Singiresu S. Rao

The Finite Element Method in Engineering, 6th Edition

Elsevier-Butterworth-Heinemann, 2018

ANSWERS TO REVIEW QUESTIONS

Chapter 1

1.1

1. The process of dividing the solution region into several pieces or finite elements.

2. To replace the unknown and complex variation of the field variable in the domain of the problem by a simpler polynomial equation within each of the finite elements.

3. Potential energy = Strain energy – Work done by external loads or forces

4. Strain energy is the energy of deformation.

5. Due to Maxwell-Betti reciprocity theorem which states that: “The load applied at node i to cause unit displacement at node j = load applied at node j to cause a unit displacement at node i”.

6. To be able to choose the transverse displacement as well as the slopes of deflection shape (due to the bending of the beam) as unknown degrees of freedom at each node (end) of the element.

7. ABAQUS, ANSYS and NASTRAN.

8. Clough

9. Turner, Clough, Martin and Topp.

10. (i) Its ability to find increasingly more accurate solutions, and (ii) Its applicability to solve problems from different areas using the same computer program.

11. Strain energy (π ) =

12 ∭

Vσ ε dV

12. Re = ρ v d

μ ; ρ = density, v = velocity, d = diameter of the pipe, and μ = viscosity of the fluid

1

13. Model the mechanism, for example, as a planar four-bar mechanism. Use one or two finite elements to represent each of the three moving links. Each finite element will have two end nodes with two translational and one rotational degrees of freedom. At each crank angle, for the known force acting on the mechanism, find the displacements of nodes and also the stresses in each element (that is, in the link) using finite element analysis procedure.

1.2

1.

A(e ) E(e )

l(e ) [ 1 − 1− 1 1 ]

2. Displacements 3. Conductivities 4. Ohm’s law

1.3

1. T 2. F 3. F

1.4

1. a 2. a 3. b 4. c 5. b

Chapter 2

2.1

1. Shape, size, number

2. When geometry, material properties and field variable of the problem are described in terms of a single spatial coordinate.

3. Temperature distribution in a rod and pressure distribution in a pipe flow

4. To be able to take the transverse displacement and slope of the beam as degrees of freedom at each node.

5. For truss analysis, the bar element is a unique choice.

6. Milling machine structure, Airplane wing

7. For a 2-dimensional element, aspect ratio is the ratio of largest dimension and the smallest dimension of the element.

8. No

9. By modeling only that part of the soil in which the load is expected to have a significant influence.

2

10. Maximum difference between the numbered degrees of freedom at the ends of any member + 1.

11. For automatic mesh generation

12. Cylindrical pressure vessel, water pipe, conical pressure vessel

2.2

1. Model 2. Automatic mesh generation 3. Tetrahedron 4. Five 5. Higher 6. Quadrilateral

7. One 8. Nodes 9. One-fourth 10. Thin slice 11. Infinite 12. Reduce or minimize

13. Shortest 14. Mesh generation

2.3

1. T

2.4

1. b

Chapter 3

3.1

1. The function used to represent the behavior of the solution within an element.

2. Corner nodes of an element

3. Improving accuracy by increasing the order of the interpolation polynomial.

4. Isoparametric element

5. Local or natural coordinate system

6. Quadratic interpolation function: φ (x )= α 1+ α 2 x + α 3 x2

7. Value of the field variable at a node

8. Temperature

9. Displacement

10. Simultaneous linear algebraic equations

3

11. The diagram that gives constant term, linear terms, quadratic terms, etc. in different lines which facilitate the implementation of geometric isotropy in the interpolation model

12. The interpolation function, its first derivative and its second derivative are continuous within an element.

13. To simplify the integration of polynomial terms (expressed in terms of natural coordinates)

14. In order to have 0 or 1 at each of the two nodes of a line element

15. L1 = 0

3.2

1.Infinity 2. One 3. Nodal 4. And/or 5. r 6. Accuracy 7. Define

8. φ (x ) = a1 + a2 x + a3 x2 9. Simplex 10. Parallel 11. Geometric 12. Satisfied

13. Representation 14. Continuous 15. Contained 16. Simplex 17. 1 18. 1 19. Volume

20. patch

3.3

1. F 2. T 3. F 4. F 5. T 6. F 7. F 8. F 9. T 10. T 11. F 12. T 13. T 14.F 15. F 16. F 17. F 18. T 19. T 20. T

3.4

1. b 2. c 3. c 4. a 5. a 6. b 7. b 8. c 9. c 10. a

3.5

1.-b 2-c 3-a

3.6

1-c 2-a 3-b

3.7

1-d 2-a 3-b 4-c

Chapter 4

4.1

4

1. When the interpolation polynomial is of order two or more, the element is called a higher order element.

2. φ (x ) = a1 + a2 x + a3 x2 + a4 x3

3. Figure:

4. Figure:

5. Figure:

6. Four primary nodes, one at each corner point and six secondary nodes, one at the middle of each edge.

7. Lagrange and Hermite8. Figure:

4.2

1. Multiplex 2. Secondary 3. Curved 4. Isoparametric 5. Gauss

4.3

1. F 2. T 3. F 4. T 5. F 6. F 7. T

4.4

1-d 2-a 3-b 4-c

4.5

1-c 2-b 3-a

Chapter 5

5.1

5

1. Using physical laws and equilibrium equations.2. It deals with the extremization of functionals in the form of integrals.3. A function of another function.4. It is the error involved in satisfying the differential equation when an approximate

solution is used in the weighted residual method.5. See book, pp 190-192.

5.2

1. Physical 2. Commutative 3. Euler or Euler-Lagrange 4. Natural or free 5. Sum 6. Domain

5.3

1. F 2. F 3. T 4. T 5. T 6. T 7. T 8. F 9. T 10. F

5.4

1. c

5.5

1-d 2-e 3-b 4-a 5-c

Chapter 6

6.1

1. To simplify the computational work involved.2. Because different elements will have different local coordinates and to derive the

equations in a common or global coordinate system, the coordinate transformation becomes necessary.

3.[ λ ] = [cos θ sin θ

− sin θ cos θ ]4. [ λ(e ) ]− 1 = [ λ( e ) ]T

5. At nodes where elements are connected, the value(s) of the unknown nodal degree(s) of freedom or variable(s) , is the same for all the elements joining at that node.

6. A boundary condition involving more than one nodal degree of freedom is called a multipoint constraint.

7. To implement the boundary conditions in a unified and simple manner.8. The displacement perpendicular to the surface must be made zero.

6

6.2

1. Physics (or physical nature) 2. Natural 3. Row column 4. Coincide

6.3

1. T 2. F 3. T 4. F 5.T 6. F 7.F 8.T 9. F

6.4

1-c 2-a 3-b 4-e 5-d

Chapter 7

7.1

1. A continuous problem involves infinite number of degrees of freedom while a discrete problem involves finite number of degrees of freedom.

2. Eq. (7.1a and b), page 257 (in book)3. Eq. (7.2a and b), page 258 (in book)4. Eq. (7.3a – 7.3d), page 258 (in book)5. In the absence of round-off and other errors, direct methods will yield the exact solution

in a finite number of arithmetic operations.6. To find the smallest eigenvalues and the associated eigenvectors.7. Any problem whose governing equation is a differential equation.8. Finite difference method, Newmark method.9. Mode superposition method.

7.2

1. Gaussian 2. Nonzero 3. Triangular 4. Singular 5. Decomposition 6. Differential

7.3

1. T 2. T 3. T 4. F 5. T 6.F 7.F 8. T 9. T 10. F

7.4

1. b

7.5

1-c 2-a 3-b

7

7.6

1-d 2-e 3-f 4-g 5-c 6-b 7-a

Chapter 8

8.1

1. Equilibrium equations, stress-strain relations, and strain-displacement relations.2. Displacements (3), stresses (6) and strains (6)3. External equilibrium equations, compatibility equations and boundary conditions.4. 35. Force per unit volume6. Plane stress and plane strain

7. σ x , σ y , and τ xy

8. A thin plate subjected to in-plane loads9. A dam, retaining wall, and buried long pipe10. A solid of revolution subjected to axisymmetric loads11. u = 0, v = 0, w = 0 at the point12. The displacement field must be continuous and single valued. This implies that (i) the

body having no cracks or holes before deformation should not have any cracks or gaps after deformation, and (ii) no point of the body should overlap on other part due to deformation.

13. 314. Potential energy = strain energy – work done by external loads15. Lagrangian = kinetic energy – potential energy

8.2

1. Moment 2. Moment 3. 2 4. Hooke’s 5. ε zz 6. σ zz 7. σ x , u

8. ε x = ∂ u

∂ x, ε y = ∂ v

∂ y, ε z = ∂ w

∂ z 9.

∂u∂ y

+ ∂ v∂ x 10. Equilibrium compatibility

11. Hamilton 12. Nodes

8.3

1. F 2. T 3. F 4. T 5.T 6. T 7. F 8. T 9. T 10. T

8.4

8

1. b 2.a 3.b 4. a 5. b 6.c 7. b 8. a 9. c 10. b

8.5

1-b 2-c 3-a

Chapter 9

9.1

1. No2. 33. 64. Members in a truss are connected by pin joints (hinges) while the members in a frame are

connected by rigid joints (fixed or welded).5. 1

6.[ k( e )] = A E

l [ 1 − 1− 1 1 ]

7.lij = cos θxX =

X j − X i

le ,mij = cos θxY =

Y j − Y i

le ,nij = cos θxX =

Z j − Z i

le

where le denotes the length of the element ij given by

le = {( X j − X i )2 + (Y j − Y i )

2 + (Z j − Zi )2}1/2

8. Coordinate transformation matrix relating the local and global degrees of freedom of an element.

9.[ k(e ) ] =∭V (e ) [ B ]T [ D ] [ B ] dV

Where [D] is the matrix relating stresses to strains and [B] is the matric relating strains to

nodal displacements, and V(e)

is the volume of element e.10. Planar frame element will have two additional degrees of freedom (axial displacements at

the end nodes)11. 12

9.2

1. Normal 2. Third 3. Torsional

9.3

1. T 2. T 3. T 4.F 5. T

9

9.4

1. c 2. a 3. a 4. b 5. b 6.a 7. b 8.c

Chapter 10

10.1

1. A membrane can carry only in-plane loads while a plate can carry transverse loads.2. Due to the assumption of small displacements.3. Constant stress (or strain) triangular element; same as a triangular membrane element.4. In-plane displacements u and v are assumed to be quadratic functions of x and y.5. Linear strain (or stress) triangle element.6. 6 x 127. See book (Section 10.9 on page 407)

10.2

1. Inplane 2. Linear 3. 6 x 9 4. 9 5. 9 x 9

10.3

1. F 2. T 3. T 4. F 5. T

10.4

1. b 2. b 3.c 4. b 5. a

10.5

1-b 2-c 3-d 4-a

Chapter 11

11.1

1. Stress analysis of the soil under the column of a high rise building, thick pressure vessel, and machine foundation.

2. Tetrahedron element with four corner nodes.3. Figure ??

10

4. Radial and axial.5. Ring element with triangular cross section6. See Fig. 11.6 on page 442 (book).

11.2

1. Linearly 2. Zienkiewicz-Irons 3. Minus Plus 4. Gaussian 5. 6 x 6

11.3

1. F 2. T 3. T 4. F

11.4

1. b 2. c 3. a 4. b 5. b 6.b

Chapter 12

12.1

1. [ m ] = ρ A l

2 [1 00 1 ]

2. [ m ] = ρ A l

6 [2 11 2 ]

3.

[ m ] = ρ A l2 [

1 0 0 0 00 000 1 0 0 0 00 0 1 0 0 00 0 0 1 0 00 0 0 0 1 00 0 0 0 0 1

]4. The vibration resulting only from the initial conditions (or disturbance).

5. [ [ K ] − ω2 [ M ] ] Q⃗ = 0⃗

6. By deleting the rows and columns corresponding to the rigid body degrees of freedom from

the matrices [K] and [M] and the corresponding elements from the displacement and load

11

vectors Q⃗ and P⃗ .

7. The property or relation Q⃗iT [M ] Q⃗ j = 0 ; i ≠ j denotes orthogonalization of the normal

modes Q⃗i and Q⃗ j .

8. Multiply Q⃗i by a constant a i whose value is chosen to satisfy the relation (ai Q⃗i )

T [M ] (ai Q⃗i )= a i2 Q⃗i

T [ M ] Q⃗ i = 1

or

a i = ( 1Q⃗i

T [ M ] Q⃗i )12

9. Gust loads on an airplane, earthquake loads on a tall building, and blast loads on a structure.

12.2

1. Time 2. Initial 3. Dynamic 4. Consistent 5. Consistent 6. Displacement 7. Shear 8. Variations 9. Invariant 10. Natural frequency 11. Numerical 12. Superposition 13. Uncoupled 14. Underdamped 15. Large

12.3

1.T 2. F 3. F 4. T 5. T 6. T 7. T 8. F 9. F 10. T 11. T 12. F 13. F 14. T

12.4

1. c 2. a 3. b 4. b 5. b 6.c 7. c 8. b 9. a 10. b 11. c 12. b 13. a

Chapter 13

13.1

1. Boilers, steam turbines and jet engines.2. Energy inflow + energy generated inside the system

= change in the internal energy of the system + energy output3. Conduction, convection, and radiation

4.α = k

ρ c

12

13.2

1. Temperature 2. k A ∂T

∂ x 3. h A (T − T∞ ) 4. σ ε A (T 4 − T∞4 ) 5. Dirichlet 6.

Neumann

7. Time 8. Differential Variational 9. Internal energy 10. Generation Heat flux 11. Electromagnetics 12. Fourth 13. Diffusivity

13.3

1. T 2. F

13.4

1. b 2. c

13.5

1-b 2-c 3-a

13.6

1-b 2-c 3-a

Chapter 14

14.1

1.k d2 T

dx2 + q̇ = 0

2. Fin

3.dTdt

|t0=

T (t 0 + Δt ) − T ( t0 )Δt

4. Element matrix associated with the unsteady state term

5. T ( x ) = a1 + a2 x + a3 x2

6. By assuming a linear taper within an element

14.2

1. P l + A 2. T4

3. Initial 4. Temperature Flux Convection

13

14.3

1. T 2. F 3. T 4. F

14.4

1. c

Chapter 15

15.1

1.k ( ∂2 T

∂ x2 + ∂2 T∂ y2 ) + q̇ = 0

2. Temperature prescribed; boundary heat flux prescribed; convection heat transfer specified.

15.2

1. Linear

15.3

1. T 2. F

15.4

1. c

Chapter 16

16.1

1.∂∂r (r kr

∂T∂r ) + ∂

∂ z (r kz∂ T∂ z ) + r q̇ = 0

2.

∂T∂n

= 0 (n is the normal direction to the surface)

3.k ( ∂2 T

∂ x2 + ∂2 T∂ y2 + ∂2 T

∂ z2 ) + q̇ = 0

14

16.2

1. Tetrahedron 2. [ K 3(e )] 3. 3 x 3

16.3

1. b 2. a 3. c

Chapter 17

17.1

1. Some plastics, colloidal suspensions and emulsions2. It is a frictionless flow characterized by zero viscosity.

3. x = f 1 ( x0 , y0 , z0 , t ) , y = f 2 ( x0 , y0 , z0 , t ), z = f 3 ( x0 , y0 , z0 , t ) where ( x0 , y0 , z0)

denotes the position at some reference time t.4. Velocities at various points in the flow are expressed as functions of time as

u = f 1 ( x , y , z , t ), v = f 2 ( x , y , z , t ) , w = f 3 ( x , y , z , t ) where (u , v , w )denote the components of velocity along (x, y, z) directions, respectively.

5. 66. Energy input = energy output + energy accumulation

7. Density, ρ = ρ( p , T ) , viscosity, μ = μ( p , T )where p = pressure and T = temperature

8.

∂ψ∂ y

= u , ∂ψ∂ x

= − v

9. Irrotational flow is one in which particles of fluid do not rotate, i.e., rotation is zero.

17.2

1.Shear 2. Absolute viscosity 3. Non-Newtonian 4. Gases 5. Viscosity 6. Eulerean 7. Momentum 8. Identical 9. Pressure

17.3

1.T 2. F 3. T 4. F 5. T 6. T 7. T

17.4.

1. b 2.a

17.5

15

1. Laminar flow: An orderly state of flow in which macroscopic fluid particles move in layers.

2. Turbulent flow: A flow in which the fluid particles have irregular fluctuating motions and erratic paths.

3. Transition flow: The flow in which laminar flow becomes unstable and tends to be a turbulent flow.

Chapter 18

18.1

1. Flow around an airfoil and flow out of an orifice

2.

∂2 φ∂ x2 + ∂2 φ

∂ y2 = 0

3.u = ∂ φ

∂ y, v = ∂ φ

∂ x

4.

∂2 ψ∂ x2 + ∂2 ψ

∂ y2 = 0

5.u = ∂ψ

∂ y, v = − ∂ψ

∂ x

6. φ = φ0 (specified value of φ on C1 (part of the boundary)

7. ψ = ψ0 (specified value of ψ on C1 (part of the boundary)

8.− ∂φ

∂ n= V 0

(specified value of velocity normal to C2 (part of the boundary)

18.2

1.Ideal 2. Stream 3. Velocity specified normal to part of boundary C2 ;

∂φ∂ n

= V 0

18.3

1. T 2. T 3. F 4. F

Chapter 19

16

19.1

1. Conservation of momentum equations along x and y directions and continuity equation.2. When convective terms (terms involving ρ ) are neglected in conservation of momentum

equations, the resulting equations are known as Stokes equations.3. The equations of motion involving the inertia, body forces, pressure and viscous forces

are called Navier-Stokes equations.4. Vorticity or fluid rotation vector is the average velocity of any two mutually

perpendicular line segments of the fluid.5. They behave as a rigid solid until a certain level of shear stress and as a Newtonian fluid

afterward.6. Green-Gauss theorem represents integration by parts in 2 and 3 dimensional spaces.

19.2

1. T 2. F 3. F

Chapter 20

20.1

1. Harmonic equation:

∂2 φ∂ x2 + ∂2 φ

∂ y2 + ∂2 φ∂ z2 = 0

Quasi-harmonic equation:

∂∂ x (k x

∂φ∂ x ) + ∂

∂ y (k y∂ φ∂ y ) + ∂

∂ z (k z∂ φ∂ z ) = 0

2. Laplace equation in two dimensions looks like ∇2 φ = ∂2 φ

∂ x2 + ∂2 φ∂ y2 = 0

while the Poisson

equation looks like∇2 φ = ∂2 φ

∂ x2 + ∂2 φ∂ y2 = C

where C is a constant.3. A quasi-harmonic equation including time differentials.4. When the coordinates x, y and z along with time t is used as the solution domain, the

finite elements are defined in the combined geometric – time space. Such elements are called space-time finite elements.

5. Galerkin method and least squares method.

20.2

1. Closed 2. Hyperbolic

17

20.3

1. F

20.4

1-c 2-e 3-a 4-b 5-d

18