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Exam (elaborations)

Solution Manual for Engineering Statics: Centroids and Distributed Loads (Chapter 5)

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This comprehensive solution manual provides detailed, step-by-step mathematical analysis for problems in Chapter 5, focusing on locating centroids and center of gravity for various geometries. The sources offer exhaustive coverage of plane areas, composite shapes, and 3D volumes, utilizing both algebraic methods and direct integration. You will find high-level solutions for finding the first moment of area (Q) and determining reactions for distributed loads on beams under equilibrium conditions. Key highlights include: • Centroids of Plane Areas: Exhaustive calculations for rectangles, triangles, circles, and semiannular areas using composite parts. • Direct Integration Techniques: Mastery of determining centroids for areas defined by functions such as parabolic curves (y=kx 2 ), trigonometric functions (y=asinπx/L), and inverse relationships (y=1/x). • 3D Bodies and Volumes: Detailed tables and analysis for composite volumes including cylinders, cones, paraboloids, and triangular prisms. • Distributed Loads: Solving for reaction forces at supports (A y ​ ,B y ​ ) when beams are subjected to linear and non-uniform loading distributions. • Specialized Applications: Analysis of sheet metal brackets, desk organizers, and the volume of gravel under oblique planes. • Mathematical Rigor: Full use of integration by parts and quadratic solutions to determine ratios (r 1 ​ to r 2 ​ ) that maximize or minimize centroid distances. This resource is an essential tool for engineering students looking to master the mechanics of rigid bodies and the geometric properties of areas and volumes

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CHAPTER 5
V

,
, PROBLEMV5.
1
LocateVtheVcentroidVofVtheVplaneVareaVsh
own.




SOLUTION


AreaV1:V RectangleV72VmmVbyV45Vm
m.
AreaV2:V TriangleVbV=V27Vmm,VhV=V4

A,Vmm2 xV,Vm yV,Vmm xVA,Vmm3 yVA,Vmm3
m
1 3240 36 22.5 116,640 72,900
2 -607.5 9 215 -5467.5 -9112.5
 2632.5 11,172.5 63,787.5



XV(2632.5Vmm2V)V=V11,172.5Vmm3



YV(2632.5Vmm2V)V=V63787.5V
mm3




CopyrightV©VMcGraw-
HillVEducation.VPermissionVrequiredVforVreproductionVorVdisplay.

, PROBLEMV5.
2
LocateVtheVcentroidVofVtheVplaneVareaVsh
own.




SOLUTION




A,Vin2 xV,Vin. yV,Vin. xVA,in3 yVA,in3
1 8 0.5 4 4 32
2 3 2.5 2.5 7.5 7.5
 11 11.5 39.5




XV(11Vin2V)V=V11.5Vin3



YV(11Vin2V)V=V39.5
Vin3




CopyrightV©VMcGraw-
HillVEducation.VPermissionVrequiredVforVreproductionVorVdisplay.

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