Resolução do Capítulo 5 - Forças Distribuídas (Centroides e Centro de Gravidade)
166 pág.

Resolução do Capítulo 5 - Forças Distribuídas (Centroides e Centro de Gravidade)

Pré-visualização18 páginas
PROBLEM 5.1 
Locate the centroid of the plane area shown. 
 
SOLUTION 
 
 2, inA , in.x , in.y 3, inxA 3, inyA 
1 8 6 48× = 4\u2212 9 192\u2212 432 
2 16 12 192× = 8 6 1536 1152 
\u3a3 240 1344 1584 
 
Then 
3
2
1344 in
240 in
xAX
A
\u3a3= =\u3a3 or 5.60 in.X = W 
and 
3
2
1584 in
240 in
yAY
A
\u3a3= =\u3a3 or 6.60 in.Y = W 
 
 
 
 
 
PROBLEM 5.2 
Locate the centroid of the plane area shown. 
 
SOLUTION 
 
 2, mmA , mmx , mmy 3, mmxA 3, mmyA 
1 
1 60 75 2250
2
× × = 40 25 90 000 56 250 
2 105 75 7875× = 112.5 37.5 885 900 295 300 
\u3a3 10 125 975 900 351 600 
 
Then 
3
2
975 900 mm
10 125 mm
xAX
A
\u3a3= =\u3a3 or 96.4 mmX = W 
and 
3
2
351 600 mm
10 125 mm
yAY
A
\u3a3= =\u3a3 or 34.7 mmY = W 
 
 
 
 
 
 
PROBLEM 5.3 
Locate the centroid of the plane area shown. 
 
SOLUTION 
 
For the area as a whole, it can be concluded by observation that 
 ( )2 24 in.
3
Y = or 16.00 in.Y = W 
 2, inA , in.x 3, inxA 
1 
1 24 10 120
2
× × = ( )2 10 6.6673 = 800 
2 
1 24 16 192
2
× × = ( )110 16 15.3333+ = 2944 
\u3a3 312 3744 
 
Then 
3
2
3744 in
312 in
xAX
A
\u3a3= =\u3a3 or 12.00 in.X = W 
 
 
 
 
 
 
PROBLEM 5.4 
Locate the centroid of the plane area shown. 
 
SOLUTION 
 
 2, mmA , mmx , mmy 3, mmxA 3, mmyA 
1 21 22 462× = 1.5 11 693 5082 
2 ( )( )1 6 9 272\u2212 = \u2212 6\u2212 2 162 54\u2212 
3 ( )( )1 6 12 362\u2212 = \u2212 8 2 288\u2212 72\u2212 
\u3a3 399 567 4956 
 
Then 
3
2
567 mm
399 mm
xAX
A
\u3a3= =\u3a3 or 1.421 mmX = W 
and 
3
2
4956 mm
399 mm
yAY
A
\u3a3= =\u3a3 or 12.42 mmY = W 
 
 
 
 
 
PROBLEM 5.5 
Locate the centroid of the plane area shown. 
 
SOLUTION 
 
 2, mmA , mmx , mmy 3, mmxA 3, mmyA 
1 120 200 24 000× = 60 120 1 440 000 2 880 000 
2 
( )260 5654.9
2
\u3c0\u2212 = \u2212 94.5 120 534 600\u2212 678 600\u2212 
\u3a3 18 345 905 400 2 201 400 
 
Then 
3
2
905 400 mm
18 345 mm
xAX
A
\u3a3= =\u3a3 or 49.4 mmX = W 
and 
3
2
2 201 400 mm
18 345 mm
yAY
A
\u3a3= =\u3a3 or 93.8 mmY = W 
 
 
 
 
 
PROBLEM 5.6 
Locate the centroid of the plane area shown. 
 
SOLUTION 
 
 2, inA , in.x , in.y 3, inxA 3, inyA 
1 
( )29 63.617
4
\u3c0 = 
( )
( )
4 9
3.8917
3\u3c0
\u2212 = \u2212 3.8917 243\u2212 243 
2 ( )( )1 15 9 67.52 = 5 3 337.5 202.5 
\u3a3 131.1 94.5 445.5 
 
Then 
3
2
94.5 in
131.1 in
xAX
A
\u3a3= =\u3a3 or 0.721 in.X = W 
and 
3
2
445.5 in
131.1 in
yAY
A
\u3a3= =\u3a3 or 3.40 in.Y = W 
 
 
 
 
 
PROBLEM 5.7 
Locate the centroid of the plane area shown. 
 
SOLUTION 
First note that symmetry implies X Y= 
 
 2, mmA , mmx 3, mmxA 
1 40 40 1600× = 20 32 000 
2 
2(40) 1257
4
\u3c0\u2212 = \u2212 16.98 21 330\u2212 
\u3a3 343 10 667 
 
Then 
3
2
10 667 mm
343 mm
xAX
A
\u3a3= =\u3a3 or 31.1 mmX = W 
 and 31.1 mmY X= = W 
 
 
 
 
 
PROBLEM 5.8 
Locate the centroid of the plane area shown. 
 
SOLUTION 
First note that symmetry implies 0X = W 
 
 2, inA , in.y 3, inyA 
1 
( )24 25.13
2
\u3c0\u2212 = \u2212 1.6977 42.67\u2212 
2 
( )26 56.55
2
\u3c0 = 2.546 144 
\u3a3 31.42 101.33 
 
Then 
3
2
101.33 in
31.42 in
yAY
A
\u3a3= =\u3a3 or 3.23 in.Y = W 
 
 
 
 
PROBLEM 5.9 
For the area of Problem 5.8, determine the ratio 2 1/r r so that 13 /4.y r= 
 
SOLUTION 
 
 A y yA 
1 
2
12
r\u3c0\u2212 143
r
\u3c0 
3
1
2
3
r\u2212 
2 
2
22
r\u3c0 243
r
\u3c0 
3
2
2
3
r 
\u3a3 ( )2 22 12 r r\u3c0 \u2212 ( )3 32 123 r r\u2212 
 
Then Y A y A\u3a3 = \u3a3 
or ( ) ( )2 2 3 31 2 1 2 13 24 2 3r r r r r\u3c0× \u2212 = \u2212 
2 3
2 2
1 1
9 1 1
16
r r
r r
\u3c0 \uf8ee \uf8f9\uf8eb \uf8f6 \uf8eb \uf8f6\uf8ef \uf8fa\u2212 = \u2212\uf8ec \uf8f7 \uf8ec \uf8f7\uf8ed \uf8f8 \uf8ed \uf8f8\uf8ef \uf8fa\uf8f0 \uf8fb
 
Let 2
1
rp
r
= 
[ ] 29 ( 1)( 1) ( 1)( 1)
16
p p p p p\u3c0 + \u2212 = \u2212 + + 
or 216 (16 9 ) (16 9 ) 0p p\u3c0 \u3c0+ \u2212 + \u2212 = 
 
 
 
 PROBLEM 5.9 CONTINUED 
Then 
2(16 9 ) (16 9 ) 4(16)(16 9 )
2(16)
p
\u3c0 \u3c0 \u3c0\u2212 \u2212 ± \u2212 \u2212 \u2212= 
or 0.5726 1.3397p p= \u2212 = 
Taking the positive root 2
1
1.340r
r
= W 
 
 
 
 
 
PROBLEM 5.10 
Show that as 1r approaches 2,r the location of the centroid approaches that 
of a circular arc of radius ( )1 2 / 2.r r+ 
 
SOLUTION 
 
First, determine the location of the centroid. 
From Fig. 5.8A: 
( )
( ) ( )2 22 2 2 222
sin2 
3
y r A r
\u3c0
\u3c0
\u3c0
\u3b1 \u3b1\u3b1
\u2212= = \u2212\u2212 
 ( )2 2
2 cos
3
r \u3c0
\u3b1
\u3b1= \u2212 
Similarly ( ) ( ) 21 1 1 122
2 cos 
3
y r A r\u3c0\u3c0
\u3b1 \u3b1\u3b1= = \u2212\u2212 
( ) ( ) ( ) ( )
( )
2 2
2 2 1 12 2
2 2
3 3
2 1
2 cos 2 cosThen
3 3
2 cos
3
yA r r r r
r r
\u3c0 \u3c0
\u3c0 \u3c0
\u3b1 \u3b1\u3b1 \u3b1\u3b1 \u3b1
\u3b1
\uf8ee \uf8f9 \uf8ee \uf8f9\u3a3 = \u2212 \u2212 \u2212\uf8f0 \uf8fb \uf8f0 \uf8fb\u2212 \u2212
= \u2212
 
( )
2 2
2 1
2 2
2 1
and 
2 2
2
A r r
r r
\u3c0 \u3c0\u3b1 \u3b1
\u3c0 \u3b1
\uf8eb \uf8f6 \uf8eb \uf8f6\u3a3 = \u2212 \u2212 \u2212\uf8ec \uf8f7 \uf8ec \uf8f7\uf8ed \uf8f8 \uf8ed \uf8f8
\uf8eb \uf8f6= \u2212 \u2212\uf8ec \uf8f7\uf8ed \uf8f8
 
( ) ( )2 2 3 32 1 2 1
3 3
2 1
2 2
2 1 2
Now
2 cos
2 3
2 cos 
3
Y A yA
Y r r r r
r rY
r r \u3c0
\u3c0 \u3b1 \u3b1
\u3b1
\u3b1
\u3a3 = \u3a3
\uf8ee \uf8f9\uf8eb \uf8f6\u2212 \u2212 = \u2212\uf8ec \uf8f7\uf8ef \uf8fa\uf8ed \uf8f8\uf8f0 \uf8fb
\u2212= \u2212\u2212
 
 
 
 PROBLEM 5.10 CONTINUED 
Using Figure 5.8B,Y of an arc of radius ( )1 21 is2 r r+ 
( ) ( )( )21 2 2
sin1
2
Y r r
\u3c0
\u3c0
\u3b1
\u3b1
\u2212= + \u2212 
 ( )1 2 2
1 cos( )
2
r r \u3c0
\u3b1
\u3b1= + \u2212 (1) 
( )( )
( )( )
2 23 3 2 1 2 1 2 12 1
2 2
2 1 2 12 1
2 2
2 1 2 1
2 1
Now 
r r r r r rr r
r r r rr r
r r r r
r r
\u2212 + +\u2212 = \u2212 +\u2212
+ += +
 
2
1
Let r r
r r
= + \u2206
= \u2212 \u2206 
Then ( )1 212r r r= + 
( ) ( )( ) ( )
( ) ( )
2 23 3
2 1
2 2
2 1
2 2
and 
3
2
r r r rr r
r rr r
r
r
+ \u2206 + + \u2206 \u2212 \u2206 + \u2212 \u2206\u2212 = + \u2206 + \u2212 \u2206\u2212
+ \u2206=
 
1 2In the limit as 0 (i.e., ), thenr r\u2206 \u2192 = 
3 3
2 1
2 2
2 1
1 2
3 
2
3 1 ( )
2 2
r r r
r r
r r
\u2212 =\u2212
= × +
 
so that ( )1 2
2
2 3 cos
3 4
Y r r \u3c0
\u3b1
\u3b1= × + \u2212 or ( )1 2 2
1 cos
2
Y r r \u3c0
\u3b1
\u3b1= + \u2212 W 
Which agrees with Eq. (1). 
 
 
 
 
 
 
PROBLEM 5.11 
Locate the centroid of the plane area shown. 
 
SOLUTION 
First note that symmetry implies 0X = W 
 
2 2 2 in., 45r \u3b1= = ° 
( ) ( )
( ) 42 4
2 2 2 sin2 sin 1.6977 in.
3 3
ry
\u3c0
\u3c0
\u3b1
\u3b1= = =\u2032 
 2, inA , in.y 3, inyA 
1 ( ) ( )1 4 3 62 = 1 6 
2 ( )22 2 6.2834\u3c0 = 2 0.3024y\u2212 =\u2032 1.8997 
3 ( ) ( )1 4 2 42\u2212 = \u2212 0.6667 2.667\u2212 
\u3a3 8.283 5.2330 
 
Then Y A yA\u3a3 = \u3a3 
 ( )2 38.283 in 5.2330 inY = or 0.632 in.Y = W 
 
 
 
 
PROBLEM 5.12 
Locate the centroid of the plane area shown. 
 
SOLUTION 
 
 2, mmA , mmx , mmy 3, mmxA 3, mmyA 
1 (40)(90) 3600= 15\u2212 20 54 000\u2212 72 000 
2 
( ) ( )40 60 2121
4
\u3c0 = 10 15\u2212 6750 10 125\u2212 
3 ( ) ( )1 30 45 6752 = 25.47\u2212 19.099\u2212 54 000\u2212 40 500\u2212 
\u3a3 6396 101 250\u2212 21 375 
 
Then XA xA= \u3a3 
 ( )2 36396 mm 101 250 mmX = \u2212 or 15.83 mmX = \u2212 W 
and YA yA= \u3a3 
 ( )2 36396 mm 21 375 mmY = or 3.34 mmY = W 
 
 
 
 
PROBLEM 5.13 
Locate the centroid of the plane area shown. 
 
SOLUTION 
 
 2, mmA , mmx , mmy 3, mmxA 3, mmyA 
1 ( ) ( )2 40 80 21333 = 48 15 102 400 32 000 
2 ( ) ( )1 40 80 16002\u2212 = \u2212 53.33 13.333 85 330\u2212 21 330\u2212 
\u3a3 533.3 17 067 10 667 
 
Then X A XA\u3a3 = \u3a3 
 ( )2 3533.3 mm 17 067 mmX = or 32.0 mmX = W 
and Y A yA\u3a3 = \u3a3 
 ( )2 3533.3 mm 10 667 mmY = or 20.0 mmY = W 
 
 
 
 
 
PROBLEM 5.14 
Locate the centroid of the plane area shown. 
 
SOLUTION 
 
 2, mmA , mmx , mmy 3, mmxA 3, mmyA 
1 ( )( )2