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Six external forces, each of magnitude F, are applied on all the faces of a unit cube. Considering its elastic modulus ,calculate the longitudinal strain and the volume strain on the unt cube. |
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Answer» Solution :Each side of a unit cube=1. `therefore` Volume of the cube =1. surface area of each of the SIX faces =1. Let x,y and z axes be chosen along three mutually perpendicular SIDES of the cube. Now at first, we choose the two opposite faces perpendicular to the x-axis. Due to the FORCES (F,F) acting on them, the side along the x-direction will have elongation=l(suppose). `therefore` LONGITUDINAL strain along the x-direction `=(elongati on)/( i nitial l eng th)=l/1=l` Also longitudinal stress `=(applied forces)/(surface area)=F/1=F` `therefore` Young modulus , `Y=(Stress)/(strai n)=F/lso,l=F/Y` Simultaneously the side along the x-direction will also suffer lateral strains due to the forces acting along y and z directions we know, Poisson.s ratio `sigma=("lateral strai n")/("longitudi nal strai n")` `thereforeLateral strai n=sigmatimeslongitu d i nal strai n=sigmal=(sigmaF)/Y` Now, an elongation is always associated with a lateral contraction. Therefore, considering the strain along all the three axes, the effective longitudinal strain along the x-axis is , `F/Y-(sigmaF)/Y-(sigmaF)/Y=F/Y(1-2sigma)` From symmetry, the effective longitudinal strain along each of y and z axes `=F/Y(1-2 sigma)` As a result the final length of each of the three sides of the cube `=1+F/Y(1-2sigma)` `therefore` Final volume of a unit cube `=[1+F/Y(1-2sigma)]^3=1+(3F)/Y(1-2sigma)` [neglecting higher order terms] `therefore` Volume EXPANSION `=1+(3F)/Y(1-2sigma)-1` `=(3F)/Y(1-2sigma)` `therefore` Volume strain `=((3F)/Y(1-2sigma))/1=(3F)/Y(1-2sigma)` |
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