Centroids Introduction • The earth exerts a gravitational force on each of the particles forming 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.

5 - 2

Centroids

• Centroid of mass – (a.k.a. Center of mass) – (a.k.a. Center of weight) – (a.k.a. Center of gravity) • For a solid, the point where the distributed mass is centered • Centroid of volume, Centroid of area

xM   xm   x dm yM   ym   y dm

Center of Gravity of a 2D Body • Center of gravity of a plate • Center of gravity of a wire

 M y xW   xW   x dW

 M y yW   yW   y dW

5 - 7 Centroids &First Moments of Areas & Lines

• Centroid of an area • Centroid of a line

xW   x dW xgM  g x dM xW   x dW M  V  tA) dM  dV  tdA x La   x adL xAt    x tdA xL   x dL xA  x dA  Q  y yL   y dL  first moment with respect to y yA  y dA  Q  x  first moment with respect to x 5 - 8 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. 5 - 9 Centroids of Common Shapes of Areas

5 - 10 Centroids of Common Shapes of Lines

5 - 11 Composite Plates and Areas • Composite weight X W   xW Y W   yW

• Composite area X  A   x A Y  A   y A

5 - 12 Sample Problem 5.1 SOLUTION: • Divide the area into a triangle, rectangle, and semicircle with a circular cutout. • Calculate the first moments of each area with respect to the axes. • Find the total area and first moments of the triangle, rectangle, and semicircle. Subtract the area and first moment of the For the plane area shown, determine circular cutout. the first moments with respect to the x • Compute the coordinates of the area and y axes and the location of the centroid by dividing the first moments by centroid. the total area.

5 - 13 Sample Problem 5.1

3 3 • Find the total area and first moments of the Qx  506.210 mm triangle, rectangle, and semicircle. Subtract the Q  757.7103 mm 3 area and first moment of the circular cutout. y 5 - 14 Sample Problem • Compute the coordinates of the area centroid by dividing the first moments by the total area. x A  757.7103 mm 3 X     A 13.828103 mm 2 X  54.8 mm

yA  506.2103 mm 3 Y     A 13.828103 mm 2 Y  36.6 mm 5 - 15 Determination of Centroids by • Double integration to find the first moment xA   xdA   x dxdy   xelIntegrationdA may be avoided by defining dA as a thin yA   ydA   y dxdy   yel dA rectangle or strip.

xA   xel dA xA   xel dA xA   xel dA   x ydx a  x 2r  1 2     a  xdx   cos  r d  2 3  2  yA   yel dA yA  y dA y  el yA   yel dA   ydx 2   ya  xdx 2r  1 2    sin  r d  3  2  5 - 16 Sample Problem 5.4 SOLUTION: • Determine the constant k. • Evaluate the total area. • Using either vertical or horizontal strips, perform a single integration to find the first moments. Determine by direct integration the location of the centroid of a parabolic • Evaluate the centroid coordinates. spandrel.

5 - 17 Sample Problem 5.4 SOLUTION: • Determine the constant k. y  k x2 b b  k a2  k  a2 b a y  x2 or x  y1 2 a2 b1 2

• Evaluate the total area. A   dA a 3 a b 2  b x   y dx  x dx      2 2 3 0 a a 0 ab  3 5 - 18 Sample Problem 5.4 • Using vertical strips, perform a single integration to find the first moments. a  b  Q  x dA  xydx  x x2 dx y  el    2  0  a  a  b x4  a2b     2 4 4 a 0 2 y a 1  b  Q  y dA  ydx  x2 dx x  el    2  2 0 2  a  a  b2 x5  ab2     4 5 10 2a 0

5 - 19 Sample Problem 5.4 • Or, using horizontal strips, perform a single integration to find the first moments.

a  x b a2  x2 Qy   xel dA   a  xdy   dy 2 0 2 1 b  a2  a2b  a2  ydy    2 0 b  4  a 1 2  Qx   yel dA   ya  xdy   ya  y dy  b1 2  b  a  ab2  ay  y3 2 dy   1 2  0 b  10

5 - 20 Sample Problem 5.4 • Evaluate the centroid coordinates.

xA  Qy ab a2b 3 x  x  a 3 4 4

yA  Qx ab ab2 3 y  y  b 3 10 10

5 - 21 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 generating curve times the distance traveled by the centroid through the rotation. A  2 yL

5 - 22 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 times the distance traveled by the centroid through the rotation.

V  2 yA

5 - 23 Sample Problem 5.7 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. 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.   7.85103 kg m3

5 - 24 Sample Problem 5.7 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.

3 3 6 3  9 3 3  m  V  7.8510 kg m 7.6510 mm 10 m mm  m  60.0 kg   W  mg  60.0 kg 9.81 m s2  W  589 N 5 - 25 Distributed Loads on Beams

L • A distributed load is represented by plotting the load W   wdx   dA  A per unit length, w (N/m) . The total load is equal to 0 the area under the load curve.

OPW   xdW • A distributed load can be replace by a concentrated L load with a magnitude equal to the area under the OPA   xdA  xA load curve and a line of action passing through the 0 area centroid.

5 - 26 Sample Problem 5.9 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 A beam supports a distributed load as summing moments about the beam shown. Determine the equivalent ends. concentrated load and the reactions at the supports.

5 - 27 Sample Problem 5.9 SOLUTION: • The magnitude of the concentrated load is equal to the total load or the area under the curve. F 18.0 kN

• The line of action of the concentrated load passes through the centroid of the area under the curve. 63 kN m X  X  3.5 m 18 kN

5 - 28 Sample Problem 5.9 • Determine the support reactions by summing moments about the beam ends.

M A  0 : By 6 m 18 kN 3.5 m  0

By 10.5 kN

M B  0 :  Ay 6 m 18 kN 6 m 3.5 m  0

Ay  7.5 kN

5 - 29 Center of Gravity of a 3D Body: Centroid of a Volume

• Center of gravity G • Results are independent of body orientation,   W j   W j xW   xdW yW   ydW zW   zdW     • For homogeneous bodies, rG W j   r  W j      W  V and dW   dV rGW  j   rW  j  xV  xdV yV  ydV zV  zdV      W   dW rGW   rdW 5 - 30 Centroids of Common 3D Shapes

5 - 31 Composite 3D Bodies • Moment of the total weight concentrated at the center of gravity G is equal to the sum of the moments of the weights of the component parts. X W   xW Y W   yW Z W   zW

• For homogeneous bodies, X V   xV Y V   yV Z V   zV

5 - 32 Sample Problem 5.12 SOLUTION: • Form the machine element from a rectangular parallelepiped and a quarter cylinder and then subtracting two 1-in. diameter cylinders.

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

5 - 33 Sample Problem 5.12

5 - 34 Sample Problem 5.12

X   xV V  3.08 in4  5.286 in3  X  0.577 in.

Y   yV V  5.047 in4  5.286 in3  Y  0.577 in.

Z   zV V  1.618 in4  5.286 in3  Z  0.577 in. 5 - 35