99

10

Thick Cylinders

When analysing thin cylinders, the assumption made is that the stress in the wall of the cylinder

will be uniform across the thickness when the cylinder is under internal pressure, and that the

radial stresses will be negligible in comparison with the circumferential (hoop) and longitudinal

stresses.

When the thickness of the cylinder is appreciable in relation to the diameter, this assumption

cannot be justied and the variation in radial and circumferential stresses across the thickness can

be deduced using Lamé’s theory.

=

σ+

σ−

−

t

d

2

P

P

1 (10.1)

where

t = thickness (m)

d = internal diameter (m)

σ = maximum hoop stress (Pa)

P = radial pressure (Pa)

Lamés theory

General formulae for shrink or force ts:

()

δ

2

=σ−σ

R

E

AB

12

(10.2)

For a cylinder subject to internal pressure (see Figure10.1):

σ=

+

−

A

P

rr

rr

Maximumstress

o

2

i

2

o

2

i

2

1

(10.3)

where

σ=

−

2Pr

rr

B

1

2

o

2

i

2

1

(10.4)

For a cylinder subject to external pressure (see Figure10.2):

σ=

−

2Pr

rr

(Maximum stress)

A

2

2

o

2

i

2

2

(10.5)

σ=−

+

−

P

rr

rr

B

o

2

i

2

o

2

i

2

2

(10.6)

100 Design Engineer's Case Studies and Examples

where

r

i

= internal radius (m)

r

o

= external radius (m)

t = thickness (r

o

– r

i

) (t).

σ = maximum hoop stress N/m

2

P = pressure (N/m

2

)

When two components are of different materials the above formula can be written as:

()

()

δ

2

=σ−ν −σ−ν

1

R

E

P

R

E

P

AB

2

12

(10.7)

E

1

and E

2

= modulus of elasticity for the two different materials

ν

1

and ν

2

= Poisson’s ratio for the two different materials

Example 10.1

A cast steel cylinder for a hydraulic press has an inside diameter of 230 mm. The internal pressure

is 300 bar. Using a maximum hoop stress of 85 MPa, determine the thickness of the cylinder.

Solution:

From Lamé’s formula:

t

d

2

P

P

1=

σ+

σ−

−

(10.8)

r

i

r

o

P (pressure N/m

2

)

FIGURE 10.1 Thick cylinder subject to internal pressure.

P (pressure N/m

2

)

r

i

r

o

FIGURE 10.2 Thick cylinder subject to external pressure.

101Thick Cylinders

t

0.230 m

2

85 x10N/m 30.0x10 N/m

85 x10N/m 30 x10N/m

1

62 62

62 62

=

+

−

−

t = 0.05129m

Example 10.2

A cast iron hydraulic cylinder has an internal diameter of 150 mm and an external diameter of

200mm. Internal pressure is 52 bar. Calculate the maximum hoop stress. Also determine the stress

at the outer surface.

Solution:

The maximum hoop stress occurs at the internal surface:

P

rr

rr

A

2

2

1

2

2

2

2

2

1

σ=

+

−

=

+

−

52 x10Pax

0.1m 0.075m

0.1m 0.75m

5

22

22

= 18.571MPa

Now stress at the outer surface:

2Pr

rr

A

1

2

2

2

1

2

1

σ=

−

=

−

2x52 x10x0.075

0.10.075

52

22

= 13.371MPa

Example 10.3

A mild steel universal coupling is shrunk on to a mild steel shaft Ø200 mm (see Figure10.3).

Theshrink allowance is 1 mm per metre on diameter.

200.0

Dimensions in mm

200.0

340.0

FIGURE 10.3 Universal coupling for Example 10.3.

102 Design Engineer's Case Studies and Examples

Calculate:

a. The stresses in the shaft and coupling.

b. For a maximum torque of 63 kNm, ﬁnd the Factor of Safety (FoS) between the actual torque

and the theoretical torque capable of being transmitted.

Solution:

a. Coupling end

where:

r

i

= 100 mm

r

o

= 170 mm

at inner radius:

P

rr

rr

A

o

2

i

2

o

2

i

2

1

σ=

+

−

P

0.170 0.100

0.170 0.100

22

22

=

+

−

= 2.058 P

at outer radius:

2Pr

rr

i

2

o

2

i

2

σ=

−

Β

1

=

−

2xPx0.100

0.170 0.170

2

22

= 1.058 P

Consider the shaft:

For a solid shaft the hoop stress at the outside radius is equal to the radial stress since r

i

= 0; then

σ

B2

= – P.

σ

B2

is used to correspond to the symbol in the formula for a hollow shaft or cylinder externally

loaded.

r

E

AB

12

()

δ

2

=σ−σ

where

δ = total shrink allowance 1 mm/m × 0.2 m

= 0.2 mm

r = 100 mm

E = 200 × 10

9

N/m

2

(for steel)

Substitute σ

A1

and σ

B2

in terms of P and write the units in terms of mm.

Hence

0.2mm

2

100 mm

200 x10N/mm

2.058 P–(–1.058 P)

32

()

=

)

(

=+

0.1mm

100 mm

200 x10N/mm

x2.0058 P1.058 P

3

32

0.1 mm = 0.0005 mm

3

/N × (3.116 P)

103Thick Cylinders

=

0.10 N

0.005 mm

3.116 P

2

200 N/mm

2

= 3.116 P

200 N/mm

3.116

P

2

=

64.185 N/mm

2

= P or 64.185 MPa (Radial Stress) 1 N/mm

2

= 1 MPa

σ=2.058 xP

A

1

= 132.093 MPa

σ=

1.058

xP

B

1

= 67.908 MPa

b. Safety Factor (SF)

These are the maximum and minimum hoop stresses generated in the coupling end due to the

shrink ﬁt on the solid shaft.

For the shaft σ

B2

= P = 64.185 MPa.

The permissible maximum hoop stress should be within the elastic limit for the material

between 432 and 494 MPa.

For this example the FoS will be:

FoS=

allowablestress

workingstress

FoS

432 MPa

132.093 MPa

=

= 3.270

The radial stress P is the gripping force between the coupling end and the shaft, if the radial

stress P = 64.185 MPa and assuming a co-efﬁcient of friction of 0.25 between the two surfaces:

The radial stress acts on a surface area:

= πDL

= π × 0.2 m × 0.2 m

= 0.1257 m

2

Therefore the tangential force:

= 64.185 MPa × 0.126 m

2

× 0.25

= 2.0164 MN

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