**3. Proposed method for the study of the cross-sectional load distribution on a girder bridge deck**

The authors propose the use of a method that allows obtaining the cross-sectional distribution of live loads on girder bridge decks without using empirical formulas of LDFs or complex structural models involving the use of specific structure calculation programs. The proposed method is based on using a virtual model that reflects the transverse stiffness of the slab deck, supported on a series of springs that provide the flexural stiffness Eq. (1) and torsional stiffness Eq. (2) of the longitudinal beams that make up the girder bridge deck (**Figure 2**).

$$K\_{v,i} = \frac{48 \cdot EI\_i}{L^3} \tag{1}$$

$$K\_{t,i} = \frac{2 \cdot G l\_i}{L} \tag{2}$$

where EIi = longitudinal bending stiffness of beam "i"; GJi = longitudinal torsional stiffness of beam "i"; L = distance between bridge supports.

The proposed method considers 2 degrees of freedom for each longitudinal beam on the bridge deck: (1) the deflection and (2) the rotation of the deck slab at the center of the beam's span. **Figure 2** represents the structural model scheme for a girder

**Figure 2.** *Proposed method model for cross-sectional distribution on a girder bridge deck.*

bridge deck composed of three longitudinal beams. The matrix approach that solves the structural problem of the cross-sectional distribution of live loads between the different beams that make up the deck is raised in matrix Eqs. (3), (4), (5). The stiffness of the springs that represents the longitudinal bending and the torsional stiffness of the longitudinal beams that make up the model corresponds to the stiffness equivalent to the center of the span.

$$K\_{\epsilon} = \frac{EI\_{\epsilon}}{L\_{\epsilon}^{3}} \begin{pmatrix} 12 & 6L\_{\epsilon} & -12 & 6L\_{\epsilon} \\ & 6L & 4L\_{\epsilon}^{2} & -6L\_{\epsilon} & 2L\_{\epsilon}^{2} \\ -12 & -6L\_{\epsilon} & 12 & -6L\_{\epsilon} \\ & 6L\_{\epsilon} & 2L\_{\epsilon}^{2} & -6L\_{\epsilon} & 4L\_{\epsilon} \end{pmatrix} = \begin{pmatrix} K\_{11,\epsilon} & K\_{12,\epsilon} \\ K\_{21,\epsilon} & K\_{22,\epsilon} \end{pmatrix} \tag{3}$$

$$K\_{Bi} = \begin{pmatrix} k\_{v,i} & \mathbf{0} \\ \mathbf{0} & k\_{t,i} \end{pmatrix} \tag{4}$$

$$
\begin{pmatrix} P\_1 \\ M\_{t,1} \\ P\_2 \\ M\_{t,2} \\ P\_3 \\ M\_{t,3} \end{pmatrix} = \begin{pmatrix} K\_{11,1} + K\_{R1} & K\_{12,1} & \mathbf{0} & \mathbf{0} \\ K\_{21,1} & K\_{22,1} + K\_{11,2} + K\_{R2} & K\_{12,2} \\ \mathbf{0} & \mathbf{0} & & K\_{21,2} + K\_{B3} \\ \mathbf{0} & \mathbf{0} & & K\_{22,2} + K\_{B3} \\ \mathbf{0} & \mathbf{0} & & \mathbf{0} \end{pmatrix} \times \begin{pmatrix} f\_{v,1} \\ \theta\_1 \\ f\_{v,2} \\ \theta\_2 \\ f\_{v,3} \\ \theta\_3 \end{pmatrix} \tag{5}
$$

Where EIe = transverse bending stiffness of element "e" of the upper deck slab; Le = length of element "e" of the upper deck slab; fv,i = vertical displacement experienced by longitudinal beam "i"; Ɵ<sup>i</sup> = transverse rotation of the bridge deck over the longitudinal beam "i."

#### **3.1 Loads applied in the center of the span of the longitudinal beams**

The maximum longitudinal bending stress in each of the longitudinal beams that make up the bridge deck corresponds to the application of a point load in the center of the beam span. The distribution of the maximum bending moment and maximum shear stress in the different longitudinal beams is obtained by Eqs. (6) and (7), respectively.

$$M\_{\text{f\'inax},n} = \frac{Q \cdot L}{4} \cdot \frac{f\_{v,n}}{\sum\_{i=1}^{N} f\_{v,i}} \tag{6}$$

$$Q\_{\text{\#ax,n}} = \frac{Q \cdot L}{2} \cdot \frac{f\_{v,n}}{\sum\_{i=1}^{N} f\_{v,i}} \tag{7}$$

where Q = point load value; L = girder bridge span length.

Example 3.1.1.

In the study of the structural behavior of the bridge deck represented in **Figure 3**, it is desired to know the distribution of the bending moment and the shear force in each of the longitudinal beams generated by the application of the following load states: (a) a point load of 300 KN in the center of the span of beam 1, and (b) a point *Perspective Chapter: Simplified Matrix Calculation for Analysis of Girder-Deck Bridge… DOI: http://dx.doi.org/10.5772/intechopen.102362*

**Figure 3.**

*Girder bridge deck: (a) side view of load states 1 and 2; (b) front view of load state 1; (c) front view of load state 2.*

load of 300 KN in the center of the span of beam 2. The girder bridge is made up of four longitudinal beams 1.05 meters deep (two end beams and two central beams) and an upper slab 0.25 meters thick and 16.83 meters wide. Considering the effective width of the upper slab in each of the longitudinal beams, the inertia to the longitudinal bending of the end and central beams takes the value of 0.1445 m<sup>4</sup> and 0.1904 m<sup>4</sup> , respectively. The longitudinal torsional inertia of the longitudinal beams takes the value of 4.3�10�<sup>3</sup> <sup>m</sup><sup>4</sup> . The spacing between longitudinal beams is 5.13 meters, while the distance between the support devices of each longitudinal beam is 25 meters. Both the upper slab and the longitudinal beams are made of structural concrete whose modulus of elasticity reaches 35,000 MPa.

The stiffness of the springs on which the upper slab of the bridge deck rests, and which simulates the bending and torsional stiffness of the longitudinal beams (**Figure 4**), is calculated as follows:

End beams

$$K\_{v,end} = \frac{48 \cdot EI\_{end}}{L^3} = \frac{48 \cdot 3.5 \cdot 10^{7} K\%\_{m^2} \cdot 0.1445 m^4}{(25 m)^3} = 15,537 \text{KN}/m$$

$$K\_{t,end} = \frac{2 \cdot G f\_{end}}{L} = \frac{2 \cdot \frac{3.5 \cdot 10^{7} K\%\_{m^2}}{2 \cdot (1 + 0.2)} \cdot 4.3 \cdot 10^3 m^4}{25 m} = 5,017 \text{ KN} \cdot m$$

$$K\_{B,end} = \begin{pmatrix} k\_{v,end} & 0\\ 0 & k\_{t,end} \end{pmatrix} = \begin{pmatrix} 15,537 \text{ kN}/\_m & 0\\ 0 & 5,017 \text{ KN} \cdot m \end{pmatrix}$$

**Figure 4.** *Proposed structural model: (a) load state 1; (b) load state 2.*

Central beams

$$K\_{\nu,\varepsilon} = \frac{48 \cdot EI\_{end}}{L^3} = \frac{48 \cdot 3.5 \cdot 10^7 \text{KN} /\_{m^2} \cdot 0.1904 m^4}{\left(25 m\right)^3} = 20,472 \text{ KN} /\_{m}$$

$$K\_{t,\varepsilon} = \frac{2 \cdot GJ\_{end}}{L} = \frac{2 \cdot \frac{3 \cdot 5 \cdot 10^7 \text{KN} /\_{m^2}}{2 \cdot (1 + 0.2)} \cdot 4.3 \cdot 10^3 m^4}{25 m} = 5,017 \text{ KN} \cdot m$$

$$K\_{B,\varepsilon} = \begin{pmatrix} k\_{v,\text{end}} & 0\\ 0 & k\_{t,\text{end}} \end{pmatrix} = \begin{pmatrix} 20,472 \text{ kN} /\_{m} & 0\\ 0 & 5,017 \text{ KN} \cdot m \end{pmatrix}$$

The mechanical characteristics of the beam elements that simulate the transverse distribution provided by the upper slab (**Figure 4**) are calculated as follows:

$$I = \frac{1}{12} \cdot 20m \cdot \left(0.25m\right)^{3} = 2.6 \cdot 10^{-2}m^{4}$$

$$L = 5, 13 \, m$$

$$K = \frac{E \cdot I}{L^{3}} \begin{pmatrix} 12 & 6L & -12 & 6L \\ 6L & 4L^{2} & -6L & 2L^{2} \\ -12 & -6L & 12 & -6L \\ 6L & 2L^{2} & -6L & 4L^{2} \end{pmatrix}$$

*Perspective Chapter: Simplified Matrix Calculation for Analysis of Girder-Deck Bridge… DOI: http://dx.doi.org/10.5772/intechopen.102362*

$$K = \begin{pmatrix} 81,0200\text{K}\% \text{m} & 207,800\text{K}\% & -81,020\text{K}\% \text{m} & 207,800\text{K}\% \text{m} \\ 207,800\text{K}\% & 710,690\text{K}\% \cdot \text{m} & -207,800\text{K}\% & 355,340\text{K}\% \cdot \text{m} \\ -81,020\text{K}\% \text{m} & -207,800\text{K}\% & 81,020\text{K}\% \text{m} & -207,800\text{K}\% \\ 207,800\text{K}\% & 355,340\text{K}\% \cdot \text{m} & -207,800\text{K}\% & 710,690\text{K}\% \cdot \text{m} \end{pmatrix}.$$

$$K = \begin{pmatrix} K\_{11} & K\_{12} \\ K\_{21} & K\_{22} \end{pmatrix}$$

The global stiffness matrix used in the proposed method for the analysis of the transverse distribution of live loads in girder bridge decks is obtained as follows:

$$K\_G = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ K\_{11} + K\_{B,md} & K\_{12} & 0 & 0 & 0 & 0 \\ K\_{21} & K\_{22} + K\_{11} + K\_{B,\varepsilon} & & & 0 & 0 \\ & & & & & K\_{12} & & \\ & & & & & & & 0 & 0 \\ & & & & & & & 0 & 0 \\ & & & & & & & K\_{21} & \\ & & & & & & & K\_{22} + K\_{11} + K\_{B,\varepsilon} & & K\_{12} \\ & & & & & & & 0 & 0 & 0 \\ & & & & & & & & 0 & 0 \\ & & & & & & & & & \\ & & & & & & & & 0 & 0 \\ \end{pmatrix}$$

The transversal distribution of each load state is obtained by planting the compatibility between loads and displacements:

Load state 1

$$\begin{pmatrix} -300KN \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ K\_{11} + K\_{B,\text{rad}} & K\_{12} & 0 & 0 & 0 & 0 \\ K\_{21} & K\_{22} + K\_{11} + K\_{B,\text{c}} & 0 & 0 & 0 \\ K\_{21} & K\_{22} + K\_{11} + K\_{B,\text{c}} & 0 & 0 & 0 \\ 0 & 0 & & K\_{22} & + K\_{11} + K\_{B,\text{c}} & K\_{12} \\ 0 & 0 & & & K\_{22} + K\_{11} + K\_{B,\text{c}} & K\_{12} \\ 0 & 0 & 0 & 0 & & K\_{21} + K\_{B,\text{c}} \\ 0 & 0 & 0 & 0 & & \\ 0 & 0 & 0 & 0 & & \\ & \begin{pmatrix} f\_{v,1} \\ g\_{1} \\ f\_{v,2} \\ g\_{2} \\ g\_{3} \\ g\_{3} \\ g\_{4} \\ \end{pmatrix} \end{pmatrix}$$

**Figure 5.** *Deflections of longitudinal beams in load state 1.*

Deflections (**Figure 5**) and rotations

$$d = \begin{pmatrix} -15.14 \, mm \\\\ 2.25 \, mRad \\\\ -4.75 \, mm \\\\ 1.55 \, mRad \\\\ 0.24 \, mm \\\\ 0.52 \, mRad \\\\ 1.79 \, mm \\\\ 0.19 \, mRad \end{pmatrix}$$

Maximum bending moments

$$M\_{\text{f\'inax.}\text{\'}\text{\'}} = \frac{\text{Q} \cdot L}{\text{4}} \cdot \frac{f\_{v,n}}{\sum\_{i=1}^{N} f\_{v,i}} = \begin{pmatrix} 1,588 \text{ KN} \cdot m \\\\ 499 \text{ KN} \cdot m \\\\ -25 \text{ KN} \cdot m \\\\ -187 \text{ KN} \cdot m \end{pmatrix}.$$

Maximum shear forces

$$\mathbf{Q}\_{\text{mix},n} = \frac{\mathbf{Q} \cdot \mathbf{L}}{2} \cdot \frac{\mathbf{f}\_{v,n}}{\sum\_{i=1}^{N} \mathbf{f}\_{v,i}} = \begin{pmatrix} 127 \text{ KN} \\\\ 40 \text{ KN} \\\\ -2 \text{ KN} \\\\ -15 \text{ KN} \end{pmatrix}.$$

*Perspective Chapter: Simplified Matrix Calculation for Analysis of Girder-Deck Bridge… DOI: http://dx.doi.org/10.5772/intechopen.102362*

Load state 2

0 0 �300*KN* 0 0 0 0 0 0 BBBBBBBBBBBBBBBBB@ 1 CCCCCCCCCCCCCCCCCA ¼ *K*<sup>11</sup> þ *KB*,*end K*<sup>12</sup> *K*<sup>21</sup> *K*<sup>22</sup> þ *K*<sup>11</sup> þ *KB*,*<sup>c</sup>* 0 0 0 0 0 0 0 0 *K*<sup>12</sup> 0 0 0 0 0 0 0 0 *K*<sup>21</sup> 0 0 0 0 0 0 0 0 *K*<sup>22</sup> þ *K*<sup>11</sup> þ *KB*,*<sup>c</sup> K*<sup>12</sup> *K*<sup>21</sup> *K*<sup>22</sup> þ *KB*,*end* 0 BBBBBBBBBBBBBBBBBBBBB@ 1 CCCCCCCCCCCCCCCCCCCCCA �*x f <sup>v</sup>*,1 *θ*1 *f <sup>v</sup>*,2 *θ*2 *f <sup>v</sup>*,3 *θ*3 *f <sup>v</sup>*,4 *θ*4 0 BBBBBBBBBBBBBBBBB@ 1 CCCCCCCCCCCCCCCCCA

Deflections (**Figure 6**) and rotations

$$d = \begin{pmatrix} -4.75 \, mm \\\\ -0.76 \, m \text{Rad} \\\\ -7.21 \, mm \\\\ 0.08 \, m \text{Rad} \\\\ -4.01 \, mm \\\\ 0.87 \, m \text{Rad} \\\\ 0.24 \, mm \\\\ 0.80 \, m \text{Rad} \end{pmatrix}$$

Maximum bending moments

$$\mathbf{M}\_{\text{f\'u\'ax},n} = \frac{\mathbf{Q} \cdot \mathbf{L}}{\mathbf{4}} \cdot \frac{\mathbf{f}\_{v,n}}{\sum\_{i=1}^{N} \mathbf{f}\_{v,i}} = \begin{pmatrix} \mathbf{566 } \text{KN} \cdot m \\ \mathbf{859 } \text{KN} \cdot m \\ \mathbf{478 } \text{KN} \cdot m \\ -\mathbf{28 } \text{KN} \cdot m \end{pmatrix}.$$

**Figure 6.** *Deflections of longitudinal beams in load state 2.*

Maximum shear forces

$$Q\_{\text{mix},n} = \frac{Q \cdot L}{2} \cdot \frac{f\_{v,n}}{\sum\_{i=1}^{N} f\_{v,i}} = \begin{pmatrix} 45 \text{ KN} \\ 69 \text{ KN} \\ 38 \text{ KN} \\ -2 \text{ KN} \end{pmatrix}.$$
