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# Statics with MATLAB Dan B. Marghitu Mihai Dupac Nels H. Madsen

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```i
LiyCi
)
, (3.15)
where yCi is the centroidal coordinate for the ith line segment Li. The generating
curve is composed of simple curves, Li and the axis of revolution is the x-axis.
Theorem 2 A generating planar surface A and an axis of revolution located in the
same plane as the surface is considered in Fig. 3.5. The volume of revolution V
developed by rotating the generating planar surface about the axis of revolution
equals the product of the area of the surface times the circumference of the circle
formed by the centroid of the surface yC in the process of generating the body of
revolution
V = 2 \u3c0yCA. (3.16)
The axis of revolution does not intersect the generating surface. It can only touch the
generating plane surface as a tangent at the boundary.
Generating plane surface
y
xAxis of
revolution
y yC
O
dV
A
C
dA
Fig. 3.5 Volume of revolution
100 3 Centers of Mass
Proof The volume generated by rotating an element dA of the plane surface, A is
shown in Fig. 3.5, about the x-axis is
dV = 2 \u3c0y dA.
The volume of the body of revolution formed from A is
V = 2 \u3c0
\u222b
A
y dA = 2 \u3c0yC A.
Thus, the volume V equals the area of the generating surface A times the circum-
ferential length of the circle of radius yC . The volume V equals 2 \u3c0 times the first
moment of the generating area A about the axis of revolution.
The areas and center of mass for some practical configurations are shown in
Fig. 3.6.
3.7 Examples
Example 3.1 Find the length and the position of the center of mass for the homoge-
neous curve given by the Cartesian equation y = b \u221axa m, where a = 3, b = 2 and
0 \u2264 x \u2264 1 m.
Solution The differential element of the curve, dl = \u221a1 + (dy/dx)2, is given in
MATLAB by:
syms x real
a = 3;
b = 2;
y = b*sqrt(x\u2c6a);
dy = diff(y,x);
dl =sqrt(1+dy\u2c62);
% dl = (1+ (dy/dx)\u2c62)\u2c60.5
% 0< x < 1
The MATLAB statement int (f, x, a, b) is the definite integral of f with
respect to its symbolic variable x from a to b. The length of the homogeneous curve
is:
L = eval(int(dl,0,1));
and the coordinates of the center of mass C are:
My=eval(int(x*dl,0,1));
xC=My/L;
3.7 Examples 101
b/2 b/2
yC
yC
xC
h
h
b
C
C
O
C
r r
r
yC
xCO
C
r
r
yC
aO
C
yC
yC
xCO
C
a
b
yC
h
a
O
C
yC
xC
C
h
a
O
yC
xC
h
a
O
C
triangle
xC = b/3
yC = h/3
A = b h/2
semicircle
xC = 0
yC = 4 r/(3\u3c0)
A = \u3c0 r2/2
quartercircle
xC = 4 r/(3\u3c0)
yC = 4 r/(3\u3c0)
A = \u3c0 r2/4
semielipse
xC = 0
yC = 4 b/(3\u3c0)
A = \u3c0 a b/2
quarterelipse
xC = 4 a/(3\u3c0)
yC = 4 b/(3\u3c0)
A = \u3c0 a b/4
parabola
xC = 0
yC = 3h/5
A = 4 a h
semiparabola
xC = 3 a/8
yC = 4h/5
A = 2 a h/3
/3
parabolic spandrel
y = k x2
xC = 3 a/4
yC = 3h/10
A = a h/3
b
y
Fig. 3.6 Coordinates of center of mass, xC and yC , and area A
Mx=eval(int(y*dl,0,1));
yC=Mx/L;
The numerical values for the length and the centroid are:
L = 2.268 (m)
xC = 0.575 (m)
102 3 Centers of Mass
yC = 0.952 (m)
The MATLAB statements for the graphical representation are:
% plot the curve and CM
xf=1;
xn = 0:xf/100:xf;
yn = b*sqrt(xn.\u2c6a);
axis ([0 1 0 1])
plot(xn,yn,\u2019-b\u2019,\u2019LineWidth\u2019,2)
hold on
plot(xC,yC,\u2019o\u2019,\u2019MarkerSize\u2019,12,...
\u2019MarkerEdgeColor\u2019,\u2019k\u2019,...
\u2019MarkerFaceColor\u2019,\u2019r\u2019)
text(xC,yC,\u2019 C\u2019,\u2019FontSize\u2019,18)
title(\u2019y=f(x)=2 x\u2c6{3/2}\u2019)
and the results are depicted in Fig. 3.7.
Example 3.2 A homogeneous circle is given by the Cartesian equation x2 +y2 = r2,
where r = 1 m. (a) Find the length of the homogeneous circle. (b) Find the length
and the position of the center of mass for the homogeneous semi-circle, \u22121 \u2264 x \u2264 1
and 0 \u2264 y \u2264 1. (c) Find the length and the position of the center of mass for the
homogeneous quarter-circle, 0 \u2264 x \u2264 1 and 0 \u2264 y \u2264 1.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
C
y=f(x)=2 x3/2
x (m)
y
(m
)
Fig. 3.7 Example 3.1
3.7 Examples 103
Solution (a) The parametric equations for the circle are:
syms r t real
x = r*cos(t);
y = r*sin(t);
% 0 < t < 2*pi
% r > 0
The differential arc length is calculated in MATLAB with:
dx = diff(x,t);
dy = diff(y,t);
% dl = ((dx/dt)\u2c62+(dy/dt)\u2c62)\u2c60.5 dt
dl = (dx\u2c62+dy\u2c62)\u2c60.5;
dl = simplify(dl);
and the result is:
dl = abs(r) dt
The length of the circle is given by:
L = int(dl,t,0,2*pi);
and the MATLAB result is:
L = 2*pi*abs(r)
(b) For the semi-circle the length and the center of mass are:
Ls = int(dl,t,0,pi);
Mys = int(x*dl,t,0,pi);
xCs = simplify(Mys/Ls);
Mxs = int(y*dl,t,0,pi);
yCs = simplify(Mxs/Ls);
The results are:
Ls = pi*abs(r)
xCs = 0
yCs = (2*r)/pi
(c) For the quarter-circle the length and the center of mass are:
Lq = int(dl,t,0,pi/2);
Myq = int(x*dl,t,0,pi/2);
xCq = simplify(Myq/Lq);
Mxq = int(y*dl,t,0,pi/2);
yCq = simplify(Mxq/Lq);
and the MATLAB results are:
104 3 Centers of Mass
Lq = (pi*abs(r))/2
xCq = (2*r)/pi
yCq = (2*r)/pi
The MATLAB statements for the semi-circle and the quarter-circle graphical repre-
sentation are:
rn=1;
% plot the semi-circle and CM
figure(1)
xCsn = subs(xCs,r,1);
yCsn = subs(yCs,r,1);
tn = 0:pi/18:pi;
xn = rn*cos(tn);
yn = rn*sin(tn);
axis manual
axis equal
hold on
grid on
sa = 1;
axis ([-sa sa -sa sa])
plot(xn,yn,\u2019-b\u2019,\u2019LineWidth\u2019,2)
text(0,0,\u2019 O\u2019,\u2019fontsize\u2019,14)
line([-sa,sa],[0,0],\u2019Color\u2019,\u2019k\u2019)
line([0,0],[0,sa],\u2019Color\u2019,\u2019k\u2019)
plot(xCsn,yCsn,\u2019o\u2019,\u2019MarkerSize\u2019,12,...
\u2019MarkerEdgeColor\u2019,\u2019k\u2019,...
\u2019MarkerFaceColor\u2019,\u2019r\u2019)
% plot the quarter-circle and CM
figure(2)
xCqn = subs(xCq,r,1);
yCqn = subs(yCq,r,1);
tn = 0:pi/18:pi/2;
xn = rn*cos(tn);
yn = rn*sin(tn);
axis manual
axis equal
hold on
grid on
sa = 1;
axis ([-sa sa -sa sa])
plot(xn,yn,\u2019-b\u2019,\u2019LineWidth\u2019,2)
text(0,0,\u2019 O\u2019,\u2019fontsize\u2019,14)
line([0,sa],[0,0],\u2019Color\u2019,\u2019k\u2019)
line([0,0],[0,sa],\u2019Color\u2019,\u2019k\u2019)
plot(xCqn,yCqn,\u2019o\u2019,\u2019MarkerSize\u2019,12,...
3.7 Examples 105
\u2019MarkerEdgeColor\u2019,\u2019k\u2019,...
\u2019MarkerFaceColor\u2019,\u2019r\u2019)
text(xCqn,yCqn,\u2019 C\u2019,\u2019FontSize\u2019,18)
The graphics are depicted in Fig. 3.8.
Example 3.3 A homogeneous quarter-astroid (one cusp) is given by the Cartesian
equation x2/3 + y2/3 = a2/3, where a = 1 m and 0 \u2264 x \u2264 1 . Find the length and
the position of the center of mass for the homogeneous given curve.
Solution The parametric equations for the astroid are:
syms t real
a = 1; % (m)
x = a*cos(t)\u2c63;
y = a*sin(t)\u2c63;
% 0 < t < pi/2 - quarter-astroid
The differential arc length, dl = \u221a(dx/dt)2 + (dy/dt)2) dt, is calculated in MATLAB
with:
dx = diff(x,t);
dy = diff(y,t);
% dl=((dx/dt)\u2c62+(dy/dt)\u2c62)\u2c60.5 dt
dl = (dx\u2c62+dy\u2c62)\u2c60.5;
dl = simplify(dl);
and the result is:
dl = (3*(sin(2*t)\u2c62)\u2c6(1/2))/2
The length of the quarter-astroid is given by:
L = int(dl,t,0,pi/2);
L = double(L);
and the MATLAB result is:
L = 6*a/4
L = 1.500 (m)
For the quarter-astroid the length and the center of mass are:
My = int(x*dl,t,0,pi/2);
xC = My/L;
xC = double(xC);
Mx = int(y*dl,t,0,pi/2);
yC = Mx/L;
yC = double(yC);
106 3 Centers of Mass
\u22121 \u22120.8 \u22120.6 \u22120.4 \u22120.2 0 0.2 0.4 0.6 0.8 1
\u22121
\u22120.8
\u22120.6
\u22120.4
\u22120.2
0
0.2
0.4
0.6
0.8
1
O
C
semi\u2212circle
x (m)
\u22121 \u22120.8 \u22120.6 \u22120.4 \u22120.2 0 0.2 0.4 0.6 0.8 1
x (m)
y
(m
)
\u22121
\u22120.8
\u22120.6
\u22120.4
\u22120.2
0
0.2
0.4
0.6
0.8
1
y
(m
)
O
C
quarter\u2212circle
Fig. 3.8 Example 3.2
3.7 Examples 107
and the MATLAB results are:
xC = 0.400 (m)
yC = 0.400 (m)
The MATLAB statements for the semi-circle and the quarter-circle graphical repre-
sentation are:
tn = 0:pi/18:pi/2;
xn = a*cos(tn).\u2c63;
yn = a*sin(tn).\u2c63;
sa = 1;
axis ([0 sa 0 sa])
plot(xn,yn,\u2019-b\u2019,\u2019LineWidth\u2019,2)
text(0,0,\u2019 O\u2019,\u2019fontsize\u2019,14)
line([0,sa],[0,0],\u2019Color\u2019,\u2019k\u2019)
line([0,0],[0,sa],\u2019Color\u2019,\u2019k\u2019)
plot(xC,yC,\u2019o\u2019,\u2019MarkerSize\u2019,12,...
\u2019MarkerEdgeColor\u2019,\u2019k\u2019,...```
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Muito bom! achei que nunca ia achar um pdf desse livro, fazia tempo que estava procurando! muito obrigado!!!
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