Hello and welcome to the workshop.
In this workshop we're going to look at
approximate methods to assess the
fundamental natural frequency of a
bridge. This bridge was designed by
Whitby and Bird - a bridge in Castlefield
in Manchester. It is curved in form.
38m span with a deck
approximately 2m wide. It is
supported by an arch to one side which
is eccentric, connected by tapered
'I' section arms, which connect to
the deck via stiffners. The deck
has plates top and bottom,
which acts rather like a torsion tube
The rise of the arch is approximately
4m and we're going to
look at some of the views of this.
This is where we measured the
worst case natural frequency which
is in the centre of the bridge using an
APP in an android phone
which was designed by Expedition Engineers. You can see in the
background the radiating arms and the
phone that took measurements of the peak
acceleration response and it gave an
output for the natural frequency which
we were going to try to assess by
approximate methods. This is the
underside of the bridge. You can see the
stiffeners and the plate to the bottom.
The tubular stringers are roughly 300mm
depth. We have assumed a
certain size for them and for the purposes
of this study we're going to ignore the
balustrading. You can see the
background the arch as it comes into the
far abutment. As we said the deck is
approximately 2m wide so we're
going to look at the sheet which is to
assist.
We're going to check two conditions
one which is a straight span, which is in
pure BENDING and we are going to assume it is
38m straight span (2m wide) and
that is is an idealized beam which is
half the rise of the arch.
Roughly 3.6m/2
which is 1.8 meters. We are going to idealize
to work out a formula for the inertia (I)
and the BENDING deflection, as you can see,
is a classic bending deflection. The
second component of the deflection in
the middle of the bridge which is by
torsion. It is essentially found from alever arm
from centreline of the bridge, based on
the centreline distance from the
abutments. The lever arm is 2.7m
2.7m and P which is half the weight
half the weight of the
bridge (dead load), which was
worked out to be 350 kN.
So that is 350kN/2 = 175kN
We are given some of the parameters like
steel modulus (210,000N/mm^2)
and G (shear modulus) which is
78,000N/mm^2 and for the TORSION
component we are going to work out the rotation....
and hence the deflection. As you can see
in the clip which follows it is really
quite a lively bridge.
Hello. Welcome to the explanation sheet
of the rough calculations for the Casltefield
bridge - the Merchants' bridge -
using the sheets which were provided
First of all we assess the deflection as
if it was a straight beam and we use a
sort of idealized beam section for this
which is a quarter (25%) of the way
along the span, which would be an
equivalent of depth of around about half
the total arch rise. We will work out
the neutral axis distance from the base
of it and this is based on the area of
the bottom section. The
combination of the plate, the box and the
CHS sections. Although you could have an
idealised box. The 'Ybar' is a standard
formula working worked out from the
distances from the base and the neutral
axis comes to 43cm or
430mm at approximately a third of the height.
Very approximately
We come to the formula for Iy in which we use
areas and distances from the neutral axis
in centimeters. And this comes to
a value of 2.42x10^6 cm^4
which if we convert this to the equivalent
mm^4 it comes
to 24.2x10^9 mm^4
And that's the intertia (Iy) which we will use.
And the deflection is given by WL^4/384EI
With equivalent load area of 4.6kN/m^2
4.6kN/m^2 from the total weight
(total dead loads) that
were given before. So coming on to the
second sheet we are now going to work
out the deflection from that formula.
The deflection this is BENDING only we
recall which is to assuming a straight
beam of 38m span. The deflection
due to dead load. First of all the formula
given by WL^4/384EI
and if w = 4.6kN/m^2
and the deck is 2m wide
this gives us equivalent line load
on the whole section of 9.2kN/m
9.2kN/m.... so putting in the numbers of
the formula with consistent units
(N and mm) we can
work out the deflection in this equation.
this deflection comes to 49mm
49mm. We can work out the
live load deflection - as a contribution
of 10% of live load, this is for the
natural frequency.
We can Pro-Rata from the dead load
And this comes to 3/4.6 (live over dead)
times 49
and divided by 10
and this gives us a value of 3mm
The total deflection, adding those two
figures together...
for the total deflection from the bending component
gives us 49+3 = 52mm
Now we go on to look at the torsion
section. This is a torsional contribution
to the deflection and this is on
sheet 3. We are going to use an
idealized section for the torsion
constant. You can refer to the hint sheets if
you like, the worksheet that came with
this. The hint sheet gives a way of
working out the torsion. Estimating
torsion which is a total load P times
the lever arm (Lever arm is the centerline of
the bridge from that straight line drawn
between the abutments) = 2.7m
and for P, you take the dead load
over half the span. From 350kN -
we take half of that which gives us 175kN
175kN. The torsion comes
simply from that the total torsion
applied to the bridge = 175 x 2.7 = 473kNm
This is owing to its
curved shape. We can work out the
rotation from this.
We take the center line of the
whole width of the bridge and you assume the
it is rotating about the middle. It is
5.3m total width.
Torsion distribution is shown in the
diagram which is T/2 goes to each
support. And the length of which is
rotating is L/2 (Span/2).
The formula for the rotation
is coming in the next sheet of
calculations....
The sheet for and the torsion continued
we're now going to work out for our
idealized box section which is 10mm
wall thickness and it is
325mm deep, and 2m -
2000mm- wide.
The torsion constant J which we can work
out. I obtained this formula from Roark (Formulas for Stress and Strain).
With Roark a thin wall box
approximates to this figure, assuming t is
the same for both the walls and the bottom
and top plates. It is
2t^2A^2B^2/(t(A+B))
taking the dimensions above - A=325mm
B=2000mm. We put in
those figures and in this particular
instance we're going to use mm.
And this gives us a figure of
3.6x10^9mm^4 - this is the
TORSION CONSTANT, J.
used to work out
the rotation. We have given it we've given it cm^4
equivalent. So from the previous
sheet we got the rotation formula
Theta (Rotation) = TL/GJ ...putting in these figures,
remembering that we are using half the applied
torsion and half the span (Length used for
the rotation) and the G value (shear modulus)
78,000 N/mm^2 (or mPa) and
torsion constant. Gives us the
rotation in radians of 0.015 radians.
This is what we can use to
work out the deflection section.
Just to recap: the section is
going to rotate about the middle - and the
total width of the bridges 5330mm and
we take half of this to obtain the
deflection which is downwards to the
left hand side of the bridge. It is on the
opposite side of the arch. The
deflection is
(total width of the bridge/2) x theta
5330/2 x 0.015 which gives us
40mm - which is approximately
equal, incidently, to the dead load
BENDING deflection. i.e. the torsion has just as large
a contribution to the TOTAL deflection compnonent
as the bending component of deflection.
Now that we've obtained both
components of the deflection and can now
add them together by the law of
superposition. This will be in our final
sheet for working out the fundamental
natural frequency. First of all we are
going to find, by PRO-RATA methods,
the deflection from 10% Live Load
for the torsion component.
Torsion continued...
We obtain 10%LL by a pro rata method
as we did for the bending component.
Torsion deflection 10%LL
by PRO-RATA from dead deflection,
is 3/4.6x40/10
which is 10% of the live contribution
3/4.6x40/10 = 2.6mm
So totaling up the torsion
component we add the dead and the
10% live deflection. And this is 40
plus 2.6 that comes approximately to
43mm. Now we obtain the
total bridge combined deflection from
the bending from sheet 4 and the torsion from
sheet 5, above. We add these two
deflections together and the total
combined deflection comes to 51+43
51mm + 43mm
which comes to 94mm. Which
is the total deflection from both components. We can now
move on to work out an estimate of the
fundamental natural frequency from the
briefing that we gave in the worksheet.
This equation gives the Fn (natural Frequency) which is
a fundamental frequency given by 18/sqrt.(dl+10%LL deflections)
dead and
and this comes to 18/sqrt.94...
which gives a fundamental natural frequency
of 1.9Hz
in other words cycles per second
and we find that the in-situ reading
using the vibrate-it app
was a very similar value
and if you look on the website you can see the output from that.
This is very close to the estimate.
And now we have an addendum
which is an addition which was not part of the brief from the original workshop.
This isn't part of the workshop.
And in this addendum we are going to look at approximate peak excitation.
The response of the bridge.
to give some idea of how it is going to be excited.
with a
forcing frequency
So we look at the response excitation which is
a comparative measure of response.
And a reference is Structural Engineering Art and Approximation
Structural Engineering Art and Approximation
136 to 7, chapter 2.12
So drawing out the bridge. Here we have a person who is forcing the response.
Weight wp
The weight of the bridge is W
So the weight or the modal mass which the mass which is
affected by the response.
We will assume is the full weight of the bridge.
Which is 350kN (3500kg)
And the force causing the resonant response is the weight of a person which is
0.75kN
The peak acceleration comes from this simple formula
a ratio of a/g which occurs at resonance
And this is
1.3 x alpha x wp / (2 x Rho. x W)
1.3 x alpha x wp / (2 x Rho. x W)
Alpha is a factor depending on the
activity. 0.5 for walking.
The damping ratio is something like 0.02 (2%). So putting the values in
We come up with a response of 0.034.
(Peak acceleration)
And if g is an acceleration of 10 m/s^2
10 m/s^2, approximately. Then the peak acceleration in this
approximate formula comes out at
comes out at 3.4%g.
Now this is a high value
for office
or sedentary activities
For a bridge it may be deemed as acceptable.
Thank you very much for listening
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