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.