## Mechanical Properties Of Materials Under Stress

A material obeys Hooke’s law up to point A and point A is known as limit of proportionality. Up to point A the material is said to be in the elastic region and the type of deformation it experiences is said to be elastic deformation. Beyond the point A Hooke’s law is not obeyed although the material remains elastic (strain completely disappears after the removal of load).

At point B elastic limit is reached and if the material is stressed up to this point it will regain its original shape on the removal of the load.

From B to C, strain increases more quickly than stress and it is this region that the material yields. On increasing the load slightly, the strain increases rapidly till point R when neck or waist is formed. After R is reached the extension continues even with lesser load and ultimately fracture occurs.

The mechanical properties shown in the stress-strain curve can be described using the following terms:

• Proportional limit – It is the maximum point at which stress remains directly proportional to strain. On the diagram, that is point A.
• Elastic limit – It is the maximum stress which the material can withstand without causing permanent deformation. On the diagram, that is point B.
• Working stress – It is the maximum calculated stress to which a material is ever subjected to. It should be well below the elastic limit.
• Factor of safety (F.O.S) – it is the ratio of ultimate stress to working stress. It is also known as factor ignorance. In America it is sometimes called as he factor of stropidity. It depends upon the following factors:
2. frequency of variation of load
3. degree of reliability required
4. decrease of section by corrosion, etc.

### Yielding and Yield Strength

Yielding point is the stress at which there first occurs a marked increase in strain without an increase in stress and that is the point when the material begins to weaken.

Yielding point can also be defined as the stress level at which plastic deformation sets in. It may be determined as the initial departure from linearity of the stress–strain curve and is soon after the proportional limit.

Tensile Yield strength is the stress at which a material exhibits a specified resistance to permanent deformation produced by tensile loads.

Yield strength of a material is difficult to measure precisely. Conventionally, it has been established by constructing a straight line parallel to the elastic portion of the stress–strain curve at some specified strain offset, usually 0.002. The stress corresponding to the intersection of this line and the stress–strain curve as it bends over in the plastic region is the yield strength.

### Ultimate Tensile Strength

Soon after yielding, the stress necessary to continue plastic deformation in metals increases to a maximum value and then decreases to the eventual tensile strength fracture.

Ultimate Tensile Strength (UTS) is the stress at the maximum point of stress–strain curve.

This is the maximum stress that can be sustained by a structure in tension. However, at this point, a small constriction or neck begins to form at some point on the material, and all subsequent deformation is confined at this neck. This is called "necking".

### Stress-strain curves for different materials

Structural steel is the only material that exhibits a marked yield point. Most of the other materials show a gradual change from linear to the nonlinear range with greater elongation for ductile materials such as copper. Brittle materials have a very low proportional limit and do not show the yield point.

## Worked Examples

### Question 1 – Determine Tensile Force and Factor Of Safety In a Tie-bar.

A tie-bar in a steel structure is of rectangular section 50mm x 30 mm. The extension measured in a 250 mm length of the bar when the load was applied to the structure was 0.1 mm. Taking E as 205 kN/mm2 and the UTS as 460 N/mm2,
find:

(i) The tensile force acting on the bar (8 marks)

(ii) The factor of safety (2 marks)

### Solution 1 (i)

From the question we can determine that:

• Cross-sectional area of the tie-bar, A = 50 × 30 = 1 500 mm2
• Extension, e = 0.1 mm
• Original length, l = 250 mm
• Young Modulus, E = 205 000 N/mm2
• Ultimate Tensile Strength, UTS, = 460 N/mm

From the given data we can calculate strain as follows:

$latex \displaystyle strain\ = \frac{extension}{original\ length}$

$latex \displaystyle \epsilon\ = \frac{e}{l}$

$latex \displaystyle \epsilon\ = \frac{0.1}{250}$

$latex \displaystyle \therefore \epsilon = 0.0004$

Now we use the Young Modulus to find the tensile force.

We know that $latex Young\ Modulus\ = \displaystyle \frac{stress}{strain}$

$latex \displaystyle E = \frac{\sigma}{\epsilon}$

Which means $latex \displaystyle E \epsilon = \sigma$

and $latex \displaystyle stress = \frac{Force}{cross-sectional\ Area}$

$latex \displaystyle \sigma = \frac{F}{A}$

Substituting everything into the formula $latex \displaystyle E \epsilon = \sigma$

We get $latex \displaystyle 205 000(0.0004) = \frac{F}{1 500}$

F = 205 000(0.0004)(1 500)

F = 123 000 N

Therefore, tensile force acting on the bar = 123 000 N

### Question 1 (ii)

$latex \displaystyle Working\ stress = \frac{F}{A}$

$latex \displaystyle Working\ stress = \frac{123 000}{1500} = 82\ N/mm^2$

$latex \displaystyle Factor\ of\ safety = \frac{UTS}{Working\ Stress}$

$latex \displaystyle Factor\ of\ safety = \frac{460}{5.6}= 5.6$

### Question 2

A brass cylinder has a diameter of 25 mm and a height of 50 mm. The cylinder supports a compressive load of 20 kN. By how much does the cylinder shorten if the load is applied? Take E for brass as 83 kN/mm2?

### Solution 2

From the question we can determine that:

• Cross-sectional Area, A = πr2 = π(12.5)2 = 156.25π
• Original length = 50 mm
• Load, F = 20 000
• Young Modulus, E = 83 000 N/mm2

We begin by calculating stress:

$latex \displaystyle stress = \frac{Force}{cross-sectional\ Area}$

$latex \displaystyle \sigma = \frac{F}{A}$

$latex \displaystyle \sigma = \frac{2 000}{156.25\pi} = 4.1\ N/mm^2$

Then, from the formula of Young Modulus:

$latex Young\ Modulus\ = \displaystyle \frac{stress}{strain}$

$latex \displaystyle E = \frac{\sigma}{\epsilon}$

Which means $latex \displaystyle \epsilon = \frac{\sigma}{E}$

and this leads us to $latex \displaystyle \frac{e}{l} = \frac{\sigma}{E}$

$latex \displaystyle e = \frac{l\sigma}{E}$

$latex \displaystyle e = \frac{50\times 4.1}{83000} = 0.00247\ mm$

Therefore, the cylinder shortens by 0.00247 mm.