Designation of the moment of inertia of the cross section with parallel transfer of the axes. Changing the moment of inertia when moving the coordinate axes in parallel Formulas for moving the axes
Come on z h, y z– central axis of pereriziv; – moments of inertia across the chodo axes. Significant moments of inertia across new axes z1, 1, parallel to the central axes and places where they are on the stand aі d. Come on dA- elementary maidan on the outskirts of the point M with coordinates yі z at the central coordinate system. 3 fig. 4.3 it can be seen that the coordinates of the point Z of the new coordinate system are updated, .
Significantly moment of inertia across the axis y 1 :
|
| z c |
| y c |
| z1 |
| y 1 |
| d |
| a |
| C |
in such a manner,
Similarly
Changing the moments of inertia of the overcut when turning the axes
We know the fallow between the moments of inertia and about the axes y, z and moments of inertia about the axes y 1, z1, turned on the cut a. Come on Jy> Jz ta positive kut a wind up in the axis y anti-year arrow. Send coordinate points M before the turn y, z, after turning - y 1, z1(Figure 4.4).
From the little one whimpers:
Now the moments of inertia are significant for the axes y 1і z1:
|
| M |
| z |
| z1 |
| y 1 |
| y |
| a |
| y |
| y 1 |
| z1 |
| z |
Similarly:
Adding term by term equal (4.13) and (4.14), we take:
tobto. the sum of the moments of inertia, if any, mutually perpendicular axes, is permanently fixed and does not change when the coordinate system is rotated.
Head axles of inertia and head moments of inertia
Zі zmіnoyu kuta turn axes a skin values change, but the sum remains unchanged. Otzhe, іsnuє the same meaning
a = a 0 , for which the moments of inertia reach extreme values, that is. one of them reaches its maximum value, and the other one reaches its minimum value. For the meaning a 0 let’s take a look at it (otherwise) and equate it to zero:
It is shown that when the axes are taken away, the center moment of inertia is equal to zero. To this right, part of the equation (4.15) is equal to zero: , stars, tobto. took the same formula for a 0 .
Axis, where some central moment of inertia is close to zero, and axis moments of inertia gain extreme values, are called head axes. Yakscho tsi osі є і central, all stinks are called head central axes. axis moments of inertia like head axes are called head moments of inertia.
Significantly headline axis through y 0і z0. Todi
If the retina can be all symmetrical, then all is one of the head central axes of inertia perezu.
Let's look at the moment of inertia of the flat figure (Fig) for the axes $(Z_1)$ and $(Y_1)$ for the given moments of inertia for the axes $X$ and $Y$.
$(I_((x_1))) = \int\limits_A (y_1^2dA) = \int\limits_A (((\left((y + a) \right))^2)dA) = \int\limits_A ( \left(((y^2) + 2ay + (a^2)) \right)dA) = \int\limits_A ((y^2)dA) + 2a\int\limits_A (ydA) + (a^2 )\int\limits_A (dA) = $
$ = (I_x) + 2a(S_x) + (a^2)A$,
de $(S_x)$ - the static moment of the figure is about the axis $X$.
Similar to the axis $(Y_1)$
$(I_((y_1))) = (I_y) + 2a(S_y) + (b^2)A$.
Central moment of inertia for axes $(X_1)$ and $(Y_1)$
$(I_((x_1)(y_1))) = \int\limits_A ((x_1)(y_1)dA) = \int\limits_A (\left((x + b) \right)\left((y + a ) \right)dA) = \int\limits_A (\left((xy + xa + by + ba) \right)dA) = \int\limits_A (xydA) + a\int\limits_A (xdA) + b\int \limits_A(ydA) + ab\int\limits_A(dA) = (I_(xy)) + a(S_x) + b(S_y) + abA$
Most often, there is a transition from the central axes (the upper axes of the flat figure) to the full, parallel ones. Then $(S_x) = 0$, $(S_y) = 0$, the shards of the axis $X$ and $Y$ are central. Remaining mayo
de, - the power moments of inertia, that is the moments of inertia according to the power of the central axes;
$a$, $b$ - vіdstanі vіd central axes to analіzovanih;
$A$ - figure area.
It should be noted that when the central moment of inertia is assigned to the quantities $a$ and $b$, the sign is to blame, so that the stench is, in fact, the coordinates of the center of gravity of the figure in the axes that are being looked at. When the axial moments of inertia and values are assigned, the values are presented behind the module (as in the standard), however, the shards of the stench rise to the square.
For help formulas parallel transfer it is possible to change the transition from the central axes to the upper ones, or navpak- in the prevіlnyh central axes The first transition is marked with a "+" sign. Another crossing is marked with a sign- ".
Apply different formulas to the transition between parallel axes
Rectangular retin
Significantly the central moment of inertia of a rectangle is proportional to the main moments of inertia around the $Z$ and $Y$ axes.
$(I_x) = \frac((b(h^3)))(3)$; $(I_y) = \frac((h(b^3)))(3)$.
.
Similarly, $(I_y) = \frac((h(b^3)))((12))$.
Trikutny Pereriz
Significantly, the central moment of inertia of the tricoutter over the given moment of inertia of the base $(I_x) = \frac((b(h^3)))((12))$.
.
If the central axis $(Y_c)$ has a different configuration, then we can also look at it. The moment of inertia of all figures along the axis $(Y_c)$ is greater than the sum of the moment of inertia of the tricot $ABD$ along the axis $(Y_c)$ and the moment of inertia of the tricot $CBD$ along the axis $(Y_c)$, tobto
.
Appointment to the moment of inertia of the folded rail
Let's put together a peratin, which is made up of okremih elements, the geometric characteristics of any of them. The area, the static moment and the moment of inertia of the warehouse figure add up to the sum of the relevant characteristics of the warehouse. Like folding the perimeters, you can make it look like a pattern of one figure from the outside, the geometric characteristics of the figure are visible. For example, moments of inertia of a warehouse figure, shown in fig. will appear like this
$I_z^() = \frac((120 \cdot ((22)^3)))((12)) - 2 \cdot \frac((50 \cdot ((16)^3)))((12 )) = 72 \, 300 $ cm 4 .
$I_y^() = \frac((22 \cdot ((120)^3)))((12)) - 2 \cdot \left((\frac((16 \cdot ((50)^3)) )((12)) + 50 \cdot 16 \cdot ((29)^2)) \right) = 1\.490\.000$cm 4
Let me see you and Ix, Iy, Ixy. Parallel to the xy axes, we draw a new line x1, y1.
І significant moment of inertia of the very cutting of the new axes.
X 1 \u003d x-a; y 1 = y-b
I x 1 = ∫ y 1 dA = ∫ (y-b) 2 dA = ∫ (y 2 - 2by + b 3) dA = ∫ y 2 dA – 2b ∫ ydA + b 2 ∫dA=
Ix - 2b Sx + b 2A.
If everything passes through the center of gravity of the cut, then the static moment Sx =0.
I x 1 = Ix + b 2 A
Similarly to the new axis y 1, we can calculate the formula I y 1 = Iy + a 2 A
Central moment of inertia for new axes
Ix 1 y 1 \u003d Ixy - b Sx -a Sy + abA.
If axis xy pass through the center of gravity of the cut, then Ix 1 y 1 = Ixy + abA
If the beam is symmetrical, if one of the central axes moves around the entire symmetry, then Ixy \u003d 0, also Ix 1 y 1 \u003d abA
Changing the moment of inertia under the hour of turning the axes.
Let us know the axial moments of inertia around the xy axes.
The new coordinate system xy is taken away by turning the old system on kut (a> 0), i.e. turning the anti-Year arrow.
Let's install the fallow between the old and new coordinates of the Maidanchik
y 1 \u003d ab \u003d ac - bc \u003d ab-de
from tricot acd:
ac/ad \u003d cos α ac \u003d ad * cos α
from tricot oed:
de/od=sinα dc=od*sinα
Let us represent the value of virase for y
y 1 \u003d ad cos α - od sin α \u003d y cos α - x sin α.
Similarly
x 1 \u003d x cos α + y sin α.
We calculate the axial moment of inertia for the new axis x 1
Ix 1 = ∫y 1 2 dA = ∫ (y cos α - x sin α) 2 dA = ∫ (y 2 cos 2 α - 2xy sin α cos α + x 2 sin 2 α) dA = = cos 2 α ∫ y 2 dA - sin2 α ∫xy dA + sin 2 α ∫x 2 dA = Ix cos 2 α - Ixy sin2 α + Iy sin 2 α .
Similarly, Iy 1 \u003d Ix sin 2 α - Ixy sin2 α + Iy cos 2 α.
We put together the left and right parts of the taken away virus:
Ix 1 + Iy 1 \u003d Ix (sin 2 α + cos 2 α) + Iy (sin 2 α + cos 2 α) + Ixy (sin2 α - cos2 α).
Ix 1 + Iy 1 = Ix + Iy
The sum of axial moments of inertia does not change when turning.
Significantly is the central moment of inertia for new axes. The value x 1 ,y 1 is visible.
Ix 1 y 1 = ∫x 1 y 1 dA = (Ix – Iy)/2*sin 2 α + Ixy cos 2 α .
Main moments and main axes of inertia.
Head moments of inertia name their extreme values.
The axes, which have some extreme values, are called the head axes of inertia. The stench is always mutually perpendicular.
Vіdtsentrovy moment іnertsії schodo head axes zavzhdі dorivnyuє 0. Oskіlki vіdomo, scho shcho have є vіs symmetry, then vіdtsentrovy moment іvіvnyuє 0, also all symmetry є head vіssyu. If we take the first line of the virus I x 1 then equate її to “0”, then we take the value of kuta = the corresponding position of the head axes of inertia.
tg2 α 0 = - 
If α 0 >0, then the old station of the head axes must be turned in the direction of the year arrow. One of the main axes is є max, and іnsha - min. With the help of the weight max, the wind blows a smaller kut tієї vypadkovoї, vyssyu schodo kakoї may have a greater axial moment of inertia. The extreme values of the axial moment of inertia are determined by the following formula:
Chapter 2. Basic understanding of the support of materials. The task of that method.
Under the hour of designing different spores, it is necessary to virishuvate different nutritional values, zhorstkost, stamina.
Mitsnist- The building of this body will showcase the difference in vanity without ruination.
Hardness- the building of the structure to take advantage of without great deformations (displacement). Forward admissible values of deformation regulate the future norms and rules (SNIP).
stamina
We can look at the grip of the gnuchka shear
If you want to increase step by step, then there will be a quick haircut on the back. When the force F reaches the critical value, the shear will bulge. - Absolutely short.
With this, the shearing does not collapse, but sharply changes its shape. Such a phenomenon is called a vtratoy stamina and leads to ruin.
Sopromat- Tse foundations of sciences about mіtsnіst, zhorstkіst, stіykіst of engineering structures. Spivpromatі vikoristovuyutsya methods theoretical mechanics, physicists, mathematicians On the vіdmіnu vіd teoreticії mekhanіki spromat vrakhovuє zminі rozmirіv i form tіl pіd ієyu navantazhennya that temperature.
Significantly fallows between different moments of inertia across two parallel axes (Fig. 6.7), connected by fallows
1. For static moments of inertia
Well,
2. For axial moments of inertia
otzhe,
Yakshcho everything z pass through the center of gravity of the cut, then
From the given moments of inertia when parallel to the axes, the axial moment of inertia can be the least important for the axis to pass through the center of gravity of the cross section.
Similarly for axis
If all y pass through the center of gravity
3. For water center moments of inertia, it is necessary to take
The rest can be written
At times, if the cob of the coordinate system yz be in the center of gravity of the cut, take it away
Have a vipadku, if one or the other offends the axis with the axes of symmetry,
6.7. Changing moments of inertia when turning axes
Let the task of the moment of inertia be cut along the coordinate axes zy.
Kut vvazhaetsya positive, like the old coordinate system for the transition to the new one, it is necessary to turn the counter-year arrow (for the right rectangular Cartesian coordinate system). New and old zy systems of coordinates po'yazanі fallows, yakі vyplyvayut іz fig. 6.8:
1. Significantly for the axial moments of inertia along the axes of the new coordinate system:
Similar to the OS
If we add up the magnitude of the moment in inertia along the axes i, then we take
i.e., when the axes are rotated, the sum of the axial moments of inertia is a constant value.
2. Let's see the formulas for the center moment of inertia.
.
6.8. Main moments of inertia. Main axes of inertia
The extreme values of the axial moments of inertia of the cut are called head moments of inertia.
Two mutually perpendicular to the axes, where such axes of moment of inertia may have extreme values, are called head axes of inertia.
For the significance of the main moments of inertia and the position of the head axes of inertia, it is significant first along the tail in the moment of inertia assigned to the formula (6.27)
Equate this result to zero:
de - Kut, on which one you need to turn the coordinate axes yі z schob stench zbіglisya z head axes.
Porіvnyuyuchi vrazi (6.30) and (6.31), you can install, scho
,
Otzhe, shdo the main axes of inertia vydtsentrovy moment of inertia to zero.
Mutually perpendicular to the axes, from which one or the other offends the axes of symmetry of the perimeter, and the head axes of inertia.
Rozv'yazhemo rivnyannya (6.31) shodo kuta:
.
If >0, then the assignment of the position of one of the head axes of inertia for the right (left) Cartesian rectangular coordinate system is necessary z turn on the kut against the course of the wrapping (along the wrapping) of the Year's arrow. Yakscho<0, то для определения положения одной из главных осей инерции для правой (левой) декартовой прямоугольной системы координат необходимо осьz turn to the kut along the wrapping (against the wrapping direction) of the Year's arrow.
Axis maximum zavzhdi skladє smaller kut z tієї osі ( y or z), so that the axial moment of inertia can be greater than the value (Fig. 6.9).
The entire maximum is straightened under the cut to the axis (), yaksho () and folded in paired (unpaired) quarters of the axes, yaksho ().
The main moments of inertia are significant. Vicorist formulas from trigonometry, which link functions, with functions, formulas (6.27) are taken
,
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