Principle of Engineering ENG2301 F Mechanics Section F Textbook: F A Foundation Course in Statics and Dynamics F Addison Wesley Longman 1997

Syllabus Overview F A Statics l B Dynamics

Units F forceNewton (N) F stressNewton per metre squared (N/m 2 ) F or Pascal, 1 Pa = 1 N/m 2 (Pa) F pressureNewton per metre squared (N/m 2 ) F or bar, 1 bar = 1x10 5 N/m 2 (bar) F moment, torque, couple F Newton. Metre (Nm)

Units F Most commonly used prefixes F microx  F millix m F kilox 10 3 k F megax 10 6 M F giga x 10 9 G F * Note Capitals and lower case letters are important

Scalars and Vector F Two kind of quantities: s Scalar s Vector F Scalar quantities have magnitude but no directional properties F can be handled by ordinary algebra, e.g. c= a+b, c= 8 if a=3, b= 5 F e.g. time, mass, speed and energy etc. etc....

Vector F Associated with directions and magnitude F e.g. Force, displacement, acceleration and velocity F Can be represented by a straight line with arrowhead and the magnitude is shown by the length  l

Vector Addition and Subtraction F By Triangle or Parallelogram laws F Addition l V = V 1 + V 2 V is called the resultant vector

Vector Addition and Subtraction F Subtraction l V’ = V 1 - V 2 can be regarded as V’ = V 1 + (- V 2 ) l - V 2 is drawn in the opposite direction V’ is the resultant vector

Vector Addition and Subtraction F Adding more than two vectors l V’ = V 1 + V 2 + V 3 + V 4

Resolution of Vectors F Any vector can be resolved into components F Commonly resolve into two components perpendicular to each other l V = V x + V y V x = V cos  V y = V sin  magnitude V =  V x 2 + V y 2 )  = tan -1 (V y /V x )

Force and Newton’s First Law F First Law F First Law - If the resultant force acting on a particle is zero, the particle will remain at rest (if originally at rest), or will move with constant speed in a straight line (if originally in motion). F State of Equilibrium F State of Equilibrium - Equilibrium exists when all the forces on a particle are in balance. The velocity of a particle does not change, if the particle is in Equilibrium.

Interpretation of First Law F A body is in Equilibrium if it moves with constant velocity. A body at rest is a special case of constant velocity i.e. v = 0 = constant. F For a body to be in Equilibrium the resultant force (meaning the vector addition of all the forces) acting on the body must be zero. F A Force can be defined as 'that which tends to cause a particle to accelerate', assuming that the force is not in Equilibrium with other forces acting on the body.

Force F A force cannot be seen, only the effect of a force on a body may be seen. F Force Units: S.I. Unit,Newton, (N) or (kN) F Force is a vector quantity. It has both magnitude and direction.

Force Vectors F Polar and Rectangular Coordinates

Example 1 F Calculate the components in rectangular coordinates of the 600 N force. F Solution

Example 2 F A force vector has the components 600 kN and 300 kN in the x and y directions respectively, calculate the components in polar coordinates. F Solution

Resultant Force F Parallelogram Method

Resultant Force F Algebraic Method

Resultant Force F Triangle of Forces Method Order is not important

Example 3 F Find the magnitude and direction of the resultant (i.e. in polar coordinates) of the two forces shown in the diagram, F a) Using the Parallelogram Method F b) Using the Triangle of Forces Method F c) Using the algebraic calculation method F Solution

Example 3 (Solution) Or from +ve x axis

Equilibrium of Concurrent forces Equilibrant E are equal and opposite to Resultant R E = -R

Conditions for Equilibrium F Coplanar: all forces being in the same plane (e.g.only x-y plane, no forces in z direction) F Concurrent: all forces acting at the same point (particle) For three forces acting on a particle

Some Definitions F Particle is a material body whose linear dimensions are small enough to be irrelevant F Rigid Body is a body that does not deform (change shape) as a result of the forces acting on it.

Polygon of Forces F Equilibrium under multiple forces Rigid body under concurrent forces Forces acting on particle

Resultant and Equilibrant Resultant = - Equilibrant R = - F5

Example 4 F The diagram shows three forces acting on a particle. l Find the equilibrant by drawing the polygon of forces.

Newton’s Third Law F The forces of action and reaction between bodies in contact have the same magnitude, but opposite in direction.

Action and Reaction

Free Body Diagram F Free Body Diagram F Free Body Diagram - used to describe the system of forces acting on a body when considered in isolation R mg R R R

Free Body Diagram

System of Particles or Bodies Two or more bodies or particles connected together are referred to as a system of bodies or particles. External Force External forces are all the forces acting on a body defined as a free body or free system of bodies, including the actions due to other bodies and the reactions due to supports.

Transmissibility of Force

Load and Reaction F Loads are forces that are applied to bodies or systems of bodies. F Reactions at points supporting bodies are a consequence of the loads applied to a body and the equilibrium of a body.

Tensile and Compressive Forces F Push on the body which is called a compressive force F Pull on a body which is called a tensile force

Procedure for drawing a free body diagram F Step 1: Imagine the particle to be isolated or cut “free from its surroundings. Draw or sketch its outlined shape. F Step 2: Indicate on this sketch all the forces that act on the particle. These forces can be applied surface forces, reaction forces and/or force of attraction.

Procedure for drawing a free body diagram

F Step 3: The forces that known should be labeled with their proper magnitudes and directions. Letters are used to represent the magnitudes and directions of forces that unknown.

Example 5

Example 6

Example 6 (Solution) F resultant R of the two forces in tow ropes No.1 & No. 2 from the components in the x and y directions:

Example 6 (Solution) Equilibrant E = - R

Example 6 (Solution) Resultant R is the sum of the actions of the tow ropes on the barge Equilibrant E is the reaction of the barge to the ropes E = - R

Moment and Couple F Moment of Force F Moment M of the force F about the point O is defined as: M = F d where d is the perpendicular distance from O to F F Moment is directional

Moment and Couple Moment = Force x Perpendicular Distance

Resultant of A System of Forces F An arbitrary body subjected to a number of forces F 1, F 2 & F 3. l Resultant R = F 1 + F 2 + F 3 l Components Rx = F 1x + F 2x + F 3x Ry = F 1y + F 2y + F 3y

Resultant of A System of Forces F Resultant moment Mo = Sum of Moments l Mo = F 1 d 1 + F 2 d 2 + F 3 d 3 l Mo = R d

Couple F For a Couple R =  F = 0 But Mo  0 l Mo = F(d+l) - Fl = Fd l Moment of couple is the same about every point in its plane

Example 7 F Calculate the total (resultant) moment on the body.

Example 7 (Solution) F Taking moments about the corner A F Note that the forces form two couples or pure moments 3.6 Nm and 3.0 Nm (resultant force =0, moment is the same about any point).

Equilibrium of Moments F The sum of all the moments is zero when the body is in moment equilibrium. or F If the body is in equilibrium the sum of the moments of all the forces on acting on a rigid body is the same for all points on the body. It does not matter at which point on a rigid body you choose for taking moments about

Example 8 F Calculate the resultant moment and the equilibrant moment.

Example 8 (Solution) Take moment about A Take moment about B

Example 8 (Solution) F Note that the body is not in vertical and horizontal equilibrium. F There is no unique value for the resultant moment. F The value depends on where the resultant force acts, ie., depends on the perpendicular distance between the resultant force and the point for taking moment. F Therefore, the moments about A and B are different.

Example 9 F Cantilever beam Find the reaction force and moment at the built in end

Example 9 (Solution) F Taking moment about A

General Equations of Equilibrium of a Plane (Two Dimensional) Rigid Body (Non-concurrent forces) For complete equilibrium, all 3 equations must be satisfied

Types of Beam Supports Simply supported beam

Types of Beam Supports F

Types of Supports and Connections Simply supported beam

Types of Loading on Beams F

F

F

F

Example 10 F Find the reactions at the supports for the beam shown in the diagram.

Example 10(Solution)

Example 11 F Express F in terms of m, a and b.

Example 11(Solution) Ratio a/b is called Mechanical Advantage

Example 12 F Find reaction forces at supports A and B.

Example 12 (Solution) F Consider the sum of vertical forces and horizontal forces are zero, and since R Ax = 0 as point B can only take up vertical force. F R Ax = 200 x sin30 o = 100 N F R Ay + R By = cos30 o F = N F = N

Example 12 (Solution) F Taking moment about A, F R By x 0.3 = 350 x cos30 o x 0.4 F R By x 0.3 = N F R By = N F Therefore the reaction at point B is 405.9N upward. F R Ay = = N

Example 12 (Solution) F The reaction at point A is N upward and 100 N to the left. F Resultant at A is: F R A =  ( ) F = N  angle  = o 

Example 13 F Find reaction forces at supports A and B.

Example 13 (Solution)

Example 14

Example 14 (Solution)

F  Fy = 0:Ry - 75 = 0Ry = 75 N F  Fx = 0:T - Rx = 0T = Rx F  M O = 0:75 x T x 100 x cos 20 0 = 0 F ThereforeRx = T = 200 N F R =  (Rx 2 + Ry 2 ) = 214 N F  = tan -1 (Ry/Rx) = 20 o 33’

Example 14 (Solution) F Reaction R is that exerted by the frame on the bellcrank, which is equal and opposite to that on the chassis.

Example 14 (Graphical Solution) F Having determined the line of reaction R, a scaled force polygon can be drawn. F By measurement, T = 200 N, R = 214 N