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5/13/2018 Ch07 Distributed Forces Centroids and Centers of Gravity 2 - slidepdf.com
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© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 1
CE 102 Statics
Chapter 7
Distributed Forces:
Centroids and Centers of Gravity
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 2
Contents
Introduction
Center of Gravity of a 2D Body
Centroids and First Moments of Areas
and Lines
Centroids of Common Shapes of Areas
Centroids of Common Shapes of Lines
Composite Plates and Areas
Sample Problem 7.1
Determination of Centroids by
Integration
Sample Problem 7.2
Theorems of Pappus-Guldinus
Sample Problem 7.3
Distributed Loads on Beams
Sample Problem 7.4
Center of Gravity of a 3D Body:Centroid of a Volume
Centroids of Common 3D Shapes
Composite 3D Bodies
Sample Problem 7.5
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 3
Introduction
The earth exerts a gravitational force on each of the particlesforming a body. These forces can be replace by a single
equivalent force equal to the weight of the body and applied
at the center of gravity for the body.
The centroid of an area is analogous to the center of gravity of a body. The concept of the first moment of an
area is used to locate the centroid.
Determination of the area of a surface of revolution and
the volume of a body of revolution are accomplished
with the Theorems of Pappus-Guldinus.
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 4
Center of Gravity of a 2D Body
Center of gravity of a plate
´
§§´§§
!
(!
!(!
dW y
W yW y M
dW x
W xW x M
y
y
Center of gravity of a wire
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 5
Centroids and First Moments of Areas and Lines
x
QdA y A y
y
QdA x A x
dAt x At x
dW xW x
x
y
respect tohmoment witfirst
respect tohmoment witfirst
!
!!
!!!
!
!
´
´
´´
KK
Centroid of an area
´´
´´
!!
!
!
dL y L y
dL x L x
dLa x La x
dW xW x
KK
Centroid of a line
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 6
First Moments of Areas and Lines
An area is symmetric with respect to an axis BB¶
if for every point P there exists a point P ¶ such
that PP ¶ is perpendicular to BB¶ and is divided
into two equal parts by BB¶.
The first moment of an area with respect to a
line of symmetry is zero.
If an area possesses a line of symmetry, its
centroid lies on that axis
If an area possesses two lines of symmetry, its
centroid lies at their intersection.
An area is symmetric with respect to a center O
if for every element dA at ( x,y) there exists an
area dA¶ of equal area at (-x,-y).
The centroid of the area coincides with the
center of symmetry.
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 7
Centroids of Common Shapes of Areas
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 8
Centroids of Common Shapes of Lines
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 9
Composite Plates and Areas
Composite plates
§§§§
!
!
W yW Y
W xW X
Composite area
§§ §§ !! A y AY A x A X
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 10
Sample Problem 7.1
For the plane area shown, determine
the first moments with respect to the
x and y axes and the location of the
centroid.
SOLUTION:
Divide the area into a triangle, rectangle,
and semicircle with a circular cutout.
Compute the coordinates of the area
centroid by dividing the first moments by
the total area.
Find the total area and first moments of
the triangle, rectangle, and semicircle.
Subtract the area and first moment of the
circular cutout.
Calculate the first moments of each area
with respect to the axes.
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 11
Sample Problem 7.1
33
33
mm107.757
mm102.506
v!
v!
y
x
Q
Q Find the total area and first moments of the
triangle, rectangle, and semicircle. Subtract the
area and first moment of the circular cutout.
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 12
Sample Problem 7.1
2333
mm1013.828
mm107.757
v
v!!§
§ A A x X
mm8.54! X
23
33
mm1013.828
mm102.506
v
v!!
§§
A
A yY
mm6.36!Y
Compute the coordinates of the areacentroid by dividing the first moments by
the total area.
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 13
Determination of Centroids by Integration
ydx y
dA y A y
ydx x
dA x A x
el
el
´
´´
´
!
!
!
!
2
? A
? Adx xa y
dA y A y
dx xa xa
dA x A x
el
el
!
!
!
!
´´
´
´
2
¹
º
¸©
ª
¨!
!
¹ º ¸©
ª¨!
!
´
´
´
´
UU
UU
d r r
dA y A y
d r r
dA x A x
el
el
2
2
2
1sin
3
2
2
1cos
3
2
´´´´ ´´´´ !!!
!!!
dA ydydx ydA y A y
dA xdydx xdA x A x
el
el Double integration to find the first moment
may be avoided by defining dA as a thinrectangle or strip.
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 14
Sample Problem 7.2
Determine by direct integration the
location of the centroid of a parabolic
spandrel.
SOLUTION:
Determine the constant k.
Evaluate the total area.
Using either vertical or horizontal
strips, perform a single integration to
find the first moments.
Evaluate the centroid coordinates.
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 15
Sample Problem 7.2
SOLUTION:
Determine the constant k.
21
21
2
2
2
2
2
yb
a xor x
a
b y
a
bk ak b
xk y
!!
!!
!
Evaluate the total area.
3
30
3
20
2
2
ab
x
a
bdx x
a
bdx y
dA A
aa
!
¼¼½
»
¬¬-
«!!!
!
´´
´
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 16
Sample Problem 7.2
Using vertical strips, perform a single integrationto find the first moments.
1052
2
1
2
44
2
0
5
4
2
0
22
2
2
0
4
2
0
2
2
ab x
a
b
dx xa
bdx y
ydA yQ
ba xab
dx xa
b xdx xydA xQ
a
a
el x
a
a
el y
!¼¼½
»
¬¬-
«!
¹ º
¸©ª
¨!!!
!¼¼½»
¬¬-«!
¹ º
¸©ª
¨!!!
´´´
´´´
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 17
Sample Problem 7.2
Or, using horizontal strips, perform a singleintegration to find the first moments.
10
421
22
2
0
2321
21
21
2
0
22
0
22
abdy yb
aay
dy yb
aa ydy xa ydA yQ
bady yb
aa
dy xa
dy xa xa
dA xQ
b
el x
b
b
el y
!¹ º ¸©
ª¨ !
¹ º
¸©ª
¨!!!
!¹¹ º ¸©©
ª¨ !
!
!!
´
´´´
´
´´´
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 18
Sample Problem 7.2
E
valuate the centroid coordinates.
43
2baab
x
Q A x y
!
!
a x4
3!
103
2abab y
Q A y x
!
!
b y10
3!
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 19
Theorems of Pappus-Guldinus
Surface of revolution is generated by rotating a
plane curve about a fixed axis.
Area of a surface of revolution is
equal to the length of the generatingcurve times the distance traveled by
the centroid through the rotation.
L y A T2!
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 20
Theorems of Pappus-Guldinus
Body of revolution is generated by rotating a plane
area about a fixed axis.
Volume of a body of revolution is
equal to the generating area timesthe distance traveled by the centroid
through the rotation.
A yV T2!
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 21
Sample Problem 7.3
The outside diameter of a pulley is 0.8
m, and the cross section of its rim is as
shown. Knowing that the pulley is
made of steel and that the density of steel is
determine the mass and weight of the
rim.
33mkg1085.7 v! V
SOLUTION:
Apply the theorem of Pappus-Guldinus
to evaluate the volumes or revolution
for the rectangular rim section and the
inner cutout section.
Multiply by density and accelerationto get the mass and acceleration.
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 22
Sample Problem 7.3
SOLUTION:
Apply the theorem of Pappus-Guldinus
to evaluate the volumes or revolution for
the rectangular rim section and the inner
cutout section.
¹ º
¸©ª
¨vv!! 3393633 mmm10mm1065.7mkg1085.7V m V kg0.60!m
2
sm81.9kg0.60!! mg W N589!W
Multiply by density and acceleration toget the mass and acceleration.
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 23
Distributed Loads on Beams
A distributed load is represented by plotting the load
per unit length, w (N/m) . The total load is equal to
the area under the load curve (dW = wdx).
´´ !!! AdAdxwW L
0
A xdA x AO P
dW xW O P L
!!
!
´
´
0
A distributed load can be replace by a concentratedload with a magnitude equal to the area under the
load curve and a line of action passing through the
area centroid.
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 24
Sample Problem 7.4
A beam supports a distributed load as
shown. Determine the equivalent
concentrated load and the reactions atthe supports.
SOLUTION:
The magnitude of the concentrated load
is equal to the total load or the area under
the curve.
The line of action of the concentrated
load passes through the centroid of the
area under the curve.
Determine the support reactions by
summing moments about the beam
ends.
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 25
Sample Problem 7.4
SOLUTION:
The magnitude of the concentrated load is equal to
the total load or the area under the curve.
kN0.18! F
The line of action of the concentrated load passesthrough the centroid of the area under the curve.
kN18
mkN63 ! X m5.3! X
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 26
Sample Problem 7.4
Determine the support reactions by summingmoments about the beam ends.
0m.53kN18m6:0 !!§ y A B M
kN5.10! y B
0m.53m6kN18m6:0 !!§ y B A M
kN5.7! y A
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 27
Center of Gravity of a 3D Body: Centroid of a Volume
Center of gravity G
§ (! jW jW TT
? A jW r jW r
jW r jW r
G
GTTTT
TTTT
v(!v
(v!v
§§
´´ !! dW r W r dW W GTT
R esults are independent of body orientation,
´´´ !!! zdW W z ydW W y xdW W x
´´´ !!! zdV V z ydV V y xdV V x
dV dW V W KK !! and
For homogeneous bodies,
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 28
Centroids of Common 3D Shapes
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 29
Composite 3D Bodies
Moment of the total weight concentrated at thecenter of gravity G is equal to the sum of the
moments of the weights of the component parts.
§§§§§§ !!! W zW Z W yW Y W xW X
For homogeneous bodies,
§§§§§§ !!! V zV Z V yV Y V xV X
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
5 - 30
Sample Problem 7.5
Locate the center of gravity of thesteel machine element. The diameter
of each hole is 1 in.
SOLUTION:
Form the machine element from a
rectangular parallelepiped and a
quarter cylinder and then subtracting
two 1-in. diameter cylinders.
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Sample Problem 7.5
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Vector Mechanics for Engineers: StaticsE i gh t h
E d i t i on
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Sample Problem 7.5
34 in.2865in08.3!! §§ V V x X
34 in.2865in5.047!!
§§V V yY
34in.2865in.6181!! §§ V V z Z
in.577.0! X
in.577.0!Y
in.577.0! Z
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33
y
x
20 mm 30 mm
Problem 7.6
36 mm
24 mm
Locate the centroid of the plane
area shown.
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34
Solving Problems on Your Own
Several points should be emphasized
when solving these types of problems.
Locate the centroid of the plane area
shown.
y
x
20 mm 30 mm
36 mm
24 mm
1. Decide how to construct the given area from common shapes.
2. It is strongly recommended that you construct a table
containing areas or length and the respective coordinates of the centroids.
3. When possible, use symmetry to help locate the centroid .
Problem 7.6
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35
Problem 7.6 Solution y
x
24 + 12
20 + 10
10
30
Decide how to construct the given
area from common shapes.C
1
C 2
Dimensions in mm
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36
Problem 7.6 Solution y
x
24 + 12
20 + 10
10
30
C 1
C 2
Dimensions in mm
C onstruct a table containing areas and
respective coordinates of thecentroids.
A, mm2 x, mm y, mm xA, mm3 yA, mm3
1 20 x 60 =1200 10 30 12,000 36,0002 (1/2) x 30 x 36 =540 30 36 16,200 19,440
7 1740 28,200 55,440
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37
Problem 7.6 Solution y
x
24 + 12
20 + 10
10
30
C 1
C 2
Dimensions in mm
X7 A = 7 xA
X(1740) = 28,200
Then
or X = 16.21 mm
and Y7 A = 7 yA
Y (1740) = 55,440
or Y = 31.9 mm
A, mm2 x, mm y, mm xA, mm3 yA, mm3
1 20 x 60 =1200 10 30 12,000 36,000
2 (1/2) x 30 x 36 =540 30 36 16,200 19,440
7 1740 28,200 55,440
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38
Problem 7.7
The beam AB supports twoconcentrated loads and
rests on soil which exerts a
linearly distributed upward
load as shown. Determine
(a) the distance a for which
w A = 20 kN/m, (b) the
corresponding value w B.
w A w
B
A B
a 0.3 m24 kN 30 kN
1.8 m
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39
The beam AB supports two
concentrated loads andrests on soil which exerts a
linearly distributed upward
load as shown. Determine
(a) the distance a for whichw A = 20 kN/m, (b) the
corresponding value w B.
w A w
B
A B
a 0.3 m24 kN 30 kN
1.8 m
Solving Problems on Your Own
1. R eplace the distributed load by a single equivalent force.
The magnitude of this force is equal to the area under the
distributed load curve and its line of action passes through
the centroid of the area.
2. When possible, complex distributed loads should be
divided into common shape areas.
Problem 7.7
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Problem 7.7 Solution
w B
A B
a 0.3 m24 kN 30 kN
20 kN/m
C
0.6 m 0.6 m
R I R II
We have RI= (1.8 m)(20 kN/m) = 18 kN
1
2
RII= (1.8 m)(w
BkN/m) = 0.9 w
BkN
1
2
R eplace the distributed
load by a pair of
equivalent forces.
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Problem 7.7 Solution
w B
A B
a 0.3 m24 kN 30 kN
C
0.6 m 0.6 m
R I = 18 kN
R II
= 0.9 w B
kN
(a) 7 M C = 0: (1.2 - a)m x 24 kN - 0.6 m x 18 kN
- 0.3m x 30 kN = 0
or a = 0.375 m
(b) 7 F y = 0: -24 kN + 18 kN + (0.9 w B
) kN - 30 kN= 0
or w B
= 40 kN/m
+
+
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Problem 7.8
y
x
0.75 in z
1 in2 in2 in
3 in
2 in2 in
r = 1.25 in
r = 1.25 in
For the machine element
shown, locate the z coordinateof the center of gravity.
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Solving Problems on Your Own
y
x
0.75 in z
1 in2 in2 in
3 in
2 in2 in
r = 1.25 in
r = 1.25 in
X 7V = 7 x V Y 7V = 7 y V Z 7V = 7 z V
where X, Y, Z and x, y, z are the coordinates of the centroid of the
body and the components, respectively.
For the machine elementshown, locate the z coordinate
of the center of gravity.
Problem 7.8
Determine the center of
gravity of composite body.For a homogeneous body
the center of gravity coincides
with the centroid of its volume. For this case the center of gravity
can be determined by
8 S
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Problem 7.8 Solution y
x
0.75 in z
1 in2 in2 in
3 in
2 in2 in
r = 1.25 in
r = 1.25 in
Determine the center of gravity
of composite body.
First assume that the machine
element is homogeneous so
that its center of gravity will
coincide with the centroid of
the corresponding volume.
y
x
z
I
IIIII
IVV
Divide the body into
five common shapes.
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y
x
z
I
IIIII
IVV
y
x
0.75 in z
1 in2 in2 in
3 in
2 in2 in
r = 1.25 in
r = 1.25 in
V, in3 z, in. z V, in4
I (4)(0.75)(7) = 21 3.5 73.5
II (T/2)(2)2 (0.75) = 4.7124 7+ [(4)(2)/(3T)] = 7.8488 36.987
III -T(11.25)2
(0.75)= -3.6816 7 -25.771IV (1)(2)(4) = 8 2 16
V -(T/2)(1.25)2 (1) = -2.4533 2 -4.9088
7 27.576 95.807
Z 7V = 7 z V : Z (27.576 in3 ) = 95.807 in4 Z = 3.47 in
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Problem 7.9
Locate the centroid of the volume
obtained by rotating the shaded area
about the x axis.
h
a
x
y y = kx1/3
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Solving Problems on Your Own
1. When possible, use symmetry to help locate the centroid .
Locate the centroid of the volume
obtained by rotating the shaded area
about the x axis.
h
a
y = kx1/3
x
y
The procedure for locating the
centroids of volumes by direct
integration can be simplified:
2. If possible, identify an element of volume dV which produces
a single or double integral , which are easier to compute.
3. After setting up an expression for dV , integrate and determine
the centroid.
Problem 7.9
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Problem 7.9 Solution
x
y
z x
dx
y = kx1/3
r
U se symmetry to help locate the
centroid . Symmetry implies
y = 0 z = 0
I dentify an element of volume dV
which produces a single or double integral.
Choose as the element of volume a disk or radius r and
thickness dx . Then
dV = Tr 2 dx xel
= x
S
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Problem 7.9 Solution
x
y
z x
dx
y = kx1/3
r
I dentify an element of volume dV
which produces a single or
double integral.
dV = Tr 2 dx xel
= x
Now r = kx 1/3 so that
dV = Tk 2 x2/3dx
At x = h, y = a : a = kh1/3 or k = a/h1/3
Then dV = T x2/3dxa2
h2/3
P bl 7 9 S l ti
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Problem 7.9 Solution
x
y
z x
dx
y = kx1/3
r
dV = T x2/3
dxa2
h2/3
I ntegrate and determine the centroid .
´0
h
V = T x2/3dxa2
h2/3
´
= T x5/3a2
h2/335[ ]
0
h
= Ta2h3
5
´0
h
xel
dV = x (T x2/3 dx) = T [ x8/3 ]a2
h2/3a2
h2/338
Also
= Ta2h23
8
P bl 7 9 S l ti
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Problem 7.9 Solution
x
y
z x
dx
y = kx1/3
r
I ntegrate and determine the centroid .
´
V = Ta2
h
3
5
xel
dV = Ta2h23
8
Now xV = xdV :´ 3
8 x ( Ta2h) = Ta2h23
5
x = h5
8
y = 0 z = 0
P bl 7 10
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Problem 7.10
The square gate AB is held in theposition shown by hinges along its
top edge A and by a shear pin at B.
For a depth of water d = 3.5 ft,
determine the force exerted on the
gate by the shear pin.
A
B
d
1.8 ft
30o
Problem 7 10
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Solving Problems on Your Own
The square gate AB is held in the
position shown by hinges along itstop edge A and by a shear pin at B.
For a depth of water d = 3.5 ft,
determine the force exerted on the
gate by the shear pin.
A
B
d
1.8 ft
30o
Assuming the submerged body has a width b, the load per unit
length is w = b Vgh, where h is the distance below the surface of
the fluid.
1. First, determine the pressure distribution acting perpendicular
the surface of the submerged body. The pressure distribution
will be either triangular or trapezoidal.
Problem 7.10
Problem 7 10
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Solving Problems on Your Own
The square gate
AB is held in theposition shown by hinges along its
top edge A and by a shear pin at B.
For a depth of water d = 3.5 ft,
determine the force exerted on the
gate by the shear pin.
A
B
d
1.8 ft
30o
2. R eplace the pressure distribution with a resultant force, and
construct the free-body diagram.
3. Write the equations of static equilibrium for the problem, and
solve them.
Problem 7.10
Problem 7 10 Solution
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Problem 7.10 Solution
A
B
Determine the pressure distribution
acting perpendicular the surface of thesubmerged body.
1.7 ft
(1.8 ft) cos 30o
P A
P B
P A
= 1.7 V g
P B = (1.7 + 1.8 cos 30o) V g
Problem 7 10 SolutionA
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Problem 7.10 Solution
A
B
(1.8 ft) cos 30o
R eplace the pressure
distribution with a
resultant force, and construct the free-body
diagram.
A y
A x
F B
1.7 V g
(1.7 + 1.8 cos 30o) V g
L AB
/3
L AB
/3
L AB
/3
P1
P2
The force of the water
on the gate is
P = Ap = A( Vgh)1
21
2
P 1 = (1.8 ft)2(62.4 lb/ft3)(1.7 ft) = 171.85 lb1
2
P 2 = (1.8 ft)2(62.4 lb/ft3)(1.7 + 1.8 cos 30o)ft = 329.43 lb1
2
Problem 7 10 SolutionA
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Problem 7.10 Solution
A
B
(1.8 ft) cos 30o
A y
A x
F B
1.7 V g
(1.7 + 1.8 cos 30o) V g
L AB
/3
L AB
/3
L AB
/3
P1
P2
P 1 = 171.85 lb P 2 = 329.43 lb
Write the equations of
static equilibrium for the
problem, and solve them.
7 M A = 0:
( L A B) P 1 + ( L A B) P 213
23
- L A B F B = 0
+
(171.85 lb) + (329.43 lb) - F B = 013
2
3 F B = 276.90 lb
30o
FB = 277 lb