Transcript ENG2000 Chapter 2 Structure of Materials

ENG2000 Chapter 6
Forces and Structures
ENG2000: R.I. Hornsey
Struct: 1
Overview
• We would like to apply our increased
understanding of the microscopic properties of
materials to larger scale structures
• This branch of engineering is called statics
 because nothing is supposed to move
• We will briefly review forces, moments and
couples
• As well as equilibrium and the free body diagram
• From there, we can consider more complex
issues, such as trusses and beams
• This is a brief overview of a large field
 and we can only touch a small part of the total subject
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Where you’ve seen some of this before
• Phys1010 covers
 forces, vectors and motion
 equilibrium and elasticity
• See Halliday, Resnick and Walker Chs 2 – 6 & 13
• While we will give a brief reminder here, it is
assumed that you know this material already
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Forces
• Force are vectors
 in order to be fully described, the magnitude and direction of
a force is required
• There are three types of vectors
 free vector – has a magnitude and direction, but its line of
action does not pass through a unique position in space (e.g.
all points in a car have the same velocity)
 sliding vector – has a unique magnitude, direction and its
line of action passes through a unique point, but the point of
application of the force can be anywhere along the line of
application (e.g. horizontally pushing or pulling a car)
 fixed vector – a vector with a specific point of application
(e.g. a load on a beam)
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Transmissability
• In all the cases considered in this course, we will
assumed that bodies are rigid
 no forces are generated within the body
• Hence, it is assumed that forces are transmitted
perfectly through the body
 “the external effect of a force on a body is the same for all
points of application of the force along its line of action”
forces have the same effect
• It should be noted that this refers to the external
effect
 internally, the microscopic response of the body might be
quite different
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Types of force
• A number of classifications of forces can be
drawn
• Contact (physical pull) versus non-contact (e.g.
gravity)
• Point forces versus distributed forces
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The lines of actions of forces due to
ropes attached to an eyelet are
considered to pass through the same
point
Such force systems are termed
concurrent
Concurrent force systems can be
replaced by a single resultant force
These forces are coplanar
These forces are collinear
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Moment
• Any force that is not through the centre of mass
of an object causes a tendency to rotation of the
object about some axis
• The moment of the force, F, about O is a measure
of the tendency to rotation about axis A–A
 which is perpendicular to the plane containing F and O
A
MO O
O
A
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MO = Fd
F
d
F
(moment is a vector
quantity)
O is the moment centre
d is the moment arm
A–A is the moment axis
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The original Starship
Enterprise has engines that
clearly do not pass through
the centre of mass.
The resulting moment would
cause the ship to rotate
This was fixed for NCC1701d
ENG2000: R.I. Hornsey
http://www.cs.umanitoba.ca/~djc/startrek/pics/BigEnPlanet.gif
http://www.research.ibm.com/people/d/dfb/trek/ncc1701.jpg
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Principle of moments
• States that the moment of the resultant of a
system of forces is equal to the vector sum of the
moments of the individual forces
 about the same point or axis
MR = Rd = R(h cos )
O
MA = Aa = A(h cos a)
a

h
MB = Bb = B(h cos b)
a
R
d
b
b
also Rcos = Acosa + bcosb
A
B
substituting for the cos’s from the
top three expressions and
multiplying through by h gives:
MR = MA + MB
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Centre of Mass
• For any object or collection of objects, there is a
location at which all the mass can be thought of
as being concentrated
 so a force acting through this centre of mass will cause no
rotation
• On earth for objects small compared with the
earth’s curvature, the centre of gravity is
essentially identical to the centre of mass
• A similar concept applies to geometrical shapes
(lines, areas, volumes), but the term is centroid
 because a shape has no physical existence
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Calculating centre of mass
• To derive the centre of mass, we make use of the
definition of centre of gravity
 the moment of the entire body (i.e. at COG) around an axis
is equal to the sum of all the moments of the elemental
masses comprising the body
 we assume that COG and centre of mass are the same, so
we can divide through by g
About y-axis:
Moment of whole body is = mgxcom
elemental mass at x, y, z
z
Moment of all elements = g∫x dm
G
y
mg
xcom gdm
zcom
Hence
ycom
Similarly:
x
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xcom = (∫x dm)/m
ycom = (∫y dm)/m
zcom = (∫z dm)/m
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Couple
• A couple consists of two equal, non-collinear
forces that are anti-parallel
 i.e. opposite directions
• The net force in any direction is zero, so a couple
gives rise to ‘pure’ rotation
F
d
O
F
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MO = Fd
provided O is in the
same plane as the couple
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Equilibrium of a particle
• We will consider only particles in equilibrium
 vector sum of all external forces = resultant force = 0
 a body is called a particle when its shape and size have no
influence on the problem
• Typically, the system to be analysed consists of a
number of interacting bodies/particles
• To isolate the one of interest, we draw a free body
diagram of just that object
 this diagram includes all the external forces on the body of
interest, whether directly applied to it or resulting from
interaction with the other bodies (not shown on the FBD)
• Constructing the FBD requires two main steps …
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Free body diagram
• 1. Decide which body to isolate and sketch the
external boundary around that object
• 2. Draw all the forces (including unknowns) with
vectors in their correct positions
 forces at interfaces between objects are assumed only to act
normal to the surface (since we assume no friction)
 ropes, chains, etc are deemed to be totally flexible (hence
transmit only axial forces)
• See examples …
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Structures
• We now have some basic tools to use on the
analysis of simple structures
• A structure can be thought of as a system of free
bodies held in equilibrium
 so a structure must be dismantled in order to examine the
internal forces
• First we will look at plane trusses
 e.g. some bridges, the frames that support the house roof
• Then we will consider concentrated loads on beams
 before moving on to distributed loading
• In all cases, the structures will be statically
determinate
 i.e. sufficient is known in order to solve for all the unknowns
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Warren truss
• For an informative look at different types of
bridge and roof trusses, see
 http://pghbridges.com/basics.htm
 http://www.trussed-rafters.co.uk/ttypes.htm
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http://www.aku.ac.ir/faculty1/aliniamm/Structural%20Slides/trusses/p/IMG0009.jpg
Struct: 18
Pratt truss
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http://www.bcsj.org/rr/bcsj/trackplan/bcsj_ver_17/MillCity/pix/pinpratt_01_m.jpg
http://www.aku.ac.ir/faculty1/aliniamm/Structural%20Slides/trusses/p/IMG0005.jpg
Struct: 19
Other trusses
ENG2000: R.I. Hornsey
http://www.kamloopsjunction.com/howe_tsa.jpg
Struct: 20
http://icampus1.mit.edu/2.973/images/Baltimore_bridge.jpg
Assumptions for truss calculations
truss members are
connected by frictionless
pins – no moment
provides vertical
& horizontal
support but no
moment
members are
weightless
provides vertical
support but no
moment or
horizontal force
loads only applied to ends of members
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Simple analysis
TAB
P
TBC
B
P
A
C
TAB
TAB
TAC
Ax
Ay
TAC
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TBC
TBC
TAC
Cy
TAC
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• It is important to identify which members are in
tension and which in compression
 long, thin truss members are relatively weak in compression
and tend to buckle
 they need to be identified and strengthened if necessary
• The roller support not only allows for expansion
of the structure
 it removes one unknown – the horizontal reaction – to make
the structure determinate
• Trusses are regarded as rigid if they maintain
their shape if the supports are removed
rigid
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non-rigid
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Method of joints
• The diagram on the previous slide was an
example of the method of joints
 the structure is ‘exploded’ and the forces on each member
and joint are identified
 the equilibrium analysis of each joint gives enough
simultaneous equations to solve for unknowns
 it’s really the ‘brute force’ approach
• There are several conventions
 initially we assume all members are in tension – if it turns out
they are in compression, the sign (direction) will tell us so
 we also draw the arrows as if members are in tension
 forces are labelled e.g. TAC regardless of whether they go
from A to C or C to A
• Choose a joint with few unknown forces to start
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1kip = 1000 pounds per
square inch (psi)
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Zero-force members
• Frequently the analysis can be simplified by
identifying members that carry no load
 two typical cases are found
• When only two members form a non-collinear
joint and there is no external force or reaction at
that joint, then both members must be zero-force
A
P
B
C
TCB
D
TCD
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E
If either TCB or TCD ≠
0, then C cannot be
in equilibrium, since
there is no restoring
C force towards the
right.
Hence both BC and
CD are zero-load
members here.
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• When three members form a truss joint for which
two members are collinear and the third is at an
angle to these, then this third member must be
zero-force
 in the absence of an external force or reaction from a support
P
A
B
D
C
Here, joint B has only
one force in the
vertical direction.
B
Hence, this force
must be zero or B
would move (provided
there are no external
loads/reactions)
TAB
E
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TBC
Also TAB = TBC
TBD
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• While zero-force members can be removed in this
configuration, care should be taken
 any change in the loading can lead to the member carrying a
load
 the stability of the truss can be degraded by removing the
zero-force member
P
A
B
C
You may think that we can
remove AD and BD to make a
triangle …
This satisfies the statics
requirements
D
E
ENG2000: R.I. Hornsey
However, this leaves a long
CE member to carry a
compressive load. This long
member is highly susceptible
to failure by buckling.
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Method of sections
• When the truss exceeds a certain size, the
method of joints becomes unnecessarily tedious
• If the joints are in equilibrium, so too is the truss
as a whole
• Hence, we can also split it into two equilibrium
part-trusses
 provided we apply the appropriate forces to the cut ends
 and we can draw an individual free body diagram for the two
parts, including the external and reaction forces
• We can cut up to three members and still have
enough equations to solve for the six unknowns
 the three reactions from the supports
 and the three forces in the members
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• If no cut can be made through three or fewer
members and which also passes through the
member of interest
 then the method of joints must be used
• The method of sections is especially useful to
find the force in a beam in the middle of a large
truss
 when lots of joints must be analysed to get the same result
 and errors made in calculations for each joint will propagate,
possibly giving an unacceptably inaccurate result
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Summary
• We have now followed a path from atomic
bonding, through material properties and basic
mechanical relations to an understanding of
complex engineering structures
• In the next step, we look in more detail at the
beams that comprise the trusses
 in particular, we are interested in how forces are transmitted
though beams
 and what effect the beam shape has on the strength
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