# Soil Mechanics Questions and Answers – Strain Components and Compatibility Equations

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This set of Soil Mechanics Multiple Choice Questions & Answers (MCQs) focuses on “Strain Components and Compatibility Equations”.

1. There are __________ strain components for a three dimensional case.
a) 2
b) 3
c) 6
d) 8

Explanation: There are six strain components for a three dimensional case and there are three linear strain components and three shearing strain components.

2. If u is the displacement in x-direction, the linear strain component in is defined by __________
a) $$ε_X=\frac{∂u}{∂x}$$
b) $$ε_X=\frac{∂x}{∂u}$$
c) $$ε_X=\frac{dx}{du}$$
d) εX=x+u

Explanation: The strain is the ratio of change in dimension of an object to its original dimension. Therefore, the linear strain component in is defined by,
$$ε_X=\frac{∂u}{∂x}$$. Where, u is the displacement in x-direction.

3. If v is the displacement in y-direction, the linear strain component in is defined by __________
a) $$ε_Y=\frac{∂v}{∂y}$$
b) $$ε_Y=\frac{∂y}{∂v}$$
c) $$ε_Y=\frac{dy}{dv}$$
d) εY=v+y

Explanation: The strain is the ratio of change in dimension of an object to its original dimension. Therefore, the linear strain component in is defined by,
$$ε_Y=\frac{∂v}{∂y}$$. Where, v is the displacement in y-direction.

4. If w is the displacement in z-direction, the linear strain component in is defined by __________
a) $$ε_Z=\frac{∂w}{∂z}$$
b) $$ε_Z=\frac{∂z}{∂w}$$
c) $$ε_Z=\frac{dz}{dw}$$
d) εZ=z+w

Explanation: The strain is the ratio of change in dimension of an object to its original dimension. Therefore, the linear strain component in is defined by,
$$ε_Z=\frac{∂w}{∂z}$$. Where, w is the displacement in z-direction.

5. The shearing strain component in X-Y plane is ________
a) $$Γ_{xy}=\frac{∂v}{∂x}+\frac{∂u}{∂y}$$
b) $$Γ_{xy}=\frac{∂v}{∂x}-\frac{∂u}{∂y}$$
c) $$Γ_{xy}=\frac{∂v}{∂y}+\frac{∂u}{∂x}$$
d) $$Γ_{xy}=\frac{∂v}{∂y}-\frac{∂u}{∂x}$$

Explanation: In order to find the shearing strain component in X-Y plane, let us consider a plane lamina of size dx, dy in X-Y plane. If u is the displacement in x-direction and v is the displacement in y-direction, the shearing strain component is,
$$Γ_{xy}=\frac{∂v}{∂x}+\frac{∂u}{∂y}$$.
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6. The shearing strain component in Y-Z plane is ________
a) $$Γ_{yz} =\frac{∂w}{∂y}+\frac{∂v}{∂z}$$
b) $$Γ_{yz} =\frac{∂v}{∂x}-\frac{∂w}{∂y}$$
c) $$Γ_{yz} =\frac{∂v}{∂y}+\frac{∂w}{∂x}$$
d) $$Γ_{yz} =\frac{∂v}{∂y}-\frac{∂w}{∂x}$$

Explanation: In order to find the shearing strain component in Y-Z plane, let us consider a plane lamina of size dy, dz in Y-Z plane. If w is the displacement in z-direction and v is the displacement in y-direction, the shearing strain component is,
$$Γ_{yz} =\frac{∂w}{∂y}+\frac{∂v}{∂z}.$$

7. The shearing strain component in Z-X plane is ________
a) $$Γ_{xz} =\frac{∂w}{∂y}+\frac{∂v}{∂z}$$
b) $$Γ_{xz} =\frac{∂v}{∂x}-\frac{∂w}{∂y}$$
c) $$Γ_{xz} =\frac{∂u}{∂z}+\frac{∂w}{∂x}$$
d) $$Γ_{xz} =\frac{∂v}{∂y}-\frac{∂w}{∂x}$$

Explanation: In order to find the shearing strain component in Z-X plane, let us consider a plane lamina of size dx, dz in Z-X plane. If w is the displacement in z-direction and u is the displacement in x-direction, the shearing strain component is,
$$Γ_{xz} =\frac{∂u}{∂z}+\frac{∂w}{∂x}.$$

8. The strain tensor is given by ____________
a) $$\begin{pmatrix} ε_{xz} & ε_{xy} & ε_{xx} \\ ε_{yx} & ε_{yz} & ε_{yy} \\ ε_{zx} & ε_{zy} & ε_{zz} \end{pmatrix}$$
b) $$\begin{pmatrix} ε_{xy} & ε_{xx} & ε_{xz}\\ ε_{yx} & ε_{yy} & ε_{yz}\\ ε_{zy} & ε_{zz} & ε_{zz} \end{pmatrix}$$
c) $$\begin{pmatrix} ε_{xx} & ε_{xy} & ε_{xz}\\ ε_{yx} & ε_{yy} & ε_{yz}\\ ε_{zx} & ε_{zy} & ε_{zz} \end{pmatrix}$$
d) $$\begin{pmatrix} ε_{xx} & ε_{xy} & ε_{xz} \\ ε_{yy} & ε_{yx} & ε_{yz} \\ ε_{zz} & ε_{zy} & ε_{zx} \end{pmatrix}$$

Explanation: The strain tensor is given by,
$$\begin{pmatrix} ε_{xx} & ε_{xy} & ε_{xz} \\ ε_{yx} & ε_{yy} & ε_{yz} \\ ε_{zx} & ε_{zy} & ε_{zz} \end{pmatrix}$$
where, the diagonal elements represent the linear strain components and the off-diagonal elements represent the shearing strain components.

9. The strain tensor is given by ____________
a) $$\begin{pmatrix} 1/2Γ_{xz} & 1/2Γ_{xy} & ε_{xx} \\ 1/2Γ_{yx} & 1/2Γ_{yz} & ε_{yy} \\ 1/2Γ_{zx} & 1/2Γ_{zy} & ε_{zz} \end{pmatrix}$$
b)$$\begin{pmatrix} 1/2Γ_{xy} & ε_{xx} & 1/2Γ_{xz} \\ 1/2Γ_{yx} & ε_{yy} & 1/2Γ_{yz} \\ 1/2Γ_{zy} & ε_{zz} & 1/2Γ_{zz} \end{pmatrix}$$
c) $$\begin{pmatrix} ε_{xx} & 1/2Γ_{xy} & 1/2Γ_{xz} \\ 1/2Γ_{yx} & ε_{yy} & 1/2Γ_{yz} \\ 1/2Γ_{zx} & 1/2Γ_{zy} & ε_{zz} \end{pmatrix}$$
d) $$\begin{pmatrix} ε_{xx} & 1/2Γ_{xy} & 1/2Γ_{xz} \\ ε_{yy} & 1/2Γ_{yx} & 1/2Γ_{yz} \\ ε_{zz} & 1/2Γ_{zy} & 1/2Γ_{zx} \end{pmatrix}$$

Explanation: The strain tensor is given by,
$$\begin{pmatrix} ε_{xx} & ε_{xy} & ε_{xz}\\ ε_{yx} & ε_{yy} & ε_{yz}\\ ε_{zx} & ε_{zy} & ε_{zz} \end{pmatrix}$$
where, the diagonal elements represent the linear strain components and the off-diagonal elements represent the shearing strain components. The linear strain components of a diagonal is equal to half the sum shearing strain components.
∴ εyx=1/2Γyx
εxz=1/2Γxz
εzy=1/2Γzy.
∴ the strain tensor is given by,
$$\begin{pmatrix} ε_{xx} & 1/2Γ_{xy} & 1/2Γ_{xz} \\ 1/2Γ_{yx} & ε_{yy} & 1/2Γ_{yz} \\ 1/2Γ_{zx} & 1/2Γ_{zy} & ε_{zz} \end{pmatrix}.$$

10. The linear strain of a diagonal is equal to half the sum shearing strain components.
a) True
b) False

Explanation: The linear strain of a diagonal is equal to half the sum shearing strain components. Thus if εyx, εxz and εzy represent the linear strains of the diagonal of a plane lamina,
∴ εyx=1/2Γyx
εxz=1/2Γxz
εzy=1/2Γzy.

11. The compatibility equations are results of application of stress equations.
a) True
b) False

Explanation: The equations resulting from the applications of the strain equations are known as the compatibility equations or the Saint-Venant’s equations.

12. ___________ is a compatibility equation.
a) $$\frac{∂^2 ε_y}{∂y^2} +\frac{∂^2 ε_y}{∂x^2} = \frac{∂^2 Γ_{xy}}{∂x∂y}$$
b) $$\frac{∂^2 ε_x}{∂y^2} +\frac{∂^2 ε_y}{∂x^2} = \frac{∂^2 Γ_{xy}}{∂x∂y}$$
c) $$\frac{∂^2 ε_x}{∂y^2} +\frac{∂^2 ε_y}{∂x^2} = \frac{∂^2 Γ_{xy}}{∂x∂y}$$
d) $$\frac{∂^2 ε_x}{∂y^2} +\frac{∂^2 ε_y}{∂x^2} = \frac{∂^2 Γ_{xy}}{∂x∂y}$$

Explanation: The strain equations in terms of displacements is given by,
$$ε_X=\frac{∂u}{∂x} —————-(1)$$
$$ε_Y=\frac{∂v}{∂y} —————- (2)$$
$$Γ_{xy}=\frac{∂v}{∂x}+\frac{∂u}{∂y} ———(3)$$
Differentiating (1) twice with respect to y, (2) twice with respect to x and (3) once with respect to x and then y,
$$\frac{∂^2 ε_x}{∂y^2} =\frac{∂^3 u}{dx∂y^2} ————-(4)$$
$$\frac{∂^2 ε_y}{∂x^2} =\frac{∂^3 v}{dy∂x^2} ————-(5)$$
$$\frac{∂^2 Γ_{xy}}{∂x∂y}=\frac{∂^3 u}{dx∂y^2}+\frac{∂^3 v}{dy∂x^2} ——-(6)$$
∴ from (4), (5) and (6), we get,
$$\frac{∂^2 ε_x}{∂y^2} +\frac{∂^2 ε_y}{∂x^2} =\frac{∂^2 Γ_{xy}}{∂x∂y}$$ which is a compatibility equation.

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