**3.2 Loads applied at a distance "x" from one of the supports of the longitudinal beams**

The proposed method is applicable to calculate the structural response of a girder bridge deck to the application of a vertical load in any cross section. Using the Maxwell-Betti reciprocity theorem [27], for the application of a point load of value "Q" at a distance "x" from one of the two support points of a longitudinal beam, the deflection at the center of the span is obtained by applying the formulation reflected in Eq. (8). Likewise, the distribution of the maximum bending moment and maximum shear stress in the different longitudinal beams is obtained by Eqs. (9) and (10), respectively.

$$\mathbf{f}\_{\rm cl,Q(x),i} = \mathbf{f}\_{\rm v,i} \cdot \sin\left(\frac{\pi \cdot x}{L}\right) \tag{8}$$

$$\mathbf{M}\_{\text{ffnax,n}} = \frac{\mathbf{Q} \cdot (\mathbf{L} - \mathbf{x}) \cdot \mathbf{x}}{\mathbf{L}} \cdot \frac{\mathbf{f}\_{\text{v,n}}}{\sum\_{i=1}^{N} \mathbf{f}\_{\text{v,i}}} \quad \text{/} \quad \mathbf{x} \le \mathbf{L}/2 \tag{9}$$

$$\mathbf{Q}\_{\text{níax,n}} = \frac{\mathbf{Q} \cdot (\mathbf{L} - \mathbf{x})}{\mathbf{L}} \cdot \frac{\mathbf{f}\_{\text{v,n}}}{\sum\_{i=1}^{N} \mathbf{f}\_{\text{v,i}}} \quad / \quad \mathbf{x} \le \mathbf{L}/2 \tag{10}$$

Example 3.2.1.

In the study of the structural behavior of the bridge deck analyzed in Example 3.1.1., it is intended to know the deflection in the center of the span of the longitudinal beams, as well as the distribution of the maximum bending moment and the maximum shear force in each of the longitudinal beams generated by the application of the following load states: (a) a point load of 300 KN at a distance equivalent to L/3 from

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


#### **Figure 7.**

*Point load of 300 KN at a distance equivalent to L/3 from one of the supports of a beam.*

one of the supports of beam 1, and (b) a point load of 300 KN at a distance equivalent to L/3 from one of the supports of beam 2 (**Figure 7**).

Load state 1

Deflection in the center of the spam of the longitudinal beams

$$f\_{cl,Q(x),i} = f\_{v,i} \cdot \sin\left(\frac{\pi \cdot x}{L}\right) = \begin{pmatrix} -13.12 \, mm \\ -4.12 \, mm \\ 0.20 \, mm \\ 1,55 \, mm \end{pmatrix}$$

Maximum bending moments

$$M\_{\text{f\'\'ax},n} = \frac{Q \cdot (L-\mathbf{x}) \cdot \mathbf{x}}{L} \cdot \frac{f\_{v,n}}{\sum\_{i=1}^{N} f\_{v,i}} = \begin{pmatrix} 1,412 \text{ KN} \cdot m \\\\ 443 \text{ KN} \cdot m \\\\ -22 \text{ KN} \cdot m \\\\ -167 \text{ KN} \cdot m \end{pmatrix}.$$

Maximum shear forces

$$Q\_{\text{mix},n} = \frac{Q \cdot (L - \infty)}{L} \cdot \frac{f\_{v,n}}{\sum\_{i=1}^{N} f\_{v,i}} = \begin{pmatrix} 169 \text{ KN} \\ 53 \text{ KN} \\ -3 \text{ KN} \\ -20 \text{ KN} \end{pmatrix}.$$

Load state 1

Deflection in the center of the spam of the longitudinal beams

$$f\_{cl,Q(x),i} = f\_{v,i} \cdot \sin\left(\frac{\pi \cdot \pi}{L}\right) = \begin{pmatrix} -4.12 \, mm \\ -6.25 \, mm \\ -3.48 \, mm \\ 0.20 \, mm \end{pmatrix}.$$

Maximum bending moments

$$M\_{\text{f\'m\'ax},n} = \frac{Q \cdot (L - \pi) \cdot \pi}{L} \cdot \frac{f\_{v,n}}{\sum\_{i=1}^{N} f\_{v,i}} = \begin{pmatrix} 503 \text{ KN} \cdot m\\ 764 \text{ KN} \cdot m\\ 425 \text{ KN} \cdot m\\ -25 \text{ KN} \cdot m \end{pmatrix}\_{\text{f\'m\'a}}$$

Maximum shear forces

$$Q\_{\text{mix},n} = \frac{Q \cdot (L - \infty)}{L} \cdot \frac{f\_{v,n}}{\sum\_{i=1}^{N} f\_{v,i}} = \begin{pmatrix} 60 \text{ KN} \\ 92 \text{ KN} \\ 51 \text{ KN} \\ -3 \text{ KN} \end{pmatrix}\_{\text{max}}$$

#### **3.3 Loads applied at any point on the bridge deck**

The proposed method is applicable to calculate the transverse response of girder bridge decks to the application of a vertical load at any point of the cross section. As in any matrix calculation, if the vertical load acts on a section of slab between longitudinal beams, the degrees of freedom of the structural model are locked and the reactions in the locked degrees of freedom are calculated (rigid step). Subsequently, the degrees of freedom are released and loaded with the reactions obtained in the previous step to obtain the vertical displacement of each of the longitudinal beams that make up the bridge deck (flexible step).

Example 3.3.1.

In the study of the structural behavior of the bridge deck analyzed in Example 3.1.1., it is intended to know the deflection in the center of the span of the longitudinal beams, as well as the distribution of the maximum bending moment and the maximum shear force in each of the longitudinal beams generated by the application of the load states described in **Figure 8**.

Rigid step (**Figure 9**).

The load generated by the action of the truck axles on longitudinal beams 1 and 2 is obtained considering the compatibility between loads and movements in the upper slab between longitudinal beams 1 and 2.

$$\begin{pmatrix} V\_{LB1} \\ M\_{LB1} \\ -Q\_i \\ 0 \\ -Q\_i \\ 0 \\ \mathbf{0} \\ V\_{LB2} \\ \mathbf{M}\_{LB2} \end{pmatrix} = \begin{pmatrix} \mathbf{0} & \mathbf{0} & \mathbf{0} & \mathbf{0} \\ K\_{11,1} & K\_{12,1} & \mathbf{0} & \mathbf{0} & \mathbf{0} \\ K\_{21,1} & K\_{22,1} + K\_{11,2} & & \mathbf{0} & \mathbf{0} \\ & & & K\_{12,2} & & \mathbf{0} \\ & \mathbf{0} & & & \mathbf{0} & \mathbf{0} \\ & \mathbf{0} & & & & K\_{21,2} + K\_{11,3} & K\_{12,3} \\ \mathbf{0} & \mathbf{0} & \mathbf{0} & \mathbf{0} & & K\_{21,3} & K\_{22,3} \\ \mathbf{0} & \mathbf{0} & \mathbf{0} & \mathbf{0} & & & K\_{22,3} \\ \mathbf{0} & \mathbf{0} & \mathbf{0} & \mathbf{0} & & & \end{pmatrix} \begin{pmatrix} \mathbf{0} \\ \mathbf{0} \\ v2 \\ \Theta \\ v3 \\ \Theta \\ \mathbf{0} \\ \mathbf{0} \\ \end{pmatrix}$$

Load on longitudinal beams 1 and 2

$$Q = \begin{pmatrix} 1.46 \cdot Q \text{ KN} \\ 1.09 \cdot Q \text{ KN} \cdot m \\ 0.54 \cdot Q \text{ KN} \\ -0.70 \cdot Q \text{ KN} \cdot m \end{pmatrix}.$$

**46**

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

**Figure 9.** *Load state generated by the actuation of a four-axle truck. Rigid step.*

#### **Figure 10.**

*Load state generated by the actuation of a four-axle truck. Flexible step.*

#### Flexible step (**Figure 10**).


Truck axle 1

Deflection in the center of the spam of the longitudinal beams

$$f\_{d,Q(x),i} = f\_{v,i} \cdot \sin\left(\frac{\pi \cdot x}{L}\right) = \begin{pmatrix} -2.39 \, mm \\ -1.20 \, mm \\ -0.21 \, mm \\ 0.30 \, mm \end{pmatrix}.$$

Maximum bending moments

$$M\_{f\text{inax},n} = \frac{Q \cdot (L-\infty) \cdot \pi}{L} \cdot \frac{f\_{v,n}}{\sum\_{i=1}^{N} f\_{v,i}} = \begin{pmatrix} 268 \text{ KN} \cdot m\\ 135 \text{ KN} \cdot m\\ 24 \text{ KN} \cdot m\\ -34 \text{ KN} \cdot m \end{pmatrix}\_{\text{if}}$$

Maximum shear forces

$$Q\_{\text{wax,n}} = \frac{Q \cdot (L - x)}{L} \cdot \frac{f\_{v,n}}{\sum\_{i=1}^{N} f\_{v,i}} = \begin{pmatrix} 31 \text{ KN} \\ 16 \text{ KN} \\ 3 \text{ KN} \\ -4 \text{ KN} \end{pmatrix}.$$

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

#### Truck axle 2

Deflection in the center of the spam of the longitudinal beams

$$f\_{cl,Q(x),i} = f\_{v,i} \cdot \sin\left(\frac{\pi \cdot \pi}{L}\right) = \begin{pmatrix} -4.27 \, mm \\ -2.15 \, mm \\ -0.38 \, mm \\ 0.53 \, mm \end{pmatrix}.$$

Maximum bending moments

$$\mathbf{M}\_{f\text{\'n\'ax},n} = \frac{\mathbf{Q} \cdot (L-\mathbf{x}) \cdot \mathbf{x}}{L} \cdot \frac{f\_{\text{\'y},n}}{\sum\_{i=1}^{N} f\_{\text{\'y},i}} = \begin{pmatrix} 468 \text{ KN} \cdot m\\ 236 \text{ KN} \cdot m\\ 41 \text{ KN} \cdot m\\ -59 \text{ KN} \cdot m \end{pmatrix}.$$

Maximum shear forces

$$Q\_{\text{wfix},\text{n}} = \frac{Q \cdot (L - x)}{L} \cdot \frac{f\_{v,\text{n}}}{\sum\_{i=1}^{N} f\_{v,i}} = \begin{pmatrix} 39 \text{ KN} \\ 20 \text{ KN} \\ 3 \text{ KN} \\ -5 \text{ KN} \end{pmatrix}.$$

Truck axle 3

Deflection in the center of the spam of the longitudinal beams

$$f\_{cl,Q(x),i} = f\_{v,i} \cdot \sin\left(\frac{\pi \cdot x}{L}\right) = \begin{pmatrix} -3.69 \, mm \\ -1.86 \, mm \\ -0.33 \, mm \\ 0.46 \, mm \\ \end{pmatrix}.$$

Maximum bending moments

$$M\_{f\text{\textquotedblleft}ax,n} = \frac{Q \cdot (L-x) \cdot x}{L} \cdot \frac{f\_{v,n}}{\sum\_{i=1}^{N} f\_{v,i}} = \begin{pmatrix} 409 \text{ KN} \cdot m\\ 206 \text{ KN} \cdot m\\ 36 \text{ KN} \cdot m\\ -51 \text{ KN} \cdot m \end{pmatrix}$$

Maximum shear forces

$$\mathbf{Q}\_{\text{\textquotedblleft}\text{ax},\text{\textquotedblright}} = \frac{\mathbf{Q} \cdot (L - \mathbf{x})}{L} \cdot \frac{f\_{v,\text{\textquotedblleft}}}{\sum\_{i=1}^{N} f\_{v,i}} = \begin{pmatrix} 41 \text{ KN} \\ 21 \text{ KN} \\ 4 \text{ KN} \\ -5 \text{ KN} \end{pmatrix}.$$

#### Truck axle 4

Deflection in the center of the spam of the longitudinal beams

$$f\_{cl,Q(x),i} = f\_{v,i} \cdot \sin\left(\frac{\pi \cdot \pi}{L}\right) = \begin{pmatrix} -3.43 \, mm \\ -1.73 \, mm \\ -0.30 \, mm \\ 0.43 \, mm \end{pmatrix}.$$

Maximum bending moments

$$M\_{\text{f\'mux},n} = \frac{Q \cdot (L - \infty) \cdot \pi}{L} \cdot \frac{f\_{v,n}}{\sum\_{i=1}^{N} f\_{v,i}} = \begin{pmatrix} \text{385 KN} \cdot m\\\text{194 KN} \cdot m\\\text{34 KN} \cdot m\\\text{-48 KN} \cdot m \end{pmatrix}\_{\text{f\'mux}}$$

Maximum shear forces

$$Q\_{\text{mixx},n} = \frac{Q \cdot (L - x)}{L} \cdot \frac{f\_{v,n}}{\sum\_{i=1}^{N} f\_{v,i}} = \begin{pmatrix} 45 \text{ KN} \\ 22 \text{ KN} \\ 4 \text{ KN} \\ -6 \text{ KN} \end{pmatrix}.$$

Deflection and efforts generated by the action of the four-axle truck

$$f\_{v,i} = \begin{pmatrix} -13.78 \text{ mm} \\ -6.94 \text{ mm} \\ -1.22 \text{ mm} \\ 1.72 \text{ mm} \end{pmatrix}$$

$$M\_{f\text{mix},i} = \begin{pmatrix} 1,323 \text{ KN} \cdot m \\ 667 \text{ KN} \cdot m \\ 117 \text{ KN} \cdot m \\ -165 \text{ KN} \cdot m \end{pmatrix}$$

$$Q\_{\text{mix},j} = \begin{pmatrix} 138 \text{ KN} \\ 70 \text{ KN} \\ 12 \text{ KN} \\ -17 \text{ KN} \end{pmatrix}$$

## **4. Conclusions**

The proposed method for the analysis of the cross-sectional distribution of live loads on girder bridge decks allows determining the cross-sectional distribution in *Perspective Chapter: Simplified Matrix Calculation for Analysis of Girder-Deck Bridge… DOI: http://dx.doi.org/10.5772/intechopen.102362*

different configurations of girder bridge decks without the need to resort to complex calculation models that involve depth computing power and excessive analysis time. The method is also applicable for modern synthetic materials, such plastic composites, self-repair. The simplicity of the method allows an easy integration into optimal bridge design strategies [28] or more heuristic approaches [29–33] to challenge today's competitive world intelligently.
