ME13A: CHAPTER FOUR

EQUILIBRIUM OF RIGID BODIES

4.1 INTRODUCTION

Equilibrium of a rigid body requires both a balance of forces, to prevent the body from translating with accelerated motion, and a balance of moments, to prevent the body from rotating.

For equilibrium: F = 0 and Mo = (r x F) = 0 In three dimensions: Fx = 0 Fy = 0

Fz = 0 Mx = 0 My = 0 Mz = 0

4.2 SOLUTION OF REAL PROBLEMS IN TWO DIMENSIONS

Step 1: Examine the space diagram and select the significant body - the free body diagram. Detach this body from the space diagram and sketch its contours

Step 2: Identify and indicate all external forces acting on the body. The external forces consist of the externally applied forces, the weight and reactions of the body at the point of contact with other external bodies.

Step 3: All the important dimensions must be included in the free body diagram.

Types of Reactions Contd.

Support Reactions

If a support prevents the translation of a body in a given direction, then a force is developed on the body in that direction.

Likewise, if rotation is prevented, a couple moment is exerted on the body.

4.4 STATICALLY DETERMINATE AND INDETERMINATE REACTIONS. TYPES OF CONSTRAINTS IN TRUSSES

Example: Find the Reactions in the Truss Below

Difference Between Improper and Partial Constraints

In improper constrained case, one equation of equilibrium has been discarded but in the partial case, there are only two unknowns so the two equations were used to determine them.

Conclusion on Supports

Apart from structures designed for movement, partial or improper constraints should be

avoided in design. A rigid body will be improperly constrained when the supports are

such that the reactions are either parallel or concurrent. An improperly constrained body

also has statically independent reactions.

P Q R

R2 B

R1

Concurrent reactions

Example

A truck mounted crane is used to lift a 3 kN compressor. The weights of the boom AB and of the truck are as shown, and the angle the boom forms with the horizontal is = 40o. Determine the reaction at each of the two (a) rear wheels (b) front wheels D

Solution

4.5 TWO AND THREE-FORCE MEMBERS

The solution of some equilibrium problems are simplified if one is able to recognise members that are subjected to only two or three forces.

  4.5.1 Two-Force Members: When a

member is subject to no couple moments and forces are applied at only two points on a member, the member is called a two-force member.

Two-Force Members

Consider a two-force member below in (a). The forces in the two points are first resolved to get their resultants, FA and FB. For

translational or force equilibrium: F3 F2

FA

F1 FA

FA FB = -FA

B

FB = - FA

FB = - FA

(a) (b) (c)

Two-Force Members Contd.

( F = 0) to be maintained, FA must be of equal magnitude and opposite direction to FB.

For rotational or moment equilibrium: ( Mo = 0) to be satisfied, FA must be

collinear with FB. See other two force members in (c)

below.

Three Force Members

Example One end of rod AB rests in the corner A and

the other is attached to cord BD. If the rod supports a 200-N load at its midpoint, C, find the reaction at A and the tension in the cord.

4.6 EQUILIBRIUM IN THREE DIMENSIONS

4.6.1 Reactions at Supports: See Fig. 4.10 of text book for possible supports and connections for a three dimensional structure.

  4.6.2 Equilbrium of a Rigid Body in Three

Dimensions Six scalar equations are useable (see

section 4.1)

Equilbrium of a Rigid Body in Three Dimensions Contd.

Problems will better be solved if the vector form of the equilbrium equations is used i.e.

F = 0 Mo = (r x F) = 0 Express the forces, F and position vectors, r

in terms of scalar components and unit vectors. Compute all vector products either by direct calculation or by means of determinants.

Solution Concluded

Recall that Equation (1) is:

Ax i + (Ay - 2 TB + Dy) j + (Az - 2 TC + DZ ) k = 0 … (1)

From (1), Ax i = 0 i.e. Ax = 0

Ay - 2 TB + Dy = 0 i.e Ay = 2 TB - DY

= (2 x 90) - 60 = 120 N

Az - 2 TC + DZ = 0 i.e. Az = 2 TC - DZ =

(2 x 150) - 166.7 = 133.3 N

A = 120 N j + 133.3 N k

Example

A 35-kg rectangular plate shown is supported by three wires. Determine the tension in each wire.