9.3 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 Provided the weight and location of the center of gravity of each of these parts are known, the need for integration to determine the center of gravity for the entire body can be neglected

9.3 Composite Bodies Accounting for finite number of weights Where represent the coordinates of the center of gravity G of the composite body represent the coordinates of the center of gravity at each composite part of the body represent the sum of the weights of all the composite parts of the body or total weight

9.3 Composite Bodies When the body has a constant density or specified weight, the center of gravity coincides with the centroid of the body The centroid for composite lines, areas, and volumes can be found using the equation However, the W’s are replaced by L’s, A’s and V’s respectively

9.3 Composite Bodies Procedure for Analysis Composite Parts Using a sketch, divide the body or object into a finite number of composite parts that have simpler shapes If a composite part has a hole, or a geometric region having no material, consider it without the hole and treat the hole as an additional composite part having negative weight or size

9.3 Composite Bodies Procedure for Analysis Moment Arms Establish the coordinate axes on the sketch and determine the coordinates of the center of gravity or centroid of each part

9.3 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

9.3 Composite Bodies Example 9.9 Locate the centroid of the wire.

9.3 Composite Bodies Solution Composite Parts Moment Arms Location of the centroid for each piece is determined and indicated in the diagram

9.3 Composite Bodies Solution Summations Segment L (mm) x (mm) y (mm) z (mm) xL (mm2) yL (mm2) zL (mm2) 1 188.5 60 -38.2 11 310 -7200 2 40 20 800 3 -10 -200 Sum 248.5 -5600

9.3 Composite Bodies Solution Summations

9.3 Composite Bodies Example 9.10 Locate the centroid of the plate area.

9.3 Composite Bodies Solution Composite Parts Plate divided into 3 segments Area of small rectangle considered “negative”

9.3 Composite Bodies Solution Moment Arm Location of the centroid for each piece is determined and indicated in the diagram

9.3 Composite Bodies Solution Summations Segment A (mm2) x (mm) y (mm) xA (mm3) yA (mm3) 1 4.5 2 9 -1.5 1.5 -13.5 13.5 3 -2 -2.5 5 -4 Sum 11.5 14

9.3 Composite Bodies Solution Summations

9.3 Composite Bodies Example 9.11 Locate the center of mass of the composite assembly. The conical frustum has a density of ρc = 8Mg/m3 and the hemisphere has a density of ρh = 4Mg/m3. There is a 25mm radius cylindrical hole in the center.

9.3 Composite Bodies Solution Composite Parts View Free Body Diagram Solution Composite Parts Assembly divided into 4 segments Area of 3 and 4 considered “negative”

9.3 Composite Bodies Solution Moment Arm Location of the centroid for each piece is determined and indicated in the diagram Summations Because of symmetry,

9.3 Composite Bodies Solution Summations Segment m (kg) z (mm) zm (kg.mm) 1 4.189 50 209.440 2 1.047 -18.75 -19.635 3 -0.524 125 -65.450 4 -1.571 -78.540 Sum 3.141 45.815

9.3 Composite Bodies Solution Summations

9.4 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 Example Line AB is rotated about fixed axis, it generates the surface area of a cone (less area of base)

9.4 Theorems of Pappus and Guldinus Example Triangular area ABC rotated about the axis would generate the volume of the cone 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

9.4 Theorems of Pappus and Guldinus 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

9.4 Theorems of Pappus and Guldinus Volume Volume of a body of revolution = product of generating area and distance traveled by the centroid in generating the volume

9.4 Theorems of Pappus and Guldinus Composite Shapes The above two mentioned theorems can be applied to lines or areas that may be composed of a series of composite parts Total surface area or volume generated is the addition of the surface areas or volumes generated by each of the composite parts

9.4 Theorems of Pappus and Guldinus Example 9.12 Show that the surface area of a sphere is A = 4πR2 and its volume V = 4/3 πR3

9.4 Theorems of Pappus and Guldinus View Free Body Diagram Solution Surface Area Generated by rotating semi-arc about the x axis For centroid, For surface area,

9.4 Theorems of Pappus and Guldinus Solution Volume Generated by rotating semicircular area about the x axis For centroid, For volume,

9.5 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

9.5 Resultant of a General Distributed Loading Pressure Distribution over a Surface This system will be simplified to a single resultant force FR acting through a unique point on the plate

9.5 Resultant of a General Distributed Loading Pressure Distribution over a Surface 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

9.5 Resultant of a General Distributed Loading Pressure Distribution over a Surface Location of Resultant Force Line of action of action of the resultant force passes through the geometric center or centroid of the volume under the distributed loading diagram