Finite Element Method Questions and Answers – Eigen Value and Time Dependent Problems – 2

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This set of Finite Element Method Questions and Answers for Entrance exams focuses on “Eigen Value and Time Dependent Problems – 2”.

1. Suppose the following eigenvalue equation represents a bar problem, then the value of the parameters a and c0 should be EA and ρA, respectively.
\(-\frac{d}{dx}(a\frac{dU}{dx})\)=λc0 U
a) True
b) False
View Answer

Answer: a
Explanation: For the given eigenvalue equation, the values of the parameters a and c0 depends upon the physical properties and phenomena involved in the problem. For a bar problem, a=EA and c0=ρA, where E is Young’s modulus, A is cross-sectional area and m is the mass density. For a heat transfer problem, a=kA and c0=ρcA.
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2. A plane wall of length L units and Cross-section area A units was initially maintained at a temperature of T units. It is subjected to an ambient temperature of T units at one surface. If the heat transfer coefficient at the surface of the wall is assumed to be h units, then what is the temperature gradient developed at the surface?
a)(T-T)\(\frac{h}{k}\)
b)(T-T)\(\frac{1}{L} \)
c) T-T
d) T-T
View Answer

Answer: a
Explanation: Let Tx be temperature gradient developed at the surface. If the heat transfer coefficient at the surfaces of a wall is assumed to be h units, then the heat interaction at the surfaces of the wall is evaluated by Equating the conduction heat transfer to the convection heat transfer, i.e.,
kATx = hA(T-T)
Tx=(T-T)\(\frac{h}{k}\).

3. A plane wall of thermal conductivity of 45\(\frac{W}{mK}\) was initially maintained at a temperature of 35°C. It is subjected to an ambient temperature of 45°C at one surface. If the heat transfer coefficient at the surface of the wall is 9\(\frac{W}{m^2K}\), then what is the temperature gradient developed at the surface?
a) 1
b) 2
c) 3
d) 4
View Answer

Answer: b
Explanation: Let Tx be temperature gradient developed at the surface. If the heat transfer coefficient at the surface of a wall is is 9\(\frac{W}{m^2K}\) then the heat interaction at the surface of the wall is evaluated by equating the conduction heat transfer to the convection heat transfer, i.e.,
45Tx = 9(35-45)
Tx = \(\frac{9}{-45}\) (35-45)
=\(\frac{1}{-5}\) (-10)
=2.
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4. A plane wall was maintained initially at a temperature of T units. It is subjected to an ambient temperature of T units at one surface. If the heat transfer coefficient at the surfaces of the wall is assumed to be infinite, then what is the new temperature at the wall?
a) T
b) T
c) T-T
d) T-T
View Answer

Answer: b
Explanation: Let X be the unknown new temperature at the wall surface. If the heat transfer coefficient at the surfaces of a wall is assumed to be h, then the heat interaction at the surfaces of the wall is evaluated by equating the conduction heat transfer to the convection heat transfer, i.e.,
kATx = hA(X – T)
\(\frac{-kT_x}{h}\) = X – T
Given h = ∞
\(\frac{-kT_x}{\infty}\) = X – T
0 = X – T
X = T.

5. A plane wall was maintained initially at a temperature of 35°C. It is subjected to an ambient temperature of 45°C at one surface. If the heat transfer coefficient at the surfaces of the wall is assumed to be infinite, then what is the new temperature at the wall surface?
a) 35°C
b) 45°C
c) 40°C
d) 50°C
View Answer

Answer: b
Explanation: Let X be the unknown new temperature at the wall surface. If the heat transfer coefficient at the surfaces of a wall is assumed to be h, then the heat interaction at the surfaces of the wall is evaluated by equating the conduction heat transfer to the convection heat transfer, i.e.,
kATx=hA(X-T)
\(\frac{-kT_x}{h}\)=X-T
Given h = ∞
\(\frac{-kT_x}{\infty}\)=X-T
0=X-T
X=T, given T=45°C
X=45°C.
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6. In thermodynamics, the following equation represents a diffusion process. If k is thermal conductivity, p is density, and c is the specific heat at constant pressure, then what is α?
\(\frac{\partial^2 T}{\partial x^2} = \frac{1}{\alpha} \frac{\partial T}{\partial t}\)
a) \(\frac{k}{pc}\)
b) \(\frac{pc}{k}\)
c) \(\frac{c}{kp}\)
d) \(\frac{c}{k}\)
View Answer

Answer: a
Explanation: The term α is called diffusion coefficient, and it is equal to \(\frac{k}{pc}\). This equation governs one-dimensional temperature distribution in a plain wall. A one-dimensional problem is solved using bar elements with one degree of freedom at each node.

7. The governing equation of an unsteady one-dimensional heat transfer problem is given below. It has a solution u(x,t) = U(x)exp(λt). What is λ appropriately called?
\(\frac{-\partial}{\partial x} (a \frac{\partial u}{\partial x}) + b \frac{\partial u}{\partial t}\) + cu = 0 for 0<x<L
a) Natural frequency
b) Eigenvalue
c) Thermal diffusivity
d) Thermal flux
View Answer

Answer: b
Explanation: The governing equation of an unsteady one-dimensional heat transfer problem is a parabolic equation. Hence, its solution is given by u(x,t) = U(x)exp(λt), where u represents temperature along a direction x at any time t, U(x) is the corresponding mode shape, and λ is the eigenvalue of the equation. The solution is periodic.
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8. The unsteady natural axial oscillations of a bar are periodic, and they are determined by assuming a solution u(x, t) = U(x) e-iwt. Which option is not correct about the solution equation?
a) w denotes the natural frequency
b) w2 denotes eigenvalue
c) U(x) denotes mode shape
d) u(x, t) denotes transverse displacements
View Answer

Answer: d
Explanation: The unsteady natural axial oscillations of a bar are periodic. They are measured by assuming a solution u(x, t) = U(x) e-iwt, where w is natural frequency, w2 is an eigenvalue, U(x) is mode shape, and u is instantaneous axial displacement. The problem is solved in FEM by employing bar elements and appropriate shape functions.

9. In matrix algebra, which option is not correct about an eigenvalue problem of the type Ax = Lx?
a) It has a discrete solution
b) It has solution only if A non-singular
c) x is called eigenvector
d) L is called eigenvalue
View Answer

Answer: b
Explanation: An eigenvalue problem of the type Ax = Lx looks as if it should have a continuous solution, but instead, it has discrete ones. The problem is to find the numbers denoted by L, called eigenvalues, and their matching vectors denoted by x, called eigenvectors. It may have a solution irrespective of whether the matrix A is singular or not.
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10. The dynamic equation of motion of a structure contains M, C and K as mass, damping and stiffness matrices of the structure, respectively. If F is an external load vector, then which option is correct about the equation?
a) M\(\ddot{x}\) + K\(\dot{x}\) + Cx = F
b) M\(\ddot{x}\) is time-dependent
c) All the forces are time-independent
d) The equation is of 3rd order
View Answer

Answer: b
Explanation: The dynamic equation of motion of a structure is a 2nd order equation. It is written as M\(\ddot{x}\) + C\(\dot{x}\) + Kx = F, where M, C and K are the mass, damping and stiffness matrices of structure, respectively. All the forces in the equation are time-dependent. M\(\ddot{x}\) is inertia force, Kx is spring force, and C\(\ddot{x}\) is damping force.

11. In matrix algebra, a matrix K equals \(\begin{pmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&3 \end{pmatrix}\). What is the value of a, if K7 = \(\begin{pmatrix} c&0&0\\ 0&b&0 \\ 0&0&a \end{pmatrix}\)?
a) 2187
b) 729
c) 6561
d) 57
View Answer

Answer: a
Explanation: Since K is a diagonal matrix, its higher powers are obtained by raising its diagonal elements to the same power. If K=\(\begin{pmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&3 \end{pmatrix}\) then K7=\(\begin{pmatrix} 1^7&0&0 \\ 0&1^7&0 \\ 0&0&3^7\end{pmatrix}\). Equating the corresponding elements of \(\begin{pmatrix} c&0&0\\ 0&b&0 \\ 0&0&a \end{pmatrix}\) and \(\begin{pmatrix} 1^7&0&0 \\ 0&1^7&0 \\ 0&0&3^7\end{pmatrix}\) we get
a=37
a=2187.

12. In matrix algebra, what is the eigenvalue of the matrix \(\begin{pmatrix} 1&1&1 \\ 1&1&1 \\ 1&1&1 \end{pmatrix}\)?
a) 1
b) 2
c) 3
d) 4
View Answer

Answer: c
Explanation: The eigenvalue, L of a matrix is equal to the root (factor) of the equation |K-LI|=0.
Let the given matrix be denoted by K then K-LI = \(\begin{pmatrix} 1&1&1 \\ 1&1&1 \\ 1&1&1 \end{pmatrix} – L \begin{pmatrix}1&0&0 \\ 0&1&0 \\0&0&1 \end{pmatrix}\)
= \(\begin{pmatrix} 1-L&1&1 \\ 1&1-L&1 \\ 1&1&1-L \end{pmatrix}\)
|K-LI| = \(\begin{vmatrix} 1-L&1&1 \\ 1&1-L&1 \\ 1&1&1-L \end{vmatrix}\)
= (1-L)((1-L)2-1)-1(-L)+1(L)
= (1-L)(L2-2L) + 2L
= -L3 + 3L2
= -L2 (L-3).
Given -L2 (L-3) = 0
On simplification L = 0, 0 and 3.

13. In matrix algebra, what is the value of a-b if the eigenvector of \(\begin{pmatrix}1&1&1 \\ 1&1&1 \\ 1&1&1 \end{pmatrix}\) corresponding to eigenvalue three is \(\begin{pmatrix}a \\ b \\ a \end{pmatrix}\)?
a) 0
b) 1
c) 2
d) 3
View Answer

Answer: a
Explanation: If X is an eigenvector corresponding to an eigenvalue L of a matrix K, then KX=LX. The eigenvector of \(\begin{pmatrix}1&1&1 \\ 1&1&1 \\ 1&1&1 \end{pmatrix}\) corresponding to eigenvalue three is \(\begin{pmatrix}1 \\ 1 \\ 1 \end{pmatrix}\). Equating the corresponding elements of \(\begin{pmatrix}a \\ b \\ a \end{pmatrix}\) and \(\begin{pmatrix}1 \\ 1 \\ 1 \end{pmatrix}\)
a=b=1
a-b=0.

14. From the Euler-Bernoulli beam theory of natural vibrations, using cubic Hermite polynomials approximation, what is the 1st element of the stiffness matrix?
a) \(\frac{12EI}{h^3}\)
b) \(\frac{12EA}{h^3}\)
c) \(\frac{12EA}{h}\)
d) \(\frac{12AI}{h^3}\)
View Answer

Answer: a
Explanation: In the formulation of the Euler-Bernoulli beam theory, there are two degrees of freedom at a point, w and \(\frac{dw}{dx}\). Typically, the finite element model of this theory uses cubic polynomial. The first element of the stiffness matrix is \(\frac{12EI}{h^3}\), where E is Young’s modulus, I is the area moment of inertia and h is the length of the element.

15. From the Timoshenko beam theory of natural vibrations, using cubic Hermite polynomials approximation, what is the 1st element of the mass matrix?
a) \(\frac{\rho A}{3}\)
b) \(\frac{\rho A}{6}\)
c) 0
d) \(\frac{\rho I}{3}\)
View Answer

Answer: a
Explanation: Using the Timoshenko beam theory applied to natural vibrations, mode shape is approximated using the cubic Hermite polynomials \(\psi_i^e\) and \(\psi_j^e\). The first element of a mass matrix is \(M_{ij}^{11} = \int_{x_a}^{x_b} \rho A \psi_i^e \psi_j^e\) dx, where x is the length of the element. For the 1st element, using appropriate values of \(\psi_i^e\) and \(\psi_j^e\), the term \(M_{ij}^{11}\) reduces to \(\frac{\rho A}{3}\), where ρ is the density of the beam material, and A is the cross-section area of the beam.

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