Copyright © 2010 Pearson Education South Asia Pte Ltd

Chapter Objectives Concept of the center of gravity, center of mass, and the centroid Determine the location of the center of gravity and centroid for a system of discrete particles and a body of arbitrary shape Theorems of Pappus and Guldinus Method for finding the resultant of a general distributed loading Copyright © 2010 Pearson Education South Asia Pte Ltd

Chapter Outline Center of Gravity and Center of Mass for a System of Particles Composite Bodies Theorems of Pappus and Guldinus Resultants of a General Distributed Loading Fluid Pressure Copyright © 2010 Pearson Education South Asia Pte Ltd

9.1 Center of Gravity and Center of Mass for a System of Particles
Locates the resultant weight of a system of particles Consider system of n particles fixed within a region of space The weights of the particles can be replaced by a single (equivalent) resultant weight having defined point G of application Copyright © 2010 Pearson Education South Asia Pte Ltd

Resultant weight = total weight of n particles Sum of moments of weights of all the particles about x, y, z axes = moment of resultant weight about these axes Summing moments about the x axis, Summing moments about y axis, Copyright © 2010 Pearson Education South Asia Pte Ltd

Although the weights do not produce a moment about z axis, by rotating the coordinate system 90° about x or y axis with the particles fixed in it and summing moments about the x axis, Generally, Copyright © 2010 Pearson Education South Asia Pte Ltd

Center Mass Provided acceleration due to gravity g for every particle is constant, then W = mg By comparison, the location of the center of gravity coincides with that of center of mass Particles have weight only when under the influence of gravitational attraction, whereas center of mass is independent of gravity Copyright © 2010 Pearson Education South Asia Pte Ltd

Centroid of a Line If the geometry of the object takes the form of a line, the balance of moments of differential elements dL about each of the coordinate system yields Copyright © 2010 Pearson Education South Asia Pte Ltd

Example 9.1 Differential element Located on the curve at the arbitrary point (x, y) Area and Moment Arms For differential length of the element dL Since x = y2 and then dx/dy = 2y The centroid is located at Copyright © 2010 Pearson Education South Asia Pte Ltd

9.2 Composite Bodies Consists of a series of connected “simpler” shaped bodies, which may be rectangular, triangular or semicircular A body can be sectioned or divided into its composite parts Accounting for finite number of weights Copyright © 2010 Pearson Education South Asia Pte Ltd

9.2 Composite Bodies Procedure for Analysis Composite Parts Divide the body or object into a finite number of composite parts that have simpler shapes Treat the hole in composite as an additional composite part having negative weight or size Moment Arms Establish the coordinate axes and determine the coordinates of the center of gravity or centroid of each part Copyright © 2010 Pearson Education South Asia Pte Ltd

9.2 Composite Bodies Procedure for Analysis Summations Determine the coordinates of the center of gravity by applying the center of gravity equations If an object is symmetrical about an axis, the centroid of the objects lies on the axis Copyright © 2010 Pearson Education South Asia Pte Ltd

9.3 Theorems of Pappus and Guldinus
A surface area of revolution is generated by revolving a plane curve about a non-intersecting fixed axis in the plane of the curve A volume of revolution is generated by revolving a plane area bout a nonintersecting fixed axis in the plane of area Copyright © 2010 Pearson Education South Asia Pte Ltd

The theorems of Pappus and Guldinus are used to find the surfaces area and volume of any object of revolution provided the generating curves and areas do not cross the axis they are rotated Surface Area Area of a surface of revolution = product of length of the curve and distance traveled by the centroid in generating the surface area Copyright © 2010 Pearson Education South Asia Pte Ltd

Example 9.12 Show that the surface area of a sphere is A = 4πR2 and its volume V = 4/3 πR3. Solution Surface Area Generated by rotating semi-arc about the x axis For centroid, For surface area, Copyright © 2010 Pearson Education South Asia Pte Ltd

9.4 Resultant of a General Distributed Loading
Pressure Distribution over a Surface Consider the flat plate subjected to the loading function ρ = ρ(x, y) Pa Determine the force dF acting on the differential area dA m2 of the plate, located at the differential point (x, y) dF = [ρ(x, y) N/m2](d A m2) = [ρ(x, y) d A]N Entire loading represented as infinite parallel forces acting on separate differential area dA Copyright © 2010 Pearson Education South Asia Pte Ltd

Magnitude of Resultant Force To determine magnitude of FR, sum the differential forces dF acting over the plate’s entire surface area dA Magnitude of resultant force = total volume under the distributed loading diagram Location of Resultant Force is Copyright © 2010 Pearson Education South Asia Pte Ltd

9.5 Fluid Pressure According to Pascal’s law, a fluid at rest creates a pressure ρ at a point that is the same in all directions Magnitude of ρ depends on the specific weight γ or mass density ρ of the fluid and the depth z of the point from the fluid surface ρ = γz = ρgz Valid for incompressible fluids Gas are compressible fluids and the above equation cannot be used Copyright © 2010 Pearson Education South Asia Pte Ltd

9.5 Fluid Pressure Flat Plate of Constant Width Consider flat rectangular plate of constant width submerged in a liquid having a specific weight γ Plane of the plate makes an angle with the horizontal as shown Copyright © 2010 Pearson Education South Asia Pte Ltd

9.5 Fluid Pressure Flat Plate of Constant Width As pressure varies linearly with depth, the distribution of pressure over the plate’s surface is represented by a trapezoidal volume having an intensity of ρ1= γz1 at depth z1 and ρ2 = γz2 at depth z2 Magnitude of the resultant force FR = volume of this loading diagram Copyright © 2010 Pearson Education South Asia Pte Ltd

9.5 Fluid Pressure Curved Plate of Constant Width When the submerged plate is curved, the pressure acting normal to the plate continuously changes direction Integration can be used to determine FR and location of center of centroid C or pressure P Copyright © 2010 Pearson Education South Asia Pte Ltd

9.5 Fluid Pressure Flat Plate of Variable Width Consider the pressure distribution acting on the surface of a submerged plate having a variable width Since uniform pressure ρ = γz (force/area) acts on dA, the magnitude of the differential force dF dF = dV = ρ dA = γz(xdy’) Copyright © 2010 Pearson Education South Asia Pte Ltd

9.5 Fluid Pressure Flat Plate of Variable Width Centroid V defines the point which FR acts The center of pressure which lies on the surface of the plate just below C has the coordinates P defined by the equations This point should not be mistaken for centroid of the plate’s area Copyright © 2010 Pearson Education South Asia Pte Ltd

Example 9.14 Determine the magnitude and location of the resultant hydrostatic force acting on the submerged rectangular plate AB. The plate has a width of 1.5m; ρw = 1000kg/m3. Copyright © 2010 Pearson Education South Asia Pte Ltd

Solution For magnitude of the resultant force FR created by the distributed load. This force acts through the centroid of the area, measured upwards from B Copyright © 2010 Pearson Education South Asia Pte Ltd

Solution Same results can be obtained by considering two components of FR defined by the triangle and rectangle. Each force acts through its associated centroid and has a magnitude of Hence Copyright © 2010 Pearson Education South Asia Pte Ltd

QUIZ The _________ is the point defining the geometric center of an object. Center of gravity B) Center of mass Centroid D) None of the above 2. To study problems concerned with the motion of matter under the influence of forces, i.e., dynamics, it is necessary to locate a point called ________. A) Center of gravity B) Center of mass C) Centroid D) None of the above Copyright © 2010 Pearson Education South Asia Pte Ltd

QUIZ 3. If a vertical rectangular strip is chosen as the differential element, then all the variables, including the integral limit, should be in terms of _____ . A) x B) y C) z D) Any of the above. 4. If a vertical rectangular strip is chosen, then what are the values of x and y? A) (x , y) B) (x / 2 , y / 2) C) (x , 0) D) (x , y / 2) ~ Copyright © 2010 Pearson Education South Asia Pte Ltd

QUIZ 5. A composite body in this section refers to a body made of ____. A) Carbon fibers and an epoxy matrix B) Steel and concrete C) A collection of “simple” shaped parts or holes A collection of “complex” shaped parts or holes 6. The composite method for determining the location of the center of gravity of a composite body requires _______. Integration B) Differentiation C) Simple arithmetic D) All of the above. Copyright © 2010 Pearson Education South Asia Pte Ltd

QUIZ 7. Based on the typical centroid information, what are the minimum number of pieces you will have to consider for determining the centroid of the area shown at the right? 1 B) C) D) 4 8. A storage box is tilted up to clean the rug underneath the box. It is tilted up by pulling the handle C, with edge A remaining on the ground. What is the maximum angle of tilt (measured between bottom AB and the ground) possible before the box tips over? A) 30° B) 45 ° C) 60 ° D) 90 ° Copyright © 2010 Pearson Education South Asia Pte Ltd

QUIZ 3cm 1 cm 7. What are the min number of pieces you will have to consider for determining the centroid of the area? 1 B) C) D) 4 8. A storage box is tilted up by pulling C, with edge A remaining on the ground. What is the max angle of tilt possible before the box tips over? A) 30° B) 45 ° C) 60 ° D) 90 ° 30º G C A B Copyright © 2010 Pearson Education South Asia Pte Ltd

QUIZ 2cm 4cm x y 9. For determining the centroid, what is the min number of pieces you can use? A) Two B) Three C) Four D) Five 10. For determining the centroid of the area, what are the coordinates (x, y ) of the centroid of square DEFG? A) (1, 1) m B) (1.25, 1.25) m C) (0.5, 0.5 ) m D) (1.5, 1.5) m A 1m y E F G C B x D Copyright © 2010 Pearson Education South Asia Pte Ltd

Copyright © 2010 Pearson Education South Asia Pte Ltd

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