Fluid Mechanics Chapter 8 – Open Channel Flow 8 8 O OP PE EN N C CH HA AN NN NE EL L F FL LO OW W
Chapter 8 Engineering Mechanics: Statics
-
Upload
tup-manila -
Category
Documents
-
view
0 -
download
0
Transcript of Chapter 8 Engineering Mechanics: Statics
9.1
Chapter 9: Center of Gravity and Centroid
Chapter Objectives
• To discuss the concept of the center of gravity, center of mass, and the centroid.
• To show how to determine the location of the centroid for a body of arbitrary
shape.
• To use the Theorems of Pappus and Guldinus for finding the surface area and
volume for a body having axial symmetry.
• To represent the method for finding the resultant of a general distributed loading.
9.1 Center of Gravity, Center of Mass, and the Centroid of a Body
Center of Gravity
Contrast the difference between the “center of gravity” and “centroid” using a
composite consisting of Styrofoam and lead.
Center of gravity Centroid
• “Center of gravity” is a function of specific weight.
• “Centroid” is a function of dimensions; that is, it is a geometric property only.
The earth (that is, gravity) exerts a force on each particle forming a rigid body.
• The summation of these forces adds up to the total weight of the rigid body W.
∑Fz : W = ∆W 1 + ∆W2 + ∆W3 + • • • + ∆Wn
9.2
• The point of application (i.e. the center of gravity) of the total weight of the
rigid body W may be found using the “principle of moments.”
∑My : x W = x 1 ∆W1 + x 2 ∆W2 + x 3 ∆W3 + • • • + x n ∆Wn
∑Mx : y W = y 1 ∆W1 + y 2 ∆W2 + y 3 ∆W3 + • • • + y n ∆Wn
x = x W/W y = y W/W
If we increase the number of elements into which the “plate” (or rigid body) is
divided and decrease the size of each element, we obtain the following equations.
W = ∫dW x W = ∫x dW y W = ∫ y dW
• The coordinates (x and y ) define the center of gravity of the plate (or of the
rigid body).
Center of Mass of a Body
Center of mass is a function of density.
Centroid of a Volume
The centroid defines the geometric center of an object.
Centroid of an Area
• In the case of a homogeneous plate of uniform thickness, the magnitude ∆W is
∆W = ∂ t ∆A
where ∂ = specific weight (weight per unit volume)
t = thickness of the plate
∆A = area of the element
• The weight of the entire plate is W = ∂ t A
where A = total area of the plate
• If we substitute these into the previous equations and divide through by the
constant “ ∂ t “:
∑My : x A = x 1 ∆A1 + x 2 ∆A2 + x 3 ∆A3 + • • • + x n ∆An
∑Mx : y A = y 1 ∆A1 + y 2 ∆A2 + y 3 ∆A3 + • • • + y n ∆An
x = x A/A y = y A/A
9.3
If we increase the number of elements
into which the “plate” (or rigid body) is
divided and decrease the size of each
element, we obtain the following
equations.
A = ∫dA
x A = ∫x dA
y A = ∫ y dA
If the plate is not homogeneous, these equations cannot be used to determine the
“center of gravity” of the plate.
• However, the equations still define the “centroid” of the area.
Centroid of a Line
For a wire of uniform cross section, the magnitude ∆W of the weight of an
element of wire may be expressed as follows.
∆W = ∂ a ∆L
where
∂ = specific weight of the material
a = cross sectional area of the wire
∆L = length of the element
and
L = ∫ dL
x L = ∫ x dL
y L = ∫ y dL
For a wire of uniform cross section and with
a constant specific weight, the center of
gravity of the wire coincides with the
centroid C.
Symmetry
• When an area or line possesses an axis of symmetry (that is, an axis that causes
a mirror image on either side of the axis) the centroid of the area or line must
be located on that axis.
9.4
• If an area or line possesses two axes of symmetry, then the centroid of that
area or line is located at the intersection of the two axes of symmetry, and the
following is true.
x = y = 0
First Moments of Areas and Lines
• The integral ∫ x dA is known as the “first moment of the area A with respect
to the y-axis” and is denoted by Qy.
Qy = ∫ x dA
• Similarly, the integral ∫ y dA is known as the “first moment of the area A with
respect to the x-axis” and is denoted by Qx.
Qx = ∫ y dA
• Using a finite summation, the first moments of the area A with respect to the
x-axis (i.e. Qx) and with respect to the y-axis (i.e. Qy) may be determined as
follows.
Qx = y A = y 1 ∆A1 + y 2 ∆A2 + y 3 ∆A3 + • • • + y n ∆An
Qy = x A = x 1 ∆A1 + x 2 ∆A2 + x 3 ∆A3 + • • • + x n ∆An
Procedure for Analysis
The centroid of an object or shape can be determined by integration using the
following equations.
A = ∫dA x A = ∫ x dA y A = ∫ y dA
• Denoting x el and y el as the coordinates of the element of dA, the last two
equations may be written as follows.
x A = ∫ x el dA and y A = ∫ y el dA
The centroid of an object or shape can be determined by finite summations using
the following equations.
A = ΣAi x A = Σx i Ai y A = Σ y i Ai
Likewise, the centroid of a line can be determined by finite summations using the
following equations.
L = Σ Li x L = Σx i Li y L = Σ y i Li
L = ∫dL x L = ∫x dL y L = ∫y dL
9.5
Examples - Centroids of Areas
Given: The triangular area shown.
Find: Centroid (x , y ), using a
horizontal element.
Applicable equations:
A = ∫ dA dA = x dy
Qy = x A = ∫ x el dA x el = x/2
Qx = y A = ∫ y el dA y el = y
To define values of “x” in terms of “y” to allow the integration with respect to y,
write an equation for the boundary of the area (i.e. line AB).
• The general form for the equation of a line is y = m x + c
• The slope “m” of the line AB is m = - h/b
• Inserting the slope, the equation may be written y = (-h/b) x + c
• To solve for the y-intercept “c”, insert the coordinates (b, 0), a point on the
line, into the equation and solve for “c”.
0 = (-h/b) (b) + c c = h
• The equation for the line is then y = (-h/b) x + h
• Rewrite the equation to define “x” in terms of “y”. x = (b/h) (h – y)
• Now define dA and x el in terms of “y”.
dA = x dy = (b/h) (h – y) dy
x el = x/2 = ½ (b/h) (h – y)
Now solve for area A and the first moments of area Qx and Qy.
A = ∫ dA = ∫ x dy = h
0(b/h ) (h – y) dy = (b/h) (h y – y2/2)|
h
0
= (b/h) [(h2 – h2/2) – 0] = (b/h) (h2/2)
A = bh/2
9.6
x A = ∫x el dA = ∫(x/2) x dy = ½ ∫x2 dy = ½ h
0(b/h)2 (h-y)2 dy
= ½ (b/h)2 h
0(h – y)2 dy = ½ (b/h)2 (1/3) (h – y)3 (-1) |
h
0
= - ½ (b/h)2 (1/3) (0 – h3)
x A = b2 h/6
y A = ∫ y el dA = ∫ y x dy = h
0y (b/h) (h – y) dy
= (b/h) h
0y (h – y) dy = (b/h)
h
0(h y – y2) dy
= (b/h) [(h y2/2 – y3/3)] |h
0 = (b/h) (h3/2 – h3/3)
y A = bh2/6
Finally, determine the location of the centroid.
x = x A/A = (b2 h/6)/(bh/2) = b/3
y = y A/A = (bh2/6)/(bh/2) = h/3
9.7
Given: Area shown.
Find: Centroid (x , y ), using a
vertical element.
Applicable equations:
A = ∫ dA dA = (y2 – y1) dx
Qy = x A = ∫ x el dA x el = x
Qx = y A = ∫ y el dA y el = ½ (y2 + y1)
Define values of “m” and “k” in terms of “a” and “b”.
For the line, when x = a and y = b: y = mx b = ma m = b/a
For the parabola, when x = a and y = b: y = kx2 b = ka2 k = b/a2
To define values of “y” in terms of “x” to allow the integration with respect to x,
write equations for the boundaries of the area (i.e. the line and the parabola).
• The equation for the line is y = (b/a) x
• The equation for the parabola is y = (b/a2) x2
• Now define dA and y el in terms of “x”.
dA = (y2 – y1) dx = [(b/a) x - (b/a2) x2] dx
y el = ½ (y2 + y1) = ½ [(b/a) x + (b/a2) x2]
Now solve for area A and the first moments of area Qx and Qy.
A = ∫ dA = ∫ x dy = a
0[(b/a) x - (b/a2) x2] dx
= [(b/a) (x2/2) – (b/a2) (x3/3)] |a
0 = (b/a)(a2/2) – (b/a2)(a3/3)
= ba/2 – ba/3
A = ba/6
9.8
x A = ∫x el dA = a
0x [(b/a) x - (b/a2) x2] dx
= a
0[(b/a) x2 – (b/a2) x3 ] dx = [(b/a)(x3/3) – (b/a2)(x4/4)] |
a
0
= (b/a)(a3/3) – (b/a2)(a4/4) = ba2/3 – ba2/4
x A = ba2/12
y A = ∫ y el dA = a
0½ [(b/a) x + (b/a2) x2] [(b/a) x - (b/a2) x2] dx
= ½ a
0[(b/a)2 x2 – (b/a2)2 x4 ] dx = ½ [(b2/a2)(x3/3) – (b2/a4)(x5/5)] |
a
0
= ½ [(b2/a2)(a3/3) – (b2/a4)(a5/5)] = ½ (b2a/3 – b2a/5)
y A = b2a/15
Finally, determine the location of the centroid.
x = x A/A = (ba2/12)/(ba/6) = a/2
y = y A/A = (b2a/15)/(ba/6) = 2b/5
9.9
Given: Area shown.
Find: Centroid (x , y ), using a horizontal
element.
Applicable equations:
A = ∫ dA dA = (x2 – x1) dy
Qy = x A = ∫ x el dA x el = ½ (x2 + x1)
Qx = y A = ∫ y el dA y el = y
To define values of “x” in terms of “y” to allow the integration with respect to y,
write an equation for the boundary of the area (i.e. the line).
• The general form for the equation of a line is y = m x + c
• The slope “m” of the line is m = (4 – 0)/(4 – 3) = 4/1 m = 4
• Inserting the slope, the equation may be written y = 4 x + c
• To solve for the y-intercept “c”, insert the coordinates (4, 4), a point on the
line, into the equation and solve for “c”.
4 = 4 (4) + c c = - 12
• The equation for the line is then y = 4 x - 12
• Rewrite the equation to define “x” in terms of “y”. x = (y + 12)/4
• Now define dA and x el in terms of “y”.
dA = (x2 – x1) dy = [(y + 12)/4 – y2/4] dy
x el = ½ (x2 + x1) = ½ [(y + 12)/4 + y2/4]
9.10
Now solve for area A and the first moments of area Qx and Qy.
A = ∫ dA = ∫ (x2 – x1) dy = 4
0[(y + 12)/4 – y2/4] dy
= ¼ 4
0(y + 12 – y2) dy = ¼ (y2/2 + 12y – y3/3) |
4
0 = ¼ (8 + 48 – 64/3)
A = 8.67 in2
x A = ∫x el dA = 4
0½ [(y + 12)/4 + y2/4] [(y + 12)/4 – y2/4] dy
= ½ 4
0{[(y + 12)/4] 2 – (y2/4)2 } dy = (1/32)
4
0[(y + 12) 2 – (y2)2] dy
= (1/32) 4
0(y2 + 24 y + 144 – y4) dy = (1/32) (y3/3 + 12 y2 + 144 y – y5/5) |
4
0
= (1/32) (64/3 + 192 + 576 – 1024/5)
x A = 18.27 in3
y A = ∫ y el dA = 4
0y [(y + 12)/4 – y2/4] dy =
4
0y {[(y + 12)/4] – y2/4} dy
= ¼ 4
0(y2 + 12y – y3) dy = ¼ [(y3/3 + 6y2 – y4/4] |
4
0 = ¼ (64/3 + 96 – 64)
y A = 13.33 in3
x = x A/A = 18.27/8.67 = 2.11”
y = y A/A = 13.33/8.67 = 1.537”
9.11
9.2 Composite Bodies
In many instances, an area may be divided into familiar shapes (rectangles,
triangles, squares, and circles).
• The centroidal distances are found by equating the first moments of area.
∑Mx : y 1 ∆A1 + y 2 ∆A2 + • • • + y n ∆An = y A
∑My : x 1 ∆A1 + x 2 ∆A2 + • • • + x n ∆An = x A
A tabular solution is often a convenient method to determine the location of the
centroid for composite areas.
9.12
Examples – Composite Areas
Given: Area shown.
Find: Centroid (x , y ).
Part Area, Ai x i y i x i Ai y i Ai
1 10,800 45.0 120.0 486,000 1,296,000
2 2,700 30.0 40.0 81,000 108,000
3 - 2,510 73.0 120.0 - 183,000 - 301,000
Totals 10,990 384,000 1,103,000
A3 = π r2/2 = π (40)2/2 = 2,510
x3 = 90 – 4r/3 π = 90 – 4(40)/3π = 73.0
x = Σx i Ai/ΣAi = 384,000/10,990
x = 34.9 mm
y = Σ y i Ai/ΣAi = 1,103,000/10,990
y = 100.4 mm
9.13
Given: The cover-plated beam shown.
Find: Neutral axis.
The “neutral axis” is an axis in the cross
section of a beam (a member resisting
bending) along which there are no
longitudinal stresses or strains.
• If the section is symmetric, then the
neutral axis is located at the geometric
centroid.
A vertical axis through the center of the web forms an axis of symmetry.
• Only the y distance is required.
Use the bottom of the bottom flange as the reference axis.
y = Σ y i Ai/ΣAi = [22.3 (18.2/2) + 1(12)(18.2 + 1.0/2)]/[22.3 + 1(12)]
= (202.93 + 224.40)/34.3
y = 12.46”
9.14
Given: Bent wire shown.
Find: Centroid (x , y ).
Segment Length, Li x i y i x i Li y i Li
1 4.00 0 5.00 0 20.00
2 6.00 3.00 7.00 18.00 42.00
3 7.00 6.00 3.50 42.00 24.50
4 6.71 3.00 1.50 20.13 10.07
Totals 23.71 80.13 96.57
x = Σx i Li/ΣLi = 80.13 / 23.71
x = 3.38”
y = Σ y i Li/ΣLi = 96.57 / 23.71
y = 4.07”
9.15
9.3 Theorems of Pappus-Guldinus
The two theorems of Pappus and Guldinus are used to find the surface area and
volume of any body of revolution.
• The theorems were first developed by Pappus of Alexandria during the 4th
century A.D.
• The theorems were restated later by the Swiss mathematician Paul Guldin (or
Guldinus) (1577 – 1643).
Surface Area
A “surface of revolution” is a surface that may be generated by rotating a plane
curve about a fixed axis.
Theorem 1: “The area of a surface of revolution is equal to the length of the
generating curve times the distance traveled by the centroid of the curve while
the surface is being generated.” (The curve must be non-intersecting.)
dL
2 π y el
A = ∫2π y el dL
= 2π ∫ y el dL
A = 2π y L
where 2π y is the distance
traveled by the centroid
of the line L
Volume
A “body of revolution” is a body that may be generated by rotating a plane area
about a fixed axis.
Theorem 2: “The volume of a body of revolution is equal to the generating area
times the distance traveled by the centroid of the area while the body is being
generated.” (The area must be non-intersecting.)
9.16
dA
2 π y el
A = ∫ 2 π y el dA
= 2 π ∫ y el dA
A = 2 π ( y A)
where 2π y is the distance
traveled by the centroid
of the area A
9.17
Examples – Theorems of Pappus and Guldinus
Given: The cone shown.
Find: The surface area and volume of
the cone.
Surface area using the Theorem
A = 2 π x L
= 2 π (1.5) 5.0 = 15.0 π
A = 47.12 in2
Surface area from formula
A = π r (r2 + h2)½
= π (3) (32 + 42)½ = 15.0 π
A = 47.12 in2
Volume using the Theorem
V = 2 π x A
= 2 π [(1/3) 3.0][½ (4.0) 3.0] = 12.0 π
V = 37.70 in3
Volume using formula
V = (π/3) r2 h
= (π/3) (32) 4 = 12.0 π
V = 37.70 in3
9.18
Given: A sphere with a radius of 12”.
Find: The surface area and volume of
the sphere.
Surface area using the Theorem
A = 2 π y L
y = 2r/π = 2(12)/π = 24/π
L = 2 π r/2 = π r = π (12) = 12 π
A = 2 π (24/π) (12 π) = 576 π
A = 1809.6 in2
Surface area from formula
A = 4 π r2
= 4 π (122) = 576 π
A = 1809.6 in2
Volume using the Theorem
V = 2 π y A
y = 4r/3π = 4 (12)/3 π = 16/π
A = π r2/2 = π (12)2/2 = 72π
V = 2π (16/π)(72 π) = 2304 π
V = 7238.2 in3
Volume from formula
V = (4π/3) r3 = (4π/3) (123) = 2304 π
V = 7238.2 in3
9.19
9.4 Resultant of a General Distributed Loading
The concept of the centroid of an area may be used to solve problems with beams
supporting a “distributed load.”
Magnitude of Resultant Force
A “distributed load” on a beam may be
replaced by an equivalent concentrated load.
• The magnitude of this single equivalent
concentrated load is equal to the area
under the load curve.
Location of the Resultant
The line of action of the resultant force
passes through the centroid of the
distributed loading.
9.5 Fluid Pressure
Pressure Distribution over a Surface
The concept of the centroid of an area is useful for this problem, whose solution is
discussed in “Fluid Mechanics.”