Chapter 5

Distributed Forces: and Centers of Gravity Application

There are many examples in engineering analysis of distributed loads. It is convenient in some cases to represent such loads as a concentrated force located at the of the distributed load.

5 - 2 Introduction

• The earth exerts a gravitational force on each of the particles forming a body – consider how your weight is distributed throughout your body. These forces can be replaced 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 is analogous to the center of gravity of a body; it is the “center of area.” The concept of the first moment of an area is used to locate the centroid.

5 - 3 Center of Mass

The center of mass of a system is the point where the system can be balanced in a uniform gravitational field. 9.2 The Center of Mass: A System of Particles

Consider a situation in which n particles are strung out along

the x axis. Let the mass of the particles are m1, m2, ….mn, and let them be located at x1, x2, …xn respectively. Then if the total mass is M = m1+ m2 + . . . + mn, then the location of the center of mass, xcom, is Center of Mass

For two objects:

The center of mass is closer to the more massive object. Center of Mass

The center of mass need not be within the object: Moment

The moment of a mass is a measure of its tendency to rotate about a point. Clearly, the greater the mass (and the greater the distance from the point), the greater will be the tendency to rotate.

The moment is defined as: Moment = mass × distance from a point

Example 1

In this case, there will be a total moment about O of: (Clockwise is regarded as positive in this work.)

M=2×1−10×3=−28 kgm Center of Mass We now aim to find the center of mass of the system and this will lead to a more general result. We have 3 masses of 10 kg, 5 kg and 7 kg at 2 m, 2 m and 1 m distance from O as shown.

We wish to replace these masses with one single mass to give an equivalent moment. Where should we place this single mass? Total moment =10×2+5×4+7×5=75 kg.m

If we put the masses together, we have: 10+5+7=22 kg For an equivalent moment, we need:

22×d =75 where d is the distance from the center of mass to the point of rotation. i.e. d =75 22 ≈3.4 m So our equivalent system (with one mass of 22 kg ) would have:

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 - 10 Centroids and First Moments of and Lines • Centroid of an area • Centroid of a

xW   x dW xW   x dW xAt    xtdA x La   x adL xA   x dA  Qy xL   x dL  first moment wit h respect to y yL   y dL yA   y dA  Qx  first moment wit h respect to x

5 - 11 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 - 12 Centroids of Common of Areas

5 - 13 Centroids of Common Shapes of Lines

5 - 14 Composite Plates and Areas

• Composite plates X W   xW Y W   yW

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

5 - 15 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 • Compute the coordinates of the area x and y axes and the location of the centroid by dividing the first moments by centroid. the total area.

5 - 16 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 - 17 Sample Problem 5.1

• 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 - 18 Theorems of Pappus-Guldinus:

The storage tanks shown are all bodies of revolution. Thus, their surface area and can be determined using Theorems of Pappus-Goldinus.

2 - 19 Theorems of Pappus-Guldinus----Surface Area Calculation

• 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 - 20 Pappus's Centroid Theorem—Surface Area Calculation

The first theorem of Pappus states that the surface area of a surface of revolution generated by the revolution of a curve about an external axis is equal to the product of the arc length of the generating curve and the distance traveled by the curve's geometric centroid ,

The following table summarizes the surface areas calculated using Pappus's centroid theorem for various surfaces of revolution.

5 - 21 Pappus's Centroid Theorem—Surface Area Calculations

solid generating curve

cone inclined

parallel line segment

sphere semicircle

5 - 22 Theorems of Pappus-Guldinus--- Calculation

• 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 Pappus's Centroid Theorem—Volume Calculations

Volume calculations:

Similarly, the second theorem of Pappus states that the volume of a solid of revolution generated by the revolution of a lamina about an external axis is equal to the product of the area of the lamina and the distance traveled by the lamina's geometric centroid ,

The following table summarizes the surface areas and volumes calculated using Pappus's centroid theorem for various solids and surfaces of revolution.

5 - 24 Pappus's Centroid Theorem---Volume Calculation

solid generating lamina

cone right triangle

cylinder rectangle

sphere semicircle

5 - 25 Sample Problem 5.7 SOLUTION: • Apply the theorem of Pappus-Guldinus to evaluate the volumes of revolution of the pulley, which we will form as a large rectangle with an inner rectangular cutout.

• Multiply by density and acceleration The outside diameter of a pulley is 0.8 to get the mass and weight. 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   7.85103 kg m3 determine the mass and weight of the rim.

5 - 26 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 weight.

m  V  7.85103 kg m37.65106mm3109m3 /mm3 m  60.0 kg W  mg 60.0 kg9.81 m s2 W  589 N 5 - 27   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 - 28 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) A beam supports a distributed load as drawing the free body diagram for the shown. Determine the equivalent beam and (b) applying the conditions concentrated load and the reactions at of equilibrium. the supports.

5 - 29 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 - 30 Sample Problem 5.9

• Determine the support reactions by applying the equilibrium conditions. For example, successively sum the moments at the two supports:

MA  0: By 6 m18 kN3.5 m 0

By 10.5 kN

 MB  0:  Ay 6 m18 kN6 m 3.5 m 0

Ay  7.5 kN

 • And by summing forces in the x-direction:

Fx  0: Bx  0

5 - 31 