consent of McGraw–Hill Education.
SOLUTION Continued
Then applying the theorems of Pappus-Guldinus for the part of the surface area generated by the lines:
( )( ) ( )( ) ( )( )
[ ]
32
31 63.906 68 52 57 95.268 10947.6 34.392 10 mm
L
A xA
ππ π
=Σ= + + = = ×
The area of the “end triangles”:
( )( )
32
1
2 52 60 3.12 10 mm
2
E
A
= = ×
Total surface area is therefore:
( )
32
34.392 3.12 10 mm
LE
AA A=+= + ×
32
or 37.5 10 mmA= ×
consent of McGraw–Hill Education.
PROBLEM 5.48
The escutcheon (a decorative plate placed on a pipe where the
pipe exits from a wall) shown is cast from brass. Knowing that
the density of brass is 8470 kg/m3, determine the mass of the
escutcheon.
SOLUTION
The mass of the escutcheon is given by
(density) ,mV=
where V is the volume. V can be generated by
rotating the area A about the x–axis.
From the figure:
22
1
2
75 12.5 73.9510 m
37.5 76.8864 mm
tan 26
L
L
=−=
= =
°
21
1
2.9324 mm
12.5
sin 9.5941
75
26 9.5941 8.2030 0.143168 rad
2
aL L
φ
a
−
=−=
= = °
°− °
= = °=
Area A can be obtained by combining the following four areas:
Applying the second theorem of Pappus-Guldinus and using Figure 5.8a, we have
22V yA yA
ππ
= = Σ
Seg.
2
, mmA
, mmy
3
, mmyA
1(76.886)(37.5) 1441.61
1(37.5) 12.5
consent of McGraw–Hill Education.
SOLUTION Continued
Then
3
3
3 63
2
2 (3599.7 mm )
22,618 mm
(density)
(8470 kg/m )(22.618 10 m )
V yA
mV
π
π
−
= Σ
=
=
=
= ×
0.191574 kg=
or
0.1916 kgm=
PROBLEM 5.49
For the beam and loading shown, determine (a) the
magnitude and location of the resultant of the distributed
load, (b) the reactions at the beam supports.
SOLUTION
(a)
I
II
I II
1(1100N/m)(6 m) 2200 N
3
(900N/m)(6m) 5400N
2200 5400 7600N
R
R
RR R
= =
= =
=+= + =
: (7600) (2200)(1.5) (5400)(3)XR x R X=Σ=+
2.5658 mX=
7.60 kN
=R
,
2.57 mX=
(b)
0: (6 m) (7600 N)(2.5658 m) 0
3250.0 N
A
MB
B
Σ= − =
=
3.25 kN=B
0: 3250.0 N 7600 N 0
4350.0 N
y
FA
A
Σ= + − =
=
4.35 kN=A
consent of McGraw–Hill Education.
PROBLEM 5.50
For the beam and loading shown, determine (a) the magnitude
and location of the resultant of the distributed load, (b) the
reactions at the beam supports.
SOLUTION
I
II
I II
1(150lb/ft)(9 ft) 675 lb
2
1(120 lb/ft)(9ft) 540lb
2675 540 1215 lb
: (1215) (3)(675) (6)(540) 4.3333 ft
R
R
RR R
XR x R X X
= =
= =
=+= + =
=Σ =+=
(a)
1215 lb=R
4.33 ftX=
(b) Reactions:
0: (9 ft) (1215 lb)(4.3333 ft) 0
A
MBΣ= − =
585.00 lbB=
585 lb=B
0: 585.00 lb 1215 lb 0
y
FAΣ= + − =
630.00 lbA=
630 lb=A
PROBLEM 5.51
Determine the reactions at the beam supports for the given
loading.
SOLUTION
I
II
(200 lb/ft)(4 ft) 800 lb
1(150 lb/ft)(3 ft) 225 lb
2
R
R
= =
= =
0: 800 lb 225 lb 0
y
FAΣ= − + =
575 lb=A
0: (800 lb)(2 ft) (225 lb)(5 ft) 0
AA
MMΣ= − + =
475 lb ft
A= ⋅M
consent of McGraw–Hill Education.
PROBLEM 5.52
Determine the reactions at the beam supports for the given
loading.
SOLUTION
First replace the given loading with the loading shown below. The two loadings are equivalent because
both are defined by a parabolic relation between load and distance and the values at the end points are the
same.
We have
I
II
(6 m)(300 N/m) 1800 N
2(6 m)(1200 N/m) 4800 N
3
R
R
= =
= =
Then
0: 0
xx
FAΣ= =
0: 1800 N 4800 N 0
yy
FAΣ= + − =
or
3000 N
y
A=
3.00 kN=A
15
0: (3 m)(1800 N) m (4800 N) 0
4
AA
MM
Σ= + − =
or
12.6 kN m
A
M= ⋅
12.60 kN m
A
= ⋅M
consent of McGraw–Hill Education.
PROBLEM 5.53
Determine the reactions at the beam supports for the given
loading.
SOLUTION
I
II
1(4 kN/m)(6 m)
2
12 kN
(2 kN/m)(10m)
20 kN
R
R
=
=
=
=
0: 12 kN 20 kN 0
y
FAΣ= − − =
32.0 kN=A
0: (12 kN)(2 m) (20 kN)(5 m) 0
AA
MMΣ= − − =
124.0 kN m
A
= ⋅
M
PROBLEM 5.54
Determine the reactions at the beam supports for the given
loading.
SOLUTION
We have
I
II
III
1(3ft)(480 lb/ft) 720 lb
2
1(6 ft)(600 lb/ft) 1800 lb
2
(2ft)(600 lb/ft) 1200 lb
R
R
R
= =
= =
= =
Then
0: 0
xx
FBΣ= =
0: (2 ft)(720 lb) (4 ft)(1800 lb) (6 ft) (7 ft)(1200 lb) 0
By
MCΣ= − + − =
or
2360 lb
y
C=
2360 lb=C
0: 720 lb 1800 lb 2360 lb 1200 lb 0
yy
FBΣ= − + − + − =
or
1360 lb
y
B=
1360 lb=B
SOLUTION Continued
Then applying the theorems of Pappus-Guldinus for the part of the surface area generated by the lines:
( )( ) ( )( ) ( )( )
[ ]
32
31 63.906 68 52 57 95.268 10947.6 34.392 10 mm
L
A xA
ππ π
=Σ= + + = = ×
The area of the “end triangles”:
( )( )
32
1
2 52 60 3.12 10 mm
2
E
A
= = ×
Total surface area is therefore:
( )
32
34.392 3.12 10 mm
LE
AA A=+= + ×
32
or 37.5 10 mmA= ×
consent of McGraw–Hill Education.
PROBLEM 5.48
The escutcheon (a decorative plate placed on a pipe where the
pipe exits from a wall) shown is cast from brass. Knowing that
the density of brass is 8470 kg/m3, determine the mass of the
escutcheon.
SOLUTION
The mass of the escutcheon is given by
(density) ,mV=
where V is the volume. V can be generated by
rotating the area A about the x–axis.
From the figure:
22
1
2
75 12.5 73.9510 m
37.5 76.8864 mm
tan 26
L
L
=−=
= =
°
21
1
2.9324 mm
12.5
sin 9.5941
75
26 9.5941 8.2030 0.143168 rad
2
aL L
φ
a
−
=−=
= = °
°− °
= = °=
Area A can be obtained by combining the following four areas:
Applying the second theorem of Pappus-Guldinus and using Figure 5.8a, we have
22V yA yA
ππ
= = Σ
Seg.
2
, mmA
, mmy
3
, mmyA
1(76.886)(37.5) 1441.61
1(37.5) 12.5
consent of McGraw–Hill Education.
SOLUTION Continued
Then
3
3
3 63
2
2 (3599.7 mm )
22,618 mm
(density)
(8470 kg/m )(22.618 10 m )
V yA
mV
π
π
−
= Σ
=
=
=
= ×
0.191574 kg=
or
0.1916 kgm=
PROBLEM 5.49
For the beam and loading shown, determine (a) the
magnitude and location of the resultant of the distributed
load, (b) the reactions at the beam supports.
SOLUTION
(a)
I
II
I II
1(1100N/m)(6 m) 2200 N
3
(900N/m)(6m) 5400N
2200 5400 7600N
R
R
RR R
= =
= =
=+= + =
: (7600) (2200)(1.5) (5400)(3)XR x R X=Σ=+
2.5658 mX=
7.60 kN
=R
,
2.57 mX=
(b)
0: (6 m) (7600 N)(2.5658 m) 0
3250.0 N
A
MB
B
Σ= − =
=
3.25 kN=B
0: 3250.0 N 7600 N 0
4350.0 N
y
FA
A
Σ= + − =
=
4.35 kN=A
consent of McGraw–Hill Education.
PROBLEM 5.50
For the beam and loading shown, determine (a) the magnitude
and location of the resultant of the distributed load, (b) the
reactions at the beam supports.
SOLUTION
I
II
I II
1(150lb/ft)(9 ft) 675 lb
2
1(120 lb/ft)(9ft) 540lb
2675 540 1215 lb
: (1215) (3)(675) (6)(540) 4.3333 ft
R
R
RR R
XR x R X X
= =
= =
=+= + =
=Σ =+=
(a)
1215 lb=R
4.33 ftX=
(b) Reactions:
0: (9 ft) (1215 lb)(4.3333 ft) 0
A
MBΣ= − =
585.00 lbB=
585 lb=B
0: 585.00 lb 1215 lb 0
y
FAΣ= + − =
630.00 lbA=
630 lb=A
PROBLEM 5.51
Determine the reactions at the beam supports for the given
loading.
SOLUTION
I
II
(200 lb/ft)(4 ft) 800 lb
1(150 lb/ft)(3 ft) 225 lb
2
R
R
= =
= =
0: 800 lb 225 lb 0
y
FAΣ= − + =
575 lb=A
0: (800 lb)(2 ft) (225 lb)(5 ft) 0
AA
MMΣ= − + =
475 lb ft
A= ⋅M
consent of McGraw–Hill Education.
PROBLEM 5.52
Determine the reactions at the beam supports for the given
loading.
SOLUTION
First replace the given loading with the loading shown below. The two loadings are equivalent because
both are defined by a parabolic relation between load and distance and the values at the end points are the
same.
We have
I
II
(6 m)(300 N/m) 1800 N
2(6 m)(1200 N/m) 4800 N
3
R
R
= =
= =
Then
0: 0
xx
FAΣ= =
0: 1800 N 4800 N 0
yy
FAΣ= + − =
or
3000 N
y
A=
3.00 kN=A
15
0: (3 m)(1800 N) m (4800 N) 0
4
AA
MM
Σ= + − =
or
12.6 kN m
A
M= ⋅
12.60 kN m
A
= ⋅M
consent of McGraw–Hill Education.
PROBLEM 5.53
Determine the reactions at the beam supports for the given
loading.
SOLUTION
I
II
1(4 kN/m)(6 m)
2
12 kN
(2 kN/m)(10m)
20 kN
R
R
=
=
=
=
0: 12 kN 20 kN 0
y
FAΣ= − − =
32.0 kN=A
0: (12 kN)(2 m) (20 kN)(5 m) 0
AA
MMΣ= − − =
124.0 kN m
A
= ⋅
M
PROBLEM 5.54
Determine the reactions at the beam supports for the given
loading.
SOLUTION
We have
I
II
III
1(3ft)(480 lb/ft) 720 lb
2
1(6 ft)(600 lb/ft) 1800 lb
2
(2ft)(600 lb/ft) 1200 lb
R
R
R
= =
= =
= =
Then
0: 0
xx
FBΣ= =
0: (2 ft)(720 lb) (4 ft)(1800 lb) (6 ft) (7 ft)(1200 lb) 0
By
MCΣ= − + − =
or
2360 lb
y
C=
2360 lb=C
0: 720 lb 1800 lb 2360 lb 1200 lb 0
yy
FBΣ= − + − + − =
or
1360 lb
y
B=
1360 lb=B