A modulus of elasticity of the wire, called Young’s modulus (E), is defined as the ratio
It should be noted that Young’s modulus , E, is calculated from the ratio stress : strain only when the wire is under ‘elastic’ conditions, that is, the load does not then exceed the elastic limit.
Dimensions of Young’s Modulus
The ‘strain’ of a wire has no dimensions of mass, length, or time, since, by definition, it is the ratio of two lengths.
Determination of Young’s Modulus
The following practical points should be specially noted in determining the Magnitude of Young’s Modulus
- The wire is made thin so that a moderate load of several kilograms produces a large tensile stress. The wire is also made long so that a measurable extension is produce.
- The use of two wires, P, Q, of the same material and length, eliminates the correction for (i) the yielding of the support when loads are added to Q, (ii) changes of temperature.
- Both wires should be free of kinks, otherwise the increase in length cannot be accurately measured. The wires are straightened by attaching weights to their ends.
- A vernier scale is necessary to measure the extension of the wire since this is always small. The ‘original length’ of the wire is measured from the B to the vernier V by a ruler. Since an error of 1 millimiter is negligible compared with an original length of several metres. For very accurate work, the extension can be measured by using a spirit level between the two wires, and adjusting a vernier screw to restore the spirit level to its original reading after a load is added.
- The diameter of the wire must be found by a micrometer screw gauge at several places, and the average value then calculated. The area of cross-section, Ai, = πr2, where r is the radius.
- The readings on the vernier are also taken when the load is gradually removed in steps of 1 kilogram; they should be very nearly the same as the readings on the vernier when the weights were added, showing that the elastic limit was not exceeded. Suppose the reading on V for loads, W, of 1 to 6 kilogram are a, b, c, d, e, f, as follows:
The average extension for 3 kilogram is found by taking the average of (d – a), (e – b), and (f -c). Young’s modulus can then be calculated from the relation stress/strain, where the stress = 3 x 9.8/πr2, and the strain = average extension/original length of wire
Magnitude of Young’s Modulus
Mild steel (0.2% carbon) has a Young’s modulus value of about 2.0 x 1011 newton m-2, copper has a value about 1.2 x 1011 newton m-2; and brass a value about 1.0 x 1011 newton m-2.
The breaking stress (tenacity) of cast-iron metal is about 1.5 x 108 newton m-2; the breaking stress of mild steel metal is about 4.5 x 108 newton m -2 .
At Royal Ordinance and other Ministry of Supply factories, tensile testing is carried out by placing a sample of the material in a machine known as an extensometer, which applies stresses of increasing value along the length of the sample and automatically measures the slight increase in length. When the elastic limit is reached, the pointer on the dial of the machine flickers, and soon after the yield point is reached the sample becomes thin at some point and then breaks. A graph showing the load v. extension is recorded automatically by a moving pen while the sample is undergoing test.
Here is an example we should look at:
Find the maximum load in kgf which may be placed on a steel wire of diameter 0.10 cm if the permitted strain must not exceed 1/100 and Young’s modulus for steel is 2.0 x 1011 newton m -2.
Force in Bar Due to Contraction or Expansion
When a bar is heated, and then prevented from contracting as it cools, a considerable force is exerted at the ends of the bar. We can derive a formula for the force if we consider a bar of Young’s modulus E, a cross-sectional area A, a linear expansivity of magnitude α and a decrease in temperature of toC. then, if the original length of the bar is l, the decrease in length e if the bar were free to contract = αlt.
Energy Stored in a Wire
Let’s assume that a wire has an original length l and is stretched by a length e when a force F is applied at one end. If the elastic limit is not exceeded, the extension is directly proportional to the applied load. Consequently, the force in the wire has increased in magnitude from zero to F, and hence the average force in the wire while stretching was F/2. Now
Work done = force x distance
Work = average force x extension
= ½ Fe – – (1)
Elasticity is the amount of energy stored in the wire. The formula ½ Fe gives the energy in joule when F is in newton and e is in metre.
it does not have exactly what i want
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