978-0073398167 Chapter 5 Solution Manual Part 1

subject Type Homework Help
subject Pages 17
subject Words 612
subject Textbook Statics and Mechanics of Materials 2nd Edition
subject Authors David Mazurek, E. Johnston, Ferdinand Beer, John DeWolf
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page-pf1
page-pf2
PROBLEM 5.2
Locate the centroid of the plane area shown.
SOLUTION
Area 1: Rectangle 6 in. by 4 in.
page-pf3
PROBLEM 5.3
Locate the centroid of the plane area shown.
SOLUTION
2
, inA
, in.x
, in.y
3
,inxA
3
,inyA
X A xAΣ=
23
(11in ) 11.5 inX=
1.045 in.X=
Y A yAΣ=Σ
23
(11 in ) 39.5 inY=
3.59 in.Y=
consent of McGraw-Hill Education.
page-pf4
PROBLEM 5.4
Locate the centroid of the plane area shown.
SOLUTION
Dimensions in mm
2
, mmA
, mmx
, mmy
3
, mmxA
3
, mmyA
6
6
page-pf5
PROBLEM 5.5
Locate the centroid of the plane area shown.
SOLUTION
By symmetry,
XY
=
Component
2
, inA
, in.x
3
, inxA
2
π
23
: (53.54 in ) 270.83 inX A xA XΣ=Σ =
5.0585 in.
X=
5.06 in.XY= =
consent of McGraw-Hill Education.
page-pf6
PROBLEM 5.6
Locate the centroid of the plane area shown.
SOLUTION
Area 1: Square 75 mm by 75 mm.
Area 2: Quarter circle radius of 75 mm.
2
, mmA
, mmx
, mmy
3
,mmxA
3
,mmyA
1
75 75 5625
×=
37.5 37.5 210,938 210,938
2
2
75 4417.9
4
π
−× =−
43.169
43.169 -190,715 −190,715
Σ
1207.10 20,223 20,223
Then by symmetry
xA
XY A
Σ
= = Σ
23
(1207.14 mm ) 20,223 mmX=
16.75 mmXY= =
consent of McGraw-Hill Education.
page-pf7
PROBLEM 5.7
Locate the centroid of the plane area shown.
SOLUTION
Area 1: Outer semicircle diameter 120 mm.
Area 2: Inner semicircle diameter 72 mm.
2
, mmA
, mm
x
, mmy
3
, mmxA
3
, mmyA
1
2
(120) 22,620
2
π
=
-50.93 0
3
1152.04 10−×
0
2
2
(72) 8143.0
2
π
−=−
-30.558 0
3
248.80 10×
0
Σ
14,477
3
903.20 10−×
0
Then
XA xA= Σ
2 33
(14,477 mm ) 903.20 10 mmX=−×
62.4 mmX= −
YA yA= Σ
0Y=
consent of McGraw-Hill Education.
page-pf8
PROBLEM 5.8
Locate the centroid of the plane area shown.
SOLUTION
2
, in.A
, in.x
, in.y
3
, in.xA
3
, in.yA
1
30 50 1500×=
15 25 22500 37500
2
2
(15) 353.43
2
π
−=
23.634 30 –8353.0 –10602.9
Σ
1146.57 14147.0 26.897
Then
14147.0
1146.57
xA
XA
Σ
= =
Σ
12.34 in.X=
26897
1146.57
yA
YA
Σ
= =
Σ
23.5 in.Y=
consent of McGraw-Hill Education.
page-pf9
PROBLEM 5.9
Locate the centroid of the plane area shown.
SOLUTION
Dimensions in in.
2
, inA
, in.x
, in.y
3
, inxA
3
, inyA
1
2(4)(8) 21.333
3=
4.8 1.5 102.398 32.000
2
1(4)(8) 16.0000
2
−=−
5.3333 1.33333 85.333 −21.333
Σ
5.3333 17.0650 10.6670
Then
X A xAΣ=Σ
23
(5.3333 in ) 17.0650 inX=
or
3.20 in.X=
and
Y A yAΣ=Σ
23
(5.3333 in ) 10.6670 inY=
or
2.00 in.Y=
consent of McGraw-Hill Education.
page-pfa
PROBLEM 5.10
Locate the centroid of the plane area shown.
SOLUTION
Area 2: Rectangle 16 in. by 3 in.
2
, in.A
, in.
x
, in.y
3
, inxA
3
, inyA
1(10)(16) 53.333
Then
XA xA= Σ
23
(101.333 in ) 1024 inX=
10.11in.X=
23
(101.333 in ) 393 in
YA yA
Y
= Σ
=
3.88 in.Y=
consent of McGraw-Hill Education.
PROBLEM 5.2
Locate the centroid of the plane area shown.
SOLUTION
Area 1: Rectangle 6 in. by 4 in.
PROBLEM 5.3
Locate the centroid of the plane area shown.
SOLUTION
2
, inA
, in.x
, in.y
3
,inxA
3
,inyA
X A xAΣ=
23
(11in ) 11.5 inX=
1.045 in.X=
Y A yAΣ=Σ
23
(11 in ) 39.5 inY=
3.59 in.Y=
consent of McGraw-Hill Education.
PROBLEM 5.4
Locate the centroid of the plane area shown.
SOLUTION
Dimensions in mm
2
, mmA
, mmx
, mmy
3
, mmxA
3
, mmyA
6
6
PROBLEM 5.5
Locate the centroid of the plane area shown.
SOLUTION
By symmetry,
XY
=
Component
2
, inA
, in.x
3
, inxA
2
π
23
: (53.54 in ) 270.83 inX A xA XΣ=Σ =
5.0585 in.
X=
5.06 in.XY= =
consent of McGraw-Hill Education.
PROBLEM 5.6
Locate the centroid of the plane area shown.
SOLUTION
Area 1: Square 75 mm by 75 mm.
Area 2: Quarter circle radius of 75 mm.
2
, mmA
, mmx
, mmy
3
,mmxA
3
,mmyA
1
75 75 5625
×=
37.5 37.5 210,938 210,938
2
2
75 4417.9
4
π
−× =−
43.169
43.169 -190,715 −190,715
Σ
1207.10 20,223 20,223
Then by symmetry
xA
XY A
Σ
= = Σ
23
(1207.14 mm ) 20,223 mmX=
16.75 mmXY= =
consent of McGraw-Hill Education.
PROBLEM 5.7
Locate the centroid of the plane area shown.
SOLUTION
Area 1: Outer semicircle diameter 120 mm.
Area 2: Inner semicircle diameter 72 mm.
2
, mmA
, mm
x
, mmy
3
, mmxA
3
, mmyA
1
2
(120) 22,620
2
π
=
-50.93 0
3
1152.04 10−×
0
2
2
(72) 8143.0
2
π
−=−
-30.558 0
3
248.80 10×
0
Σ
14,477
3
903.20 10−×
0
Then
XA xA= Σ
2 33
(14,477 mm ) 903.20 10 mmX=−×
62.4 mmX= −
YA yA= Σ
0Y=
consent of McGraw-Hill Education.
PROBLEM 5.8
Locate the centroid of the plane area shown.
SOLUTION
2
, in.A
, in.x
, in.y
3
, in.xA
3
, in.yA
1
30 50 1500×=
15 25 22500 37500
2
2
(15) 353.43
2
π
−=
23.634 30 –8353.0 –10602.9
Σ
1146.57 14147.0 26.897
Then
14147.0
1146.57
xA
XA
Σ
= =
Σ
12.34 in.X=
26897
1146.57
yA
YA
Σ
= =
Σ
23.5 in.Y=
consent of McGraw-Hill Education.
PROBLEM 5.9
Locate the centroid of the plane area shown.
SOLUTION
Dimensions in in.
2
, inA
, in.x
, in.y
3
, inxA
3
, inyA
1
2(4)(8) 21.333
3=
4.8 1.5 102.398 32.000
2
1(4)(8) 16.0000
2
−=−
5.3333 1.33333 85.333 −21.333
Σ
5.3333 17.0650 10.6670
Then
X A xAΣ=Σ
23
(5.3333 in ) 17.0650 inX=
or
3.20 in.X=
and
Y A yAΣ=Σ
23
(5.3333 in ) 10.6670 inY=
or
2.00 in.Y=
consent of McGraw-Hill Education.
PROBLEM 5.10
Locate the centroid of the plane area shown.
SOLUTION
Area 2: Rectangle 16 in. by 3 in.
2
, in.A
, in.
x
, in.y
3
, inxA
3
, inyA
1(10)(16) 53.333
Then
XA xA= Σ
23
(101.333 in ) 1024 inX=
10.11in.X=
23
(101.333 in ) 393 in
YA yA
Y
= Σ
=
3.88 in.Y=
consent of McGraw-Hill Education.

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