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  • ruRussian
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00:00:01
Aleksandrovich Makeev, we are continuing the school of a
00:00:04
young or novice designer,
00:00:06
today we will talk about calculating a score. We
00:00:09
will be interested in, say, the
00:00:10
calculation of platform beams, the calculation of
00:00:13
beams allowed for roofs with the presence of
00:00:16
snowman bags and the calculation of statically
00:00:18
indeterminate beams with any number of
00:00:20
intermediate supports, and so go to the
00:00:23
first tab
00:00:24
calculation beams and consider the operation of this
00:00:28
automated template using the example of
00:00:30
calculation: platform beams made of schiller 20
00:00:33
t steel 245 design resistance 240
00:00:36
megapascals this template is intended
00:00:38
for the calculation of cut beams with an
00:00:41
environment uniformly distributed
00:00:43
load along the entire length and any
00:00:45
concentrated forces
00:00:47
of limitation that there can only be
00:00:48
six of them placed on given the
00:00:51
distance between each support,
00:00:52
let's look at how this template works, using
00:00:55
our example,
00:00:56
platform beams mean what kind of
00:00:58
element this is, in this case we have a
00:01:01
20 p schiller, and in the template a solid section can be used,
00:01:04
any rolled profile, a
00:01:06
composite section, which means the length must be
00:01:09
known with us you length and
00:01:11
consists of drawing the distance between the
00:01:14
applied forces n who lived known
00:01:16
from the drawing a1 a2 a3 a4
00:01:19
values ​​n given to us we took from the
00:01:23
calculation of the stringer for strength and
00:01:25
rigidity of the last video, here we found the
00:01:27
full arrow the exact force acting on the
00:01:30
stringer here it is 1732 kilogram means
00:01:35
this load falls on one and the
00:01:37
second raincoat summer beam, so let's
00:01:39
divide this value in half, the result is
00:01:41
exactly the value n
00:01:42
they are all the same 4 values, so
00:01:45
let's fill in the initial data so
00:01:47
uniformly distributed load
00:01:49
concentrated clear uniform
00:01:51
stable load is taken from two
00:01:54
components constant component from
00:01:56
weight half of the platform slab
00:01:59
in this way, knowing the density of the volume,
00:02:01
we can calculate all this weight in
00:02:04
kilograms or days tones.
00:02:05
I need data, we will distribute it along the entire
00:02:08
length of the beam,
00:02:09
this will be a component of its own
00:02:11
weight, but you can also include sleeping and a
00:02:13
uniformly distributed load will also be
00:02:15
added from the temporary load
00:02:18
temporary We also know the load, but
00:02:20
if we collect a temporary load,
00:02:22
or the easiest way is to calculate the temporary load
00:02:24
on our stairs, we
00:02:27
found out in the last classes 300 kilograms per
00:02:29
square meter, so if we take 300
00:02:31
kilograms per square meter, multiply the
00:02:33
platform design by half the width and but we’ll
00:02:36
just get the time component,
00:02:38
add it to the constant, this will be the
00:02:41
value q, so let’s fill out the table of
00:02:43
initial data, you should know that
00:02:46
green checks need to be filled out
00:02:48
when we put pink checks,
00:02:50
this template is filled in automatically and so the length is
00:02:53
3 meters three 300 millimeters due to the fact
00:02:55
that
00:02:56
in template, you can calculate both
00:02:57
solid section and
00:02:59
rolled alimony, then here in this case we’ll
00:03:02
consider another option for an important
00:03:04
rectangular section, let’s say it’s a wooden
00:03:07
structure 300 by 150 with a rolled
00:03:10
section bh, but you can enter either 0 and or
00:03:13
leave
00:03:14
any other numbers, then it goes with a
00:03:16
uniformly distributed load
00:03:17
total from permanent and temporary,
00:03:21
it is clear that this is our
00:03:23
total design load,
00:03:26
let’s put the letter in pairs, how can I get it, we have
00:03:28
already talked to you
00:03:30
now p1 p2 p3 and p4 here they are 4 loads,
00:03:35
they are equal to half of
00:03:36
our vertical load on the
00:03:39
Kosovo is 86 64. 80 you can
00:03:43
paste this number here, then copy it and
00:03:47
put it here 4 by ordinary
00:03:52
copying, the distance is exactly the same a1
00:03:55
a2 a3 a4 direct measurements from the
00:03:58
drawing means 1 350 enter and two
00:04:05
1400 enter a 3900
00:04:10
a 4 2 950 enter and since we have
00:04:14
all the tasks with you - after all, the channel is 20 p, then
00:04:18
here in the initial data you should enter the
00:04:20
calculated resistance steel 245 fish
00:04:25
240r y n 245 gamma im 10 25
00:04:29
we determined these parameters in the last video,
00:04:31
well, here we just enter the
00:04:34
calculated resistance on the
00:04:36
shear tool is considered automatic, also the
00:04:39
review talked about this the modulus of elasticity of
00:04:41
steel
00:04:42
and these cells are filled in automatically
00:04:44
if we do not have a rolled section and
00:04:48
here let’s say we see b in a cube at 12
00:04:51
bash square at 6
00:04:54
bash square at 8 s this is the thickness of our
00:04:58
element, here it is 150 and the permissible
00:05:02
deflection as a length is 200 narrower It’s
00:05:06
not important that you can enter
00:05:07
any numbers, but I’ll repeat once again due to
00:05:10
the fact that we still have
00:05:12
a rolled channel, especially for rolled
00:05:15
elements, here
00:05:16
you need to manually enter its parameters and
00:05:19
where to get the value of the moment nr for
00:05:21
the resistance itself is very simple, we
00:05:24
have a copy here he is our Schiller 20 p
00:05:28
GOST some kind of steel with 245, but the parameters are the
00:05:33
moment of inertia, the moment of resistance, the
00:05:36
static moment and the wall thickness of 5 and
00:05:40
2 millimeters, these numbers are with you here, we
00:05:46
enter here sequentially in millimeters of the fourth power, do not forget
00:05:48
that in the centimeter table we entered four
00:05:50
names By and large, the calculation
00:05:54
is prepared, you just need to run the
00:05:57
automatic solution, for this there
00:05:59
is such a procedure as searching for a
00:06:02
solution, then if, when you activate the button, the
00:06:05
data does not pop up, this is the
00:06:08
search for a solution icon, which means you need to
00:06:10
revive it in a very
00:06:13
simple way, for example, here are the
00:06:15
excel
00:06:16
add-in parameters further here we find the
00:06:19
excel settings, go to and
00:06:21
here you need to check the box check the box
00:06:24
and say ok then after
00:06:26
a while when you activate
00:06:28
the data you will see a search for a solution here you
00:06:30
need
00:06:31
this is all already configured in the template
00:06:34
you need to click the button execute the
00:06:36
solution found all the restrictions
00:06:38
optimality layer to fulfilled do not forget
00:06:40
you need to click ok and check our graphs,
00:06:43
it means a diagram of transverse forces,
00:06:46
here we have a break in us, I have chiseled forces and are
00:06:50
constantly increasing due to the fact that
00:06:53
we have a constantly distributed load,
00:06:55
so here is a diagram of the transverse force,
00:06:58
then bending moments, here it is broken
00:07:02
and the shape of the deflection is a
00:07:05
reminder summary data plates
00:07:06
should run the search procedure
00:07:08
decided
00:07:09
and the results are displayed means
00:07:12
maximum deflection
00:07:13
maximum with borscht maximum and
00:07:15
normal stresses
00:07:16
maximum transverse force
00:07:18
maximum shear stresses and
00:07:20
three conditions for fulfillment or failure
00:07:23
strength for normal stresses
00:07:25
is satisfied strength for shear
00:07:28
stresses is satisfied stiffness
00:07:30
is satisfied and the coefficients are shown here
00:07:33
using the creature's
00:07:36
ability according to these parameters,
00:07:37
if you want to change something here, then
00:07:39
this is done very simply, let's say we
00:07:41
want to increase the load, well, it’s clear that
00:07:45
we have a minus here because the
00:07:46
value is directed downwards against the y-axis,
00:07:50
just like with a uniformly distributed
00:07:52
load,
00:07:53
let’s try to increase uniformly
00:07:55
distributed load -7 and if so, then
00:08:00
let’s also increase the black force to 10
00:08:03
thousand newtons minus 10 times two three
00:08:07
center we’ll
00:08:09
leave them in the same places,
00:08:14
we’ll just cure the loads with you, it’s clear that all the
00:08:18
graphs are wrong, but here we have the
00:08:21
search data again for now to find solutions solutions
00:08:25
execute solution found ok
00:08:29
stress increased a little deflection
00:08:30
increased tangential stress
00:08:32
a little eternity and here are the
00:08:35
coefficients of load-bearing capacity utilization
00:08:39
so I have the initial data and
00:08:42
periodically run the procedure for finding a
00:08:45
solution you can vary the solution to swap
00:08:48
forces increase me flow
00:08:51
for comparison just below is the exact
00:08:54
same beam with the same dimensions, the
00:08:56
same loads but with rigidly fixed
00:08:58
ends, and the same results are given:
00:09:01
maximum deflection,
00:09:03
maximum stresses, load-
00:09:05
bearing capacity utilization factor, this
00:09:08
calculation is given here automatically, well, at
00:09:10
least to compare the
00:09:11
difference between a split hinge design
00:09:14
and a continuous one, let's compare
00:09:17
briefly, which means transverse forces are the
00:09:20
same values ​​are the same because the
00:09:23
river load and reactions on the supports have not
00:09:25
changed the bending moment in the hinged
00:09:28
diagram of the zeros on the supports the maximum moment in
00:09:32
the span is
00:09:33
equal to 27 million, but let's
00:09:36
broaden our horizons a little and look at
00:09:39
classic examples of connecting beams with
00:09:43
decks, here is an example of a
00:09:45
hinged connection girder columns
00:09:48
I-beam I-beam I-column
00:09:51
corner welded to flange
00:09:54
2 by connecting bolts and mounting table
00:09:57
welding welding classic hinge
00:10:02
joint
00:10:03
in this case the load is transferred from the
00:10:05
beam
00:10:06
through the bolts and through welding the corner to the
00:10:12
column flange and then to the foundation
00:10:14
another classic example of a hinged
00:10:17
connection mounting table or the support
00:10:20
table is made, the
00:10:22
support rib is installed, the beam is installed and
00:10:24
secured with mounting bolts, the bolts are
00:10:27
here only for safety reasons, it holds the
00:10:32
entire load, this support table,
00:10:35
due to welding, is
00:10:36
also considered a classic hinge
00:10:39
unit, let's consider the option of a
00:10:42
rigid connection between the beam and the column, this is
00:10:44
how this unit is organized
00:10:46
plate plate mounting table and
00:10:50
we see the plate with welding
00:10:52
is connected to the upper chord and with the lower
00:10:55
chords additional stiffening ribs are also installed,
00:11:00
such a unit is like this, and as we see, the
00:11:03
central part is also tacked; in
00:11:06
this case, the plate is welded to the shelf
00:11:10
and welded with mounting bolts with welding
00:11:13
to the wall here are different types of plates and
00:11:17
different options for connections, which is also
00:11:21
considered a classic rigid node
00:11:23
connecting
00:11:24
steel beams with the rest of the column, but
00:11:27
here I would also add, if it is
00:11:28
possible structurally, of course, the
00:11:31
same plate on the supporting one on
00:11:34
top to increase the rigidity of the
00:11:37
connection with these, let’s say an explanation and
00:11:41
now the following question appears,
00:11:43
look, here we are
00:11:46
when we calculated the beam using shear
00:11:49
stresses,
00:11:50
remember here we took the static
00:11:53
moment of the flange in this section
00:11:56
and I said, please note that this
00:11:58
calculation is approximate because we do not
00:12:01
know the design of the
00:12:03
beam-column connection unit,
00:12:06
now if Let's say we consider
00:12:09
this option, then you see that here the
00:12:12
vertical load or reaction of the support
00:12:15
is transmitted to the bottom as if through the section of the
00:12:18
beam, and here the weaker flow is the
00:12:21
flow of this corner,
00:12:23
so when calculating such a node and
00:12:28
these beams according to the condition of strength according to
00:12:31
tangential stresses,
00:12:33
you should take the moment inertia and
00:12:37
static moment, cross-section and thickness c in the
00:12:41
Zhuravsky formula, it is this part of the
00:12:45
support, that is, this corner, and let’s
00:12:49
do the calculation directly, that is,
00:12:52
if we have such a support there and
00:12:54
correctly, almost
00:12:55
if there is such a support, then it is necessary to take the
00:12:59
sections in this place and as a
00:13:03
comparison, let's calculate the
00:13:05
tangential stresses using the
00:13:07
Zhuravsky formula in the case of the node where the
00:13:12
beam is attached to the column using a corner, let's
00:13:17
take this section in yellow
00:13:20
numbers
00:13:21
150, let the height of the corner be 6, its
00:13:24
thickness is commensurate with the thickness of the section of the
00:13:28
beam of the wall, and here the yellow
00:13:31
cells will also be determined the moment of inertia of
00:13:35
this section is bh cubed by 12, the
00:13:40
static moment of this
00:13:46
section bh squared by 8 is determined and the
00:13:51
tangential stresses are determined as the transverse
00:13:54
force multiplied by the static moment of
00:13:56
the section divided by the moment of inertia and by the
00:14:02
wall thickness of this element and we see
00:14:07
that in our case if you considered the
00:14:12
support node as a whole, like a whole beam,
00:14:15
then these are the stresses; if we take
00:14:19
into account the structural elements and
00:14:21
consider just such a technical
00:14:24
solution, then the stresses can reach up to
00:14:29
these values; therefore, when calculating the
00:14:32
shear stresses, you should understand
00:14:36
which node
00:14:38
connecting the beams with the column is being discussed
00:14:41
a few words about collecting loads on a beam,
00:14:45
beam cages, here are a lot of
00:14:47
photographs showing the
00:14:50
designs of
00:14:51
beams that can be calculated using these
00:14:54
templates, well, let’s say we have a
00:14:57
main beam, a
00:14:58
main beam, secondary beams and
00:15:02
they are also scary shelves, so how to assemble
00:15:05
let’s say the load is on this beam, yes
00:15:07
very simply, half the
00:15:09
load distributed over the area on
00:15:12
this side is half on this side, that
00:15:15
is, we multiply the distributed load over the
00:15:16
area by this
00:15:19
distance, then we get a uniformly distributed
00:15:21
load on this beam in the same way
00:15:24
here, if we calculated this beam,
00:15:27
then we could do the next thing you can do is
00:15:30
this load from this area
00:15:33
transform the concentrated force to get
00:15:37
one two three 4 5 six concentrated
00:15:41
forces and let’s
00:15:46
take the own weight of the beam as a uniformly distributed load, then
00:15:49
in this beam the same thing
00:15:50
can be divided either the load
00:15:56
area transform this area or this
00:16:00
load distributed over the area into a
00:16:02
concentrated force
00:16:05
one-two-three-four are scientists force and
00:16:07
let's assume a uniformly distributed
00:16:09
load as the beam's own weight,
00:16:13
here the load
00:16:15
areas of the columns are also shown the load areas of other
00:16:21
beams in general, everything is just for reference
00:16:24
and here are the data parameters
00:16:26
from the EU for a wooden structure, the modulus of
00:16:28
elasticity of wood, depending on the
00:16:31
strength class and calculated resistances in
00:16:35
order to be able to perform the calculation of a
00:16:38
specific beam, a few words
00:16:41
about the calculation of welds,
00:16:44
we are now preparing an automated software package
00:16:48
where
00:16:51
calculations of any welds will be performed,
00:16:53
here we will only touch on questions
00:16:55
related to bur loads on welds
00:16:59
and drawing up calculation diagrams of
00:17:02
welds, if we were to take
00:17:06
this weld that is made along the
00:17:10
perimeter of the corner in
00:17:11
this way and we understand that the
00:17:14
load from the stringer is transmitted through
00:17:18
this bolted connection, there is
00:17:20
no permissibility of welding there, then like this
00:17:23
in this way, a transverse force is applied here
00:17:26
to y, which we find from the calculation of
00:17:32
the stringer, this load is applied here in
00:17:35
this plane
00:17:36
of the plane, here the leaves can be seen that I am
00:17:39
showing, that is, this load is known, the
00:17:43
known distance a and b and in
00:17:46
order to go to the design diagram of the
00:17:48
weld we need to bring this
00:17:51
load to the center of gravity of our
00:17:55
welds, the center of gravity will be in the middle,
00:17:56
so I
00:17:58
corrected the mechanics, this force is transferred
00:18:00
here with the addition of y moment, in
00:18:03
this case it will be a torque for
00:18:06
our weld,
00:18:08
that is, or a moment that we will call
00:18:12
if we have y and x this is how it is
00:18:14
oriented z is looking at us, then let’s
00:18:17
apply a torque moment q
00:18:21
y multiplied by arm b in half and a
00:18:24
transverse force this is how the
00:18:26
design diagram of
00:18:28
welds looks like in this connection
00:18:32
let’s look at the beam-column support unit
00:18:35
which we looked at a little higher
00:18:39
and here are two views welds, a
00:18:42
weld connecting directly the beam
00:18:44
to the shelf during the period of operation and a
00:18:48
weld connecting the installation or table to the
00:18:51
shelf of the column, the
00:18:53
installation period, that is, installation
00:18:55
loads must be used, so if
00:19:00
we have this force and we want to
00:19:03
draw up a design diagram for these
00:19:05
welds,
00:19:06
this angle, then you need to do
00:19:09
the following,
00:19:10
you need to transfer this transverse force
00:19:12
to the center of gravity of the section,
00:19:14
transfer it to the center of gravity as y and
00:19:17
add a torque relative to the
00:19:19
z axis, which look at us sat in
00:19:24
half as much as this is the distance between the
00:19:28
welds if we talk about the design diagram of the
00:19:31
assembly welds table, the
00:19:33
seams are made in this way,
00:19:35
here they are depicted at a distance f3, then
00:19:40
in this case there is a transverse
00:19:43
force q,
00:19:45
which manifests itself at the moment of installation,
00:19:48
it is clear that this is not the case, which
00:19:51
acts before operation,
00:19:54
this is probably its own weight in half, so
00:20:00
this is the shoulder, well, here you will have some
00:20:03
shoulder to accept, there is a possibility that
00:20:05
such a calculation makes sense to carry out when
00:20:07
n is equal to the full height of this corner,
00:20:12
that is, adding here, the
00:20:14
bending moment relative to the x axis is
00:20:18
equal, which means this is q on this shoulder,
00:20:22
this is the design diagram of the weld seam of the
00:20:26
assembly table, and if we
00:20:28
say more and a bolted connection then
00:20:31
requires calculation and they themselves require calculation
00:20:35
or selection of the diameter of the bolts that will
00:20:39
work together with the wall from
00:20:44
this corner, these elements should be
00:20:47
calculated for crushing and the bolt on the estimate and
00:20:51
for shearing, in short,
00:20:54
we will talk in more detail about the calculation of welds in a
00:20:57
separate section trainings and now let's go further in terms of
00:21:00
calculation of beams, let's consider also the
00:21:02
automated calculation of beams, this
00:21:04
tab
00:21:05
go here 2 support beam with
00:21:08
the possibility of organizing a snow bear,
00:21:11
or in this case, one two three about
00:21:14
any
00:21:15
linearly varying loads along the length
00:21:18
can be entered and also up to 6 any
00:21:22
concentrated strength,
00:21:23
let's try to solve the problem in this
00:21:25
template,
00:21:26
let the length of the element be 3,300 in this
00:21:30
case the same principles work
00:21:32
either yours for excision and the characteristics are
00:21:36
entered automatically geometrically
00:21:38
or here we enter from the table
00:21:41
I left the same profile here these are the
00:21:44
characteristics as you remember the goals are
00:21:47
joy correct and shelves p20 moment of
00:21:49
inertia moment resistance
00:21:51
static impala section
00:21:53
wall thickness here distributed
00:21:56
loads are also very simple and
00:21:59
intuitive q1 start to one end q2 start
00:22:04
q2 end here it is and 3 load q3 start
00:22:08
q3 end can be evenly
00:22:11
distributed and in any
00:22:12
sequence in any combination, the
00:22:14
beginning and end are set here, the beginning and
00:22:17
end of the corresponding loads, well, let’s
00:22:19
say let 1 load start from 0
00:22:23
from minus 5 to minus 2 and up to 1300 times it
00:22:27
is checked, it automatically builds the whole
00:22:30
here, here we see up to from 5 to 2
00:22:34
by 1300 then the load since a
00:22:36
small spelling error crept in here, here
00:22:39
they are, the goal is for a distributed load,
00:22:42
so I corrected you in the template,
00:22:44
and this is for concentrated forces, then
00:22:47
constant load 2 from 1300 to the end of
00:22:52
3,300 madara
00:22:53
and the third load, we simply don’t have clay,
00:22:58
you can leave this unimportantly at
00:23:00
replacing
00:23:02
any load, the
00:23:06
diagram of the
00:23:08
distributed load is automatically rearranged, you can also
00:23:10
change the lengths of the concentrated forces,
00:23:13
let's enter the same numbers that
00:23:16
we had in the last template,
00:23:18
here are 350400 1902 402 600 and 3 meters, but then
00:23:26
this does not work out 10 thousand newtons
00:23:30
1000 kilograms or this is a ton minus languidly,
00:23:34
let's bring David here and
00:23:39
at 1400 they are also automatically viewed here,
00:23:47
if we have fewer of them, then we enter a
00:23:49
smaller number of loads, then the
00:23:53
calculated resistance is
00:23:54
steel 240 schiller mountains of fungus at 245 to the
00:24:00
reliability center for steel 10 25
00:24:04
calculated shear resistance
00:24:08
The modulus of elasticity is calculated automatically; the limiting deflection is calculated automatically;
00:24:12
we have a hundred steps of integration; the integra step
00:24:15
is also automatic for me as
00:24:17
soon as we have entered the initial data; here we
00:24:20
need to run the search for a
00:24:23
solution procedure; this search for a solution is all
00:24:27
set up; okay, here is our
00:24:30
load, and here is our distribution of the
00:24:34
shear force, bending moment the
00:24:37
angle of rotation of the section and the deflection and,
00:24:41
accordingly, the results are given in the
00:24:43
form of such a compact list whether the
00:24:48
strength conditions for normal
00:24:50
tangential stresses are satisfied or not for rigidity and the
00:24:52
corresponding utilization coefficients are given
00:24:54
for these three parameters,
00:24:57
here’s a convenient template, let’s
00:25:01
test it under a uniformly
00:25:03
distributed load,
00:25:04
that is, here let's take minus 6 from 0 to the
00:25:12
end and the rest we enter or and the environment we'll
00:25:17
introduce the chiseled forces so we'll execute the solution
00:25:27
okay and check for example two numbers
00:25:30
maximum moment and maximum
00:25:32
deflection maximum moment let's here we
00:25:35
calculate the stake squared by 8 equals k
00:25:40
multiply this we have b in square
00:25:49
divided by
00:25:55
8,816,520 and there is a maximum deflection of 5
00:26:04
multiplied so there is 5 golf 4 divided by 384
00:26:11
and p5q
00:26:16
multiplied by 4 divided 384 multiplied by a and
00:26:31
multiplied by up to 88 36 but 88
00:26:42
but hardly quite acceptable accuracy of
00:26:46
calculations and in the beams section we’ll see more
00:26:50
one template with the possibility of placing an
00:26:55
unlimited number of intermediate
00:26:57
supports, we go to this template, here it is
00:27:02
statically indeterminate and the beam here
00:27:06
retains all the conditions of the previous template,
00:27:09
that is, up to 6 concentrated forces of
00:27:12
various sizes, up to three
00:27:14
distributed loads with any angle of
00:27:17
inclination and let's see
00:27:19
how to organize it,
00:27:21
get to know briefly the technique
00:27:23
of forming a boundary conditions in the
00:27:26
procedure for finding solutions
00:27:28
initial parameters or boundary
00:27:32
conditions what we know at the ends of the beam
00:27:36
we know that here the moment is zero
00:27:38
here the moment is zero the deflection is
00:27:41
zero prague correct and the personal force
00:27:45
and angle of rotation are
00:27:47
not known neither here nor here means
00:27:50
at the origin of coordinates,
00:27:51
we know that the moment is zero, so we
00:27:56
fix it here 0 and the deflection is equal to zero,
00:28:00
this is known, the initial parameters are the
00:28:04
angle of rotation at the origin of coordinates and the
00:28:07
lateral force is unknown, they are in
00:28:10
orange, they will be varying by
00:28:14
parameter, then what do we know on the
00:28:16
intermediate supports
00:28:18
on the intermediate support the reaction is manifested, the
00:28:22
magnitude of this reaction is such that
00:28:26
it delivers a deflection equal to zero in this
00:28:29
place, that is, r1 and r2 or there will be several of them
00:28:33
will also be included in the varying
00:28:37
parameters and the goal function deflection and deflection
00:28:41
will be equal to zero at the end, we said that
00:28:45
we should get the moment equal
00:28:47
zero and the deflection is equal to zero,
00:28:49
these are the conditions that have been spoken, we will now
00:28:53
try to formulate them in our
00:28:56
search for solutions, which means that the initial
00:28:58
parameters we have entered are 00, this is
00:29:01
known, here you can enter any
00:29:03
values, either zero or
00:29:05
ones, it doesn’t matter, we
00:29:08
varied the parameters now we said
00:29:13
let’s let's agree that
00:29:15
we will put this reaction just well, which one is here, but it
00:29:19
is also drawn, that is, at the end of the first and
00:29:21
started the last one, that is, at 1300 and
00:29:26
4,500, let's select these cells, so we
00:29:29
have highlighted them, these lines are suitable for example
00:29:33
for a thousand
00:29:34
rice, which means it is marked in green light and
00:29:39
thousands and 4,500
00:29:42
you 4,500 are also highlighted in
00:29:46
green now we have a reaction column
00:29:52
here we must also put
00:29:56
the wine box and make it varied as a cell
00:30:02
and the second reaction must also put
00:30:08
one
00:30:09
0 and 0 does not matter and make it also varying
00:30:18
May chayka
00:30:21
function of the goal we must ensure that
00:30:24
here at the end the moment is
00:30:27
equal to zero and the deflection is equal to this.
00:30:30
We will form this in the search for solutions and
00:30:32
the goal we have 2 more goals is that the
00:30:37
deflection is equal to zero here, let’s also
00:30:41
indicate it with the same property and the deflection
00:30:55
together of this reaction should also be
00:30:58
equal to zero, now take a closer look at the data, search for a
00:31:03
solution, now it’s as if we have
00:31:06
nothing, delete, delete, delete everything
00:31:17
zero, search for a solution, once again
00:31:24
vary the parameters 1 2 3
00:31:31
and it should be in this
00:31:35
cell, in this cell and in these
00:31:40
two cells it
00:31:43
means searching for a solution, well, let’s have the first
00:31:48
target seagull, well, let the deflection at the end
00:31:50
be equal to the value will and
00:31:56
we’ll form the rest, which means the rest
00:31:58
just need to be added because
00:32:00
we have a goal function, target cell 1
00:32:02
constraint, you can add as much as
00:32:04
you want,
00:32:05
we add the moment at the end should be equal to
00:32:10
zero,
00:32:11
add deflection at the second reaction
00:32:17
should be equal to 0 and add deflection
00:32:21
at the railing reaction should be equal to
00:32:24
zero ok and form varying draws
00:32:29
valera and we and the seagulls you and I, this is the
00:32:36
first one formed by semicolon
00:32:42
enumeration 2 and 2 reactions 3 4
00:32:52
all if I have formed everything correctly
00:32:55
click execute solution solution
00:32:59
found obtained values ​​obtained
00:33:01
graphs here we have deflection 0 deflection 0
00:33:05
graphs all converge and the
00:33:10
corresponding
00:33:13
utilization coefficients are obtained for normal tangents
00:33:16
for stiffness in the same way you can
00:33:19
add a few more pores let's
00:33:22
try let's add terms by adding a
00:33:25
time exactly to the middle of the span, it is clear
00:33:28
that the reaction will also manifest itself here, so
00:33:33
the middle of the span, the middle of the span, you and
00:33:36
I have a length of 6 meters,
00:33:38
which means at three meters we
00:33:51
will have an additional cell that varies,
00:33:54
but in this cell we will ask to
00:33:59
find a solution to ensure equality to zero
00:34:04
then yes, he will now also look for
00:34:06
the addition of
00:34:07
this reaction, which delivers
00:34:10
zero here, how to add this to the search for a solution,
00:34:13
add varying seagulls separated by a
00:34:16
semicolon, this cell will
00:34:20
vary and in this cell we
00:34:23
need to add its equality to zero in poker
00:34:33
execute the solution solution found here
00:34:41
she is the reaction, here it is 0 and here is our support 0
00:34:51
meter thirty-three meters four and a
00:34:55
half and 6 meters, so with
00:34:59
this template you can solve almost any
00:35:02
problems of beams of this type with any
00:35:08
loads up to the snow bear and
00:35:11
any combinations of distributed and
00:35:15
concentrated forces with any amount
00:35:18
intermediate supports and so we
00:35:20
have worked enough on which beams and
00:35:23
beam cages
00:35:25
let's move on to calculating the railings

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