Lecture 1 - Mech 351 Statics & Dynamics of Rigid BodiesPage | 14Engr. Rejie C. Magnaye

INTRODUCTION

ENGINEERING MECHANICS Science that deals/considers the effects of forces on rigid bodies. A branch of physical sciences concerned with the state of rest or motion of bodies that are subjected to the actions of forces. Mechanics is the study of forces that act on bodies and the resultant motion that those bodies experience. With roots in physics and mathematics, Engineering Mechanics is the basis of all the mechanical sciences: civil engineering, materials science and engineering, mechanical engineering and aeronautical and aerospace engineering. Engineering Mechanics provides the "building blocks" of statics, dynamics, strength of materials, and fluid dynamics. Engineering mechanics is the discipline devoted to the solution of mechanics problems through the integrated application of mathematical, scientific, and engineering principles. Special emphasis is placed on the physical principles underlying modern engineering design.

PARALLELSTATICSNON CONCURRENTFORCE SYSTEMCONCURRENTDYNAMICSAPPLICATIONSTRUSSESFRICTIONCENTROIDSKINEMATICSKINETICSTRANSLATIONROTATIONROTATIONTRANSLATIONPLANE MOTIONPLANE MOTIONPrinciples of Statics Fundamental Concepts and Definition

Rigid BodyDefinite amount of matter the part of which are fixed in position relative to each other.

Basic Quantities Length Time Mass ForceForce Considered as a push or a pull. As that which changes or tending to change the state of motion of the body (external effect) As that which produce stress and deformation on the body (internal effect of force) Mutual interaction between two bodies that produces deformation of the bodies or affects the motion of the bodiesContact forces direct contact of bodies.Field forces gravitational, electrical and magnetic forces.

Characteristic of ForcesMagnitude- Position of its line of action- Direction in which the force acts along its line of action

Principle of Transmissibility- states that the external effect of a force on a body is the same for all points of application along its line of action (i.e independent of the point of application)

Illustration:

Considered three identical bars subjected to the loading shown. The bar is loaded at its ends by two forces that have equal magnitude and are oppositely directed along the same action.

Scalars and Vector QuantitiesScalar Quantities which possess magnitude only can be added arithmetically Can be added arithmeticallyVector Quantities which possess both magnitude and direction A vector of quantity can be represented geometrically (graphically) by drawing a line acting, the length of the line representing to some scale the magnitude of the quantity. An arrow is placed on the line usually at the end to denote the sense of the direction.

Axioms of MechanicsThe Parallelogram Law the resultant of two forces is the diagonal formed on the vectors of these forces.

Triangle Law (Corollary to Parallelogram Law) If two forces are represented by their free vectors placed tip to tail, their resultant vector is the third side of the triangle, the direction of the resultant being from tip to tail of the first vector to the tip of the last vector.

Special Case: if the angle between two forces becomes zero or 180, the forces act along the same line (i.e forces are collinear). By taking one direction as positive and the other direction as negative, resultant therefore is their algebraic sum.

1. Two forces are in equilibrium only when equal in magnitude, opposite in direction and collinear in action.2. A set system of forces in equilibrium may be added to any system of forces without changing the effect of the original system.3. Action and reaction forces are equal but oppositely directed.

Introduction to Free Body Diagram One of the most important concepts in engineering mechanics is that of free body diagram.

Free Body Diagram Sketch of the isolated body which shows only the forces acting on the body It may consist of the entire assembled structure or a isolated part of it

Numerical CalculationsOne of the first things a student should is the ability to organize his work in a neat and orderly manner. Properly arranged work helps to eliminate personal errors but also permits checking by another person a frequent occurrence in engineering offices.

Procedures:1. After identifying the problem, start by constructing a neat diagram of the quantities involved.2. State as concisely as possible what data are given and what information is required.3. Errors are frequently caused by mental substitution in equations and subsequent failure to include the term in the equation. For this reason, write down the equation you want to use before substituting in it. If a equation is not used, write a short note indicating the principle used or the operation performed.

Dimensional HomogeneityThe terms of any equation used to describe a physical process must be dimensionally homogeneous (i.e each term must be express in the same units)

Resultant of Forces Resultant The effect of a system of forces on a body is usually expressed in terms of a resultant The value of this resultant determines the motion of the body (i.e if R = 0, R = 0, in motion)

Forces systems Any arrangement where two or more forces act on a body or group of related bodies Coplanar Force System When line of action of all forces in system lie on one plane Concurrent Force system Lines of action of forces passes through a common point (point of concurrency) Parallel Force System Lines of action of the forces are parallel Non concurrent Force System Also termed as the general coplanar force system Lines of action of the forces neither are parallel nor intersect at a common point Non Coplanar Force System (Force System on Space) Line of action of the forces not on the same plane

Forces and components Resolution of VectorsA vector may be resolved into two component having known lines of action by using parallelogram law.For the figure shown, if R is to be resolve into components acting along line and b, one starts at the head of R and extend a line parallel to a until it intersect b. likewise a line parallel to b is drawn from head of R to the point of intersection with a.

Example Problem1. The tension T in the vertical cable equals the weight of the crate. Calculate the components Tt and Tn along and normal to the boom, respectively of the force T applied to the boom A by the crate.

2. The 10 kN vertical force is to be replaced by two forces F1 directed along 45 line a a and F2 which has a magnitude of F1 and the counter clockwise angle made by x axis. (F1= 10.81 kN with = 17.1; F1= 3.33 kN with 72.9)

3. At what maximum angle should the force F be directed so that the magnitude of its components along CA does not exceed 80percent of the magnitude of its component along BC? (ans. 53)

4. The hydraulic cylinder exerts a force of 40 kN in the direction of its shaft against the load that is hoisting. Determine the components Fn and F1 normal and tangent to AB for the position =30

Resolution of Force in Mutually Perpendicular ComponentsIn engineering, forces are not generally at right angles. While the resultant of such forces may be found graphically (parallelogram law), it is not convenient to do so. It is frequently desirable to resolve each force into a pair of right angled components for analytical computation.

Direction of F given in terms of Slope

By ratio and proportion

Fy = F ()Fx = ()

Components Given or KnownF =

tan x = Fy/Fx; the direction of F is determined by sign of its componentsExample Problem 1. At what angle must the 400N force be applied in order that the resultant R of the forces will have a magnitude of 1000N? for this condition will be the angle between R and the horizontal?

2. It is desired to remove the spike from the timber by applying a force along its horizontal axis. An obstruction A prevents direct access so that two forces one 1.6kN and the other P, are applied by cables as shown. Compute the magnitude of P necessary ensure tension T along the axis of the spike. Also find T.

3. Calculate the magnitude of single force R is equivalent to the two forces shown. Also calculate the angle measured counter clockwise from the positive axis.

Resultant of Three or More Concurrent Force System- requires the sum of three or more vectors

Graphical method Analytical methodRx=FxRy=Fy

R=

Tan x=

Example Problems:

1. Calculate the magnitude of tension T and the angle for which the eyebolt will be under a resultant downward force of 15kN.

2. Determine the resultant R of the four acting on the gusset plate. Also find the magnitude of R and the angle x which resultant makes with the x-axis.

Moment of a Force Moment of a force about an axis or line is the measure of its ability to produce turning or twisting about the axis. The magnitude of the moment about an axis which is perpendicular to the plane containing the line of action of the force is defined as the product of the force and the perpendicular distance from the axis to the line of the force moment is twice the area of the triangle formed by joining the center of moments with the ends of the force.M=F*d

Signs and Units Unit (kN-m, N-m, lb-ft, lb-in, kip-ft, kip-in) Sign is arbitrarily chosen whichever is more convenient

The principle of Varignons TheoremThe moment of the resultant forces is equal to the sum of the moment of its components ApplicationsIn some cases it is more convenient to determine the moment of the force from the sum of the moments of its components rather than from the force itself.

Mo+ = F*dMo+= Fx *y - Fy*x (using components at pt. A)

Mo+= Fx*Fy (using components at pt. B)

Ly = (y intercept; distance from the origin where the force intersects the y-axis)

Mo+= Fy* Ix (using components at pt. C)

Lx = (y intercept; distance from the origin where the force intersects the x axis)

Example Problems:1. Calculate the moment about point A exerted by the 160 kN force supported by the hoisting cable of the tractor crane for the position shown.2. In raising the flagpole from the position shown, the tension T in the cable must supply a moment about 0 of 72kN m. Determine T?3. Determine the angle , which will maximize the moment MO of the 200N force about the shaft axis at O. also compute MO

Resultant of Parallel Force SystemResultant is determined when it is known in: Magnitude Direction Position

MagnitudeR = FR = T + S P Q[ (+); (-)]Position of the Resultant from any point (apply Varignons Theorem)MO = R* drMO = T* dt + P * dt + Q * dq S * ds (using components)

Example Problems1. Determine the height h above the base B at which the resultant of the three forces acts.

2. Determine the magnitude and position of the resultant from A of the forces acting on the cantilever beam showed. (R = 1.5kN, downward; d= 11m)

3. Determine the magnitude of the resultant and its position from O.

The special case in which the resultant has zero magnitude but does have a moment is said to consists of a couple. Define as made up of two equal, parallel, oppositely directed forces The perpendicular distance between the action lines of the forces is called moment arm of the couple Theirsum is constant and independent of the moment armC = F * d

Since the only effect of a couple is to produce a moment that is independent of the moment center, the effect of a couple is unchanged if (equivalent couple):

Original Couple The couple is rotated through any angle in its plane The couple is shifted to any other position in its plane The couple is shifted to a parallel plane Changing the magnitude of F of each force and the perpendicular distance d while keeping the product F * d constantC = F*dC = 10 (2)C = 20kN m y

Example Problems:1. To close a gate valve it is necessary to exert two forces of 60 lbs at opposite sides of a hand wheel 3 ft. diameter. Through an accident the wheel is broken and the valve must be closed by thrusting a bar through a slot in the valve stem and exerting a force 4 ft out from the center. Determine the force required.

2. A vertical force P at A and another vertical force F at V in the figure shown, produce a resultant of 100 lbs down at D and counter clockwise couple C of 200 lbs - ft. find the magnitude and direction of forces P and F.