Finite Element Method Questions and Answers – Penalty – Finite Element Model – 1

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This set of Finite Element Method Multiple Choice Questions & Answers (MCQs) focuses on “Penalty – Finite Element Model – 1”.

1. Which type of problem can be obtained by reformulating a problem with differential constraints by using the penalty method?
a) A problem with no constraints
b) A problem with variable constraints
c) A problem with fixed constraints
d) A problem with structural constraints
View Answer

Answer: a
Explanation: Penalty method is introduced in connection with constraint equations. It can also be used to reformulate a problem with differential constraints as one without constraints. In the viscous fluid flow problem, the constraints are created on velocity components by the elimination of pressure term, and thus, the Penalty method is applied.
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2. In the interest of the simple formulation of viscous flows, which case does not involve time derivative terms?
a) Static case
b) Transient case
c) Unsteady case
d) Non-periodic
View Answer

Answer: a
Explanation: The equations governing flows of viscous incompressible fluids can be viewed as equivalent to minimizing a quadratic functional with a constraint. In the penalty method, in the interest of simplicity, for the static case, the constraint condition does not involve time derivative terms. Then, we add time derivative terms to study transient problems.

3. In the penalty formulation of the fluid flow model, if the velocity field (vx, vy ) satisfies the continuity equation, then the weight functions (w1, w2) also satisfy the continuity equation.
a) True
b) False
View Answer

Answer: b
Explanation: In the penalty formulation, we begin with the unconstrained problem described by the weak forms of the mixed model, without the time—derivative terms. Now, suppose that the velocity field (vx, vy ) is such that the continuity equation is satisfied identically. Then the weight functions (w1, w2) being (virtual) variations of the velocity components, also satisfy the continuity equation.
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4. In the weak forms of the fluid flow model, as the weight functions (w1, w2) are virtual variations of the velocity components(vx, vy ),respectively, which relation is satisfied by the weight functions?
a) \(\frac{\partial w1}{\partial x}+\frac{\partial w2}{\partial y}\)=0
b) \(\frac{\partial w2}{\partial x}+\frac{\partial w1}{\partial y}\)=0
c) \(\frac{\partial w1}{\partial x}-\frac{\partial w2}{\partial y}\)=0
d) \(\frac{\partial w2}{\partial x}-\frac{\partial w1}{\partial y}\)=0
View Answer

Answer: a
Explanation: In the weak forms of the fluid flow model, suppose that the velocity field (vx, vy ) is such that the continuity equation \(\frac{\partial v_x}{\partial x}+\frac{\partial v_y}{\partial y}\)=0 is satisfied identically. Then the weight functions (w1, w2) being (virtual) variations of the velocity components, also satisfy the continuity equation. Thus the relation possessed by the weight functions is \(\frac{\partial w1}{\partial x}+\frac{\partial w2}{\partial y}=0\).

5. In finite element modeling, which formulation introduces constraints on variables and satisfies them in an approximate sense?
a) Velocity-pressure formulation
b) Penalty formulation
c) Mixed formulation
d) Lagrange multiplier formulation
View Answer

Answer: b
Explanation: There are two different finite element models of momentum and continuity equations. The first one is a natural and direct formulation of momentum and continuity equations. The second model, penalty formulation, is based on the interpretation that the continuity equation is a new relationship among the velocity components, and it eliminates the pressure term leading to a constraint on the velocity components. The constraint is satisfied in a least-squares (i.e., approximate) sense.
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6. In the formulation of the finite element model, which option is the complete restated form of the weak forms of viscous fluids flow equations?
a) Only Bt(w,v)+Bv(w,v)-B̅p(w,P)=l(w)
b) Bt(w,v)+Bv(w,v)-B̅p(w,P)=l(w) and –Bp(w3, v)=0
c) Only –Bp(w3, v)=0
d) Bt(w,v)+Bv(w,v)-B̅p(w,P)=0
View Answer

Answer: b
Explanation: The problem described by weak forms of viscous fluids flow equations can be restated as a variational problem of finding (vx, vy, P) such that Bt(w,v)+Bv(w,v)-B̅p(w,P)=l(w) and –Bp(w3, v)=0 holds for all weight functions (w1, w2, w3) and t>0. Here, we have used the notation w=\(\begin{Bmatrix}w_1\\w_2\end{Bmatrix}, \)v=\(\begin{Bmatrix}v_x\\v_y\end{Bmatrix},\) f=\(\begin{Bmatrix}f_x\\f_y\end{Bmatrix}\) and t=\(\begin{Bmatrix}t_x\\t_y\end{Bmatrix}\).

7. In the following variational problem of finding velocity components and pressure, which bilinear form includes time-derivative terms?
Bt(w,v)+Bv(w,v)-B̅p(w,P)=l(w); –Bp(w3, v)=0
a) Bt(w,v)
b) Bv(w,v)
c) B̅p(w,P)
d) Bp(w3,v)
View Answer

Answer: a
Explanation: The problem described by weak forms of viscous fluids flow equations can be restated as a variational problem of finding (vx, vy, P) such that Bt(w,v)+Bv(w,v)-B̅p(w,P)=l(w) and –Bp(w3, v)=0 holds for all weight functions (w1, w2, w3) and t>0. The term Bt(w,v)=∫ΩeρwTvdx contains the time derivative term v. Further more, in the penalty method, we add time derivative terms to study transient problems.
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8. In the weak forms of the fluid flow model, since the weight functions are linearly dependent on each other, the sum of the three weak forms is the same as the three individual equations.
a) True
b) False
View Answer

Answer: b
Explanation: if x1, x2 and x3 represent three weak forms, then x1=ax2+bx3 is true only if the weight functions are linearly dependent. Since the weight functions are linearly independent of each other, the sum of the three weak forms is the same as the three individual equations i.e, the sum must be the expression x1+x2+x3 only.

9. Which option is not correct concerning the bilinear term B(v,w) in the variational problem of the viscous fluid flow equation?
a) It is symmetric
b) It contains the viscosity matrix
c) Bv(w,v)= Bv(v,w)
d) It is bilinear in both w and v
View Answer

Answer: d
Explanation: The complete variational problem of the viscous fluid flow equation is Bt(w,v)+Bv(w,v)-B̅p(w,P)=l(w) and –Bp(w3, v)=0. The term, Bv(v,w) is linear in both v and w, it is symmetric because it contains the symmetric matrix, C. Thus Bv(w,v)= B(v,w). The variational problem Bv(v,w)=l(w) is a constrained problem.
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10. In the variational problem of fluid flow, what is the correct matrix form of C in the bilinear form Bv(w,v)=∫Ωe(Dw)TC(Dv)dx?
a) \(\begin{pmatrix}2&0&0\\0&2&0\\0&0&1\end{pmatrix}\)
b) μ\(\begin{pmatrix}2&0&0\\0&1&0\\0&0&2\end{pmatrix}\)
c) μ\(\begin{pmatrix}2&0&0\\0&2&0\\0&0&1\end{pmatrix}\)
d) μ\(\begin{pmatrix}1&0&0\\0&2&0\\0&0&2\end{pmatrix}\)
View Answer

Answer: c
Explanation: The complete restated form of the weak forms of viscous fluids flow equations is given by Bt(w,v)+Bv(w,v)-B̅p(w,P)=l(w) and –Bp(w3,v)=0. The bilinear form, Bv(w,v)=∫Ωe(Dw)TC(Dv)dx is symmetric because it contains the symmetric matrix μ \(\begin{pmatrix}2&0&0\\0&2&0\\0&0&1\end{pmatrix}\), denoted by the letter C. C is the viscosity matrix since it contains the viscosity term μ.

11. For the given weight functions [2x 3y] and applied pressure of 3pa, what is the value of K if B̅p(w,P)=∫ΩeKdx denotes the bilinear term in the variational problem of fluid flow?
a) 5
b) 10
c) 15
d) 20
View Answer

Answer: c
Explanation: Since the bilinear term has the form B̅p(w,P)=∫Ωe\((D_1^Tw)^T\)Pdx; K=\((D_1^Tw)^T\)p.
We have D1=\(\begin{bmatrix}\frac{\partial}{\partial x}\\\frac{\partial}{\partial y}\end{bmatrix}\) and w=\(\begin{bmatrix}2x\\3y\end{bmatrix}\)
\(D_1^T w=\begin{bmatrix}\frac{\partial}{\partial x}&\frac{\partial}{\partial y}\end{bmatrix}\begin{bmatrix}2x\\3y\end{bmatrix}\)
= \(\frac{\partial (3y)}{\partial y}+\frac{\partial (2x)}{\partial x}\)
=2+3
=5.
K=(D1T w)Tp
=5*3
=15.
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