calucations

Yield stress= 20.74 Mpa

Yield stress = applied load at yield point /original cross section area

Tensile strength= 31.85 Mpa

Tensile strength = maximum load applied/original cross section area

Sample of CalculationsEngineering stress

Engineering strain

DuctilityPercentage elongation

= (290-66/66)*100%= 339.3%

Percentage reduction in area

=(0.1348-0.0307/0.1348)= 77%

True stress = applied load / current area (N/mm2)

= 0.1 N/ 0.134617 = 0.742851 (N/mm2)

True strain = Ln(current length/original length) = Ln(66.188/ 66) = 0.002844

Tensile stress at fracture = 12.59 Mpa

Discussion:In the beginning we calculated the original cross sectional area to calculate

the engineering stress and engineering strain, we make this equation:

Engineering stress= given load (N)/ original cross sectional area (5mm

*0.027mm)

Engineering Strain= given deflection / original length (66mm)

Then by using excel program, we plot the graph of engineering stress-strain.

After that we determine the yield strength from the graph, which is located on the

point between the elastic (linear) and plastic (non-linear) region . The tensile

strength is the highest point in the graph

Also continue in calucation and determine the values of the current length

and area in order to get the results of true stress and strain, that will be used for the

plotting of the true stress-strain graph and we calucated the ductility represented by

the elongation percentage and the reduction in area percentage

Conclusion and recommendation:The remarkable differenece between the true stress-strain in compare to the

engineering stress-strain graphs, which is represented by the engineering stress

strain graph, does not reveal the real relationship between the stress and strain

beyond the elastic region because it is divided on the original area, it is recognizable

that in the true stress strain graph , it reflects the real relationship between them

and the reason is because of dividing on the current area.

The material we used in experiment is of plastic nature, that get affected

relatively by the temperature factor in directiononal propation, in way that it affects

on the elongation when we increase the temperature, in addition to the effect on the

force we need, that will decrease as well as increasing the strain, so the strength of

the speciman will increase

the temperature will affect the force that we need, when we increase the

temperature the needed force will decrease and also when we increase the strain

the strength of the piece work will increase.

It is recommended to consider the temperature factor before doing the

temperature, and it is of course related to the nature of the material if it is plastic or

metal.