Questions and Answers

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

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 Gen.

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 Gen.

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 Gen does not take into account the unloading and reloading. It is recommended that proportional monolithic loading is tried with the Isotropic option.

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.

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.

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):

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. This is the verification of accidental eccentricity of earthquake load by the response spectrum analysis.

For the test purpose, the model is one story building which has only one column at the center.

In midas Gen, the effects of accidental eccentricity due to the X direction and Y direction of the response spectrum load case can be checked separately by RX(ES) and RY(ES) respectively.

Accidental eccentricity will be verifed by checking the torsional moment of the column for the following two cases.

Case 1) Rigid diaphragm: Floor diaphragm is considered in the Roof floor.

Case 2) Flexible diaphragm: Floor diaphragm is not considered in the Roof floor.

Mass distribution

Spectral acceleration

Only one mode is considered for the test purpose.

Case 1) Floor diaphragm is considered.

Case 2) Floor diaphragm is not considered.

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. 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 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 Gen, 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 as per Cl.6.3 in EN1993-1-1 during Steel Code Checking in midas Gen. It is according to the Cl.5.2.2(3)b. In Steel Code Checking, LTB check is also verified for steel I-shape section.

However, if the user wants to consider the local second order effect and imperfection as per Cl.5.2.2(3)a, we can use buckling analysis. In midas Gen, 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, which considers Cl.5.2.2(3)b.

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.

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.

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 Gen 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. 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. 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. 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. In serviceability verification, deflection of beam considers only (b) in the figure below where as analysis results display total displacements (a+b). If a beam member is assigned with support at both ends or connected with vertical members, it will be considered as simple/fixed beam condition instead of cantilever beam condition. If a member has no support condition and no connected elements in i-end or j-end, it will be determined as cantilever condition.

[Example 1. Cantilever beam condition]

Calculate the initial deflection “a” using tangential line.

Tangential line is determined by dividing the element into 199 segments.

In this example, deflection of node no. 5 (D1): -16.734108 mm

Displacement of 1st segment among 199 segments (D2) : -16.759696mm

Displacement of a:

a = D1 + (D2-D1)*199*5 = -42.194168 mm

b = -42.529 – (-42.1942) = -0.3348 mm

[Example 2. Fixed/Simple beam condition]

Divide a member into 18 segments and calculate global displacement.

Local displacement is also calculated considering tangential displacement between i-end and j-end.

Local Displacement = Di + (Dj-Di) * Position Number/Number of Division (=18)

ex. Local Displacement in position 8 = 3.19270+(-3.91826-3.19270)*8/18 = 3.51517 mm

A1. When I run the steel optimal design for your model, I get the following table.

As you can see that the section 1 has Combined Stress Ratio of 1.242 >1 (Allowable ratio). Thus you would go for optimization of this section. D1 value shown buy the software for this section is 0.656. When you click on Design & Analysis the software would look for sections with the depth of 0.656(+/- 5%)m only and it will present the optimal sections by iterative analysis and iterative design. Since there is no other section with this depth which can have a combined stress ratio within limits, the software can find the most optimum section of W8x40 in your model as shown below.

But this section doesn’t has the Combined Stress Ratio <1. Thus I would recommend you to change the D1 value to zero. Doing so the software will look all the sections as per the database of the code.

You observe a message stating “Section is not changed” when the ‘Number of Repetitions’ under Analysis Option are specified more than required. Getting this message is not a concern since it just indicates that the software has already found the most optimum section in previous iteration and thus the section is not changed. See the image below to view the number of repetitions I am talking about.

A1. I looked into your model and I can say that the modelling seems to be fine. The reason for obtainment of the warning message stating the inability to find the performance point is the definition of the 5% damped response spectrum.

You are trying to perform pushover analysis using Procedure-A. Below is the explanation for the procedure adopted by the software for this procedure.

1) Transform the capacity curve into a capacity spectrum. This is done in software by clicking "For Performance Point (FEMA)". Clicking this, the Base shear vs displacement curve is converted into Sa by Sd curve.

2) Develop the 5% damped (elastic) response spectrum appropriate for the site. This is done in the software by clicking onto "Define Elastic Spectrum". Before you click this option the software automatically finds out one response spectrum function for which it shows the performance point.

3) Software selects a trial performance point api, dpi (on the capacity spectrum) and for that a bilinear representation is developed internally. In your model file, you are not getting any performance point for the automatically defined spectrum and not for some other spectrums from the database as well because, the software is unable to find a trial performance point for which the demand spectrum would intersect the capacity spectrum (shown in the image below).

4) Software calculates the reduction factors using which it obtains the demand spectrum. If this demand spectrum intersects the capacity spectrum within acceptable tolerance of 5% then that point is considered to be the performance point. Demand spectrum can be obtained for various inherent damping of the structure (shown in the image below for response spectrum defined as per IS standards and inherent damping of 2%).

The point where the green line (demand spectrum) intersects the yellow line (capacity spectrum) is the performance point.