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 justied 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
R
E
AB
12
(10.2)
For a cylinder subject to internal pressure (see Figure10.1):
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 Figure10.2):
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
−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
= Poissons 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 = 0.05129m
Example 10.2
A cast iron hydraulic cylinder has an internal diameter of 150 mm and an external diameter of
200mm. 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
= 18.571MPa
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
= 13.371MPa
Example 10.3
A mild steel universal coupling is shrunk on to a mild steel shaft Ø200 mm (see Figure10.3).
Theshrink 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
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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