This set of Finite Element Method Multiple Choice Questions & Answers focuses on “Boundary Value Problems – 2”.

1. For A1=5, A2=10, A3=5, what is the value of the shape function at node 1 of the element shown?

a) 0.15

b) 0.5

c) 0.35

d) 0.25

View Answer

Explanation:

Total area, A=A1+A2+A3

A=5+10+5

=20.

The shape function at node 2 is given by (A2/A)

=10/20

=0.5.

2. In a solid of revolution, if the geometry, support conditions, loads, and material properties are all symmetric about the axis and are independent of θ, then the problem can be treated as a ____

a) two-dimensional one

b) one-dimensional one

c) three-dimensional one

d) plane strain

View Answer

Explanation: In a solid of revolution, if the geometry, support conditions, loads, and material properties are all symmetric about the axis and are independent of θ, then the problem can be treated as a two-dimensional problem. Moreover, due to the absence of stress variation in the third dimension, such a problem is treated as a plain stress problem.

3. A function Q is evaluated at boundary 1-2 by boundary integral Q=∮q(s)*S(s)ds where q(s)=q_{0} and shape functions S(s) are S_{1}, S_{2}.S_{1}=1-(s/l) and S_{2}=1-S_{1} then Q_{1} is given by expression ____

a) \((\frac{1}{2})\)*q_{0}*l

b) q_{0}*l

c) \((\frac{1}{3})\)*q_{0}*l

d) \((\frac{1}{6})\)*q_{0}*l

View Answer

Explanation: Given Q=∮q*S(s)ds

Q

_{1}=\(\int_{0}^l\)q(s)*S1*ds

=\(\int_{0}^l\)q

_{0}*(1-\((\frac{s}{l}))\)*ds

=[q

_{0}*s*(1-\( (\frac{s}{2*1})\))]

Putting limits of s from zero to l

Q

_{1}=[(1-\(\frac{1}{2}\))*q₀*l]-0

=\((\frac{1}{2})\)*q₀*l.

4. In a static structural type Boundary Value Problem, at any fixed support, How many non-zero Degrees Of Freedom exist?

a) 0

b) 1

c) 2

d) 3

View Answer

Explanation: In a static structural type Boundary Value Problem, three types of supports exist. They are roller, fixed and hinged support. A fixed support has zero degrees of freedom where as a roller and a hinged support have two and one degree of freedom respectively.

5. A function Q is evaluated at boundary 1-2 by boundary integral Q=∮q(s)*S(s)ds where q(s)=q_{0} and shape functions S(s) are S_{1}, S_{2}.S_{1}=1-(s/l) and S_{2}=1-S_{1} then Q_{3} is given by the value ____

a) \((\frac{1}{2})\)

b) 1

c) \((\frac{1}{3})\)

d) 0

View Answer

Explanation: Given Q=∮q*S(s)ds

Since there is no q(s) defined on sides 2-3 and 3-1 we take q(s)=0.

Q

_{3}=\(\int_{0}^l\)0*S1*ds

=0.

6. In a static structural type Boundary Value Problem, at any roller support, How many non-zero Degrees Of Freedom exist?

a) 0

b) 1

c) 2

d) 3

View Answer

Explanation: In a static structural type Boundary Value Problem, three types of supports exist. They are roller, fixed and hinged support. A fixed support has zero degrees of freedom where as a roller and a hinged support have two and one degree of freedom respectively.

7. A function Q is evaluated at boundary 1-2 by boundary integral Q=∮q(s)*S(s)ds where q(s)=q_{0}*(s/l) and shape functions S(s) are S_{1}, S_{2}.S_{1}=1-(s/l) and S_{2}=1-S_{1} then Q_{1} is given by expression ___

a) \((\frac{1}{2})\)*q₀*l

b) q₀*l

c) \((\frac{1}{3})\)*q₀*l

d) \((\frac{1}{6})\)*q₀*l

View Answer

Explanation: Given Q=∮q*S(s)ds

Along line 1-2, Q

_{1}=\(\int_{0}^l\)q(s)*S1*ds

=\(\int_{0}^l\)q

_{0}*\((\frac{s}{l})\)*(1-\((\frac{s}{l}))\)*ds

\(\int_{0}^l\)q

_{0}*\((\frac{s}{l})*ds-\int_{0}^l\)q

_{0}\(((\frac{s}{l})^2)\)*ds

Putting limits of s from zero to l

Q

_{1}=\((\frac{1}{2})\)*q

_{0}*l-\((\frac{1}{3})\)*q

_{0}*l

=q₀*l*\(((\frac{1}{2})-(\frac{1}{3}))\)

=\((\frac{1}{6})\)*q₀*l.

8. In a static structural type Boundary Value Problem, at any hinged support, How many non-zero Degrees Of Freedom exist?

a) 0

b) 1

c) 2

d) 3

View Answer

Explanation: In a static structural type Boundary Value Problem, three types of supports exist. They are roller, fixed and hinged support. A fixed support has zero degrees of freedom where as a roller and a hinged support have two and one non-zero degree of freedom respectively.

9. A function Q is evaluated at boundary 1-2 by boundary integral Q=∮q(s)*S(s)ds where q(s)=q₀*(s/l) and shape functions S(s) are S_{1}, S_{2}.S_{1}=1-(s/l) and S_{2}=1-S_{1} then Q_{2} is given by expression ___

a) \((\frac{1}{2})\)*q₀*l

b) q₀*l

c) \((\frac{1}{3})\)*q₀*l

d) \((\frac{1}{6})\)*q₀*l

View Answer

Explanation: Given Q=∮q*S(s)ds

Along line 1-2, Q

_{2}=\(\int_{0}^l\) q(s)*S2*ds

=\(\int_{0}^l q_0*(\frac{s}{l})*(\frac{s}{l})*ds\)

=\(\int_{0}^l*((\frac{s}{l})^2)*ds\)

Putting limits of s from zero to l

= \((\frac{1}{3})\)*q₀*l.

10. For a linear triangular element with (xi, yi) as the coordinates of the ith node of the element, which option denotes twice the Area of the triangle?

a) (x1y2 − x2y1) + (x2y3 − x3y2) + (x3y1 − x1y3)

b) (x1y2 – x3y1) + (x2y3 – x1y2) + (x3y1 – x2y3)

c) (x1y2 − x2y1) + (x2y3 − x3y2)

d) (x1y1 − x2y2) + (x2y2 − x3y3) + (x3y3 − x1y1)

View Answer

Explanation: A linear triangular element has 3 nodes. With (xi, yi) as coordinates of i

^{th}node, the twice of area is given by determinant of the matrix \(\begin{pmatrix}1&x1&y1\\1&x2&y2\\1&x3&y3\end{pmatrix}\)which equals to (x1y2 − x2y1) + (x2y3 − x3y2) + (x3y1 − x1y3).

11. For a linear triangular element with (xi, yi) as the coordinates of the i^{th} node of the element the area=10units, the value of ∑αi from the standard relation αi+βiX+γiY=(2/3)*Area where X=∑xi, Y=∑yi is ___

a) 10

b) 20

c) 30

d) 40

View Answer

Explanation: A linear triangular element has 3 nodes. With (xi, yi) as coordinates of ith node, the twice of area is given by determinant of the matrix \(\begin{pmatrix}1&x1&y1\\1&x2&y2\\1&x3&y3\end{pmatrix}\) which equals to (x1y2 − x2y1) + (x2y3 − x3y2) + (x3y1 − x1y3). Then from the standard relation we have ∑αi = (x2y3 − x3y2) + (x3y1 − x1y3) + (x1y2 − x2y1)

=2*Area

=2*10

=20.

12. For a linear triangular element with (xi, yi) as the coordinates of the ith node of the element the area=10units, the value of ∑βi from the standard relation αi+βiX+γiY=(2/3)*Area where X=∑xi, Y=∑yi is ___

a) 0

b) 10

c) 20

d) 30

View Answer

Explanation: A linear triangular element has 3 nodes. With (xi, yi) as coordinates of ith node, the twice of area is given by determinant of the matrix \(\begin{pmatrix}1&x1&y1\\1&x2&y2\\1&x3&y3&\end{pmatrix}\) which equals to (x1y2−x2y1)+(x2y3−x3y2)+(x3y1−x1y3). Then from the standard relation we have ∑βi=(y2−y3)+(y3−y1)+(y1−y2)

=y2−y3+y3−y1+y1–y2

=0.

13. In a 3D axisymmetric solid, because of symmetry about the longitudinal axis, the stresses do not vary along ___ coordinate.

a) x

b) y

c) z

d) θ

View Answer

Explanation: In a 3D axisymmetric solid, because of the symmetry about the longitudinal z-axis, the stresses does not vary along circumferential direction i.e. along θ coordinate and such a problem can be treated as a two-dimensional problem.

14. For a linear triangular element with (xi, yi) as the coordinates of the ith node of the element the area=10units, the value of ∑γi from the standard relation αi+βiX+γiY=(2/3)*Area where X=∑xi, Y=∑yi is ___

a) 0

b) 10

c) 20

d) 30

View Answer

Explanation: A linear triangular element has 3 nodes. With (xi, yi) as coordinates of ith node, the twice of area is given by determinant of the matrix\(\begin{pmatrix}1&x1&y1\\1&x2&y2\\1&x3&y3\end{pmatrix}\) which equals to (x1y2−x2y1)+(x2y3−x3y2)+(x3y1−x1y3).Then from the standard relation we have ∑γi=−(x2−x3)−(x3−x1)−(x1 − x2)

=−x2+x3−x3+x1−x1+x2.

=0.

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