TEST BANK FOR Introduction to Finite Element Analysis 3rd Edition By Reddy J.N.

Problem 1.1: Newton’s second law can be expressed as
F = ma (1)
where F is the net force acting on the body, m mass of the body, and a the
acceleration of the body in the direction of the net force. Use Eq. (1) to determine
the mathematical model, i.e., governing equation of a free-falling body. Consider
only the forces due to gravity and the air resistance. Assume that the air resistance
is linearly proportional to the velocity of the falling body.
Solution: From the free-body-diagram it follows that
m
dv
dt
= Fg − Fd, Fg = mg, Fd = cv
where v is the downward velocity (m/s) of the body, Fg is the downward force (N or
kg m/s2) due to gravity, Fd is the upward drag force, m is the mass (kg) of the body,
g the acceleration (m/s2) due to gravity, and c is the proportionality constant (drag
coefficient, kg/s). The equation of motion is
dv
dt
+ αv = g, α =
c
m
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2 AN INTRODUCTION TO THE FINITE ELEMENT METHOD
Problem 1.2: A cylindrical storage tank of diameter D contains a liquid at depth
(or head) h(x, t). Liquid is supplied to the tank at a rate of qi (m3/day) and drained
at a rate of q0 (m3/day). Use the principle of conservation of mass to arrive at the
governing equation of the flow problem.
Solution: The conservation of mass requires
time rate of change in mass = mass inflow - mass outflow
The above equation for the problem at hand becomes
d
dt
(ρAh) = ρqi − ρq0 or
d(Ah)
dt
= qi − q0
where A is the area of cross section of the tank (A = πD2/4) and ρ is the mass density
of the liquid.
Problem 1.3: Consider the simple pendulum of Example 1.3.1. Write a computer
program to numerically solve the nonlinear equation (1.2.3) using the Euler method.
Tabulate the numerical results for two different time steps Δt = 0.05 and Δt = 0.025
along with the exact linear solution.
Solution: In order to use the finite difference scheme of Eq. (1.3.3), we rewrite
(1.2.3) as a pair of first-order equations
dθ
dt
= v,
dv
dt
= −λ2 sin θ
Applying the scheme of Eq. (1.3.3) to the two equations at hand, we obtain
θi+1 = θi + Δt vi; vi+1 = vi − Δt λ2 sin θi
The above equations can be programmed to solve for (θi, vi). Table P1.3 contains
representative numerical results.
Problem 1.4: An improvement of Euler’s method is provided by Heun’s method,
which uses the average of the derivatives at the two ends of the interval to estimate
the slope. Applied to the equation
du
dt
= f(t, u) (1)
Heun’s scheme has the form
ui+1 = ui +
Δt
2
h
f(ti, ui) + f(ti+1, u0i
+1)
i
, u0i
+1 = ui + Δt f(ti, ui) (2)
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SOLUTIONS MANUAL 3
Table P1.3: Comparison of various approximate solutions of the equation
(d2θ/dt2) + λ2 sin θ = 0 with its exact linear solution.
Exact Approx. solution θ Exact Approx. solution v
t θ Δt = .05 Δt = .025 v Δt = .05 Δt = .025
0.00 0.78540 0.78540 0.78540 -0.00000 -0.00000 -0.00000
0.05 0.76965 0.78540 0.77828 -0.62801 -0.56922 -0.56922
0.10 0.72302 0.75694 0.74276 -1.23083 -1.13844 -1.13027
0.15 0.64739 0.70002 0.67944 -1.78428 -1.69123 -1.66622
0.20 0.54578 0.58980 0.56482 -2.26615 -2.20984 -2.15879
0.25 0.42229 0.50496 0.47627 -2.65711 -2.67459 -2.58816
0.30 0.28185 0.37123 0.34225 -2.94148 -3.06403 -2.93371
0.35 0.13011 0.21803 0.19218 -3.10785 -3.35605 -3.17573
0.40 -0.02685 0.05023 0.03148 -3.14955 -3.53018 -3.29791
0.45 -0.18274 -0.12628 -0.13374 -3.06491 -3.57060 -3.29007
0.50 -0.33129 -0.30481 -0.29690 -2.85732 -3.46921 -3.15014
0.60 -0.58310 -0.63965 -0.59131 -2.11119 -2.85712 -2.50787
0.80 -0.78356 -1.05068 -0.91171 0.21536 -0.50399 -0.28356
1.00 -0.50591 -0.94062 -0.74672 2.41051 2.29398 2.19765
In books on numerical analysis, the second equation in (2) is called the predictor
equation and the first equation is called the corrector equation. Apply Heun’s method
to Eqs. (1.3.4) and obtain the numerical solution for Δt = 0.05.
Solution: Heun’s method applied to the pair
dθ
dt
= v,
dv
dt
= −λ2 sin θ
yields the following discrete equations:
θ0
i+1 = θi + Δt vi
vi+1 = vi − λ2 Δt
2
³
sin θi + sinθ0
i+1
´
θi+1 = θi +
Δt
2
(vi + vi+1)
The numereical results obtained with the Heun’s method and Euler’s method are
presented in Table P1.4.
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4 AN INTRODUCTION TO THE FINITE ELEMENT METHOD
Table P1.4: Numerical solutions of the nonlinear equation d2θ/dt2 + λ2 sin θ = 0
along with the exact solution of the linear equation d2θ/dt2+λ2θ = 0.
Exact Approx. solution θ Exact Approx. solution v
t θ Euler’s Heun’s v Euler’s Heun’s
0.00 0.785398 0.785398 0.785398 -0.000000 -0.000000 -0.000000
0.05 0.769645 0.785398 0.771168 -0.628013 -0.569221 -0.569221
0.10 0.723017 0.756937 0.728680 -1.230833 -1.138442 -1.121957
0.20 0.545784 0.615453 0.564818 -2.266146 -2.209838 -1.121957
0.40 -0.026852 0.050228 0.015246 -3.149552 -3.530178 -3.073095
0.60 -0.583104 -0.639652 -0.544352 -2.111190 -2.857121 -2.194398
0.80 -0.783562 -1.050679 -0.787095 0.215362 -0.503993 -0.114453
1.00 -0.505912 -0.940622 -0.587339 2.410506 2.293983 2.023807
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SOLUTIONS MANUAL 5
Chapter 2
MATHEMATICAL PRELIMINARIES,
INTEGRAL FORMULATIONS, AND
VARIATIONAL METHODS
In Problem 2.1—2.5, construct the weak form and, whenever possible, quadratic
functionals.
Problem 2.1: A nonlinear equation:
−
d
dx
μ
u
du
dx
¶
+ f = 0 for 0 < x < L
μ
u
du
dx
¶¯¯¯¯
x=0
= 0 u(1) = √2
Solution: Following the three-step procedure, we write the weak form:
0 =
Z 1
0
v
∙
−
d
dx
(u
du
dx
) + f
¸
dx (1)
=
Z 1
0
∙
u
dv
dx
du
dx
+ vf
¸
dx −
∙
v(u
du
dx
)
¸1
0
(2)
Using the boundary conditions, v(1) = 0 (because u is specified at x = 1) and
(du/dx) = 0 at x = 0, we obtain
0 =
Z 1
0
∙
u
dv
dx
du
dx
+ vf
¸
dx (3)
For this problem, the weak form does not contain an expression that is linear in both
u and v; the expression is linear in v but not linear in u. Therefore, a quadratic
functional does not exist for this case. The expressions for B(·, ·) and `(·) are given
by
B(v, u) =
Z 1
0
u
dv
dx
du
dx
dx (not linear in u and not symmetric in u and v)
`(v) = −
Z 1
0
vfdx (4)
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6 AN INTRODUCTION TO THE FINITE ELEMENT METHOD
♠ New Problem 2.1:
The instructor may assign the following problem:
−
d
dx
∙
(1 + 2x2)
du
dx
¸
+ u = x2 (1a)
u(0) = 1 ,
μ
du
dx
¶
x=1
= 2 (1b)
The answer is
B(v, u) =
Z 1
0
∙
(1 + 2x2)
dv
dx
du
dx
+ vu
¸
dx (symmetric)
`(v) =
Z 1
0
v x2 dx + 6v(1) (2)
I(u) =
1
2
B(u, u) − `(u) =
1
2
Z 1
0
"
(1 + 2x2)
μ
du
dx
¶2
+ u2
#
dx
−
Z 1
0
u x2 dx − 6u(1)
Problem 2.2: The Euler-Bernoulli-von K´arm´an nonlinear beam theory [7]:
−
d
dx
(
EA
"
du
dx
+
1
2
μ
dw
dx
¶2
#)
= f for 0 < x < L
d2
dx2
Ã
EI
d2w
dx2
!
−
d
dx
(
EA
dw
dx
"
du
dx
+
1
2
μ
dw
dx
¶2
#)
= q
u = w = 0 at x = 0, L;
μ
dw
dx
¶¯¯¯¯
x=0
= 0;
Ã
EI
d2w
dx2
!¯¯¯¯
x=L
= M0
where EA, EI, f, and q are functions of x, and M0 is a constant. Here u denotes the
axial displacement and w the transverse deflection of the beam.
Solution: The first step of the formulation is to multiply each equation with a weight
function, say v1 for the first equation and v2 for the second equation, and integrate
over the interval (0, L). In the second step, carry out the integration-by-parts once
in the first equation, twice in the first term of the second equation, and once in the
second part of the second equation. Then use the fact that v1(0) = v1(L) = 0 (because
u is specified there), v2(0) = v2(L) = 0 (because w is specified), and (dv2/dx)(0) = 0
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SOLUTIONS