Question

Strain-displacement relationship) Consider a unit cube of a solid occupying the region 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1 After loads are applied, the displacements are

Answer 1

please see answers are as in the explanation.

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`

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

`\epsilon_(xx)=(\partial u)/(\partial x)=(\partial (\alpha x))/(\partial x)=\alpha =0.03\\\epsilon_(yy)=(\partial v)/(\partial y)=(\partial ( \beta y))/(\partial y)=\beta =-0.01\\\epsilon_(zz)=(\partial w)/(\partial z)=(\partial (0))/(\partial z)=0\\`

Shear Strain Components

`\gamma_(xy)=\gamma_(yx)=(\partial u)/(\partial y)+(\partial v)/(\partial x)=0\\\gamma_(xz)=\gamma_(zx)=(\partial u)/(\partial z)+(\partial w)/(\partial x)=0\\\gamma_(yz)=\gamma_(zy)=(\partial w)/(\partial y)+(\partial v)/(\partial z)=0`

Part c: Find the volume change

`\Delta V=(1.03 * 0.99 * 1)-(1 * 1 * 1)\\\Delta V=(1.0197)-(1)\\\Delta V=0.0197\\`

Also the change in volume is 0.0197

For the unit cube, the change in terms of strains is given as

`\Delta V={V_0}[(1+\epsilon_(xx))]*[(1+\epsilon_(yy))]* [(1+\epsilon_(zz))]-[1 * 1 * 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)]\\`

As the strain values are small second and higher order values are ignored so

`\Delta V\approx {V_0}[\epsilon_(xx)+\epsilon_(yy)+\epsilon_(zz)]\\ \Delta V\approx [\epsilon_(xx)+\epsilon_(yy)+\epsilon_(zz)]\\`

As the initial volume of cube is unitary so this result can be proved.