Statics of Rigid Bodies: Equilibrium, Free-Body Diagrams, and Truss Analysis, Study notes of Mathematics

A comprehensive introduction to the principles of statics of rigid bodies, focusing on equilibrium of force systems, free-body diagrams, and truss analysis. It includes numerous examples and exercises to illustrate the concepts and their application in real-world scenarios. Suitable for students studying engineering or physics, providing a solid foundation for understanding the behavior of rigid bodies under static loads.

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2023/2024

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Statics of Rigid Bodies
Prepared by: Franklin Joven Pasilbas, RCE
Lecture 4:
Equilibrium of force system
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Statics of Rigid Bodies

Prepared by: Franklin Joven Pasilbas, RCE Lecture 4:

  • Equilibrium of force system

Equilibrium

  • A body is said to be in equilibrium if the resultant of the force system that acts on the body vanishes.

Equilibrium means that both the resultant force and the resultant couple are zero.

Free-Body Diagram of a Body

  • The first step in equilibrium analysis is to identify all the forces that act on the body. This is

accomplished by means of a free-body diagram.

  • The free-body diagram (FBD) of a body is a sketch of the body showing all forces that act on it. The term

free implies that all supports have been removed and replaced by the forces (reactions) that they exert

on the body.

The homogeneous 6-m bar AB in figure is supported in the vertical plane by rollers at A and B and by a cable at C .The mass of the bar is 50kg. Draw the FBD of bar AB. Determine the number of unknowns on the FBD.

Example 1:

The homogeneous, 250-kg triangular plate in figure is supported by a pin at A and a roller at C. Draw the FBD of the plate and determine the number of unknowns.

Example 2:

An 80-N box is placed on a folding table as shown in figure. Neglecting friction and the weights of the members, determine all forces acting on member ABCDE, EFG and ACFH

Example 4:

From the figure shown, two spheres having equal weights of 360 N each is placed on top of the three identical spheres. The radius of the upper two spheres is 0.80m while that at the bottom is 1.0 m having equal weights of 420 N. Determine the:

  1. reaction at A
  2. reaction at C
  3. reaction at B

Example 5:

If the intensity of the distributed load acting on the beam is w = 3 kN/m, determine the reactions at the roller A and pin B.

Example 7:

Determine the reaction at the supports A and B.

Example 8:

The towline exerts a force P = 4 kN at the end of the 20-m-long crane boom. If θ = 30°

  1. determine the placement x of the hook at A so that this force creates a maximum moment about point O.
  2. What is this maximum moment?

Example 10:

The structure consists of two identical bars joined by a pin at B. Neglecting the weights of the bars, find the magnitude of the pin reaction at C.

Example 11:

An 80-N box is placed on a folding table as shown in figure. Neglecting friction and the weights of the members, Determine the tension in the cable connecting points B and D.

Example 13:

Statics of Rigid Bodies

Prepared by: Franklin Joven Pasilbas, RCE Lecture 5:

  • Truss

The analysis of trusses is based on the following three assumptions:

  1. The weights of the members are negligible. A truss can be classified as a lightweight structure, meaning that the weights of its members are generally much smaller than the loads that it is designed to carry.
  2. All joints are pins. In practice, the members at each joint are usually riveted or welded to a plate, called a gusset plate. However, if the members at a joint are aligned so that their centroidal axes (axes that pass through the centroids of the cross-sectional areas of the members) intersect at a common point, advanced methods of analysis indicate that the assumption of pins is justified.
  3. The applied forces act at the joints. Because the members of a truss are slender, they may fail in bending when subjected to loads applied at locations other than the joints. Therefore, trusses are designed so that the major applied loads act at the joints.
  • each member of a truss is a two-force body
  • tension and compression.

Zero-force members

There is a special case that occurs frequently enough to warrant special attention. shows the FBD

for joint G of the truss. Because no external loads are applied at G, the joint equilibrium equations

∑Fx = 0 and ∑Fy = 0 yield PGH = PGF and PGC = 0. Because member GC does not carry a force, it is

called a zero-force member. It is easily verified that the results remain unchanged if member GC is

inclined to GH and GF, When analyzing a truss, it is often advantageous to begin by identifying

zero-force members, there by simplifying the solution.