Settlement of foundations can occur due to loads from the superstructure of buildings. Geotechnical engineers are usually interested in seeing that the settlements are uniform and limited to a certain depth. In some cases however, it is anticipated that differential settlement may occur.
The geotechnical engineer has a responsibility of determining the magnitude of the settlement, while the structural engineer investigates the effect of the settlement on the structural members. If the structure is statically determinate, internal stresses are not induced due to differential settlement. However, if it is statically indeterminate, then internal stresses due to the settlement are induced. This is usually treated as a special load case in design of structures, and this is what this article explores.
Example
Let us consider a 6m cantilever that is supported on a roller at the free end. The roller support undergoes a vertical settlement of 25mm (downwards). We are to determine the additional bending moment at the fixed support due to the settlement. We will approach this problem using force and stiffness methods.
Solution (a) By force method
A propped cantilever (roller support) is indeterminate to the 1st order. The adopted basic system for the structure is a simple 6m cantilever. We will now replace the redundant vertical reaction with a unit load, and the resulting bending moment diagram is as shown below.
The appropriate cannonical equation is given by;
δ11X1 + δ1∆ = 0
Where;
δ11 = Deformation at point 1 (free end) due to unit load at point 1
δ1∆ = Deformation at point 1, due to settlement of Support = – EI(∆.S)
On the other hand;
δ1∆ = – EI(∆.S) = -22500 (1 × -0.025) = 562.5
Substituting into the canonical equation and solving;
72X1 + 562.5 = 0
On solving;
X1 = – 7.8125 kN (downward reaction)
With this reaction now known, the moment at support A due to the sinking of support;
MA = – 7.8125 × 6 = -46.875 kNm (hogging moment)
(b) By stiffness method
With support A fully fixed and support B simply supported, the general equation for bending moment at support A due to external load and settlement of support is given by;
MAB = [(3EIθA/L) – (3EI∆/L2)] + FAB
Where FAB is the fixed end moment due to externally applied load = 0
Also θA = 0 since support A is fully fixed.
∆ = 25mm = -0.025m (negative due to downward settlement).
MAB = [((3 × 22500)/6) × (-0.025/6)] = -46.875 kNm
Now, we obtained the answer above expressly because of very known facts. Since the bending moment at B zero, we may not pass through the stress of obtaining the rotation at support B. However for the sake of knowledge, let us apply a more general method which a lot of people might be more familiar with. Here, we will obtain the rotation (slope) at support B, and use it to obtain the bending moment at support A. We are initially assuming that all supports are fixed in the development of the slope deflection equations.
MAB = EI/L[(4θA + 2θB – 6∆/L)] + FAB
MBA = EI/L[(2θA + 4θB – 6∆/L)] + FBA
For equilibrium and compatibility;
MBA = 0
We also know that θA = 0 (fixed support), and there is no external load being considered;
The general arrangement (GA) or structural layout is a drawing that clearly specifies the disposition of the structural elements in a building such as the columns, beams, panelling of the floor slabs etc, on which the design of the structure is based. By looking at the GA of a building, other engineers can identify the model of the building, the shape and type of structural elements, and the possible assumptions made in the design. In order to carry out a design properly, the design engineer should be able to adequately idealise the structure in order to obtain the closest theoretical and practical behaviour of the structure under load. This interpretation is usually made from standard and well-prepared architectural drawings of the proposed building.
The architectural drawings enable the engineer to prepare what is normally referred to as the ‘general arrangement’ of the building, popularly called the ‘GA’ or the ‘Structural Layout’. The GA also contains the labelling of the axes and members, unique grid lines, building structural levels, etc. After completing the general arrangement drawing, the engineer makes preliminary sizing of the structural elements which may be governed by past experience or by deflection requirements based on the code of practice. After the sizing, the engineer is faced with the challenge of loading the structure. But let us briefly review how we go about the GA.
General Ideas on the Preparation of General Arrangement (GA)
There are no hard and fast rules on how to select the appropriate general arrangement of a structure. To the best of our knowledge, adequate presentation of the general arrangement drawing has more to do with the years of design and construction experience.
However, let us highlight some important guidelines which are very necessary.
(1) Respect the architect’s original disposition: In the preparation of a general arrangement drawing, try as much as possible to respect the architect’s design. Architectural drawings supersede structural drawings for building projects, therefore, the structural drawing must conform to the architectural drawing – not the other way around.
For instance, when arranging your columns, do not place columns where the architect desires a free/uninterrupted space, no matter how well the column placement will positively affect the structural behaviour. By implication, none of your structural elements should interrupt the interaction of spaces. In addition to that, your columns and beams should not project out or drop where the architect has intended plain walls or flush ceilings, etc. Your arrangement of the structural elements should be consistent with the original form of the building.
(2) Select a stable model: The model or general arrangement you are adopting should be statically stable, and fully representative of the behaviour of the structure. There should be no stability or equilibrium problem at any joint or location in the building.
(3) Clearly define the load path The load path is the way through which loads of the building travel through the connected parts of the structure before being transmitted to the foundation. In a conventional building, the load path is normally from the floor slab to the beams, then to the columns, and finally to the foundations. All special situations such as the use of tension columns or transfer structures should be carefully planned out.
(4) Consider buildability and construction consequences: The structural model adopted should be buildable. This starts by considering the technical capacities of the contractors who will execute the design from your model. For instance, in a region where reinforcement bending machines and cranes are not readily available, you should not provide models and structural arrangements that will require the provision of high-yield 32mm reinforcement bars, or recommend the use of precast or prestressed elements. This comes in handy by preventing very large spans as practically as possible.
Every construction project has a budget, and your designs should reflect that. For instance, it is very normal to limit the sizes of the reinforcement bars to a maximum of 16mm for simple residential duplex design, unless the building model cannot help it. This can be achieved by moderating the span of the structural elements, knowing full well that the live load in residential buildings are not so heavy.
(5) Know the economic and structural consequences of your arrangement: Between more steel reinforcements and more concrete, which one is more economical in the region? For instance, let us consider an external beam that should have been 8m long under simply supported assumptions. If there are no openings at the wall panels under the mid-span of the beam, you can comfortably hide an intermediate column there, thereby having two spans of 4m each.
The original 8m beam span would have required more steel reinforcements, deeper sections, or both to satisfy ultimate and serviceability limit state requirements. However, if you introduce a column at the mid-span, there will be a redistribution of stresses, with a hogging moment at the propped mid-span, and hence, generally lesser reinforcements, concrete section, and deflection.
However, you should note that you are now going to construct a new column and new isolated base (requiring concrete, reinforcement, and additional excavation cost). Between the two options, which one gave you the most economical solution? Is the cost of constructing a foundation in that area cheap or expensive? Are there groundwater problems, etc? These are all influencing factors, and as highlighted earlier, years of design experience counts.
Case Study on General Arrangement Drawing Preparation
In this article, a small residential building on a 10m x 15m plot of land has been presented for the purpose of preparing the structural general arrangement. The ground and first-floor plans of the building are shown in the figures below. From the architectural disposition, the building is a two-family occupancy arrangement, with each family occupying a floor level. The 1st floor of the building has a balcony in the kitchen area, a cantilever sit-out, and a little balcony by the staircase area. Apart from that, the general arrangement is fairly the same.
Ground floor plan
First floor plan
How to prepare the general arrangement of a building
The architectural drawing of a building can come to a structural engineer in many formats such as hard copy or soft copy (pdf or cad drawing) or both. If it comes as a soft copy of a CAD drawing, then the work is made easier and more accurate. If it comes as a hard copy, then kindly request a soft copy, otherwise, you will have a lengthier job to do.
So let us quickly run down through Structville’s style of preparing the general arrangement drawing of a building, when the drawing comes in AUTOCAD format.
(1) It is advisable to place the ground floor plan side by side with the floor plans of the subsequent storeys on the graphical user interface window of your AUTOCAD (see Figure below).
Note that architectural drawings come with their own unique grid lines. In the building we are trying to design here, some details (like gridlines) have been removed from the architectural drawing for the purpose of clarity of very necessary details like dimension lines. After placing the floor plans side by side on the window, you can start noticing a few different things about the floor plans immediately. For ease of construction setting out, it is important that the architectural and structural drawings have consistent gridline labels.
(2) Copy the floor plans to another location on the window (still leaving the ones you placed side by side for quick reference).
(3) Copy the plan of the first floor, and paste it on the ground floor plan, so that the dimensions and axes are matching perfectly. Just like I hinted earlier, grid lines are unique. Grid line (axis) A-A on the ground floor still represents axis A-A on the 4th floor whether they contain the same elements or not.
HINT: You should choose a prominent corner of the building as a pick-up point for your copy-and-paste operation (this is to make your axis match more properly). For more clarity, you can change the colour or thickness (or both) of the ground-floor elements and first-floor elements and gridlines, so that you can rightly distinguish between the two, but do not mess with the drawing layers on your AUTOCAD during this process.
(4) After you have superimposed the first floor on the ground floor, you should see the interaction between the two floors. All axes with elements that are coincidental should be visible, and all axes with members that are not coincidental should also become visible.
HINT: Architectural drawings can be clouded with lots of details that are not important to a structural engineer. To help you see more clearly and make your decisions faster, you can turn off the layers of irrelevant details such as furniture, sanitary fittings, etc as appropriate. Whenever you get confused, look at the sections of the building for more information, and when the details provided are still not clear, you can contact the architect for more clarification.
At this point, you can also see the outline of blockwork on the first floor, and this can be more critical for the general arrangement of the first-floor slab. It will properly guide you on the selection of the floor beam axis. Engineers usually prefer to have their major block works sit directly on beams unless they cannot help it. This is to prevent major block wall loads from sitting directly on slab panels (note that this can be designed for)
On the other hand, the blockwork axis of the ground floor will aid you in the design of the foundation layout, especially for the strips. Do not work on the ground floor alone without looking at the first floor – the last thing you will want to happen is to place a column somewhere on the ground floor, and helplessly see it popping out through the lobby of the first floor (unless the column will be terminated at the first floor).
So carefully make your selections based on matching axes, and fair uniformity. And as hinted earlier, your arrangement must be consistent with what the architect has in mind. So this is much like art, and you have to use your ingenuity here.
“My supervisor during my industrial training once told me that preparation of structural layout is the MAJOR WORK to do in structural design. Five engineers can prepare the GENERAL ARRANGEMENT of a building using the same drawing, and come up with five different layouts that are plausible. But WHEN EVALUATED critically, some solutions may be better than the others.”
(5) After studying the two floors, the next thing to do is to create a rectangular or square box (say 230 x 230mm) on AUTOCAD, and hatch it with any pattern appealing to you (I normally use SOLID). This represents your columns on the floor plan. Now carefully copy this element and start pasting it at the locations where you have decided to place your columns (this usually occurs at intersections between axes). Personally, I normally start at the corners of the building because more often than not, columns must be there irrespective of the arrangement. After that, you can move to the interiors and place your columns as desired.
(6) After you are satisfied with what you have done, carefully check the interaction of the arrangement, and make sure that they are reasonable. At this point, you can start seeing how your floor beams will connect. Areas, where primary and secondary beams will interact, will now become visible, and this is another stage of critical thinking to see if there are better solutions and alternatives.
Once you connect your floor beams as appropriate, the general arrangement drawing work is basically done. You can now ‘fine tune’ it, and add other relevant details.
Sample thought process in general arrangement creation
To make some points clearer, let us look at a portion of the plan we are considering in this text (see figure below).
We wish to place columns along gridline A.
A little consideration will show that we can place columns at points A1, A3, and A5. Also, we can alternatively place columns at points A1, A2, A4, and A5. Without reading further, ponder on that arrangement and see the alternative that you will prefer.
If I should choose to place a column at points A1, A3, , and A5 (neglecting A2 and A4), these are some of the implications;
(1) I will have a larger span for A:1-3 and A:3-5. But note that the spans are considerably moderate for such RC structure in our case study. (2) I will have a floor beam running along gridline 3 (the beam will probably have to run down to gridline B or C before encountering another support. (3) I may need to have secondary beams along gridlines 2 and 4 to support the wall load above. (4) If I ignore the use of secondary beams along gridlines 2 and 4, the floor slabs on the bedroom and kitchen will be subjected to block wall load from the walls on axis 2 and 4 which must be designed for.
If I should choose the second alternative (placing columns at points A1, A2, A4, and A5);
(1) I will have shorter spans and of course there will be three spans instead of two. The bending moment on Span A:2 – 4 will probably be hogging due to its short span relative to others. (2) I will have a wall along axis 3, but by proximity and considering load sharing implication, it will not be critical, and will not affect my designs like the previous alternative. (3) My floor beam at axis 2 will stop at axis B, and my floor beam at axis 4 will stop at the wall close to axis B. So I will not have a complex arrangement to deal with.
Considering all these consequences, I preferred the second alternative. I feel it gives the building more robustness.
However, if the building is to be located in an area where the soil is so bad that constructing foundations will be very expensive, we can settle for alternative 1 since we want as fewer foundation points as possible. The final GA I adopted for the whole model is shown in the figure below.
However, if the drawing comes in form of a paper work, then nothing changes in the approach. Place the floor plans side by side, and manually make your decisions as highlighted above, before you go into drafting.
I will be dropping my pen here guys. Note that this post is an excerpt from my highly interactive design book called ‘Structural Analysis and Design of Residential Buildings using Staad Pro, CSC Orion, and Manual Calculations’
It is a very affordable piece of information, and if it interests you, contact; E-mail: info@structville.com WhatsApp: +2347053638996
Thank you for visiting today, and keep checking back.
Clause 5.3.2.1 of EN 1992-1-1:2004 covers the calculation of the effective flange width of beams for all limit states. For T-beams, the effective flange width, over which uniform conditions of stress can be assumed, depends on the web and flange dimensions, the type of loading, the span, the support conditions and the transverse reinforcement. The effective width of the flange should be based on the distance lo between points of zero moments, which is shown in the Figure below (Figure 5.2 EN 1992-1-1:2004).
According to EC2, the length of the cantilever, l3, should be less than half the adjacent span and the ratio of adjacent spans should lie between 0.667 and 1.5.
The flange width for T-beams and L-beams can be derived as shown below. The notations are shown in the figure below and above (Figure 5.3 EN 1992-1-:2004).
beff = Σbeff,i + bw ≤ b ———– (1)
where; beff,i = 0.2bi + 0.1lo ≤ 0.2lo ———– (2) and beff,i ≤ bi
Solved Example Consider the floor plan shown below; We are going to calculate the flange width of the various floor beams in the general arrangement. All floor beams are 230mm x 450mm
External Beams (1) Beams A:1-3 and D:1-3
b1 = (6000 – 230)/2 = 2885 mm (there will be no b2 since it is an L-beam) lo = (0.85 × 5000) = 4250 mm (assumed point of zero moment)
beff,1 = 0.2bi + 0.1lo = 0.2(2885) + 0.1(4250) = 577 + 425 = 1002 mm < (0.2 × 4250 = 850 mm) Therefore take beff,i = 850 mm
beff = Σbeff,i + bw ≤ b = 850 mm + 230 mm = 1080 mm < 3000 mm Therefore effective flange width = 1080 mm
Comparing this answer with BS 8110-1:1997 (clause 3.4.1.5) beff = lz/10 + bw
Where lz is the distance between points of zero moment (take as 0.85L = 4250 mm) in this case. beff = 4250/10 + 230 = 655 mm
(2) Beams 1:A-B
b1 = (5000 – 230)/2 = 2385mm (there will be no b2 since it is an L-beam) lo = (0.85 × 6000) = 5100 mm (assumed point of zero moment)
beff,1 = 0.2bi + 0.1lo = 0.2(2385) + 0.1(5100) = 477 + 510 = 987 mm < (0.2 × 5100 = 1020 mm) Therefore take beff,i = 987 mm
beff = Σbeff,i + bw ≤ b = 987 mm + 230 mm = 1217mm < 2500 mm Therefore effective flange width = 1080 mm
Comparing this answer with BS 8110-1:1997 (clause 3.4.1.5) beff = lz/10 + bw
Where lz is the distance between points of zero moment (take as 0.85L = 5100 mm) in this case. beff = 5100/10 + 230 = 740 mm
(3) Beams 1:B-C
b1 = (5000 – 230)/2 = 2385mm (there will be no b2 since it is an L-beam) lo = (0.7 × 6000) = 4200 mm (assumed point of zero moment)
beff,1 = 0.2bi + 0.1lo = 0.2(2385) + 0.1(4200) = 477 + 420 = 897 mm < (0.2 × 4200 = 840 mm) Therefore take beff,i = 840 mm
In our last post, I highlighted the different methods of loading sub-frames such as to obtain the maximum design moment on the columns (Follow the link below to see it). In this post, I am going to show with a practical solved example how to analyse sub-frames for obtaining the maximum moment.
We are going to pick column A1 as a case study, to obtain the design moment in the z direction. Due to the location of this column (corner column), we are going to consider one span (Beam A:1-2) only fully loaded at ultimate limit state 1.35gk + 1.5qk.
In our previous post following the link above, this has been determined as 41.752 kN/m .
Therefore, the loading is as shown below;
We know that the preliminary dimensions of the columns are 230 x 230mm, while the beams are 230 x 450mm. However, we know that the beams are technically not rectangular sections (they are L- beams) by virtue of their location. Therefore, we have a little job of calculating the beams flange width, and moment of inertia. Note that at this stage, we do not consider the effect of reinforcement in determining the moment of inertia, and the main aim of this process is to determine the value of k in the picture above so that our calculation will be as accurate as possible.
For an L-beam, the flange width is as given below;
Beff = 0.2(5770/2) + 0.1 (0.85 × 50000 + 230 = 1232mm
To learn how to calculate flange width, click here;
Therefore, the section is as shown below;
We can use our knowledge of statics to find the moment of inertia of the shape above. I believe there are charts that can help us compute this faster based on the bw/beff and h/hf ratio or thereabout. But let us just go through this the old school way.
We can now calculate the moment of inertia using the parallel axis theorem;
We have now gotten the moment of inertia for the beam
We can quickly verify that for the columns, the moment of inertia can be obtained by;
IC = bh3/12 = (230 × 2303)/12 = 2.332 × 108 mm4
Summarily, we can now say that;
k = IB/IC = (3.4074 × 109)/(2.332 × 108) = 14.611
Having obtained our k, the diagram of the structure is as shown below. But because we are actually overestimating the stiffness of the structure by assuming all ends to be fixed, we have to reduce this value by half. Therefore k/2 = 7.3055
We are going to tackle this problem using the stiffness approach. Kinematically, the structure is indeterminate to the first order, being the rotation at node 1.
We are going to fix the node up, so as to restrain it from any form of rotation. This forms the basic system of the structure. Now, we will apply a unit rotation at the node, and the resulting bending moment on the basic system is as given below;
The appropriate cannonical equation is given by;
k11Z1 + k1P = 0
The bending moment at point 1 due to unit rotation at point 1 (k11) is given by;
k11 = (4EI/3) + (4EI/3) + (29.222EI/5) = 8.511
k1P = fixed end moment due to external load at that point = -ql2/12 = (-41.752 × 52)/12 = -86.983 kNm
8.511Z1 = 86.983
On solving;
Z1 = (10.22/EI) radians
Therefore the moment at the top of the column;
M1B = (10.22/EI) × (4EI/3) = 13.627 kNm
So by implication, the ground floor column A1 is subjected to a moment of 13.627 kNm in the z-direction at the top. Can you model this structure on Orion or Staad and see what your answer will be?
You can also follow the same procedure and obtain the design moment in the x-direction.
Just in case you are not very familiar with stiffness method of structural analysis, this post below will help.
Previously, we have made a submission on how to calculate column axial loads by considering beam support reactions. Just in case you missed it, you can read it by following the link below. But this post highlights the methods of loading columns in a building in order to obtain the maximum design moment.
Columns in buildings may be axially, uniaxially, or biaxially loaded. An axially loaded column is a column that is subjected to axial load only. This type of column usually exists in the centres of a building where bending moments from floor beams will normally neutralise each other. When a column is subjected to a bending moment in one direction, then it is uniaxially loaded, and when it is subjected to bending in both directions, it is biaxially loaded.
Column A1 is a biaxially loaded column because the load from beam 1:A-B is generating a bending moment in the x-direction, while the load from beam A:1-2 is generating a moment on the column in the z-direction. This moment is usually in the form of a fixed end moment, which is finally distributed to the other members meeting at that node based on their stiffness.
Column B1 is a typical example of a uniaxially loaded column. The bending moments from beams 1:A-B and 1:B-C will typically neutralise each other, while the load from beam B:1-2 will exert a bending moment in the z-direction. Typical examples of axially loaded columns in the building are columns B2 and C2.
When a structural design is to be carried out manually, the design engineer has to determine the bending moment on columns as appropriate, alongside the axial loads.
However, while it has been pointed out above that column B1 is uniaxially loaded due to the beams on gridline 1-1 neutralising each other, have you considered a situation whereby PANEL 1 is fully loaded with stored materials, while PANEL 2 is empty?
This situation is very possible during the service life of the structure, and as a result, the column at that instance will not behave as uniaxially loaded, but as biaxial. This is just to give us a little idea of what may influence our loading while carrying out such an analysis. The consideration to be made during designs is usually the worst possible load regime.
Reynolds and Steedman (2005) gave us a good idea of how to approach some aspects of this issue. Loading all the spans at the ultimate limit state (say 1.35gk + 1.5qk) will seldom give us the maximum design moment. However, alternating the loads (say one span 1.35gk + 1.5qk and the other span 1.0gk) will give us an unbalanced moment, which is a possible worst scenario in practice.
So I am going to give us a rundown of the methods that we can use to load structures (sub-frames) in order to determine the maximum bending moment on columns.
Method 1: Full Subframe
Let us consider the image above showing the sub-frame of a building with the intent of determining the maximum design moment of the lower column at joint C. What we have to do is to load 1.0gk and (1.35gk + 1.5qk) on the spans adjoining span C such that the unbalanced moment is maximised.
In this case, it is preferable to apply 1.35gk + 1.5qk on the longer span which is CD, and 1.0gk on the shorter span BC. While this is more representative, the setback is that you will have to solve a 5 x 5 simultaneous equation using the stiffness method of structural analysis before arriving at your answer and hence may not be very handy for a simple scientific calculator process.
Method 2
In this case, the sub-frame is simplified as shown above, but it works best if span BC is longer than the adjoining spans. If span BC is shorter than the adjoining span, you can switch to the use of method 3. Here, we apply a load of 1.35gk + 1.5qk on span BC, and 1.0gk on span AB and CD.
We only solve a 2 x 2 simultaneous equation when using this method, and it is very convenient for classroom and simple design purposes. Realise also that we have to reduce the stiffness of the adjoining beams by half because we are actually overestimating the actual stiffness of the beam by considering all ends to be fully fixed.
Method 3
In this case, we are going to load (1.35gk + 1.5qk) on the longer span, and 1.0gk on the shorter span such that the unbalanced moment at C is maximum. We are still going to reduce the stiffness of the beam by half for the same reasons stated in method 2 (see how to calculate the stiffness of beams).
The advantage of this method is that we are going to solve for just one unknown (rotation at point C) in order to obtain the design moment in question. This is the most popular approach that is taught in most classrooms and utilised for simple designs or checks. I have written some MATLAB programs for carrying out such analysis.
After loading the column, we can now analyse the structure completely by using the stiffness method (my recommendation). An example of this has been done in the next post, and you can take a look by clicking the link below;
Grillage analysis presents a sufficiently accurate method of analysing bridge decks for estimation of design bending moment, torsion, shears force etc, and has been adapted for use in most computer software around the world. Basically, the grillage analogy method uses the stiffness approach for analyzing the bridge decks (Jaggerwal and Bajpai, 2014). The whole bridge deck is divided into a number of longitudinal and transverse beams (planar grids).
In grillage analysis, the elements of a grid are assumed to be rigidly connected, so that the original angles between elements connected together at a node remain unchanged. Both torsional and bending moment continuity then exist at the node point of a grid (Qaqish et al, 2008). The method has proved to be reliable and versatile for a wide variety of bridge decks. Conceptually, the grillage analogy method attempts to discretize the continuous or dispersed stiffness of the bridge and concentrates it into discrete longitudinal and transverse members.
Fig 1: Equivalent grillage of a bridge deck (Source: Kalyanshetti and Bhosale, 2015)
According to Shreedhar and Kharde (2013), the basic steps in carrying out grillage analysis are as follows;
Idealization of the physical deck into equivalent grillage
Evaluation of equivalent elastic inertia of members of grillage
Application and transfer of loads to various nodes of grillage
Determination of force responses and design envelopes and
Interpretation of results.
Some basic guidelines in carrying out grillage analysis are as follows;
Gridlines are placed along the centre line of the existing beams, if any and along the centre line of the leftover slab, as in the case of T-girder decking.
Longitudinal gridlines at either edge are placed at 0.3D from the edge for slab bridges, where D is the depth of the deck.
Gridlines should be placed along lines joining bearings.
A minimum of five grid lines are generally adopted in each direction.
Gridlines are ordinarily taken at right angles.
Gridlines in general should coincide with the centre of gravity of the section. Some shifts, if it simplifies the idealisation, can be made.
Over continuous supports, closer transverse grids may be adopted. This is so because the change is more dependent upon the bending moment profile.
For better results, the side ratios i.e. the ratio of the grid spacing in the longitudinal and transverse directions should preferably lie between 1.0 to 2.0.
Methodology
While it is easy to visualise a grillage model of a bridge deck, there must be a careful thought process especially when you want to model on software like StaadPro. In this post, we are going to carry out a grillage analysis for the bridge deck model shown below.
The intermediate transverse beams are assumed to be 700mm deep, and 300mm wide. The profile of the bridge can be assumed to be of the form shown below;
To adequately present the grillage analogy of the bridge deck, we will discretize the deck as shown in the image below. If you have AUTOCAD, it is advisable to carefully map out your grids and verify all dimensions and sections before proceeding to Staad Pro for modelling. In the image shown below, I have used AUTOCAD to show the disposition of my gridlines, and the sections they are representing are given in the legends that follow.
The legend that gives the description of the lines is given below;
Having known these, we can now go over to Staad Pro for the modelling;
Step 1: Launch Staad and identify the model as a space structure
Step 2: Create the nodes on the plan view to reflect the general arrangement
Step 3: Connect all the nodes as adequate
This lands us somewhere here as shown below. Make sure you utilise the ‘connect along’ command for speed of execution.
Step 4: Assign supports
Since only the main girders will be sitting on bearings, we will model this by placing one support on a pin, and the other on a roller. You can obtain roller support by using the ‘enforced but’ command. This gives you the option of releasing other support constraints and leaving only the vertical reaction. What we obtain is shown below.
Step 5: Create properties and assign them to where they belong
By looking at the legends above, you can create the section properties, and assign them to where they belong. If any section is not available in Staad by default. You just have to input the section properties manually. Note that for the slab sections, you can use the rectangle option to define it, we have a T-beam by default in Staad, while we will have to define the L-section manually.
An example for the T-section model is shown below;
Step 6: Apply the traffic loads and analyse
In grillage analysis, uniformly distributed loads are applied on the main girders, while tandem loads are better applied as moving loads. For instance using the Eurocodes, the tandem load of 300kN is applied on notional lane 1 (150 kN per axle at disatance of 1.2m) with a width of 2.0m. You can define the moving load as you wish based on the code of practice you are using. Staad automatically distributes this load as appropriate to the members.
For our model above, this only has been applied (other loads have been neglected for brevity), and deflection of the grillage at different locations is as shown in the images below;
The deflection profile when the load is at another location is as shown below;
In such a similar way, you can view the internal forces when the load is at different locations. You can verify that we are interested in the location of the load that will produce the worst effect on the structure.
N/B: Some dummy members from the initial model were removed.
So we will stop this post here for now, see you later…..
References
[1] Jaggerwal H. And Bajpai Y (2014): Effects of skewness on Three Span Reinforced Concrete T-Girder Bridges. International Journal of Computational Engineering Research. ISSN(e): 2250-3005 Vol 04 Issue 8 pp 1-9 [2] Qaqish M., Fadda E., Akawwi E. (2008): Design of T-beam Bridge by Finite Element Method and AASHTO specification. KMITL Science Journal Vol 8 No 1 (January- June 2008) pp24-34 [3] Shreedhar R., Rashmi Kharde (2013): Comparative study of Grillage method and Finite Element Method of RCC Bridge Deck. International Journal of Scientific & Engineering Research Volume 4, Issue 2, February-2013 1 ISSN 2229-5518 [4] Kalyanshetti M.G., Bhosale (2013): Alternative Forms of Two Lane T-Beam Bridge Superstructure – Study by Grillage Analogy. International Journal of Latest Trends in Engineering and Technology (IJLTET) Vol. 5 Issue 3 ISSN: 2278-621X pp 88-96
Overhead steel tank is one of the most adopted water storage system due to lots of reasons. The speed and ease of construction, flexibility in supporting different types of storage facilities, and diverse options are some of the most profound reasons. Furthermore, it can be easily dismantled and removed unlike their concrete counterpart that will offer more difficulties if it were to be removed at a later time.
This post succinctly explains the process of carrying out the analysis of overhead tank using the popular Staadpro software. The tank to be modelled is as shown below. The stanchions are braced at intervals of 2.5m, and forms a square profile of 3m x 3m in plan. There are two primary beams sitting directly on top of the stanchions, and six secondary beams sitting on the primary beams. The secondary beams receive the tank directly, and are spaced at distance of 1m centre to centre at the middle span. The cantilever area has 0.8m projection (see image below). Note that this arrangement is just to briefly show modelling steps. I am convinced this structural arrangement could be made better especially when you look at the final result. That is what structural engineering is all about, you can treat this as a preliminary analysis that has been carried out, prior to final and more refined modelling, analysis and design.
Step 1: Create your nodes to form the base and the location of the stanchion.
To make this step easier, change your view to plan view. To switch to plan view, click on the icon circled in red at the top left corner of the screenshot below. Make sure that your UCS icon is as shown in the bottom left corner of the screen shot. You can create the first node, and use the copy and paste command to create the other nodes.
Step 2:Create other nodes to capture the exact geometry of the frame
Using the ‘copy and paste’ command can make this step very fast for you. The ‘translational repeat’ command can also be used. For instance to create the next level using the copy and paste command, you can highlight the entire nodes, right click, select the copy command, and paste in the positive Y direction.
At the end of the process, you should be able to obtain something like this as shown below. It may not be so clear looking at the 3D view, but I hope it makes some sense to you.
Step 3:Connect the nodes to represent the form of the structure
You can use the ‘connect geometry’ command (go to geometry, and select ‘connect along’) but note that the members to be connected must be selected or highlighted. Alternatively, you can use the ‘add beam’ command.
Step 4:Assign the member properties
The properties that we are going to assign is as follows;
Stanchions – UB 254 x 102 x 25
Tie beams – UB 127 x 76 x 13
Top beams – UB 203 x 102 x 23
Bracings – UA 50 x 50 x 6
(Note that for practical purposes, it is very advisable to keep the variability of sections to a minimum).
To create properties, click on general, go to properties, and click on section database (circled on the screen shot below). Select the properties above, and click add.
Assign the properties to the designated members. Your rendered view should look as shown below;
Step 5: Model the steel tank
We want the frame shown above to carry a cube steel tank of dimension 3m. This can be achieved by using the mesh tool. We are going to use a steel plate of thickness 15mm to model the square tank. Just in case the steel frame is going to carry plastic tanks as usually obtainable in Nigeria, you can model the floor plate only, and apply pressure load on the plate to represent the weight of the tank and water.
As you can see, the storage tank has been modelled. You can notice from the 3D rendering above that the centroid of the plates coincide with the centroid of the main beams. We can model it to look as realistic as possible by offsetting the beams such the plates will rest on the beams (the way it appears in construction). However, the only difference is that Staad will see this as a form of geometric imperfection, and high magnitude of unrealistic axial forces will be induced in the beams. But for all design and practical purposes, leaving it this way is very okay, and gives a representation of the loads transferred so that appropriate sections can be selected.
However, we know that the primary beams are not receiving the loads from the tank directly. The load from the tank goes first to the secondary beams, which now transfer the loads to the primary beams, and to the stanchions. You can achieve this by merging the primary beams where necessary.
Step 6: Support
Assign pinned support to the structure. This shouldn’t be challenging to you at all.
Step 7: Loading
The height of the main storage tank is 3m, but the maximum water level is 2.8m before it is discharged through a spill way.
The loading of the structure is as given below;
(1) Hydrostatic pressure on the walls of the tank (trianular load of 28 kN/m2)
(2) Full pressure load of 28 kN/m2 on the bottom of the tank
(3) Dynamic wind pressure of 2.5 kN/m2 acting on one side of the tank
(4) A live load of 1.5 kN/m2 acting on the exterior perimeter of the tank (maintenance area)
(4) Self weight of structural members.
We will treat the weight of water as dead load in this case, since we know the maximum level it can get to.
Step 8: Analysis and results
When the structure is analysed, the following results are obtained;
Stanchions: Maximum Axial loads:
Dead load – 82.2 kN (compression)
Live load – 4.39 kN (compression)
Wind load – 36.3 kN (compression) and 36.3 kN (tension)
Maximum bending moment
Dead load – 0.73 kNm
Live Load – 0.33 kNm
Wind Load – 4.21 kNm
Main beams Bending Moment
Dead load = 28.6 kNm
Live load = 0.913 kN
A sample bending moment of the structure is as given in the image below;
The deflected profile of the structure due to wind load is given below;
After such analysis, you can take your time to view the results of your model. This was just to show some salient approaches that can be used to model overhead tank. I will keep improving on the post terms of delineating the steps, and reporting the results as appropriate as time goes bye……
Your reactions are highly welcome.
See you later……..
Raft foundation is a type of shallow foundation that is mostly used on soils of low bearing capacity, where the foundation pressures need to be spread over a large area. They are also used in areas where the foundation soils are of varying compressibility and the foundation has to bridge over them. The geotechnical design of raft foundation ensures that there is no bearing capacity failure and that the settlement is kept to a minimum, while the structural design ensures that adequate thickness and reinforcements are provided to avoid structural failure of the raft plate.
The types of raft foundation commonly encountered in practice are;
A flat raft slab can be adopted and analysed using the ACI empirical method called the direct design method (DDM) when there is a symmetrical arrangement of spacing of columns and loadings that leads to uniform ground pressure. The DDM method involves analysing the flat raft foundation as flat slab (similar to the analysis and design of an inverted roof flat slab).
When there is uniform ground pressure but with unequal spans which do not satisfy the conditions for DDM, a flat slab can be adopted, but it has to be analyzed by the equivalent frame method (EFM). However, beam and slab rafts are generally preferred in such cases.
A beam and slab raft construction should be adopted when there is a symmetrical arrangement of unequally loaded columns but with uniform ground pressure.
On soils of very low bearing capacity, the loads on the raft can be compensated by excavation of soil, and the raft designed as a cellular raft.
When there is non-symmetrical loading with the centre of gravity of loads not coinciding with the centre of gravity of the area, the ground pressures will vary from place to place. In such cases, it is advisable to adopt a beam and slab construction for such cases. To some extent, the space around the loading area should be adjusted to give the least eccentricity. The slab should then be designed for the maximum pressures. The beam loading must be for the maximum values of the varying pressure in its location.
When the bearing capacity is satisfactory but there is excessive settlement, some parts of the load on the raft can be relieved by installing a few piles so that raft settlement can be reduced. This is known as piled raft foundation.
Raft foundations can be analysed using the rigid approach or flexible approach.
The conventional rigid combined footing approach is a method of analysing raft foundation using simple statics without any consideration of the elastic properties of the raft and the soil and their interaction. Here, the raft is analysed as a large beam member independently in both directions. The row of column loads perpendicular to the length of the beam is coupled together in single column load. Then for these column loads acting on the beam, the upward soil pressure is calculated and the moments and the shears at any section is determined by simple statics. Hence, the moment per unit width of the raft is determined by dividing the moment values by the corresponding width of the section (Gupta, 1997).
In order to obtain the upper bound values of the stresses, the raft is divided into strips bounded on the centre line of the column bays in each direction. Each of these strips is then analysed as independent combined footing by simple statics. Using the column loads on each strip the soil pressure under each strip is determined without reference to the planar distribution determined for the raft as a whole.
The eccentricity of the load and the pressure distribution below the raft which is considered to be linearly varying are taken into account in this analysis.
Approximate Rigid Analysis of Flat Raft Foundation with Eccentricity
In theory, the flat slab analysis is restricted to uniformly distributed load. However, if the eccentricity is small, we may proceed as follows. We adopt the following steps to analyze such a flat slab:
Step 1: Check eccentricity of the resultant load and find ex and ey with service loads.
Step 2: Find Mx and My with factored loads of all loads about XX- and YY-axes to calculate the varying base pressures.
Step 3: Find the moment of inertia of the raft slab in the x-x direction (Ixx) and y-y direction (Iyy).
Step 4: A flat slab has to be analyzed in frames formed by cutting the slab along mid-spans in the XX- and YY-directions. Find the ground pressures due to the eccentric loading at the end and mid points of the slab at the top and along the bottom.
Step 5: Check whether the maximum pressure exceeds the safe bearing capacity or not.
Step 6: Take each strip along YY. As the pressure in the strip varies, find the average pressures over the top edge and also at the bottom edge. (This average pressure for the interior frame will be pressure along the column line.) Draw the load diagram with the column loads P1, P2, etc. and base reaction in the strip along its length.
Step 7: Check for balance of downward and upward forces. If they do not match, modify the ground pressures for equilibrium.
Step 8: Draw the shear force and bending moment for the strip for design. As the loads from the top and the base pressure from below are known, we may use any of the following:
(a) Simple statics or neglecting colulmn loads used (b) Direct design Methd (DDM), or (c) Equivalent Frame Method (EFM) (we may use the transverse distribution percentage in detailing of steel).
(As a quick and short procedure, we first find the pressures along the boundary lines of the spans take and design each span for the average pressure in the span. We will also assume for simplicity that the load from the apron below the peripheral column is transmitted directly to the beams.)
Step 9: Distribute the moments using transverse distribution.
Design Example of a Flat Raft Foundation
It is desired to design the raft slab shown below to support the column loads for a building as given below. All columns are 230 x 450mm, the grade of concrete (fck) is 30 N/mm2, and the yield strength of the reinforcement (fyk) is 500 N/mm2. The allowable bearing capacity for the supporting soil is 60 kN/m2.
The service loads on the columns are given in the table below;
Total axial load = 18296 kN
Eccentricity along the x-direction This is obtained by taking moment about grid 5 x = [22.4(770 + 1050 + 776 + 350) + 16.8(870 + 1450 + 860 + 660) + 11.2(1211 + 1850 + 1000 + 779) + 5.6(875 + 1400 + 865 + 660)] / 18296 = 11.258 m
ex = x – (L/2) = 11.258 – (22.4/2) = 0.058 m
Eccentricity along the y-direction This is obtained by taking moment about grid D y = [15.1(770 + 870 + 1211 + 875 + 770) + 9.1(1050 + 1450 + 1850 + 1400 + 1050) + 3.1(776 + 860 + 1000 + 865 + 700) ] / 18296 = 7.8045 m
Width of strip = 4 m Length of strip = 24.4 m Taking the average load = (38.7 + 49.975) / 2 = 44.3375 kN/m2 Summation of upward force = (44.3375 × 4m × 24.4m) = 4327.34 kN
Check; 4496 kN – 4327.34 kN = 169 kN Now the average ground pressure can be increased by [169 kN/(4m × 24.4m) = 1.73 kN/m2] which is 44.3375 + 1.73 = 46 kN/m2. This has been however ignored in this analysis.
Multiplying by the width of the strip = 44.3375 kN/m2 × 4m = 177.35 kN/m On factoring at ultimate limit state = 1.37 × 177.35 = 242.9695 kN/m
Using simple statics and analysing as a continuous beam as shown below, the bending moment and shear forces along the strip can be determined.
This analysis is repeated using the same procedure for all the strips in the raft foundation.
Structural Design Check for punching shear at column perimeter We can determine the thickness of the raft slab by considering the punching shear at the column perimeter. At the column perimeter, the maximum punching shear stress should not be exceeded.
VEd < VRd,max
Where; VEd = βVEd /u0d
VRd,max = 0.5vfcd
Considering Column B3 bearing the maximum axial load, VEd = 1.37 × 1850 kN = 2534.5 kN β = 1.15 (an approximate value from clause 6.4.3(6)) u0 = column perimeter = 2(230) + 2(450) = 1360mm v = 0.6[1 – fck/250] (strength reduction factor for concrete cracked in shear) v = 0.6[1 – 30/250] = 0.528 fcd = αccfck/γc fcd = (1.0 × 30)/1.5 = 20 N/mm2 VRd,max = 0.5 × 0.528 × 20 = 5.28 N/mm2 VEd = (1.15 × 2534.5 × 1000) / (1360 mm × d)
Therefore; 2914675/1360d = 5.28 On solving; dmin = 405.89 mm
The basic control perimeter for punching shear check is normally taken at 2d, but when the concentrated force is resisted by high pressure as can be found in foundations, the punching control perimeter is taken at less than 2d.
Let us consider a trial footing depth of 700mm.
Effective depth (d) = 700 – 70 – (20/2) = 620 mm (concrete cover is taken as 70mm and assumed diameter of bar is 20mm)
Minimum area of reinforcement = 0.0013bd = 0.0013 × 1000 × 620 = 806 mm2/m (clause 9.2.1.1(1)) Also provide 18H16 @ 225mm c/c TOP (ASprov = 893 mm2/m or 3216 mm2) along the strip.
Beam shear Check critical section d away from column face considering the highest force in the shear force diagram. The point of contraflexure for shear between column A1 and A2 is 2.315 m.
The shear force at d from column A2 (towards the left) can therefore be calculated using similar triangle. The length d from column face is 0.15 m + 0.62 = 0.77 m (width of column is 230 mm). VEd = 611 kN vEd = VEd/bd = (611 × 1000) /(4000 × 620) = 0.246 N/mm2
vRd, c = CRd, c × k × (100 × ρ1 × fck) 0.3333 CRd, c = 0.12 k = 1 + √ (200/d) = 1 + √ (200/620) = 1.568 ρ = As/bd = 3216/(4000 × 620) = 0.00129 vRd, c = 0.12 × 1.568 × (100 × 0.001296 × 30)0.333 = 0.295 N/mm2 Vmin = 0.035k(3/2) fck0.5 Vmin = 0.035 × (1.568)1.5 × 300.5 = 0.376 N/mm2 => vEd (0.246 N/mm2) < vRd,c (0.376 N/mm2) beam shear is ok
Punching Shear (Use column A3) Punching shear: Basic control perimeter at 2d from face of column vEd = βVEd/uid < vRd,c
β = 1, a = 2d = 2(620) = 1240 mm Perimeter length, u = 2(c1 + c2 + π × a) c1 = 230 mm; c2 = 450 mm ui = 2(230 + 450 + π × 1240) = 9151 mm
VEd = load minus net upward force within the area of the control perimeter) Area inside the perimeter A = c1 × c2 + 2 × (c1 + c2) × a + π × a2 = 0.23 × 0.45 + 2 × (0.23 + 0.45) × 1.24 + π × 1.242 = 6.62 m2 Average earth pressure at ultimate limit state = 1.37 x 44.3357 = 60.733 kN/m2 (note that the SLS actual soil pressure under the column is44.288 kN/m2). Since this is meant to beneficial, the lowest value can be used.
VEd = 1.37 × 1211 kN = 1659.07 kN
VEd,red = 1659.07 – (60.733 x 6.62) = 1257 kN vEd = (1257 x 103)/(9151 x 620) = 0.221 N/mm2
This shows that shear is ok. Using this approach, the entire reinforcement for the mat foundation can be obtained in the longitudinal and transverse directions.
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A combined footing is a type of shallow foundation where a common base is provided for two closely spaced columns. When a common base in form of a strip is provided for three or more rows of columns, it is better described as a continuous footing. Combined footings are normally employed in cases where two or more columns are closely spaced such that their individual pad footings would overlap each other. They can also be used where property boundaries would not permit the design of separate bases.
The design of a combined footing involves the geotechnical and structural design, where the proper proportioning of the dimensions, thickness of the base, and proper reinforcement is provided. The foundation of a combined footing must not undergo excessive settlement or shear failure, and the footing itself must be strong enough to resist the bending moment and shear forces as a result of the superstructure load. The thickness of combined footings is usually governed by shear force considerations.
In the design of combined footings, the centre of gravity of the two column loads should as practically as possible coincide with the centre of area of the base. The base should preferably be rectangular in plan and symmetrically disposed about the line of loads. If it is not practicable to proportion the base as described above, the load will be eccentric, and the centre of pressure of the higher ground pressure will have the same eccentricity relative to the centre of the base.
If the base is thick enough, the pressure distribution diagram will be trapezoidal for eccentric loads, but uniform for concentric loads. If the base is relatively thin, the pressure distribution will be variable, with the maximum pressure occurring directly under each column.
The most common method of design of combined footings is the rigid approach, which makes the following assumptions.
The footing is infinitely rigid, and therefore, the deflection of the footing does not influence the pressure distribution.
The soil pressure is distributed in a straight line or a plane surface such that the centroid of the soil pressure coincides with the line of action of the resultant force of all the loads acting on the foundation.
However, flexible method can also be used for the design of combined footings, taking into account soil-structure interaction.
Solved example on the design of combined footing
Two (300 x 300)mm square columns spaced at a distance of 2.45 m c/c are loaded as shown below. The foundation is founded on soil of a bearing capacity of 125 kN/m2. It is desired to design the footing to satisfy all requirements using the concrete grade of 30 N/mm2 and steel of yield strength 500 N/mm2. The concrete cover is 50mm.
Combined footing
At serviceability limit state; PEd = 1.0Gk + 1.0Qk Total service load on column P1 = 665 + 122 = 787 kN Total service load on column P2 = 825 + 145 = 970 kN Total load on both columns = 787 + 970 = 1757 kN
Assuming 10% of the service load to account for the self-weight of the footing; Sw = 0.1 × 1757 = 175.7 kN
Area of footing required = Total service load/Allowable bearing capacity = (1757 + 175.7)/125 = 15.46 m2
Adopt a rectangular base = 6.5m x 2.5m (Area provided = 16.25 m2)
To locate the centroid of the footing, let us take moment about column P1; (970 × 2.45) – 1757x = 0 On solving, x = 1.352m from column P1
Let the projection of the footing from the lighter column be L1 Therefore; L1 + 1.352m = (Total length of footing)/2 = 6.5/2 = 3.25 m Therefore, L1 = 3.25 – 1.352 = 1.898 m (say 1.9 m) Hence L3 = 6.5 – (1.9 + 2.45) = 2.15 m
Alternatively, the footing can be proportioned using a straight-forward formula based on the calculations done above;
The final disposition of the footing is given below assuming a trial footing depth of 600 mm.
We can check the preliminary thickness of 600 mm by considering the punching shear at the column perimeter. At the column perimeter, the maximum punching shear stress should not be exceeded.
Considering Column P2 bearing the maximum axial load, VEd = 1331.25 kN
β = 1.5 (an approximate value from clause 6.4.3(6) of EN 1992-1-1:2004) u0 = column perimeter = 2(300) + 2(300) = 1200 mm v = 0.6[1 – fck/250] (strength reduction factor for concrete cracked in shear) v = 0.6[1 – 30/250] = 0.528 fcd = αccfck/γcfcd = (1.0 × 30)/1.5 = 20 N/mm2 VRd,max = 0.5 × 0.528 × 20 = 5.28 N/mm2
VEd = (1.5 × 1331.25 × 1000) / (1200mm × d)
Therefore; 1996874/1200d = 5.28
On solving; The minimum thickness of footing dmin = 315.263 mm
Hence the depth provided exceeds the minimum required.
Earth pressure intensity at ultimate limit state = (1080.75 + 1331.25)/16.25 = 148.43 kN/m2
For a width of 2.5m, q = 148.43 × 2.5 = 371.075 kN/m
Bending moment at column P1 = (371 × 1.92)/2 = 669.65 kNm Bending moment at column P2 = (371 × 2.152)/2 = 857.47 kNm
The maximum bending moment in the span will occur at the point of zero shear; Mx = 371x2/2 – 1080.75(x – 1.9) = 185.5x2 – 1080.75x + 2055.425 ∂Mx/∂x = 371x – 1080.75 = 0 On solving; x = 1080.75/371 = 2.91 m
The shear force just to the left of column P1 = (371 × 1.9) = 704.9 kN The shear force just to the right of column P1 = 704.9 – 1080.75 = -375.85 kN The shear force just to the left of column P2 = -375.85 + (371 × 2.45) = 533.1 kN The shear force just to the right of column P2 = 533.1 – 1331.25 = -798.15 kN
As an be seen, there is no hogging moment in the combined footing.
Bottom Reinforcement We are supposed to take our maximum moment from the column face which can be easily determined as shown below.
MEd = 857.47 kNm Effective depth (d) = 600 – 50 – 10 = 540 mm (assuming 20mm bars). b = 2000mm
k = MEd/(fckbd2 ) k = (857.47 × 106)/(30 × 2500 × 5402 ) = 0.039
Since k < 0.167 No compression reinforcement required z = d[0.5 + √(0.25 – 0.882k)] = z = d[0.5 + √(0.25 – (0.882 × 0.039)] = 0.95d
Top reinforcement There is no hogging moment on the footing as shown in the bending moment diagram. However, we can provide minimum reinforcement at the top.
Check for shear Clause 6.4.2(2) of EN 1992-1-1:2004 states that control perimeters at a distance less than 2d should be considered where the concentrated force is opposed by a high pressure (e.g. soil pressure on a base), or by the effects of a load or reaction within a distance 2d of the periphery of the area of application of the force.
Beam shear Check critical section d away from the most loaded column face VEd = 148.43 x (2 – 0.54) x 2.5 = 541.7 kN vEd = (541.7 x 103)/(2500 x 540) = 0.401 N/mm2
vRd, c = CRd, c × k × (100 × ρ1 × fck) 0.3333 CRd, c = 0.12 k = 1 + √ (200/d) = 1 + √ (200/540) = 1.608 ρ = 4082/(540 × 2500) = 0.00302 vRd, c = 0.12 × 1.608 × (100 × 0.00302 × 30)0.333 = 0.402 N/mm2 => vEd (0.401 N/mm2) < vRd,c (0.402 N/mm2) Beam shear ok
Punching Shear Punching shear: Basic control perimeter at 2d from face of column vEd = βVEd/uid < vRd,c
β = 1, ui = (300 x 4 + 540 x 2 x 2 x π) = 7986 mm
VEd = load minus net upward force within the area of the control perimeter) VEd = 1331.25 – 148.43 x (0.302 + π x 1.0802 + 1.080 x 0.30 x 4) = 581.6 kN vEd = (581.6 x 103)/(7986 x 540) = 0.1348 N/mm2
Punching shear resistance of the section VRd,c = [CRd,c.k.(100ρ1 fck)(1/3) ] × 2d/a ≥ (Vmin × 2d/a) a = 540 + 540 = 1080 mm
Introduction
Indeterminate trusses are analysed usually analysed using force method or direct stiffness method. In this post, we are going to analyse step by step, the analysis of the truss loaded as shown below.
(EA = Constant)
Solution
Step 1: Determine the degree of static indeterminacy
For a truss to be determinate;
m + r = 2j
Where;
m = number of members = 13
r = number of support reactions = 4
j = number of joints = 8
Therefore;
13 + 4 – 2(8) = 1
Therefore, the truss is indeterminate to the 1st order at the supports.
Step 2: Select redundant and remove constraint
To solve for the single degree of indeterminacy, the structure has to be reduced to a statically determinate and stable structure. This can be achieved by removing a redundant support, and a little consideration will show that for the structure to be stable, only a horizontal redundant will have to be removed. In this case, let us remove the horizontal support at H. This will give us the basic system that is given below.
Step 3: Analyse the basic system completely and obtain the internal forces
We now have to obtain the support reactions and internal forces using the principles of statics.
Step 4:Calculate the deformation at the redundant
Having obtained the internal forces at the redundant, we can now use virtual work method to calculate the horizontal translation at support H which corresponds to the removed redundant Hx. This is done by removing the external load on the basic system and applying a unit horizontal force in the direction of the removed force.
By looking at what is happening at the above structure, you will agree with me that the entire top chord is in a uniform compression of -1.0 kN.
Hence;
FAB = FBD = FDF = FFH = -1.0 (Compression)
All other members of the truss have zero forces.
Since the horizontal displacement at H in the original structure is equal to zero, it means that the horizontal reaction at support H must produce a deflection in the opposite direction that will counter the deflection.
Deformation of the basic system due to externally applied load can be obtained by using the relationship below;
Where;
m = number of members
n = Internal forces due to virtual load
N = Internal forces due to externally applied load
L = Length of member
A = Cross sectional area of member
E = Modulus of Elasticity of member
To analyse it, we normally present this in tabular form. Due to the fact that the force in the members in the virtual load state are zero in all members except the top chord, we will just focus on the top chord.
δ1P = 2(65.08/AE) + 2(74.96/AE) = 280.08/AE
The deflection at point H due to the virtual load can be obtained using the same relationship, and this is shown in the table below;
δ11 = 4(4/AE)) = 16/AE
The appropriate cannonical equation is therefore given by;
