In the original project of the viaduct [20], the torque moment diagram in Figure 8 was obtained using Equation (1), which provides a strong approximation of the real torsional structural behavior of the viaduct under study: the bending rotation at the beam ends is released, providing the maximum torsional stresses at the supports, according to ref. [21] and also considering that the bending moment is the first derivative of the torque ( d T / d ϕ = M ):

T ( ϕ ) = q ⋅ R 2 ⋅ [ cos ( ϕ ) sen ( ϕ 0 ) − ϕ ] (1)

whereas the shear diagram is given by Equation (2):

V ( ϕ ) = q ⋅ R ⋅ ϕ (2)

where R is the radius of curvature; ϕ is the angle defined by a generic point with respect to the mid-span of the beam; q is the uniformly distributed load; T ( ϕ ) is torque moment and V ( ϕ ) is the shear force.

In the original design, the bending moment was calculated by considering the bridge deck as a curved continuous beam on four supports (see Figure 9): this design choice turns out to be inconsistent with the computation of the torque moment diagram, but more adherent to reality with regard to the distribution of the bending moment. Based on the Principle of Virtual Works [20] [22] [23] [24] [25] [26] and starting from the straight continuous beam on n + 2supports, the main isostatic structure must be identified by eliminating n constraints of the real structure [27] [28] [29] [30]: the bending rotation of the internal supports can be released and, consequently, the n hyperstatic unknowns are the moments of continuity at the internal supports.

In Figure 10 N is the number of spans; n=N – 1 represents the number of flexural hyperstatic unknowns; L_i is the span length and q is the uniformly distributed load applied to the single span. In the case of structural system constrained with rigid supports, the resolving system is made by n-linear algebraic equations of congruence t [20] which allow to derive the response of the structure in terms of internal stress diagrams.

[ ∫ S M ( 1 ) 2 E I d s ∫ S M ( 1 ) M ( 2 ) E I d s 0 ∫ S M ( 2 ) M ( 1 ) E I d s ∫ S M ( 2 ) 2 E I d s ⋱ ⋱ ⋱ ∫ S M ( n − 1 ) M ( n ) E I d s 0 ∫ S M ( n ) M ( n − 1 ) E I d s ∫ S M ( n ) 2 E I d s ] [ X 1 , 1 ⋯ X 1 , n X 2 , 1 ⋯ X 2 , n ⋮ ⋱ ⋮ X n , 1 ⋯ X n , n ] = [ P 1 , 1 ⋯ P 1 , n P 2 , 1 ⋯ P 2 , n ⋮ ⋱ ⋮ P n , 1 ⋯ P n , n ] (3)

[ E ] [ X ] = [ P ] → [ X ] = [ E ] − 1 [ P ]

where: [ E ] is the flexibility matrix made of influence coefficients of dimension n∙n (in the case of continuous rectilinear HCB the matrix [ E ] is tri-diagonal and symmetrical); [ X ] is matrix of the hyperstatic unknowns, of dimension n∙n and [ P ] is the matrix of the load conditions, of dimension n∙n:

P i , j = ∫ s M ( i ) ⋅ M ( 0 ) d s = 1 24 ⋅ ( q i , j ⋅ L i 3 I i + q i + 1 , j ⋅ L i + 1 3 I i + 1 ) (4)

By changing the terms of Equation (5) for the straight beam the terms of flexibility matrix are calculated from the following formulations:

· main diagonal ( i = 1 , 2 , ⋯ , n ):

E i , i = R i 4 ⋅ E i ⋅ I i { ϕ 0 , i [ cos ( ϕ 0 , i ) ⋅ sen ( ϕ 0 , i ) ] 2 − cos 2 ( ϕ 0 , i ) − sen 2 ( ϕ 0 , i ) cos ( ϕ 0 , i ) ⋅ sen ( ϕ 0 , i ) }   + R i + 1 4 ⋅ E i + 1 ⋅ I i + 1 { ϕ 0 , i + 1 [ cos ( ϕ 0 , i + 1 ) ⋅ sen ( ϕ 0 , i + 1 ) ] 2 − cos 2 ( ϕ 0 , i + 1 ) − sen 2 ( ϕ 0 , i + 1 ) cos ( ϕ 0 , i + 1 ) ⋅ sen ( ϕ 0 , i + 1 ) } (5)

· lower diagonal ( i = 2 , 3 , ⋯ , n ):

E i , i − 1 = R i 4 ⋅ E i ⋅ I i { 1 cos ( ϕ 0 , i + 1 ) ⋅ sen ( ϕ 0 , i + 1 ) − ϕ 0 , i + 1 sen 2 ( ϕ 0 , i + 1 ) + ϕ 0 , i + 1 cos 2 ( ϕ 0 , i + 1 ) } (6)

· upper diagonal ( i = 1 , 2 , ⋯ , n − 1 ):

E i , i + 1 = R i + 1 4 ⋅ E i + 1 ⋅ I i + 1 { 1 cos ( ϕ 0 , i + 1 ) ⋅ sen ( ϕ 0 , i + 1 ) − ϕ 0 , i + 1 sen 2 ( ϕ 0 , i + 1 ) + ϕ 0 , i + 1 cos 2 ( ϕ 0 , i + 1 ) } (7)

By adding the contributions of the hyperstatic unknowns and of the applied loads, in Equations (8) and (9), the overall one bending moment is M i t o t = M i X + M i K :

M i X ( ϕ i ) = X i + X i + 1 2 ⋅ cos ( ϕ 0 , i ) ⋅ cos ( ϕ 0 , i ) + X i + 1 − X i 2 ⋅ sen ( ϕ 0 , i ) ⋅ sen ( ϕ 0 , i ) (8)

M i K ( ϕ i ) = k i , j ( cos ( ϕ i ) cos ( ϕ 0 , i ) − 1 ) (9)

Following the discussion in refs. [20] and [21], on the other hand the shear ad torque diagrams follow that of the isostatic HCB as reported in Equations ((1), (2)).

The span that best lends itself to the validation of the FEM results is the central span of 120 m, between piers P10 and P11, as the piers provide a bending constraint. The maximum dimensions for the box-girder section taken as reference are deducible from Table 4 and Figure 12. From the theoretical treatment reported in ref. [32], the bending moment M_B and the torque moment T_B in correspondence with the piers axis are obtained (see Table 5) using Castigliano’s Theorem [22].

Using the values in Table 5, it is then possible to define the bending moment and torque diagrams using Equations (10) to (13) as the angle θ subtended by the generic point of the deck axis varies.

M ( θ ) = q ⋅ R 2 ⋅ [ sen ( θ ) ⋅ ( c t − θ 2 ) + cos ( θ ) ⋅ ( c m − 1 ) + 1 ] (10)

T ( θ ) = q ⋅ R 2 ⋅ [ sen ( θ ) ⋅ ( θ − c m ) − cos ( θ ) ⋅ ( c t − θ 2 ) − θ 2 ] (11)

where: θ is the angle defined by a generic point on the axis of the beam, from the node j; m is the ratio between flexural and pure torsional rigidities of the main girder as defined in Equation (12):

m = E ⋅ I G ⋅ J = 2 ( 1 + υ ) ⋅ I J (12)

E is the Young Modulus; I is the moment of inertia of the cross section; G is the shear modulus, J is the torsional stiffness factor, υ is the Poisson’s ratio, and c m and c t are the flexural and torsional constants as defined in Equation (13).

c m = 2 ( m + 1 ) ⋅ sen ( ϕ ) − m ⋅ ϕ ⋅ [ 1 − cos ( ϕ ) ] ϕ ⋅ ( m + 1 ) − sen ( ϕ ) ⋅ ( m + 1 ) − 1 c t = 2 ( m + 1 ) ⋅ [ 1 − cos ( ϕ ) ] − m ⋅ ϕ ⋅ sen ( ϕ ) ϕ ⋅ ( m + 1 ) − sen ( ϕ ) ⋅ ( m + 1 ) − ϕ 2 (13)

The diagrams of the bending and torque moments of the hyperstatic HCB, reported in Figure 13, are strongly different from those of the isostatic case: in the former case torsion assumes low values at the clamped-ends and the maximum values at about a quarter and three quarter of the span, whereas in the latter case the maximum torsion is located at the ends and assumes null values in the center of the span.

The comparison of the results that can be obtained from the two analytical calculation methods described in the previous section for the determination of the torsional structural response, in particular of the torque stress characteristic, is now proposed for the side spans.

The torsion problem of the 76.33 m long lateral span of the South carriageway is solved extrapolating it from the structural context [21] [23] and solving the static scheme of the HCB clamped in pier P10 and hinged in pier P9 (refer to Figure 3). The maximum dimensions for the box-girder section taken into account are summarized are deducible from Figure 14 and Figure 15.

The vertical reaction R_A and the torque moment T_A at the supported end have been computed following the procedure proposed in ref. [23]. Their values are reported in Table 6. The bending and torque moments due to the distributed load uniformly applied to the circular cantilever beam are given in Equation (14):

M q ( ω ) = ∫ 0 ω − q ⋅ R 2 ⋅ sen ( ω − ε ) d ε = − q ⋅ R 2 [ 1 − cos ( ω ) ] T q ( ω ) = ∫ 0 ω q ⋅ R 2 ⋅ [ 1 − cos ( ω − ε ) ] d ε = q ⋅ R 2 [ ω − sen ( ω ) ] (14)

The Hinged-Clamped static scheme can be analysed from the known problem of the isostatic cantilever curved beam, by assuming the boundary reactions at the free-end (R_v,A, M_A and T_A shown in Figure 16) as unknowns and imposing the corresponding boundary constraints through the displacement and rotational congruence equations at the extreme A, given in Equation (15).

The problem is statically determined, as it consists of a system of three equations in three unknowns.

{ ζ A R v , A + ζ A q = 0 → R v , A                                                                                                 ( 15 -a ) ϑ A R v , A + ϑ A q + ϑ A T , A + ϑ A M , A = 0 → T A                                                                 ( 15 -b ) φ A R v , A + φ A q + φ A T , A + φ A M , A ≠ 0 → M A = 0                                                   ( 15 -c )

Equation (15-a) shows that imposing the vertical displacement equal to zero [24] by superposing the effect of the vertical constraint reaction ζ A R v , A and of the uniformly distributed load ζ A q , see Equations (16) and (17), it is possible to derive the vertical constraint reaction in A.

ζ A R v , A = 1 E I ∫ 0 ϕ R v , A ⋅ R 3 ⋅ sen 2 ( ω ) d ω + 1 G J ∫ 0 ϕ R v , A ⋅ R 3 ⋅ [ 1 − cos ( ω ) ] 2 d ω (16)

ζ A q = 1 E I ∫ 0 ϕ q ⋅ R 4 ⋅ [ 1 − cos ( ω ) ] ⋅ sen ( ω ) d ω     + 1 G J ∫ 0 ϕ q ⋅ R 4 ⋅ [ ω − sen ( ω ) ] ⋅ [ 1 − cos ( ω ) ] d ω (17)

Similarly, Equation (15-b) shows that cancelling the torsional rotation at the node A due to the load ϑ A q , the vertical reaction ϑ A R v , A , the torque ϑ A T , A and bending moment ϑ A M , A the torque offered by the additional support constraint T A is obtained.

ϑ A R v , A = 1 E I ∫ 0 ϕ R v , A ⋅ R 2 ⋅ sen 2 ( ω ) d ω − 1 G J ∫ 0 ϕ R v , A ⋅ R 2 ⋅ [ 1 − cos ( ω ) ] ⋅ cos ( ω ) d ω (18)

ϑ A q = − 1 E I ∫ 0 ϕ q ⋅ R 3 ⋅ [ 1 − cos ( ω ) ] ⋅ sen ( ω ) d ω     + 1 G J ∫ 0 ϕ q ⋅ R 3 ⋅ [ ω − sen ( ω ) ] ⋅ cos ( ω ) d ω (19)

(3) (20)

ϑ A M , A = 1 E I ∫ 0 ϕ M A ⋅ R ⋅ sen ( ω ) ⋅ cos ( ω ) d ω − 1 G J ∫ 0 ϕ M A ⋅ R ⋅ sen ( ω ) ⋅ cos ( ω ) d ω (21)

The bending moment M A can be neglected as there is a flexural hinge in the extremity A that provide free rotation φ A M , A . Finally, bending and torque moments in a generic point along the axis of the beam are provided by Equations (22) and (23) as a function of the boundary reactions applied to the support A of the beam.

M ( ω ) = M R v , A + M q + M T , A + M M , A = R v , A ⋅ R ⋅ sen ( ω ) − q ⋅ R 2 ⋅ [ 1 − cos ( ω ) ] ⋅ sen ( ω )     + M A ⋅ cos ( ω ) + T A ⋅ sen ( ω ) (22)

T ( ω ) = T R v , A + T q + T T , A + T M , A = − R v , A ⋅ R ⋅ [ 1 − cos ( ω ) ] + q ⋅ R ⋅ [ ω − cos ( ω ) ]     − M A ⋅ sen ( ω ) + T A ⋅ cos ( ω ) (23)

By replacing the values of the reactions provided in Table 6 in Equations (22) and (23) it is possible to derive the diagrams of the bending and torque moment along the deck axis. The comparison between the analytical calculation of the hyperstatic Hinged-Clamped HCB and the isostatic Hinged-Hinged case, is reported in Figure 17.

For the side spans, the redistribution of the torque and bending moments, compared to the isostatic case, is affected only by the presence of a clamped-end that leads to maximum torsion where the bending moment is null and vice-versa for the minimum values. At the annulment of the bending moment there is the maximum torsion at about three quarters of the span.