Download Engineering Science - Mechanics Statics – Course Content

Engineering Science - Mechanics
Statics – Course Content
1. Introduction
Units and Dimensions
Scalars and Vectors
Forces, Couples and Moments
Resolution of Forces
2 Lectures
2. Equilibrium
Newton’s Laws
Equilibrium equations
Free body diagrams
Trusses
4 Lectures
3. Friction
Friction on horizontal and inclined Planes
Friction Problems
11/2 Lectures
4. Distributed Forces
Centre of Area
Body forces
Pressure
Hydrostatics
Second moments of area
Buoyancy and stability
41/2 Lectures
1
Objectives
What will you be able to do?
Vectors
• Use vectors competently
• Add, subtract and multiply vectors
• Resolve vectors
• Calculate moments due to vectors
Equilibrium
• Calculate the support reactions for statically
determinate plane frame structures.
• Use principles of equilibrium to draw free body
diagrams and hence calculate the forces within
members of a pin-jointed truss.
Friction
• Define the static and dynamic coefficients of friction
• Define angle of slip
• Tackle static equilibrium problems involving friction
Pressure and Body forces
• Solve problems involving pressure and body forces
Centroids and 2nd Moments of Area
• Calculate the centroid of a cross-section
• Calculate the centre of gravity of a lamina
• Calculate the 2nd moment of Area of a cross-section
2
Hydrostatics
• Solve buoyancy and stability problems
• Calculate the hydrostatic thrust on dams and other
liquid retaining structures
Recommended Text:
Engineering Mechanics – Vol. 1 (SI Version)
By J.L. Meriam & L.G. Craig
John Wiley & Sons (1998)
Recommended Reading
I recommend that all students read “Structures – or why
things don’t fall down”, by J.E. Gorgon, published by
Penguin.
Tutorials
All tutorials are held in the Drawing Office (at the top of
the museum building)
Tuesday 9:00 – 10:00
Groups 25-48
Thursday 17:00-18:00
Groups 1-24
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Lecture I
What is statics and why study it?
Statics involves studying the forces acting on bodies that
are in equilibrium. Dynamics concerns the motion of
bodies in response to forces.
ANSWER………… DESIGN
Although this course in statics is concerned with analysis
the long-term objective is to develop the analytical skills
required for design.
Remember the Master Masons who built Notre Dame did
so with only an intuitive, or qualitative, understanding of
statics.
Conventions and Units
We will use SI units
Length…………...meter m
Time…………..…second s
Mass…………..…Kilogram kg
Force…………….Newton N mass x length / time2
Work, Energy…...Joule J force x length
4
Vectors & Scalars
A vector quantity, such as force, has magnitude and
direction and can be represented graphically by a line of
given length and direction. Forces and some other vector
quantities, in addition to these two properties also have a
line of action that is most important.
Examples: displacement, velocity, acceleration and
force
A scalar quantity has magnitude only.
Examples: mass, length, time and temperature
Properties of Vectors
•
•
•
•
•
Can be represented graphically
Can be resolved
Can be represented in Cartesian form i,j,k along x,y,z
Can be added and subtracted
Can be multiplied
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Resolution of vectors
Y
F
Fy
0
Fx
X
Fx = F cosθ
Fy = F sin θ
F = Fx 2 + Fy 2
Cartesian Representation
F
Y
Z
Fk
Fj
Fi
X
6
Vector Addition
A
C=A+B
C=B+A
A
B
Vector Subtraction
-B
A=C-B
C
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Vector Multiplication
Y
d
0
F
d
F
X
Dot (or scalar) product
Example - work done by a force
F.d = Fd cosθ
F = Fx i + Fy j
d = d xi + d y j
F.d = Fx d x + Fy .d y
Note: F.d is zero if F and d are mutually at right angles
8
Cross product
Y
0
F
r
(x,y)
X
M = r×F
The cross (vector) product arises, for example, in the
calculation of a moment (which is a vector quantity). The
moment M due to the Force F (a vector) about the point
(x,y) is at right angles to the plane of the paper and into
the page (right hand screw rule). The magnitude of the
moment is the force by the perpendicular distance from
the line of action of F to (x,y). If r is some space vector
from (x,y) to the line of action then the magnitude of the
moment about (x,y) is;
M = r×F
M = Fr sin θ
It is convenient, in general three dimensional terms, to
write the result as a vector determinant in terms of the
unit vectors i,j,k.
i
j
k
M = r×F = x
Fx
y
Fy
z
Fz
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where r = xi + yj + zk
in the two dimensional case both z and Fz are zero so that
r × F = k (xFy − yFx )
and it can be shown from simple geometry that
(xFy − yFx ) = (Fr sin θ )
Right hand screw rule
Care has to be taken in defining the set of cartesian axes
to be used because from the above definition
r × F = −(F × r ). In the usual convention the set of axes
used is a right-hand set of axes, that is, the x,y,z axes are
along the directions of the first three digits, respectively,
of the right hand when these digits are held mutually at
right angles.
10
Lecture II
Forces, Moments & Couples
Force
A force is a vector and has the properties;
• Magnitude
• Direction
• Point of Application
When two forces P and Q act on a particle, they are
together equivalent to a single force F=P+Q the vector
sum being defined by the parallelogram construction.
P+Q is called the resultant of P and Q; P and Q are
called the vector components of P+Q
Moment
Moments are also vectors and have the following
properties.
• Magnitude
• Axis – direction
• Line of action coincides with axis
Moments have dimensions Force by length (units of
Nm). The cross (vector) product of a displacement vector
and a force vector is a moment vector.
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Moment vectors can be added and resolved.
Varignon’s Theorem
The moment of a force about any point is equal to the
sum of the moments of the components of the force
about the same point.
Couple
The moment produced by two equal and opposite and
non-colinear forces is known as a couple.
Examples
Forces
Direct force - Hydraulic jack
Gravity – weight
w = mg
Wind load – quasi-static
g ≅ 9.81m s 2
Moments
Diving board
Wrench
Lever
Couple
Cork screw
Cross brace – hand auger
Hero’s Engine
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Resultants
Consider a body acted upon by a series of forces. Can
these forces be replaced by a single force?
FC
FB
FA
O
The forces can be added to find their vector sum.
FD
FC
FA
FB
∑ F = FA + FB + FC = FD
This resultant force is the net force which acts upon the
body.
13
The forces FA, FB and FC also cause moments. The
magnitude of their resultant moment does of course
depend upon the point about which the moments are
calculated.
FC
FB
FA
O
The point at which FD acts must be chosen so that the
moment due to FD equals the moments due to the three
forces FA, FB and FC.
FD
O
Thus the line of action of the resultant force is not
arbitrary.
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The Fundamental Laws of
Newtonian Mechanics
First Law
A body continues in a state of rest or constant velocity
unless acted upon by a force.
Second Law
A body on which a force acts experiences an acceleration
in the direction of the force and proportional to it: and,
more generally; the time rate of change of the momentum
vector is equal to the applied force vector.
F = ma
Or more generally F =
(in general F = kma )
d (mv )
dt
Third Law
The mutual interaction (force) between two bodies is
always equal and directed in the opposite sense on each
body (often briefly stated as “Action and reaction are
equal and opposite”).
Rigid Bodies
Newton’s laws were formulated to account for motions
of “particles” and rectilinear forces but there are also a
corresponding and analogous set of laws which govern
rotational motion and moments and these two sets must
be taken together to solve many engineering problems.
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Equilibrium
First Law
A body continues in a state of rest or constant velocity
unless acted upon by a force.
The corollary of this law is that if a body is in a state of
rest or constant velocity then the sum of the forces and
moments (which are akin to forces) must be zero.
This is the cornerstone of STATICS !!!
Consider some examples
•
•
•
•
•
Ice skaters
Ship coasting into harbour
Voussoir Arch
Pulley assembly
See-saw
The vector sum of all external forces is zero. Hence the
sum of all horizontal and vertical forces is zero.
∑ FH
= 0,
∑ FV
=0
The vector sum of the moments of all external forces
about any line is zero
∑M = 0
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Force polygons and Free-body
diagrams
mg
If a rigid body is in equilibrium then
∑ FH
= 0,
∑ FV
= 0 and
∑M = 0
If a rigid body is in equilibrium then so too is any
component part of this rigid body and the sum of the
forces and moments which act on any sub-section of the
body must also satisfy the condition,
∑ FH
= 0,
∑ FV
=0
and
∑M = 0
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Static Analysis of Plane Trusses
Supports
Graphical Conventions
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Calculating the Support Reactions
H
W
RA
x
L-x
RB
If a rigid body is in equilibrium then
∑ FH
∑ FV
= 0,
= 0 and
∑M = 0
Therefore the sum of the horizontal forces acting on the
truss is zero.
∑ FH
= 0 ⇒ HA = 0
Similarly the sum of the vertical forces on the truss is
also zero.
∑ Fv = 0 ⇒ RA + RB − W = 0
This equation is not sufficient on its own to solve for RA
and RB.
However the third equation of static equilibrium must
also be satisfied.
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∑ M = 0 ⇒ W × x − RB × L = 0
⇒ RB =
W×x
L
Therefore putting this value for RB back into the previous
equation yields
RA = W −
W×x
x

= W 1 − 
L
 L
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Joint Equilibrium
The forces in a statically determinate truss can be
determined using the equations of statics.
When a rigid body is in equilibrium the sum of the
external forces and moments acting on the body must
sum to zero. In the same way if any section of a rigid
body is in equilibrium then the sum of the forces and
moments acting on that section of the body must also
sum to zero.
For example,
FA
FB
RB
Since this section of the truss is in equilibrium we have;
∑ FH
= 0,
∑ FV
=0
Thus we have two unknown variables and two constraint
equations the hence we can solve for FA and FB.
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Method of Sections
Joint equilibrium can be used to find the forces in all the
members of the truss shown below. However if all that is
required is to calculate the force in a number of particular
members then it may be more efficient to use the method
of sections.
If a rigid body is in equilibrium then so too are all
sections of the rigid body. By carefully choosing the
section of the rigid body we can calculate the forces in
particular members of a truss without the necessity to
solve for the forces in all members.
Consider the example shown below. All that is required
is to calculate the forces in members FDF, FDG and FEG.
F
D
H
E
RA
x
G
W
L-x
RB
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Consider the free body diagram of the section of the truss
shown below. The free body diagram has been selected
such that the three members whose forces we require
have been cut.
FDF
FDG
H
E
RA
L/6
FEG
L/2
Since the truss is in equilibrium therefore the section of
the truss shown above is in equilibrium and hence the
following static equations must be satisfied.
∑ FH
= 0,
∑ FV
= 0 and
∑M = 0
The value of the reactions at the left hand support for this
problem were calculated previously and are,
RA = W −
Since
∑ FV
W×x
x

= W 1 −  and H=0
L
 L
=0


L


6
⇒ RA = FDG 

2
2
 L +L 
4
6 

⇒ FDG = 1.803 × RA
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Since
∑ FH
=0
⇒ FDF = FEG


L


4
+ FDG × 

2
2
 L +L 
4
6 

However, there are two unknowns in this equation and
hence this condition is not sufficient to establish the
magnitude of these forces. However there is one more
equilibrium condition; ∑ M = 0 .
The point which moments are taken about is arbitrary.
The equilibrium equation will be satisfied regardless of
which point is chosen. However an intelligent choice will
make the calculations easier. Remember if the point
chosen lies on the line of action of a force then that force
will not cause a moment about the point. For this
problem the ideal point to take moments about is joint D.
This joint lies along the line of action of FDF and FDG
therefore, ∑ M = 0 becomes
FEG × L = RA × L
6
2
⇒ FEG = 3RA
⇒ FDF


L


4
= 3RA − FDG × 

2
2
L
L


+
4
6 

⇒ FDF


L


4
= RA ×  3 − 1.803 ×

2
2
L +L 

4
6 

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Friction – Tribology
Consider a mass resting on a plane and with a horizontal
force being exerted on it.
mg
P
Consider two possible situations. First consider the
situation where the horizontal force P is not sufficiently
large to cause the force to move. The free body diagram
below shows the forces acting on the block.
mg
F
P
N
N
F
If the block is in equilibrium then the friction corce F
must be in equilibrium with the force P and hence F must
have the same magnitude and act in the opposite
direction to P.
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F
R
N
If the force P is increased then at some value the block
will start to move in the direction of application of P.
F
P
The load P at which the block starts to slide is the
maximum frictional force which can develop between the
block and the surface and is termed the static friction.
Once the body has started to move then a lesser force P is
required to keep the body moving (without accelerating)
the friction force which is being overcome is referred to
as the dynamic friction. The dynamic friction force is
less than the static friction force.
26
Coefficients of Friction
The maximum static friction force which can develop
between two bodies is a function of the normal force
acting at right angles to the contact surfaces and the
material from which the materials are made. The contact
area does not have a significant effect. As the magnitude
of the friction force F increases the angle of the resultant
vector R relative to the normal vector N increases.
F
R
N
For two given materials the maximum angle which R can
make with N is a constant and does not vary with
changing values for N. The maximum friction force is
given by
Fmax = µ S N
where µ S is called the static coefficient of friction. The
maximum friction force occurs when the angle between
the normal vector and the resultant vector is a maximum.
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This angle is referred to as the angle of static friction.
The angle of static friction and the coefficient of friction
are related;
tan φ S = µ S
A similar expression can be developed for the dynamic
condition. Where µ K is the dynamic coefficient of
friction and φ K is the angle of kenetic friction.
FK = µ K N
tan φ K = µ K
Note I:
The dynamic coefficient of friction is more complex than
this presentation suggests.
Note II:
The friction force must always be less than the maximum
static friction force.
F ≤ µS N
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Distributed Forces
In the physical world one does not encounter discrete
forces which act at a single point. Similarly three
dimensional objects do not have their masses
concentrated in a single point.
In many cases distributed forces can be represented by
their resultants.
29
Centre of Area- Centre of mass
The co-ordinates of the centroid of an area of a two
dimensional plane shape are given by the following
formulae.
x=
∫ xdA
A
y=
∫ ydA
A
The resultant of a uniform distributed force acting on a
flat surface will act at the centroid.
The centre of mass of a body is the point which the
resultant gravitational and inertial forces act. For this
reason it is often referred to as the centre of gravity.
The co-ordinates of the centre of mass or a body are
given by the following formulae.
x=
∫ xdm
m
y=
∫ ydm
m
z=
∫ zdm
m
The resultant of a uniform body force (gravity for
example) will act through the centre of mass of the body.
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Hydrostatics
A fluid at rest exerts a pressure that is everywhere
normal to any surface immersed in it.
The pressure intensity at a point P in a liquid is equal to
that at the free surface of the liquid together with ρgh ,
where h is the depth of P below the free surface and ρ is
the density of the liquid.
Force on an area
Taking atmospheric pressure as a datum, the total force
on a plane area A = ρghA where h is the vertical depth of
the centroid of the whole area. The resultant force will
act through a point in the immersed area A known as its
centre of pressure. The inclined depth of the centre of
pressure for a plane area is given by
∫z
Z=
2
dA
zA
Buoyancy
A liquid of density ρ exerts a vertical upwards force Vρ
on an immersed body of volume V. If the weight of the
body is greater than Vρ it will sink. If the weight of the
body is less than Vρ the body will float in such a way
that the portion immersed has a volume V ' which
satisfies the following equation
V ' ρ g = total weight of the body
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