MECHANICAL PRINCIPLES H2

TUTORIAL 4

SELF ASSESSMENT EXERCISE No.1

1. A thin walled cylinder is 80 mm mean diameter with a wall 1 mm thick. Calculate the
longitudinal and circumferential stresses when the inside pressure is 500 kPa larger than on the
outside.
pD 500 x 10 3 x 90 x 10 −3
σC = = = 20x10 6 N/m 2
2t 2 x 0.001
pD 500 x 10 3 x 90 x 10 −3
σL = = = 10x10 6 N/m 2
4t 4 x 0.001

2. Calculate the wall thickness required for a thin walled cylinder which must withstand a pressure
difference of 1.5 MPa between the inside and outside. The mean diameter is 200 mm and the stress
must not exceed 60 MPa.
pD 1.5 x 10 6 x 200 x 10 −3
t= = = 2.5 x 10 -3 m
4σ C 4 x 1.5x10 6

3. Calculate the stress in a thin walled sphere 100 mm mean diameter with a wall 2 mm thick when
the outside pressure is 2 MPa greater than the inside. (Answer -25 MPa).
pD - 2 x 10 6 x 100 x 10 −3
σ= = = −25x10 6 N/m 2 (compressive)
4t 4 x 0.002

SELF ASSESSMENT EXERCISE No. 2

1. A cylinder is 200 mm mean diameter and 1 m long with a wall 2.5 mm thick. It has an inside
pressure 2 MPa greater than the outside pressure. Calculate the change in diameter and change in
volume. Take E = 180 GPa and ν = 0.3
pD(2 - ν) 2 x 10 6 x 200 x 10 −3 (1 − 2x0.3)
εC = = 9
= 377.87x10 −6
4tE 4 x 0.002 x 180x10
∆D ∆D
εC = = = 377.87 x 10 −6 ∆D = 0.075 mm
D 200
pD(5 - 4ν )
ε v = 2ε C + ε L = 844x10 −6 or = 844x10 −6
4tE
Volume = (π x 2002/4) x 1000 = 31.416 x 106 mm3
∆V = εv V = 26529 mm3 Final Volume = 31.416 x106 + 26529 = 31.442 x 106 mm3

2. A sphere is 50 mm mean diameter with a wall 0.5 mm thick. It has an inside pressure 0.5 MPa
greater than the outside pressure. Calculate the change in diameter and change in volume.
Take E = 212 GPa and ν= 0.25
(Answers 0.0022 mm and 8.68mm3)

pD(1 - ν) 0.5 x 10 6 x 50 x 10 −3 (1 − 0.25)

εC = = = 44.22 x10 −6
4tE 4 x 0.0005 x 212 x10 9
∆D ∆D
εC = = = 44.22 x 10 −6 ∆D = 0.0022 mm
D 50
ε v = 3ε C = 132.6
Volume = (π x 503/6) = 64450 mm3 ∆V = εv V = 8.68 mm3
3a. A thin walled cylinder of mean diameter D and length L has a wall thickness of t. It is subjected to
an internal pressure of p. Show that the change in length ∆L and change in diameter ∆D are
∆L=(pDL/4tE)(1 -2ν) and ∆D =(pD2/4tE)(2 - ν)

The derivation is given in the tutorial

b. A steel cylinder 2 m long and 0.5 m mean diameter has a wall 8 mm thick. It is filled and
pressurised with water to a pressure of 3 MPa gauge. The outside is atmosphere. For steel E= 210
GPa and ν=0.3. For water K = 2.9 GPa.
Calculate the following.
i. The maximum stress.
ii. The increase in volume of the cylinder.
iii. The volume of water at atmospheric pressure required.
pD 3 x 10 6 x 0.5
σC = = = 93.75x10 6 N/m 2
2t 2 x 0.008
pD(5 - 4νν 3x10 6 x 0.5 x (5 − 4 x 0.3)
εv = = 9
= 848.2 x 10 −6
4tE 4 x 0.008 x 210x10
Initial volume of water.= π(500)2(2000)/4 = 392699100 mm3
∆V = εvV = 848.2 x 106 x 392699100 = 333087 mm3
Volume of compressed water = 392699100 + 333087 = 393032187 mm3
K = pV/∆V V = k∆V/p = 2.9 x109 x ∆V/(3 x 106) = 966.7∆V
966.7∆V = 393032187 + ∆V
∆V = 406992 mm3
Volume of uncompressed water = = 393032187 + ∆V(water) = 393439179 mm3
Volume of water added to the vessel is 393439179 - 392699100 = 740079 mm3

4a. A thin walled sphere of mean diameter D has a wall thickness of t. It is subjected to an internal
pressure of p. Show that the change in volume ∆V and change in diameter ∆D are
∆V =(3pDV/4tE)(1 - ν) where V is the initial volume.
The derivation is given in the tutorial

b. A steel sphere 2m mean diameter has a wall 20 mm thick. It is filled and pressurised with water so
that the stress in steel 200 MPa. The outside is atmosphere. For steel E= 206 GPa and ν = 0.3.
For water K = 2.1 GPa. Calculate the following.
i. The gauge pressure
ii. The volume water required.

pD 4tσ 4 x 0.02 x 200x10 6

σ= p= = = 8 x 10 6 N/m 2
4t D 2
3pD(1 - ν) 3 x 8 x 10 6 x 2 x (1 − 0.3)
δV = V= 9
= 2.03 x 10 −3 V
4tE 4 x 0.02 x 206x10
Initial volume of water.= πD3/6 = 4.18888 x 109 mm3
δV = 8.5403 x 106 mm3
Final Volume = 4.18888 x 109 + 8.5403 x 106 = 4.1973 x 109 mm3
Volume of uncompressed water = K ∆V/p = 2.1 x 109x ∆V/(8 x 106)
4.1973 x 109 = 2.1 x 109x ∆V/(8 x 106) – ∆V = 261.5 ∆V
∆V (water) = 16.0509 x 103 mm3
V + ∆V = 4.1973 x 109 + 16.0509 x 103 = 4.2134 x 109 mm3
Volume added to vessel = 4.2134 x 109 - 4.18888 x 109 = 24.55 x 106 mm3
 SELF ASSESSMENT EXERCISE No.3

1. A thick cylinder has an outer diameter of 150 mm and an inner diameter of 50 mm. The OUTSIDE is
pressurised to 200 bar greater than the inside.

Calculate the following.

• The circumferential stress on the inside layer.

• The circumferential stress on the outside layer.

The boundary conditions are

Inner surface ri = 25 mm σR = 0 MPa (compressive)
Outer surface ro = 75 mm σR = 200 x 105 Pa (compressive)
Substituting into Lame's equation we have
σR = 0 = a - b/r2 = a - b/0.0252 = a – 1600b a = 1600b
σR = -20 x 106 = a - b/r2 = a - b/0.0752 = a – 177.8 b
-20 x 106 = a – 177.8 b = 1600b – 177.8 b = 1422.2b
b = -20 x 106/1422.2 = -14062.5
a = 1600b = -22.5 x 106

Now solve the circumferential stress on the inside. σc = a + b/ri2 = -45 MPa
Now solve the circumferential stress on the outside. σc = a + b/ro2 = -25 MPa

2. A thick cylinder has an outside diameter of 100 mm and an inside diameter of 60 mm. It is
pressurised until internally until the outer layer has a circumferential stress of 300 MPa.

Calculate the pressure difference between the inside and outside. (266.6 MPa)

The boundary conditions are

Inner surface ri = 30 mm σR = -pi
Outer surface ro = 50 mm σR = -po and take this as zero and solve pi relative to it.
Substituting into Lame's equation we have
σR = -po = 0 = a - b/ro2 a = b/ro2 = 400 b

Outer surface ro = 50 mm σc = 300 x106

Substituting into Lame's equation we have
σC = 300 x106 = a + b/ro2 = 400b + 400b = 800b
b = 375000 hence a = 150 x 106

On the inner surface ri = 30 mm

σR = -pi = a - b/ro2 = 150 x 106- 375000/0.032
pi = 266.7 MPa
The pressure difference is hence 266.7 MPa
3. A thick cylinder is 100 mm outer diameter and 50 mm inner diameter. It is pressurised to 112 MPa
gauge on the inside. Calculate the following.

• The circumferential stress on the outside layer

• The circumferential stress on the inside layer
• The longitudinal stress
• The circumferential strain in the outside layer
• The circumferential strain in the inside layer
• The change in the inner diameter
• The change in the outer diameter

Take E = 205 GPa and ν = 0.27

The boundary conditions are

Inner surface ri = 25 mm σR = -112 MPa (compressive)
Outer surface ro = 50 mm σR = 0
Substituting into Lame's equation we have
σR = 0 = a - b/ro2 = a - b/0.052 = a – 400b a = 400b
σR = -112 x 106 = a - b/ri2 = a - b/0.0252 = a – 1600 b
-112 x 106 = 400b – 1600 b = -1200b
b = 112 x 106/1200 = 93333.3
a = 400b = 37.33 x 106

Now solve the circumferential stress on the inside. σc = a + b/ri2 = 186.7 MPa
Now solve the circumferential stress on the outside. σc = a + b/ro2 = 74.64 MPa

The longitudinal stress is Force/Area

The force is p x πri2 = 219.9 kN
Wall section area = π(ro2 - ri2) = 5.89 x 10-3 m2
σL = F/A = 37.33 MPa

Circumferential strain at outside

εC = (1/E){σC – ν(σR +σL)}
εC = (1/205 x 109 ){74.64 – 0.27(0 + 37.33)} x 106 = 315 µε
∆D = Do εC = 100 x 315 x10-6 = 0.031 mm

Circumferential strain at inside

(1/205 x 109 ){186.7 – 0.27(-112 +37.33)} x 106 == 1.009 µε
∆D = Di εC = 50 x 1.009 x10-6 = 0.05 mm
4. A shaft has a diameter of 45.08 mm and is an interference fit with a sleeve 60 mm outer diameter,
45 mm inner diameter and 80 mm long. Calculate the force needed to slide the sleeve on the shaft
if the coefficient of friction is 0.25. The elastic properties for both parts are the same with E = 200
GPa and Poisson’s ratio = 0.3

Calculate the change in radius of the shaft and sleeve at the inside.

δ
p=
R ⎡ R + R o2
2 ⎤
⎢ 2 2
+ 2ν -1 ⎥
E ⎣Ro − R ⎦
0.08
p= = 111.2 x 10 6 N/m 2
45.08 ⎡ 45.08 + 60
2 2 ⎤
9 ⎢ 2
+ (2 x 0.3) - 1 ⎥
200 x 10 ⎣ 60 2 − 45.08 ⎦

Surface Area = 2πRL = 2π x 0.04508 x 0.08 = 0.0226 m2

Normal Force = N = pA = 2.519 MN

Friction force = µ N = 0.25 x 2.519 = 0.63 MN

pR ⎛ R 2 + R o2 ⎞
For the sleeve ∆R = ⎜ ⎟
E ⎜ R2 − R2 + ν ⎟
⎝ o ⎠
112x10 6 x 0.04508 ⎛ 45.08 2 + 60 2 ⎞
∆R = ⎜ + 0.3 ⎟ = 97.5 x 10 −6 m
9 ⎜ 2 2 ⎟
200 x 10 ⎝ 60 − 45.08 ⎠
pR
For the shaft ∆R = (− 1 + ν )
E
112x10 6 x 0.04508
∆R = 9
(− 1 + 0.3 ) = −17.5 x 10 −6 m
200x10
Check δ = 97.5 – 17.5 = 80 µm which was the original interference fit.