Bendable reinforced concrete structures rectangular cross-sections are not effective from an economic point of view. This is due to the fact that normal stress the height of the sections when bending the element are distributed unevenly. Compared to rectangular sections, T-sections are much more profitable, because at the same bearing capacity Concrete consumption in T-profile elements is less.
The T-section, as a rule, has single reinforcement.
In strength calculations of normal sections of bending T-profile elements, there are two design cases.
The algorithm for the first design case is based on the assumption that the neutral axis of the bending element is located within the compressed flange.
The algorithm for the second design case is based on the assumption that the neutral axis of the bending element is located outside the compressed flange (passes along the edge of the T-section of the element).
Calculation of the strength of the normal section of a bending reinforced concrete element with single reinforcement in the case when the neutral axis is located within the compressed flange is identical to the calculation algorithm rectangular section with single reinforcement with a section width equal to the width of the brand flange.
The design diagram for this case is presented in Fig. 3.3.
Rice. 3.3. To calculate the strength of the normal section of a bending reinforced concrete element in the case when the neutral axis is located within the compressed flange.
Geometrically, the case when the neutral axis is located within the compressed flange means that the height of the compressed zone of the section of the tee () is not greater than the height of the compressed flange and is expressed by the condition: .
In terms of ongoing efforts from external load and internal forces, this condition means that the strength of the section is ensured if the calculated value of the bending moment from the external load (M ) will not exceed the calculated value of the moment of internal forces relative to the center of gravity of the section of tensile reinforcement at values .
M 
(3.25)
If condition (3.25) is satisfied, then the neutral axis is indeed located within the compressed flange. In this case, it is necessary to clarify what size width of the compressed flange should be taken into account in the calculation. The norms establish the following rules:
Meaning b " f , entered into the calculation; taken from the condition that the width of the shelf overhang in each direction from the rib should be no more 1 / 6 element span and no more:
a) in the presence of transverse ribs or when h " f ≥ 0,1 h - 1 / 2 clear distances between longitudinal ribs;
b) in the absence of transverse ribs (or when the distances between them are greater than the distances between the longitudinal ribs) and h " f < 0,1 h - 6 h " f
c) with cantilever overhangs of the shelf:
at h " f ≥ 0,1 h - 6 h " f ;
at 0,05 h ≤ h " f < 0,1 h - 3 h " f ;
at h " f < 0,05 h - overhangs are not taken into account.
Let us write down the strength condition relative to the center of gravity of the tensile longitudinal reinforcement
M 
(3.26)
Let us transform equation (3.26) similarly to the transformations of expressions (3.3). (3.4) we obtain the expression
M (3.27)
From here we determine the value
= (3.28)
By value from table 
Let's determine the values of 𝛈.
Let's compare the value . element sections. If condition 𝛏 is satisfied, then it constitutes a strength condition relative to the center of gravity of the compressed zone of the tee.
M 
(3.29)
Having carried out the transformation of expression (3.29) similar to the transformation of expression (3.12), we obtain:
= (3.30)
it is necessary to select the area values of the stretched longitudinal working reinforcement.
Calculation of the strength of the normal section of a bending reinforced concrete element with single reinforcement in the case where the neutral axis is located outside the compressed flange (passes along the edge of the tee) is somewhat different from that discussed above.
The design diagram for this case is presented in Fig. 3.4.
Rice. 3.4. To the calculation of the strength of the normal section of a bending reinforced concrete element in the case when the neutral axis is located outside the compressed flange.
Let us consider the cross-section of the compressed zone of the tee as a sum consisting of two rectangles (flange overhangs) and a rectangle related to the compressed part of the rib.
Strength condition relative to the center of gravity of tensile reinforcement.
M + (3.31)
Where – force in compressed shelf overhangs;
Shoulder from the center of gravity of the tensioned reinforcement to the center of gravity of the shelf overhangs;
– force in the compressed part of the tee rib;
- shoulder from the center of gravity of the tension reinforcement to the center of gravity of the compressed part of the rib.
= (3.32)
= (3.33)
= b (3.34)
= (3.35)
Let's substitute expressions (3.32 – 3.35) into formula (3.31).
M + b (3.36)
Let us transform the second term on the right side of the equation in expression (3.36) similarly to the transformations performed above (formulas 3.3; 3.4; 3.5)
We get the following expression:
M + (3.37)
From here we define numerical value .
=
(3.38)
By value from table 
Let's determine the values of 𝛈.
Let's compare the value with the limit value of the relative height of the compressed zone . element sections. If condition 𝛏 is satisfied, then the equilibrium condition for the projections of forces on the longitudinal axis of the element is created. Σ N=0
--=0 (3.39)
=+ b (3.40)
From here we define required area sections of tensile longitudinal working reinforcement.
=
(3.41)
By assortment of rod reinforcement 
it is necessary to select the area values of the stretched longitudinal working reinforcement.
The calculations are the same as for a rectangular beam. They cover the determination of forces in the beam and at the corners of the slab. The forces then lead to the center of gravity of the new T-section.
The axis passes through the center of gravity of the slab.
A simplified approach to accounting for slab forces is to multiply the forces at slab nodes (common slab and beam nodes) by the design width of the slab. When positioning a beam relative to a slab, displacements (also relative displacements) are taken into account. The resulting abbreviated results are the same as if the T-section was raised from the plane of the slab by an amount of displacement equal to the distance from the center of gravity of the slab to the center of gravity of the T-section (see figure below).
Bringing forces to the center of gravity of the T-section occurs as follows:
M = Mb + Mp * B + Np * B * e1 + Nb * e2
B = beff1+b+beff2
Determining the center of gravity of a T-section
Static moment calculated at the center of gravity of the slab
S = b*h*(offset)
A = (beff1+b+beff2)*hpl + b*h
Center of gravity raised relative to the center of gravity of the slab:
b - beam width;
h - beam height;
beff1, beff2 - calculated slab widths;
hpl - slab height (slab thickness);
displacement is the displacement of the beam relative to the slab.
NOTE.
- It is necessary to take into account that there may be common areas of the slab and beam, which, unfortunately, will be calculated twice, which will lead to an increase in the rigidity of the T-beam. As a result, forces and deflections are reduced.
- The slab results are read from the finite element nodes; mesh refinement affects the results.
- In the model, the axis of the T-section passes through the center of gravity of the slab.
- Multiplying the corresponding forces by the accepted design width of the slab is a simplification, which leads to approximate results.
A feature of the center of gravity is that this force does not act on the body at any one point, but is distributed throughout the entire volume of the body. The forces of gravity that act on individual elements bodies (which can be considered material points) are directed towards the center of the Earth and are not strictly parallel. But since the sizes of most bodies on Earth are much smaller than its radius, therefore these forces are considered parallel.
Determining the center of gravity
Definition
The point through which the resultant of all parallel forces of gravity, affecting the elements of the body at any location of the body in space, passes is called center of gravity.
In other words: the center of gravity is the point to which the force of gravity is applied at any position of the body in space. If the position of the center of gravity is known, then we can assume that the force of gravity is one force, and it is applied at the center of gravity.
The task of finding the center of gravity is a significant task in technology, since the stability of all structures depends on the position of the center of gravity.
Method for finding the center of gravity of a body
Determining the position of the body's center of gravity complex shape You can first mentally break the body into parts of a simple shape and find the centers of gravity for them. For bodies of simple shape, the center of gravity can be immediately determined from considerations of symmetry. The force of gravity of a homogeneous disk and ball is at their center, of a homogeneous cylinder at a point in the middle of its axis; a homogeneous parallelepiped at the intersection of its diagonals, etc. For all homogeneous bodies, the center of gravity coincides with the center of symmetry. The center of gravity may be outside the body, such as a ring.
Let's find out the location of the centers of gravity of parts of the body, find the location of the center of gravity of the body as a whole. To do this, the body is represented as a set material points. Each such point is located at the center of gravity of its part of the body and has the mass of this part.
Center of gravity coordinates
In three-dimensional space, the coordinates of the point of application of the resultant of all parallel forces of gravity (coordinates of the center of gravity), for solid are calculated as:
\[\left\( \begin(array)(c) x_c=\frac(\sum\limits_i(\Delta m_ix_i))(m);; \\ y_c=\frac(\sum\limits_i(\Delta m_iy_i) )(m);; \\ z_c=\frac(\sum\limits_i(\Delta m_iz_i))(m) \end(array) \right.\left(1\right),\]
where $m$ is the mass of the body.$;;x_i$ is the coordinate on the X axis of the elementary mass $\Delta m_i$; $y_i$ - coordinate on the Y axis of the elementary mass $\Delta m_i$; ; $z_i$ is the coordinate on the Z axis of the elementary mass $\Delta m_i$.
In vector notation, a system of three equations (1) is written as:
\[(\overline(r))_c=\frac(1)(m)\sum\limits_i(m_i(\overline(r))_i\left(2\right),)\]
$(\overline(r))_c$ - radius - a vector that determines the position of the center of gravity; $(\overline(r))_i$ are radius vectors that determine the positions of elementary masses.
Center of gravity, center of mass and center of inertia of the body
Formula (2) coincides with the expressions that determine the center of mass of the body. If the dimensions of the body are small compared to the distance to the center of the Earth, the center of gravity is considered to coincide with the center of mass of the body. In most problems, the center of gravity coincides with the center of mass of the body.
The force of inertia in non-inertial reference systems moving translationally is applied to the center of gravity of the body.
But it should be taken into account that the centrifugal force of inertia (in the general case) is not applied to the center of gravity, since in a non-inertial reference frame different centrifugal forces of inertia act on the elements of the body (even if the masses of the elements are equal), since the distances to the axis of rotation are different.
Examples of problems with solutions
Example 1
Exercise. The system is made up of four small balls (Fig. 1). What are the coordinates of its center of gravity?
Solution. Let's look at Fig. 1. The center of gravity in this case will have one coordinate $x_c$, which we define as:
The body mass in our case is equal to:
The numerator of the fraction on the right side of expression (1.1) in case (1(a)) takes the form:
\[\sum\limits_(i=4)(\Delta m_ix_i=m\cdot 0+2m\cdot a+3m\cdot 2a+4m\cdot 3a=20m\cdot a).\]
We get:
Answer.$x_c=2a;$
Example 2
Exercise. The system is made up of four small balls (Fig. 2). What are the coordinates of its center of gravity?
Solution. Let's look at Fig. 2. The center of gravity of the system is on the plane, therefore, it has two coordinates ($x_c,y_c$). Let's find them using the formulas:
\[\left\( \begin(array)(c) x_c=\frac(\sum\limits_i(\Delta m_ix_i))(m);; \\ y_с=\frac(\sum\limits_i(\Delta m_iy_i) )(m).\end(array)\right.\]
System weight:
Let's find the coordinate $x_c$:
Coordinate $y_с$:
Answer.$x_c=0.5\ a$; $y_с=0.3\ a$
