Saturday, October 31, 2009

How to Calculate Hoop Stress, Internal Pressure Only

It is often desirable to calculate how much a cavity, hydraulic cylinder, or some other cylindrically shaped tube will expand due to internal pressure. This is applicable when designing molds for thin walled parts, with cavities that are mounted on the face of a plate. If the cavity expands more than the wall thickness of the plastic part will shrink, when the injections pressure stops, the strained cavity walls will contract, transmitting the force through the plastic to the core. This has been seen to be so severe, at times, that a mold could not be opened by the molding machine.

There are other reasons to minimize cavity expansion, including controlling the size of the molding and preventing flash in certain mold designs.

I am supplying 4 choices of calculations:

1. Stress at the outside of the hoop

2. Stress at the inside of the hoop

3. Increase of the inner radius of the hoop

4. Increase of the outer radius of the hoop

All calculations assume zero external pressure. The stress calculated is the tangential stress. The stress must be kept below the fatigue stress endurance limit for the material chosen to get the maximum number of cycles.

One of the material properties that you need to know is the Modulus of Elasticity.

The Machinery Handbook defines Modulus of Elasticity as follows: "Modulus of Elasticity, E, (also called Young' s modulus) is the ratio of unit stress to unit strain within the proportional limit of a material in tension or compression". In general, the Modulus of Elasticity, for steel is 30,000,000 p.s.i. Some steels are as low as 28,000,000 and some as high as 34,000,000 p.s.i.

The shear Modulus of Elasticity, G, also known as the Modulus of Rigidity, is also the ratio of stress to strain within the proportional limit of a material, but in shear. It is usually expressed as a percentage of the Modulus of Elasticity, E, in tension. The Modulus of Elasticity in shear, G, is usually about 38% of the Modulus of Elasticity in tension, E.

Some values for Modulus of Elasticity for some common metals are as follows:

Aluminum, Alcoa QC-7: 10,300,000 p.s.i.

Brush-Wellman's Moldmax XL Beryllium Copper: 17,000,000 p.s.i.

Brush-Wellman's Protherm Beryllium Copper: 20,000,000 p.s.i.

Ampco MoldMATE 90: 22,000,000 p.s.i.

Steel: 30,000,000 p.s.i.

The stress, due to internal pressure only, calculations require only the outer and inner radii, and the internal pressure. The increase in outer wall requires that you also use the Modulus of Elasticity quantity, which is described in step 2.

The calculation for increase of inner radius requires an additional piece of information, Poisson's ratio. For steel this is approximately .30, for copper it is .34. These can be looked up in the Machinery's Handbook, Mark's Standard Handbook for Mechanical Engineer's, or a variety of other places, including web searches.

The calculation for increase of inner radius requires an additional piece of information, Poisson's ratio. For steel this is approximately .30, for copper it is .34. These can be looked up in the Machinery Handbook, Mark's Standard Handbook for Mechanical Engineer's, or a variety of other places, including web searches.

The nomenclature for the following equations is as follows:

P = Pressure inside hoop
R = Outer radius
r = Inner Radius
E = Modulus of Elasticity
Po = Poisson's Ratio

Stress at Inside of Hoop = (P * ((R ^ 2 + r^ 2) / (R ^ 2 - r^ 2)))

Stress at Outside of Hoop = ((P * 2) * ((r ^ 2) / (R ^ 2 - r ^ 2)))

Increase of Inner Radius = (P * (r / E)) * (((R ^ 2 + r ^ 2) / (R ^ 2 - r ^ 2)) - Po)

Increase of Outer Radius = (P * (R / E)) * ((2 * (r ^ 2)) / (R ^ 2 - r ^ 2))

Simply substitute the numbers from your problem, or application, into the appropriate equation, and do the calculations. Obviously, it's much quicker and easier to do this with software, or a spreadsheet.

It is advisable to start any engineering calculation with a sketch, or diagram, of the problem that you want to solve. It helps you to visualize the relationships involved.

Sunday, October 18, 2009

How to Calculate Thermal Expansion

Calculating linear thermal expansion and contraction, or change of size due to change in temperature, is a common engineering calculation. Materials that will be welded, glued, or rigidly fastened together should have similar coefficients of thermal expansion, or breakage and cracking may occur. Knowing the expansion due to temperature change is extremely importnat in mold design, and can cause problems with running fits, location misalignment, and more.

Fortunately, the calculation for size change due to temperature change is fairly simple:

Size Change = initial size x temperature change x coefficient of expansion.

The important thing to remember, as with all calculations, is to keep all of the terms of the equation in the same units.

In order to use this equation, you need to know the coefficient of thermal expansion for the material in question. Here is a list of some common coefficients of thermal expansion, for different materials, in inches per inch per degree Fahrenheit:

Aluminum - .000013 or 12.8 x 10^-6 in scientific notation
Beryllium Copper - .0000093 or 9.3 x 10^-6
Brass - .000014 or 10.4 x 10^-6
Brick - .0000032 or 3.2 x 10^-6
Cement - .000006 or 6.0 x 10^-6
Concrete - .000008 or 8.0 x 10^-6
Glass, hard - .0000033 or 3.3 x 10^-6
Glass, Plate - .000005 or 5.0 x 10^-6
Glass, Pyrex - .0000022 or 2.2 x 10^-6
Gold - .0000084 or 8.4 x 10^-6
Graphite - .0000044 or 4.4 x 10^-6
Lead - .0000151 or 15.1 x 10^-6
Marble - .0000065 or 6.5 x 10^-6
Porcelain - .0000017 or 1.7 x 10^-6
Rubber - .0000428 or 42.8 x 10^-6
Silver - .0000107 or 10.7 x 10^-6
Steel - .0000064 or 6.4 x 10^-6
Tin - .000013 or 13.0 x 10^-6
Titanium - .0000048 or 4.8 x 10^-6
Wood, oak parallel to grain - .0000027 or 2.7 x 10^-6
Wood, oak across the grain - .000003 or 3.0 x 10^-6
Zinc - .0000165 or 16.5 x 10^-6

You can use the internet, the Machinists Handbook, Marks Standard Handbook for Mechanical Engineers, and many other resources to find the coefficient of thermal expansion for materials not listed here.

OK, suppose you have a piece of steel that is 12.0000 inches long, and has a 100 degree increase in temperature. What will the new length be?

The change in length = 12.0000 x 100 x .0000064 = .00768

The new length will be 12.0000 = .00768 = 12.00768

As shown in the list, .0000064 inch/inch/degree F is the coefficient of thermal expansion for steel. This can also be represented as 6.4 x 10^-6 in scientific notation.

If you want software to automate this calculation, go to www.dzynsource.com