Chapter - 1. Equilibrium of a Rigid Body

Transcription

1 Chapter - 1 Equilibrium of a Rigid Body Dr. Rajesh Sathiyamoorthy Department of Civil Engineering, IIT Kanpur hsrajesh@iitk.ac.in; Condition for Rigid-Body Equilibrium A rigid body is subjected to an external force and couple moment system that is the result of the effects of gravitational, electrical, magnetic, or contact forces caused by adjacent bodies. The force and couple moment system acting on a body can be reduced to an equivalent resultant force and resultant couple moment at any arbitrary point O on or off the body If this resultant force and couple moment are both equal to zero, then the body is said to be in equilibrium In statics, we deal primarily with bodies at rest (i.e. they are in the state of static equilibrium ). 1

2 Equilibrium of a system All physical bodies are inherently 3D, but many may be treated as 2D or sometimes 1D the heart () of mechanics R M Fi 0 M 0 i System Idealisation R F i 0 Equilibrium of Objects Particles (2D & 3D) 2D Rigid Bodies 3D Rigid Bodies Particle Ideal particle can not rotate. (no couple acting on it) 3 Equilibrium conditions A body is in equilibrium if all forces and moments applied to it are in balance Category 1: equilibrium of collinear forces, clearly requires only the one force equation in the direction of the forces (x- direction), since all other equations are automatically satisfied. Category 2: equilibrium of forces which lie in a plane (x-y plane) and are concurrent at a point O, requires the two force equations only, since the moment sum about O, that is, about a z-axis through O, is necessarily zero. Category 3: equilibrium of parallel forces in a plane, requires the one force equation in the direction of the forces (x-direction) and one moment equation about an axis (z-axis) normal to the plane of the forces. Category 4: equilibrium of a general system of forces in a plane (x-y), requires the two force equations in the plane and one moment equation about an axis (z-axis) normal to the plane. 2D Eq. system 3D Eq. system 2

3 Idealisation of a System A complex system can be simplified by enforcing some simplifying assumption. This process is called Idealisation of system Member Loading condition Support reactions Idealisation of a System Internal forces are not essential because these forces cancel each other. They will not create an external effect on the body Suitable idealisation of support and loading conditions, and materials used 3

4 External Loads A body is subjected to only two types of external loads; namely, surface forces or body forces. A body force is developed when one body exerts a force on another body without direct physical contact between the bodies. Surface forces are caused by the direct contact of one body with the surface of another. In all cases these forces are distributed over the area of contact between the bodies. If this area is small in comparison with the total surface area of the body, then the surface force can be idealized d as a single concentrated force, which is applied to a point on the body. If the surface loading is applied along a narrow strip of area, the loading can be idealized as a linear distributed load. 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. If rotation is prevented, a couple moment is exerted on the body. 1) Roller supports N - Roller, rocker, or ball transmits a compressive force normal to the supporting surface N 4

5 Support Reactions 2) Pin connection Pin free to turn Pin not free to turn R R x R R x M R y R y 3) Built-in or fixed support F M V Weld A A A 9 4) Flexible cable, belt, chain, or rope Support Reactions T always away from the body T tangent to the cable 5) Spring action x F For linear springs, F = kx Normal distance 10 5

6 6) Contact surfaces Smooth surfaces N - Contact force at contact point normal to the surface/contact plane N - always compressive Rough surfaces F R N -A rough surface can produce a tangential ti force (F, friction) as well as a normal force (N) - direction of F depend on situations 11 Support Reactions Summary of 2D support reactions 6

7 Support Reactions Summary of 3D support reactions Support Reactions 7

8 System Isolation - All physical bodies are inherently 3D, but many may be treated as 2D; e.g. when all forces are on the same plane. Mechanical system isolation R F i 0 - Before we apply the equilibrium conditions M M 0 we need to know what force or moment are involved. F F F F Isolate body i - Draw Free Body Diagram (FBD) FBD is used to isolate body (or bodies / system of bodies) so that force/couple acting on it can be identified. 15 Construction of FBD 1) Pick body/combination of bodies to be isolated 2) Isolating the body. Draw complete external boundary of the isolated body 3) Add all forces and moments Label all the known and unknown force magnitudes and directions on the diagram The sense of a force having an unknown magnitude can be assumed 4) Indicate a coordinate system 5) Indicate necessary dimensions y F x 150 cm 50 cm body of interest Free Body Diagram N mg Most important step is solving problems in mechanics. *** If an FBD is not drawn (when it is needed), you will get no credit (0 point) for the whole problem!!!! *** 8

9 Free Body Diagram W FBD of A 17 F 1 F 2 F 3 y F V F 1 F 2 F 3 P P A mass m B mass m A M No internal force is shown in FBD y x x P F P V M W=mg W=mg y x N M Correct? A B m N F V T not always mg 18 9

10 Free Body Diagram - Examples The sphere has a mass of 6 kg and is supported as shown. Draw a free-body diagram of the sphere, the cord CE, and the knot at C. 19 How many unknowns? How many Equations? o Fx 0 FCBAcos60 FCD 0 o Fy 0 FCBAsin 60 F CE 0 F CE 58.9 FCBA o o sin 60 sin 60 o o F F cos cot 60 CD CE 58.9 F F N F CE F CE Fy 0 FCE F CE 0 y x 0 F F y CE F CE 58.9 N F F 58.9 N CE CE Action-reaction pair - use same symbol - opposite direction 10

11 Example Determine the tension in cables BA and BC necessary to support the 60-kg cylinder 3D -> 2D Force Equilibrium is sufficient to solve the problem! Example Determine the tension in each cord used to support the 100-kg crate Force Equilibrium is sufficient to solve the problem! 3D 11

12 Free Body Diagram - Examples Two smooth pipes, each having a mass of 300 kg, are supported by the forks of the tractor. Draw the FBDs for each pipe and both pipes together. 23 Two- and Three Force Members The solutions to some equilibrium problems can be simplified by recognizing members that are subjected to only two or three forces. Two-Force Members: If a member is subjected to only two forces, it is called a two-force member.. Two force member Due to Force equilibrium Due to Moment equilibrium For any two-force member to be in equilibrium, the two forces acting on the member must have the same magnitude, act in opposite directions, and have the same line of action, directed along the line joining the two points where these forces act. 12

13 Two- and Three Force Members The solutions to some equilibrium problems can be simplified by recognizing members that are subjected to only two or three forces. Three-Force Members: If a member is subjected to only three forces, it is called a threeforce member. Due to Moment equilibrium i Moment equilibrium can be satisfied only if the three forces form a concurrent or parallel force system. Example The lever ABC is pin supported at A and connected to a short link BD. If the weight of the members is negligible, determine the force of the pin on the lever at A. Two Force Member Three Force Member 26 13

14 Example The cord shown in figure supports a force of 100 N and wraps over the frictionless pulley. Determine the tension in the cord at C and the horizontal and vertical components of reaction at pin A. 0.5m 0.5m 100 N 100 N FBD of Cord FBD of Pulley 100 N FBD of combined cord and pulley 27 Example The wing of the jet aircraft is subjected to a thrust of T = 8 kn from its engine and the resultant lift force L = 45 kn. If the mass of the wing is 2.1 Mg and the mass center is at G, determine the x, y, z components of reaction where the wing is fixed to the fuselage A. 14