Quiz questions for fundamental theorem for finite difference
Answers
Answer:
Multiple-Choice Test
Finite Difference Method
1.
The exact solution to the boundary value problem
2
2
2 6x 0.5x
dx
d y = − , y(0) = 0 , y(12) = 0
for y(4) is
(A) − 234.67
(B) 0.00
(C) 16.000
(D) 37.333
2.
Given
2
2
2 6x 0.5x
dx
d y = − , y(0) = 0 , y(12) = 0
the value of
2
2
dx
d y
at y(4) using the finite difference method and a step size of h = 4
can be approximated by
(A) ( )
( )
8
y 8 −
y 0
(B) ( )
( ) ( )
16
y 8 − 2
y 4 + y 0
(C) ( )
( ) ( )
16
y 12 − 2
y 8 + y 4
(D) ( )
( )
4
y 4 −
y 0
3.
Given
2
2
2 6x 0.5x
dx
d y = − , y(0) = 0 , y(12) = 0 ,
the value of y(4) using the finite difference method with a second order accurate
central divided difference method and a step size of h = 4 is
(A) 0.000
(B) 37.333
(C) − 234.67
(D) − 256.00
4.
The transverse deflection u of a cable of length L that is fixed at both ends, is given
as a solution to
( )
R
qx x L
R
Tu
dx
d u
2
2
2 − = +
where
T = tension in cable
R = flexural stiffness
q = distributed transverse load
u(x)
T
T
q
x
Given L = 50", T = 2000 lbs,
in
lbs
q = 75
, and 6 2 R = 75×10 lbs⋅in
Using finite difference method modeling with second order central divided difference
accuracy and a step size of h = 12.5", the value of the deflection at the center of the
cable most nearly is
(A) 0.072737″
(B) 0.080832″
(C) 0.081380″
(D) 0.084843″
5.
The radial displacement u of a pressurized hollow thick cylinder (inner radius = 5
″,
outer radius = 8″) is given at different radial locations.
Radius
(in)
Radial
Displacement
(in)
5.0
0.0038731
5.6
0.0036165
6.2
0.0034222
6.8
0.0032743
7.4
0.0031618
8.0
0.0030769
The maximum normal stress, in psi, on the cylinder is given by
(
)
( )
= ×
+ 0.3
5
5
5
3.2967 106
max
dr
u
du
σ
The maximum stress, in psi, with second order accuracy is
(A) 2079.6
(B) 2104.5
(C) 2130.7
(D)2182.0
6.
For a simply supported beam (at x = 0 and x = L ) with a uniform load q, the vertical
deflection v(x) is described by the boundary value ordinary differential equation as
( )
EI
qx x L
dx
d v
2
2
2 − = , 0 ≤ x ≤ L
where
E = Young’s modulus of the beam
I = second moment of area
This ordinary differential equation is based on assuming that
dx
dv
is small. If
dx
dv
is
not small, then the ordinary differential equation is given by
L
(A)
( )
EI
qx x L
dx
dv
dx
d v
2
1
2
2
2
− =
+
(B)
( )
EI
qx x L
dx
dv
dx
d v
2
1
23
2
2
2
− =
+
(C)
( )
EI
qx x L
dx
dv
dx
d v
2
1
2
2
− =
+
(D)
(
)
EI
qx
x
L
dx
dv
dx
d v
2
1
2
2
−
=
+
F