PROBLEM 5.56
Determine the reactions at the beam supports for the given loading.
SOLUTION
First replace the given loading by the loadings shown below. Both loadings are equivalent since they are
both defined by a linear relation between load and distance and have the same values at the end points.
1
2
1(900 N/m)(1.5 m) 675 N
2
1(400 N/m)(1.5 m) 300 N
2
R
R
= =
= =
0: (675 N)(1.4 m) (300 N)(0.9 m) (2.5 m)
A
M BCΣ=− + + =
270 NB=
270 N=B
0: 675 N 300 N 270 N 0
y
FAΣ= − + + =
105.0 NA=
105.0 N=A
PROBLEM 5.57
Consider the composite body shown. Determine (a) the
value of
x
when
/2,hL=
(b) the ratio h/L for which
.xL=
SOLUTION
V
x
xV
Rectangular
prism Lab
1
2L
2
1
2L ab
Pyramid
1
32
b
ah
1
4
Lh+
11
64
abh L h
+
Then
2
1
6
11
3
64
V ab L h
xV ab L h L h
Σ= +
Σ= + +
Now
X V xVΣ=Σ
so that
22
11 1
3
66 4
X ab L h ab L hL h
+ = ++
or
2
2
11 1
13
66 4
h hh
XL
LL
L
+ = ++
(1)
(a)
?X=
when
1.
2
hL
=
Substituting
1into Eq. (1),
2
h
L=
2
11 1 1 11
13
62 6 2 42
XL
+ = ++
or
57
104
XL=
0.548XL=
consent of McGraw–Hill Education.
SOLUTION Continued
h
PROBLEM 5.58
Determine the location of the centroid of the composite body shown
when (a)
2,hb=
(b)
2.5 .hb
=
SOLUTION
V
x
xV
Cylinder I
2
ab
π
1
2b
22
1
2ab
π
Cone II
2
1
3ah
π
1
4
bh+
2
11
34
ahb h
π
+
2
22 2
1
3
11 1
2 3 12
V ab h
xV a b hb h
π
π
= +
Σ= + +
(a) For
2,hb=
22
15
(2 )
33
V a b b ab
ππ
=+=
22 2
22 22
11 1
(2 ) (2 )
2 3 12
121 3
233 2
xV a b b b b
ab ab
π
ππ
Σ= + +
= ++ =
2 22
53 9
:3 2 10
XV xV X ab ab X b
ππ
=Σ==
Centroid is
1
10
b
to the left of base of cone.
consent of McGraw–Hill Education.
SOLUTION Continued
(b) For
2.5 ,hb=
22
1(2.5 ) 1.8333
3
V a b b ab
ππ
=+=
22 2
22
22
11 1
(2.5 ) (2.5 )
2 3 12
[0.5 0.8333 0.52083]
1.85416
xV a b b b b
ab
ab
π
π
π
Σ= + +
= ++
=
2 22
: (1.8333 ) 1.85416 1.01136XV xV X ab ab X b
ππ
=Σ==
Centroid is 0.01136b to the right of base of cone.
Note: Centroid is at base of cone for
6 2.449 .
hb b= =
consent of McGraw–Hill Education.
PROBLEM 5.59
The composite body shown is formed by removing a semiellipsoid
of revolution of semimajor axis h and semiminor axis a/2 from a
hemisphere of radius a. Determine (a) the y coordinate of the
centroid when h = a/2, (b) the ratio h/a for which
y
= −0.4a.
SOLUTION
V
y
yV
Hemisphere 3
2
3
a
π
3
8a−
4
1
4a
π
−
Semiellipsoid
22
21
32 6
ah ah
ππ
−=−
3
8h−
22
1
16 ah
π
+
Then
2
222
(4 )
6
(4 )
16
V a ah
yV a a h
π
π
Σ= −
Σ=− −
Now
Y V yVΣ=Σ
So that
2 222
(4 ) (4 )
6 16
Y aah aah
ππ
−=− −
or
2
3
44
8
hh
Ya
aa
−=− −
(1)
(a)
? when 2
a
Yh
= =
Substituting
1
2
h
a=
into Eq. (1)
2
13 1
44
28 2
Ya
−=− −
or
45
112
Ya= −
0.402Ya= −
consent of McGraw–Hill Education.
SOLUTION Continued
(b)
? when 0.4
hYa
a= = −
PROBLEM 5.60
Locate the centroid of the frustum of a right circular cone when r1 = 40 mm,
r2 = 50 mm, and h = 60 mm.
SOLUTION
By similar triangles:
11 12
60 240 mm, therefore 240 60 300 mm
40 50
hh hh
+
= = = +=
( ) ( ) ( ) ( )
12 2 1
22
33
12
1 300 1 240
75 mm 60 60 120 mm
44 4 4
50 300 250 10 40 240 128 10
33
yh y h
VV
ππ
ππ
= == =+=+=
==×==×
3
mmV
mmy
4
mmyV
3
250 10
π
×
6
40 mm
50 mm
60 mm
h
1
2
h = 300 mm
50 mm
y
1
40 mm
60 mm
y
2
240 mm
consent of McGraw–Hill Education.
PROBLEM 5.61
For the machine element shown, locate the y coordinate of
the center of gravity.
SOLUTION
For half–cylindrical hole,
III
0.95 in.
4(0.95)
1.5 3
1.097 in.
r
y
π
=
= −
=
For half–cylindrical plate,
IV
1.5 in.
4(1.5)
5.25 5.887 in.
3
r
z
π
=
=+=
3
, inV
, in.y
, in.z
4
, inyV
4
, inzV
PROBLEM 5.56
Determine the reactions at the beam supports for the given loading.
SOLUTION
First replace the given loading by the loadings shown below. Both loadings are equivalent since they are
both defined by a linear relation between load and distance and have the same values at the end points.
1
2
1(900 N/m)(1.5 m) 675 N
2
1(400 N/m)(1.5 m) 300 N
2
R
R
= =
= =
0: (675 N)(1.4 m) (300 N)(0.9 m) (2.5 m)
A
M BCΣ=− + + =
270 NB=
270 N=B
0: 675 N 300 N 270 N 0
y
FAΣ= − + + =
105.0 NA=
105.0 N=A
PROBLEM 5.57
Consider the composite body shown. Determine (a) the
value of
x
when
/2,hL=
(b) the ratio h/L for which
.xL=
SOLUTION
V
x
xV
Rectangular
prism Lab
1
2L
2
1
2L ab
Pyramid
1
32
b
ah
1
4
Lh+
11
64
abh L h
+
Then
2
1
6
11
3
64
V ab L h
xV ab L h L h
Σ= +
Σ= + +
Now
X V xVΣ=Σ
so that
22
11 1
3
66 4
X ab L h ab L hL h
+ = ++
or
2
2
11 1
13
66 4
h hh
XL
LL
L
+ = ++
(1)
(a)
?X=
when
1.
2
hL
=
Substituting
1into Eq. (1),
2
h
L=
2
11 1 1 11
13
62 6 2 42
XL
+ = ++
or
57
104
XL=
0.548XL=
consent of McGraw–Hill Education.
SOLUTION Continued
h
PROBLEM 5.58
Determine the location of the centroid of the composite body shown
when (a)
2,hb=
(b)
2.5 .hb
=
SOLUTION
V
x
xV
Cylinder I
2
ab
π
1
2b
22
1
2ab
π
Cone II
2
1
3ah
π
1
4
bh+
2
11
34
ahb h
π
+
2
22 2
1
3
11 1
2 3 12
V ab h
xV a b hb h
π
π
= +
Σ= + +
(a) For
2,hb=
22
15
(2 )
33
V a b b ab
ππ
=+=
22 2
22 22
11 1
(2 ) (2 )
2 3 12
121 3
233 2
xV a b b b b
ab ab
π
ππ
Σ= + +
= ++ =
2 22
53 9
:3 2 10
XV xV X ab ab X b
ππ
=Σ==
Centroid is
1
10
b
to the left of base of cone.
consent of McGraw–Hill Education.
SOLUTION Continued
(b) For
2.5 ,hb=
22
1(2.5 ) 1.8333
3
V a b b ab
ππ
=+=
22 2
22
22
11 1
(2.5 ) (2.5 )
2 3 12
[0.5 0.8333 0.52083]
1.85416
xV a b b b b
ab
ab
π
π
π
Σ= + +
= ++
=
2 22
: (1.8333 ) 1.85416 1.01136XV xV X ab ab X b
ππ
=Σ==
Centroid is 0.01136b to the right of base of cone.
Note: Centroid is at base of cone for
6 2.449 .
hb b= =
consent of McGraw–Hill Education.
PROBLEM 5.59
The composite body shown is formed by removing a semiellipsoid
of revolution of semimajor axis h and semiminor axis a/2 from a
hemisphere of radius a. Determine (a) the y coordinate of the
centroid when h = a/2, (b) the ratio h/a for which
y
= −0.4a.
SOLUTION
V
y
yV
Hemisphere 3
2
3
a
π
3
8a−
4
1
4a
π
−
Semiellipsoid
22
21
32 6
ah ah
ππ
−=−
3
8h−
22
1
16 ah
π
+
Then
2
222
(4 )
6
(4 )
16
V a ah
yV a a h
π
π
Σ= −
Σ=− −
Now
Y V yVΣ=Σ
So that
2 222
(4 ) (4 )
6 16
Y aah aah
ππ
−=− −
or
2
3
44
8
hh
Ya
aa
−=− −
(1)
(a)
? when 2
a
Yh
= =
Substituting
1
2
h
a=
into Eq. (1)
2
13 1
44
28 2
Ya
−=− −
or
45
112
Ya= −
0.402Ya= −
consent of McGraw–Hill Education.
SOLUTION Continued
(b)
? when 0.4
hYa
a= = −
PROBLEM 5.60
Locate the centroid of the frustum of a right circular cone when r1 = 40 mm,
r2 = 50 mm, and h = 60 mm.
SOLUTION
By similar triangles:
11 12
60 240 mm, therefore 240 60 300 mm
40 50
hh hh
+
= = = +=
( ) ( ) ( ) ( )
12 2 1
22
33
12
1 300 1 240
75 mm 60 60 120 mm
44 4 4
50 300 250 10 40 240 128 10
33
yh y h
VV
ππ
ππ
= == =+=+=
==×==×
3
mmV
mmy
4
mmyV
3
250 10
π
×
6
40 mm
50 mm
60 mm
h
1
2
h = 300 mm
50 mm
y
1
40 mm
60 mm
y
2
240 mm
consent of McGraw–Hill Education.
PROBLEM 5.61
For the machine element shown, locate the y coordinate of
the center of gravity.
SOLUTION
For half–cylindrical hole,
III
0.95 in.
4(0.95)
1.5 3
1.097 in.
r
y
π
=
= −
=
For half–cylindrical plate,
IV
1.5 in.
4(1.5)
5.25 5.887 in.
3
r
z
π
=
=+=
3
, inV
, in.y
, in.z
4
, inyV
4
, inzV