Appunti completi corso Bridge Theory and Design

A POSTERIORI VERIFICATION OF THE GUYON THEOREM. CASE 1

Let's try to proof this result by considering this 2 span continuous beam with a linearized cable:

Let's proof that center of pressure depend only at the eccentricity at the end:

To do this, we solve this statically indeterminate problem by using force method:

We remove the support in B. The line pressure is coincident with the cable profile. The displacement in B is (Mohr's Theorem):

Now how is the line pressure? Is coincident with the cable profile since it's a statically determinate structure. Computing the displacement now at point B due to the prestressing:

(Note that the eccentricity vary along the beam. V is found by applying the PVW. If Xb=1 the reactions at the ends are = 1/2)

This is the displacement if we don't have the intermediate support, and this deflection depend on a, b and the prestressing force.

For displacement compatibility in B results:

So now we have:

(Reaction depends on eccentricity at the end and)

eccentricity at the central support).The system of V ,V ,V are SELF EQUILIBRATED FORCES SYSTEMA B C

The redundant moment is given by the moment due to prestressing (primary moment) + contribution of the reaction: ‘’a’’.

This is the TOTAL MOMENT and depends only on A POSTERIORI VERIFICATION OF THE GUYON THEOREM. CASE 2 LINEAR TRANSFORMATION OF THE CABLE

Also in this case we proceed as in the previous case (we have zero eccentricity at the end because external loading doesn’t create any moment at the middle). Following the same procedure by changing the function:

What we get is:

The reactions depend on where the cable is in the middle span. The secondary effect of prestressing in the middle span in terms of moments is the same, considering the tendon sag. So also the center of pressure in the middle span depends only on the tendon sag, not on eccentricity.

HANDLING PRESTRESSING WITH FORCE AND DISPLACEMENT METHODS

In the principal system, we have a statically determinate beam with those configurations,

where we compute the rotations which change according to the prestressing force. Usually we have the schemes:

What change with prestressing are the rotations. To compute them we can apply PVW:

So with those rotations we can apply force method in the standard procedure.

COEFFICIENTS FOR THE DISPLACEMENT METHOD SOLUTIONS

With the same way we can see the displacement method where we are interested in the reactions which are fixed under the prestressing actions

Direct flexibility coefficients: phi11, phi22

In order to understand this we see the case of BALANCING LOADING: we can fix the end of a beam and compute the reactions. We have to apply those forces in opposite verse in order to talk about EQUIVALENT MODAL FORCES.

SUMMARY: END ROTATIONS AND FIXED-END MOMENTS DUE TO PRESTRESSING (helpful for exercises)

We have the geometrical parameter that define the tendon layout, the 2 fixed end moments and the rotations to solve a statically indeterminate problem.

EXAMPLES:

WORKED EXAMPLE 1. TWO-SPAN BEAM. STRAIGHT TENDON.

FORCE METHOD:

We want to solve this beam and find the bending moment diagram, by ex. use the force method, by computing the rotation in B due to prestressing.

Writing now compatibility:

In this case due to continuity, the actual rotation is zero.

So we will get the solution:

Given X we can find the reactions in A and C:

The systems are self equilibrated, so since prestressing is already equilibrated maybe reactions could be not satisfied. So we have a parasitic effect.

WORKED EXAMPLE 2. TWO-SPAN BEAM. STRAIGHT TENDON. DISPLACEMENT METHOD.

In this case:

1st diagram represent primary effect and 2nd the secondary effect. Prestressing effect are involved into the middle support.

In 1st column we have the different tendons layout.

Looking at the 2nd column we have the moments at the ends. Ex for the 1st row we have a parabolic cable with at end 1 an eccentricity e1 and on the other end an eccentricity e2.

3rd column give us the TOTAL MOMENTS.

4th column give us the REDUNDANT COMPONENTS. We have 2 schemes.

Clamped supported and supported clamped. They are not the same unless the cable layout is symmetrical.

SKEW BRIDGES:

Are necessary when we have to comply with some BC with some angles between longitudinal girder and some different components.

INTRODUCTION The structural behavior of skew girder bridges is characterized by the coupling between flexural and torsional behavior. An intuitive interpretation of the structural behavior of a skew plate under uniformly distributed load is shown in the Figure (redrawn after C. Menn, 1986).

We consider two sections, cut out from the obtuse corners and normal to the longitudinal axis: the opposite rotations of the left and right sections give rise to torsional effects.

In the following, we can distinguish between:

• skew girder bridges with beam-like behavior;
• skew girder bridges with plate-like behavior.

SKEW GIRDER BRIDGES BEAM-LIKE BEHAVIOUR

To evaluate equilibrium of a static scheme as the previous one we must take into account the inclination of the system.

We consider RS according to either the support and the beam: Eta = Oriented along the support line.

In order to impose the BC at the supports we have to do a transformation of coordinates. This can be done by making projection along the 2 axis. The denote as s= sin and c=cos.

stLet’s consider 1 :

In matrix form:

In terms of moments:

In terms of moments we get the TRANSFORMATION MATRIX.

Also we can see: stIn this transformation matrix the 1 row provide the moment around the psi axis and this must be =0.

Let’s make an example:

SINGLE-SPAN SKEW BEAM SUBJECT TO UNIFORM LOAD

To solve this problem we remove the rotation in B and we consider a simple support The beam is one time statically indeterminate. We assume the moment X , directed as the axis as the -unknown and draw the diagrams of the internal forces for the 2 structures (b) and (c).- (a) Released Structure- (b) Structure subject to X moment only

Using the PVW we can determine the unknown:

The value of the unknown X is determined

through the virtual work principle. We consider an auxiliary structure, subject to the unitary moment X =1 and set the internal work equal to the external one:

We have 3 contributions: the contribution of the external loading, and the contribution due to the application of X=1 but in the 2 directions x and y (so 2 contributions). The last integral is the TORSIONAL CONTRIBUTIONS (in addiction to the bending contribution).

By solving with respect to X we obtain:

By letting:

We have:

We can express the same results also in terms of those functions f1,f2,f3, which can be plotted for different value of rho, to see:

If alpha is = 90 degree we have no redundancy. At 45 degree we have the maximum effect of torsional stiffness. (in this case the beam is simply supported while when alpha is zero the rotations are prevented and – the beam become clamped clamped).

This aspect have a strong impact in the design of the reinforcement.

SKEW GIRDERS BRIDGES. PLATE-LIKE BEHAVIOUR

Are skew decks made by plates

where the distribution of the reinforcement will be more important. Is fundamental how to design the supports.

LEZIONE 22/11/22

By recap: The main girder in the longitudinal direction of the bridge have not a straight angle and also with stiffening beams. This emphasize the flexural and torsional behavior with important effect on the supports (in particular on corners).

The torsional moment is TOTALLY ASSOCIATED to SKNEWNESS.

SKEW GIRDERS BRIDGES. PLATE-LIKE BEHAVIOUR

Let's consider a solid slab, but using plate theory instead of beam theory.. We could have singularities that may arise into the corners.

So we consider this skew plate, and with 2 RSS (one orthogonal x-y and a skewed one x bar- y bar).

We can relate the 2 coordinate systems using the equations:

Those new RS can be used to introduce the governing equations for plates, where we remember:

So this is the same equation in the new RS xbar-ybar. So in the inclined RS we have:

Is important to change the coordinate cause now the BC's

are associated to xbar-ybar.This advantage in BC’s is paid by a more complex governing differential equations (which is complex tosolve).In the design stage is important to have an immediate value of reference to design (for ex. the depth of thedeck, the minimum amount of reinforcements..). We have also other aspects. When we create a complexmodel we have problems associated to SKEWNESS.BEARING SUPPORTS. TYPICAL LAYOUTSWhen the structure is subjected to loading that leading to singularities. In general the load path over the platewill create internal stress distribution which will be convenient on the lateral supports.Looking at what happen into those angles, the reactions are not all upward (it can be a mix, and so whenreaction are downward the supports tends to uplift)When we have unilateral support, since we have static indeterminacy (since a portion of the stresses isequilibrated by the tensile stresses) this will be an advantage and so we reduce the amount of stresses in thesupports.

For this reason we NEVER use continuous supports. This is what happen as follow:

At the bottom of the photo, all supports not allow displ in the longitudinal direction, so the force that arise in the longitudinal direction are taken by the supports on one of the 2 sides.

This scheme is not optimal cause the main flue of forces is developed along the shortest diagonal. If we want an optimal configurations for the supports we want a configuration that take also those forces. This can be achieved by using spherical supports:

This figure shows a different choice that makes use of spherical supports.

In this case the displ is along the diagonal so that the reaction along the diagonal is taken by the support on the other side.

Near the lower left corner, a fixed spherical bearing support restrains vertical, longitudinal and transversal displacements. The other three adjacent supports restrain the longitudinal displacements, but allow transversal ones. On the opposite side, the support near the upper