YT YouTube2Text

YouTube2Text → Transcripts → PLAXIS tutorial for the calculation of modulus of subgrade reaction(ks)

PLAXIS tutorial for the calculation of modulus of subgrade reaction(ks) — Transcript

by puttatt jayarajan · 3,296 words · 600 segments · language en · Watch on YouTube

Open in transcriber Summarize with AI Download SRT/VTT

Full transcript

0:02Hello, a great welcome to this
0:03presentation on a Plaxis tutorial that
0:06demonstrates the calculation of modulus
0:08of subgrade reaction KS.
0:10So, as usual, let me first take you to
0:12the basics of KS.
0:17So, what is modulus of subgrade
0:18reaction? It is simply
0:20a relationship between the soil pressure
0:22and deflection.
0:25The parameter KS is widely used by
0:27design engineers
0:28in the analysis of design of
0:30foundations.
0:32So, mathematically, we can write KS is
0:34equal to Q by delta, where Q is the
0:37applied pressure,
0:38and delta is the associated settlement.
0:42So, remember that KS
0:45is not a constant. It essentially
0:47depends on the load level.
0:49To demonstrate,
0:50here is a shown
0:52the pressure versus settlement curve
0:54that is obtained from
0:55a plate load test.
0:59As expected, you can find that the curve
1:01is non-linear, which implies that
1:04the value of KS depends upon the load
1:07level.
1:08Referring to this plot, we can say that
1:10at lower load levels, we have
1:13higher KS,
1:15and at higher load levels, we have got a
1:17lower KS value.
1:20So, that is the objective of the
1:22presentation is to evaluate the modulus
1:24of subgrade reaction KS for a raft
1:26foundation
1:27supported on a multi-layered soil
1:29deposit using the geotechnical FM tool
1:31Plaxis.
1:34Now, let me provide you
1:35some details on the input parameters.
1:39As usual, the problem consists of a raft
1:41of size 12 m by 8 m
1:43and 0.5 m thick
1:45that is supported on a three-layered
1:47soil deposit
1:49for a depth of 20 m.
1:51The top 9 m is a clay with C equal to 20
1:55kPa, phi equal to 20 degrees,
1:57and a modulus of elasticity ESC to 8,000
1:59kPa. The second one is a medium sand
2:03that extends for another 5 m with a C
2:05equal to 0 kPa, 5° to 35°, and ES is
2:08100,000 kPa. And finally, the third
2:11layer
2:12that extends for 6 m is a silty sand
2:14strata with a C equal to 0 kPa,
2:185° to 28°, and ES is equal to 12,000
2:21kPa.
2:23The other parameters includes
2:25the unit weight of the soil taken for
2:27all the three strata as unsaturated at
2:2916 kN/m³, the saturated weight of 18
2:32kN/m³, and the Poisson's ratio for the
2:35soil is taken as 0.3.
2:37So, now let me directly take you to the
2:39Plaxis presentation.
2:43Okay, fine. So, this is the Plaxis 3D.
2:45So, we'll start defining the project
2:47properties first.
2:48So, we have provided all the title for
2:51the project and it's also saved
2:54uh in the info folder. So, let us start
2:57defining the model dimension.
3:00So, here
3:01the the units are acceptable to us, so
3:03we'll not make any changes.
3:05So, the model dimension that is used for
3:08this project will be 60 m length and 40
3:10m wide.
3:12So, accordingly we'll define X minimum,
3:15X max, and Y minimum and Y max as
3:18So, this is minus 30,
3:23plus 30,
3:26minus 20,
3:30and plus 20.
3:32So, means that we have defined the
3:34contour limits for the model, minus 30
3:36plus 30, and minus 20 plus 20.
3:39Now, we'll proceed to the next.
3:42So,
3:43now we are ready to define
3:45the borehole.
3:48Press this button.
3:51And
3:52we'll create a borehole at the origin
3:54for the model.
3:58So, as already defined, we have three
4:00layers.
4:01So, let us start defining
4:04the levels or the elevations of the
4:06various layers.
4:09So, the first layer, obviously, we know
4:10that it is from
4:120 to -9.
4:15So, this is -9.
4:18And we'll add
4:21one more layer.
4:22That is from -9 to -14.
4:28Add one more.
4:30That is from -14 to -20.
4:34-20.
4:35So, we have defined the elevations of
4:38the various layers.
4:39And here, as
4:41as you can see that currently the water
4:43table is set at the ground level. So,
4:45we'll change it to the bottom of the
4:46borehole, same height to -20, so that
4:48the borehole so that the water
4:51won't be having any effect on the
4:52calculations.
4:54So, now as you can see that here uh
4:56uh all the layers are defined, but the
4:58material is not assigned. So, we are uh
5:01uh now ready to define the materials.
5:03So, we have three soil layers, means we
5:04have to define three soil properties.
5:07Press this button, materials.
5:10So, here as you can see that there are
5:11various types. Here, if we select the
5:14soil and interfaces,
5:16press the new button.
5:19And the top layer, we know that it is a
5:20clay. So, let us name it as a clay.
5:24And we'll use the out of the various
5:26uh
5:27the soil characteristic models available
5:28in Plaxis, we'll use them simpler one,
5:31that's the elastoplastic Mohr-Coulomb
5:32model.
5:34And the drainage type will kept as a
5:36drained.
5:37And here
5:38for all the
5:40soil strata, we'll define the unit
5:42weights as unsaturated as 16
5:4516 and the saturated is 18.
5:5118, okay.
5:52And now we are ready to define the
5:54mechanical properties where we need to
5:56specify the stiffness as well as the
5:58strength parameters.
5:59And the stiffness for the first top
6:01layer
6:02is basically the soft clay.
6:04So, we define this 8,000 KPA.
6:08And the mu mu will keep the same for all
6:11the soil layers.
6:15And now we can proceed for defining the
6:18shear strength parameters. The C value
6:20will set it as a 20 KPA.
6:2420 KPA and the phi value will keep it as
6:2720°.
6:30And the dilatancy is kept as a zero.
6:33And all other parameters will kept as
6:35will keep the same as the interfaces,
6:37the initial conditions. For example, if
6:39you talk about the initial will
6:42use the automatic method for the
6:43calculation of the K0.
6:45So, which means that we have defined the
6:47clay materials and the color is also
6:49assigned.
6:50So, now let us proceed for defining
6:54the second layer that is our medium
6:57sand.
6:58So, we'll identify this as medium sand.
7:01Medium sand.
7:03Okay, we'll use the same Mohr-Coulomb
7:05model.
7:07And the unit weights will
7:09keep it as a 16
7:11and 18.
7:15Now, let us again define the stiffness
7:16as well as the strength parameters.
7:19And this is as a medium sand, so we have
7:22got a
7:23a higher
7:24the stiffness that is almost 100,000
7:28uh KPA.
7:30And we have it as the mu will keep it as
7:33the same 0.30.
7:41And now coming to the C value, since
7:43it's medium sand, we'll keep it as zero.
7:45And the phi, it is 35°.
7:48And the dilatancy, using the standard
7:50formula, we'll keep it as five.
7:53And all our other parameters are
7:54acceptable to us.
8:07So, we have defined the medium sand as
8:10So, now we have one more layer that's
8:12the bottom-most layer, that is the silty
8:16layer. So, let us define a new material
8:18again. So, we'll call it as silt.
8:24So, use our Mohr-Coulomb again.
8:29We use the same unit weights.
8:36Define the stiffness as well as the
8:38shear strength parameters.
8:40The stiffness is 12,000.
8:44And the Poisson's ratio we'll keep it as
8:460.30.
8:49And coming to the
8:52shear strength parameters,
8:53we have a C value of a zero
8:56and a phi value of 28.
9:00And the dilatancy we'll keep it as zero.
9:04So, all other parameters we'll keep it
9:05as default because it is not going to
9:07affect our calculations.
9:09So, that is okay for us. Okay, fine.
9:12Now, what we need to do is that we need
9:14to assign all this So, we have defined,
9:16as you can see, the three layers.
9:20So, this is okay for us. Now, let us
9:22assign start assigning these materials.
9:25So, the top one is clay, we know.
9:28And the second one we know that it is a
9:30medium sand.
9:33And the bottom-most we know that it is
9:36silt.
9:37Silt.
9:39Okay, fine. So, we have defined
9:44all our soil strata. And this is the
9:493D model for the soil as you can see
9:51that there are three
9:53soil layers
9:54each identified by soil mechanical
9:56properties.
9:58Now, we are ready to define
10:00a surface.
10:02Surface means we'll convert this later
10:04into a raft plate.
10:07That's a plate element. So, let us
10:09define
10:10the surface.
10:11So, for the surface, we need to move to
10:14the structure mode.
10:15So, this uh is a create surface, I
10:17think. Yes.
10:19Create surface.
10:20Obviously, we know that our uh
10:23raft foundation is of dimensions 12 m by
10:268 m, which means that this coordinate
10:28should be kept as
10:33-6 and 4 is okay for us.
10:37So, press the button there.
10:39And then this should be +6 and 4.
10:44Okay for us.
10:48Now, we'll keep it as
10:506 and -4. That's okay.
10:54And this again -6 and -4. So, that's
10:57okay for us.
10:59So, as you can see that all the
11:00coordinates are shown on
11:02the right side
11:04and it is acceptable to us.
11:07So, means that we have defined right now
11:10only a surface.
11:11So, now let us
11:15convert this into a first a plate
11:18element.
11:20So, we'll say it's a create plate.
11:23So, fine. So, we have created a plate
11:24out of it.
11:25Now, what we need to do is that we need
11:27to define the raft properties, that's
11:30the mechanical properties, and assign to
11:32this plate.
11:33So, let us proceed again for the
11:35materials here. You can see that we are
11:36right now under the structures.
11:38So, here we can see that there's a icon
11:40called to show materials.
11:42So, let us define a new one.
11:46So, obviously we'll select the plate
11:48here.
11:50And the new one we'll select. And let us
11:53call it as a raft.
11:56And the material type is elastic.
12:00And then we shall go for uh
12:04the mechanical properties.
12:06And the E value,
12:08uh remember that it is a isotropic, that
12:10is all the properties are the same in
12:12both directions.
12:13And this we'll set as 25 E6 KPA.
12:18And the thickness
12:20and this we'll keep it as a 0.20.
12:24And the thickness we'll keep the
12:26thickness of the raft as 0.50.
12:30So, that's okay. So, we defined E value,
12:33we have defined the thickness, also we
12:34have defined the mu value, that's a
12:36Poisson's ratio. So, that's okay for us.
12:39Okay.
12:41Now, we'll select this uh
12:43plate element, and we need to set its
12:46property also.
12:49So, here you can see that there's an
12:51option. We'll keep it as a raft.
12:54So, which means that we have defined
12:57the soil model, we have defined the raft
13:00element. Now, we need to apply a
13:03prescribed displacement. So, here the
13:05philosophy is a simple. We apply a
13:07prescribed displacement,
13:09and then calculate using Plaxis what
13:11will be the load required to cause this
13:14displacement.
13:17So, here you can see that here we have a
13:19created surface of prescribed
13:21displacement. Just press it.
13:24And here if you see on the left-hand
13:25side here, you can see that
13:27a surface displacement
13:30of user is equal to minus one meter is
13:33applied over here. Obviously we will
13:35change this to 0.5 meters during our
13:40uh definition of the so-called
13:43staged construction phases.
13:45So, having defined to all the
13:49basic elements such as a soil as well as
13:50a raft, now we are ready to move over to
13:54the so-called machine.
13:56So, for the machine
13:58uh
13:59uh we'll keep the default mesh
14:01everywhere except the raft.
14:03And where in currently we'll find that
14:05the coarseness factor is 0.5 that will
14:08little bit uh
14:10we'll refine the mesh.
14:17So, we have made it to say 0.25. So,
14:19that's okay for us.
14:21So, let us just
14:23review the so-called
14:26mesh. So, press the generate mesh
14:28button.
14:29So, we'll keep for this problem element
14:31distribution as medium that is good
14:33enough for us. So, that's okay.
14:41So, let us uh
14:45view the generated mesh.
14:58Okay, fine. So, you can see that this is
15:01acceptable to us because considering
15:04that we require
15:05somewhat a finer mesh
15:08uh near the raft and at other locations
15:11this coarser distribution is acceptable
15:12to us.
15:14So, this is how the meshed model look
15:16like. So, let us just close this.
15:19So, having said that,
15:22uh okay, for this problem mesh is
15:23complete and the flow conditions we
15:25don't have any other parameters to
15:27mention over here. We'll
15:29go directly to the stage of
15:31construction.
15:32And here as you can see that the Plaxis
15:34starts with an initial phase wherein it
15:36will compute all the initial stresses
15:38due to the sulfate of the soil. So, just
15:41double click it.
15:42We see whether it is everything is
15:44acceptable to you. Okay, we are using
15:45the calculation type as K0 procedures.
15:48And
15:50the rest of everything is seems to be
15:51okay for us. All the thing that you need
15:53you need to check is that
15:55what are the elements that should be
15:57made active and what are the loading to
15:59be made active corresponding to each
16:01phase. For example, in the initial phase
16:03you can
16:04see that here
16:06all the soils are made active and your
16:08plates as well as the surface
16:11displacements are deactivated because
16:13our purpose in the initial phase is just
16:15to calculate the stresses due to the
16:16sulfate of the soil.
16:18So, which means that we are now ready to
16:21create another phase.
16:23So, phase one. So, just double click it
16:26and here you'll find that we'll start
16:28this phase from the initial phase and we
16:31are using a calculation type known as
16:33the plastic.
16:34And uh
16:36all of the parameters are acceptable to
16:38us.
16:39There is nothing to be changed over
16:40here.
16:41So, that is okay.
16:43And so, only thing what we need to do is
16:45that
16:46during this phase
16:49uh components such as the plates and the
16:51associated surface displacements need to
16:53be made active.
16:55So, just make
16:57this in phase one. So, right now you can
17:00see that we are in phase one. So, keep
17:02the surface as
17:04the plates as that's our foundation as
17:06active.
17:07And also keep
17:09the surface displacements also
17:11activated. If you remember, we have
17:14earlier
17:15applied a load of displacement load of 1
17:18m. Now that we would like to change to
17:24-1 -0.5
17:27so -0.50
17:33Okay.
17:34And that's that must be good enough for
17:36us because uh
17:38the all other um
17:40parameters are
17:41okay for us.
17:43So the model conditions the soil surface
17:45displacements, plates all are active.
17:48So it means that we are now ready to
17:53run the analysis.
17:54So go for this calculate button.
17:58Okay, fine. So we need to select uh
18:00uh one node for the calculations.
18:04So select a points
18:07for monitoring
18:11the load displacement behavior.
18:18Okay, so let us just
18:21go over here line.
18:24So here you can see that uh the point of
18:27interest for the coordinates is taken as
18:29the origin 0 0 0. So let us say our sub
18:32classes to select the various points.
18:43Okay, fine. So you will find that
18:44there's a point with a coordinate of
18:460.030
18:48uh 0.030 that is acceptable to us. So
18:51just select this.
18:53So just update it.
18:56Which means that uh the plots will be
18:59the curves and the plots can be easily
19:00now monitored using this
19:02control point.
19:05So proceed for this calculate button.
19:12Okay, it's a very simple model So, I
19:14think that it do hardly takes uh
19:16uh not more than say 30 40 30 to 45
19:19seconds.
19:22As you can see that the calculations are
19:23in progress.
19:44Okay, fine. So, what does this mean? The
19:46analysis fine.
19:48So, we can straight to go further the
19:50viewing the calculation reports.
19:52So, here uh
19:54the only thing that we would like to see
19:55is the curve for the monitored point.
19:59That is a curve that is a connecting the
20:00force applied to the deformation.
20:03So, we can go straight to
20:06the curves.
20:08Go to manager.
20:09And uh we can go for a new curve.
20:13And here uh along the x-axis obviously
20:15we would like to
20:17plot the displacements. So, we'll select
20:19this control point which we have.
20:21And then uh
20:24we would like to
20:26find the displacement that is the UZ.
20:32UZ and uh
20:36and we will need to
20:38plot this
20:40along the y-axis as a corresponding
20:42force FZ.
20:44Okay.
20:46So, that's okay for us.
20:49So, you can see that here uh this is the
20:51plot. So, if you want I think that uh we
20:54can just invert the axis.
20:58Settings, I would like to
21:02flip the both of
21:04once.
21:06Apply. Okay.
21:08So, obviously, so this is the so-called
21:10uh
21:12the response of the raft foundation
21:14under the prescribed displacement. And
21:17along the x-axis,
21:19uh you'll find that uh it's a
21:20displacement that's 0.1 m, 0.2, 0.3,
21:230.4, 0.5 because 0.5 is the maximum
21:26applied displacement. And along the
21:28y-axis is the corresponding load that
21:30that need to apply it on the raft in
21:32order to cause this displacement.
21:34And for example, here we find that
21:36the load is as high as
21:3945,000 kN.
21:41So, now using this
21:43uh this is the basis curve uh or the
21:46basis
21:47basic response curve that we'll use in
21:50the evaluation of the modulus of
21:53subgrade reaction for the raft
21:55foundation.
21:56So, let us go a little bit deeper into
21:59how the modulus of subgrade reaction is
22:01evaluated from this plot.
22:05Okay, fine. So, this is the response
22:07plot that we obtained from a Plaxis. And
22:10as you can see that along the x-axis is
22:12the settlement,
22:13which is taken up to 500 mm because we
22:15know that
22:16this is the maximum settlement that is
22:19displacement level that is applied to
22:20the Plaxis model.
22:22And along the y-axis is the pressure
22:24kPa.
22:25In fact, Plaxis provide us the
22:28corresponding loads which are converted
22:29into the pressures by dividing these
22:30loads by the area of the raft.
22:33And now we will proceed for the
22:35calculation of the Ks using this
22:37response plot. And this Ks evaluation
22:39will be done at a pressure that is
22:41corresponding to the safe bearing
22:43capacity.
22:45So, now assuming a permissible
22:46settlement of 75 mm, we can directly
22:49read the corresponding SBC from the
22:51graph. That is 75 mm means
22:54from this level. And that evaluates to
22:56be 120 kPa.
22:58Now, let us also obtain
23:00what would be the SBC for the
23:02consideration of ultimate bearing
23:03capacity?
23:06So, from the graph, as you can see that
23:08the response is non-linear. So, what we
23:11do is that we draw two tangents
23:13that
23:14and the point of intersection of these
23:16two tangents represents a transition in
23:18the behavior of the raft foundation
23:20and corresponds to an ultimate bearing
23:22capacity level. And this pressure is
23:25obtained as indicated in the arrows as
23:27275 kPa. Applying a factor of safety of
23:302.5, we evaluate SBC as 275 / 2.5,
23:35that's 110 kPa.
23:37Which means that the SBC SBC, that's the
23:40safe bearing capacity for the design,
23:41will be taken as the lower of this 120
23:44and 110 kPa, that's 110 kPa.
23:47Now, at corresponding to 110 kPa, we
23:49obtain
23:51This is the 110 kPa. We obtain the
23:53corresponding settlement as 68.1 mm.
23:57Which means that at the SBC level,
24:01the value of the modulus of subgrade
24:02reaction works out to be Ks is equal to
24:04110, which is the SBC level, divided by
24:07the associated settlement 68.1 mm.
24:10And that is equal to 1650 kN/m³, so
24:12remember that here the Ks is evaluated
24:14at a pressure level corresponding to
24:16SBC.
24:17In a similar way, if you are proceeding
24:19for the design of the raft foundation,
24:22you can similarly evaluate
24:24the Ks corresponding to the ultimate
24:26bearing capacity as well.
24:28So, that's all regarding this
24:29presentation.
24:31So, as usual, please go ahead and
24:32subscribe to this channel. If you have
24:34any comments, please let me know.
24:36Thank you very much and have a nice day.

About this transcript

This page contains the full transcript of PLAXIS tutorial for the calculation of modulus of subgrade reaction(ks) by puttatt jayarajan, generated from the public captions YouTube serves with the video. The transcript has 3,296 words across 600 segments, with the original timestamps preserved so you can click any line to jump to that moment in the embedded player.

What you can do with it

Use the transcript to take notes, quote the speaker, build a study guide, generate a summary with ChatGPT or Claude via the YouTube Summary tool, or export it as a timed subtitle file with YouTube to SRT. You can also re-open it in the transcriber to choose a caption track available on the video.

Free YouTube transcript tool

YouTube2Text is a free YouTube transcript generator — no signup, no daily limit. Paste any YouTube link and get the full transcript instantly, with timestamps, click-to-jump, caption language selection, AI prompts for ChatGPT, Claude, and Gemini, and exports to TXT, SRT, VTT, or Markdown.