Chapter 5 Distributed Forces: Centroids and Centers of Gravity PDF

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EGR 240: Solid Mechanics - Statics
David L...

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EGR 240: Solid Mechanics Statics

Chapter 5 Distributed Forces: Centroids and Centers of Gravity

Center of gravity • Any rigid body is essentially made up of many infinitesimally small particles. • The earth exerts a gravitational force on each particle. The sum of all these forces equal the weight of the object. • The center of gravity or center of mass of the object is the point where a single equivalent force equal to the weight and replacing all the minute forces can be applied. • The weight applied at the center of gravity must not produce any net moments.

Derivation • Consider a rigid body of arbitrary shape. Gravity acts in the negative y direction.

• Divide the body into small volume elements, of weight ΔWi, i = 1, 2, 3, … n.

• The distances of each element from the x, y, and z axes are xi, yi, and zi, i = 1, 2, 3, … n. • In the example, the weights produce no moments about the y axis since they are all parallel to the y axis. They do, however, produce moments about the x and z axes.

• Recall the requirement that net moments about the center of gravity have to be zero. The center of gravity has unknown coordinates (x, y, z). • Since there are no moments about the y axis, we can’t deal with y yet.

• We can calculate moments about the center of gravity, G, as follows:

"M z = (x1 # x )$W1 + (x 2 # x )$W 2 + (x 3 # x )$W 3 n

+ ...+ (x n # x )$W n = % (x i # x )W i = 0 i=1

"M x = (z1 # z )$W1 + (z 2 # z )$W 2 + (z 3 # z )$W 3 n

+ ...+ (z n # z )$W n = % (z i # z )W i = 0 i=1

• Manipulating and rearranging the Mz equation,

(x1 " x )#W1 + (x2 " x )#W 2 + (x3 " x )#W 3 + ...+ (xn " x )#W n = 0 x1#W1 + x2 #W 2 + x3 #W 3 + ...+ xn #W n = x #W1 + x #W 2 + x #W 3 + ...+ x #W n $xi #W i = x $#W i Letting the size of each element approach 0:

% xdW = x W • Similarly for the Mx equation,

" zdW = z W

• How do we find the y coordinate? Rotate the body and the coordinate axes so that gravity now acts in the negative x or negative z direction. • Say we rotate such that gravity acts in the negative z direction. We now can sum up moments about the x and y axes. The derivation for x remains the same. We get for y,

" ydW = yW

• The integral equations just derived are for cases where the weight is not distributed uniformly (i.e., density is not the same over the body). • We will consider only cases where density is constant (i.e., the bodies are homogeneous). We start with the 2-D case. • Consider a plate with uniform thickness (t) in the z direction and lying on the xy plane. For each volume element of weight dW, we can express the weight in terms of its specific weight γ (weight per unit volume) and volume (t dA, where dA is area): dW = γt dA

• Likewise, the total weight of the plate is W = γtA • We can now go back to the derived integral equations and rewrite them:

" xdW = xW # " x$t dA = x$tA " ydW = yW # " y$t dA = y$tA • We get:

" xdA yA = " ydA xA =

In 2-D, x and y are the coordinates of the centroid of the area.

• There’s an analogy for deriving the centroid of a wire or line of length L. We get:

" xdL yL = " ydL xL =

• Note that the centroid does not have to lie on the line.

• In our equations for the centroid, we define the integral terms as the first moment with respect to the y axis (Qy) and the first moment with respect to the x axis (Qx):

# xdA = xA " # ydA = yA

Qy " Qx

• To find the centroids of common shapes, there is no need to calculate the integrals. Equations are given in Fig. 5.8A and Fig. 5.8B in the text book. These tables are also on the inside back cover.

Centroids of common areas

Centroids of common lines

Rules and shortcuts for finding centroids: • If an area has a line of symmetry, its first moment with respect to that line is zero and its centroid lies on that line.

• If an area has two lines of symmetry, its centroid is at the intersection of those two lines.

• 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). • If such a center of symmetry exists, the centroid of the area coincides with its location.

Centroids of composite areas • A composite area is an area that’s made up of several “standard” shapes. For example, the following area can be considered a composite made up of two triangles and a rectangle:

• Each sub-area has a centroid (xi, yi). • Designate the centroid of the composite area by (X, Y)

• Similar to our derivation before, the net moments about the centroid must be zero:

"M y = (x1 # X )W1 + (x 2 # X )W 2 + (x 3 # X )W 3 + ...+ (x n # X )W n = 0 x1W1 + x 2W 2 + x 3W 3 + ...+ x nW n = X (W1 + W 2 + W 3 + ... + W ) "xW = X"W • Again assuming constant specific weight,

X"A = "(xA) • Note that the above fits our definition of first moment, Qy.

• The derivation is obviously similar for the y coordinate of the centroid. • To find the centroid of a composite area, we therefore use:

X"A = "(xA) = Qy Y "A = "(yA) = Qx • Note that a composite area can contain subareas that are “negative,” such as holes. • The use of the above equations will be demonstrated by an example.

Sample Problem 5.1

For the plane area shown, determine the first moments with respect to the x and y axes and the location of the centroid.

Sample Problem 5.1 (cont’d)

Notice that the circular hole is a negative area. It is a good idea to make a table like this one.

Qx = +506.2 ! 10 3 mm 3 Q y = +757.7 ! 103 mm3

Sample Problem 5.1 (cont’d) From our equations, compute the coordinates of the area centroid by dividing the first moments by the total area.

x A + 757.7 ! 10 3 mm 3 " X = = " A 13.828 !10 3 mm 2

X = 54.8 mm

y A + 506.2 !10 3 mm 3 " Y = = " A 13.828 ! 10 3 mm 2 Y = 36.6 mm

Finding centroids by integration • To find centroids of non-standard areas not included in the tables or not comprised of standard shapes, we must apply the integral definition. Recall:

" xdA yA = " ydA xA =

• The integrals suggest that we break the area into infinitesimal area elements. Although we can use square elements of lengths dx and dy, such an approach requires double integration. There are simpler alternatives.

• Two such alternatives are to use rectangular strips, either vertical or horizontal:

x A = ! xel dA = ! x ( ydx) yA = ! y el dA =!

y (ydx ) 2

Better if writing the curve as y = f (x) yields a simpler equation.

x A = " xel dA a+ x [(a ! x)dx] 2 yA = " y el dA ="

= " y [(a ! x)dx ]

Better if writing the curve as x = f (y) yields a simpler equation.

• For areas that are defined using polar coordinates:

x A = ' xel dA ='

2r 1 cos( &$ r 2 d( #! 3 %2 "

yA = ' y el dA ='

2r 1 sin ( &$ r 2 d ( #! 3 " %2

Sample Problem 5.4

Determine by direct integration the location of the centroid of a parabolic spandrel.

Sample Problem 5.4 (cont’d) SOLUTION: • Determine the constant k.

y = k x2 b = k a2 ! k = y=

b a2

x2

or

b a2 x=

a b1 2

y1 2

• Evaluate the total area. A = ' dA a

a

& b x3 # 2 = ' y dx = ' 2 x dx = $ 2 ! $% a 3 !" 0 0a ab = 3 b

Sample Problem 5.4 (cont’d) Using vertical strips, perform a single integration to find the first moments. a

, b 2) Q y = - xel dA = - xydx = - x* x ' dx 2 a ( 0 + a & b x4 # a2b =$ 2 ! = 4 4 a !" 0 %$ 2

a y 1, b ) Qx = - yel dA = - ydx = - * 2 x 2 ' dx 2 ( 02+a a & b 2 x5 # ab2 =$ 4 = ! $% 2a 5 !" 0 10

Sample Problem 5.4 (cont’d) Or, using horizontal strips, perform a single integration to find the first moments. b 2 a+ x a ' x2 Q y = ( xel dA = ( dy (a ' x)dy = ( 2 2 0

1 b &$ 2 a2 = ( $a ' 20% b

# a 2b ! y dy = ! 4 "

a & # Qx = ( yel dA = ( y(a ' x)dy = ( y$ a ' 1 2 y 1 2 !dy % " b a 3 2# ab 2 & = ( $ ay ' 1 2 y ! dy = 10 " b 0% b

Sample Problem 5.4 (cont’d) Evaluate the centroid coordinates.

xA = Q y ab a 2b x = 3 4

x=

3 a 4

y=

3 b 10

yA = Q x ab ab 2 = y 3 10

Theorems of Pappus-Guldinus • A surface of revolution is generated by rotating a plane curve about a fixed axis:

• A volume of revolution is generated by rotating a plane area about a fixed axis:

• Theorem I. The area of a surface of revolution is equal to the length of the generating curve times the distance traveled by the centroid through the rotation. • Theorem II. The volume of a body of revolution is equal to the generating area times the distance traveled by the centroid through the rotation.

A = 2! yL

V = 2! y A

Sample Problem 5.7

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 ρ = 7.85 × 103 kg/m3, determine the mass and weight of the rim.

Sample Problem 5.7 (cont’d) 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 acceleration to get the mass and acceleration.

(

m = "V = (7.85 # 10 3 kg m 3 )(7.65 # 10 6 mm 3 ) 10 $9 m 3 mm 2 W = mg = (60.0 kg)(9.81 m s )

3

)

m = 60.0 kg W = 589 N

Center of gravity of a 3-D body or centroid of a volume • The derivation of the center of gravity or centroid in 3-D is very similar to the 2-D case. • We continue from our derivation of the weight integrals:

" xdW = xW " ydW = yW " zdW = z W • Again, we will assume that the bodies we deal with are homogeneous.

• Each weight dW can then be replaced by dW = γ dV where γ is specific weight (weight per unit volume) and dV is volume. • We end up with:

" xdV yV = " ydV z V = " zdV xV =

• To use the above integrals, we need to use volume elements. Centroids for common volumes are tabulated, just like for areas.

Centroids of common 3-D volumes

Centroids of composite volumes • To find the centroid (X, Y, Z) of a composite volume, the procedure and derivation are very similar to the 2-D case.

X"V = "(xV ) Y "V = "(yV ) Z "V = "(z V ) • Note again that we can have negative volumes. • The procedure is illustrated by Sample Problem 5.12.

Sample Problem 5.12

Locate the center of gravity of the steel machine element. The diameter of each hole is 1 in.

Sample Problem 5.12 (cont’d)

Create a table similar to the ones used to find centroids of 2-D areas.

Sample Problem 5.12 (cont’d)

X = " xV " V = ( 3.08 in 4 ) (5.286 in 3 ) X = 0.577 in.

Y = " y V " V = (#5.047 in 4 ) (5.286 in 3 ) Y = !0.955 in.

Z = " z V " V = ( 1.618 in 4 ) ( 5.286 in 3 ) Z = 01.618 in.

Finding centroids of volumes by integration • To find centroids of non-standard volumes not included in the table or not comprised of standard shapes, we apply the integral equations as in the 2-D case. Recall:

" xdV yV = " ydV z V = " zdV xV =

• Although we can use cube elements of lengths dx, dy,and dz, such an approach requires triple integration. It’s usually possible to avoid triple integrals.

• The more general case is to use rectangular block elements that span one dimension of the volume (i.e., one of dx, dy, or dz is replaced with a constant or with a function of the other two variables. For example: z

z is constant

x el = x, y el = y, zel = z /2 dV = z dx dy

z

" x dV yV = " y dV z V = " z dV xV =

y

zel x

yel

dy

xel dx

el

el

el

• Even easier is if we have a body of rotation. We then use disks or partial disks as our volume elements. We have to align our coordinate axes with the axis of rotation. y

x el = x, y el = 0, zel = 0

xel

dV = " r 2 dx r

xV = x

dx z

#x

el

dV

• What if we have a partial body of rotation? We will consider rotations of 180°. We then use half disks as our volume elements. We again align our coordinate axes with the axis of rotation. Note now that yel ≠ 0. y

4r x el = x, y el = , zel = 0 3" dV = 21 " r 2 dx

y = f (x) xel dx

x z

#x yV = # y xV =

r

el

dV

el

dV

See Sample Problem 5.13...