 N/mm2
J = Polar moment of inertia of the shaft

— D= x 4 for a solid shaft. (shear stress is maximum on the surface of the shaft)
32
Total strain energy in the shaft due to torsion, Modulus of rigidity, C = 8 x 104 N/mm2

ULet = Shear strain energy stored in the shaft.

U — Dx2 .l it 4 Using equation (16.20), we get

= CA2 x’ 52- 32

(v D = 2R)

= 76699 Nmm. Ans.

Problem 16.29. The external and internal diameters of a hollow shaft are 40 cm and
N/mm N/mm20 cm. Determine the maximum strain energy stored in the hollow shaft if the maximum
allowable shear stress is 50 2 and length of the shaft is 5 m. Take C = 8 x 104 2

.

VVolume, xR= 2 ...(16.20)

.l)

Total strain energy in the hollow shaft due to torsion Sol. Given : D = 40 cm = 400 mm
External dia., d = 20 cm = 200 mm
DLet = Outer dia. of shaft Internal dia.,

d = Inner dia. of shaft A — mmArea of cross-section, = (402 — 202 ) = 942.47 cm2 — 94247 2

J = Polar moment of inertia of hollow shaft

Maximum allowable shear stress ( i.e ., shear stress on the surface of the shaft),

x = 50 N/mm2

Substituting the value of J in equation (i), we get Length of shaft, ! = 5m = 500 cm

Total strain energy in the hollow shaft due to torsion, .-. Volume of hollow shaft,
mmV - A x l = 942.47 x 500 = 471235 cm3 = 471235 x 103
3

2 [D 4 - 4 Modulus of rigidity, C - 8 x 104 N/mm2
<f l
CR2
32

—^-5-= ULet = Strain energy stored

JD. X -5. [D2 + d2 l[D2 - d2 Using equation (16.21), we have
]
2
t 32 L 2J

A= x [£)« + d2][D 2 - d2 ] U =^tf iD2 + d2) * V
2C.D 2 32
50 2 X (400 2 + 2002 ) X 471235 x 10 3
u=
4 x 8 x 10 4 x 400 2

= ————— x - x (D 2 - d2 )i(D2 + d2 ) = 4601900 Nmm = 4601.9 Nm. Ans.

4CD 2 4 Problem 16.30. Calculate the diameters of a hollow shaft of the same length and same

= x Vx(D2 + d2) V -d••• = ^(IZD) 2 - 2 )).dIl cross-sectional area as a solid shaft of 15 cm diameter if the strain energy in the hollow shaft is
L4 25% greater than that of solid shaft transmitting the same torque at the same maximum shear
4CZ) 2 . J

= _e!_ (D 2 + d2 X V ...(16.21) stress.
)

4CD 2 Sol. Given :
Dia. of solid shaft,
Problem 16.28. Determine the maximum strain energy stored in a solid shaft of D = 15 cm
diameter 10 cm and of length 1.25 m, if the maximum allowable shear stress is 50 N/mm2 .
A —Area of solid shaft, = x 15 2 = 56.25 it cm 2
Take C = 8 x 104 N/mm2.
4

Sol. Given : Area of solid shaft = Area of hollow shaft

Dia. of shaft, D = 10 cm Length of solid shaft = Length of hollow shaft

.-. Area of shaft, mmA = — x 102 = 78.54 cm2 = 7854 2 .'. Volume of solid shaft = Volume of hollow shaft

4

——:

STRENGTH OF MATERIALS TORSION OF SHAFTS AND SPRINGS

Strain energy in solid shaft = 1.25 of strain energy in solid shaft DSubstituting this value of 1 in equation (if), we get

Maximum shear stress in solid shaft Z?2 = 0.5 x 17.32 = 8.66 cm. Ans.

= Maximum shear stress in hollow shaft A mProblem 16.31. solid circular shaft of 10 cm diameter of Length 4 is transmitting

DLet = External dia. of hollow shaft kW112.5 power at 150 r.p.m. Determine : (i) the maximum shear stress induced in the shaft
1
D2 - Internal dia. of hollow shaft and (ii) Strain energy stored in the shaft. Take C = 8 x 104 N/mni3 .

Cross-sectional area of hollow shaft Sol. Given : D = 10 cm = 100 mm
Dia. of shaft,
= f[ZV-D22]. Length of shaft, l = 4 m = 4000 mm
Power,
But cross-sectional area of hollow shaft WkWP = 112.5
= Area of solid shaft (given) = 112.5 x 10 3

Speed of shaft, IV = 150 r.p.m.
Modulus of rigidity,
C = 8 x 104 N/mm2
D~ (Dj 2 -
2 = 56.25n Let t = Maximum shear stress induced in the shaft and
2)
V = Strain energy stored in the shaft.
DDj 2 - 2 = 56.25 x 4 = 225
2

Let t = Max. shear stress in solid as well as in hollow shafts, We, IT ,know ——P„ = 2jzNT
60
U, = Strain energy of solid shaft, and
——or 112.5 x 103, = 2ji x 150 x T
Uh = Strain energy of hollow shaft.

Using equation (16.20) for the strain energy in solid shaft, we get

U ——T2 112,5 x 10 3 x 60

= x Volume of solid shaft. : 7159 Nm = 7159000 Nmm
s 4C
Using equation (16.21) for the strain energy in hollow shaft, 2n x 150

U t2 But we know, T* = —— xxx D3
16
h=
D(Dj 2 - 2 x Volume of hollow shaft —7159000 =
2}

16 x x x 10G 3

D(v External dia. = v Internal dia. = D„) 7159000 x 16

1.25 x U = 36.5 N/mm2

71 X 10®

DD( Using equation (16.20) for strain energy,
2+ 2 ) x Volume of hollow shaft
4CD, 2
1

—= 1.25 x — x Volume of solid shaft U = x Volume of shaft

4C 2

ACDy D+ 2) ~ 1-25 36.5

2 4C x Volume of shaft
4 x 8 x 10'4
('.'
Volume of hollow shaft = Volume of solid shaft) 5J“ Volume = — x D2 xl
x T x 100 2 x 4000
x4 8a x 4
10 4

—Cancelling — to both sides = 130793 Nmm. Ans.
*4C
AProblem 16.32. hollow shaft of internal diameter 10 cm, is subjected to pure torque
DDD 2 + 2 = 1.25 2
and attains a maximum shear stress q on the outer surface of the shaft. If the strain energy
D D DD, 2 = 1.25 2 - 2 = 0.25 2
——T 2 x V, determine the external diameter of the shaft.
stored in the hollow shaft is given by

D D2 = 70.25D! 2 = 0.5 l Sol. Given
Internal dia.,
DSubstituting this value of 2 in equation (i), we get d = 10 cm
Maximum shear stress =x
-— D20.25 Dj 2 = 225 or 0.75 2 = 225

£>l
D,,.= 225 .300

D- 7300 = 17-32 cm. Ans. *PIease see equation 16.4 on page 675. Here q - x.
l

yD v

718 STRENGTH OF MATERIALS torsion of shafts and springs 719

The shaft is subjected to a constant torque T.

Strain energy stored, T x wV where V - Volume H*^ir =
3C
=3
-*2

DLet = External diameter of the hollow shaft. 2- t x

2
D D Dor 3 = 2
x 2 tx x
.(j
Using equation (16.21) for the strain energy in hollow shaft
Let us find the total angle of twist for this tapering shaft of length ‘L\

—U = 4CD2 ( V2 + d2) x Consider a small length dx of this shaft.

Equating the two values of strain energy, we get > Let d% = Angle of twist of small length dx

6 = Total angle of twist of total length ‘L’

{D2 + d2)xV=~ x V tUtsi• ng equation, ——T = C9
oG
-J

D 2 + d 2 __ 1 — VCancelling x to both sides For a small length dx, the angle of twist is rf6. Hence the above equation becomes as

4D 2 3 t c x de

ZD2 + 3 d2 = 4Z>2 J dx

3d2 = AD2 - ZD2 = D2 Dwhere J = Polar moment of inertia corresponding to diameter x

— —=3or or ** -v/3 = 1.732 — D= 4

r

D = 1.732 x d = 1.732 x 10 = 17.32 cm. Ans. T x dx _ T x dx 32T x dx
DJtC X
JxC *_ D * xC
16.13.1. Torsion of tapering shafts. Fig. 16.10 (a) shows a shaft which tapers uniformly 32

from radius R to radius R and the shaft is subjected to a twisting moment T. Due to the 32T dx
x2
C71 X X.
twisting moment, the shear stress will be developed in the shaft. The shear stress is directly

proportional to radius. Hence shear stress will be

different at the surface of the two ends of the shaft

(as the two ends are having different radius). ||

Let Tj = Shear stress on the surface of the " ' D D2 - { = k (some constant)
shaft at left end where radius is ft, "
g
R|
t = Shear stress on the surface of the || —~~ |
' - ]
2 ~\\
T"
shaft at right end where radius Ris 2 x H r~" dx Z2Tdx
r
xx = Shear stress on the surface of the —l Cn x x [D + kx 4
shaft which is at a distance of x from |« t ]

p, g ' The total angle of twist for the total length of the shaft is obtained by integrating the

the left end above equation as given by

L = Length of the shaft

D = Dia. of shaft at left end = 2R I = J dQ
x1

D - Dia. of shaft at right end = 2R2 L 32T dx ( x varies from 0 to L)
?
=f (v T is constant)

Dx = Dia. of shaft at a distance x from the left end = 2RX where R is the radius at CJo re x x [Dj + kxf
x

that section. Z2T rt. dx

RThe radius x at a distance x from left end is given by, ~

rexCJo D[ x + kxf

Dor diameter at a distance x from left end is given by * x c Jo jxC (- 3) X k

D = D + = _, 32T
x l
( ^

Cre x 3k

We know that, ^-^c ^ ^QOT 1
x KD
Vk
kLrS ~ {D ]

) :

720 STRENGTH OF MATERIALS TORSION OF SHAFTS AND SPRINGS 721

48 x 10 s = —x x 0.163

ixC 3k !(£>! + kLY D{ 48 x 10 3 x 16

D Lk D DD. = + kL 7t x 0.16 3 = 59683103 N/m2 = 59.683103 MN/m2. Ans.
2 l ;
= k hence = + or 1

Now above equation becomes as 16.14. SPRINGS

32T 1 1 Springs are the elastic bodies which absorb energy due to resilience. The absorbed en-

) Aergy may be released as and when required. spring which is capable of absorbing the great-

1 jtC *3k DY est amount of energy for the given stress, without getting permanently distorted, is known as
the best spring. The two important types of springs are :
32T 1 1 ...[16.21(A)]
C"it * 3k D, 3 1. Laminated ot leaf springs and

Problem 16.32(A). Determine the angle of twist and maximum shear stress developed 2. Helical springs.

mmin a shaft which tapers uniformly from a diameter of 160 to a diameter of 240 mm. The 16.14.1. Laminated or leaf spring. The laminated springs are used to absorb shocks
in railway wagons, coaches and road vehicles (such as cars, lorries etc.).
mlength of shaft is 2 and transmitts a torque of 48 kNm. Take the value of modulus of rigidity

GN mfor shaft material as = 80
z Fig. 16.11 shows a laminated spring which consists of a number of parallel strips of a
/.

Sol. Given metal having different lengths and same width, placed
one over the other. Initially all the plates are bent to
mm mmDj = 160 the same radius and are free ro slide one over the other. w/2 w/2
m H= 0.16 ; 2 = 240 = 0.24 m, L = 2 m, T = 48 kNm = 48 x 103 Nm, and Fig. 16.11 shows the initial position of the spring, ^
which is having some central deflection 6. The spring ^
C = 80 GN/m2 = 80 x 109 N/m2 rests on the axis of the vehicle and its top plate is j -ypy
pinned at the ends to the chassis of the vehicle. H
Find : (i) Angle of twist 6 and //; 5
\V
(ii) Maximum shear stress developed. y

(i) Angle of twist, 0 i

The angle of twist, 0 is given by equation 16.21(A), as When the spring is loaded to the designed load ^"7-—_ !

32T _lj_l 1_' W, all the plates becomes flat and the central deflec- |
X
9~ 3 D3 tion (6) disappears. w
jtC 3k 2
[Zlj Let b = Width of each plate

where DDi) - n = Number of plates Fig. 16.11
Dk - x
l = Span of spring
= 0.240-0.160 = 0.08 = 0.04
22 a = Maximum bending stress developed in the plates

32 x 48 x io 3 i ri i__; t = Thickness of each plates

__ W = Point load acting at the centre of the spring and

ax80 x 10 9 * 3x0.04 3 0.24 3 .

L0.16

[v T = 48 x 103 Nm and C = 80 x 10 6 N/m2 ] 6 = Original deflection of the top spring.

= 0.00005093 [244.14- 72.34] WExpression for maximum bending stress developed in the plate. The load

= 0.00875 radians = .00875 x = 0.501. Ans. acting at the centre of the lowermost plate, will be shared equally on the two ends of the top
plate as shown in Fig. 16.11.

n

(ii Maximum shear stress developed B.M. at the centre = Load at one end x —

We know —T = x x x D3 WM= l W.l
lb 2 *2- 4
The moment of inertia of each plate,
For a given torque, the shear stress on the surface of the shaft will be maximum where

diameter is minimum. Hence at smaller diameter, the shear stress will be maximum.

— mT = xmax x
Jg x D, 3 where D, = smaller diameter = 0.160
1 1

1 I

STRENGTH OF MATERIALS TORSION OF SHAFTS AND SPRINGS
722
Substituting this value of R in equation (Hi), we get
But the relation among bending stress (a), bending moment (M) and moment of inertia
5_ 2 x 2ct aj}_ ..,(16.23)
(I) is given by l 4Et

— fllere y = E8 x x t

Equation (16.23) gives the central deflection of the spring.

AProblem 16.33. leaf spring carries a central load of 3000 N, The leaf spring is to be
mm Nmade of 10 steel plates 5 cm wide and 6 2
M = —° x /t = ° * o.bt 2 !mmthick. If the bending stress is limited to 150
12 =

y t_ 6 determine :

2 (i) Length of the spring and

.-. Total resisting moment by n plates (ii) Deflection at the centre of the spring.

n x a 2 Take E = 2 x 10s N/mm2.

M= n x . 6£

= Sol. Given :
Central load,
As the maximum B.M. due to load is equal to the total resisting moment, therefore No. of plates, W = 3000 N
Width of each plates,
equating (t) and (ii), Thickness, n = 10
Bending stress,
W.l no. 2 Modulus of elasticity, b = 5 cm = 50 mm
bt /t = 6 mm

4^ 6 a - 150 N/mm2

6W.Z 3WZ ...(16.22) E = 2 x 105 N/mm2.

° 4. n.b.t 2 2nbt 2 Let l = Length of spring
8 = Deflection at the centre of spring.
Equation (16.22) gives the maximum stress developed in the plate of the spring.

Expression for central deflection of the leaf spring Using equation (16.22),

Now R = Radius of the plate to which they are Q/1 x o= 3 Wl
-

2nbt 25

From triangle ACO of Fig. 16.12, we have 3 x 3000 x l
AO2 = AC2 + CO2 °r 160 = 2 x 10 x 50 x 6 2

——=, 10 x 50 x 2

I 6
150 x 2 x = 6B0n0n mm. A. ns.

R2 = - +(R- 6)2 3 x 3000

\2/ Using equation (16.23) for deflection,

R+ 2 + 82 - 21? 6 150 x 600 2

,
„2

8„ = a. =
= 11.25 mm. Anso.

4 Et 4 x 2 x 10 5 x 6

+ R2 - 2R6 (Neglecting 82 which Fig. 16.12 Problem 16.34. A laminated spring 1 m long is made up ofplates each 5 cm wide and
is a small quantity)
N/mm1 cm thick. If the bending stress in the plate is limited to 100 2 how many plates would
,

—2 N mmbe required to enable the spring to carry a central point load of 2 k ? If E = 2.1 x 10s N/ 2

l ,

2Rb = what is the deflection under the load ? (AMIE, Summer 1982)
4
Sol. Given :
— i-,2 ,2
5= - = ...(iii) Length of spring, mmm/ = 1 = 1000

4x21? 81? Width of each plate, 6 = 5 cm = 50 mm

But the relation between bending stress, modulus of elasticity and radius of curvature mmThickness of each plate, t - 1 cm = 10

(R) is given by Bending stress, a = 100 N/mm2

o_£ WCentral load on spring, = 2 kN = 2000 N

y~R Young’s modulus, E - 2.1 x 105 N/mm2

£xy £xt ( Here y = — Let n = Number of plates and
=

8 = Deflection under the load.

— :

STRENGTH OF MATERIALS TORSION OF SHAFTS AND SPRINGS 725

Using the equation (16.22), Expression for deflection of spring

100 = 3 x 2000 x 1000 Now length of one coil = xDor 2nR
“
or .-. Total length of the wire = Length of one coil x No. of coils or l - 2xR x n.
2o x n x 5707 x 1770(2 As the every section of the wire is subjected to torsion, hence the strain energy stored by
the spring due to torsion is given by equation (16.20).
n= 3 x 2000 x 1000 = „ .
6. .-. Strain energy stored by the spring,
Ans.

100 x 2 x 50 x 100

Deflection under load Volume . Volume
Using equation (16.23),
.

a x 2 ~ 100 x 1000" = 11.9 nun. Ans. 161V. R —x x ltd 2 x2xR.n

l s
jxd
4E x t 4 x 2.1 x 10 s x 10

16.14.2. Helical Springs. Helical springs are the thick spring wires coiled into a helix. - and Volume = — or x Total length of wire

4

They are of two types : 32W 2rR—2 32W 2 R3
t
Cd 4
—— .n
1. Close-coiled helical springs and n =
Cd 4 ...(16.25)
.R.n
( v Deflection = 6)
2. Open coiled helical springs. Work done on the spring = Average load x Deflection

Close-coiled helical springs. Close-coiled helical springs //////// = Iff* 5
are the springs in which helix angle is very small or in other words
|! p Equating the work done on spring to the energy stored, we get
Athe pitch between two adjacent turns is small. close-coiled helical |

spring carrying an axial load is shown in Fig. 16.13. As the helix JL q---
angle in case of close-coiled helical springs are small, hence the
bending effect on the spring is ignored and we assume that the D 32W 2 i? 3 . n

coils of a close-coiled helical springs are to stand purely torsional ,

mVR„

8
= sn ...(16.26)
...(16.27)
stresses. Cd4

Expression for max. shear stress induced in wire. H- R_H Expression for stiffness of spring
Fig. 16.13 shows a close-coiled helical spring subjected to an axial
nJ
load.
T^ The stiffness of spring,
Let d = Diameter of spring wire
s = Load per unit deflection

Flg' 16,1

_W W Cd 4

p = Pitch of the helical spring 6 ~ 64.1VR 3 .n 64 . 3
n - Number of coils f?

R = Mean radius of spring coil Note. The solid length of the spring means the distance between the coils when the coils are

IV = Axial load on spring touching each other. There is no gap between the coils. The solid length is given by

C = Modulus of rigidity Solid length = Number of coils x Dia. of wire = n x d ...(16.28)

x = Max. shear stress induced in the wire AProblem 16.35. closely coiled helical spring is to carry a load of 500 N. Its mean
0 = Angle of twist in spring wire, and
6 = Deflection of spring due to axial load coil diameter is to be 10 times that of the wire diameter. Calculate these diameteis if the
N mmmaximum, shear stress in the material of the spring is to be 80 /
1 = Length of wire. 2

Now twisting moment on the wire, .

T=WxR (AMIE, Summer 1985)

-(*) Sol. Given : W = 500 N
Load on spring,
But twisting moment is also given by Max. shear stress. x = 80 N/mm2

Let d - Diameter of wire

D ~ Mean diameter of coil

Equating equations (i) and (ii ), we get D = 10 d.

VV A XL — LLi, UL T — ,3 —Using equation (16.24), t = -3-
Jta
16

Equation (16.24) gives the max. shear stress induced in the wire.

—

726 STRENGTH OF MATERIALS TORSION OF SHAFTS AND SPRINGS 727
16 x 500 x
No. of turns, n = 10
*(¥)
Mean dia. of coil, D = 12 cm = 120 mm
R = D/2 = 60 mm
Radius of coil. W = 200 N
Axial load.
Modulus of rigidity, C = 8 x 10 4 N/mm2

Let 5 : Deflection of the spring

or 80 x rod3 = 8000 x 5d x = Max. shear stress in the wire

or **™»** m mXi s = Stiffness of the spring.

•
80 x 71

mmd = -J 159.15 = 12.6 (,i ) Using equation (16.26),

= 1.26 cm. Ans. — — —6 = 60 3 x
x t1t0t4
D = 10 x d = 10 x 1.26 = 12.6 cm. Ans. 64WR3 xn = 64—x 200 x 10 = 34.5 mm.
x 10 4
Cd 4 8 Ans.

N mmProblem 16.36. In problem 16.35, if the stiffness of the spring is 20 per deflection

mmand modulus of rigidity - 8.6 x 104 N/ 2 find the number of coils in the closely coiled helical (ii) Using equation (16.24),

,

spring. WRx = 16 x 200 x 60
16 x— = = 61.1 N/mm-,. AAns.
Sol. Given : 35
3
rod rex 10

Stiffness, s = 20 N/mm (iii) Stiffness of the spring,

Modulus of rigidity, C = 8.4 x 104 N/mm2 —W 200

From problem 16.35, s = “t” = 7T7T = 5.8 N/mm. Ans.
o 34.5
W = 500 N, x = 80 N/mm2
d - 12.6 mm and D = 126 mm Problem 16.38. A close coiled helical spring of 10 cm mean diameter is made up of 1 cm
mmR - D/2 - 126/2 = 63
diameter rod and has 20 turns. The spring carries an axial load of 200 N. Determine the shear-

mming stress. Taking the value of modulus of rigidity = 8.4 x 104 N/ 2 determine the deflection
,

Let n = Number of coils in the spring when carrying this load. Also calculate the stiffness of the spring and the frequency of free

We know, —= a<^ vibration for a mass hanging from it. (AMIE, Winter 1982)

stiffness Sol. Given :

mmDMean diameter of coil, = 10 cm = 100

R mm.'. Mean radius of coil,
- 5 cm = 50

= 25 mm Diameter of rod, d = 1 cm = 10 mm

Using equation (16.26), Number of turns, n = 20

WR65 n3 Axial load, W=200N
.
Modulus of rigidity., C - 8.4 x 104 N/mm2

Let x == Shear stress in the material of the spring

64 x 500 x 8 x n ( v R = 63 mm) 8 = Deflection of the spring due to axial load
(63)

8.4 x 10 4 x 4 s = Stiffness of spring

12.6

n ~_ 25 x 8.4 x 10 4 x 4 = 6.6 say 7.0 x = Frequency of free vibration.

12.6

64 x 500 x 3 Using equation (16.24),

(63)

n - 7. Ans. 1 6WR 16x200x50 50.93 N/mm2„. Ans.

mmProblem 16.37. A closely coiled helical spring of round steel wire 10 in diameter

having 10 complete turns with a mean diameter of 12 cm is subjected to an axial load of200 N. Deflection of the spring
Using equation (16.26),
Determine : (i) the deflection of the spring (ii) maximum shear stress in the wire, (iii) stiffness of

the spring. Take C = 8 x 104 N/mm2
.

WR64
Sol. Given : 3 xn _ 64 x 200 x 50 3 x 20 = 38.095 mm. Ans.
Dia. of wire, d = 10 mm Cd 4 " 8.4 x 10 4 x 10 4

: d:

strength of materials TORSION OF SHAFTS AND SPRINGS

Stiffness of the spring spring is 125 N/mm2. The solid length of the spring (when the coils are touching) is given as

—Stiffness = 5 cm. Find : (i) diameter ofwire, (ii) mean diameter ofthe coils and (Hi) number ofcoils required.
Load on spring 200 Ans.
Take C = 4.5 x 104 N/mm2.
~. 7 : = = 5.25 N/mm,
Deflection of spring 38.095
Sol. Given
frequency of free vibration s = 1.5 N/mm
Stiffness of spring.
mm5 = 38.095 = 3.8096 cm W = 60 N
Load on spring,
Using the relation, 2.55 cycles/sec. Ans. x = 125 N/mm2
2% 11 3.8095 Maximum shear stress
= 5 cm = 50 mm
AProblem 16.39. closely coiled helical spring of mean diameter 20 cm is made of 3 cm Solid length of spring,
C~ 4.5 x 104 N/mm2.
Modulus of rigidity.

diameter rod and has 16 turns. A weight of 3 kN is dropped on this spring. Find the height by Let d = Diametter of wire,

which the weight should be dropped before striking the spring so that the spring may be com- D = Mean dia. of coil, and

pressed by 18 cm. Take C = 8 x 10 4 N/mm2 —DR =
.

Sol. Given Mean , ,t =

Mean dia. of coil, radius of coil

D = 20 cm = 200 mm n = Number of coils.

Mean radius of coil, mm200 Using equation (16.27),

R = —jjj- = 100

Dia. of spring rod, d = 3 cm = 30 mm ——s = Cd4 ,- = 4.5 x 10 4 x d4
64.E s .n 1.5
or 3
ii
Number oTturns, n = 16 64 x x n

Weight dropped, W = 3 kN = 3000 N Rd =,4, 1.5 x 64 x 3 x n = 0.002 133f? s x n
Compression of the spring, 5=18 cm = 180 mm
4.5 xlO 4

Modulus of rigidity, C = 8 x 104 N/mm2 Using equation (16.24),

WLet h = Height through which the weight is dropped —Rt = 16 x 60 x ER
W = Gradually applied load which produces the compression of spring equal to 16W x or 125 =
-z
180 mm. 5 xd 3
3
ltd

Now using equation (16.26), 125 x n 3

R = 11f6l x 6cn0 = 0.40906d3

„ 64W.E 3 .n. RSubstituting the value of in equation (i), we get

W64 x x 100 J x 16 d4 = 0.002133 x (0.40906d3 )3 x n
= 0.002133 x (0.409063 ) x d9 x n - 0.00014599 x d9 x n

°r 180 = 8 x 104 x 30 4 d 9 .n _

1 or ds n_ 1
0.00014599
W= .lg0 -8x,iyx30l = 11390N d 4 0.00014599
64 x 1003 x 16
Now using equation (16.28),

Work done by the falling weight on spring Solid Length = n.x. d or 50 = n x <7

= Weight falling (h + 6) = 3000 (h + 180) N-mm 50
n~ d
WxEnergy stored in the spring = | 5
Substituting this value of n in equation (Hi), we get

= | x 11390 x 180 = 1025100 N-mm. ,s 50 1
Equating the work done by the falling weight on the spring to the energy stored in the
spring, we get d 0.00014599

3000(ft + 180) = 1025100 d* = X = 136 99

0.00014599 50

1025100 d = (136.99) 174 = 3.42 mm. Ans.

— — mm>r h + 180 = 3QQQ = 341.7 Substituting this value in equation (iv)

h = 341.7 - 180 = 161.7 mm. Ans.

N/mmProblem 16.40. The stiffness of a close-coiled helical spring is 1.5 ofcompression 14.62 say 15. Ans.

under a maximum load of 60 N. The maximum shearing stress produced in the wire of the

\ : 3

730 STRENGTH OF MATERIALS TORSION OF SHAFTS AND SPRINGS
But mean coil radius,
Also from equation (it),

R mm= 0.40906 d3 = 0.40906 x (3.42)3 = 16.36

Meani.e., dia. of coil, D = 2R = 2 x 16.36 = 32.72 mm. Ans. DD3 = 3 2 *

Problem 16.41. A close-coiled helical spring has a stiffness of 10 N/mm. Its length p]d5 = 3.2 x = 0.4 A

when fully compressed, with adjacent coils touching each other is 40 cm. The modulus of rigid-

N/mmity of the material of the spring is 0.8 x 10s 2

.

—(i) Determine the wire diameter and mean coil diameter if their ratio is = 0.4 or d2 = 0.4

,

. 3

(ii) If the gap between any two adjacent coil is 0.2 cm, what maximum load can be applied / 1\

before the spring becomes solid, i.e., adjacent coils touch ? UJ

d2 = 0.4 x 103 = 400

(Hi) What is the corresponding maximum shear stress in the spring ? (AMIE, May 1976) d = V400 = 20 mm = 2 cm. Ans.

Sol. Given : s = 10 N/mm B„ut d 1
~
Stiffness of spring, D 10 *

Length of spring when fully compressed i.e., solid length D - 10 x d = 10 x 2 = 20.0 cm. Ans.

= 40 cm = 400 mm Let us find first number of turns.

Modulus of rigidity, C = 0.8 x 105 N/mm2 Prom equation (ii), we have

Let d — Diameter of wire of spring 400 400 =20 ( v d = 20)
n= *
D - Mean coil diameter d 20

n = Number of turns mm6 = 2 x n = 2 x 20 = 40

W = Maximum load applied when spring becomes solid We know, stiffness of spring is given by

x = Maximum shear stress induced in the wire. WW

xt d 1 Ts= 0r i1n0 =
40
D 10
W = 10 X 40 = 400 N. Ans.
mmGap between any two adjacent coil = 0.2 cm = 2.0
Using equation (16.24), we have
.-. Total gap in coils = Gap between two adjacent coil x Number of turns
W16 . . i?
= 2 x n mm.
When spring is fully compressed, there is no gap in the coils and hence maximum dX 3
it
compression of the coil will be equal to the total gap in the coil.
_ 16x400x 100 f,. fi = ^,20_ 100ccmm = 110000mmmml
mm.-. Maximum compression, 6 = 2 x n
It x 20 3 V 22 )
Now using equation (16.27),
= 25.465 N/mm2. Ans.

Problem 16.42. Two close-coiled concentric helical springs of the same length, are

Cd 4 0.8 x 10 s x d 4 wound out of the same wire, circular in cross-section and supports a compressive load 'P7 The

S~ °r 10 “ a inner spring consists of 20 turns of mean diameter 16 cm and the outer spring has 18 turns of
64 .R B .n i?
644.. ..n mean diameter 20 cm. Calculate the maximum stress produced in each spring if the diameter of

^ a 10x 64 8 ' R3xra wire - 1 cm and P - 1000 N. (AMIE, Summer 1989)
0.8 x10s U03 J
But from equation (16.28), Sol. Given

NTotal load supported, P = 1000

Solid length = n x d or 400 = n x d. Both the springs are of the same length of the same material and having same dia. of

400 wire. Hence values of L, C and ‘d’ will be same.
n~ d
For inner spring
Substituting the value of n in equation (£),
No. of turns, n = 20
t

——R 400 Mean dia., mmD = 16 cm = 160 mm .-. R, = ——— = 80
x 3x = 3.2 x
d
Dia. of wire, d, = 1 cm = 10 mm
Bdor
5 = 3;2 x 3

.

W- '

732 STRENGTH OF MATERIALS TORSIOty OF SHAFTS AND SPRINGS 733

For outer spring For outer spring, the maximum shear stress produced is given by,
No. of turns,
n = 18 64W xRq 16 x 362.7 x 100
0 0 a x 10 3

Mean dia., mmD R = —.-. 100 mm 0 jTd^
0
= 20 cm = 200 Afl

cm mm = 184.72 N/mm2. Ans.

Dia. of wire, d = 1 = 10 Similarly for inner spring, the maximum shear stress produced is given by,
Q

WLet = Load carried by inner spring W R16 x x
i

W0 = Load carried by outer spring {t 16 x 637.3 x 80

x = Max. shear stress produced in inner spring 1 nx d z n x 10 3
f t

t = Max. shear stress produced in outer spring. = 259.66 N/mm2. Ans.

0 mmProblem 16.43. A closely coiled helical spring made of 10

WNow W; + 0 = Total load carried = 1000 .,,(/) diameter steel wire has

Since there are two close-coiled concentric helical springs, hence deflection of both the mm15 coils of 100 mean diameter. The spring is subjected to an axial load of 100 N. Calculate :
springs will be same.
(i) The maximum shear stress induced,

S = 8j where 8 = deflection of outer spring (ii) The deflection, and

0 0

6 = Deflection of inner spring. (iii) Stiffness of the spring.
;
N/mmTake modulus of rigidity, C = 8.16 x 104 2.
The deflection of close-coiled spring is given by equation (16.26) as

64Wxi?3 xn (AMIE, Winter 1990 Converted to S.I. units)
;
_
Sol. Given : d = 10 mm
^ C xd4 Dia. of wire,

Hence for outer spring, we have

3 64 0 x 100 3 x 18 Number of coils, n = 15
flD
64W nfl x x 0 mmDMean dia. of coil, - 100

Cx 4 ) = 100, d0 - 10)
cf0

Similarly for inner spring, we have mmR.-. Mean radius of coil, = - = 50
^
W64 x R x n 64W x 80s x 20
t tt £ W = 100 N

Axial load,

(Material of wires is same. Hence value of C will be same.) Modulus of rigidity, C = 8.16 x 10 4 N/mm2.

But = (i) Maximum shear stress induced

60 S 16x 100x50
;
3
64W x 100 3 x 18 64Wj- x 80 3 x 20 Using equation (16.24), x = 16W-i—? = = 24.46 N/mm02. A,ns.
0 5
jid it x 10
CxlO 4 " CxlO4 .

W W0 x 1003 x 18 = x 80a x 20 (ii) The deflection (8)
Using equation (16.26),
t

wWj x 80 3 x 20 64W xR3 xn 64 x 100 x 50 3 x 15

= = Q g6g
100 3 x 18 *
•

Substituting the value of W„ in equation (i), we get Cxd4 ~ 8.16 x 10 4 x 10 4

W W+ 0.569 = 1000 or 1.569 W. = 1000 = 14.7 mm. Ans.
;

1000 (iii) Stiffness of the spring

W WBut from equation (i), 0 = 1000 ‘ Load on spring
+ Stiffness =

; Deflection of spring

W ~W0 = 1000 = 1000 - 637.3 = 362.7 N. Load on spring _ 100
Deflection of spring 14.7
t

The maximum shear stress produced is given by equation (16.24) as 6.802 N/mm. Ans.

16 Wi?

STRENGTH OF MATERIALS TORSION OF SHAFTS AND SPRINGS 735

HIGHLIGHTS mStrength of a shaft means the maximum torque or axim um power the shaft can transmit.

A shaft is in torsion, when equal and opposite torques are applied at the two ends of a shaft. The product of modulus of rigidity and polar moment of inertia of a shaft is known as torsional

The relation of maximum shear stress induced in a shaft subjected to twisting moment is given rigidity or stiffness of the shaft. Mathematically,

by C JTorsional rigidity = x

r__ce '-llL 3£.(: . CxJ-Zl±)

RL 0 { JL 0J
where x = Maximum shear stress,
= T if L = 1 metre and 0 = one radian.
R = Radius of shaft,
C = Modulus of rigidity, The power transmitted by a shaft is given by

0 = Angle of twist in radian, and 2nNT

L = Length of the shaft. The strain energy stored in a shaft due to torsion is given by,
When a circular shaft is subjected to torsion, the shear stress at any point varies linearly from
Um lC* V ... For a solid shaft
the axis to the surface i.e..
(D2 + d2 x V ... For a hollow shaft
4 C.D 2 )

Dwhere = External diameter of shaft,

6. x = Maximum shear stress on the' surface of the shaft d - Internal diameter of shaft,

where

R = Radius of surface T = Modulus of rigidity, and

s q = Shear stress at a point which is at a radius ‘r\ x = Shear stress on the surface of the shaft.
4. The shear stress is maximum on the surface of the shaft and is zero at the axis of the shaft.
.

13. Polar moment of inertia (J) is given by

The torque, transmitted by a solid shaft is given by Jr D4

32 ... For a solid shaft

T= xD3 —= (D4 - d*) ... For hollow shaft.

16 32..

Dwhere - Dia. of solid shaft and

x = Max. shear stress. 14. Springs are the elastic bodies which absorb energy due to resilience. Two important types of

The torque transmitted by a hollow circular shaft is given by springs are :

1. Laminated or leaf springs and

Do* -Dr 2. Helical springs.

16. The maximum stress developed in the plates of a leaf spring is given by,

O= 3.W.J

D = Internal diameter. 2n b 2s-
i t. .

7. Relation between torque, polar moment of inertia and shear stress is given as where W = Point load acting at the centre of leaf spring,

t x ce l = Span of leaf spring or length of leaf spring,

J~ R~ L n = Number of plates,

where I = Polar moment of inertia b = Width of each plate, and

—= D4 ... For a solid shaft t - Thickness of each plate.

32 16. The central deflection (6) of the laminated spring is given by,

—= [D 4 - IX 4 ... For a hollow shaft.

j 6 ~ 4E.t
where a = Maximum stress developed in the plates,
32
E = Modulus of elasticity,
8. The ratio of the polar moment of inertia to the radius of the shaft is known as polar modulus or
L = Length of leaf spring, and
Ztorsional section modulus. It is denoted by . T Thickness of each plate.
p
vj 17. Helical springs are the thick spring wires coiled into a helix. They are of two types :
1. Close-coiled helical springs and
—= D3 ... For a solid shaft
2. Open-coiled helical spring.
D 4 - UTj ... For a hollow shaft.

(0

STRENGTH OF MATERIALS TORSION OF SHAFTS AND SPRINGS 737

18. The maximum shear stress induced in the wire of a close-coiled helical spring which carries an 7. Define the term ‘Polar modulus’. Find the expressions for polar modulus for a solid shaft and for
a hollow shaft.
axial load is given by,
8. What do you mean by ‘strength of a shaft’ ?
where 16.W.jR
X= o 9. Define torsional rigidity of a shaft. Prove that the torsional rigidity is the torque required to
produce a twist of one radian in a unit length of the shaft.
nd 3
10. Prove that the strain energy stored in a body due to shear stress is given by,
W = Axial load on the spring,

R = Mean radius of spring coil, and

d = Diameter of spring wire. where x = Shear stress,
19. For a close-coiled helical spring which carries an axial load, we have
C = Modulus of rigidity, and
(i) Strain energy stored, V = Volume of the body.

W R n32 .2 3 1 1. Find an expression for strain energy stored in a body which due to torsion or
Prove that the strain energy stored in a body due to torsion is given by.
.
U ~ Cdi

(ii) The deflection of the spring at the centre due to axial load is given as

—6=
64 ,W.R3 .n and

4

(hi) The stiffness of the spring, where x = Shear stress on the surface of the shaft,

——Cd4 C = Modulus of rigidity, and V = Volume of the body.
A Dhollow shaft of external diameter and internal diameter d is subjected to torsion, prove that
s= =
R64 . n3 the strain energy stored is given by,
.

where W = Axial load on the spring,

Vi = Number of coils, —-2 / V
[/= (D2 + 2 )X
C = Modulus of rigidity, and 4C-Dn2r
c/

R = Mean radius of spring coil. where V = Volume of the hollow shaft and

EXERCISE 16 t = Shear stress on the surface of the shaft.

What is a spring ? Name the two important types of spring.
Prove that the maximum stress developed in the plates of a leaf spring is given by

(A) Theoretical Problems 3W.l

Define the terms : Torsion, torsional rigidity and polar moment of inertia. 2n b 2
t. .
Derive an expression for the shear stress produced in a circular shaft which is subject to torsion.
Wwhere = Point load acting at the centre ofleaf spring,
What are the assumptions made in the derivation ?
Prove that the torque transmitted by a solid shaft when subjected to torsion is given by l = Span of leaf spring or length of the leaf spring,
n = Number of plates,
—T= Dx 3
16 b - Width of each plate, and
t - Thickness of each plate.
Dwhere = Dia. of solid shaft and x = Max. shear stress. Prove that the central deflection of the laminated spring (or leaf spring) is given by

When a circular shaft is subjected to torsion show that the shear stress varies linearty from the

axis to the surface. IAMIE, Winter 1982 )

Derive the relation for a circular shaft when subjected to torsion as given below where a = Maximum stress developed in plates,

T_x^C6 E - Modulus of elasticity,

J ~ R~ L l = Length of leaf spring, and

where T = Torque transmitted, t = Thickness of each plate.

J Polar moment of inertia, Define helical springs. Name the two important types of helical springs.

x = Max. shear stress, Prove that the maximum shear stress induced in the wire of a close-coiled helical spring is given

R = Radius of the shaft, by

C = Modulus rigidity, W R16 . .

6 = Angle of twist, and

L - Length of the shaft. x = Maximum shear stress induced in the wire,

Find an expression for the torque transmitted by a hollow circular shaft of external diameter W = Axial load on spring,

D D= R = Mean radius of spring coil, and d = Diameter of spring wire.
and internal diameter - v

0



•:W

thin cylinders and SPHERES 741

17

Thin Cylinders and Spheres

17.1. INTRODUCTION (a)

J The vessels such as boilers, compressed air receivers etc., are of cylindrical and spheri- Cross-section perpendicular
cal forms. These vessels are generally used for storing fluids (liquid or gas) under pressure to the axis of vessel
The walls of such vessels are thin as compared to their diameters. If the thickness of the wall
(b)
^ iof the cylindrical vessel is less than to of its internal diameter, the cylindrical vessel is Fig. 17.2
known as a thin, cylinder. In case of thin cylinders, the stress distribution is assumed uniform
17.3. STRESSES IN A THIN CYLINDRICAL VESSEL SUBJECTED TO INTERNAL
over the thickness of the wall. PRESSURE

17.2. THIN CYLINDRICAL VESSEL SUBJECTED TO INTERNAL PRESSURE When a thin cylindrical vessel is subjected to internal fluid pressure, the stresses in the

Fig. 17.1 shows a thin cylindrical vessel in which a fluid under pressure is stored. wall of the cylinder on the cross-section along the axis and on the cross-section perpendicular

Fig. 17.1 to the axis are set up. These stresses are tensile and are known as :
1. Circumferential stress (or hoop stress) and
Let d = Internal diameter of the thin cylinder
t = Thickness of the wall of the cylinder 2. longitudinal stress.
p - Internal pressure of the fluid
L = Length of the cylinder. The name of the stress is given according to the direction in which the stress is acting.
The stress acting along the circumference of the cylinder is called circumferential stress whereas
On account of the internal pressure p, the cylindrical vessel may fail by splitting up in the stress acting along the length of the cylinder (i.e., in the longitudinal direction) is known as
longitudinal stress. The circumferential stress is also known as hoop stress. The stress set up
any one of the two ways as shown in Fig. 17 .2 (a) and 17.2 (6). in Fig. 17.2 (a) is circumferential stress whereas the stress set up in Fig. 17.2 (6) is longitudi-
The forces, due to pressure of the fluid acting vertically upwards and downwards on the
nal stress.
thin cylinder, tend to burst the cylinder as shown in Fig. 17.2 (a).
The forces, due to pressure of the fluid, acting at the ends of the thin cylinder, tend to 17.4. EXPRESSION FOR CIRCUMFERENTIAL STRESS (OR HOOP STRESS)

burst the thin cylinder as shown in Fig. 17.2 (6). Consider a thin cylindrical vessel subjected to an internal fluid pressure. The circumfer-
ential stress will be set up in the material of the cylinder, if the bursting of the cylinder takes
740 place as shown in Fig. 17.3 (a).

The expression for hoop stress or circumferential stress' (Oj) is obtained as given below.
Let p = Internal pressure of fluid

d = Internal diameter of the cylinder
t = Thickness of the wall of the cylinder

a, = Circumferential or hoop stress in the material.

)

STRENGTH OF MATERIALS THIN CYLINDERS AND SPHERES 743

742

(b)

Longitudinal stress (o2 )
develop

r Fig. 17.4

(a) (b Resisting force = x Area on which a acting
2
cj2 is

Fig. 17.3 = a x nd x t
2

The bursting will take place if the force due to fluid pressure is more than the resisting Hence in the limiting case
force due to circumferential stress set up in the material. In the limiting case, the two force
Force due to fluid pressure = Resisting force
should be equal.
p x — d2 = a x nd x t
Force due to fluid pressure = p x Area on which p is acting 2
=px(dxL)
P*4 d72 pd
(v p is acting on projected area d x L)

Force due to circumferential stress

= Oj x Area on which a-, is acting The stress a is also tensile.
2

= Oj x (L x t + L x t) Equation (17.2) can be written as

= Oj x ZLt = 20j xLxi •••(*»)

Equating (t) and (it), we get =CT2 2 x 21

p x d x L - 2a x L x t or Longitudinal stress = Half of circumferential stress.
1

This also means that circumferential stress (Oj) is two times the longitudinal stress (o ).

Oj = •—• (cancelling L) ...(17.1) 2

Hence in the material of the cylinder the permissible stress should be less than the circumfer-

This stress is tensile as shown in Fig. 17.3 (6). ential stress. Or in other words, the circumferential stress should not be greater than the

17.5. EXPRESSION FOR LONGITUDINAL STRESS permissible stress.

Maximum shear stress. At any point in the material of the cylindrical shell, there are

Consider a thin cylindrical vessel subjected to internal fluid pressure. The longitudinal two principal stresses, namely a circumferential stress of magnitude o = acting circum-
L

stress will be set up in the material of the cylinder, if the bursting of the cylinder takes place -—ferentially and a longitudinal stress of magnitude ct2 =

ABalong the section of Fig. 17.4 (u). acting parallel to the axis of the

The longitudinal stress (o2) developed in the material is obtained as : shell. These two stresses are tensile and perpendicular to each other.
Let p = Internal pressure of fluid stored in thin cylinder
pd pd

d = Internal diameter of cylinder Maximum shear stress %max = 1 ^ ...[17.2 (A)]
^^
t = Thickness of the cylinder ^

0 = Longitudinal stress in the material. Note, (t) If the thickness of the thin cylinder is to be determined then equation (17.1) should be
2
(used.
The bursting will take place if the force due to fluid pressure acting on the ends of the
(ii) If maximum permissible stress in the material is given. This stress should be taken circum-
cylinder is more than the resisting force due to longitudinal stress (<J2 ) developed in the mate-
ferential stress (Oj).

rial as shown in Fig. 17.4 (b). In the limiting case, both the forces should be equal. (iti) While using equations (17.1) and (17.2), the units of p, <7 and ct should be same. They should
2
Force due to fluid pressure = p x Area on which p is acting be expressed either in N/mm2 or N/m2 . Also the units of d and t should be same. They may be in metre

(m) or millimetre (mm).

= P*7<*2

4

~ :

:

STRENGTH OF MATERIALS THIN CYLINDERS AND SPHERES

mProblem 17.1. A cylindrical pipe of diameter 1.5 and thickness 1.5 cm is subjected to m N/mmAProblem 17.3. cylinder ofinternal diameter 0.50 contains air at a pressure of 7 2

an internal fluid pressure of 1.2 N/mm2. Determine : N mm(gauge). If the maximum permissible stress induced in the material in 80 / 2 find the

(i) Longitudinal stress developed in the pipe, and ,

thickness of the cylinder.

(ii) Circumferential stress developed in the pipe. Sol. Given

Sol. Given mInternal dia. of cylinder, d - 0.50

Dia. of pipe, md = 1.5 Internal pressure of air, p = 7 N/mm2
Thickness, mt = 1.5 cm = 1.5 x 10~2 Maximum permissible stress in the material means the circumferential stress (a,).

Internal fluid pressure, p = 1.2 N/mm2 As stated earlier that the circumferential stress should not be greater than the maxi-

mum permissible stress. Hence take circumferential stress equal to maximum permissible

— — —As the ratio stress.

= 1-5 X 10 = which is less than -^r, hence this is a case of thin Circumferential stress, a, = 80 N/mm 2
,
d 1.5 20 Let t = Thickness of the cylinder
100

cylinder.

Here unit of pressure ip) is in N/mm2 . Hence the unit of a, and a will also be in N/mm2 . Using equation (17.1),
2

(i) The longitudinal stress (a2) is given by equation (17.2) as, „°i“ Pd

pxd 21
——— mt= - pd
41 2xo,
°2 ~ = 7 x 0.50 = 0.021875

— N/mm1.2 x 1.5 2*80

= j4 x - = 30 2„ Ans. m(Here jo and o, are in N/mm2 d is in hence t will be in m)
2 ,
.

1.5 x 10 = 2.188 cm. Ans.

(ii) The circumferential stress (a,) is given by equation (17.1) as If the value of t is taken 2.1875 cm, the stress induced will be 80 N/mm2 . If the value of
t is less than 2.1875 cm, the stress induced will be more than 80 N/mm2 . But the stress induced
pd should not be more than 80 N/mm2 . If the value of t is taken more than 2.1875 cm (say t -
2.188 cm), the stress induced will be less than 80 N/mm2 .
=

ltd

N/mm1.2 x 1.5 „2
2 x 1.5 x 10'2 - go , Ans. Hence take t = 2.188 cm or say 2.2 cm. Ans.
.
A mProblem 17.4. thin cylinder of internal diameter 1.25 contains a fluid at an internal
Problem 17.2. A cylinder of internal diameter 2.5 rn and of thickness 5 cm contains a
mmgas. If the tensile stress in the material is not to exceed 80 N/ pressure of 2 N/mm2. Determine the maximum thickness of the cylinder if:
2 determine the internal
,

pressure of the gas. (i) The longitudinal stress is not to exceed 30 N/mm 2
.

Sol. Given : N/mm(ii) The circumferential stress is not to exceed 45 2
Internal dia. of cylinder,
Thickness of cylinder, .

md = 2.5 Sol. Given : md = 1.25
mt = 5 cm = 5 x 10~2 Internal dia. of cylinder,

Maximum permissible stress = 80 N/mm2 Internal pressure of fluid, p - 2 N/mm2

As maximum permissible stress is given. Hence this should be equal to circumferential Longitudinal stress, a = 30 N/mm2
Circumferential stress, 2
stress (a ).
x .a, = 45 N/mm2
We know that the circumferential stress should not be greater than the maximum per-
Using equation (17.1),
missible stress. Hence take circumferential stress equal to maximum permissible stress.
pd
= 80 N/mm2
—— m—pxd 2 x 1.25
Let p = Internal pressure of the gas
Using equation (17.1), t = -r- r = 0.0277

pd = 2.77 cm.

Using equation (17.2),

— N/mmor pd

p- X 2 x 5 x 10 xj?0 (HHeerre uunnijtt 0of a, is in 2 it
-?JL =
, pd
d . 2.5
2 x 1.25 m: 0.0208
hence unit of p will also be in N/mm2)
4 x o2
= 3.2 N/mm2. Ans.
= 2.08 cm.

::

746 STRENGTH OF MATERIALS THIN CYLINDERS AND SPHERES

The longitudinal or circumferential stresses induced in the material are inversely pro- and the longitudinal stress (a2 ) is given as
portional to the thickness (t) of the cylinder. Hence the stress induced will be less if the value
pxd
of‘<’ is more. Hence take the maximum value of 7’ calculated in equations (i) and (ii)
a2~4:txr) —(17,4)
c

From the equations (i) and (ii) it is clear that t should not be less than 2.77 cm. Note, (i) In longitudinal joint, the circumferential stress is developed whereas in circumferen-

Take t = 2.80 cm. Ans. tial joint, the longitudinal stress is developed.

AProblem 17.5. water main 80 cm diameter contains water at a pressure head of (ii) Efficiency of a joint means the efficiency of a longitudinal joint.

100 m. If the weight density of water is 9810 N/m3 find the thickness of the metal required for (Hi) If efficiencies of a joint are given, the thickness of the thin shell is determined from
,

the water main. Given the permissible stress as 20 N/mm2 . (AMIE, Summer 1974) equation (17.3).

Sol. Given AProblem 17.6. boiler is subjected to an internal steam pressure of2 N/mm2. The thick-

Dia. of main, d = 80 cm ness of boiler plate is 2.6 cm and permissible tensile stress is 120 N/mm 2. Find out the maxi-
mum diameter, when efficiency of longitudinal joint is 90% and that of circumferential joint is
mPressure head of water, h = 100
40%. (AMIE, 1976)

Weight density of water, w-.py.g- 1,000 x 9.81 = 9810 N/m3 Sol. Given :

Permissible stress = 20 N/mm2 Internal steam pressure, p =2 N/mm2

Permissible stress is equal to circumferential stress (af) Thickness of boiler plates, t - 2.0 cm

or CTj = 20 N/mm2 Permissible tensile stress = 120 N/mm2

Pressure of water inside the water main, In case of a joints, the permissible stress may be circumferential stress or longitudinal

p=pxgxh -wh - 9810 x 100 N/m2 stress.

Here o is in N/mm2 hence pressure (p) should also be N/mm2. The value ofp in N/mm2 Efficiency of longitudinal joint, = 90% = 0.90
x ,

is given as Efficiency of circumferential joint, qc = 40% = 0.40.
Max. diameter for circumferential stress is given by equation (17.3).
|

9810x100 (v lm= 1000 mm)

mmP = iooo2 2 ^/mm2

= 0.981 N/mm2 .'. Using equation (17.3),

Let t = Thickness of the metal required. 0 = —£pxd
1 2 X T|; X /
Using equation (17.1),
where o = Given permissible stress = 120 N/mm2
pxd t

= V(Here ‘d’ is in cm hence -will also be in cm) 120 = —2 x d —— (Here p and a are in same units. Thickness
z2xx0u..9y0uxx2.0 1
2 xt 0.981x80
is in cm hence ‘d’ will be in cm)
pxd 2x20 = 2 cm. Ans.
120 x 2 x 0.9 x 2.0
* ~ 2 x ct = 216.0 cm.
x

17.6. EFFICIENCY OF A JOINT Max. diameter for longitudinal stress is given by equation (17.4).

The cylindrical shells such as boilers are having two types ofjoints namely longitudinal .-. Using equation (17.4),
joint and circumferential joint. In case of a joint, holes are made in the material of the shell for
the rivets. Due to the holes, the area offering resistance decreases. Due to the decrease in area, pxd
the stress (which is the equal to the force divided by the area) developed in the material of the
2 4 x t]c x t
shell will be more.
where a = Given permissible stress
2

Hence in case of rivetted shell the circumferential and longitudinal stresses are greater = 120 N/mm2
than what are given by equations (17.1) and (17.2). If the efficiency of a longitudinal joint and .
circumferential joint are given then the circumferential and longitudinal stresses are
120 = -
obtained as 4 x 0.4 x 2.0

Let n, = Efficiency of a longitudinal joint, and 120x4x0.4x2.0 ...(ii)
Tf. = Efficiency of the circumferential joint. a, = = 192 cm.

Then the circumferential stress (Oj) is given as z

pxd The longitudinal or circumferential stresses induced in the material are directly propor-

ai~ 2t x q, tional to diameter (d). Hence the stress induced will be less if the value of ‘d’ is less. Hence take

the minimum value of ‘d’ calculated from equations (i) and (ii).

...(17.3)

:

74g STRENGTH OF MATEFIJALS THIN CYLINDERS AND SPHERES

Maximum diameter of the boiler is equal to the minimum* value of diameter given by 120 x 4 x 0.3 x 15
2
mm“, -
aquation (i) and (ii). = 1080 r^y

Hence maximum diameter, d = 192 cm. Ans. SLT lh^'»“‘"“m *h* •** <» ** m«y b. satisfied

(Please note that if d is taken as equal to 216.0 cm, the longitudinal stress (a ) will be „ 1080

2

more than the given permissible value as shown below : (n) Permissible intensity of internal pressure when the shell diameter is l S
d = 1.5 = 1500 mm.
„ = P xd = - 2x216 - = 135 N/mm2.) m mKil or

2 4xric xt 4x 0.4 x 2.0 i (a) Taking limiting tensile stress = Circumferential Stress (of

mmProblem 17.7. A boiler shell is to be made of 15 thick plate having a limiting tensile = 120 N/mm 2

N/mmstress of 120 2 If the efficiencies of the longitudinal and circumferential joints are 70% - Using equation (17.3),

.

and 30% respectively determine :

(i) The maximum permissible diameter of the shell for an internal pressure of 2 N/mm2 1 2 X r); X t
,

and Por x 1500

(ii) Permissible intensity of internal pressure when the shell diameter is 1.5 m. ioa_. .. _
(AMIE, Winter 1981) IS ln N/mm hence ‘p’ will be in N/mm 2)
“ (Here
2~x 0.7 x 15

.Sol. Given " 120 x 2 x 0.7 x 15

mmThickness of boiler shell, t - 15 = i1.68 N/mm2 ...(f)

Limiting tensile stress = 120 N/mm2 (b) Takmg limiting tensile stress = Longitudinal stress (a,)
= 120 N/mm2
Limiting tensile stress may be circumferential stress or langitudinal stress.
.

Using equation (17,4),

Efficiency of longitudinal joint, r\ = 70% = 0.70 pxd
t

Efficiency of circumferential joint, r) = 30% = 0.30. On = 4 x rj
c c
xt
(i) Maximum Permissible diameter for an internal pressure,
-^i120=
p = 2 N/mm2. 500
The boiler shell should be designed for the limiting tensile stress of 120 N/mmz. First
4 x 0.30 x 15

consider the limiting tensile stress as circumferential stress and then as longitudinal stress. 120 x 4 x 0.30 x 15

The minimum diameter of the two case will satisfy the condition. m” 1500 =11-.4444 N/mm2 ...(«)

(a) Taking limiting tensile stress = Circumferential stress Hence . order both the conditions may be satisfied the mavirriim permissible internal
pressure is equal to the minimum* value (i) and (ii).
= 120 N/mm2 ofpressure given by
.
Maximum permissible internal pressure = 1.44 N/mm2
0l = 120 N/mm2
pxdN/mm(*Ifp is taken equal to 1.68 2 then longitudinal
But Oj is also given by equation (17.3) as stress (o2 ) will.be,
,

.pxd _ 1.68x1500
2 4xnc xr4^0.30xl5 =i4° N/mm -
0l “ 2 X T|( x /
This value is more than the given limiting tensile stress
2xd ^ ^ ^sure
120 =
Zffaacctor
2x0.7x15
mm" hence ‘d’ will be in mm)
(Here t is in to withstand maximum internal pres-
material of the cylinder is 300 N/mm 2
d, = 120x2x0.7x15 = 1260 mm ••a•"wi ooff s<affetty 3.0 *1 U 'tlr> lte ten8lle stress Ln the
and joint efficiency 80%, determine the diameter of the cylinder.
2•
(b) Taking limiting tensile stress = Longitudinal stress Sol. Given :

(o2) = 120 N/mm2 . Thickness of cylinder, t = 1.5 cm
o = 120 N/mm2
Internal pressure, p = 1.5 N/mm 2
2 Ultimate tensile stress = 300 N/mm 2

Using equation (17.4), Factor of safety = 3.0

° .'. Working stress or o _
2 4 x qc x t (
T ™W-
Ultimate tensile stress = 300 = 10<

or 4 x 0.30 x 15 Factor of safety

Joint efficiency, n = 80% = 0.80

..V- ;
..v&cK-i

d

THIN CYLINDERS AND SPHERES

means the efficiency of longitudinal joint (or q ), pd ppd
;
Joint efficiency

T = 0.80. 2tE 4tE 4t J
);
pd 1_ Ji ...(17.6):
The stress corresponding to longitudinal joint is given by equation (17.3).
= 2.
Using equation (17.3), 2tE

and longitudinal strain.

a_ 2 xr) xt Oz poi ...(17.7)
i

( EE

1,5 xd pd \tpd (substituting values of a and a2 )
= x
^ 00 = 2 x 0.80 x 1.5
4tE 2tE
100 x 2 x 0.80 x 1.5
d= = 160 cm = 1.6 m. Ans. pd fl I
L5
= ...(17.8)
2 tE

17.7. EFFECT OF INTERNAL PRESSURE ON TIDE DIMENSIONS OF A THIN But circumferential strain is also given as,

CYLINDRICAL SHELL 01 - Change in circumference due to pressure

When a fluid having internal pressure (p) is stored in a thin cylindrical shell, due to Original circumference

internal pressure of the fluid the stresses set up at any point of the material of the shell are . Final circumference - Original circumference
(i) Hoop or circumferential stress (Oj), acting on longitudinal section.
Original circumference
mitnhethster(Tteiihhs)sieLrsidoennptgsrhitiertneuctsdihsipieanrsladlapprrlseiatnrnpceerisipsinasc(lizappezal)rlaoanscaettsriienstsghrseeaosdnt,ihatailhsceskttnchreieesrsyscsua(wmrth)feieoarcfcehttnihitsneigavcleoyrnslyeipcnsrtdmiieaonrlnc.lipfaovlrerptlyhaisnnmecasylllTi.nhAdecetrssutaralenlsdsy
n(d + 5d) - nd

nd

7t + JtSd - ltd _ nbd

nd nd

can be neglected. M Change in diameter \
f

Let p = Internal pressure of fluid d[ Original diameter J

L = Length of cylindrical shell Equating the two values of e given by equations (17.6) and (17.9), we get

d = Diameter of the cylindrical shell x

t = Thickness of the cylindrical shell 5d Pd L pi

E = Modulus of Elasticity for the material of the shell d 2tE L 2J

.•. Change in diameter.

Oj = Hoop stress in the material

o = Longitudinal stress in the material
2

u = Poisson’s ratio Similarly longitudinal strain is also given as.

6d = Change in diameter due to stresses set up in the material Change in length due to pressure

6L = Change in length 2 Original length

SV = Change in volume. L
The values of a, and o2 are given by equations (17.1) and (17.2) as
Equating the two values of e given by equations (17.8) and (17.12).
pd
2
CTi= 2t
UbL pdf 1 ^^
pxd
L 2tE
°=

2 4f

Let e = Circumferential strain, Change in length,
1
pxixl
e = Longitudinal strain.
2
“ 2 tE (.2 )
Then circumferential strain, Volumetric strains. It is defined as change in volume divided by original volume.

...(17.5)

8V
VVolumetric strain =
~

) —: 2

STRENGTH OF MATERIALS THIN CYLINDERS AND SPHERES

But change in volume (8V) = Find volume - Original volume 5m=5xLength of shell, L = 100 = 500 cm
Original volume (V) = Area of cylindrical shell x Length
Internal pressure, p = 3 N/mm2
Young’s modulus, E = 2x 10s N/mm2

= 74 d2 x L Poisson’s ratio, p = 0.30

Final volume = (Final area of cross-section) x Final length (0 Change in diameter (8d) is given by equation (17.11) as

= 7 [d + 6d]2 x [L + 8L] pd
1 4.
fir/ Cl

7= [d2 +(8d)2 + 2d 6d] x [L + 6L] 3 x 100 2 l-|x0.30j

4 2 x 1 x 2 x 10 5

= 74 [d2 L + L(8d)2 + 2d Lbd, + 5Ld2 + bL (bd 2 + 2d 8d5L] —= [1 ~ 0.15] = 0.06375 cm. Ans.
40
Neglecting the smaller quantities such as (5d)2L, 8L(6d)2 and 2d bdbL, we get
(ii) Change in length (bL) is given by equation (17.14) as

Final volume = - [d2 L + 2d Lbd + bL d2] 6L = pdL f 1 "1

4 2tE L2

Change in volume (6V)

= 74 [d2 L + 2 dL8d + 8Ld2]- 74 d2 xL . 3 x 100 x 500 fl Q 3q1
2x1x2x 105 L2
j

= 74 [2d L8d + 6Ld2] —= x 0.20 = 0.075 cm. Ans.
40

—ov — [2d £8d + 2 (iii) Change in volume (6V) is given by equation '17. 18) as
SZ-d ]
8V= V&ei+ej)
F.= 4

Volumetric strains — d2 x L fv bd

26d 6L dL [ ei = T’*2 = -

+ ...(17.15) Substituting the values of bd, bL, d and L, we get

dL —bV=V |"
Tr 20
0.06375 0.075] .

= 2 c +£ = e2 ...(17.16) x +
2
1 T“ ’T6l 100 500 j

= V [0.001275 + 0.00015] = 0.001425 V.

2®2J5£ L 2 1,2 But V = Original volume = — d2 L

(Substituting the values of e and e 2)

x

2®P± f2 _2!i + l = 74 x 1002 x 500 cm3 = 3926990.817 cm3
i, 2 SV= 0.001425x3926990.817 = 5595.96 cm3. Ans.
2

^r2+ i-,-gi mProblem 17.10. A cylindrical thin drum 80 cm in diameter and 3 long has a shell

2Ei l 2 J thickness of 1 cm. If the drum is subjected to an internal pressure of 2.5 N/mm2 determine
,

(i) change in diameter, (ii) change in length and (iii) change in volume.

U= ...(17.17) Take E = 2 x 10s N/mm2 : Poisson’s ratio = 0.25. (Annamalai University 1990)
2Et
j

J

VAlso change in volume (6V) = {2e + e 2 ). ...(17.18) Sol. Given

l

Problem 17.9. Calculate : (i) the change in diameter, (ii) change in length and Diameter of drum, d = 80 cm
m(iii) change in volume ofa thin cylindrical shell 1 00 cm diameter, 1 cm thick and 5 long when Length of drum,
i = 3m = 3x 100 = 300 cm

subjected to internal pressure of 3 N/mm2. Take the value of E = 2 x 10s N/mm2 and Poisson’s Thickness of drum, t = 1 cm
Internal pressure,
ratio, p = 0.3. Young’s modulus, p = 2.5 N/mm2
E - 2 x 105 N/mm2
Sol. Given :

Diameter of shell, d = 100 cm Poisson’s ratio, p = 0.25

Thickness of shell, t = 1 cm '\

1

THIN CYLINDERS AND SPHERES

(i) Change in diameter (bd) is given by equation (17.11) as Now using equation(17.16), volumetric strain is given as

6V =

-y 2 el+ e

2

25x8°2 20

2 'x lx 2 x 10 5 °r 2827433 = 2 *i + e2 -

= [ ! _ 1 x 0.25 :. But e and e are circumferential, and longitudinal strains and are given by equation

L2 t 2

(17,6) and (17.8) respectively as
.

= 0.04 [1 - 0.125] = 0.035 cm. Ans 1
Ti 1
(ii) Change in length (6L) is given by equation (17.14) as
HL 2
pdL r 1 ]
pd (
2,5 X 80x30
2tE{2 H-

Substituting these values in equation (i), we get

= 1 x 2 x 10 5 [1- 0.2sl = 0.0357 cm. Ans. 2pd r 1 1 pd r 1 1

2x !_2 J 28274.33 ~ 2Et\} Xfl + -M

2 J 2tEL 2 J

(iii) Using equation (17.15) for volumetric strain I • we have - 2px20 r^.l^J px20 [-1 1

W 02 6d 6L 2x2 x 10s x 0.8 L 2 O.8x2xl0 5 |_2 03
T*T-ir-
J

0.000707 = —2— x 0.85 + x 0.20 = 1‘05p

(,. bd- 0.035, 6L = 0.0375^ 8000 8000 8000

0 X. 0.035 +. 0.0375 V ^ = 80, £ = 300 J p= 0.000707 x 8000 = 5.386 N/mm2.
80 300
. Ans.
= 0.000875 + 0.000125 = 0.001
(ii) Hoop stress (Oj) is given by equation (17.1) as
6V = 0.001 x V
pd = 5.386 x 20
°i = 2T 2 x 0.8 = 67‘33 N/mm2- Ana.

where volume V= — d2 x i = 74 x 80 2 x 300 = 1507964.473 cm3 Problem 17.12. A cylindrical vessel whose ends are closed by means of rigid flange

4

Change in volume, 6V = 0.001 x 1507964.473 = 1507.96 cm3. Ans. mmplates, is made of steel plate 3 thick. The length and the internal diameter of the vessel are

AProblem 17.11. cylindrical shell 90 cm long 20 cm internal diameter having thick- 50 cm and 25 cm respectively. Determine the longitudinal and hoop stresses in the cylindrical

mmness of metal as 8 is filled with fluid at atmospheric pressure. If an additional 20 cm3 of N/mmshell due to an internal fluid pressure of 3 2 Also calculate the increase in length, diam-

.

fluid is pumped into the cylinder, find (i) the pressure exerted by the fluid on the cylinder and Eeter and volume of the vessel. Take = 2 x 10s N/mm2 and p = 0.3. (AMIE, Winter 1984)

E N/mm(ii) the hoop stress induced. Take - 2 x 10s 2 and p = 0.3. (AMIE, Summer 1977) SoL Given :

Sol. Given : Thiekness, t =-33mmim = 0.3 cm

Length of cylinder., L = 90 cm Length of the cylindrical vessel, L = 50 ccjm

Diameter of cylinder, d = 20 cm Internal diameter, d = 25 ccim
Thickness of cylinder,
t - 8 mm = 0.8 cm Internal fluid pressure, p = 3 NN//i mm2
Young’s modulus, E£' = 2 x 11 0s N/mm2
Volume of additional fluid = 20 cm3

V= ~ d2 *L Poisson’s ratio, p = 0.3

Volume of cylinder, 4 = t4 *202 x 90 Let Oj = Hoop stress and

= 28274.33 cm3 o = Longitudinal stress.
2

Increase in volume, V6 = Volume of additional fluid Using equation (17.1) for hoop stress,

= 20 cm3 p xd 3 x 25 N/mm2.
~2 2x0.3
(t) Let p ~ Pressure exerted by fluid on the cylinder T= = = 1125 Ans.

E = 2 x 10s N/mm2 Using equation (17.2) for longitudinal stress,

p = 0.3. 4x0.3 = 62.5 N/mm2. Ans.

= 21 2

STRENGTH OF MATERIALS THIN CYLINDERS AND SPHERES

Using equation (17.5) for circumferential strain, Sol. Given : mmd = 1.5 ni = 1500
L = 4 m = 4000 mm
Ee _ Qi |x x 02 Dia.,
L
E Length,

Internal pressure, p = 3 N/mm2

Max. principal stress = 150 N/mm2

—= =• [125 - 62.5 x 0.3] ( v p = 0.3, o = 125 and a = 62.5) Max. principal stress means the circumferential stress
2 x 10 6 1 2
.'. Circumferential stress, a = 150 N/mm2
I^ ^^1 106.25 x

= (125 ' 18 ' 75) = Value of E = 2 x 10s N/mm2.

= 53.125 x 10-5 Poisson’s ratio, g = 0.25
. Let t = thickness of the shell,

... .

But circumferential strain is also given by equation (17.9) as bd - change in diameter,

6d bL = change in length, and
6l ~ d
Equating the two values of circumferential strain ev we get SV = change in volume.

~ = 53.125 x 10-5 (i) Using equation (17.1),

bd = 53.125 x 10-® x d = 53.125 x 10"5 x 25 = 0.0133 cm p xd (Here p and o, are in same units, ‘d’
/. Increase in diameter, 5d = 0.0133 cm. Ans.
Longitudinal strain is given by equation (17.7) as 01 " 21 mmis in hence 't' will be in mm)

bL = ct2 (x x o p xd _ 3 x 1500
~E 1 2xa 2x150

x

= 15 mm. Ans.

T Ee = ~ Hi) Using equation (17.11),
2

=~ .-5 [Oa~-iht*xuqiJJ 2txE{ 2 J

jjj

= 1 - x = [62.5 -- 37.5]

-g5 [62.5 125 0.3] - ;10r“sg 3 x 1500 z
10 2 x 15 x 2 x 105 ^1 -
2x 2 x x 0.25 = 0.984 mm. Ans.

— —= j
—— = Xx 10- 5
: 12.5
2 x10s
(Hi) Using equation (17.14),

.-. Increase in length, bL ~ 12.5 x 0~® x L bL = —pxdxL (
2t x E 1
1

= 12.5 x 10"s x 50 = 0.00625 cm. Ans. u

Volumetric strain is given by equation (17.16), as 3 x 1500 x 4000 f 1

2x 15x2 x 10 s
7'I6V ' !2n bd + bl U(rH~
mm= 0.75
V dl 10-® . Ans.

= 2e + e = 2 x 53.125 x 10~® + 12.5 x

t 2 (iv) Using equation (17.17),

= 106.25 x 10-5 + 12.5 x HP® = 118.75 x 10"5

.-. Increase in volume, V 2Ext\2 )

bV= 118.75 x 10-5 x V

volume - j d‘ x Lj ~ 3 x 1500 (5 „ 33 xx1155000xx 22
2 x 2 x 10 s x 15 11 2
. 116.75 x KTS x j . 25s x 50 ) 4 x 10s x 15

= 29.13 cm2. Ans. UV = -?— xf-xd2 xil

m mAProblem 17.13. cylindrical vessel is 1.5 diameter and 4 long is closed at ends by 2000 )

rigid plates. It is subjected to an internal pressure of3 N/mm2. If the maximum principal stress mmx 1500 2 x 4000 = 10602875 3 Ans.
find the thickness of the shell. Assume E = 2 x 105 N/mm and .
N/mrn2
is not to exceed 150 , j

Poisson’s ratio - 0.25. Find the changes in diameter, length and volume of the shell.

(AMIE, Winter 1988)

: J

THIN CYLINDERS AND SPHERES

A mmProblem 17.14. closed cylindrical vessel made of steel plates 4 thick with plane 750 750 x 1.928
(2.5 - 0.572) =
ends, carries fluid under a pressure of 3 N/mm2. The dia. of cylinder is 25 cm and length is 16.8 x 10 s ' 16.8 x 10s

calculate the longitudinal and hoop stresses in the cylinder wall and determine the 750 x 1.928
in diameter, length and volume of the cylinder.
Take = 2.1* 10s N/mm and p = 0.286.
75 cm, 16.8 x 10 s
m U-
Echange

(AMIE, Summer 1990)

750 x x f*x 2502 x 760

Sol. Given : 16.8 xlO 5 )
Thickness, J
Fluid pressure, t- 4 mm
Diameter, V = ^d 2 xL-^x 250 2 x
Length, p — 3 N/mm2 750
Value of
• d = 25 cm = 250 mm mm= 31680
L = 75 cm = 750 mm
3 Ans.
E = 2.1 x 105 N/mm2
.

AProblem 17.15. cylindrical shell 3 metres long which is closed as the ends has an
minternal diameter of 1 and a wall thickness of 15 mm. Calculate the circumferential and

Poisson’s ratio, p = 0.286 longitudinal stresses induced and also changes in the dimensions of the shell, if it is subjected

Let cr, = Hoop stress, to an internal pressure of 1.5 N/mm2. Take E = 2 x 10s N/mm2 and p = 0.3.

o = Longitudinal stress, (AMIE, Summer 1983 Annamalai University 1991)
2 ;

bd = Change in diameter, Sol. Given

6L = Change in length, and Length of shell, mZ, = 3 = 300 cm
Internal diameter, d=lm = 100 cm
5V = Change in volume.

(i) Longitudinal stress is given by equation (17.2) as Wall thickness, mmi = 15 = 1.5 cm

° pxd Internal pressure, p - 1.5 N/mm2
2 Young’s modulus,
~ 4xt E = 2 x 105 N/mm2

= 3 * 2 -- == 46.875 N/mm2. Ans. Poisson’s ratio, p = 0.3
4x4
Let CTj = Circumferential (or Hoop) stress, and

{ii) Hoop stress is given by equation (17.1) as o = Longitudinal stress.
2
pxd
Using equation (17.1) for hoop stress,
=
0l 2 x t pd

= 3 x 250 N/mm= 9„3„._7_5 . .. o2 Aans.

.

2x4 1.5 x 100

(Hi) The change in diameter is given by equation (17.11) as = - = 50 N/mm2. Anss.
2 x 1.5

bd=P^(l-^) Using equation (17.2) for longitudinal stress,
2txE\ f
pxd
2J
°2= ^r
3 x 250 2 1 1 - -1 x 0„.286 I = 0.0956 mm.
Ans. 1.5x100
N= /mm2.
2 x 4 x 2. lx 10 s v. 2 J . x _ = 25 Annss.
,1.5
4

(to) The change in length is given by equation (17.14) as Changes in the dimensions

bL - f--.nl Using equation (17.11) for the change in diameter (8<f),

2Ext l2 *)

— 3x250x750 f—l_ U0i2M8uuq| _— 0.0716 mm. Ans.

:1 1

2 x 2.1 x 10 5 x 4 {2 J 1.5 x 10Q Z M-|-
2 x 1.5x2 x 10 s
(u) The change in volume is given by equation (17.17) as ) (v p = 0.3)

V 2txE \2 J - 1 (1-0.15)= °' 85

3 x 250 4 x 10 3 4 x 10¥3
2 x 4 x 2.1 x 10 5
- 2 x 0.286 = 0.2125 x 10-3 cm. Ans.

- .

STRENGTH OF MATERIALS THIN CYLINDERS AND SPHERES 761

Using equation (17.14) for change in length, we get 25000 --2x0.33
31908528
^- -^p x d x L ( 1 2 x 2.1 x lO 3 x 8 .2 J
8L = 2
11 —25000 x 2 x 2.1 x 10” x 8

N/mmFp
1.5 x 100 x 300 f1 = VA’ A = 7.77 2 Ans.
(2.5
2x 1.5x2 x 10 8 .
U ^= -
Q 31908528 x 184 x 0.66)

' (it) Using equation (17.1),

J

x 100 x 300
4 x 10 s6
—= 10 0.06
x ._ = ~ = 0.015 cm. Ans. —pxd 7.77 x 184
°' 2 Oj = “ = 89.42 N/mm2. Ans.
7
4 2x8

Using equation (17.17) for volumetric strain, we get mm mProblem 17.17. A hollow cylindrical drum 600
in diameter and 3 long, has a

“L.psiri.sj shell thickness of 10 mm. If the drum is subjected to an internal air pressure of 3 N/mm2
,

V 2 Et |_2 J Edetermine the increase in its volume. Take = 2 x 105 N/mm2 and Poisson’s ratio = 0.3 for the

5 x 100 _ material. (AMIE, Winter 1986)

= = [2.5- 2x0.3] (v p = 0.3) Sol. Given :
External diameter,
2 x 2 x 105 x 1.5 Length of drum, D = 600 mm
Thickness of drum, L = 3 m = 3000 mm
= 0.25 x 10-3 x [2.5 - 0.6] Internal pressure, t - 10 mm
Young’s modulus,
= 0.25 x 10 3 x 1.9 = 0.475 x lO^3 p - 3 N/mm2
Poisson’s ratio, E = 2 x 105 N/mm3
V.-. Change in volume, 6V = 0.475 x 10 3 x Internal dia.,
p = 0.3
’"here V = Original volume
mmd = D — 2 x f = 600 — 2 x 10 = 580
= ^d2 xL=:4x 1002 x 300 = 2356194.49 cm3.
44 Using equation (17.17),
6V = 0.475 x 10-3 x 2356194.49 = 1119.19 cm3. Ans.

AProblem 17.16. thin cylindrical shell with following dimensions is filled with a liquid

'w atmospheric pressure : Length = 1.2 m, external diameter = 20 cm, thickness of metal = 8 mm.

Find the value of the pressure exerted by the liquid on the walls of the cylinder and the' V 2E x t\2 J

op stress induced if an additional volume of 25 cm3 of liquid is pumped into the cylinder.- — 3 x 580 . - 2 x 0.3 = 0.000435 x 1.9 = 0.0008265

Take E =2.1 x 10s N/mm2 and Poisson’s ratio = 0.33. (AMIE, Summer 1989) =- X 11(0 j
2 x 2 x 10 s

Sol. Given : 8V= 0.0008265 x V
Length,
External dia. L = 1.2 m = 1200 mm d2 * 580 2 x 3000^
Thickness,
D = 20 cm = 200 mm x x Lj == 0.0008265 x
Internal dia.,
Additional volume, j
t - 8 mm ^= 0.0008265 x
Value of mm= 792623000
3 Ans.

.

dd == D--22xxtt- 200 -- 22 x 8 = 184 mm

mm mmb5V -= 25 cm3 = 25 x 103 n 3 = 25000 3 A17.8. THIN CYLINDRICAL VESSEL SUBJECTED TO INTERNAL FLUID PRES-
SURE AND A TORQUE
E = 2.1 x 105s N/mm2

Poisson’s ratio, p = 0.33 When a thin cylindrical vessel is subjected to internal fluid pressure (p), the stresses set

Let p = Pressure exerted, and up in the material of the vessel are circumferential stress and longitudinal stress o . These
2

ct, = hoop stress produced. two stresses are tensile and are acting perpendicular to each other. If the cylindrical vessel is

Volume of liquid or inside volume of cylinder, subjected to a torque, shear stresses will also be set up in the material of the vessel.

V= - d*xL Hence at any point in the material of the cylindrical vessel, there will be two tensile
4 stresses mutually perpendicular to each other accompanied by a shear stress. The major prin-
mmx 1842 x 1200 = 31908528 3
cipal stress, the minor principal stress and maximum shear stress will be obtained as given in

Art. 3.4.4 on page 106 and 108.

(i) Using equation (17.17), Let Oj = Circumferential stress (tensile)

V6 = p x < o = Longitudinal stress (tensile)
Y~ 2E~x 2

x = Shear stress due to torque.

762 STRENGTH OF MATERIALS THIN CYLINDERS AND SPHERES 763

Hence the material of the tube is subjected to two tensile stresses (i.e., a = 48 N/mm2
t

The major principal stress and a = 24 N/mm2 ) accompanied by a shear stress (x = 40 N/mm2 ).
2

.-. Maximum principal stress* is given by equation (17.19) as

Minor principal stress £1 +02 °I ~ °2 ...(17.20) Max. principal stress =
2
2

and maximum shear stress = — [Major principal stress - Minor principal stress]

A

...(17.21)

mm mmAProblem 17.18. thin cylindrical tube 80
internal diameter and 5 thick, is

A Nmmclosed at the ends and is subjected to an internal pressure of 6 N/mm2 . torque of2009600

is also applied to the tube. Find the hoop stress, longitudinal stress, maximum and minimum = 36 + 41.76 = 77.7 N/mm2 (tensile). Ans.

principal stresses and the maximum shear stress. (AMIE, Summer 1984) Minimum principal stress is given by equation (17.20) as

Sol. Given : d = 80 mm Min. principal stress —= Oj + a2
Internal diameter, t ~ 5 mm
Thickness of tube,
Internal pressure, p = 6 N/mm8
Torque applied,
T = 2009600 Nmm

Let a = Hoop stress and = 36-41.76
t
a, = Longitudinal stress. = - 5.76 N/mm2 (compressive). Ans.

Using equation (17.1) for hoop stress, (- ve sign shows compressive stress)
Max. shear stress is given by equation (17.21) as
o, = —pxd - 6 x 80 = 48 N/mm2„. Ans.
1 2t 2x5 —Maximum shear stress = Max. principal stress - Min. principal stress

Using equation (17.2) for longitudinal stress,

— — —o,2 =
pxd = —6 x 8—0 - 24 N/mm2„. Ans. _ 77.76 - (- 5.76) — 77.76 + 5.76 _ 83.52
- ~ 2 2 2
4t 4x5

Maximum and minimum principal stresses = 41.76 N/mm2. Ans.

The stresses <ij and a are tensile stresses. But the cylindrical tube is also subjected to Problem 17.19. A copper cylinder, 90 cm long, 40 cm external diameter and wall thickness
2
mm6 has its both ends closed by rigid blank flanges. It is initially full of oil at atmospheric
torque. Due to torque, shear stress will be produced in the tube. As the thickness of the wall of
pressure. Calculate the additional volume of oil which must be pumped into it in order to raise
the tube is small, the shear stress is assumed to be uniform throughout the thickness.
the oil pressure to 5N/mm2 above atmospheric pressure. For copper assume E = 1.0 x 10s N/mm2
Let x = Shear stress in the wall of the tube and Poisson’s ratio - 1/3. Take bulk modulus ofoil as 2.6 x 103 N/mm2. (AMIE, Winter 1983)

Shear force = Shear stress x Area Sol. Given : L = 90 cm
Length of cylinder,
= x x (nd x t) (• Area = Circumference x Thickness = itdx t) External diameter,
= txnx80x5 Wall thickness,

= 400irx

—torque, T = Shear force x cL = 400it d D = 40 cm

xxx-and mmt = 6 = 0.6 cm.

: 400lt XXX Nmm16000it x x .-. Internal diameter, d = External dia. - 2 x Wall thickness

= 40 - 2 x 0.6 = 38.8 cm.

But torque applied Nmm= 2009600 (given) VInitial volume of oil, = Internal volume of cylinder

.'. Equating the two values of the torque, we get =7X^x1
160 003t X X = 2009600 4

2009600 N/mm2
.
= 40

16000H *See equation 3.15 on page 106.

STRENGTH OF MATERIALS Ti-IIM PVI ihincoc A tin Phucnrn

Increase in oil pressure. = - x 38.8 x 90 = 106413 cm3 Resultant additional space created in the cylinder
Young’s modulus for copper, = Increase in volume of cylinder + Decrease in volume of oil
4
= SVj + 6V
p ~ 5 N/mm!. 2

E = 1.0 x 10s N/mm2 = 314.98 + 204.64 = 519.62 cm3 .
.'. Additional quantity of oil which must be pumped in order to raise the oil pressure to
Poisson’s ratio,
5 N/mm2
*Bulk modulus of oil, k = 2.6 x 10s N/mm2 = 519.62 cm3. Ans.

Due to internal pressure of fluid inside the cylinder, there will be a change in the dimen- 17.9. WIRE WINDING OF THIN CYLINDERS
sions of the cylinder. Due to this, there will be a increase in the volume of the cylinder. Let us
We have seen in previous articles that hoop stress (or circumferential stress) is two
first calculate this increase in volume of the cylinder.
times the longitudinal stress in a thin cylinder, when the cylinder is subjected to internal fluid
Let 6V, = Increase in volume of cylinder. pressure. Hence the failure of a thin cylinder will be due to hoop stress. Also the hoop stress
which is tensile in nature (for a given internal diameter and thickness of a pipe) is directly
Then volumetric strain = . proportional to the fluid pressure inside the cylinder. Hence the maximum fluid pressure in-

But volumetric strain due to fluid pressure is given by equation (17.17), as side the cylinder is limited corresponding to the condition that the hoop stress reaches the
permissible value. In case of cylinders which have to carry high internal fluid pressures, some
— J—gy pci -f 5 - ^ (Here SV, = Increase of volume instead of 6V) methods of reducing the hoop stresses have to be devised.
1= 2p.
V 2Et\2 ) 1

-

5 x 38.8 i-2xi

2 x 1.0'x 10 s x 0.6 2 3

5 x 38.8 (v V = 106413 cm 3

2 x 105 x 0.6 (2.5 - 0.667) = 0.00296. )

bVj = 0.00296 x V

= 0.00296 x 106413

= 314.98 cm3 .

As bulk modulus of oil is given, then due to increase of fluid pressure on the oil, the ! Unit U

1 length 1

original volume of oil will decrease. Let us find this decrease in volume of the oil. -.

Let SV2 = Decrease in volume of oil due to increase of pressure. °w = Winding stress in wire (tensile)
Bulk modulus is given as
o - Compressive stress exerted by wire on cylinder
c

Fig. 17.5

Increase in pressure of oil One method is to wind strong steel wire under tension on the walls of the cylinders. The
effect of the wire is to put the cylinder wall under an initial compressive stress. If now this
( Decrease in volume of oil

^ Original volume of oil cylinder is subjected to internal fluid pressure, the walls of the cylinder will be subjected to
hoop tensile stress. The net effect of the initial compressive stress due to wire winding and
pxV
those due to internal fluid pressure is to make resultant stress less. The resultant stress in the
6V2
material of the cylinder will be the hoop (tensile) stress due to internal fluid pressure minus

the initial compressive stress. Whereas the stress (tensile) in the wire will be equal to the sum

of the tensile stress due to internal pressure in the cylinder and initial tensile winding stress.

k If ow = Initial winding stress in wire (tensile)

o = Compressive circumferential stress exerted by wire on cylinder
€

5 x 106413 = 204.64 cm3 . The relation between aw and a is obtained by considering a length ‘L’ of cylinder [Refer
c
3
2.6 x to Fig. 17.6 (a) and (6)].
10

Increase of pressure
* Bulk modulus, k = ' Decrease in volume )

, Volume )

THIN CYLINDERS AND SPHERES 767

= L x — x dw x aw
J4

2xLxixaCompressive force exerted by wire on cylinder for length L =
c

For equilibrium

Initial tensile force in wire = Compressive force on cylinder

or L x —it dw x a -2xLxtx a
x

2

or a = ———it x dw x ow ...[17.21 (A)]
c

a* = Circumferential stress developed in the cylinder due to fluid pressure only

(tensile)

ow* = Stress developed in the wire due to fluid pressure only (tensile)

Then resultant stress in the cylinder = (a* - a.)

and resultant stress in wire = ( aw + a *)

Hence wire-winding of thin cylinder is used :

(i) to increase the pressure-carrying capacity of the cylinder, and

(ii ) to reduce the Chances of bursting of the cylinder in longitudinal direction.
The hoop stress (or circumferential stress) will be set up in the material of the cylinder
if the bursting of the cylinder takes place in logitudinal direction. In wire winding of thin

cylinders, the bursting of the cylinder should be considered only in longitudinal direction.

In such cases the bursting force (due to internal pressure) per cm length will be resisted

by the pipe as well as the wires, offering tensile stresses.

The bursting force due to fluid along longitudinal section per cm length = p x d x L

(Here p, d and L should be in consistant units i.e., if p is in N/mm 2 d and L should be in mm.
,

Hence L = 1 cm = 10 mm.)

Resisting force of cylinder along longitudinal section per cm length

- a* x2L x t
C

where a * = circumferential stress in cylinder due to fluid pressure.
c
(Here L and t are in mm. L = 1 cm = 10 mm)

Resisting force of wire per cm length

= No. of turns of wire x (2 x Area of cross-section of wire)

x Stress in wire due to fluid pressure

_ n .„ ... (wh, ere n = L and, d, w = Dia. of wire
=nx2x dw

= L x —n x dw 2i, x aaw** =L x —n xdw x am*

~z—• 2 2

dw

Bursting force due to fluid pressure

= Resisting force of cylinder + resisting force of wire

p x d x L = a* x 2L x t + L x — x dw + a

or p x d = a* x 2t + — x dw x ow* ...[17.21 CB)]

The circumferential strain in the pipe is also equal to the strain in the steel wire.

Since the wire and cylinder remain in contact, the circumferential strain in the cylinder
should be equal to the strain in the steel wire. Due to fluid pressure, the stresses set up in the

cylinder are circumferential stress and longitudinal stress. But in the wire, there is only one

stress.

9

STRENGTH OF MATERIALS THIN CYLINDERS AND SPHERES

Circumferential strain in cylinder = Strain in wire mm mm- 2 x 10 x 12
2 (v Here / = 1 cm = 10 and / = 12 mm)

Initial compressive stress in the material of the cylinder due to wire winding,

Ee = Young’s modulus for cylinder _ Total compressive force on the cylinder
Ew = Young’s modulus for wire
,vhere Sectional area of cylinder
a* = Circumferential stress in cylinder due to fluid pressure
—= 471 = 19.63 N/mm2. Ans.
CT;* = Longitudinal stress in cylinder due to fluid pressure
2 x 10 x 20

pxd Hence before the water is admitted into the pipe, stresses in the pipe and wire are :
~ 4 xt
In the pipe = 19.63 N/mm2 (compressive)

In the wire = 60 N/mm 2 (tensile) '•

ow* = Stress developed in wire due to fluid pressure. (ii) Stresses due to fluid pressure alone

mm mmProblem A17.20. cast iron pipe of200 Let a* = Stress in the pipe due to fluid pressure 3.5 N/mm2 alone
internal diameter and 12 thick is wound aw * - Stress in the wires due to pressure 3.5 N/mm2 alone.

mmclosely with a single layer of circular steel wire of 5 diameter, under a tension of 60 N/mm2. Consider 1 cm length of the pipe.

Find the initial compressive stess in the pipe section. Also find the stresses set up in the pipe

Eand steel wire, when water under a pressure of 3.5 N/mm2 is admitted into the pipe. Take for The force of fluid which tends to burst the cylinder along longitudinal section

N/mmlast iron as 1 x 10s N/mm2 and for steel as 2 x 105 2 Poisson’s ratio is given as 0.3.

.

Sol. Given: = p.d.l = 3.5 x 200 x 10 (v /=lcm=10mm)

mmInternal diameter of pipe, d = 200 -wire = 7000 N ...(i)

/ Cylinclor This bursting force is resisted by the cylinder as well as wire.

Pipe thickness, t- 12 mm

Diameter of wire = 5 mm Resisting force of cylinder

Tension in wire = 60 N/mm2 Js. _ = Stress in the cylinder x Area of cylinder resisting
FFiigg.’ 17.,7
Water pressure, p = 3.5 N/mm2 t0 a T] = c * x 21 x i = a* x 2 x 10 x 12
i
E for C.I., £ = 3xlOs N/mm2 '1 = 240 ct *
£ for steel, f aw c
CTw
Poisson’s ratio, ' E =2 x 105 N/mm2
s ~Resisting force of wire = No. of turns x ^2 x x 5 2

p = 0.3 j

(i) Before the fluid under pressure is admitted in the cylinder x Stress in wire due to fluid pressure

Let aw = Winding stress in wire = 2 x ^2 x -j x 5 2 x aw* = 78.54 aw*

= 60 N/mm2 j

P,, = Compressive force exerted by wire on cylinder .-. Total resisting force

a = Compressive stress exerted by wire on cylinder. = 240 a * + 78.54 aj ...(ii)
c c

Consider one cm length of the pipe. Number of turns of the wire of 1 cm pipe length Equating the resisting force to the bursting force given by equations (t) and (it), we get

_ Length of pipe _ 1cm = 240 a* + 78.54 aw* = 7000 „.(tit)
Diameter of wire 0.5 cm
Now we know that circumferential strain in cylinder is equal to the strain in wire.
The compressive force exerted by one turn of the wire on the cylinder (see Pig. 17.7)
But circumferential strain in cylinder
= 2 x Area of cross-section of wire x a
~-Circumferential stress Longitudinal stress x p

= 2 x - x 52 x 60 N ('pxd'l [t <" For pipe, E = E)

4 = MtT- x ¥-

.-. Total compressive force exerted by the wire on the cylinder per cm length of the pipe 3.5 x 200
= No. of turns x Force exerted by one turn

N~= 2 X {% x X 5 Z X 60 = 4712 (a* - 4.375) ...(to)

j

Sectional area of the cylinder which takes this compressive force

=2 xl xt

_

THIN CYLINDERS AND SPHERES 771

770 STRENGTH OF MATERIALS The stress is tensile in nature.
The fluid inside the shell is also having tendency to split the shell into two hemispheres
Strain in wire E E(v For wire = ) ...(u)

Equating equations (iv ) and (v), we get along Y-Y axis. Then it can be shown that the tensile hoop stress will also be equal to Let

T (V- 4.375)=^ this stress is a2 .

The stress cj2 will be at right angles to or internal diameter

A mProblem 17.21. vessel in the shape of a spherical shell of 1.20
and 12 mm. shell thickness is subjected to pressure of 1.6 N/mm2. Determine the stress induced
2 x 10“ <o * - 4.375) = 2(o * - 4.375)
'
= 1 x l1r0tS -I in the material of the vessel.

Substituting the above value in equation {iit), we get Sol. Given : m mmd = 1.2 = 1.2 x 10s
Internal diameter,
240 a* + 78.54 x [2(0,,* - 4.375)] = 7000
£ - 12 mm
' 397.08a * = 7687.225 Shell thickness,
c
or p = 1.6 N/mm2

7687.225 N/mm19.36 _ r , ... Fluid pressure,
(tensile)
a* = = 2 The stress induced in the material of spherical shell is given by,

39? Q8

Substituting the value of a* in equation (oi), we get £—„ - p.d = 1.6x1.2x10* = 40 N/mm2„ . .

aw * = 2(19.36 - 4.375) = 29.97 N/mm2 (tensile). Ans.

(its) Resultant stresses in pipe and wire 1 At 4 x 12

Resultant stress in pipe A mProblem 17.22. spherical vessel 1.5 diameter is subjected to an internal pressure of
= Initial stress in pipe + Stress due to fluid pressure alone
2 N/mm2 . Find the thickness of the plate required if maximum stress is not to exceed 150 N/mm2

and joint efficiency is 75%.

= 19.63 (compressive) + 19.36 (tensile) Sol. Given : m mmd = 1.5 == 1.5 x 103
Diameter of shell,
= 0.27 N/mm2 (compressive). Ans. Fluid pressure, p -~ 2 NN//mmim2
Stress in material, o0 = 150 NN//; mm2
Resultant stress in wire Joint efficiency,
= Initial stress in ware + Stress due to fluid pressure alone !1

= 60 (tensile) + 29.97 (tensile) q = 75% or 0.75

= 89.97 N/mm2 (tensile). Ans. Let £ = thickness of the plate.

The stress induced is given by,

17.10.THIN SPHERICAL SHELLS ^//' p.d
/' JA °
Fig. 17.8 shows a thin spherical shell of internal diam- j
eter 'd' and thickness '( and subjected to an internal fluid pres- l 4£.q
sure ‘p\ The fluid inside the shell has a tendency to split the
shell into two hemispheres along x-x axis. p.d 3

The force (P) which has a tendency to split the shell 4.T1.0! 2 x 1.5 x 10

: 6.67 mm. Ans.

4 x 0.75 x 150

Resisting 17.11. CHANGE IN DIMENSIONS OF A THIN SPHERICAL SHELL DUE TO AN
area
INTERNAL PRESSURE
M-t) t
In previous article, we have seen that the stresses 0 and 0 at any point are equal
j 2
The area resisting this force = K.d.t
Hoop or circumferential stress (Oj) induced in the (each equal to p.d/At) and like. There is no shear stress at any point in the shell.

material of the shell is given by. —Maximum shear stress = jZi 12—P x _ P * d = 0 o2 are acting at
2 , and
Force P . 0 . The stresses
0l Area resisting the force P
]

At At )

Fig. 17.8 right angles to each other.

yp x —J£ d ... Strain in any one direction is given by,

4 , a p x q
e= 1 t
p.d
~E
“ E

O; HXO[ STRENGTH OF MATERIALS
°i = <r 2

We know that strain in any direction is also

_ 6d
d

—Volumetric strain ^
[ I
v)l

The ratio of change of volume to the original volume is known as volumetric strain. If
—V= original volume and dV = change in volume. Then
= volumetric strain.

VTLervtf- T 7 —= Origina_li voilume

= -d 3 v — —VFor a sphere,
= 3 - d3
jxr

Taking the differential of the above equation, we get 36

dV= ~ x 3d2 x d(d)

Hen. V ~ *xd3

6

did)

_3

d —Hi)

But from equation (i), we have
— WIT 0*
&d d(d) p x d

<!-•*>

Substituting this value in equation Hi), we get

dV 3 x pxd
.

V “ 4tE 1_|X)
-

XM Pm WW. &AProblem 17.23.
D^muna m mmU,o an i. rnal
spherical shell of internal diameter 0.9 and of thickness 10 is

of ,.4 incr^e in

in volume. Take E = 2* 10s N/mm2 and u = -

3'

Sol. Given
:

Internal diameter, m mmd = 0.9 = 0.9 x 103
Thickness of shell, t = 10 mm

Fluid pressure, p = 1.4 N/mm2

Value of E = 2 x 105 N/mm2

Value of jw = —

Fig. 17.9