Questions and Answers

The following Image will help clarify this  :

At the moment, solid models cannot be imported from AutoCAD .

 

A1. There are mainly three methods for performing forward Construction Stage analysis. All three with their procedure and limitation are mentioned below.

Depending on the requirements and suitability you can choose your analysis method.

Thank you for the fast replay, the material sent is helpful but a kind of confusing with the terms "effective width", "longitudinal length", "W", "Distribution width", "Longitudinal width". These terms are used and mixed up in a way that the explanation results tricky.

In the program, both "W" and "Distribution width" represent the effective width of slab for the concentrated load. "Longitudinal length" represents the width of section of the transverse model.

Anyway, following your instructions I created the tranverse section model which I attach in this e-mail, where the LM1 according to Eurocode was applied (The detail of this Load Model is also attached in this e-mail. Note that the loads of each lane are differents, how to manage it?),

The different loads in the different lanes cannot be considered in the transverse moving load analysis.

please check the input data because the max. and min. moments obtained by Midas is too high in comparation with the reality. For instance, the min. moment at node 5 should have been around -250 [kN-m] (value calculated applying Homberg charts and FEM); but Midas gives a value equal to 

In your model, the wheel load is wrong. The axle load of LM1 is 300 kN, so the wheel load should be 150 kN. If you fix this in your model, the moment will be reduced by half, which is -382.5 kN-m. This value corresponds to the moment per 1.2 m because the longitudinal width is 1.2 m in your model. The moment, -250 kN-m obtained from Homberg charts and FEM will be the moment per unit width (1m). Therefore, if you divide -382.5 by 1.2, the moment per unit width becomes -318.3 kN-m.

Still there is some difference between the chart and the transverse analysis model. I think this results from the effective width (1.2m), I don’t know how you obtained this value but after all the formula of effective width recommended by the code is approximate. If you apply 1.5m as effective width, you will get -254 kN-m per unit width.

Now, in the case of the LM2, where there is just one axle load, the longitudinal lenght and "W" should be equal to 0? (Load Model 2 is also attached)

The effective width (W) of slab for the concentrated load does not change between LM1 and LM2. You can use same value. And longitudinal length can be entered as the same value as effective width .

On the other hand, I have doubts about the scale factor, I used the diagram shown in the Fig.11 Scale Factor and Element Number from the PDF you sent me to determine my own scale factors; but I find the information not enough to clearly understand how to manage it. 

The effective width of slab for the concentrated load varies along the span (transverse direction). It is the smallest at the web position and becomes larger at the tip of cantilever and at the center point of middle slab as shown in the image. It is recommended the effective width calculated for the web position is applied in the Transverse Analysis Model Wizard and the different effective widths for other locations are taken into account by the scale factor.

In reference to the load combinations, Midas does not allow me to create them, I dont know what the problem is, as I mensioned, I am working with the Eurocode.

In your model, there is just one load case, moving load. If you add more load cases like self weight, temperature, etc, you can create load combination.

During the transverse analysis, does MidasCivil find the most unfavorable location of traffic loads in the transversal direction to define the max. and min. moments?

Yes.

Some results seem strange to me. In the attached model, left side is without drilling DOF, and the right is with. Yet still, both produce bending moment (I was expecting the left to be 0 at the connection with the plates, thus, beam behaving as simply supported). Where is this bending moment coming from? Also, notice that this diagram is only obtained if the section of the beam is very small (2mm x 2mm). For any decent size section, the stiffness of the drilling degree of freedom appears to be negligible, as the bending moment diagram is almost perfectly 'simply supported'. Is it supposed to provide any adequate stiffness, or is it more of a 'no singularity problem' tool, just so there is some minimum stiffness provided'?

A1. A: As you observed, the plate element without drilling dof is assigned a very small rotational stiffness so that no singularity problem can occur. There is additional information below.

Plate element without drilling vs. plate element with drilling
- Plate element without drilling
Typical plate elements have three translational and two rotational dof at a node. These element types produce very reliable results as long as we use undistorted and square-shaped elements. But when we use relatively long rectangular elements, shear locking problem can occur, which cause very stiff response. So, incompatible mode was added. This is the explanation about plate element without drilling in Gen.

- Plate element with drilling
The above element types may give inaccurate results for the distorted mesh. In order to reduce the mesh distortion sensitivity and enhance the coarse mesh accuracy, we have introduced plate element with drilling degrees of freedom. This element type is particularly convenient in the analysis of shell structure and available at some sophisticated CAE software e.g. Abaqus, Nastran. The deficiencies of this element type are the zero energy mode (hourglass deformation) and that the node rotations are not true rotations. The boundary conditions of these elements have a high influence in the results. The rotations should be fixed at the boundaries because the node rotations are not true rotations. 

A1. Here is the explanation about plasticity in concrete with midas Civil.

 

One parameter models

Under tensile stresses, concrete will fail by a brittle fracture with very little plastic before failure.

Associated fracture criterion: Rankine (Maximum tensile stress criterion)

This is not supported in midas Civil.

 

Under high hydrostatic pressure (compression), concrete can yield and flow like a ductile metal on the failure (or yield) surface.

 

Hydrostatic pressure

 

For metals and for concrete in the high pressure range, the effect of hydrostatic pressure on the yield value of material may be neglected. It follows that shearing stress must be the major cause of yielding.

Associated yield criterion:

-          Tresca yield criterion (Shear stress criterion)

-          Von Mises yield criterion (Shear stress criterion)

This can be simulated with midas Civil.

 

Two parameter models

In the intermediate stress range, the failure criteria of concrete are sensitive to hydrostatic pressure.

Associated yield criterion:

-          Mohr-Coulomb criterion

-          Drucker-Prager criterion

 

Mohr-Coulomb criterion

 

The failure is governed by the equation below.

c (cohesion) and φ (internal-friction angle) are determined by experiment. This equation is referred to as the Mohr-Coulomb criterion.

c (cohesion) becomes equal to the yield stress in pure shear. c = fc’/sqrt(3) = 4000 psi/sqrt(3) = 2309.4 psi

From the test, m = (1+sin φ)/(1-sin φ) = 4.1. Thus, φ=37.4 degree

The stiffness after yield point can be defined by Hardening Coefficient. The zero value of coefficient represents perfectly plastic or zero stiffness.

The option of hardening rule: Isotropic, Kinematic, Mixes. Hardening rules describe how this yield surface changes in shape as yielding occurs.

 

Isotropic: Expansion of the yield surface. This is meant for proportional loading only.

 

 

 

Kinematic: Translation of the yield surface. This is meant for cyclic loading.

 

Mixed: Translation and expansion of the yield surface

 

Note that material nonlinear analysis in midas Civil does not take into account the unloading and reloading. It is recommended that proportional monolithic loading is tried with the Isotropic option.

Answer:

Let me first discuss the results for the section generated using the Composite Steel-I option in Civil. The torsion of such sections is calculated using a method where the section is broken down into rectangles, the torsional constant of each of them is calculated and then the torsion of the combined section is a sum of all the parts. In ‘Bridge Deck Behavior’ by E.C. Hambly we can find detailed information on this approach and how to calculate the torsional constant for a single rectangle:

And complex sections can be divided in the following manner to be calculated as rectangles:

Using this approach we can break your section into 4 parts – bottom flange, web, top flange and slab. Each of those is a thin rectangle and we can apply (2.18) to obtain the torsional constants and then combine them to get the resistance of the composite section (using a modular ratio to convert the contribution of the concrete). In case of composite girders where the slab is continuous and only part of it is effective for the composite action with the steel girder, one more consideration has to be taken into account. The normal distribution of torsion has the following pattern:

However, for continuous slabs the vertical components at the two ends of the slab are not present for the general case of intermediate girders and the torsional resistance is better approximated by:

Now using (2.18) for the beam parts and (3.15) for the slab we can carry out a manual calculation.

Before composite:  

and after composite:  

Now if we look at the values provided by the software you will see these match very well.

Now let’s move on to the SPC section. The first thing that needs to be pointed out is that SPC is a tool for calculation of section properties for arbitrary sections, however, as such it has to work for the general case. As I said in my previous email the way SPC calculates section properties is by performing finite element analysis applying unit loads. But when we analyse a composite beam which has a part of a continuous slab acting as flange, the software cannot take into account the reduced torsional resistance of the slab (formula 3.15) as it analyses the beam as a separate unit. So to check the results manually, we have to use (2.18) for the slab as well:

Manual calculation after composite:

SPC:  

Again, we observe a good match between the values for Ixx after composite. The only significant discrepancy is between the ‘before composite’ values. The Composite Steel – I section gives 1.00E7, while SPC gives 1.2E7. This difference is purely due to the accuracy of the approach using the formulae. After all, this is an approximate method and some inaccuracy is normal. For example, some torsional resistance is lost depending on the division of the section into  rectangles as we lose resistance along the dividing lines. Hambly provides a good illustration of this:

You can easily see how splitting the section in this case reduces the value of the resistance obtained.

Now that it is clear where the difference in the values is coming from, it is up to engineers to choose what approach they want to use to. As a solution for most accurate calculation I can suggest to use the SPC feature to generate the section with the exact properties and then apply a factor of 0.5 for Ixx for the slab in the Composite Section for Construction Stage definition.

However, as I-beams have generally poor torsional performance, we can simply use the Composite Steel-I section and neglect the inaccuracy of the torsion of the I section as its contribution to the composite properties is negligible (two orders of magnitude lower than the one of the slab). However, if the structure is such that the effects of torsion before composite are important it may be reasonable to go for the more accurate solution with SPC which I suggested above.

I hope this clarifies the topic and you will find this information useful. If there is anything else we could help you with, please let us know.

Kind regards,

MIDAS Support

Answer:

The difference is coming from the way torsional stiffness is calculated using approximating methods (that is, the equation option). In midas Civil the section properties are calculated using approximating simplified methods, while SPC can run an FEM analysis of the section to obtain the section properties. Before continuing with this question please read the answer to Q1.

Now that it is clear how the two methods work, it is easy to see why the equation option provides much lower results.

In this case, as SPC does not support curved lines, the circular parts of the section are represented by a number of straight lines. When we calculate the torsional constant using ‘equation’ the software simply applies the formula for each rectangle (line with thickness) and then sums those up. Obviously, this is incorrect as it completely ignores the connectivity of the lines and the global enclosed shape which the section has and the torsional stiffness which we get is very low. For this reason we have provided an extra function to define the correct stress flow path in such enclosed polygons so that the section properties are calculated correctly. This is done using closed loops (Model > Curve > Closed Loop > Register in SPC). Select the lines on the outer perimeter of the steel section and click apply:

Once this is defined, the torsional constant will be correctly calculated using the equation method as well:

Answer:

There could be two scenarios:

1) Section Properties cannot be calculated for the defined tapered section: If this is the case then there could be two possible reasons as mentioned below.
a) The two pre-defined sections had different number of nodes in the Section Property Calculator. If the number of nodes/divisions are different in between the two sections then the software can't generate a tapered section between them.
b) When the sections are defined in Section Property Calculator, the software assigns a number to the user defined nodes in accordance with the sequence of generation. If the sequence of generation in between the two sections defined in SPC is different, the software may fail to generate a tapered section between them because while generating the tapered section, midas Civil joins the node "x" of one section to the node "x" to another. Here, "x" is the node number defined in SPC. In this case, redefining the section in SPC with same sequence for both the sections is the solution.

2) Section properties can be calculated but section cannot be defined: This could occur when one of the two predefined sections has zero value for some section property and the other section has non-zero value for the same section property. To define the tapered section successfully, either the section property should be zero for both the sections or non-zero.

For the calculation of torsional constant before composite action, the program assumes that tub section is closed before slab concrete is hardened. During erection in practice, lateral bracings are attached to top flanges in tub girder as shown below.

                                                      

Top flange lateral bracing creates a quasi-closed section, which increases the torsional stiffness of  tub  girder  sections  during  erection,  handling,  and  deck  casting. The top flange lateral bracing forms a quasi-closed section resisting shear flow from the non composite loading. This reflects the practice in the US. If you need to apply any different values of torsional constant, you need to use Section Stiffness Scale Factor. Figure below shows the one of the example comparing tub section Ixx and SPC calculation by modeling it as closed box. When we consider it as closed box, torsional stiffness is almost same.

     

After composite action, the value Ixx is calculated based on the shear modulus ratio (Gs/Gc). When Es/Ec is entered from DB, Poisson's Ratios from DB are used for the calculation of Gs and Gc. When Es/Ec is inputted by the user, Poisson's Ratio is set to 0.

                                                            

In this case, Ixx_girder is calculated by assuming it as closed section. The torsional constant for closed section used in midas Civil can be found in the following online help.

http://manual.midasuser.com/en_common/civil/845/Start/04_Model/05_Properties/Section.htm

 

The models attached herewith are very puzzling and raise some doubts about the link connectivity. One of the models is a base model. The models with horizontal link and inclined link were later made for our verification. The support nodes (at 3 support location) were shifted in our test models and placed equidistant from the central support node. That brought down the abnormal reaction values. However, with everything else being the same, if the horizontal and vertical link assembly is changed to inclined link assembly, the vertical reactions change! Why is that? Supporting nodes and support conditions remain same. Just the change in flow of force due to links is changing vertical reactions drastically. This shouldn't be happening. A simple test model made showed that. Why is that not the case with the attached files? Please verify and let me know.

 

A1.

 

A1.

 

Algorithm of Beam’s semi-rigid release boundary by relative ratio or specified values

In case of 2-Dimensional stiffness of Beam element, if we disregarding shear deformation, Beam element stiffness matrix can be shown below equation.

In order to incorporate semi-rigid boundary condition, we add additional node A and 2node spring

stiffness matrix which is constituted by rigid stiffness and specified stiffness. Rigid stiffness value is

defined penalty method and specified stiffness is defined by user’s input: ratios or values.

The total dofs size of K  matrix can be expanded to 18 or 24 dofs. If we condense beam end dofs(uB,ΘB), the dofs size will be reduced to 12 dofs. In this condensation, including load vectors is needed.

In case of a relative type semi-rigid connection, the Ksp is defined by multiplying the factor to the related element stiffness value.

The other unchecked(in the dialog) components’ stiffness are defined by penalty method.

BIG/(BIG+1) = 0.999990000999… : There is no reference. This is from a kind of engineering experiences.

A1. For the specified displacement, midas uses the penalty method. According to this method, the solver internally creates a very stiff spring at the node to which displacement is specified. The stiffness of the spring is determined as about 10⁴ times the largest value of stiffness matrix of structure. And then the solver applies a load to the spring, the amount of which is equal to the product of spring stiffness and the specified displacement. Therefore, the same displacement as the specified displacement will occur at the node.

This method has an advantage that the calculation is effective when there are many load cases. But it may be that the sum of reaction is not exactly zero for the specified displacement load case, which is generally negligible. If the stiffness of spring is further increased more than 10⁴ times, the summation of reaction will be closer to zero. On the other hand, the increase of the stiffness of spring might cause some other problem in the FEM analysis e.g. numerical error.

There is no setting in which this can be controlled by the user.

A2: Whenever we deactivate a load in midas Civil, an opposite loading with the same magnitude is applied on the structure to nullify the effects of previously applied loading. In your model file, the support at the pier base is fixed. Due to this reason, when the software applies the load opposite to the specified displacement, the node is unmoved. Thus the deformations in CS3 are same as those in stage 2.

Instead of deactivating the specified displacement, it is recommended that you activate another specified displacement with the sign opposite to the previously specified displacement. The reasons behind this recommendation are discussed below.

Specified Displacement is considered as the relative displacement with reference to undeformed nodal coordinates. This means that if node 1 (say) has dz deformation of -2.0m at stage 1 and you activate a specified displacement of +0.5m at stage 2, then the resultant displacement at node 1 will be +0.5m and not -1.5m. Likewise, if node 1 has dz deformation of +3.0m at stage 1 and you activate a specified displacement of +0.4m at stage 2, the resultant displacement at node 1 will be +0.4m.

Some more information on specified displacement application in construction stages:

Specified Displacement is applied to CS: Dead Load so that the displacement due to CS:Summation is equal to specified displacement.
(Assumption: The load case of Specified Displacement is not distinguished from CS: Dead Load.)

Example:

 

A1. This is the case for the temperature load. Let us consider a cantilever whose length is 10m at temperature of 0 °C. If temperature is increased to 20 °C, the length of a member will be increased to let’s say 10.2m. Note that axial force in the member remains zero although the length is increased. We should not think that this member should be under tension because the length is increased. The axial stress will be zero because the lengthening of member is not restrained in the cantilever.

Let me take another case. Simple beam at temperature of 0 °C.

If temperature gradient of 50 °C (25 °C at the top and -25 °C at the bottom, linear variation in between) is applied, the deformed shape of the member will be hogging (upward) as shown below. Similarly the bending moment remains zero along the member.

This time, we change the boundary condition to fully-fixed support at both ends and apply temperature gradient of 50 °C. Then, the deformed shape of the member will be straight (no deflection) but the bending moment diagram will be sagging as shown below.

Actually, the length or shape of member at which the internal stresses are zero is determined based on the temperature.

For a beam with temperature gradient of 50 °C, the stress-free shape is as shown below.

If both ends are fixed before applying temperature gradient, the sagging bending moment will occur due to temperature gradient, which will cause the beam to deflect in sagging (downwards).

The above explanation can be applied to your question.

In the figure below, the first image from the left shows the stress-free shape of horizontal member for the temperature gradient of 50 °C. The rotation of horizontal member is resisted by the bending stiffness of columns at both ends. So, the sagging bending moment will occur due to temperature gradient, which will cause the beam to deflect in sagging (downwards) from stress-free shape. The bending stiffness of columns will be less than fully-fixed support. Therefore the final deformed shape will not be straight in this case. Instead, it will be above the straight line.

 

A1. When we input the temperature load, we need to input the temperature value based on the middle of the tapered section. However, in this case, the stress at the larger size end will be less estimated whereas smaller size end will be over-estimated as shown in the figure below. It means, middle of the section has correct stress value.

In order to resolve the problem, we can use equivalent uniform beam elements by dividing them into several elements as shown below.

A1. External tendons are the same as unbonded internal tendons (post-tension) except for the followings:

- External tendons attach to concrete at several locations, i.e. anchorages and deviator blocks.

- Load transfers to the concrete occur at anchorages and deviator blocks. 

- Losses due to wobble friction will not be counted.

- Losses due to curvature friction occur only at deviator blocks.

- No relative displacements occur at the deviation points.

- Creep & Shrinkage losses are constant over the length between anchors or deviators.

When defining tendon property in prestressed structures, one shall enter the wobble friction factor (among other factors) in the add/modify tendon property-window. The units of the wobble friction factor (K) used in MIDAS is [1/m]. For the post-tensioning system I am using, I can only find the "unintentional angular displacement"-factor (k) given in the units of [rad/m]. Is it correct, that the value which I should enter in MIDAS then is K=k*µ, where µ is the curvature friction factor [1/rad].

A2.

In general, yes, in MIDAS you should enter K=k*µ.

If you look at the Analysis Reference manual of MIDAS you will see the formula used to account for friction losses (Px is the force in the tendon after losses):

In the same time if you look for the standard definition of the loss due to friction you will find the following formula:

So in MIDAS the expression for the power of ‘e’ has been defined without the brackets and the wobble friction factor which we define is the product of the coefficient of friction (µ) and the unintentional angular displacement per unit length (k).

A3. Software uses following consideration from EC2 for the relaxation loss calculation:

A1. Theoretically, the maximum possible moment due to a group of wheel loads does not occur at the center of simple span. We can find the location of maximum possible moment in the simple span by applying two things.

1. The maximum possible moment will occur at the wheel load which is closest to the resultant load of group of wheel loads. In case where the amount of the wheel load closest to the resultant load is small, the maximum possible moment can occur at the second closest wheel load. In your case, the group of loads consists of just two wheel loads with the same amount. If we denote it as P1 and P2, the closest wheel load can be either P1 or P2. If we take the closest wheel load as P1, the maximum possible moment will occur at P1.

2. If we denote the distance between the closest wheel load (P1) and the resultant load (R) as ‘d’,  the location of P1 which causes the maximum possible moment will be d/2 away from the center of span as shown below.

A1. The surface lane can be only applied only if the plate is within 15 degrees from the horizontal plane. If the angle between the plate element and the horizontal plane is more than 15 degrees, the surface lane cannot be applied.

 

 

A1. The purpose of ‘periodic’ analysis is to find the steady-state response of vibration. The typical application is the vibration analysis of a floor due to vibrating machine. If we measure the vibration of floor, it will be as shown below. At the beginning there is some fluctuation but after some time, the response becomes stable and exactly periodic. The former is called transient response and the latter steady-state response. The peak response is larger in the transient response than steady-state response. The steady-state response can be obtained by the ‘Periodic’ analysis. The whole response (transient plus steady-state) can be obtained by the ‘Transient’ analysis.

 

In terms of your analysis results, both results are not correct. the followings should be met in order to obtain correct result in the ‘periodic’ analysis:

- The end time should be ‘period of forcing function’ multiplied by N (= integer).

- The period of forcing function should be ‘Time Increment’ multiplied by N (= integer).

In this model, the above two conditions are not satisfied.

End time (= 13.4 sec.) divided by ‘period of forcing function’ (= 0.33557 sec.) equals 39.932, which is not an integer.

Period of forcing function (= 0.33557 sec.) divided by ‘Time Increment’ (0.01 sec.) equals 33.557, which is not an integer.

End time should be 13.4228 sec. = 0.33557 x 40. Also time increment should be, for example, 0.0033557 sec. = 0.33557 / 100. Normally time increment is taken less than one tenth of period of the highest vibration mode of structure.

From the above change, the maximum displacement becomes 2.133e-4 m. The same result is obtained with end time of 0.33557 sec. = 0.33557 x 1, 0.671141 sec. = 0.33557 x 2, …, 0.33557 x N as long as identical time increment (0.003357 sec.) is used.

This is what is happening in the program in case where the end time is not ‘period of forcing function’ multiplied by N (= integer).

 

Therefore, the two results were different in the user’s model.

A1. The vertical (or gravitational) load cases should be added in “P-Delta Combination” so that the program can find the modified stiffness matrix of the structure based on the geometric stiffness matrix due to P-Delta load combination. Even if the lateral load case is added here, the results will remain the same because the lateral load will not make any axial forces in case of a cantilever column.

For example, if we make a test model as below, all the three cases give the same results for P-Delta analysis.

According to the information as mentioned above, in midas, P-Delta analysis will find the geometric stiffness matrix of structure due to the entered load cases in “P-Delta Combination”. This geometric stiffness matrix will be added to the elastic stiffness matrix to generate the modified stiffness matrix of structure, which will be applied to the analysis for all the load cases including lateral loads.

Additionally, when the user wants to see the 2nd order effect for multiple load combinations which have different load factors, the user has to modify the load case factor in “P-Delta Combination” and perform analysis again.

A1. In midas Civil, both flexural buckling and lateral torsional buckling of the beam elements are considered in buckling analysis. After that, local second order effect and imperfection can be considered during Steel Code Checking. In Steel Code Checking, LTB check is verified for steel I-shape section.

However, if the user wants to consider the local second order effect and imperfection, we can use buckling analysis. In midas Civil, both flexural buckling and lateral torsional buckling are considered in Buckling Analysis.

For this verification, I had a test as below.

The report below is generated from "LTB.mgb" model file.

By performing buckling analysis, we can see the 1st buckling factor, which is LTB, is determined as 1.3618. "My" value is 250kNm. Therefore lateral torsional buckling moment is 1.3618*250 = 340.45kNm.

We can see this value is very similar to Mcr above.

After checking buckling load, we can update the geometry of model with the buckling mode 1 in order to consider local imperfection using the Results > Buckling Mode Shape function. This model can be used for the verification of buckling force by performing geometric nonlinear analysis. Here, maximum eccentricity was entered as 0.015m.

   *Why 0.015m is determined?

       e0/L for buckling curve a  = 1/300

       L=10m

       e=0.03m

       For LTB, k=0.5

      Therefore, eccentricity is 0.5*0.03 = 0.015m

In the updated model with 1st mode imperfection, if we check the value of My, it is shown as 362kNm which is also similar to Mcr.

Eigenvalue analysis of a cable structure

For the calculation of accurate natural frequency of a cable structure, the following two things should be considered in the analysis.

- The effect of cable tension on the axial stiffness of cable by Ernst equation (horizontal direction in the verification example below)

- The effect of cable tension on the geometric stiffness of cable (vertical direction in the verification example below)

 

Natural frequency

The natural frequency of single degree of freedom system is determined as the equation below.

  

Ernst equation

The axial stiffness of truss element is determined as below.

k (truss) = EA/L

 

In order to calculate the axial stiffness of cable, first we calculate the equivalent modulus of elasticity using Ernst equation.

where E indicates the elastic modulus of the cable, L the horizontal projection length of the cable element, w the cable weight per unit length, A the cross-sectional area of the cable, and T the axial force of the cable element.

 

Then, the axial stiffness of cable can be determined as below. Refer to midas analysis manual.

Geometric stiffness

The element stiffness matrix may be separated into an elastic part, k_E, plus a geometric part, k_G that accounts for the effects of finite deformation.

where the first matrix is the elastic stiffness matrix and the second matrix is the geometric element stiffness matrix. The approximation (T/L) = (T/L_o) is sufficiently accurate in most applications.

 

How Ernst equation and Geometric stiffness can be taken into account for the eigenvalue analysis?

1) Ernst equation

Use cable element type.

Create construction stage model.

Check on the 'Change Cable Element to Equivalent Truss Element for PostCS' option in the 'Construction Stage Analysis Control Data' dialog as below.

 

2) Geometric stiffness

Check on the 'Convert Final Stage Member Forces to Initial Forces for PostCS' option in the 'Construction Stage Analysis Control Data' dialog as below.

Perform eigenvalue analysis.

 

Verification

Title: Eigenvalue analysis of a cable

Description: Obtain natural frequencies of vibration of a cable in the horizontal and vertical direction.

Structural geometry and analysis model

 

Analysis type: 2D eigenvalue analysis (X-Z plane)

 

Unit system: m, kN

 

Dimension: Length = 5 m

 

Element: Cable element

 

Material

Modulus of elasticity: E = 2.10 x 10⁸ kN/m²

Poisson's ratio: ν = 0.3

 

Section property

Area: A = 0.001963 m²

 

Boundary condition

Node 1: Restrain DX, DY & DZ

Node 2: Negligible stiffness in the Z-axis (or vertical direction) in order to avoid singular condition.

 

Load Case

A concentrated load, 10 kN is applied to the node 2 in the (+)X direction.

This load will generate cable tension of 10 kN.

 

Mass

MX = 100 kN/(m/sec²)

MZ = 100 kN/(m/sec²)

 

Results: Eigenvalue analysis results

Mode shape (mode 1): Vertical vibration

 

Mode shape (mode 2): Horizontal vibration

 

Frequency

 

Hand-calculation of frequency

Mode 1 (vertical vibration)

- Elastic stiffness = 0

- Geometric stiffness

 

- Frequency

 

Mode 2 (horizontal vibration)

Axial stiffness with equivalent modulus of elasticity

 

Equivalent modulus of elasticity (Ernst equation)

 

 

    

 

Comparison of results

A1.  Tendon Primary represents member forces caused by Tendon Prestress forces. Tendon Secondary represents member forces resulting from Tendon Prestress forces acting in an indeterminate structure. To check analysis results, Primary and Secondary can be regarded as internal forces and external forces respectively. For design, however, the program internally recalculates member forces due to Primary considering the translation of neutral axis so as to use them as internal forces, and member forces due to Secondary are used as external forces.

Likewise in creep, we have both primary as well as secondary effects. Creep Primary Effects refers to deformation and forces required to cause creep strain. Whereas creep secondary refers to effects caused by creep in indeterminate structure.

A2.

Primary forces are the imaginary forces whereas, the secondary forces are the actual forces developing inside the member. So for design, only the consideration of secondary forces is adequate. But especially for shrinkage, there are few more considerations as discussed below. There are three effects of shrinkage in the steel composite girder as follows:

1.        Auto-equilibrated stresses and an imposed rotation in a determinate structure.

2.        Deflection due to the imposed deformation.

3.        Member forces in a indeterminate structure.

Often, item 1 and 2 are denoted primary effects and item 3 secondary effects of shrinkage. In midas, the item 1 and 3 can be checked with CS: Shrinkage Secondary. The item 2 can be checked with CS: Shrinkage primary.

For the serviceability stress check, the item 1 and stresses caused by item 3 should be considered. For the ultimate limit state check, the item 3 should be considered. Since item 2 is not checked as per the code’s provisions, the auto generated load combinations in midas Civil do not include the Shrinkage primary.

A1. There are some mistakes to be corrected in the SAR file. Please change the file according to the following explanation.

1. Unit must be divided using “,”. Also “KIPS” should be used instead of “KIP”.

2. The number of load cases for reaction is 6 in the model file, whereas there is only one load case in the SAR file.

3. At the end of the file, it should be “*ENDDATA” instead of “ENDDATA(TAB)(TAB)”.

A1. For using Import feature for the load combinations in midas Civil you need to manually create *.lcb file which will later be imported. The format of the *.lcb file is as follows.

Sequential number, combination method, unit load case i, load factor i, unit load case j, load factor j, ..., load combination k, load factor k, load combination l, load factor l

Examples)

1, , 1, 1.0, 2, 1.0

2, , 1, 1.4, 2, 1.7

Note Combination methods

blank,0	: Add
1	: Envelope
2	: ABS
3	: SRSS

Rather than importing using the import option we also have two other alternatives for importing the load combinations from one file to another. You might be knowing them already but I believe its worth mentioning.

1) Copy and paste feature is supported for the load combinations in the Spread Sheet Form. In this form by correctly copying the cells from one model file you can paste it into other model file.

2) You can export the model as MGT file and from there you can copy the load combinations. By pasting the load combinations in another model file's MGT command Shell and clicking run, you can get the load combinations imported.

A1. If you want to check the results of “CS:Live Load + CS:Tendon Primary” in a certain stage, you need to manually calculate them using table. If you sort the table with Node as top priority, you can check the table as below.

If you are dealing with the results in the final construction stage, you can use “Load combination” in PostCS as shown below.

A1. Major reason causing the difference between program results and your hand-calculation is as follows:

The consideration of forces in the tendons is missing in your hand-calculation in which only geometric properties and modulus of elasticity are included to calculate the eccentricity. Different forces in the different tendons affect the eccentricity obtained using axial force and moment in the section. In the schematic drawing below, for example, if tendon forces of two tendons are equal, then the eccentricity obtained from axial force and moment( or e=M/P) will be the same as your hand-calculation (or geometric eccentricity). If the force in the upper tendon is higher than the lower tendon, the eccentricity obtained from axial force and moment will move up from the centroid of tendons.

 

 

 

 

A1. In Response Spectrum analysis, nodal results obtained from modal combination are not coexisting results, while the nodal results of individual mode are coexisting results. This is because of the modal combination, i.e. SRSS, CQC. Thus, in order to get accurate results with response spectrum analysis, any results related to more than one node such as total reaction, story displacement, Local Direction Force Sum etc. must be calculated in such a way that they are first calculated for each mode and then combined by SRSS or CQC. It is wrong to first calculate the results of each node by modal combination and then, based on the nodal results, calculate total reaction, story displacement, Local Direction Force Sum. This is your hand calculation. This method only gives correct results when only one mode is considered because the nodal results obtained from any single mode are coexisting results. The reason why your hand calculation matched the LDFS results in case of two modes is that the contribution of second mode was very small, so the LDFS results were obtained from the first mode alone. If we consider only one mode, then your hand calculation has no problem.

Case 1) Six elements are activated in the model view.

Case 2) One element is activated in the model view.

A2. First, let me show you how the LDFS result is obtained in case when six elements are activated. The LDFS results are obtained using nodal forces (Result Tables>Plate: Global) of blue dots shown in the image below.

How to determine elements to be considered in the LDFS calculation:

Step 1. Find the elements whose node(s) is connected to the dotted line (or Cutting line of LDFS). In this model, the element 5737, 5747, 5757, 5738, 5748, 5758 are selected.

Step 2. Among the elements which are connected to the dotted line only at one node, i.e. element 5738, 5737, 5758, 5757, exclude elements whose edge line is located at the same straight line as dotted line. Element 5738 and 5737 are excluded because edge line connecting node 5839 and 5850 is parallel to dotted line.

Step 3. Find elements whose node(s) is located below the white line. The element 5747, 5757, 5758 are selected. Element 5758 is selected because node 5872 is located below the white line.

How to determine nodes to be considered in the LDFS calculation:

Step 1. For the elements whose edge line is located directly on the dotted line, i.e. element 5747, only select the nodes which are located on the dotted line. Node 5850, 5861 of element 5747 are selected and node 5849, 5860 are not selected.

Step 2. For the elements whose edge line is not parallel to the dotted line, i.e. element 5757, 5758, only select the nodes which are located below the white line. Node 5860, 5871, 5872 of element 5757 are selected and Node 5872 of element 5758 are selected. Note that node 5861 is not selected which are on the dotted line.

From step 1 and step 2 above, the following nodes are selected in the end.

How to calculate resultant forces using the nodes selected above:

Basically, the resultant forces are obtained by summing the plate global forces of the total selected nodes. But, the sign of the plate global forces are taken as follows: For the nodes which belong to the elements whose edge line is located directly on the dotted line, i.e. node 5850, 5861 of element 5747, the same sign as plate global forces are taken. For the nodes which belong to the elements whose edge line is not parallel to the dotted line, i.e. node 5860, 5871, 5872 of element 5757 and node 5872 of element 5758, the opposite sign as plate global forces are taken. The above rule contributes to the following sign convention: (-) for compression, (+) for tension

After all, this is the algorithm of Local Direction Force Sum, which is generally applied to the unstructured mesh as well as structured mesh. Note that this algorithm is only applied to the activated elements. Thus, if you activate element 5747 only, then the final selected nodes are as shown below.

The user might think that even when all the elements are activated, the LDFS should consider only one element (element 5747). It may be true for the structured mesh. But, there are many possible complex cases of unstructured mesh. In this case, it is not so simple which elements and nodes should be included in the calculation of LDFS.

       

Concluding remarks:

For the practical project, first we recommend using finer meshes, which will reduce the error resulting from which elements are selected. Secondly, you can select elements to be included in the calculation using Activation / Deactivation of elements.

A1. The modal mass participation shown by the software is correct. You can get a clear idea by looking at the vibration mode shapes. The vibration mode shapes for the model having the point springs clearly show that the piles with point springs don't contribute significantly to the effective mass for deformations in x and y direction.

Let me explain this using the formula given in our analysis reference manual.

If certain degrees of freedom of a given mass become constrained, the mass will be included in the total mass but excluded from the effective modal mass due to the restraints on the corresponding mode vectors. Accordingly, when you attempt to compare the effective modal mass with the total mass, the degrees of freedom pertaining to the mass components must not be constrained.

While converting load to mass you have not kept this in mind at the time of defining the point springs. Since the point springs have constraint in x and y direction, their mass is excluded from the effective mass for a and y deformations but not from the total mass. This is why the mass participation factor is not reaching 90% value with the constraint.

Answer:

The spacing between the cross beams is inappropriate. The curved bridge is modelled using linear elements closely representing the curved profile. Due to the curvature, there is slight gap between the cross beams on the outer side of the curvature and there is overlap on the inner side of curvature.

If the cross beams are not placed at appropriate spacing then there is no intersection on the outer edge and the load distribution is inaccurate which in turn leads to reduced values of moving load results.

The solution to this problem can be to introduce more intermediate cross beams or use 'lane element' type lane definition. But for grillage model, cross beams are recommended, thus reducing the spacing between the cross beams can be a better option.

A1. The snapshot below should provide a good explanation for the concurrent reaction table:

A1. In order to obtain the results of Plate Force (Unit Length), we need to  understand the following diagram.

The program has the stresses at the nodes of the plate elements. These stresses are then integrated in the following manner to obtain the desired results.

A1. As the tendon group can be composed of a number of tendons, the program calculates the inclination of individual tendon from their profile. And then their average sine and cosine can be calculated.

Reference axis for calculating theta is determined as the neutral axis of an element. Also tendon inclination is determined based on the variation of “Zp” in the table. Due to this reason, the value between i-end and j-end is always identical. Please refer to the image below for detail calculation.

 As the average stresses and forces due to prestress are provided in the table, the users can calculate the vertical component and horizontal component of prestressing force. Therefore the sin and cos theta can be applied for the verification of result values.

Q1. How the value 0.01172 for Load Case type PO is calculated.

The total rotation at node consists of elastic rotation and plastic rotation.

The yield moment of this hinge is 87.275 kN-m and the applied moment is 90 kN-m. So, in this case plastic rotation occurs.

1)      Elastic rotation at node 2: ML/EI = 87.275 x 5 / (32308000 x 0.002133) = 6.331e-3

2)      Plastic rotation at node 1: 0.002683

3)      Plastic rotation at node 2: 0.002703

 

              

          Total rotation at node 2:     1) + 2) + 3) = 0.01172

This is how the value 0.01172 for Load Case type PO is calculated.

 

Q2. If this is the rotation of the node, how it is possible that it is less than the rotation of the hinge in J-end, which is 0.015129 ?

         

As I mentioned in the web board, the “Pushover Hinge(Deformation)” table shows the rotation of pushover hinge,  which is different from the rotation of the node. The hinge rotation is calculated based on the hinge property curve. The elastic rotation of hinge at the yielding moment of 87.275 kN-m is calculated by dividing moment by initial stiffness of the curve.

Moment=87.275 kN-m, Initial stiffness=7023.3957 kN-m

Thus, elastic hinge rotation = 87.275/7023.3957 = 0.012426 rad.

Total hinge rotation = elastic hinge rotation + plastic hinge rotation

     i-end: 0.012426 + 0.002683 = 0.015109

     j-end: 0.012426 + 0.002703 = 0.015129

Hinge rotation of Eurocode corresponds to the chord rotation like FEMA as shown below.

                                   

 

So, the initial stiffness of hinge in case of cantilever subjecting uniform moment along the member should be 2EI/L.                   

You can do this using “User Input” in the hinge property dialog or using FEMA type hinge as shown below.

               

Then, the total rotation for the pushover load case is 6.739e-003 which is slightly higher than the rotation (6.529) of static load case as expected.

                     

 

Total hinge rotation:

              

 

Plastic hinge rotation:    

  

 

 

 

A1. Deflection Amplification Factor (C_d) amplifies the drift computed using elastic analysis in order to account for the drift due to inelastic response of the lateral force resisting system (LFRS). As general philosophy of earthquake resistant design, the structure is designed for the lower amount than the actual force which it will be subjected to during actual earthquake. Because we rely on nonlinearity hence we reduce the actual force by dividing Response modification factor to get the design force in Response Spectrum Analysis or Static Seismic Analysis. But the displacement of the structure will be according to the actual force. Hence, the displacements obtained are multiplied because of reduced forces by that response modification factor. This factor is specified as per ASCE/SEI 7-2010 (American Society of Civil Engineers/Structural Engineers Institute). Therefore, the factor was included in the serviceability parameters. If user is using design code other than ACI and he wants to amplify the deflection as per the above reason, then the deflection amplification factor may be used. Otherwise, it may be used as 1 by default.

A1. The governing Load Combination for Shear Design is not decided on the basis of the magnitude of the shear force. The governing load combination is decided on the basis of the ratio of the shear force and shear capacity.

Now, if there are two shear forces V_Ed1 and V_Ed2 such that V_Ed1>V_Ed2. V_RDs (Shear Strength provided by steel) and V_Rd,c (Shear Strength provided by concrete) will be calculated. If V_ED1/V_Rd,s is less than the V_ED2/V_Rd,c, the V_Ed2 will be used for shear design, even if V_Ed2 is smaller of the two shear forces.

A1. There is no change in any parameter or formula due to the consideration of the transverse stiffener. When the user specifies the pitch for the stiffener, the software does the calculations to check the shear capacity of the section. For checking the shear capacity, the shear stiffener width, moment of inertia of the stiffener and area of transverse stiffener are checked to satisfy standard constraints.

A1. Analysis terminates if the maximum number of increments is reached, or negative values are encountered in the stiffness matrix. Negative values may be encountered in the stiffness matrix due to excessive force development or large deformations making the structure unstable.

In the user's model file, the forces are exceeding the limits on nodes connected to the multi-linear point springs (numbered as 111 and 112). If you delete these multi-linear point spring definitions or reduce the force limits then the program doesn't terminate. FYI, 2215 indicates the point spring definition at the node which is inappropriate for the Pushover analysis.

A1. The reason for hanging of GSD is the large section size (diameter = 8ft) and very small mesh size (mesh size = default mesh size = 0.1in). In order to perform the design using GSD, please increase the mesh size. Mesh size greatly affects the time required for design. Larger is the mesh size lesser will be the time needed for the design.

In order to change the mesh size go to Model tab and click on Section property. Software will take time but finally you will get the option for inputting the mesh size. By increasing the mesh size to 1in, I could obtain the design results successfully.

1. Compression-only bearing

General Link with Gap type can be used to simulate the compression-only bearing in the nonlinear time-history analysis.

For example, the properties of Gap are as shown below.

   

2. Vehicle or train running on the bridge.

 

3. Nonlinear time-history load case

Note that nonlinear time-history analysis requires an additional option: Inelastic Time History.

Make sure that you have purchased this option separately.

 

 

 

 

 

 

 

 

4. Results

Answer:

Software uses following consideration from EC2 for the relaxation loss calculation:

User question:

When defining tendon property in prestressed structures, one shall enter the wobble friction factor (among other factors) in the add/modify tendon property-window. The units of the wobble friction factor (K) used in MIDAS is [1/m]. For the post-tensioning system I am using, I can only find the "unintentional angular displacement"-factor (k) given in the units of [rad/m]. Is it correct, that the value which I should enter in MIDAS then is K=k*µ, where µ is the curvature friction factor [1/rad].

Answer:

In general, yes, in MIDAS you should enter K=k*µ.

If you look at the Analysis Reference manual of MIDAS you will see the formula used to account for friction losses (Px is the force in the tendon after losses):

In the same time if you look for the standard definition of the loss due to friction you will find the following formula:

So in MIDAS the expression for the power of ‘e’ has been defined without the brackets and the wobble friction factor which we define is the product of the coefficient of friction (µ) and the unintentional angular displacement per unit length (k).

bES: This parameter is used for the load combinations generated as per Japanese code required for the 2nd design i.e.
pushover. For all other codes, the value is zero.

iSERV-TYPE: This parameter is used to classify the Serviceability load combinations generated as per Eurocode and categorized under the Serviceability load combination type. The corresponding values for this parameter are 1, 2 and 3 for Quasi-permanent, Frequent and Characteristic combinations respectively.

nLCOMBTYPE: This is used to internally assign the auto-generated load combinations to the Load Combination Type.
This is used for several codes like:

Steel
         - SNiP 2.05.03-84*, SP 35.13330.2011, SNiP 2.05.03-84*(MKS), SP 35.13330.2011(MKS), KSCE-LSD12
Concrte (PSC)
         - SNiP 2.05.03-84*, SP 35.13330.2011, SNiP 2.05.03-84*(MKS), SP 35.13330.2011(MKS), KSCE-LSD12
Steel Composite
         - KSCE-LSD12, KSCE-ASD10, KSCE-RAIL-ASD11, AASHTO-LRFD12, AASHTO-LRFD07

Lets consider that the load combinations are auto-generated for AASHTO-LRFD as shown in the image below:

 

Now to internally classify these load combinations under strength, service and Fatigue combinations so that they are automatically available under Design > Composite Design (as per AASHTO-LRFD12) > Load Combination Type, the software assigns the number from 0 to 3 increasing downwards, i.e. Strength Limit State = 0, Service = 1, Fatigue 1 = 2 & Fatigue 2 = 3.

This is how the auto generated load combinations are automatically assigned internally.