Answer:
please see answers are as in the explanation.
Step-by-step explanation:
As from the data of complete question,
![0\leq x\leq 1\\0\leq y\leq 1\\0\leq z\leq 1\\u= \alpha x\\v=\beta y\\w=0](https://tex.z-dn.net/?f=0%5Cleq%20x%5Cleq%201%5C%5C0%5Cleq%20y%5Cleq%201%5C%5C0%5Cleq%20z%5Cleq%201%5C%5Cu%3D%20%5Calpha%20x%5C%5Cv%3D%5Cbeta%20y%5C%5Cw%3D0)
The question also has 3 parts given as
<em>Part a: Sketch the deformed shape for α=0.03, β=-0.01 .</em>
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+<em>α</em><em>x,y+</em><em>β</em><em>y) </em>
Point A'(0+<em>(0.03)</em><em>(0),0+</em><em>(-0.01)</em><em>(0))</em>
Point A'(0<em>,0)</em>
Point B(x=1,y=0)
Point B'(x+<em>α</em><em>x,y+</em><em>β</em><em>y) </em>
Point B'(1+<em>(0.03)</em><em>(1),0+</em><em>(-0.01)</em><em>(0))</em>
Point <em>B</em>'(1.03<em>,0)</em>
Point C(x=1,y=1)
Point C'(x+<em>α</em><em>x,y+</em><em>β</em><em>y) </em>
Point C'(1+<em>(0.03)</em><em>(1),1+</em><em>(-0.01)</em><em>(1))</em>
Point <em>C</em>'(1.03<em>,0.99)</em>
Point D(x=0,y=1)
Point D'(x+<em>α</em><em>x,y+</em><em>β</em><em>y) </em>
Point D'(0+<em>(0.03)</em><em>(0),1+</em><em>(-0.01)</em><em>(1))</em>
Point <em>D</em>'(0<em>,0.99)</em>
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.
<em>Part b: Calculate the six strain components.</em>
Solution
Normal Strain Components
![\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\\](https://tex.z-dn.net/?f=%5Cepsilon_%7Bxx%7D%3D%5Cfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20x%7D%3D%5Cfrac%7B%5Cpartial%20%28%5Calpha%20x%29%7D%7B%5Cpartial%20x%7D%3D%5Calpha%20%3D0.03%5C%5C%5Cepsilon_%7Byy%7D%3D%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20y%7D%3D%5Cfrac%7B%5Cpartial%20%28%20%5Cbeta%20y%29%7D%7B%5Cpartial%20y%7D%3D%5Cbeta%20%3D-0.01%5C%5C%5Cepsilon_%7Bzz%7D%3D%5Cfrac%7B%5Cpartial%20w%7D%7B%5Cpartial%20z%7D%3D%5Cfrac%7B%5Cpartial%20%280%29%7D%7B%5Cpartial%20z%7D%3D0%5C%5C)
Shear Strain Components
![\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](https://tex.z-dn.net/?f=%5Cgamma_%7Bxy%7D%3D%5Cgamma_%7Byx%7D%3D%5Cfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20y%7D%2B%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20x%7D%3D0%5C%5C%5Cgamma_%7Bxz%7D%3D%5Cgamma_%7Bzx%7D%3D%5Cfrac%7B%5Cpartial%20u%7D%7B%5Cpartial%20z%7D%2B%5Cfrac%7B%5Cpartial%20w%7D%7B%5Cpartial%20x%7D%3D0%5C%5C%5Cgamma_%7Byz%7D%3D%5Cgamma_%7Bzy%7D%3D%5Cfrac%7B%5Cpartial%20w%7D%7B%5Cpartial%20y%7D%2B%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20z%7D%3D0)
Part c: <em>Find the volume change</em>
<em></em>
<em></em>
<em>Also the change in volume is 0.0197</em>
For the unit cube, the change in terms of strains is given as
![\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}]\\](https://tex.z-dn.net/?f=%5CDelta%20V%3D%7BV_0%7D%5B%281%2B%5Cepsilon_%7Bxx%7D%29%5D%5Ctimes%5B%281%2B%5Cepsilon_%7Byy%7D%29%5D%5Ctimes%20%5B%281%2B%5Cepsilon_%7Bzz%7D%29%5D-%5B1%20%5Ctimes%201%20%5Ctimes%201%5D%5C%5C%5CDelta%20V%3D%7BV_0%7D%5B1%2B%5Cepsilon_%7Bxx%7D%2B%5Cepsilon_%7Byy%7D%2B%5Cepsilon_%7Bzz%7D%2B%5Cepsilon_%7Bxx%7D%5Cepsilon_%7Byy%7D%2B%5Cepsilon_%7Bxx%7D%5Cepsilon_%7Bzz%7D%2B%5Cepsilon_%7Byy%7D%5Cepsilon_%7Bzz%7D%2B%5Cepsilon_%7Bxx%7D%5Cepsilon_%7Byy%7D%5Cepsilon_%7Bzz%7D-1%5D%5C%5C%5CDelta%20V%3D%7BV_0%7D%5B%5Cepsilon_%7Bxx%7D%2B%5Cepsilon_%7Byy%7D%2B%5Cepsilon_%7Bzz%7D%5D%5C%5C)
As the strain values are small second and higher order values are ignored so
As the initial volume of cube is unitary so this result can be proved.