Answer:
please see answers are as in the explanation.
Step-by-step explanation:
As from the data of complete question,
[tex]0\leq x\leq 1\\0\leq y\leq 1\\0\leq z\leq 1\\u= \alpha x\\v=\beta y\\w=0[/tex]
The question also has 3 parts given as
Part a: Sketch the deformed shape for α=0.03, β=-0.01 .
Solution
As w is 0 so the deflection is only in the x and y plane and thus can be sketched in xy plane.
the new points are calculated as follows
Point A(x=0,y=0)
Point A'(x+αx,y+βy)
Point A'(0+(0.03)(0),0+(-0.01)(0))
Point A'(0,0)
Point B(x=1,y=0)
Point B'(x+αx,y+βy)
Point B'(1+(0.03)(1),0+(-0.01)(0))
Point B'(1.03,0)
Point C(x=1,y=1)
Point C'(x+αx,y+βy)
Point C'(1+(0.03)(1),1+(-0.01)(1))
Point C'(1.03,0.99)
Point D(x=0,y=1)
Point D'(x+αx,y+βy)
Point D'(0+(0.03)(0),1+(-0.01)(1))
Point D'(0,0.99)
So the new points are A'(0,0), B'(1.03,0), C'(1.03,0.99) and D'(0,0.99)
The plot is attached with the solution.
Part b: Calculate the six strain components.
Solution
Normal Strain Components
[tex]\epsilon_{xx}=\frac{\partial u}{\partial x}=\frac{\partial (\alpha x)}{\partial x}=\alpha =0.03\\\epsilon_{yy}=\frac{\partial v}{\partial y}=\frac{\partial ( \beta y)}{\partial y}=\beta =-0.01\\\epsilon_{zz}=\frac{\partial w}{\partial z}=\frac{\partial (0)}{\partial z}=0\\[/tex]
Shear Strain Components
[tex]\gamma_{xy}=\gamma_{yx}=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}=0\\\gamma_{xz}=\gamma_{zx}=\frac{\partial u}{\partial z}+\frac{\partial w}{\partial x}=0\\\gamma_{yz}=\gamma_{zy}=\frac{\partial w}{\partial y}+\frac{\partial v}{\partial z}=0[/tex]
Part c: Find the volume change
[tex]\Delta V=(1.03 \times 0.99 \times 1)-(1 \times 1 \times 1)\\\Delta V=(1.0197)-(1)\\\Delta V=0.0197\\[/tex]
Also the change in volume is 0.0197
For the unit cube, the change in terms of strains is given as
[tex]\Delta V={V_0}[(1+\epsilon_{xx})]\times[(1+\epsilon_{yy})]\times [(1+\epsilon_{zz})]-[1 \times 1 \times 1]\\\Delta V={V_0}[1+\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}+\epsilon_{xx}\epsilon_{yy}+\epsilon_{xx}\epsilon_{zz}+\epsilon_{yy}\epsilon_{zz}+\epsilon_{xx}\epsilon_{yy}\epsilon_{zz}-1]\\\Delta V={V_0}[\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\[/tex]
As the strain values are small second and higher order values are ignored so
[tex]\Delta V\approx {V_0}[\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\ \Delta V\approx [\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\[/tex]
As the initial volume of cube is unitary so this result can be proved.
