CE 102 Statics. Chapter 7 Distributed Forces: Centroids and Centers of Gravity. 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
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Chapter 7
Distributed Forces:
Centroids and Centers of Gravity
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
For the plane area shown, determine the first moments with respect to the x and y axes and the location of the centroid.
Determine by direct integration the location of the centroid of a parabolic spandrel.
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.
A beam supports a distributed load as shown. Determine the equivalent concentrated load and the reactions at the supports.
Locate the center of gravity of the steel machine element. The diameter of each hole is 1 in.
Problem 7.6
y
20 mm
30 mm
Locate the centroid of the plane
area shown.
36 mm
24 mm
x
y
Problem 7.6
20 mm
30 mm
Solving Problems on Your Own
Locate the centroid of the plane area shown.
36 mm
Several points should be emphasized when solving these types of problems.
24 mm
x
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 Solution
y
20 + 10
Decide how to construct the given area from common shapes.
C1
C2
24 + 12
30
x
10
Dimensions in mm
Problem 7.6 Solution
y
20 + 10
Construct a table containing areas and respective coordinates of the centroids.
C1
C2
24 + 12
30
x
10
Dimensions in 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
S 1740 28,200 55,440
Problem 7.6 Solution
y
20 + 10
XSA = S xA
Then
X (1740) = 28,200
X = 16.21 mm
or
C1
C2
YSA = S yA
and
24 + 12
30
Y (1740) = 55,440
x
10
Y = 31.9 mm
or
Dimensions in 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
S 1740 28,200 55,440
Problem 7.7
The beam AB supports two concentrated loads and rests on soil which exerts a linearly distributed upward load as shown. Determine (a) the distance a for which wA = 20 kN/m, (b) the corresponding value wB.
24 kN
30 kN
a
0.3 m
A
B
wA
wB
1.8 m
Problem 7.7
Solving Problems on Your Own
24 kN
30 kN
a
0.3 m
The beam AB supports two concentrated loads and rests on soil which exerts a linearly distributed upward load as shown. Determine (a) the distance a for which wA = 20 kN/m, (b) the corresponding value wB.
A
B
wA
wB
1.8 m
1. Replace 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.
1
2
1
2
Problem 7.7 Solution
24 kN
30 kN
a
0.3 m
Replace the distributed
load by a pair of
equivalent forces.
C
A
B
20 kN/m
wB
0.6 m
0.6 m
RI
RII
RI = (1.8 m)(20 kN/m) = 18 kN
We have
RII = (1.8 m)(wB kN/m) = 0.9 wB kN
+
+
Problem 7.7 Solution
24 kN
30 kN
a
0.3 m
C
A
B
wB
0.6 m
0.6 m
RII = 0.9 wB kN
RI = 18 kN
(a)
SMC = 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)
SFy = 0: -24 kN + 18 kN + (0.9 wB) kN - 30 kN= 0
or wB = 40 kN/m
y
2 in
1 in
2 in
3 in
r = 1.25 in
x
z
0.75 in
2 in
r = 1.25 in
2 in
Problem 7.8
For the machine element
shown, locate the z coordinate
of the center of gravity.
y
Problem 7.8
2 in
1 in
2 in
3 in
r = 1.25 in
x
z
0.75 in
2 in
r = 1.25 in
2 in
Solving Problems on Your Own
For the machine element
shown, locate the z coordinate
of the center of gravity.
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
X S V = S x V Y S V = S y V Z S V = S z V
where X, Y, Z and x, y, z are the coordinates of the centroid of the body and the components, respectively.
y
Problem 7.8 Solution
2 in
1 in
2 in
3 in
r = 1.25 in
x
z
0.75 in
2 in
r = 1.25 in
2 in
y
V
IV
I
III
x
II
z
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.
Divide the body into five common shapes.
y
y
2 in
1 in
V
2 in
IV
3 in
r = 1.25 in
I
III
x
II
x
z
0.75 in
z
2 in
r = 1.25 in
2 in
V, in3z, in. z V, in4
I (4)(0.75)(7) = 213.5 73.5
II (p/2)(2)2 (0.75) = 4.71247+ [(4)(2)/(3p)] = 7.8488 36.987
III -p(11.25)2 (0.75)= -3.6816 7 -25.771
IV (1)(2)(4) = 8 2 16
V -(p/2)(1.25)2 (1) = -2.4533 2 -4.9088
S27.576 95.807
Z S V = S z V : Z (27.576 in3 ) = 95.807 in4Z = 3.47 in
Problem 7.9
y
y = kx1/3
Locate the centroid of the volume obtained by rotating the shaded area about the x axis.
a
x
h
y
Problem 7.9
y = kx1/3
Solving Problems on Your Own
Locate the centroid of the volume obtained by rotating the shaded area about the x axis.
a
The procedure for locating the
centroids of volumes by direct
integration can be simplified:
x
h
1. When possible, use symmetry to help locate the centroid.
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 Solution
y
Use symmetry to help locate the centroid. Symmetry implies
x
dx
y = 0 z = 0
x
z
r
Identify an element of volume dV which produces a single or double integral.
y = kx1/3
Choose as the element of volume a disk or radius r and thickness dx. Then
dV = p r2 dx xel = x
Problem 7.9 Solution
y
x
Identify an element of volume dV which produces a single or double integral.
dx
dV = p r2 dx xel = x
x
z
r
r = kx1/3
so that
Now
dV = p k2 x2/3dx
y = kx1/3
a = kh1/3
k = a/h1/3
or
At x = h, y = a :
a2
h2/3
dV = p x2/3dx
Then
a2
h2/3
dV = p x2/3dx
3
5
h
ò
0
3
8
= p a2h2
Problem 7.9 Solution
y
Integrate and determine the centroid.
x
dx
a2
h2/3
h
ò
V = p x2/3dx
x
z
r
0
a2
h2/3
h
[ ]
3
5
= p x5/3
y = kx1/3
0
= p a2h
a2
h2/3
a2
h2/3
3
8
ò
xel dV = x (p x2/3 dx) = p [ x8/3 ]
Also
3
5
3
5
5
8
x = h
Problem 7.9 Solution
y
Integrate and determine the centroid.
x
dx
V = p a2h
3
8
ò
xel dV = p a2h2
x
z
r
y = kx1/3
3
8
ò
xV = xdV:
x ( p a2h) = p a2h2
Now
y = 0 z = 0
Problem 7.10
The square gate AB is held in the position 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
d
1.8 ft
B
30o
Problem 7.10
Solving Problems on Your Own
The square gate AB is held in the position 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
d
1.8 ft
B
30o
Assuming the submerged body has a width b, the load per unit length is w = brgh, where h is the distance below the surface of the fluid.
1. First, determine the pressure distributionacting perpendicular
the surface of the submerged body. The pressure distribution
will be either triangular or trapezoidal.
Problem 7.10
Solving Problems on Your Own
The square gate AB is held in the position 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
d
1.8 ft
B
30o
2. Replace 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 Solution
Determine the pressure distribution
acting perpendicular the surface of the submerged body.
1.7 ft
A
PA
PA = 1.7 rg
PB = (1.7 + 1.8 cos 30o)rg
(1.8 ft) cos 30o
B
PB
1
2
1
2
1
2
1
2
Problem 7.10 Solution
Ay
A
Replace the pressure distribution with a resultant force, and construct the free-body diagram.
Ax
1.7 rg
(1.8 ft) cos 30o
LAB/3
P1
FB
The force of the water on the gate is
LAB/3
P2
B
LAB/3
(1.7 + 1.8 cos 30o)rg
P = Ap = A(rgh)
P1 = (1.8 ft)2(62.4 lb/ft3)(1.7 ft) = 171.85 lb
P2 = (1.8 ft)2(62.4 lb/ft3)(1.7 + 1.8 cos 30o)ft = 329.43 lb
30o
Problem 7.10 Solution
Ay
A
Ax
Write the equations of
static equilibrium for the
problem, and solve them.
1.7 rg
(1.8 ft) cos 30o
LAB/3
P1
FB
S MA = 0:
+
LAB/3
P2
B
2
3
1
3
( LAB)P1 + ( LAB)P2
LAB/3
(1.7 + 1.8 cos 30o)rg
- LABFB = 0
P1 = 171.85 lb
P2 = 329.43 lb
2
3
1
3
(171.85 lb) + (329.43 lb) - FB = 0
FB = 276.90 lb
FB = 277 lb