Unit III Thick CylindersDifference between Thick cylinder and Thin cylinder:

Thick Cylinder Thin Cylinder1.Circumferential stress varies along the thickness of the shell.2. Radial stress is no longer negligible since a thick cylinder is required to have a heavy internal pressure.

1. Circumferential stress ‘f’ is constant throughout the thickness of the shell.2. Radial stress ‘p’ is negligible in comparison of ‘f’ and ‘f0’.

Lame’s theory:

Assumptions:

1. The material is homogeneous and isotropic2. Plane sections of the cylinder perpendicular to the longitudinal axis remain plane under

pressure.i.e., longitudinal strain is the same at all points in the cylinder wall. (i.e., it is independent of the radius)Hence to satisfy the requirements of uniform longitudinal strain, we have

e0 = 1E ( f 0−

f xm

+pxm )=¿ Constant

f x−px = Constant = 2A (say)f x = Circumferential stress (tensile)px = radial pressure.

Figure 1a shows a thick cylinder subjected to internal and external radial pressure. Consider an angular ring of the cylinder, of internal radius x and thickness δx. Let the internal radial pressure on this ring be px and external pressure (px + δpx). On any small element of this ring, fx is circumferential stress. The condition for equilibrium of one-half of thin ring are similar to those in the case of thin cylinder. Thus we have,the bursting force = ( p¿¿ x2 xl)−¿¿) 2 (x+ δx¿ l.

= 2l [– pxδx−xδ px−δx . δ px ]

= −2 l [ pxδx+ xδ px ] ( Neglecting the products of small quantities)

The resisting force = 2f x lδx .

For equilibrium we have, 2f x lδx = −2 l [ pxδx+ xδ px ] Or, f x =−[ px+x

δpxδx

]

In the limit when thickness of the element is reduced indefinitely,f x+ px+x dpxdx =0------------(i)Another relation is obtained from the assumption that the longitudinal strain is independent of x . Thus from the equation f x−px=2 A---------(ii)

And hence, f x=px+2 A and by substituting it in equation (i), one can obtain the following relation;

( px+2 A)+ px+x dpxdx =0

Or dpxdx

= −2 ( px+A )x

dpx(px+A )

=−2 dxx

Integrating, log e ( px+A ) = −loge x2+ logeB

Where log eB is a constant of integration.Therefore, log e ( px+A ) = log e

Bx2

px = Bx2 – A

From (ii), f x = Bx2 +AThus we have Lame’s equations,

Problems:

1. The internal and external diameter of a thick hollow cylinder are 80 mm and 120 mm respectively. It is subjected to an external pressure of 40 N/mm2 and an internal pressure of 120 N/mm2. Calculate the circumferential and radial stresses at the mean radius. (Oct’97)

Solution:

At x = r, pr = 120 N/mm2

¿ f x = Bx2 +A

At x=R , pR = 40 N/mm2

i.e., At r = 40mm, pr=120N /mm2

At R= 60 mm, pR=40N /mm2

Lame’s equations are:

Where px = radial stress at a radius x from the centre of the cylinder.

f x=¿ circumferential stress at a radius ‘x’ from the centre of the cylinder

From equation (1),

At x=r , pr = Br2 – A-----------(1a)

At x=R , pR = BR2 – A ----------------(1b)

From (1a), 120 = B402 −A

From (1b), 40 = B602 −A

------------------------------(1a) – (1b), 80 = B ( 1

402 −1

602 )B= 80∗402∗602

602−402 = 230400From (1a), 120 = 230400

402 −A

120=144−A

A= 24Radial stress at the mean radius ‘50 mm’ is:p50 = B

502 – A

px = Bx2 – A−−−(1)

¿ f x = Bx2 +A−−−(2)

= 230400

502 −24 = 68.16 N/mm2

Circumferential stress at mean radius ’50 mm’ is,

f 50 = B502 +A

= 230400

502 +24 = 116.16 N/mm2

2. A cylinder has an internal radius of 200 mm and external radius of 300 mm. Permissible stress for the material is 15.5 N/mm2. If the cylinder is subjected to an external pressure of 4 N/mm2, find the internal pressure that can be applied. (Apr’97)

Solution: r =20 mm; R= 300mm,

f r = 15.5 N/mm2 ; pR=4N /mm2

Lame’s equations are:

From (1), pR = BR2 – A

4 = B3002 −A ---------------(1a)

From (2), ¿ f r =Br2 +A 15.5 = B

2002 +A ----------------(2a) 4 = B

3002 −A ---------------(1a) 15.5 = B

2002 +A ----------------(2a)________________________________

px = Bx2 – A−−−(1)

(1a) + (2a), 19.5 = B ( 13002 +

12002 )

B= 19.5∗30 02∗20 02

2002+3002 = 540000From (1a), 4= 540000

3002 −A

A = 6-4 = 2Internal pressure, pr = B

r2 – A = 540000

2002 −2=11.5N /mm2

3. A pipe with internal diameter 400 mm is to carry a fluid pressure of 12 MPa. If the maximum stress in the material of the pipe is restricted to 110 MPa, calculate the minimum thickness of the pipe required. (Apr’96)

Solution:

d= 400 mm, r = 200 mm

pr=12MPa=12N /mm2

f r=110MPa=140N /mm2

Lame’s equations:

From (1), pr = Br2 – A

12 = B2002 −A ---------------(1a)

From (2), ¿ f r =Br2 +A 110 = B

2002 +A ----------------(2a)

px = Bx2 – A−−−(1)

12 = B2002 −A ---------------(1a)

_________________________________________________(2a) +(1a) is, 122=

2B2002

B = 122×2002

2 = 2440000

From (1a), 12 = 2440000

2002 −A

A = 49.

pR = BR2 – A =0

2440000R2 −49=0

R2=244000049 = 49795.9

R= 223.15 mm.

Thickness of the pipe is R-r = 223.15-200 = 23.15 mm.

4. A pipe with internal diameter 400 mm is to carry a fluid at a pressure of 10 MPa. If the maximum stress in the material of the pipe is restricted to 150 MPa, calculate the minimum thickness of the pipe required. (Apr’97, Prior to 93-94 batch)

Solution:

d= 400 mm

r = 200 mm

pr=10MPa=10N /mm2

f r=1 5 0MPa=15 0N /mm2

Lame’s equations: pr = B

r2 – A−−−(1)

400 mm

Pr=10 MPa

fr = 150 MPa

From (1), 10 = B2002 −A ---------------(1a)

From (2), 150 = B2002 +A ----------------(2a)

__________________________________________(1a) + (2a), 160 = 2B

2002

And hence, B = 160×2002

2 = 320×104

From (1a), 10 = 320×104

2002 −A

A = 70.

Since PR = 0,

BR2 – A =0

320×104

R2 −70=0

R2 = 320×104

R2 =45714.2

R=213.8Thickness of pipe = R-r = 213.8-200 = 13.8 mm.

5. A thick cylinder of external diameter 40 cm and internal diameter 30 cm is shrunk on to another cylinder of external diameter 30 cm and 5 cm thick. If the radial pressure at the junction due to shrink fit is 15 MPa, calculate the initial difference in radii at the junction. (Oct’96)