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0:00

начало

0:46

условие задачи

1:23

кинематический анализ

4:37

расчет простой рамы

9:46

расчет средене части рамы

14:08

построение эпюры М для всей рамы

14:33

обобщение

Расчет составной многопролетной балки (начало) / строительная механика

Расчет составной многопролетной балки (начало) / строительная механика

Channel: Строительная механика с Татьяной

Метод перемещений. Угловое перемещение

Метод перемещений. Угловое перемещение

Channel: Строительная механика с Татьяной

Расчет рам методом сил. Обобщение

Расчет рам методом сил. Обобщение

Channel: Строительная механика с Татьяной

Расчет симметричных систем методом перемещений (ч.2)/строительная механика

Расчет симметричных систем методом перемещений (ч.2)/строительная механика

Channel: Строительная механика с Татьяной

расчет симметричной рамы комбинированным методом / строительная механика

расчет симметричной рамы комбинированным методом / строительная механика

Channel: Строительная механика с Татьяной

Определение коэффициентов и свободных членов кононических уравнений м.п. кинематическим способом

Определение коэффициентов и свободных членов кононических уравнений м.п. кинематическим способом

Channel: Строительная механика с Татьяной

Расчет симметричных систем методом перемещений (ч 1)

Расчет симметричных систем методом перемещений (ч 1)

Channel: Строительная механика с Татьяной

Расчет рамы методом сил с двумя неизвестными (ч. 2)

Расчет рамы методом сил с двумя неизвестными (ч. 2)

Channel: Строительная механика с Татьяной

основная система и канонические уравнения метода перемещений

основная система и канонические уравнения метода перемещений

Channel: Строительная механика с Татьяной

метод перемещений

симметричные рамы

степень кинематической неопределимости

основная система

степень свободы

кинематический анализ

эпюра

изгибающиий момент

грузовая эпюра

00:00:03

[applause]

00:00:04

[music] [applause]

00:00:09

[music]

00:00:12

hello dear subscribers

00:00:14

listeners we continue to study

00:00:16

Tatyana’s construction mechanics today

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I want to offer you another, in my

00:00:23

opinion, interesting problem and before you

00:00:25

start solving this problem I want to ask

00:00:28

you to subscribe to my channel who not

00:00:31

subscribed please like if

00:00:33

you like the video and also comment on

00:00:35

my videos, thereby you will help me

00:00:39

promote it in the comments you can

00:00:42

ask me questions, I will be happy to

00:00:44

answer them so we have a symmetrical frame in front of us,

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it’s easy to notice here is

00:00:52

our axis of symmetry, a

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uniformly distributed load acts on the frame

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and for this we

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need to construct only a diagram of

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bending moments, I want to draw your

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attention to the fact that sometimes

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problems are not written with what method to solve,

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but are simply asked to construct a diagram,

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therefore, before solving the problem, you

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must conduct a

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kinematic analysis

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and what it includes first is

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to determine the degree of freedom and make sure

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that this is the system in front of us,

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it is statically determinable and or indeterminate and so the

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degree of freedom of the system is 3d minus 2x

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minus t zero d is the number of hard

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drives, here we have a hinge and here there is no ball,

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so we have one disk, the second disk is

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this disk 3 4 5 disk 3 multiplied by 5

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minus 2 multiplied by shh this is the number of

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simple hinges connecting the disks

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regarding this question in the degree of

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freedom, what is it about your hinge, I

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have a separate video,

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please, you can watch it, the

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link is at the top,

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since our hinge connects more of two

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disks it is a multiple of connecting three

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disks 3 minus 1 will be 2 and here is exactly the

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same hinge therefore 4 hinges and

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minus c the supporting number of supporting

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links here are 3 each and here there is one in total 11

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total 15 minus 19 we get 4

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which means the number of extra links if we

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we get a negative number, we get a

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statically indeterminate system,

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therefore there are extra connections in this system,

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if you calculate this system using the

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force method, then there are extra connections and there will be

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minus w, that is, there will be 4 extra connections, but

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I want to draw your attention once again to the fact

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that kinematic analysis does not include

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only

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determining the degree of freedom and also you

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need to be able to

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analyze the formation diagram of our

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system, I want to draw your attention

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to this hinge and to this ball, not

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this hinge, it connects these 2 disks, but

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through this same hinge we have

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this disk adjacent, that is, this

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hinge not a simple multiple and we considered

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that it is for two hinges, that is, we have

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the right,

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instead of

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2 hinges, to draw here

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such a support, a

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hinge on a fixed support, that is, ours,

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this disk is attached to this

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group of disks by means of a hinge, instead of

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this hinge, we have the right to show

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the hinge on a fixed support,

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we have the right to do the same thing on the right with

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a hinge,

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as a result, what we get if we

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consider this part of the frame

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is that it is

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statically

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definable and a frame

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in which there are no extra connections w it is

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equal to zero in the same way in this part

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frames, so we can construct any diagrams in this frame without any

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problems, be

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given and boutique hotel, so let’s

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try to determine the value of the support

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reactions here in this frame and construct a diagram of the bending

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moments two kilo newtons no here a

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vertical reaction occurs here a

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vertical and horizontal

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reaction occurs ordinary simple frame I will designate

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this as support a this is support b if I draw up

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an equation for the projection of forces onto the x axis yes

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this horizontal reaction is already equal to

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zero since we have nothing projected onto the x axis

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except this reaction and

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therefore schools of us but to

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determine these two reactions in the past,

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we must make up our moment

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around point 2, multiply at a distance of 4,

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this will be the force to which we apply

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evenly in the middle and multiply by 2

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minus v.b. multiply by 4 equals zero,

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which means the reaction in b will be equal to 4

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kilonewtons, if we add up the sum of

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the moments around point b, we also

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get that the reaction in will also be 4

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kilonewtons, and

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4 kilonewtons, and well, puree diagrams of

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bending moments, I have a

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separate video,

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you can watch the construction diagrams in

00:06:44

frames and

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so we assign a section here, two sections

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here two sections are

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not enough since there is a

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uniformly distributed load here, so in the

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middle we need to

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assign an additional section, I will start

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with this section, considering the equilibrium of the

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lower part, the force in the shoulder would be zero,

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which means the moment will be equal to zero, well and

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I remind you that the moment at the hinge rave the gauze,

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then the moment at this point, again, it is more profitable for me

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to consider the equilibrium of the lower

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part and in this force relative to this

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section also the shoulder is zero, which means the moment is

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0 due to the equilibrium of the node if here

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we have 0. that point will also be 0,

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we can say the same thing about this

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point and about this point here we

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will also have 0 since this is a hinge and

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we have this one left.

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I’m looking at the equilibrium of the left side,

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we have a force of 4 kilonewtons acting on the left,

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and a uniformly

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distributed load of 2 kilonewtons per

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meter is acting, the

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length of this section is two meters, we

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take only half the force, I

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’ll duplicate this force here again, 4, I have to

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multiply by 2 since it acts

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from below stretches the lower fibers 4

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multiply by 2 I get 8 but since we

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also have a uniformly

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distributed load 2 multiply by 2

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will be 4 force 4 act from above but

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her shoulder is already 1 meter acting from above on

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will stretch the upper fibers 4 by 1

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we get 4 that as a result

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m8 will stretch from below a4 from above 8

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minus 4 will be 4, which means at this point

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the moment will be equal to 4 and we must

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draw a smooth curve for the parabola,

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I want to draw your attention to the fact that

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the parabola will always bend in the

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direction in which it acts uniformly

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distributed load,

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if at this place there was an unevenly

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distributed load, the

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concentrated force here and the feather

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looked like a triangle, but

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again the ether would bend in the direction of the

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action of the force not towards, be

00:09:35

careful, the same applies to this

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frame, it will be symmetrical relative to the

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axis of symmetry this means that the diagram will not be

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exactly the same, let's start calculating the

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middle part of the frame, that is,

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I drew this part of the frame with a load,

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it is also symmetrical

00:09:58

in this frame, we have counted 4

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extra connections using the force method,

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let's see how many unknowns

00:10:08

there will be according to the method of displacements n is equal to n

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angular + r Lenin

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angular displacements only one

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linear movement we will also have

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one in the horizontal direction + 1

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we get 2 as you can see by the method of

00:10:30

displacements away the next stage in the

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method of moving the construction of the main

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system on rigid nodes we impose

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connections or rigid floating ones seals that

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causes the system from linear

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movement we install a

00:10:48

hinge support

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angular movement we have for this day

00:10:56

linear movement we have

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z2 but I want to draw attention to the fact

00:11:05

that z2 linear movement in the

00:11:10

movement method is considered inversely

00:11:14

symmetrical unknown and if we have a

00:11:18

symmetrical frame and our

00:11:21

symmetrical frame acts

00:11:24

symmetrically load then the reverse

00:11:27

symmetrical unknowns will be equal to

00:11:30

zero, please watch the video

00:11:33

link at the top

00:11:37

z1 is our angular displacement,

00:11:41

but there is also one small

00:11:46

pleasant surprise since we have this node

00:11:50

located along the axis of symmetry, then this

00:11:54

angular displacement will also be

00:11:57

equal to zero since this part of

00:12:01

the structure will try to

00:12:03

bend like this, and this part of the

00:12:05

design contract tries to bend like this,

00:12:07

and that is, this node

00:12:11

will not rotate for us, therefore, what conclusion

00:12:14

can we draw since both of our

00:12:16

unknowns will be equal to zero, which means

00:12:20

we won’t have to solve anything, but the

00:12:23

bending diagrams moments the

00:12:26

final will be equal to simply and puree I

00:12:30

remind you how a load diagram is built in the

00:12:33

displacement method, we build it only on

00:12:37

those rods where there is a given load,

00:12:40

namely on this rods and on this rods

00:12:43

what this rod is,

00:12:46

rigid on the left, fastening on the right we have

00:12:50

a hinge, I remind you that we will rebuild

00:12:53

using these signs, these signs I

00:12:56

have repeatedly dropped in the playlist using the

00:13:00

moving method in several videos,

00:13:04

please look for links to these

00:13:07

signs in 1 plate we have all the diagrams

00:13:11

for the beam, rigid on one side,

00:13:15

fastened on the other side, hinged,

00:13:18

ours this shell has

00:13:21

rigid on the left hinge on the right therefore

00:13:26

since a uniformly distributed load acts on it,

00:13:28

it means that we are

00:13:31

building the pen here in this 5th scheme in the 5th scheme

00:13:36

of pressure we have Mr. one gel square for 8

00:13:40

in the middle jule square for 16

00:13:44

here we will have a large ordinate k

00:13:48

here less

00:13:53

than r square by 8 here forces r-square

00:13:59

by 16 in this section our diagram will be

00:14:04

absolutely symmetrical

00:14:08

so dear listeners,

00:14:11

for this part of the frame, that is,

00:14:14

here on the floor diagram we have a diagram, here it is for

00:14:18

this part of the frame, we also built it

00:14:22

for the left side of the frame, our diagrams

00:14:26

will be mirrored, then symmetrically, here we

00:14:29

get the final pen of bending

00:14:32

moments,

00:14:33

dear listeners, as usual, I’ll move on

00:14:36

to generalization and the first thing I want to

00:14:39

draw your attention to when calculating

00:14:42

symmetrical systems is

00:14:44

if the stand is located along the axis of symmetry,

00:14:48

I didn’t talk about this earlier then the angle of

00:14:52

rotation of the rigid assembly

00:14:55

will always be equal to zero if this

00:14:59

symmetrical frame is acted upon by a symmetrical

00:15:03

load, if it is a load that is inversely

00:15:06

symmetrical, then there will be no zeros

00:15:10

involved.

00:15:11

The following is what I want to pay attention to

00:15:14

when solving problems,

00:15:16

pay attention to the formation diagram of

00:15:20

our system, like in this case,

00:15:24

by selecting this part of the frame, we got a

00:15:27

statically definable rom, and if we had

00:15:30

not paid attention to this, we would have had to

00:15:33

add one more rigid node here, that’s

00:15:36

all I wanted to tell you today

00:15:39

about taking you to a tirade who

00:15:42

wants to build diagrams of transverse and

00:15:46

longitudinal forces so that you are sure of the

00:15:50

correctness, I will drop them in the link in the

00:15:54

description and for today I will finish

00:15:57

my video, do not forget to please like

00:15:59

if you liked the video,

00:16:02

comment please help me

00:16:05

promote my videos all the best to you, good

00:16:08

luck in your exams, be

00:16:12

healthy bye bye

Description:

Задача о том, как упростить расчет рамы, зная некоторые особенности проведения кинематического анализа и расчета симметричных систем 00:00 - начало 00:46 - условие задачи 01:23 - кинематический анализ 04:37 - расчет простой рамы 09:46 - расчет средене части рамы 14:08 - построение эпюры М для всей рамы 14:33 - обобщение эпюры Q и N: https://drive.google.com/file/d/1Z14bwu0ku2fQLUAT2zHJqn4TLDfdNvdB/view?usp=drivesdk

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