1. Introduction

2. Materials and Methods

2.2. Methods

Some beams with intermediate support can be studied as statically indeterminate continuous beams. It is considered as a statically indeterminate continuous beam, articulate at the left edge, with simple support at the right edge and with intermediate supports, as in Figure 1. The calculus is done with TMM.

2.2.1. Work Hypotheses

The beam with intermediate support that will be studied is considered as a statically indeterminate continuous beam, articulate at the left edge, with a simple support at the right edge and with i (i=1,n-1) intermediate simple supports (Figure 1).

The lengths of all parts of the beam are known.

The constant inertia of the entire beam is also considered.

It is considered a reference system with the origin in the left edge, the edge 0.

External vertical forces are considered to act on the beam, which can be characterized by a charge density q’(x) for the vertical concentrated loads and q”(x) is the charge density for the force uniformly distributed over the entire beam.

Due to the external forces, unknown vertical reactions arise at each edge, that are (n+1) vertical reactions. Further, these reactions will be considered as external forces with the charge density as q’”(x) .

2.2.2. Total Charge of Density for a Continuous Beam

It is considered as a statically indeterminate continuous beam, articulate at the left edge, with a simple support at the right edge and with n intermediate supports (Figure 1).

Thus, the total charge density q(x) can be written as (1):

$$q\left(x\right)=q\text{'}\left(x\right)+q"\left(x\right)+q\text{'}"\left(x\right),$$

(1)

where:

• q(x) is the total charge density;
• q’(x) is the charge density for the exterior vertical concentrated loads;
• q”(x) is the charge density for the force uniformly distributed over the entire beam;
• q’”(x) is the density of charge corresponding of the vertical reaction in the edges.
• The density of charge corresponding of the exterior concentrated loads q’(x), for a number j of exterior loads F_j, j=1, k, acting in the point a_j, j=1, k, can be written, with Dirac’s and Heaviside’s functions and operators, as (2):

$$q^{\text{'}}\left(x\right)=-\sum _{j=1}^k{F}_j\delta \left(x-a_j\right).$$

(2)

• The charge density for the force uniformly distributed over the entire beam q”(x) can be written, with Dirac’s and Heaviside’s functions and operators, as (3):

$$q"\left(x\right)=-p\left(x\right)\delta \left(x\right),$$

(3)

• and the density of charge corresponding for the Vertical Reaction, (VR_i, i=0, n) in the edges, q’”(x), with Dirac’s and Heaviside’s functions and operators too, can be written as (4):

$$q"\left(x\right)=\sum _{i=0}^n{VR}_i\delta \left(x-l_i\right),$$

(4)

when:

• j refers to the point where act the concentrated vertical force F_j;
• F_j, j=1, k, are the exterior vertical concentrated loads;
• k is the total number of the concentrated vertical force;
• i refers to the edge with number i;
• n+1 is the total number of edges;
• p(x) is the force uniformly distributed over the entire beam;
• VR_i is the Vertical Reaction in the edge i.

Applying the mathematical calculus with Dirac’s and Heaviside’s functions and operators for (2), we can obtain successively the relations (5):

$$\left\{\begin{array}{c}{q}_1^{\text{'}}\left(x\right)=-{\sum }_{j=1}^k{F}_{j}Y\left(x-a_j\right)\\ q_2^{\text{'}}\left(x\right)=-{\sum }_{j=1}^k{F}_j\left(x-a_j\right)Y\left(x-a_j\right)\\ q_3^{\text{'}}\left(x\right)=-{\sum }_{j=1}^k{F}_j{\displaystyle \frac{{\left(x-a_j\right)}^2}{2}}Y\left(x-a_j\right)\\ q_4^{\text{'}}\left(x\right)=-{\sum }_{j=1}^k{F}_j{\displaystyle \frac{{\left(x-a_j\right)}^3}{6}}Y\left(x-a_j\right)\end{array}\right.,$$

(5)

In the same way, the mathematical formalism will be applied to expressions (1) and relations (6) are obtained:

$$\left\{\begin{array}{c}{q}_1\left(x\right)=-{\displaystyle \sum _{j=1}^k}{F}_{j}Y\left(x-a_j\right)-p\left(x\right)xY\left(x\right)+{\displaystyle \sum _{i=1}^n}{VR}_{i}Y\left(x-l_i\right)\\ q_2\left(x\right)=-{\displaystyle \sum _{j=1}^k}{F}_j\left(x-a_j\right)Y\left(x-a_j\right)-p\left(x\right){\displaystyle \frac{x^2}{2}}Y\left(x\right)+{\displaystyle \sum _{i=1}^n}{VR}_i\left(x-l_i\right)Y\left(x-l_i\right)\\ q_3\left(x\right)=-{\displaystyle \sum _{j=1}^k}{F}_j{\displaystyle \frac{{\left(x-a_j\right)}^2}{2}}Y\left(x-a_j\right)-p\left(x\right){\displaystyle \frac{x^3}{6}}Y\left(x\right)+{\displaystyle \sum _{i=1}^n}{VR}_i{\displaystyle \frac{{\left(x-l_j\right)}^2}{2}}Y\left(x-l_i\right)\\ q_4\left(x\right)=-{\displaystyle \sum _{j=1}^k}{F}_j{\displaystyle \frac{{\left(x-a_j\right)}^3}{6}}Y\left(x-a_j\right)-p\left(x\right){\displaystyle \frac{x^4}{24}}Y\left(x\right)+{\displaystyle \sum _{i=1}^n}{VR}_i{\displaystyle \frac{{\left(x-l_j\right)}^3}{6}}Y\left(x-l_i\right)\end{array}\right.,$$

(6)

2.2.3. The State Vectors for Different Sections of the Continuous Beam

The state vectors for different sections of the continuous beam can be defined as follows.

It is noted with {SV(x)} the State Vector corresponding at the section x, at a point on the x-axis, the abscissa measured from the origin of the reference system, which is in the left edge 0, as (7):

$$\left\{SV\left(x\right)\right\}={\left\{y\left(x\right), \omega \left(x\right), M\left(x\right),T\left(x\right)\right\}}^{-1},$$

(7)

Elements of the state vector (7) are:

• y(x) is the arrow in the section x;
• ω(x) is the rotating in the section x;
• M(x) is the bending moment in the section x;
• T(x) is the cutting force in section x.

The state vector at the origin {SV(0)}=${\left\{SV\right\}}_0$, in the section 0 is as (8):

$$\left\{SV\left(0\right)\right\}={\left\{y\left(0\right), \omega \left(0\right), M\left(0\right),T\left(0\right)\right\}}^{-1}={\left\{SV\right\}}_0={\left\{y_0,{\omega }_0,M_0,T_0\right\}}^{-1},$$

(8)

The total length l of continuous beam is as (9):

$$l=\sum _{i=1}^n{a}_i,$$

(9)

when a_i, i=1, n is the distance between two consecutive supports.

The state vector at the last section, in the section n, {SV(l)} =${\left\{SV\right\}}_l,$for x=l, is as (10):

$$\left\{SV\left(l\right)\right\}={\left\{y\left(l\right), \omega \left(l\right), M\left(l\right),T\left(l\right)\right\}}^{-1}={\left\{SV\right\}}_l={\left\{y_l,{\omega }_l,M_l,T_l\right\}}^{-1},$$

(10)

2.2.4. The Transfer-Matrix for the Section x of the Continuous Beam

The connection between the state vector from the origin, the section 0, and the state vector from some section x, is made with relation (11):

$$\left\{\begin{array}{c}y\left(x\right)\\ \omega \left(x\right)\\ M\left(x\right)\\ T\left(x\right)\end{array}\right\}={\left[TM\right]}_x\left\{\begin{array}{c}y\left(0\right)\\ \omega \left(0\right)\\ M\left(0\right)\\ T\left(0\right)\end{array}\right\}+\left\{VEF\left(x\right)\right\},$$

(11)

when:

• [TM]_x is the Transfer-Matrix corresponding at section x;
• {VEF(x)} is the Vector for Exterior Forces at section x.

The Transfer-Matrix [TM]_x is as (12), after [1]:

$${\left[TM\right]}_x=\left[\begin{array}{cccc}1& x& {\displaystyle \frac{x^2}{2EI}}& -{\displaystyle \frac{x^3}{6EI}}\\ 0& 1& {\displaystyle \frac{x}{EI}}& -{\displaystyle \frac{x^2}{2EI}}\\ 0& 0& 1& -x\\ 0& 0& 0& 1\end{array}\right],$$

(12)

The vector for exterior forces at the section x is as (13):

$$\left\{VEF\left(x\right)\right\}=\left\{\begin{array}{c}-{\displaystyle \sum _{j=1}^k}{F}_j{\displaystyle \frac{{\left(x-a_j\right)}^3}{6}}Y\left(x-a_j\right)-p\left(x\right){\displaystyle \frac{x^4}{24}}Y\left(x\right)+{\displaystyle \sum _{i=1}^n}{VR}_i{\displaystyle \frac{{\left(x-l_j\right)}^3}{6}}Y\left(x-l_i\right)\\ -{\displaystyle \sum _{j=1}^k}{F}_j{\displaystyle \frac{{\left(x-a_j\right)}^2}{2}}Y\left(x-a_j\right)-p\left(x\right){\displaystyle \frac{x^3}{6}}Y\left(x\right)+{\displaystyle \sum _{i=1}^n}{VR}_i{\displaystyle \frac{{\left(x-l_j\right)}^2}{2}}Y\left(x-l_i\right)\\ -{\displaystyle \sum _{j=1}^k}{F}_j\left(x-a_j\right)Y\left(x-a_j\right)-p\left(x\right){\displaystyle \frac{x^2}{2}}Y\left(x\right)+{\displaystyle \sum _{i=1}^n}{VR}_i\left(x-l_i\right)Y\left(x-l_i\right)\\ -{\displaystyle \sum _{j=1}^k}{F}_{j}Y\left(x-a_j\right)-p\left(x\right)xY\left(x\right)+{\displaystyle \sum _{i=1}^n}{VR}_{i}Y\left(x-l_i\right)\end{array}\right\},$$

(13)

The matrix relation (11) can be developed as (14):

$$\begin{gathered} \left\{\begin{array}{c}y\left(x\right)\\ \omega \left(x\right)\\ M\left(x\right)\\ T\left(x\right)\end{array}\right\}=\left[\begin{array}{cccc}1& x& {\displaystyle \frac{x^2}{2EI}}& -{\displaystyle \frac{x^3}{6EI}}\\ 0& 1& {\displaystyle \frac{x}{EI}}& -{\displaystyle \frac{x^2}{2EI}}\\ 0& 0& 1& -x\\ 0& 0& 0& 1\end{array}\right]\left\{\begin{array}{c}y\left(0\right)\\ \omega \left(0\right)\\ M\left(0\right)\\ T\left(0\right)\end{array}\right\}+ \\ +\left\{\begin{array}{c}-{\displaystyle \sum _{j=1}^k}{F}_j{\displaystyle \frac{{\left(x-a_j\right)}^3}{6}}Y\left(x-a_j\right)-p\left(x\right){\displaystyle \frac{x^4}{24}}Y\left(x\right)+{\displaystyle \sum _{i=1}^n}{VR}_i{\displaystyle \frac{{\left(x-l_j\right)}^3}{6}}Y\left(x-l_i\right)\\ -{\displaystyle \sum _{j=1}^k}{F}_j{\displaystyle \frac{{\left(x-a_j\right)}^2}{2}}Y\left(x-a_j\right)-p\left(x\right){\displaystyle \frac{x^3}{6}}Y\left(x\right)+{\displaystyle \sum _{i=1}^n}{VR}_i{\displaystyle \frac{{\left(x-l_j\right)}^2}{2}}Y\left(x-l_i\right)\\ -{\displaystyle \sum _{j=1}^k}{F}_j\left(x-a_j\right)Y\left(x-a_j\right)-p\left(x\right){\displaystyle \frac{x^2}{2}}Y\left(x\right)+{\displaystyle \sum _{i=1}^n}{VR}_i\left(x-l_i\right)Y\left(x-l_i\right)\\ -{\displaystyle \sum _{j=1}^k}{F}_{j}Y\left(x-a_j\right)-p\left(x\right)xY\left(x\right)+{\displaystyle \sum _{i=1}^n}{VR}_{i}Y\left(x-l_i\right)\end{array}\right\}, \end{gathered}$$

(14)

For the last section, for x=l, the expression (12) can be written as (15):

$$\left\{\begin{array}{c}y\left(l\right)\\ \omega \left(l\right)\\ M\left(l\right)\\ T\left(l\right)\end{array}\right\}={\left[TM\right]}_l\left\{\begin{array}{c}y\left(0\right)\\ \omega \left(0\right)\\ M\left(0\right)\\ T\left(0\right)\end{array}\right\}+\left\{VEF\left(l\right)\right\},$$

(15)

The relation (15) with the developed Transfer-Matrix can be written as (16):

$$\begin{gathered} \left\{\begin{array}{c}y\left(l\right)\\ \omega \left(l\right)\\ M\left(l\right)\\ T\left(l\right)\end{array}\right\}=\left[\begin{array}{cccc}1& l& {\displaystyle \frac{l^2}{2EI}}& -{\displaystyle \frac{l^3}{6EI}}\\ 0& 1& {\displaystyle \frac{l}{EI}}& -{\displaystyle \frac{l^2}{2EI}}\\ 0& 0& 1& -l\\ 0& 0& 0& 1\end{array}\right]\left\{\begin{array}{c}y\left(0\right)\\ \omega \left(0\right)\\ M\left(0\right)\\ T\left(0\right)\end{array}\right\}+ \\ +\left\{\begin{array}{c}-{\displaystyle \sum _{j=1}^k}{F}_j{\displaystyle \frac{{\left(l-a_j\right)}^3}{6}}Y\left(l-a_j\right)-p\left(x\right){\displaystyle \frac{l^4}{24}}+{\displaystyle \sum _{i=1}^n}{VR}_i{\displaystyle \frac{{\left(l-l_j\right)}^3}{6}}Y\left(l-l_i\right)\\ -{\displaystyle \sum _{j=1}^k}{F}_j{\displaystyle \frac{{\left(l-a_j\right)}^2}{2}}Y\left(l-a_j\right)-p\left(x\right){\displaystyle \frac{l^3}{6}}+{\displaystyle \sum _{i=1}^n}{VR}_i{\displaystyle \frac{{\left(l-l_j\right)}^2}{2}}Y\left(l-l_i\right)\\ -{\displaystyle \sum _{j=1}^k}{F}_j\left(l-a_j\right)Y\left(l-a_j\right)-p\left(x\right){\displaystyle \frac{l^2}{2}}+{\displaystyle \sum _{i=1}^n}{VR}_i\left(l-l_i\right)Y\left(l-l_i\right)\\ -{\displaystyle \sum _{j=1}^k}{F}_{j}Y\left(l-a_j\right)-p\left(x\right)l+{\displaystyle \sum _{i=1}^n}{VR}_{i}Y\left(l-l_i\right)\end{array}\right\}, \end{gathered}$$

(16)

In matrix relation (16), the conditions on the extreme edges can be placed, in section 0, as (17):

$$\left\{\begin{array}{c}{y}_0=0\\ M_0=0\end{array}\right. ,$$

(17)

and in section n for x=l, conditions are as (18):

$$\left\{\begin{array}{c}y\left(l\right)=0\\ M\left(l\right)=0\end{array}\right. ,$$

(18)

With the conditions (17) and (18), the matrix relation (16) can be written as (19):

$$\begin{gathered} \left\{\begin{array}{c}0\\ \omega \left(l\right)\\ 0\\ T\left(l\right)\end{array}\right\}=\left[\begin{array}{cccc}1& l& {\displaystyle \frac{l^2}{2EI}}& -{\displaystyle \frac{l^3}{6EI}}\\ 0& 1& {\displaystyle \frac{l}{EI}}& -{\displaystyle \frac{l^2}{2EI}}\\ 0& 0& 1& -l\\ 0& 0& 0& 1\end{array}\right]\left\{\begin{array}{c}0\\ {\omega }_0\\ 0\\ T_0\end{array}\right\}+ \\ +\left\{\begin{array}{c}-{\displaystyle \sum _{j=1}^k}{F}_j{\displaystyle \frac{{\left(l-a_j\right)}^3}{6}}Y\left(l-a_j\right)-p\left(x\right){\displaystyle \frac{l^4}{24}}+{\displaystyle \sum _{i=1}^n}{VR}_i{\displaystyle \frac{{\left(l-l_j\right)}^3}{6}}Y\left(l-l_i\right)\\ -{\displaystyle \sum _{j=1}^k}{F}_j{\displaystyle \frac{{\left(l-a_j\right)}^2}{2}}Y\left(l-a_j\right)-p\left(x\right){\displaystyle \frac{l^3}{6}}+{\displaystyle \sum _{i=1}^n}{VR}_i{\displaystyle \frac{{\left(l-l_j\right)}^2}{2}}Y\left(l-l_i\right)\\ -{\displaystyle \sum _{j=1}^k}{F}_j\left(l-a_j\right)Y\left(l-a_j\right)-p\left(x\right){\displaystyle \frac{l^2}{2}}+{\displaystyle \sum _{i=1}^n}{VR}_i\left(l-l_i\right)Y\left(l-l_i\right)\\ -{\displaystyle \sum _{j=1}^k}{F}_{j}Y\left(l-a_j\right)-p\left(x\right)l+{\displaystyle \sum _{i=1}^n}{VR}_{i}Y\left(l-l_i\right)\end{array}\right\}, \end{gathered}$$

(19)

Also, the conditions for the arrows in each intermediate supports must be set, they must be null in each intermediate supports as (20):

$$y_i=0, \mathrm{f}\mathrm{o}\mathrm{r} i=1,n-1 ,$$

(20)

or, (21):

The conditions for the right end of the beam, for x=l, are (18) and give (22):

The relations (21) and (22) can be written in a matrix relation like (23), considering that ω_0, ${\displaystyle \frac{T_0}{6EI }}$ and $\sum _{i=1}^{n-1}{\displaystyle \frac{{VR}_i}{EI}}$ are the unknowns:

$$\left[\begin{array}{ccccccc}{l}_1& -l_1^3& 0& 0& 0& \cdots & 0\\ l_2& -l_2^3& ({l_2-l_1)}^3& 0& 0& \cdots & 0\\ l_3& -l_3^3& ({l_3-l_1)}^3& ({l_3-l_2)}^3& 0& \cdots & 0\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ l_i& -l_i^3& ({l_i-l_1)}^3& ({l_i-l_2)}^3& ({l_i-l_3)}^3& \cdots & 0\\ ⋮& ⋮& ⋮& ⋮& ⋮& ⋮& ⋮\\ l& -l^3& ({l-l_1)}^3& ({l-l_2)}^3& ({l-l_3)}^3& \cdots & ({l-l_{n-1})}^3\\ 0& -l& l-l_1& l-l_2& l-l_3& \cdots & l-l_{n-1}\end{array}\right]\left\{\begin{array}{c}{\omega }_0\\ {\displaystyle \frac{T_0}{6EI}}\\ \begin{array}{c}{\displaystyle \frac{{VR}_1}{6EI}}\\ ⋮\\ \begin{array}{c}⋮\\ {\displaystyle \frac{{VR}_{n-2}}{6EI}}\\ {\displaystyle \frac{{VR}_{n-1}}{6EI}}\end{array}\end{array}\end{array}\right\}\left\{\begin{array}{c}-{\displaystyle \frac{q_4^{\text{'}}\left(l_1\right)}{6EI}}-{\displaystyle \frac{q_4^{"}\left(l_1\right)}{24EI}}\\ -{\displaystyle \frac{q_4^{\text{'}}\left(l_2-l_1\right)}{6EI}}-{\displaystyle \frac{q_4^{"}\left(l_2-l_1\right)}{24EI}}\\ \begin{array}{c}⋮\\ -\sum _{i=1}^{n-1}{\displaystyle \frac{q_4^{\text{'}}\left({l( seq EquationSeq \backslash * \backslash Arabic )}_i-l_1\right)}{6EI}}-\sum _{i=1}^{n-1}{\displaystyle \frac{q_4^{"}\left(l_i-l_1\right)}{24EI}}\\ \begin{array}{c}⋮\\ -{\displaystyle \frac{q_4^{\text{'}}\left(l\right)}{6EI}}-{\displaystyle \frac{q_4^{"}\left(l\right)}{24EI}}\\ {-q}_2^{\text{'}}\left(l\right)-q_2^{"}\left(l\right)\begin{array}{c}\\ \end{array}\end{array}\end{array}\end{array}\right\}$$

(23)

Solving the system of equations given by the matrix relation (23) leads to the calculus of all unknowns. It can be also programmed very easily, thus quickly obtaining the values ​​of the unknowns, and then, all the efforts and deformations in any section of the beam can be calculated.

3. Application and Results

4. Discussion

5. Conclusions

References

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