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
View Answer
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
View Answer
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
View Answer
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
View Answer
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) \(Γ_{xy}=\frac{∂v}{∂y}-\frac{∂u}{∂x} \)
View Answer
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,
\(Γ_{xy}=\frac{∂v}{∂x}+\frac{∂u}{∂y} \).
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) \(Γ_{yz} =\frac{∂v}{∂y}-\frac{∂w}{∂x} \)
View Answer
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) \(Γ_{xz} =\frac{∂v}{∂y}-\frac{∂w}{∂x} \)
View Answer
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) \(\begin{pmatrix}
ε_{xx} & ε_{xy} & ε_{xz} \\
ε_{yy} & ε_{yx} & ε_{yz} \\
ε_{zz} & ε_{zy} & ε_{zx}
\end{pmatrix}\)
View Answer
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) \(\begin{pmatrix}
ε_{xx} & 1/2Γ_{xy} & 1/2Γ_{xz} \\
ε_{yy} & 1/2Γ_{yx} & 1/2Γ_{yz} \\
ε_{zz} & 1/2Γ_{zy} & 1/2Γ_{zx}
\end{pmatrix}\)
View Answer
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
View Answer
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
View Answer
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) \(\frac{∂^2 ε_x}{∂y^2} +\frac{∂^2 ε_y}{∂x^2} = \frac{∂^2 Γ_{xy}}{∂x∂y}\)
View Answer
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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