1 Rubber specimens, having an initial length of 5 cm, are tested, one in compression and one in tension. If the engineering strains are βˆ’1.5 and +1.5, respectively, what will be the final lengths of the specimens? What are the true strains, and why are they numerically different?

Accepted Solution

Answer:We know that[tex]Strain=\frac{L_{f}-L_{o}}{L_{o}}[/tex]For specimen in compression we have engineering strain is -1.5 which is not possible since a item cannot compress more than it's initial length2)For item in tension we have[tex]1.5=\frac{L_{f}-5}{5}\\\\\therefore L_{f}=1.5\times 5+5\\L_{f}=12.5cm[/tex]3) the true strain are obtained when we use the actual area of the specimen in calculating stress in the specimen which decreases due to poission effect and necking Β and not the initial nominal area. The true strain is related to engineering strain as [tex]\epsilon _{true}=ln(1+\epsilon _{engg})[/tex]Applying values For specimen in tension we have [tex]\epsilon _{true}=ln(1+1.5)[/tex][tex]\epsilon _{true}=0.916[/tex]