Category: Finite Element Objective Questions

Finite Element Method Multiple Choice Questions on “One Dimensional Problems – Temperature Effects”.

1. With temperature effect which will vary linearly?
a) Horizontal stress load
b) Potential energy
c) Vertical stress load
d) Kinematic energy
Answer: c
Clarification: Temperature is a variant which varies from one point to another point. It has adverse effects on different structures. By temperature effect Vertical stress load vary linearly.

2. α means ____
a) Co-efficient of thermal expansion
b) Co-efficient of linear expansion
c) Thermal expansion
d) Thermal effect
Answer: a
Clarification: The co-efficient of thermal expansion describes how the size of an object changes with a change in temperature. Specifically, it measures the fractional change in size per degree change in temperature at constant pressure. It is denoted by symbol α.

3. In temperature effect, initial strain, ε₀= ____
a) α ΔT
b) α+ΔT
c) α-ΔT
d) Load
Answer: a
Clarification: Strain is relative change in shape or size of an object due to externally applied forces. Temperature is a variant which varies from one point to another point. In temperature effect of FEM, Initial strain ε₀=α ΔT.

4. In a structure, a crack is formed as a result of ______
a) Thermal expansion
b) Thermo couple
c) Thermal strain
d) Thermal stress
Answer: d
Clarification: Thermal stress is caused by differences in temperature or by differences in thermal expansion. A crack formed as a result of Thermal stress produced by rapid cooling from a high temperature.

5. In the given diagram, the line indicates ____________

a) Stress relation
b) Strain relation
c) Stress – strain relation
d) Undefined
Answer: c
Clarification: Stress is physical quantity that expresses the internal forces that neighboring particles of a continues material exert on each other. While, strain is the measure of deformation of the material. In the given diagram, the line indicates Stress-strain relation.

6. Stress – strain relation is given as _____
a) σ=E
b) σ=Eε
c) σ=E(ε-ε₀)
d) Undefined
Answer: c
Clarification: Stress is a physical quantity that expresses the internal forces that neighboring particles of a continuous material exert on each other, while strain is the measure of the deformation of the material. Stress- strain relation is given as
σ=E(ε-ε₀).

7. Strain energy per unit volume is ___
a) u₀=E(ε-ε₀)
b) u₀=(frac{1}{2})E(ε-ε₀)
c) Non symmetric
d) Specified displacement
Answer: b
Clarification: Energy stored in a body due to deformation is called Strain energy. The strain energy per unit volume is called strain energy density and the area under the stress strain curve towards the point is deformation. Strain energy per unit volume is
u₀=(frac{1}{2})E(ε-ε₀)

8. Temperature change is denoted as_____
a) ΔT=(T₂-T₁)
b) θ^e
c) l_e
d) A_e
Answer: b
Clarification: Temperature is the measure of average kinetic energy of the particles in a system. Adding heat to system causes temperature rise. In Finite element analysis temperature is denoted as θ^e.

9. θ^e=(frac{E_eA_el_ealphaDelta T}{x_2-x_1}begin{Bmatrix}-1 \ 1 end{Bmatrix}) is the temperature effect.
a) True
b) False
Answer: a
Clarification: Temperature is the measure of average kinetic energy of the particles in a system. Adding heat to system causes its temperature to rise. Temperature effect describes that how much of temperature is rised in body when load is applied. Temperature effect formula is as shown
θ^e=(frac{E_eA_el_ealphaDelta T}{x_2-x_1}begin{Bmatrix}-1 \ 1 end{Bmatrix}).

10. Stress in each element is ____
a) Eliminated
b) σ=EBq
c) σ=αΔT
d) σ=E(Bq-αΔT)
Answer: d
Clarification: Stress is a physical quantity that expresses the internal forces that neighboring particles of a continuous material exert on each other. After solving Finite element equations KQ=F for the displacements Q, the stress in each element can be obtained from equation.
σ=E(Bq-αΔT).

Finite Element Method Multiple Choice Questions on “Beams and Frames – Load Vector”.

1. On a simply supported beam with uniformly distributed load (p) over length (l) the value of reaction force at one support is _______
a) p*l
b) (p*l)/2
c) (p*l)²
d) (p*l)/3
Answer: b
Clarification: Overall force acting on the beam=p*l
The reaction force at one of the two supports=(p*l)/2.

2. In Galerkin method we convert a continuous operator problem into a discrete problem.
a) True
b) False
Answer: a
Clarification: In Galerkin method we convert a continuous operator problem into a discrete problem. In beams the problem is formulated into finite elements using Galerkin Method.

3. Which of the following is not a one dimensional element?
a) bar
b) brick
c) beam
d) rod
Answer: b
Clarification: Brick is not a one dimensional element. One dimensional element is used when one dimension of the model geometry is significantly greater than the other two dimensions.

4. Three geometrically identical beams made out of steel, aluminum, and titanium are axially loaded. Which of the following statements is correct?
a) Stress in titanium is the least
b) Stress in cast iron is the highest
c) Stress in steel is the least
d) Stress in all three beams is the same
Answer: d
Clarification: The stress in all three beams will be induced equally. Stress in not dependent on the material, but the geometry and cross section of the element.

5. The application of force at which of the following point on a beam will nullify the effect of torsion?
a) Centroid
b) Center of mass
c) Extreme fiber
d) Shear centre
Answer: d
Clarification: The application of force on shear centre nullifies the effect of torsion on a beam element. This is useful when the cross section of beam is asymmetric and creates a twisting effect on application of force.

6. For a taper beam element two cross sections are necessary to define the geometry.
a) True
b) False
Answer: a
Clarification: A taper beam element requires two cross sections to define the geometry. Regular beam element cannot take into account the variation in cross section required to define the geometry.

7. Which of the following statements is correct?
a) Beam elements are recommended for unsymmetrical cross sections
b) Bar elements are recommended for unsymmetrical cross sections
c) Beam and bar elements can both be used for unsymmetrical cross sections
d) Neither beam nor bar elements can be used for unsymmetrical cross sections
Answer: a
Clarification: Beam elements are recommended for unsymmetrical cross sections due to their ability to take into account the shear centre. Thus shear centre can nullify effects of torsion acting on unsymmetrical cross section which is limited in case of bar elements.

8. The equivalent load (f) acting on a simply supported beam element loaded with uniformly distributed load (p) over an element of length (l) is given by which of the following expressions?
a) f=([frac{pl}{2},frac{pl^2}{12},frac{pl}{2},-frac{pl^2}{12}])
b) f=([frac{pl}{2},frac{pl^2}{12},frac{pl}{2},frac{pl^2}{12}])
c) f=([frac{pl}{2},-frac{pl^2}{12},frac{pl}{2},-frac{pl^2}{12}])
d) f=([frac{pl}{2},frac{pl^2}{12},frac{-pl}{2},-frac{pl^2}{12}])
Answer: a
Clarification: The value of equivalent load is given by
f=([frac{pl}{2},frac{pl^2}{12},frac{pl}{2},-frac{pl^2}{12}])
Here l is length of beam element, and p is the uniformly distributed load. Here (frac{pl}{2}) represents the reaction force on the supports, and (frac{pl^2}{12}) represents the moment acting on the supports.

Finite Element Method Multiple Choice Questions on “Plane Elasticity – Finite Element Model”.

1. If only the first derivatives of u_x and u_y appear in the weak forms, then their interpolation must be at least bilinear.
a) True
b) False
Answer: a
Clarification: An examination of the weak form, 0=∫_Ωeh_e([frac{partial w_1}{partial x}(c_{11}frac{partial u_x}{partial x}+c_{12}frac{partial u_y}{partial y}) + c_{66}frac{partial w_1}{partial y} (frac{partial u_x}{partial y} + frac{partial u_y}{partial x}))+ρw₁u_x]dxdy-∫_Ωeh_ew₁f_xdxdy-∮_┌dh_ew₁t_xds reveals that only first derivatives of u_x and u_y with respect to x and y appear respectively. Therefore, u_x and u_y must be approximated by the Lagrange family of interpolation functions, and at least a bilinear (i.e., linear both in x and y) interpolation is required.

2. In FEM, if two independent variables are components of the same vector, then they can be approximated by two different types of interpolations.
a) True
b) False
Answer: b
Clarification: u_x and u_y are the primary variables in the expanded weak forms of the plane elasticity problems. Although u_x and u_y are independent of each other, they are the components of the displacement vector. Therefore, both components should be approximated using the same type and degree of interpolation.

3. What can one conclude about the displacement components u_x and u_y in the finite element model of the plane elasticity equations?
a) They are primary variables and must be carried as primary nodal degrees of freedom
b) They are secondary variables and must be carried as primary nodal degrees of freedom
c) They are primary variables and must be carried as secondary nodal degrees of freedom
d) They are secondary variables and must be carried as secondary nodal degrees of freedom
Answer: b
Clarification: An examination of the weak form, 0=∫_Ωeh_e([frac{partial w_1}{partial x}(c_{11}frac{partial u_x}{partial x}+c_{12}frac{partial u_y}{partial y}) + c_{66}frac{partial w_1}{partial y} (frac{partial u_x}{partial y} + frac{partial u_y}{partial x}))+ρw₁u_x]dxdy-∫_Ωeh_ew₁f_xdxdy-∮_┌dh_ew₁t_xds reveals the following:
(i) u_x and u_y are the primary variables, which must be carried as the primary nodal degrees of freedom.
(ii) only first derivatives of u_x and u_ywith respect to x and y, respectively, appear.

4. Which interpolation functions must be used for the primary variables in weak forms of plane elasticity equations?
a) Hermite family interpolation function
b) Lagrange family interpolation function
c) Hierarchical interpolation function
d) Quadratic interpolation function
Answer: b
Clarification: An examination of expanded weak forms of the plane elasticity problems reveals that the variables u_x and u_y are the primary variables, and only first derivatives of u_x and u_y with respect to x and y appear, respectively. Therefore, u_x and u_y must be approximated by the Lagrange family of interpolation functions, and at least bilinear (i.e., linear both in x and y) interpolation is required.

5. What are the simplest elements that fit for the finite element model of the plane elasticity equations?
a) Linear triangular and quadratic quadrilateral elements
b) Quadratic triangular and linear quadrilateral elements
c) Linear triangular and linear quadrilateral elements
d) Quadratic triangular and quadratic quadrilateral elements
Answer: c
Clarification: Because the first-derivatives of the primary variables, u_x and u_y with respect to x and y, respectively, appear in the expanded weak forms of the plane elasticity problems, they must be approximated by the Lagrange family of interpolation functions, with at least a bilinear interpolation. The simplest elements that satisfy those requirements are the linear triangular and linear quadrilateral elements.

6. What is the correct statement regarding the shape function S of a linear triangular element?
a) The first derivatives of S and hence, all the strains are element-wise constant
b) The first derivatives of S are linear, but the strains are element-wise constant
c) The first derivatives of S and hence the strains are linear functions
d) The first derivatives of S are element-wise constant, but the strains are linear
Answer: a
Clarification: A linear triangular element (number of nodes=3) has two degrees of freedom, u_x and u_y per node and a total of six nodal displacements per element. Since the shape functions are linear, their first derivatives are element-wise constant, and hence, all the strains computed for the linear triangular element are element-wise constant.

7. What is the other name of a linear triangular element for plane elasticity problems?
a) Constant-strain triangle
b) Linear-strain triangle
c) Quadratic-strain triangle
d) Variable-strain triangle
Answer: a
Clarification: In a linear triangular element (number of nodes=3), there are two degrees of freedom, u_x and u_y per node and a total of six nodal displacements per element. Since the first derivatives of shape functions for a triangular element are element-wise constant, all the strains computed for the linear triangular element are element-wise constant. Therefore, the linear triangular element for a plane elasticity problem is known as the constant-strain triangular (CST) element.

8. What is the function on (frac{partial s}{partial n}) if S (n, e) denotes the shape function of a linear quadrilateral element?
a) Linear in e and constant in n
b) Constant in e and Linear in n
c) Linear in both e and n
d) Constant in both e and n
Answer: a
Clarification: For a quadrilateral element, the first derivatives of the shape function are not constant. (frac{partial s}{partial n}) is linear in e and constant in n whereas (frac{partial s}{partial e}) is linear in n and constant in e. Also, the first derivatives of shape functions for a triangular element are constant element wise, and hence all the strains computed are constant element wise.

9. For a linear triangular element, what is the order of matrix B in the strain-displacement relation ε=BD, where D denotes the displacement matrix?
a) 6×3
b) 3×6
c) 3×8
d) 8×3
Answer: b
Clarification: For plane elasticity problems the strain-displacement relation is given by ε=BD, where ε=[ε_xxε_yyε_xy]^T, B=(begin{pmatrix}frac{partial psi_1}{partial x}&0&frac{partial psi_2}{partial y}&0&…&frac{partial psi_n}{partial y}&0\0&frac{partial psi_1}{partial y}&0&frac{partial psi_2}{partial y}&…&0&frac{partial psi_n}{partial y}\frac{partial psi_1}{partial y}&frac{partial psi_1}{partial x}&frac{partial psi_2}{partial y}&frac{partial psi_2}{partial x}&…&frac{partial psi_n}{partial y}&frac{partial psi_n}{partial x}end{pmatrix}) and D=(begin{bmatrix}u_x^1&u_y^1&u_x^2&u_y^2&u_x^3&u_y^3end{bmatrix})^T. The order of matrix B is 3x2n, where n is the number of nodes in the element. A linear triangular element has three nodes, thus n=3.
Order of B is 3×2*3
=3×6.

10. For a linear quadrilateral element, what is the order of matrix B in the strain-displacement relation ε=BD, where D denotes the displacement matrix?
a) 6×3
b) 3×6
c) 3×8
d) 8×3
Answer: c
Clarification: For plane elasticity problems the strain-displacement relation is given by ε=BD, where ε=[ε_xxε_yyε_xy]^T, B=(begin{pmatrix}frac{partial psi_1}{partial x}&0&frac{partial psi_2}{partial y}&0&…&frac{partial psi_n}{partial y}&0\0&frac{partial psi_1}{partial y}&0&frac{partial psi_2}{partial y}&…&0&frac{partial psi_n}{partial y}\frac{partial psi_1}{partial y}&frac{partial psi_1}{partial x}&frac{partial psi_2}{partial y}&frac{partial psi_2}{partial x}&…&frac{partial psi_n}{partial y}&frac{partial psi_n}{partial x}end{pmatrix}) and D=(begin{bmatrix}u_x^1&u_y^1&u_x^2&u_y^2&u_x^3&u_y^3end{bmatrix})^T. The order of matrix B is 3x2n, where n is the number of nodes in the element. A linear quadrilateral element has four nodes, thus n=4.
Order of B is 3×2*4
=3×8.

11. Which option is not correct with respect to the orders of the matrices in the following finite element model of plane elastic equations?
M^e(ddot{Delta}^e)+K^eΔ^e=F^e+Q^e
a) Mass matrix M has order 2n x 2n
b) Stiffness matrix K has order 2n x n
c) The element load vector F has order 2n x 1
d) The vector of internal forces Q has order 2n x 1
Answer: b
Clarification: For the given vector form of finite element model of plane elastic equations, the element mass matrix M and stiffness matrix K are of the order 2n x 2n, the element load vector F and the vector of internal forces Q are of the order 2n x 1, where n is the number of nodes in a Lagrange finite element.

12. Which form of a periodic solution is sought for the natural vibration study of plane elastic bodies?
a) {Δ}={Δ₀}e^iωt
b) {Δ}={Δ₀}e^-iω
c) {Δ}={Δ₀}e^-iωt
d) {Δ}={Δ₀}e^-iω/t
Answer: c
Clarification: For natural vibration study of plane elastic bodies, we seek a periodic solution of the form {Δ}={Δ₀}e^-iωt, where Δ denotes the displacements, ω is the frequency of natural vibration and i=(sqrt{-1}). With this, the finite element models of plane elastic problems reduce to an eigen value problem (-ω²M^e+K^e)(Delta_0^e)=Q^e.

13. Which option is correct for the first derivative of the shape function S in the study of plane elasticity problems?
a) It is element-wise constant for triangular element whereas not a constant for quadrilateral element
b) It is element-wise constant for both the triangular element as well as a quadrilateral element
c) It is not a constant for triangular element whereas element-wise constant for quadrilateral element
d) It is a combination of a linear function and constant for both the triangular element and a quadrilateral element
Answer: a
Clarification: In the study of plane elasticity problems, the first derivatives of the shape function S for a triangular element are element-wise constant, whereas, for a quadrilateral element, they are not constant. Notably, (frac{partial s (n, e)}{partial n}) is linear in e and constant in n whereas (frac{partial s (n, e)}{partial e}) is linear in n and constant in e.

Finite Element Method Multiple Choice Questions on “Plane Trusses”.

1. Plane trusses are also known as _____
a) One–dimensional trusses
b) Two-dimensional trusses
c) Three-dimensional trusses
d) Poly dimensional trusses
Answer: b
Clarification: Truss elements are two- node members which allow arbitrary orientation in XYZ co-ordinate system. Truss transmits axial force only. Planar truss is one where all members and nodes lie within Two dimensional plane.

2. A truss structure consists only ___ force members.
a) Only one
b) Two
c) Three
d) Poly
Answer: b
Clarification: Truss elements are two node members which allow arbitrary orientation in XYZ co-ordinate system. Truss transmits axial force only, in general, three degree of freedom element. A truss structure consists only 2 truss members.

3. Plane truss element can be shown in _____
a) Local coordinate system
b) Global coordinate system
c) Local and global coordinate systems
d) Dimensional structure
Answer: c
Clarification: Truss is one where all members and nodes lie within two dimensional plane. Truss is that elements of a truss have various orientation. These different orientations can be shown in local and global coordinate system.

4. The truss element is a _____ when we see it in a local co-ordinate system.
a) Three dimensional
b) One dimensional
c) Two dimensional
d) Thermal component
Answer: b
Clarification: Truss is one where all members and nodes lie within two dimensional plane. Truss is that elements of a truss have various orientations. When we see a truss in local co-ordinate system, the element of a truss can be seen as one dimensional element.

5. Where l and m are direct cosines, then transformation matrix L is given by ___
a) (begin{bmatrix}l & 0 \ m & 0\0 & l\0 & m end{bmatrix})
b) (begin{bmatrix}l & 0 \0 & m end{bmatrix})
c) (begin{bmatrix}l & m end{bmatrix})
d) (begin{bmatrix}l & m & 0 & 0\0 & 0 & l & m end{bmatrix})
Answer: d
Clarification: The direct cosines l and m are introduced as l = cos θ and m=cos ф. These direction cosines are the cosines of the angles that the local x’-axis makes with the global x-, y- axes.

6. Formula for direct cosine l=cosθ= ______
a) x₂-x₁
b) (frac{x_2}{l_e})
c) (frac{x_1}{l_e})
d) (frac{x_2-x_1}{l_e})
Answer: d
Clarification: A truss is a two node element which allows arbitrary orientation in XYZ co-ordinate system. Planar truss is one where all the members and nodes lie within two dimensional plane.

7. The direct cosine m is given by formula cosф= ____
a) (frac{x_2-x_1}{l_e})
b) (frac{y_2-y_1}{l_e})
c) (frac{x_1-y_1}{l_e})
d) (frac{x_2-y_2}{l_e})
Answer: b
Clarification: The direct cosines l and m are introduced as l=cosθ and m=cosф. These direction cosines are cosines of the angles of that local x’- axis makes with global x- and y- axes. Let (x₁,y₁) and (x₂,y₂) be the co-ordinates of nodes 1 and 2.

8. Strain energy (U) in global co-ordinates can be written as ____
a) q^{’ T} k^’q^’
b) q^{’ T} k^’
c) k^’ q^’
d) (frac{1}{2}) q^{’ T} k^’q^’
Answer: d
Clarification: Strain energy is defined as the energy stored in the body due to deformation. Strain energy per unit volume is known as strain energy density and the area under stress- strain curve towards the point of deformation.

9. The stress σ in a truss element is given by ____
a) σ=E
b) σ=ε
c) σ=E_eε
d) σ=α
Answer: c
Clarification: Stress is a physical quantity that expresses the internal forces that neighboring the particles of a continuous material exert on each other. Expression for element stresses can be obtained by noting that a truss element in local co-ordinates is simple two- force member.

10. In a truss element, element temperature load is ___
a) θ^e=E_eA_e
b) θ^e=E_eA_eε₀
c) θ^e=E_eA_eε₀(begin{Bmatrix}l \ m end{Bmatrix})
d) θ^e=E_eA_eε₀(begin{Bmatrix}-l \ -m \l\mend{Bmatrix})
Answer: d
Clarification: In a truss, temperature effect is arised then thermal stress problem is considered here. Since the element is simply a one dimensional element when viewed in local co-ordinate system, the element temperature load in the local co-ordinate system.

Finite Element Method Multiple Choice Questions on “ Beams and Frames – Boundary Conditions”.

1. Symmetry in application of boundary conditions should be avoided in which of the following type of analysis?
a) Linear static analysis
b) Modal analysis
c) Thermal analysis
d) Nonlinear static analysis
Answer: b
Clarification: Symmetric boundary conditions should not be used for modal analysis. Symmetric model would miss some of the modes of modal analysis or out of phase modes.

2. Boundary conditions are applied to simulate the physical constraints on the finite element model.
a) True
b) False
Answer: a
Clarification: Boundary conditions simulate the physical constraints on the finite element model. Application of boundary conditions is a crucial preprocessing step to yield accurate solution.

3. Which of the following is the correct equation for stiffness (K) of an element given value of force (F) and displacement (Q)?
a) FQ=K
b) KQ=F
c) KF=Q
d) KFQ=1
Answer: b
Clarification: The correct equation is given by KQ=F. The value of force (F) is the product of stiffness (K) and displacement (Q). The value of stiffness (K) of the element determines the displacement of the node.

4. Which of the following conditions must be fulfilled to apply symmetry in a finite element model?
a) Geometry of the model is symmetric
b) Boundary conditions to be applied are symmetric
c) Geometry model has large number of nodes
d) Geometry of the model is symmetric and boundary conditions to be applied are symmetric
Answer: d
Clarification: The geometry and boundary conditions both have to be symmetric to apply any kind of symmetry. Half or quarter portions of a model can be used to reduce computational cost.

5. Which of the following boundary conditions cannot be directly applied on solid elements?
a) Force
b) Pressure
c) Support
d) Torque
Answer: d
Clarification: Torque cannot be directly applied on solid element in finite element model. Since solid elements have three translational degrees of freedom and no rotational degrees of freedom torque cannot be directly applied on solid elements.

6. Traction is force acting on an area in any direction other than normal.
a) True
b) False
Answer: a
Clarification: Traction is force acting on an area in any direction other than normal. The force acting on an area in normal direction is called as pressure. Traction is boundary condition applied where force acting on a surface is not normal to the surface such as friction and drag.

7. Which of the following statements is correct?
a) Reaction force at supports is equal to the sum of the product of stiffness and displacement
b) Reaction force at supports is equal to the product of sum of stiffness and displacement
c) Reaction force at supports is equal to the sum of stiffness’s
d) Reaction force at supports is equal to the sum of displacements
Answer: a
Clarification: Reaction force at supports is equal to the sum of the product of stiffness and displacement. The stiffness and displacement matrices are multiplied for each element and then cumulated to find the reaction force at supports.

8. Which of the following equations give the relation between material properties like modulus of elasticity (E), modulus of rigidity (G), and Poisson’s ratio (u)?
a) E = 2*G*(1+u)
b) E = 3*G*(1+u)
c) E = 2*G*(1-u)
d) E = 3*G*(1-u)
Answer: a
Clarification: The relation between the material properties is given by
E = 2*G*(1+u)
Here E is the ratio of normal stress to normal strain. G is the ratio of shear stress to shear strain and u is the ratio of lateral strain to longitudinal strain.

Finite Element Method Multiple Choice Questions on “Plane Elasticity – Evaluation of Integrals”.

1. In the Finite Element Method, which expression is correct for a linear triangular element if S is the shape function, A_e is its area, and K is a constant?
a) (frac{partial S}{partial x}=frac{K}{A_e})
b) (frac{partial S}{partial y}=frac{K}{A_e^2})
c) (frac{partial S}{partial x})=KA_e
d) (frac{partial S}{partial y})=KA_e²
Answer: a
Clarification: For a linear triangular (i.e., constant-strain triangle) element, the shape function (psi_i^e) and its derivatives are given by (psi_i^e=frac{1}{2A_e}(alpha_i^e+beta_i^e x+gamma_i^e y)), (frac{partial psi_i^e}{partial x}=frac{beta_i^e}{2A_e}) and (frac{partial psi_i^e}{partial y}=frac{gamma_i^e}{2A_e})where A_e is the area of the element, α, β and γ are constants. Note that the derivatives of the shape function are constants.

2. In Finite Element Analysis, what is the correct load vector for a linear triangular element with area A_e, thickness h_e and uniform body force vector f?
a) (frac{A_e h_e}{4})f
b) (frac{A_e h_e}{3})f
c) (frac{h_e}{3A_e})f
d) (frac{h_e}{4A_e})f
Answer: b
Clarification: For a linear triangular (i.e., constant-strain triangle) element, for the case in which the body force is uniform and thus the body force components fx and fy are element-wise constant (say, equal to, (f_{x0}^e) and (f_{y0}^e) respectively), the load vector F has the form F=∫_Ωch_e(ψ^e)^T(f_0^e)dx
=(frac{A_e h_e}{4}begin{bmatrix}f_{x0}^e\f_{y0}^e\f_{x0}^e\f_{y0}^e\f_{x0}^e\f_{y0}^eend{bmatrix}). The internal load vector Q is computed only when the element falls on the boundary of the domain on which tractions are specified.

3. In Finite Element Analysis, what is the correct load vector for the linear quadrilateral element with area A_e, thickness h_e and uniform body force vector f?
a) (frac{A_e h_e}{4} )f
b) (frac{A_e h_e}{3})f
c) (frac{h_e}{3A_e})f
d) (frac{h_e}{4A_e})f
Answer: a
Clarification: For a linear quadrilateral element,for the case in which the body force is uniform and thus the body force components are element-wise constant (say, equal to, (f_{x0}^e) and (f_{y0}^e) respectively), the load vector F has the form F=(int_{Omega_c}h_e(psi^e)^T f_0^edx)
=(frac{A_e h_e}{4}begin{bmatrix}f_{x0}^e\f_{y0}^e\f_{x0}^e\f_{y0}^e\f_{x0}^e\f_{y0}^e\f_{x0}^e\f_{y0}^eend{bmatrix}). The internal load vector Q is computed only when the element falls on the boundary of the domain on which tractions are specified.

4. In the Finite Element Method, the vector of internal forces is computed only when the element falls on the boundary of the domain on which tractions are absent.
a) True
b) False
Answer: b
Clarification: In Finite Element Analysis, internal load vector Q is computed only when the element falls on the boundary of the domain on which tractions are specified (i.e., known). Computation of Q involves the evaluation of line integrals (for any type of element). In practice, it is convenient to express the surface traction t in the element coordinates. In that case, Q can be evaluated in the element coordinates and then transformed to the global coordinates for assembly.

5. Which option is not correct concerning the internal load vector in the finite element model of plane elasticity problems?
a) It is computed at all the nodes interior of the element
b) It is computed only when the element falls on the boundary of the domain on which tractions are known
c) Its computation doesn’t involve evaluation of line integrals for any type of element
d) It is evaluated in global coordinates but not in element coordinates
Answer: b
Clarification: In Finite Element Analysis, internal load vector Q is computed only when the element falls on the boundary of the domain on which tractions are specified (i.e., known). Computation of Q involves the evaluation of line integrals (for any type of element). In practice, it is convenient to express the surface traction t in the element coordinates. In that case, Q can be evaluated in the element coordinates and then transformed to the global coordinates using a transformation matrix.

6. In transformations, what is the transformation matrix R in the relation F=RQ if the load vector in global coordinates is F and the load vector in element coordinates is Q?
a) (begin{bmatrix}
cos alpha & sin alpha & 0 & 0 & \-sin alpha & cos alpha & 0 & 0 & \0 & 0 & cos alpha & sin alpha & \0 & 0 & -sin alpha & cos alpha & \& & & & ddotsend{bmatrix})
b) (begin{bmatrix}cos alpha & -sin alpha & 0 & 0 & \
sin alpha & cos alpha & 0 & 0 & \0 & 0 & cos alpha & -sin alpha & \0 & 0 & sin alpha & cos alpha & \& & & & ddotsend{bmatrix})
c) (begin{bmatrix}cos alpha & sin alpha & 0 & 0 & \-sin alpha & cos alpha & 0 & 0 & \0 & 0 & cos alpha & -sin alpha & \0 & 0 & sin alpha & cos alpha & \& & & & ddotsend{bmatrix})
d) (begin{bmatrix}cos alpha & -sin alpha & 0 & 0 & \
sin alpha & cos alpha & 0 & 0 & \0 & 0 & cos alpha & sin alpha & \0 & 0 & -sin alpha & cos alpha & \& & & & ddotsend{bmatrix})
Answer: a
Clarification: In practice, it is convenient to express the surface traction T in the element coordinates. In that case, the element load vector can be evaluated in the element coordinates and then transformed to the global coordinates for assembly. If Q denotes the element load vector referred to the element coordinates, then the corresponding load vector referred to the global coordinates is given by
F=R^TQ, where R is the transformation matrix R=(begin{bmatrix}cos alpha & sin alpha & 0 & 0 & \-sin alpha & cos alpha & 0 & 0 & \0 & 0 & cos alpha & -sin alpha & \0 & 0 & sin alpha & cos alpha & \& & & & ddotsend{bmatrix}) and α is the angle between the global x-axis and the traction vector T.

7. What is the global load vector in Finite Element Analysis of the following structure if the local load vector is (begin{bmatrix}0\0\2\0\1\0end{bmatrix}) and θ=0?

a) (begin{bmatrix}0\0\0\2\0\1end{bmatrix})
b) (begin{bmatrix}0\0\2\0\0\1end{bmatrix})
c) (begin{bmatrix}0\0\2\0\1\0end{bmatrix})
d) (begin{bmatrix}0\0\0\2\1\0end{bmatrix})
Answer: a
Clarification: If Q denotes the element load vector referred to the element coordinates then the corresponding load vector referred to the global coordinates is given by F=R^TQ where R is the transformation matrix R=(begin{bmatrix}cos alpha & sin alpha & 0 & 0 & \
-sin alpha & cos alpha & 0 & 0 & \0 & 0 & cos alpha & sin alpha & \0 & 0 & -sin alpha & cos alpha & \& & & & ddotsend{bmatrix}) and α is the angle between the global x-axis and the traction vector T.
Here α=90-θ, given θ=0
α=90-0
=90.
Cos90=0 and sin90=1.
R=(begin{bmatrix}0&1&0&0&0&0\-1&0&0&0&0&0\0&0&0&1&0&0\0&0&-1&0&0&0\0&0&0&0&0&1\-1&0&0&0-1&0end{bmatrix})
F=(begin{bmatrix}0&1&0&0&0&0\ -1&0&0&0&0&0\0&0&0&1&0&0\0&0&-1&0&0&0\0&0&0&0&0&1\-1&0&0&0-1&0end{bmatrix}^T)x(begin{bmatrix}0\0\2\0\1\0end{bmatrix})
=(begin{bmatrix}0\0\0\2\0\1end{bmatrix}).

8. What is the expression for the traction term t_n in the element load vector Q^e=∮_┌ch_eψ^Ttds of the following figure where L₂₃ is the length of the line 2-3?

a) t_n=-T(-1+(frac{s}{L_{23}}))
b) t_n=T(-1+(frac{s}{L_{23}}))
c) t_n=-T(1+(frac{s}{L_{23}}))
d) t_n=T(1+(frac{s}{L_{23}}))
Answer: b
Clarification: For the element shown in the given figure, the side 2-3 is subjected to linearly varying normal force t_n. The traction term on the side 2-3 of the element is t_n=-T(1- (frac{s}{L_{23}})), where the minus sign for T accounts for the direction of the applied traction. Traction is acting towards the body in the present case. The local coordinate system s used in the above expression is chosen along the side connecting node 2 to node 3, with its origin at node 2. We are not restricted to this choice.

9. In Finite Element Analysis, what are the values of nodal forces in the following element if the line 2-4 is 160 in long?

a) 1600 along both the DOF 3 and 7
b) 800 and 0 along the DOF 3 and 4 respectively
c) 0 and 800 along the DOF 7 and 8 respectively
d) 0 and 800 along the DOF 3 and 4 respectively
Answer: b
Clarification: Consider the thin elastic plate subjected to a uniformly distributed edge load, as shown in the given figure. The specified displacement degrees of freedom for the problem are U₁=U₂=0, and the known forces are F₃= (frac{pbh}{2}), F₄=0, F₇=(frac{pbh}{2}) and F₈=0.
Given p*h=10 and b=160 thus (frac{pbh}{2})
=(frac{10×160}{2})
=800.
Thus, the forces along the DOF 3 and 4 are 800 and 0, respectively.

10. In vibration and transient analysis of beams, if the linear acceleration scheme predicts the solution,then it is unstable for the first several time steps, but it eventually becomes stable.
a) True
b) False
Answer: b
Clarification: In the determination of natural frequencies and transient response using plane elements, the time approximation scheme is used. Note that the solution predicted by the linear acceleration scheme is stable for the first several time steps, but it eventually becomes unstable.

11. In Finite Element Analysis, which option is correct for computation of load due to specified boundary stress?
a) Can be computed using a local coordinate system and one-dimensional interpolation functions
b) Can be computed using a local coordinate system but not one-dimensional interpolation functions
c) Cannot be computed using a local coordinate system but one-dimensional interpolation functions can be used
d) Neither a local coordinate system nor one-dimensional interpolation functions can be used
Answer: a
Clarification: In general, the loads due to specified boundary stresses can be computed using an appropriate local coordinate system and one-dimensional interpolation functions. When higher-order elements are involved, the corresponding order of one-dimensional interpolation functions must be used.

12. In the Finite Element Method, which element is known for the slowest convergence?
a) Linear triangular element
b) Quadratic triangular element
c) Linear rectangular elements
d) Quadratic rectangular elements
Answer: a
Clarification: Mesh convergence determines how many elements are required in a finite element model to ensure that the results of an analysis are not affected by varying the size of the mesh. Once a mesh is converged, no change is observed in the results even after changing its density. The linear triangular element mesh has the slowest convergence compared to the quadratic triangular element, linear and quadratic rectangular elements.