# Finite Element Method Questions and Answers – Library of Elements and Interpolation Functions – 2

This set of Finite Element Method Questions and Answers for Campus interviews focuses on “Library of Elements and Interpolation Functions – 2”.

1. In the FEM element library, what is the correct name for a three noded triangular element?
a) Linear strain triangular element
b) Constant strain triangular element
c) Variable strain triangular element
d) Higher-order triangular element

Explanation: A Constant Strain Triangular (CST) element is the simplest triangular element with only three nodes that are located at its ends. A Linear Strain Triangular element (LST) is a six-noded triangular element with three intermediate nodes in addition to three end nodes. For plane stress applications, LST gives an accurate result compare to the three-noded CST element. The variable strain triangular element is a higher-order triangular element with more than six nodes.

2. If the geometry and other parameters of an element are defined in terms of only one spatial coordinate, then the element is a one-dimensional element?
a) True
b) False

Explanation: A one-dimensional element possesses one degree of freedom at each node. It is also known as a bar element or line element. Geometry and other parameters of a bar element are defined in terms of one spatial coordinate only. If a one-dimensional element has two nodes with corresponding displacements u1, u2 and corresponding shape functions N1, N2, then the displacement function is given by N1u1+N2u2.

3. In FEM, what is the name of the element specified by a polynomial of order two or more?
a) Nonlinear element
b) Higher-order element
c) Linear element
d) Master element

Explanation: Simplex and linear elements contain nodes only at endpoints but not at the interior. They have linear polynomials as interpolation functions. Higher-order elements can be created easily from simplex elements by adding additional intermediate nodes to each element. They are also called complex elements.

4. In FEM, what is the name of the shape function of an Euler-Bernoulli beam element?
a) Hermite cubic interpolation function
b) Lagrange cubic interpolation function
c) Consistent element functions

Explanation: Interpolation function of a beam element is continuous with nonzero derivatives up to order two. It is derived by interpolating the displacement polynomial as well as its derivative at the nodes. Such interpolation functions are called as Hermite cubic interpolation (or cubic spline) function. The Lagrange cubic interpolation Functions are derived by interpolating the displacement polynomial but not its derivatives.

5. In FEM, which option is used to develop the Higher-order rectangular elements (i.e., rectangular elements with interpolation functions of higher degree) systematically?
a) A rectangular array of binomial coefficients
b) Galerkin method
c) Jacobi method
d) Delaunay triangulation

Explanation: Analogous to the Lagrange family of triangular elements, the Lagrange family of rectangular elements can be developed from a rectangular array of binomial coefficients. Since a linear rectangular element has four corners (hence, four nodes), the polynomial should have the first four terms 1, x, y, and xy(which form a parallelogram in Pascal’s triangle and a rectangle in the rectangular array of binomial coefficients).
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6. In FEM, what are the elements in which the same shape functions describe the geometry and field displacement variables?
a) Iso-parametric
b) Axi-Symmetric
c) Super-parametric
d) Sub-parametric

Explanation: In sub-parametric formulations, the geometry is represented by elements of a lower order than those used to approximate the dependent variable. An example of this category is provided by the beam element. In iso-parametric formulations (which are the most common in practice), the same element is used to approximate the geometry as well as the dependent unknowns. In super-parametric formulations, the geometry is represented by elements of a higher order than those used to approximate the dependent variables.

7. In the Finite Element Method (FEM), if the geometry is represented by elements of a higher order than those used to approximate the field displacement variables, then it is called super-parametric formulation.
a) True
b) False

Explanation: In super-parametric formulations, the geometry is represented by elements of a higher order than those used to approximate the dependent variables. In sub-parametric formulations, the geometry is represented by elements of a lower order than those used to approximate the dependent variable. The beam element provides an example of this category. In iso-parametric formulations (which are the most common in practice), the same element is used to approximate the geometry as well as the dependent unknowns.

8. Which option is not correct about the Lagrange rectangular element used in FEM?
a) Second-order Lagrange element has nine nodes
b) Zero-order Lagrange element has one node
c) First-order Lagrange element has four nodes
d) Third-order Lagrange element has fifteen nodes

Explanation: In general a pth-order Lagrange rectangular element has n nodes with n= (p+1)2, where p=0,1,2 …
For p=0,
n=1.
For p=1,
n=22,
=4.
For p=2,
n=32
=9.
For p=3,
n=42
=16.

9. Which option is not correct about the Lagrange family of triangular elements used in FEM?
a) 2nd-degree polynomial corresponds to 6 noded triangle
b) 0th-degree polynomial corresponds to 1 noded triangle
c) 1st-degree polynomial corresponds to 3 noded triangle
d) 3rd-degree polynomial corresponds to 9 noded triangle

Explanation: A pth degree polynomial corresponds to n noded triangular element with n=0.5(p+2)(p+1), where n=0, 1, 2 …
For p=0,
n=1.
For p=1,
n=0.5*3*2,
=3.
For p=2,
n=0.5*3*4
=6.
For p=3,
n=0.5*4*5
=10.

10. In FEM, if x, y represents Cartesian coordinates, then which term of Pascal’s triangle corresponds to the position of the interior node of the following element?

a) 1
b) xy
c) x2y
d) xy2

Explanation: The interior node of the nine noded Lagrange rectangular element is the central term in the corresponding parallelogram on Pascal’s triangle. The central term is the middle term of the 3rd row. As the 3rd row contains three terms viz. x2, xy and y2 in the same order, the interior node corresponds to the term xy.

11. What is the number of nodes present on the boundary of the Lagrange quadratic rectangular element used in FEM?
a) 1
b) 4
c) 8
d) 9

Explanation: The Lagrange quadratic rectangular element has nine regularly spaced nodes. Four nodes are paced at the four corners, four at midpoints of the sides, and one at the center of the element. Thus, a total of 8 boundary nodes are present. Its associated polynomial has a total of nine terms, including the second degree and third-degree terms.

12. Which nodes of the higher-order elements of the Lagrange family do not contribute to the inter-element connectivity?
a) Interior nodes
b) All nodes
c) Corner nodes
d) Intermediate nodes

Explanation: Since the Interior nodes of the higher-order elements of the Lagrange family do not contribute to the inter-element connectivity, they can be condensed out at the element level so that the size of the element matrices is reduced. The elements formed after removing the Interior nodes are called serendipity elements.

13. What is the reason for an element in the Serendipity family to have a smaller size of stiffness matrix compare to a similar element in the Lagrange family?
a) Absence of interior nodes
b) Modified element connectivity
c) Lesser interpolation functions
d) Irregular arrangements of nodes

Explanation: The internal nodes of the higher-order elements of the Lagrange family do not contribute to inter-element connectivity, and hence they are condensed out at element level; as a result, the size of the element matrices is reduced. However, the element connectivity remains unaffected. The elements formed after removing the internal nodes are called serendipity elements.

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