mechanical principles - complex loaded cylinders and system

Steve Goddard

Mechanical Principles – Assignment 1

Complex Loaded Systems & Cylinders

1. Using a diagram define Poisson’s ratio and give typical values for steel, Titanium and Aluminum Alloy.

When a sample of material is stretched in one direction, it tends to get thinner in the other two directions. Poisson's ratio (v - Greek letter Nu), named after Simeon Poisson, is a measure of this tendency. Poisson's ratio is the ratio of the relative contraction strain, or transverse strain (normal to the applied load), divided by the relative extension strain, or axial strain (in the direction of the applied load). For a perfectly incompressible material deformed elastically at small strains, the Poisson's ratio would be exactly 0.5. Most practical engineering materials have ν between 0.0 and 0.5. Cork is close to 0.0, most steels are around 0.3, and rubber is almost 0.5.

Some materials, mostly polymer foams, have a negative Poisson's ratio; if these auxetic materials are stretched in one direction, they become thicker in perpendicular directions.

For the materials in question Poisons ratio is as follows:

Steel – 0.27 – 0.30

Titanium – 0.34

Aluminium Alloy – 0.33

Poisson's ratio values for different materials

material poisson's ratio

concrete 0.20

cast iron 0.21-0.26

glass 0.24

clay 0.30-0.45

saturated clay

0.40-0.50

copper 0.33

cork ca. 0.00

magnesium 0.35

stainless steel

0.30-0.31

rubber 0.50

foam 0.10 to 0.40

sand 0.20-0.45

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auxetics negative

Values based on www.answers.com

2. A bar 300mm long and 30mm diameter is pulled by a tensile load of 40kN. If Poisson’s ratio is 0.3 and the Modulus of Elasticity is 100GN/m2, find the reduction in diameter of the bar.

Firstly I drew out the question:

Next I had to work out the stress applied using the formula:

Stress σ =

I worked out the area:

Strain in the x direction

Using

Change in Diameter =

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40kNØ 30mm

300mm

E = 100GN/m2

v = 0.3

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Steve Goddard

3. A brass plate shown is subjected to stresses as shown below. Find the changes in the 160mm length and in the 7-mm width if for brass E = 105 GN/m2 and poisson’s ratio is 0.3.

X Direction

Y Direction

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10 MN/m2

20 MN/m220 MN/m2

10 MN/m2

180

mm

70mm

E = 105 GN/m2

V = 0.3

Steve Goddard

4. The plate shown is made from a metal for which Poisson’s ratio is 0.29 and E=180 GN/m2. Find the changes in the 100mm and 45mm direction.

Firstly I need to convert the forces into stresses with the equation:

Now that I have the stresses I can work out the changes in direction.

5. Find the change in volume of a square cross section 80mm by 80mm and 1.2m long when subjected to an axial load of 20kN.

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5mm

10

0m

m

45mm

10 kN

15 kN

10 kN

15 kN

X Direction Y Direction

Steve Goddard

Take R as 200 GN/m2 and Poisson’s ratio as 0.3.

Firstly I drew out the problem so it was easier to interpret.

Volumetric Strain =

Linear Stresses

Change in volume = Volumetric Strain x Original Volume

Original Volume =

Volumetric Strain =

6. A round steel bar 25mm diameter and 250mm long is loaded with an axial tensile force of 200kN.Calculate the change in volume if E=200 GN/m2 and Poisson’s ratio is 0.32.

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80mm

80

mm

1200mm

20kN

Z

X

Y

E = 200 GN/m2

V = 0.3

Steve Goddard

Converting force to stress:

Linear Strains:

Volumetric Strain =

Change in volume = Volumetric Strain x Original Volume

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Ø 25mm

250mm

200 kN

E = 200 GN/m2

V = 0.32

Steve Goddard

7. A Component in a machine which is submerged in water is subjected, as a result, to a hydrostatic pressure of 5 bar. The Modulus of Elasticity of the material of the component is 200GN/m2.If poisson’s ratio is 0.25 and the original volume was 20 x 10-6 m3, find:

7.1 The volumetric strain

7.2 Change in volume

8. Find the value of the Shear Modulus G of a material if the Young’s Modulus of Elasticity is 210GN/m2 and poisson’s ratio is 0.3.

9. Find the value of the Bulk Modulus of a material if it has an Elastic Modulus of 200GN/m2 and poisson’s ratio of 0.3.

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1 Bar = 100 KN/m2

Steve Goddard

10. A Cylinder 1.5m in length has an internal diameter of 75mm and a wall thickness of 1.5mm.

Determine

10.1 The Hoop Stress

Hoop stress =

10.2 The Longitudinal Stress

I know longitudinal stress is equal to

To check this I can use the equation:

10.3 The percentage increase in internal volume of the tube.

I can calculate this using the following equation

Percentage increase =

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Original Volume

Steve Goddard

11. A cylindrical air compressor is 2m internal diameter and made of plate 15mm thick. If the hoop stress is not to exceed 90MN/m2 and the axial stress is not to exceed 60MN/m3, find the maximum safe air pressure.

Axial Stress =

Therefore the maximum safe pressure would be 1.35 Mpa

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12. A closed cylinder has an internal diameter of 60mm and an external diameter of 100mm. It is subject to an internal pressure of 20Mpa and an external pressure of 5Mpa. Determine the radial and hoop stresses at the inner and outer wall surfaces and the longitudinal stress experienced at the cylinder wall.

Subtract equation 1 from equation 2

Using b in equation 1

Now I can work out the radial and hoop stresses by substituting a and b into the original equations:

Longitudinal Stress:

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Equation 1

Equation 2

Steve Goddard

13. A thick cylinder having an external diameter of 230mm and an internal diameter of 120mm is subjected to an internal pressure of 48MPa and an external pressure of 9MPa.Find: 13.1 the maximum direct stress in the cylinder

13.2 the change in the external diameterGiven that E is 200Gpa and Poisson’s ratio is 0.28

13.1

Inner Radius

Outer Radius

Subtract Equation 1 from Equation 2

Substitute b into equation 1

Inner Radius – Circumferential Stress

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I will call this Equation 1

I will call this Equation 2

Steve Goddard

Outer Radius – Circumferential Stress

Radial Stress at outer radius

Change in Diameter

Change in Diameter

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Maximum Direct Stress

Steve Goddard

Bibliography

http://www.answers.com/poisons%20ratio

Lecture Notes

HNC/HND Book – Tooley & Dingle

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http://www.answers.com/poisons%20ratio

mechanical principles - complex loaded cylinders and system

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