This should be the first section in the blog, but I cannot
work out how to rearrange things. I suggest that you read the first part of the
Introduction to see where I am coming from and then carry on with this section
Dinosaurs rool OK!
The Whinge
Recently, to my surprise, I have been let loose interviewing
graduates. My manager is not totally daft and, rather than doing the
professional serious bit, I got to give the poor little dears what we thought
was a simple test and then to put them through a short coaching session to see
how they react when we discuss the bits that they got wrong. Finding out how
someone reacts WHEN they get something wrong is an important thing to find out:
it also allows a bit of coaching, and it is useful for prospective employees to
find out how we would treat them in a working environment.
I was surprised to find that a large proportion of
candidates could not do calculations that they should have been able to solve
whilst at school. As part of the coaching sessions, however, I found out that
this was often because they had been taught methods that, while rigorously
correct, were counter-intuitive and mathematically difficult.
One of the problems is that there are often a number of
different ways that a problem can be solved but some of the graduates had been
taught methods that caused great amusement/concern to working engineers.
Simple
or complex ?
As usual, academics insist on using complicated terms that
make little sense if you were looking out of the window during the one lecture
that they briefly explained it.
Aside: perhaps lecturers need flash cards, like Sheldon
Cooper in The Big Bang Theory, but rather than flashing up “Sarcasm” they need
to have a rarely used “Fundamental” flash card, as well as a more widely used
“Useful example” as well as the inevitable “This is my speciality and I am”, “
going to teach this obscure bit of research”,
“as though the world revolves around it” (three flash cards probably
required for that last one).
Further aside: I am quite worried about the number of people
who suggested that I watch Big Bang Theory, because they thought that I had
similar character traits to Sheldon.
The terms that are commonly used are “Statically
determinate” and “Statically indeterminate” – partial translation - “Solvable
using statics” and “Not solvable using statics”.
And
in English ?
On a Plane structure (i.e. one where you can draw the entire
structure and plot all of the deflections in a single view on a single piece of
paper) solving a problem using statics just means that to find the unknowns
(usually the support reactions) the only techniques that you need are resolving
forces in two directions (usually horizontally and vertically) and taking
moments ONCE. A shorter term to use, and one that I will use from now on, would
be “Simple”.
Aside: I have never managed to work out why, but taking moments
TWICE will just give you another equation that could be found by rearranging
the other three equations: this would not give you any extra information.
Statically indeterminate just means that you cannot work out
the unknowns using statics alone and need to use a more complicated method. I
will just refer to these structures as “Complex”.
You will be pleased to know that a large proportion of
structures are simple, or can be treated as simple through a bit of fast
talking. After a bit more explaining I will take you through easy methods that
include everything that you need to know to solve every Simple structure. These
methods will also demonstrate some principles that will allow you to check the
results for more complex structures, although that is all I will cover here for
complex structures.
Simple ??????
So how do we decide whether something is simple ?
I seem to recall being given some very complicated rules
about degrees of freedom (how many joints there are combined with how many ways
they can move or rotate) numbers of members and numbers of supports: but it did
not make any sense to me at the time and I have completely forgotten it. In
thirty plus years I have never needed to use the rule and it does not seem to
be necessary. The rule was written to cover all sorts of eventualities, and was
probably rigorously correct, but common sense can cover most situations. My
method is a combination of simple ideas, but requires you to picture the
behaviour of the structure. To keep things simple, although the principles
apply to all linear structures, I will mainly talk about structures where all
of the loads, and all of the movements of the structure, are in a single vertical
Plane: as I mentioned above (but it never hurts to mention the important things
twice – flash card : fundamental) these structures are called Plane structures.
The first thing to look at is the structure as an overall
rigid body: by this I mean that a structure, even if it has many elements like
a truss, is treated as a single body. Using statics, there are only three
equations that can be written so you can only solve for three unknowns. Looking
at this the other way round, if there are more than three unknowns then you
cannot solve using statics and it is Complex.
Looking at the simplest structure, i.e. a single beam, the
most common form is to have two supports, distantly followed by the form where
one end is clamped to form something like a diving board. For a plane
structure, supports can typically be fixed against either horizontal or
vertical movement in the plane, or against rotation about an axis perpendicular
to the plane. These fixities are the unknowns that we need to calculate using
statics.
If we consider the beam with two supports, then one will
typically provide both vertical and horizontal support and the other will only
provide vertical support. There are three unknowns and they can be calculated
using the three equations.
If there were an extra fixity, such as fixing the second
support horizontally, then you could get one support working against the other
and you cannot simply resolve forces horizontally to split how much of any
horizontal load is resisted by either end. This is a statically indeterminate
structure, i.e. one that I refer to as Complex.
Aside: if you remove one of the fixities then the structure will
not be stable: removing the horizontal fixity would allow the beam to slide
freely and removing one vertical support would allow the structure to rotate
freely: both of these are referred to as mechanisms and cannot be analysed as
they are not stable.
If we look at the springboard then one end is fixed to
prevent both vertical and horizontal movement and is also fixed against
rotation. Again, there are three unknowns and three equations allowing you to
solve the problem, meaning that it is Simple. Adding another simple support at
the far end (propped cantilever) makes the structure complex, requiring special
treatment, and removing any of the fixities would create a mechanism.
If we extend the form of structure by adding an extra span,
with the two spans acting continuously (i.e. the shared ends of the two spans,
over the support, are rigidly linked horizontally, vertically and against
rotation) then you will need at least four supports to keep it stable. You will
always need one horizontal fixity to stop the structure from sliding off, and
adding three vertical supports would make four in total. Since this is more than
the three equations it is a Complex structure. That is the numerical way of
deciding that it is Complex, but a better way is to decide whether or not you
can move, or rotate, any one of the supports without any of the structure being
stressed in any way.
The structure below (Example the First) is a simply
supported structure, with one horizontal fixity.
The original location is shown behind in grey for reference.
If you move the left support to the right, the right support
is free to slide so there is no problem.
If you lift either the left or right support then the
structure is free to move as a rigid body. There are no axial forces, shears or
bending moments under any of these movements.
Conclusion: the structure is Simple.
The structure below (Example the Second) is a continuous
structure with four pinned vertical supports and one horizontal fixity.
If you move the left support to the right the other supports
are free to slide so there is no problem.Conclusion: the structure is Complex.
Note that in this problem there are 5 fixities so we would
not expect the problem to be Simple.
I will no longer labour the point about which forces etc are
occurring since it will (hopefully) be clear by now what I am talking about.
Aside: Being a mean little Adrian, who loves winding up his
grads, one of my favourite recent course developments has been to get them all
standing by the projection screen, showing a series of structures, and then
getting them to stand in front of markers saying Simple, Complex or Something
funny. After the initial problem where they all waited, until one of my brighter
grads moved first, and then followed like sheep (I made him stand still until
they had all chosen for later problems) it was an amusing (and mildly sadistic)
way of checking what they were learning.
I will come back later to some more interesting structures,
and discuss whether they are Simple or Complex, but the next stage is to work
out how to analyse a Simple structure.
Stage
the First
The first operation is to calculate the forces and moments
in the structure supports. This is simply done by taking moments about one of
the supports. Taking moments about that one point allows you to calculate the
reaction in the other support. There are some little variations that sometimes
come up, but I will confuse you with those at the very end of this part of the
guide.
Once you have calculated one reaction then the sum of the
two vertical reactions will equal the total applied vertical load: this is
resolving vertically and allows you to calculate the second reaction.
Finally, if there are any horizontal forces then these will
be resisted by the horizontal fixity: this is resolving horizontally.
OK, you have done the hard bit since you have calculated all
of the unknowns.
Stage
the Second:
What are bending moments and axial and shear forces ?
A bending moment is similar
to a normal moment (i.e. force x lever arm) but it is an internal moment that is acting on a cross section as a
result of all of the loads (including reactions) acting on that structure.
Axial force is just the force acting along the member,
stretching or compressing it, with shear force being the force acting to try to
shear the member.
Fundamental: Bending moments are the internal moments that would
be required to keep the two sides of a structure in equilibrium if you were to
take a saw and cut through that structure at the point in question. The oddity
is that there are actually two moments required at any section, one for each
cut face, and these moments are of opposite sign, as shown above, because one
will be acting in a clockwise direction and the other will be
anti-clockwise. Axial force is the
internal in-line force required to keep the structure in equilibrium and the
shear force is the internal normal force required to do the same: again, both
of these will be of opposite sign on the two cut faces.
Fundamental: if the term “Shearing” is unfamiliar, try the
following: hold a ruler horizontally, but with the wide face set vertically.
Grasp the ruler in both hands, with your hands touching each other. Try to move
your left hand upwards, and your right hand down, but using your grip to clamp
the ruler so that the ends stay level. You will not be able to do much to the
ruler, but if you were to cut the ruler vertically, between your hands (I am
not sure how many hands you would need for that in total, and I am sure that
the Elf ‘n Safety guys would complain about a blade so close to flesh) then the
left side of the cut would move up, past the other side moving down. Shear
force is the internal force resisting this movement in the uncut beam.
Aside: before I go on any further, I have an important thing to
say …..
I am sorry. I am really very, very sorry. Please accept my most
heartfelt apologies. I am really, really, really, incredibly, really, massively
sorry.I am not apologising for something that I have done, but am apologising on behalf of the engineering community over the past 150 years. The sign conventions for moments and shears are rigorous but very confusing. It does not actually matter what sign convention you use, provided that you are consistent. But I will use the convention used by my favourite analysis programme, STAAD (as well as Superstress, which is widely used for simpler analyses in good old Blighty).
Fundamental: in the diagram above, the sign convention is
that sagging moments are positive and are plotted DOWNWARDS. Although this is
very odd, it means that the sign of the moments for simply supported beams,
which used to make up the vast majority of structures, did not require the
engineer to put a negative sign in front of every number, but the moment
diagram was drawn on the tension side, and was drawn downwards, which seemed
more logical when the beam was deflecting down (but please do not get confused
that the moment diagram will always follow deflection because it does not do so
over internal supports on multi-span continuous structures, as you can see in
the diagram above).
The shear force diagram is defined positive upwards, when
looking at the left hand end of a cut structure, and that is plotted UPWARDS.
Grrrrrrhhh !! This is consistent with what I will discuss later, about
gradients, but only because the first red section on the left is going
downwards which is actually upwards when you consider the sign convention.
Confused much ?There is a very simple way to calculate the values. Any structure is subject to a series of applied moments and forces, and also to a series of reaction forces and moments. I will demonstrate this later for simple structures (but all that more complex methods do is calculate reactions on Complex structures). Once you know all of these reactions, applying them back onto the structure will result in a structure that is in equilibrium with the applied loads.
In the earlier diagram, the bending moment at mid span is
(as we all know) wL2/8. The reactions are both wL/2. The method for
working out the bending moment at mid span is to cut the structure in half and
remove the left hand side.
Taking moments about the cut, M = wL/2 . L/2 – wL/2 . L/4 =
wL2/8. In a similar manner, the shear force is the vertical force
required to keep the cut structure in equilibrium and in his case it is wL/2 –
wL/2 = 0. Finally, the axial force P is simply F since there is no horizontal
reaction at the sliding support.
And that is all there is to it.
Aside: recently, whilst interviewing, I was surprised to find
that most of the graduates tried to solve a series of examples by calculating
the shear force diagram and then integrating the shear force to get the moment.
Although this is a perfectly legitimate method it is ridiculously complex and
relies on getting the directions of the reactions correct, which is not always
easy to do: it also carries an error made in one part of the calculation
through to the end (although at least it allows you to check that everything sums
to zero for a pin end). As part of the interview, I spend time coaching them on
the bits that they got wrong, partially as part of my relentless war to
educate, but partially to test their character when they get something wrong
(as we all do). I am often flattered/amazed/worried by the number of grads who
say “But your method is really easy. I wish that we had you lecturing us
instead”. Much as I love being mean to lecturers, I have to add that coaching
one to one is a lot easier than lecturing a roomful of hung over students, but
teaching a more simple method has got to be better.
Example
the First:
This is a typical example that covers the majority of simply
supported beams that you can analyse using statics.
The
Method:
First off you need to calculate the reactions, either as
detailed earlier or by using some more advanced method for a complex structure.
The beam has three fixities and can therefore be analysed
using statics.
Taking moments about A:
10.RB = 10.2+20.5 = 120
Hence RB = 120/10 = 12kN
Resolving vertically: RA+RB = 10+20 = 30
Hence RA = 30-RB = 30-12 = 18kN
To show that taking a second set of moments does not produce
extra info (although it does provide a check:
Taking moments about B:
10.RA = 10.8+20.5 = 180Hence RA = 180/10 = 18kN (amazing)
So now we have both unknowns that we are interested in
(there are no horizontal forces).
Next, you get two pieces of paper. On one piece you draw the
structure and plot all of the loads and reactions. Be careful that the loads
are plotted in the right direction, with gravity loads acting downwards and
reactions generally (BUT NOT ALWAYS, AS WE WILL SEE IN AN EXAMPLE AT THE END)
acting upwards.
The force acting on a support will go in one direction but
we are interested in the forces on the structure (i.e. the reaction) which act
in the opposite direction. On the second piece of paper, you draw a dot on the
left hand side of the page, a moment arrow, going in a clockwise direction and
a vertical arrow.
The final stage is to cover all of the structure, except for
the right hand end, and then calculate the moment (bending moment) and total
vertical force (shear) for all of the forces NOT covered up by the sliding
sheet at any point of interest on the structure. What you are effectively doing
is to cut the structure and throw one side away: calculating the equilibrium
moment and vertical force gives you the bending moment and shear force.
Plotting out these results for all points will give you the bending moment and
shear force diagrams for the structure under the applied load case.
The final little niggle is the sign conventions. It is so
blindingly obvious that it is really hard to describe exactly what to do, but I
will try anyway. In the diagram, the curved moment arrow drawn is a positive
moment, which is a sagging moment. This will result in a tensile stress in the
bottom of the beam (as shown) and that is the side that you plot the bending
moment diagram.
The translucent bit is supposed to represent a piece of
paper that has been laid over the drawing. The dot represents the point that we
are taking moments about, as well as the location where we are resolving
forces. In theory we are setting it an infinitesimally small distance either
side of supports or load points. This makes no difference for the moments,
since a tiny change in location will make no difference in the length of the
lever arms and having a force appearing with a zero lever arm will contribute
no moment. It makes a massive difference for the shear forces since the shear
force is the total of all of the forces that have been revealed and moving to
one side of the load or another means that the force gets either included or
not. Hence for the figure above:
Moment = 12.0 = 0kNmShear force = -12kN
The 20kN force has not appeared so is not included.Shear force = -12kN
Now the 20kN force has appeared so it is included.
Moment = 12.5 = 60kNm (no change from previous calculation)
Shear force = -12 + 20 = 8kN (step change from previous
calculation)
Moment = 12.8 - 20.3 = 96 - 60 = 36kNm
Shear force = -12 + 20 = 8kN
Moment = 12.8 - 20.3 = 96 - 60 = 36kNm
Moment = 12.10-20.5-10.2 = 120-100-20 = 0kNm
Shear force = -12 + 20 + 10 = 18kN
Simples !
The above method can be used to analyse absolutely any
Simple structure.
Fundamental: you will often hear me banging on about us all
making mistakes and looking for ways to check things. Having started at the
right hand end to uncover the structure, and using the sign convention that I
have used, you will notice that the step changes in the shear force diagram, at
the concentrated forces, follow the direction and magnitude of the reactions
and applied forces. UDL loads actually do the same thing, but you need to
visualise the load as a series of small discrete loads and the sloping line as
a series of small steps.
This example also shows another way that you can check any
complex structure that you analyse as well as showing the overly complicated
method that I was so rude about earlier. If you look at the slope of the moment
diagram between A and the 10kN load, you can see that it is sloping downwards.
Actually, it is sloping upwards since the moment is plotted positive downwards.
The slope is (36-0)/2 = 18, which is also the shear force over that length.
Between the 10kN and 20kN loads, the slope is (60-36)/3 = 8, which is the shear
force between those points. Finally, the change between the 20kN load and B is
(0-60)/5 = -12, which is …. You get the idea. It is a fundamental that the
shear force is always equal to the slope (differential) of the moment diagram.
The complicated method (about which I was so rude earlier) is to calculate the
shear force along the entire structure and then to integrate the shear force to
calculate the bending moments (which is the reverse operation). This is
technically correct but relies on getting the reactions correct and, more
importantly, in the right direction. It is also easy to integrate uniform shear
patterns, but more complex for uniformly distributed loads, and worse for
variable distributed loads.
The shear force diagram for a structure with point loads
will always be a series of rectangles, since there are step changes at the load
points and no other loads in between. Since the shear force diagram is always
the slope of the bending moment diagram, the series of shear force rectangles
will result in the diagram being a series of straight lines. Likewise a UDL
will give a uniformly changing shear force diagram, which results in the
gradient of the BM diagram uniformly changing, resulting in a parabolic moment
diagram.
Below is the bending moment diagram for a three span
continuous beam with a single point load on the first span and a UDL on the
third span.
The resulting shear force on the first span has a stepped
shear force diagram resulting in two straight lines for the moment diagram. The
middle span has a uniform shear force resulting in a single sloping moment
diagram. The shear force in the final span has a uniform slope, giving a
parabolic bending moment diagram. The rule works for every single structure and
is a good check of what you work out, no matter what method you used to work it
out.
Fundamental: although we have been using statics to analyse
simple structures, they are also vital for complex structures, especially if
you are analysing something on a computer. Although you will not be able to
actually analyse a complex structure by resolving forces, if you think of the
structure as a single body, with reactions being just another set of forces,
then you can pick up a lot of mistakes just by adding all of the applied
loads in a direction (usually downwards but it also works horizontally) and comparing
the applied total to the total of all of the reactions. If they are not
equal then you have a problem, and this is usually caused by you applying the
loads incorrectly (although it can also be because your check calcs are wrong).
Shear
= slope of moment diagram
In case you are interested, it is very easy to show why this
is. The diagram below (which may look familiar from above) shows a very short
length (cut out of beam) of length δL.
At one end there is Bending Moment M and Shear Force V: at
the other end these have changed to M+δM and V+δV. Taking moments about the
right end M – (M+δM) + V.δL = 0, hence V = δM/δL which is the first differential
of the bending moment diagram and also the slope.
By the way, if you took moments about the other end then
there would be a term δV.δL which would tend to zero.
Fundamental: please bear in mind that, when taking moments about
a point, the moment due to a force is the product of force x perpendicular
distance, but the “Moment” due to an applied moment is just the moment (i.e. it
does not matter where that moment is actually applied).
Time
to test yourself
Before you can try to apply statics, you need to be able to
determine if statics will work. My method is to check whether or not you could
move all of the restraints,
one at a time, without causing any stress in the structure e.g. by trying to
bend or shorten it. If you can then it is Simple, if you cannot then it is
Complex.
To check your logic, and to test the results, move down to
the end of the examples. Try hard not to move too far because it will be hard
not to see the answers unintentionally.
Example the Third – two fixed pins
Example the Fourth – two sliding pins
Example the Fifth – encastré cantilever
Example the Sixth – pinned cantilever
Example the Seventh – propped cantilever
Example the Eighth – three span with drop in beam
The
Answers
Example the Third – two fixed pins
If you move the left support to the right, the right support
is fixed so the beam is squashed (bad).
If you lift either the left or right support then the
structure is free to move as a rigid body (fine).
Conclusion: the structure is Complex.
If the structure gets hot or cold then it will be stressed
because it will not be able to expand or contract as it wants. Also, since the
structure has depth, the bottom fibre will get longer under bending which will
also be resisted.
HOWEVER a lot of the old structures that I work with are
beams sitting on padstones (stone bearings) so they are not free to move. We
do, however, cheat and assume that the walls will flex enough to allow for the
relatively small movements required under bending or temperature effects.
Example the Fourth – two sliding pins
If you move the left support to the right, the right support
is free to slide so there is no problem with the beam. Unfortunately the
structure has no way to restrain any horizontal load and is a skate board –
good to let off some stream – bad as a structure.
Conclusion: the structure is a mechanism and cannot be
analysed.
Example the Fifth – encastré cantilever
If you lift the support then the structure follows as a
rigid body.
If you move the support to the right then the structure
follows as a rigid body.
If you rotate the support then the structure rotates without
restraint.
Conclusion: the structure is Simple.
Example the Sixth – pinned cantilever
Bad case of brewer’s droop (if English is not your first
language, then good luck translating that J). There is nothing to stop the structure from just
dropping.
Conclusion: the structure is a mechanism.
Example the Seventh – propped cantilever
Lifting the encastré left support will bend the structure.
Rotating the support will bend the structure.
Dropping the pinned support will bend the structure.
Conclusion: the structure is Complex.
Example the Eighth – three span with drop in beam
If you move the left support to the right, the other
supports are free to slide so there is no problem.
Conclusion: the structure is Simple (Fairly).
Surprise ! There are 5 fixities so we would not expect the
problem to be Simple but because of the releases this actually is (Fairly)
Simple. In order to analyse this structure (which is a quite common construction
form from the 70’s, but rarely used now due to problems of maintaining the
bearings in the hinges as well as hidden corrosion) you have to work from the
simple bit and work out. The drop in span is a simply supported structure and
can be easily solved. Then just take the span out whilst applying the end
reactions from the drop in span as loads on the remaining two structures.
Principle
of superposition
This is one of those lovely phrases that does not leap out
as either a) English, or b) useful: it is however both of these.
All it means is that for certain types of structures you can
analyse a series of load cases and get a series of results. If you add the
series of results together then you will get the same answer as if you analysed
the structure for one load case with all of the loads applied at once.
This means that if you have a structure with lots of
complicated loads then you can break it down into little bits and solve them
one at a time.
The relevant type of structure is called linear and covers probably
99% of all structures that we have to deal with. Oh, and to be honest, if you
are reading this then you probably are not going to be working on non-linear
structures J.
Linear structures are those structures that do not change
their shape very much under load, so suspension bridges and structures that are
on the way to buckling are GEOMETRICALLY non-linear, whilst most cable stayed
structures, as well as practically everything else, are geometrically linear.
There is also another form of analysis where the structures that are loaded so
heavily that they go beyond elastic behaviour and these are MATERIALLY
non-linear. We generally try to avoid pushing structures that far, so material
non-linearity is rare.
Hence linear structures are anything that are neither
geometrically or materially non-linear, and it is for these structures that
superposition works.
Simple
and Complex Truss structures
The structures that I was discussing before were all beam
structures but there are similar rules for truss structures. The first thing to
do is to treat the truss as a rigid body and to think of it in the same way as
a beam. Hence a truss with two fixed supports is likely to be Complex whilst
one with one fixed support and one sliding one is likely to be Simple.
But trusses also have a further set of guidelines that
determine whether or not they are Simple, and which can break the 3 fixity
guidance. If the structure is made out of a series of triangulated pin jointed
truss members (i.e. ones where the members can take tension or compression but
are free to rotate at the joints) then it is very likely to be Simple. If,
however, any of the members are continuous and can carry bending moments then
the structure will be Complex. Most real world truss structures are a hybrid of
pin ended truss members and bending members. The primary support mechanism for
most trusses is the pin ended behaviour, but most trusses will also have parts
that are continuous that will attract bending moments whether you want them to
or not.
Aside: there is a form of “Truss” where there are no triangles,
and everything is a rectangular grid of members that resist bending as well as
axial forces, and these are called Vierendeel Trusses. They are much beloved by
Architects (and I hope that you are impressed by the fact that I did not stick
a rude adjective before the A word) because they are open structures that look
pretty. Unfortunately they are massively inefficient and one small area of
corrosion at a joint can cause progressive collapse of the entire structure.
Bless ! Winning awards and keeping structures up are two different things,
obviously.
Back to the real world. Typically the top and bottom chords
(as well as the end diagonal if sloped) will be continuous with each other and
relatively stiff in bending, whilst the internal diagonals and verticals will
typically be more slender, be connected more lightly and may often be treated
as pin ended. These structures are Complex and require complex analysis.
Aside: If you are looking into the primary support mechanism,
then the above truss can be treated as pinned, allowing Simple analysis, but I
have concerns about this method since it is fine as a plastic analysis (i.e. if
the local sections are so chunky that they can form plastic hinges and rotate
if necessary) but a lot of trusses have cross sections that are slender
locally, and I would only consider this if all relevant sections were compact
i.e. chunky enough that they could form a full plastic hinge without local
buckling.
If you add one diagonal joining one support to the opposite
corner then it becomes a triangulated structure and is therefore stable.
In a more practical arrangement, in the example above, the
two diagonals would take the same compression and tension, IF they were the
same section. But if one were a lot stiffer than the other then it would take a
lot more load (this would require complex analysis and is not covered here).
Back to my simple rule, in a Simple truss you can change the
length of any of the members by some amount without the rest of the structure
being stressed.
Back in the good old days (more memories from when dinosaurs roamed the Earth) trusses were analysed by calculating the reactions in the normal way, followed by a large series of resolutions of forces, horizontally and vertically, at each joint in turn. This allowed the tension and compression forces to be calculated in every member. This was only possible if the truss were treated as Simple, but would have given reasonable results if there were some continuity because the pin jointed truss was the primary support mechanism.In the truss above, one top chord member has been lengthened and the structure has rearranged itself to suit. The structure is therefore Simple. Notice that the sliding bearing has slid sideways: if the bearing had been fixed then the structure would not have been free to rearrange itself in an unstressed way and the structure would have been complex.
Members
at an angle
If the free end allows movement towards the other end then
it does not matter which end you take moments about to calculate reactions. If
the free movement does line up with the fixed end then you need to take moments
about the fixed end. If you take moments about the top reaction, in the problem
above, then you have two unknowns, which makes life a little too interesting.
But if you take moments about the lower point then two of the unknown reactions
drop out of the equation and the equations are much simpler.
This is the sort of problem that I learnt at school when I
was fourteen and we had to show whether or not the ladder would slip, by
comparing the friction required to the maximum available of μ.RV.
The only thing that we never worked out was whether or not the ladder would
snap under the load, by comparing the bending moment to the bending strength:
the principles for calculating reactions were, however, exactly the same as we
use in structural analysis.
Aside: at school we, of course, became obsessed by the
impossibility of the frictionless wall. The teachers told us that it could not
be solved if there was friction on both surfaces. If only they had told us, as
someone with an engineering PhD would, that it was statically indeterminate
then all would have become clear. For those who do not know me, the last bit
was written with an ironic tone in my head.
Further aside: just to clear up one point – I have met several
PhD’s for whom I have a lot of respect as engineers. They just tend to be the
ones who talk a recognisable language rather than dropping “But of course it is
just a negative second order differential” to describe a hog curve.
Final
example
I will now demonstrate a solution to one problem to
demonstrate most of what we have been looking at, as well as to point out a
common mistake that people make.
The beam is simply supported, with a fixed pin at one end, a
sliding pin inset from the other end, self-weight applied over the entire length
and two point loads, one in the middle of the main span and one at the end of
the overhanging section.
1) Taking moments about A:
8.RB = 6.5.12 + 10.4 + 20.12 = 640
Hence RB = 640/8 = 80kN
Resolving vertically: RA + RB = 5.12 + 10 + 20 = 90
Hence RA = 90 - RB = 90 - 80 = 10kN
At right hand end:
Moment = 0kNm
Shear force = 20kN
At B (RHS):
Moment = -4.5.2 - 20.4 = -120kNm
Shear force = 4.5 + 20 = 40kN
At B (LHS):
Moment = -4.5.2 - 20.4 = -120kNm
Shear force = 4.5 + 20 - 80 = -40kN
At midspan (RHS):
Moment = -8.5.4 - 20.8 + 4.80 = 0kNm
Shear force = 8.5 +20 - 80 = -20kN
At midspan (LHS):
Moment = -8.5.4 - 20.8 + 4.80 = 0kNm
Shear force = 8.5 +20 +10 - 80 = -10kN
At A:
Moment = -5.12.6 – 10.4 – 20.12 + 80.8 = 0kNm
Shear force = 5.12 + 10 + 20 - 80 = 10kN
I will now split the above into three separate loads to
demonstrate superposition.
2) Taking moments about A:
8.RB = 5.12.6 = 360
Hence RB = 360/8 = 45kN
Resolving vertically: RA + RB = 5.12 = 60
Hence RA = 60 - RB = 60 - 45 = 15kN
At right hand end:
Moment = 0kNm
Shear force = 0kN
At B (RHS):
Moment = -4.5.2 = -40kNm
Shear force = 4.5 = 20kN
At B (LHS):
Moment = -4.5.2 = -40kNm
Shear force = 4.5 - 45 = -25kN
At midspan (LHS and RHS):
Moment = -8.5.4 + 4.45 = 20kNm
Shear force = 8.5 - 45 = -5kN
At A:
Moment = -5.12.6 + 45.8 = 0kNm
Shear force = 5.12 - 45 = 15kN
3) Taking moments about A:
8.RB = 10.4 = 40
Hence RB = 40/8 = 5kN
Resolving vertically: RA + RB = 10
Hence RA = 10 - RB = 10 - 5 = 5kN
At right hand end:
Moment = 0kNm
Shear force = 0kN
At B (RHS):
Moment = 0kNm
Shear force = 0kN
At B (LHS):
Moment = 0kNm
Shear force = -5kN
At midspan (RHS):
Moment = 5.4 = 20kNm
Shear force = -5kN
At midspan (LHS):
Moment = 5.4 = 20kNm
Shear force = 10 - 5 = 5kN
At A:
Moment = 5.8 – 10.4 = 0kNm
Shear force = 10 - 5 = 5kN
4) Taking moments about A:
8.RB = 20.12 = 240
Hence RB = 240/8 = 30kN
Resolving vertically: RA + RB = 20
Hence RA = 20 - RB = 20 - 30 = -10kN
Please notice the sign of the reaction. If there were
nothing else holding down Bearing A then the beam would lift off and the flip
over. In this case, the loads from the other two parts of the overall load case
are there at the same time and will hold the beam down overall.
At right hand end:
Moment = 0kNm
Shear force = 20kN
At B (RHS):
Moment = -20.4 = -80kNm
Shear force = 20kN
At B (LHS):
Moment = -20.4 = -80kNm
Shear force = -30 + 20 = -10kN
At midspan (LHS and RHS):
Moment = 30.4 - 20.8 = -40kNm
Shear force = -30 + 20 = -10kN
At A:
Moment = 30.8 - 20.12 = 0kNm
Shear force = -30 + 20 = -10kN
If you add all of the results together, you get 2) + 3) +
4). If you look you can see that these match the result when you apply all
cases at once. This may seem trivial, and if you can spare the time then you
can split up all of the reactions into their individual components in case 1
and it should be fairly obvious that the results will add together.
RA
|
A
|
LHS
MS
|
RHS
MS
|
LHS
B
|
RB
|
RHS
B
|
R
End
|
|
1)
Reaction
|
10
|
80
|
||||||
Bending mom.
|
0
|
0
|
0
|
-120
|
-120
|
0
|
||
Shear force
|
10
|
-10
|
-20
|
-40
|
40
|
20
|
||
2)
Reaction
|
15
|
45
|
||||||
Bending mom.
|
0
|
20
|
20
|
-40
|
-40
|
0
|
||
Shear force
|
15
|
-5
|
-5
|
-25
|
20
|
0
|
||
3)
Reaction
|
5
|
5
|
||||||
Bending mom.
|
0
|
20
|
20
|
0
|
0
|
0
|
||
Shear force
|
5
|
5
|
-5
|
-5
|
0
|
0
|
||
4)
Reaction
|
-10
|
30
|
||||||
Bending mom.
|
0
|
-40
|
-40
|
-80
|
-80
|
0
|
||
Shear force
|
-10
|
-10
|
-10
|
-10
|
20
|
20
|
||
2+3+4)
Reaction
|
10
|
80
|
||||||
Bending mom.
|
0
|
0
|
0
|
-120
|
-120
|
0
|
||
Shear force
|
10
|
-10
|
-20
|
-40
|
40
|
20
|
Conclusion
Well I could go on and on for a few more hours, but I hope
that you have the idea by now. These methods are used for a large proportion of
analyses and are very simple when you understand the basics. If you have a
structure that is more complex than the simple structures then you will most
likely be using a computer analysis programme.
The problem is that these simple methods are some of the
first things that you will have been taught at University. Unfortunately, these
methods are not interesting for researcher/lecturer types and you will probably
have been taught a lot of special cases that you will never use.
There will be some of you who wonder why I have spent so
much time producing something so basic. But if you are one of the majority who
has learnt something then please be aware that you are not alone.
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