(1)

Why do we define stress and strain?

To model the way a material will act when a force is exerted upon it. Ensuring that material can perform in its desired application.

(2)

Fatigue accounts for ~90% of mechanical engineering failures. Please list 3 measures that may be taken to increase the resistance to fatigue of metallic materials. Please also briefly explain how each of these measures improves fatigue performance

Shot Peening

Suppresses surface cracks by imposing a compressive surface stress.

Surface Design

Increasing the radius of edges under stress, reduce's the stress concentration factor.

Surface Polishing

Reduces surface defacts that could become crack sites.

(3)

From one of the previous Materials 1 lectures, we learnt that Hooke’s law can be described as $\sigma = E\epsilon$, in which $\sigma$ is the stress applied to a specimen, E is the Young’s modulus of the specimen, and $\epsilon$ is the resulting strain. In some other sources, Hooke’s law is expressed as $F=kx$, F where is the force applied to the specimen, k is a constant, and x is the displacement caused by the force. Please discuss the similarities and differences between these two types of description.

Looking at the expansion of Hooke's Law $\cfrac{F}{A} = E\cfrac{x}{l_0}$ where $x = \Delta l = displacement$ equating the two descriptions shows the second description assumes the specimen size is constant
$k = \cfrac{A}{l_0}E$ It can be seen that both expressions show the relationship between an applied force and the deformation it causes in a given material. The expression $F = kx$ assumes a constant specimen size (as k is constant) where as Hooke's law shows deformation with respect to the specimen size ($A$, $l_0$).

(4)

A long cylindrical rod with a diameter of 10 mm is cut into two cylindrical rods. The length of one rod (Rod 1) is a half of the length of the other one (Rod 2). Tensile stress of 600 MPa is applied to both rods. Rod 1 experiences only elastic deformation with elongation of 1 mm and strain of 0.5%, what are the elongation and strain values of Rod 2.

$\epsilon_{r1} = 0.5\%$ $\Delta l_{r1} = 1mm$ $\sigma_{r1} = \sigma_{r2}$ $E = \cfrac{\sigma_{r1}}{\epsilon_{r1}} = \cfrac{\sigma_{r2}}{\epsilon_{r2}}$
$$\epsilon_{r1} = \epsilon_{r2} = 0.5\%$$
$e_{r1} = \cfrac{\Delta l_{r1}}{\epsilon_{r1}}$ $l_{r2} = 2l_{r1} = 400mm$ $\Delta l_{r2} = \epsilon_{r2} l_{r2}$
$$\Delta l_{r2} = 2mm$$

(5)

For the engineering stress–strain curve shown on the right, the values of the engineering strain at point M (just before necking) and point F (the fracture point) are 35% and 53%, respectively. Can we calculate the true strain values at these two points using formula $\epsilon_T = ln (1 + \epsilon)$ if yes, provide the true strain values. If no, provide the reason.

As point M occurs just before necking we can use the formula to calculate true strain.
$$\epsilon_{T_M} = ln(1.35) = 0.3$$
The true strain formula is only valid before necking, hence we cannot calculate the true strain at point F.

(6)

Please explain what mechanical properties of materials that we can acquire from a tensile engineering stress–strain curve (shown to the right) and how. Please also provide the definition of these mechanical properties.

Young's Modulus (E)

Represented on Graph: The gradient of the linear region of the stress-strain ($\sigma - \epsilon$) curve.
Definition: Young's modulus is a measure of a materials ability to resist elastic deformation.

Tensile strength

Represented on Graph: The maximum stress on the $\sigma - \epsilon$ curve
Definition: The stress a material can withstand before necking occurs.

Yield strength

Represented on Graph: The stress at which a line running through the point (0.002, 0) parallel with the linear (elastic) region of the $\sigma - \epsilon$ curve, intersects the $\sigma - \epsilon$ curve.
Definition: The stress at which noticeable plastic deformation occurrs.

Ductility

Represented on Graph: The strain at which a line running trough the fracture point on the $\sigma - \epsilon$ curve, parallel to the elastic region, intersects strain axis (plastic strain).
Definition: The plastic strain a material can take before fracturing.

(7)

Based on your reading and knowledge that you have learnt from Chapter 6, please provide detailed explanation of

(i)

if change in specimen dimensions affects the measured Young’s modulus, yield strength, and tensile strength.

Young's Modulus

From the reading, specimen size does not apear to have any affect on the gradient of the linear region of the $\sigma - \epsilon$ curves. In elastic elongation atomic bonds are streatch and will return. The number of atomic bonds should be propertional to the specimen size. Hence change in specimen dimension will not affect the measured Young's modulus.

Yield Strength

Yield strength is dependant on the plastic region of the $\sigma - \epsilon$ curve. From the reading, specimen size affected the plastic elongation of the material they tested. Plastic deformation causes shearing within the structure of the material. This changes the geometry of the specimen, engineering strain and stress doe's not account for this. Hence speciman size will affect the measured yield strength.

Tensile Strength

Similarly to yield strength, tensile strength is dependant on the plastic strain region of $\sigma - \epsilon$ curve. From the reading and the definition of strain, speciman size will affect the measured tensile strength.

(ii)

the reason for using specimens with standard dimensions for mechanical property testing.

It is important when performing scientific experiments that all variables not being tested are kept constant in order to maintain validity and repeatability of the experiment. Without standardisation we assume that a specimen's size is independant of it's mechanical properties. From the evidence provided in the reading, and a general knowledge of materials, it would be naive to assume that a specimen's dimensions have neglible affect on it's mechanical properties. Hence using a standard specimen size for mechanical property testing ensures validity of comparing data of one material with another.

(8)

A material presents S–N fatigue behaviour shown on the right with the fatigue limit being 100 MPa. If a cylindrical bar made from the material has a diameter of 20 mm is subjected to repeated tensile stress along its axis, calculate the maximum allowable tensile load to ensure no occurrence of fatigue failure.

$\sigma_{max} = \sigma_a$(reversed stress cycles) $\sigma_{max} = 100MPa$ $\cfrac{L_{max}}{A} = \sigma_{max}$
$$L_{max} = 100MPa \times \pi10^2mm^2 = 31.4kN$$