33
CHAPTER 5
5.1 The function to evaluate is
tanh t
v t
m gc
c gm
c f
d d
d
− ⎟
⎟ ⎠
⎞ ⎜
⎜ ⎝
⎛ =
or substituting the given values
36 4
80 81
. 9
tanh 80
81 .
9 −
⎟ ⎟
⎠ ⎞
⎜ ⎜
⎝ ⎛
=
d d
d
c c
c f
The first iteration is
15 .
2 2
. 1
. =
+ =
r
x 175944
. 204516
. 860291
. 15
. 1
. −
= −
= f
f Therefore, the root is in the first interval and the upper guess is redefined as x
u
= 0.15. The second iteration is
125 .
2 15
. 1
. =
+ =
r
x
20 100
125 .
15 .
125 .
= −
=
a
ε
273923 .
318407 .
860291 .
125 .
1 .
= =
f f
Therefore, the root is in the second interval and the lower guess is redefined as x
u
= 0.125. The remainder of the iterations are displayed in the following table:
i x
l
fx
l
x
u
fx
u
x
r
fx
r
|
ε
a
| 1 0.1
0.86029 0.2 −1.19738
0.15 −0.20452
2 0.1 0.86029 0.15
−0.20452 0.125
0.31841 20.00 3 0.125
0.31841 0.15
−0.20452 0.1375
0.05464 9.09 4 0.1375 0.05464
0.15 −0.20452
0.14375 −0.07551
4.35 5 0.1375 0.05464 0.14375
−0.07551 0.140625
−0.01058 2.22
6 0.1375 0.05464 0.140625 −0.01058 0.1390625
0.02199 1.12
Thus, after six iterations, we obtain a root estimate of 0.1390625 with an approximate error of 1.12.
5.2 function root = bisectnewfunc,xl,xu,Ead
bisectnewxl,xu,es,maxit:
34 uses bisection method to find the root of a function
with a fixed number of iterations to attain a prespecified tolerance
input: func = name of function
xl, xu = lower and upper guesses Ead = optional desired tolerance default = 0.000001
output: root = real root
if funcxlfuncxu0 if guesses do not bracket a sign change errorno bracket display an error message and terminate
end if necessary, assign default values
if nargin4, Ead = 0.000001; end if Ead blank set to 0.000001 bisection
xr = xl; compute n and round up to next highest integer
n = round1 + log2xu - xlEad + 0.5; for i = 1:n
xrold = xr; xr = xl + xu2;
if xr ~= 0, ea = absxr - xroldxr 100; end test = funcxlfuncxr;
if test 0 xu = xr;
elseif test 0 xl = xr;
else ea = 0;
end end
root = xr;
The following is a MATLAB session that uses the function to solve Prob. 5.1 with E
a,d
= 0.0001.
fcd = inlinesqrt9.8180cdtanhsqrt9.81cd804-36,cd fcd =
Inline function: fcdcd = sqrt9.8180cdtanhsqrt9.81cd804-36
format long bisectnewfcd,0.1,0.2,0.0001
ans = 0.14008789062500
5.3
The function to evaluate is 36
4 80
81 .
9 tanh
80 81
. 9
− ⎟
⎟ ⎠
⎞ ⎜
⎜ ⎝
⎛ =
d d
d
c c
c f
35 The first iteration is
141809 .
19738 .
1 86029
. 2
. 1
. 19738
. 1
2 .
= −
− −
− −
=
r
x 030292
. 03521
. 860291
. 141809
. 1
. −
= −
= f
f Therefore, the root is in the first interval and the upper guess is redefined as x
u
= 0.141809. The second iteration is
140165 .
03521 .
86029 .
141809 .
1 .
03521 .
141809 .
= −
− −
− −
=
r
x
17 .
1 100
140165 .
141809 .
140165 .
= −
=
a
ε
Therefore, after only two iterations we obtain a root estimate of 0.140165 with an approximate error of 1.17 which is below the stopping criterion of 2.
5.4 function root = falseposfunc,xl,xu,es,maxit
falseposxl,xu,es,maxit: uses the false position method to find the root
of the function func input:
func = name of function xl, xu = lower and upper guesses
es = optional stopping criterion default = 0.001 maxit = optional maximum allowable iterations default = 50
output: root = real root
if funcxlfuncxu0 if guesses do not bracket a sign change errorno bracket display an error message and terminate
end default values
if nargin5, maxit=50; end if nargin4, es=0.001; end
false position iter = 0;
xr = xl; while 1
xrold = xr; xr = xu - funcxuxl - xufuncxl - funcxu;
iter = iter + 1; if xr ~= 0, ea = absxr - xroldxr 100; end
test = funcxlfuncxr; if test 0
xu = xr; elseif test 0
xl = xr; else
ea = 0;
36 end
if ea = es | iter = maxit, break, end end
root = xr;
The following is a MATLAB session that uses the function to solve Prob. 5.1: fcd = inlinesqrt9.8180cdtanhsqrt9.81cd804-36,cd
fcd = Inline function:
fcdcd = sqrt9.8180cdtanhsqrt9.81cd804-36 format long
falseposfcd,0.1,0.2,2 ans =
0.14016503741282 5.5
Solve for the reactions:
R
1
= 265 lbs. R
2
= 285 lbs.
Write beam equations:
0x3
3 2
55 .
5 265
1 265
3 667
. 16
x M
x x
x M
− =
= −
+
3x6
150 415
50 2
265 3
3 2
150 2
3 3
100
2
− +
− =
= −
− +
− −
+ x
x M
x x
x x
M
6x10
1650 185
3 265
5 .
4 300
3 3
2 150
+ −
= −
− +
− =
x M
x x
x M
10x12
1200 100
4 12
100 −
= =
− +
x M
x M
Combining Equations: Because the curve crosses the axis between 6 and 10, use 3.
3
1650 185
+ −
= x
M
Set
10 ;
6 =
=
U L
x x
37
200 540
L
− =
=
U
x M
x M
8 2
= +
=
U L
r
x x
x
L R
x replaces
x M
→ = 170
200 170
L
− =
=
U
x M
x M
9 2
10 8
= +
=
r
x
U R
x replaces
x M
→ −
= 15
15 170
L
− =
=
U
x M
x M
5 .
8 2
9 8
= +
=
r
x
L R
x replaces
x M
→ =
5 .
77
15 5
. 77
L
− =
=
U
x M
x M
75 .
8 2
9 5
. 8
= +
=
r
x
L R
x replaces
x M
→ =
25 .
31
15 25
. 31
L
− =
=
U
x M
x M
875 .
8 2
9 75
. 8
= +
=
r
x
L R
x replaces
x M
→ = 125
. 8
15 125
. 8
L
− =
=
U
x M
x M
9375 .
8 2
9 875
. 8
= +
=
r
x
U R
x replaces
x M
→ −
= 4375
. 3
4375 .
3 125
. 8
L
− =
=
U
x M
x M
90625 .
8 2
9375 .
8 875
. 8
= +
=
r
x
L R
x replaces
x M
→ = 34375
. 2
4375 .
3 34375
. 2
L
− =
=
U
x M
x M
921875 .
8 2
9375 .
8 90625
. 8
= +
=
r
x
U R
x replaces
x M
→ −
= 546875
.
546875 .
34375 .
2
L
− =
=
U
x M
x M
9140625 .
8 2
921875 .
8 90625
. 8
= +
=
r
x
8984 .
R
= x
M
Therefore,
feet x
91 .
8 =
5.6 a The graph can be generated with MATLAB
