Pile caps are concrete mats that rests on piles with adequate rigidity to transfer the column loads to the piles. Piles are provided as alternatives to shallow foundations when a firm and stable soil to carry column load is too deep below the surface, or when high lateral load is anticipated. More often than not, pile caps are usually so rigid that they make the entire group of piles to behave like one unit.
EN 1992-1-1:2004 permits us to use strut and tie models to analyse structures where non-linear strain distribution exists (e.g. pile caps, deep beams and corbels).
In strut and tie models, trusses are used with the following components:
• Struts (concrete)
• Ties (reinforcement)
• Nodes (intersections of struts and ties)
Eurocode 2 gives guidance for each of these.
Design Example
A 500mm x 500mm column is carrying an ultimate limit state load of 2140 kN. We are to design the pile cap using the following data;
Grade of concrete fck = 30 N/mm2
Fyk = 500 N/mm2
Concrete cover = 75mm
Spacing of pile = 1800mm
Diameter of piles = 600mm
Solution
The strut and tie model is as given below;
Angle of inclination of strut γ = tan-1 (1100/900) = 50.710°
cos γ = 0.633
sin γ = 0.7739
Since pile spacing is less than three times pile diameter, the bars may be spread uniformly across the cap. Check for Shear
Consider the critical section for shear to be located at 20% of the pile diameter inside the pile cap.
Distance of this section from the column face;
av = 0.5(Spacing between piles – width of column) – 0.3(pile diameter)
av = 0.5(1800 – 500) – 0.3(600) = 470 mm
Length of corresponding perimeter for punching shear
u = 2(900 + 1440) = 4680mm
Perimeter of pile cap = 2(900 + 2700) = 7200mm
Since the perimeter of the pile cap is less than 2u, normal shear extending across the full width of the pile cap is more critical than punching shear.
The contribution of the column load to the shear force may be reduced by applying a factor β = av/2d, where 0.5d ≤ av ≤ 2d
But a little consideration will show that av(470 mm) < 0.5d(550 mm), therefore, take av as 0.5d (550)
Therefore β = 550 / 2(1100) = 0.25
v = βV/bd
V = 1070 KN + (Self weight of pile cap/2)
Self weight of pile cap = 1.35(25 × 2.7 × 0.9 × 1.2) = 98.415 KN v = (0.25 × 1119.2 × 1000)/(2700 × 1100) = 0.0942 N/mm2
v = V/ud v = (2140 × 1000)/(2000 × 1100) = 0.972 N/mm2
VRd,max = 0.2(1 – fck/250)fck
VRd,max = 0.2 (1 – 30/250)30 = 5.28 N/mm2
This shows that the punching shear around column perimeter is ok.
Extra
This pile cap has been modelled on Staad Pro and the results obtained are given below.The set back in the model is the piles were modelled as columns that could sway under the load. What do you think?
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1.0 Introduction
The fin plate connection is very popular in the construction industry due to its ease of erection, and the absence of shared bolts in two sided connections. It consists of a length of plate welded to the supporting member (eg column or primary beam), and the supported member bolted to the fin plate. Technically, fin plates derive their rotational capacity from shear deformation of the bolts, and hole distortion in the fin plate and/or the beam web.
2.0 Practical Considerations and Recommended Geometry
With 10mm thick fin plate in S275 steel, 8mm fillet weld to the supporting member will guard against any possibility of weld failure (CSI, 2011). Fin plates may be classified as short or long as follows;
Short tp/zp ≥ 0.15
Long tp/zp < 0.15
Where zp is the distance between the face of the support and the first line of bolts. For short fin plates, erection on site is usually more difficult, but with long fin plates, care must be taken against lateral torsional buckling, especially if the beam is laterally unrestrained.
According to SCI (2011), when detailing the joint, the following recommendations should be followed;
(1) Full strength fillet welds are provided
(2) The fin plate is positioned close to the top flange in order to provide positional restraint
(3) The depth of the fin plate is at least 0.6 times the supported beam depth in order to provide the beam with adequate torsional restraint
(4) The thickness of the fin plate or the beam web is ≤ 0.50d (for S275 steel)
(5) Property class 8.8 bolts non-preloaded are used
(6) All end and edge distances on the plate and the beam web are at least 2d
3.0 Solved Example
A secondary beam UKB 305 x 165 x 40 is supported on a primary beam of UKB 533 x 210 x 101 with an ultimate shear force of 100.25 kN. We are to design a fin plate connection for this joint using grade 8.8 non preloaded bolts.
Initial Considerations
Since the depth of beam is less than 610mm, let us adopt fin plate size of 100 x 10mm
Gap gh = 10mm
Bolt rows = 3
Length of plate = 0.6 × 303.4 = 182.04mm (provide plate length = 220mm)
Right now, we wish to check the result obtained from the manual analysis above with result from Prokon software. This present design has been carried out according to BS 5950:2000, and in this post, we will highlight all the input variables that are necessary to carry out the design, as well as the results.
Axial Load on Column
Characteristic force from permanent action Gk = 620 kN
Characteristic force from leading variable action Qk = 132 kN
At ultimate limit state;
N = 1.4Gk + 1.6Qk
N = 1.4(620) + 1.6(132) = 1079.2 kN
Column Dimensions and Properties(UC 203 x 203 x 60)
Depth (d) = 209.6mm
Width (b) = 205.8mm
Thickness of web (tw) = 9.4mm
Thickness of flange = (tf) = 14.2mm
Root radius (r) = 10.2mm
Perimeter of section = 1206.4mm
Area of section = 76.4 cm2
Base Plate Bedding Details
Strength of concrete/grout = 25 N/mm2
Design Using Prokon
Step 1: Getting stared
Launch the Prokon Software, go to ‘Connections Module’, and select ‘Base Plate’. The window below comes up.
This is a simple interactive interface that allows you to carry out your design in a very simple manner. As you input the design variables, the graphical representation of what you are doing comes up on the window, so that you will remain properly guided.
Step 2:Verify Design Code
At the upper left hand side of the screen, go to ‘File’, and make sure that the design code is set to BS 5950
Step 3:Select the Section Size
At the column dialog box at the upper left hand side of the screen, check the universal column symbol (see screenshot above). Now, go to the drop down list and select the appropriate column section. In this example, we are dealing with UC 203x203x60.
Step 4:Input Dimensions
As a guide, the area of base plate required should be equal to or greater than;
N/0.6fcu = (1079.2 x 1000)/(0.6 x 25) = 71946.667 mm2
It is always a good practice to allow at least 100mm from the edge of the column to the base plate.
Let us adopt base plate 350mm x 350mm (Aprov = 122500 mm2).
Now input the base plate dimensions as given below;
You will discover that the drawing below automatically appears on the window;
What this requires is for you to input the offset, in order to position the column on your desired location on the base plate. However, we want our column to centralised, and Prokon gives us that option by the click of a button. Go to the area where you input loads, and by the side of the table, click on ‘Centralise Column’. This brings the column to the centre of the base plate, and the offset value is calculated automatically for us. See the new image below;
Now, by imputing the rest of the variables, we can position the bolts to suit the design appropriately. We are offsetting the bolts 35mm from the edge of the plate. See the input and the result below;
Step 5: Define materials properties (general parameters)
From the dialog box below, define the properties of the materials to be used in the construction.
We are saying no to the use of studs because we want the column load to be transferred directly to the concrete bedding and not to the bolts. The grade of concrete and yield stress of steel selected is as given above.
Step 6: Define the loads
Since we have already factored our loads, we can input the ultimate load directly and set the load factor as unity (1.0). Alternatively, we can input the dead load and live load separately with their appropriate factor of safety, but note that they must be presented as the same load case. See below;
Step 7: Design
If you have gotten to this point without any errors, we can then click on ‘Design’ function, and the window below comes up with the design completely done;
When the plate is unstiffened, the design result above holds good, with base plate thickness of 18mm.
When the plate is stiffened, the design result below holds good, which also includes the thickness of the stiffeners, and weld designs. The plate thickness in this case reduced to 10mm.
Step 8: Detail Drawing
Click on drawing, and you will be able to see the shop drawing for the design. The images below gives the detailing for the unstiffened and stiffened plates respectively.
Conclusion
I think the result above needs no interpretation. The details are quite crystal clear and I must say that it is very impressive. It is very interesting to note that when this same base plate was designed manually using EC3 (unstiffened), we obtained a thickness of 18.447mm, and then provided a base plate of 20mm. This shows reliability of both methods for all practical purposes. You can read the manual design below just in case you missed it.
1.0 Introduction
Industrialization is one of the major keys to development and sustainable economy. Industrial structures are usually very easy to identify because of their unique features that are quite different from residential or commercial buildings. Engineers are often tasked with analysing and designing industrial frames, and to simplify the analysis for manual calculations, they are usually idealised as 2D plane frames.
2.0 Solved Example
In this post, a gable industrial frame structure is subjected to a load regime as shown below. The frame is hinged at point D, and it is desired to obtain the internal forces (bending moment, axial forces, and shear forces) due to the externally applied load.
N/B: In practical construction, it is not advisable to hinge the structure at the apex due to the problem of excessive deflection and stability of the frame.
Solution
(a) Support Reactions
Geometrical Properties
Angle of inclination of the rafter
θ = tan-1(2.5/6) = 22.619°
cos θ = 0.923
sin θ = 0.385
Length of rafter (z) = sqrt(62 + 2.52) = 6.5m
Notations: Example: NCR = Axial Load at point C, just to the Right
(b) Internal Stresses Section A – BB (0 ≤ y ≤ 4.0m)
(i) Bending moment
My = -Ax.y = -17.939y
At y = 0; MA = 0
At y = 4.0m; MBB = (-17.939 × 4) = -71.756 kNm
(ii) Shear Force
Qy = ∂My/∂y = -17.939 kN
(iii) Axial Force
Ny + Ay = 0
Ny = -Ay = -81.917 kN (compression)
Section 1 – BR (0 ≤ x ≤ 1.0m)
(i) Bending moment
Mx = -25.x
At x = 0; M1 = 0
At x = 1.0m; MBR = (-25 × 1) = -25 kNm
(ii) Shear force
Qx = ∂Mx/∂x = -25 kN
(iii) Axial Force
Nx – 7kN = 0
Nx = 7kN (tension)
Section BUP – CB (4 ≤ y ≤ 5.5m)
(i) Bending moment
My = -Ax.y + (25 × 1) – 7(y – 4)
At y = 4m; MBUP = -(17.939 × 4) + 25 = -46.756 kNm
At y = 5.5m; MCB = -(17.939 × 5.5) + 25 – 7(1.5) = -84.1645 kNm
(ii) Shear force
Qx + Ax + 7kN = 0
Qx = -17.939 – 7 = -24.939 kN
(iii) Axial force
Ny + Ay – 25kN = 0
Ny = -81.917 + 25 = -56.917 kN (compression)
Section CR – DL (0 ≤ z ≤ 6.5m) (Pay careful attention here)
(i) Bending moment
The bending moment transferred transferred to the rafter at node C is -84.1645 kNm
Summation of vertical force transferred ∑V = Ay – 25 = 81.917 – 25 = 56.917 kN
Summation of horizontal force transferred ∑H = Ax + 7 = 17.939 + 7 = 24.939 kN
Coming from the right Section G – FB (0 ≤ y ≤ 4.0m)
(i) Bending moment
My = -Gx.y = -24.939y
At y = 0; MG = 0
At y = 4.0m; MFB = (-24.939 × 4) = -99.756 kNm
(ii) Shear Force
Qy = ∂My/∂y = +24.939 kN (note that we are coming from right to left)
(iii) Axial Force Ny + Gy = 0
Ny = -Gy = -79.083 kN (compression)
Section 2 – FR (0 ≤ x ≤ 1.0m)
(i) Bending moment
Mx = -16.x
At x = 0; M2 = 0
At x = 1.0m; MFR = (-16 × 1) = -16 kNm
(ii) Shear force
Qx = ∂Mx/∂x = -16 kN
(iii) Axial Force
Nx – 0 = 0
Nx = 0 (no force)
Section FUP – EB (4 ≤ y ≤ 5.5m)
(i) Bending moment
My = -Gx.y + (16 × 1)
At y = 4m; MFUP = -(24.939 × 4) + 16 = -83.756 kNm
At y = 5.5m; MEB = -(24.939 × 5.5) + 16 = -121.1645 kNm
(ii) Shear force
Qx – Gx = 0
Qx = 24.939 kN
(iii) Axial force
Ny + Ay – 16kN = 0
Ny = -79.083 + 16 = -63.083 kN (compression)
Section EUP – DR (0 ≤ z ≤ 6.5m)
(i) Bending moment
The bending moment transferred transferred to the rafter at node C is -121.1645 kNm
Summation of vertical force transferred ∑V = Gy – 16 = 79.083 – 16 = 63.083 kN
Summation of horizontal force transferred ∑H = Gx = 24.939 kN
Construction experts in Lagos Island, Nigeria are no strangers to pile foundation. The poor soil profile in the area means that foundations are more economically founded at depths way beyond the weak upper soil strata. Faseki et al (2016) carried out assessment of sub-soil properties of some parts of Lekki Peninsula, Lagos.
The result of their findings form the basis of this post, which aims to explore the design of pile foundations on sand, using Eurocode 7, and Lekki, Lagos State Nigeria, as case study. The summary of their findings regarding geotechnical properties of the soil in the area is given in the Table below;
The reports from soil investigation in most parts of Lekki indicate poor bearing capacity for shallow foundations. In a research work by Warmate and Nwankwoala (2019), an average bearing capacity of 55 kN/m2 was reported within 1m depth of the soil using direct shear analysis and SPT. However, in an independent soil investigation report carried out within Abijo Village, Ibeju Lekki, fibrous peat was observed within 0 – 5m depth of the area, necessitating the adoption of pile foundation in such areas.
Theoretical Background of Pile Foundation Design
When the static formulae is applied in the geotechnical design of piles, the ultimate load carrying capacity is calculated from the soil properties obtained from site investigation. This tends to be a more natural approach since it is the soil that actually carries the load, but more often than not, pile load tests are employed for validation of results. The ultimate capacity of piles is assumed to be reached when;
Qu = Qb + Qs + Ws ——- (1)
Where: Qu = Ultimate capacity of pile Qb = Base resistance = quAb Qs = Shaft resistance = fuAs Ws = Weight of displaced soil
Quite often, the weight of the pile and the weight of the soil removed is assumed to be equal, such that the last term of the equation is eliminated.
qu = ultimate value of the resistance per unit area of the base due to shear strength of the soil fu = ultimate value of the tangential force per unit area of the shaft due to adhesion and skin friction As = surface area of the pile shaft Ab = base area of the pile L = Length of pile below ground.
Meyerhof (1951) expressed the base resistance per unit area as;
qu = cNc + ksPo‘Nq + 0.5BγNγ ———(2)
In the case of a pile of normal proportions, L/B is about 30 or more, the term containing B in the above equation is so small that is usually ignored. The above equation now reduces to;
qu = cNc + ksPo‘Nq———-(3)
Where; ks = the coefficient of earth pressure on the shaft within the failure zone Nc, Nq = Bearing capacity factors dependent on the embedment ratio and angle of internal friction
In a soil where both adhesion and friction can be mobilised on the pile shaft, Meyerhof (1951) expressed the tangential force per unit area as;
fu = Ca + ksPo‘tanδ ——— (4)
Where; Ca = adhesion per unit area δ = angle of friction between the soil and the pile material For a purely cohesive soil such as clay, δ is zero and for non-cohesive soil such as sand, Ca is zero.
Pile Foundation in Sand
Knowing full well that in purely cohesionless soil (sand), fu = ks‘Po‘tanδ. What this equation means is that the skin friction continues to increase linearly with increasing depth. But for piles founded on sand, it has been determined that the overburden pressure of the soil adjacent to the pile does not increase without limits. When a certain depth of penetration is reached, the overburden pressure remains more or less constant, and this is called the critical depth Dc.
The critical depth depends on the field condition and the dimensions of the pile. It is generally agreed that at a penetration depth between 10 and 20 pile diameters, a peak value of skin friction is reached which cannot be exceeded at greater penetration depth (see figure below). Therefore, the equation fu = ks‘Po‘tanδ gives increasingly unsafe values as the penetration depth increases and exceeds about 20 pile diameters. The assumptions usually made for piles as reported by Wrana (2015) are;
Pile in loose sand (10D)
Pile in medium dense sand (15D)
Pile in dense sand (20D)
Figure 1: Critical Depth For Pile Founded on Sand (Picked from Wrana, 2015)
Therefore, for piles in a homogenous dense sand for example; Qu = Qb + Qs = qu.Ab + fu.As
Total skin friction = fu.As = ks‘tanδ × Area of PV diagram × Circumference of pile Area of PV diagram = 0.5γ.Dc2 But Dc = 20B qu = Pv. Nq = γ.Dc
The working load or design load for all pile types is equal to the sum of the base resistance and the shaft friction divided by a suitable factor of safety.
Pile Design to Eurocode 7 (EN 1997-1:2004)
EN 1997-1 clause 7.4(1)P states that the design of piles shall be based on one of the following approaches:
(1) The results of static load tests, which have been demonstrated by means of calculations or otherwise, to be consistent with other relevant experience, (2) Empirical or analytical calculation methods whose validity has been demonstrated by static load tests in comparable situations, (3) The results of dynamic load tests whose validity has been demonstrated by static load tests in comparable situations, (4) The observed performance of a comparable piles foundation, provided that this approach is supported by the results of site investigation and ground testing.
Pile Resistance from ground parameters
According to clause 7.6.2.3(8) pf EC7, the characteristic base and shaft resistances may also be determined directly from the ground parameters using the following equations;
Rb;k = Ab qb;k Rs;k = ∑As;i qs;i;k
Where; qb;k is the characteristic unit base resistance qs;i;k is the characteristic unit shaft resistance
The design compressive resistance of a pile Rc,d may be obtained either by treating the pile resistance as a total resistance;
Rc,d = Rc;k / γt (where Rc;k = Rb;k + Rs;k)
or by separating it into base and shaft components Rb;k and Rs;k using the relevant partial factors, γb and γs
Rc,d = [Rb;k / γb + Rs;k / γs]
The partial resistance factors in the UK National Annex have been modified to take account of the type of pile and whether the serviceability behaviour is to be determined either by load test or a rigorous and reliable calculation (Raison 2017).
The table below gives the partial factors for bored piles (culled from Raison, 2017)
The equilibrium equation to be satisfied in the ultimate limit state design of axially loaded piles in compression is Fc,d ≤ Rc,d .
Exampleof Pile Foundation Design for Lekki Peninsula
A structure to be built at Lekki Peninsula, Lagos has a group of piles to be subjected to a characteristic permanent load (Gk) of 665 kN and characteristic variable load Qk of 275 kN each. It has been decided to use piles of diameter 600mm, and soil investigation report from Faseki et al (2016) is to be used for the design. The design involves determining the length of embedment of the piles.
Solution
For details of this calculation and approach, go to clause 7.6.2.3 of EN 1997-1:2004.
Let us make an initial trial of 20m depth.
For this particular design, we will ignore the effect of shaft resistance on the first and second layers. Our interest is to determine the appropriate length of penetration into the fourth layer or possibly fifth layer. We are assuming that the water table lies at 1m below the ground.
For simplicity, let me summarise the properties of the layers below as reported by Faseki et al (2016);
First layer (3m thick) – Loose to Medium Silty Sand, Cu = 0, ϕ = 32°, γ = 18.5 kN/m2 Second layer (1.5m thick) – Soft silty clay, Cu = 24 kN/m2 , ϕ = 28°, γ = 15 kN/m2 Third layer (4.5m thick) – Very loose sand, Cu = 0, ϕ = 30°, γ = 18.5 kN/m2 Fourth layer (11m thick) – Medium dense sand, Cu = 0, ϕ = 33°, γ = 19.5 kN/m2
The decision to ignore the shaft (friction) resistance of the first two layers is conservative. I am not very confident about the behaviour of the 1st layer, and the clay layer (2nd layer) sandwiched between the upper and lower sandy formations. But a little consideration shows that the 3rd and 4th layer gives more confidence for design purposes.
(1) Calculation of the Base Resistance Base resistance Rb;k = qb:k.Ab = (Po‘)(Nq)(πD2/4)
From Faseki’s report, the angle of internal friction at layer of interest φ = 33°
N/B: Table A.4 of EC7 recommends that we apply material factor of safety to some soil properties, but UK Annex to EC7 does not recommend application of partial safety factors except in the case of negative skin friction.
Values of bearing capacity factor Nq according to NAVFAC DM 7.2(1984), is given below;
From the table above; Nq at the receiving layer = 17 (pile will be bored) Effective pressure at the base of the pile = (18.5 × 3) + (15.5 × 1.5) + (18.5 × 4.5) + (19.5 × 11) = 376.5 kN/m2 Pore water pressure = 10kN/m3 × 19m = 190 kN/m2 Effective Stress at foundation depth = 376.5 – 190 = 186.5 kN/m2
Base resistance = qb:k × Ab = (186.5 × 17) × (π × 0.62/4) = 896.553 kN
(2) Shaft Friction We will ignore the shaft resistance of the first two layers as earlier stated, but typical calculation values are shown.
fu.As = ks‘ tanδ × 0.5Po‘ × πDL
Typical values for δ and ks‘ as suggested by Broms (1966) is given in the table below.
Therefore, Design Approach 1 Combination 1 (DA1.C1) is more critical, and pile length will need to extend beyond 20m length. Therefore from Design Approach 1, Combination 1, the length of 20m is rejected.
Design Approach 2
Combinations of sets of partial factors DA2 —– A1 + M1 + R2
Hence Length of 20m is rejected using Design Approach 2
Conclusion
We have seen from the design that the pile length of 20m at a diameter of 600mm is inadequate to support the design load specified. An alternative is to increase the diameter of the pile, but the reader should pay close attention to the assumptions made in arriving at the ultimate values.
The probable shaft resistance of the rejected two layers was shown, so that it could serve as a guide for judgement and further decision making. However, the closeness of the answers offer very intellectually stimulating ideas about the assumptions made However, I will also be happy if I can get an alternative design based on the soil properties, which might include the SPT and CPT values reported for the various layers.
As a matter of fact, it took a lot of time to arrive at some decisions especially regarding the implementation of the Eurocode 7. I more or less agree with the conclusions of Raisons (2017) regarding Eurocode 7 as follows;
EC7 does not tell the Designer how to design piles but does give rules and procedures to be followed
EC7 has complicated pile design with the introduction of numerous partial factors; load factors, combination factors, material factors, resistance factors, model factors and correlation factors
More design effort is required in EC7, and ultimately,
Engineering judgement cannot be suspended
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References [1] Wrana B. (2015): Pile Load Capacity – Calculation Methods. Studia Geotechnica et Mechanica, Vol. 37, No. 4, 2015 DOI: 10.1515/sgem-2015-0048 [2] Faseki O.E., Olatinpo O.A., Oladimeji, A.R. (2016): Assessment of Sub-Soil Geotechnical Properties for Foundation Design in Part of Reclaimed Lekki Pennisula, Lagos, Nigeria. International Journal of Advanced Structures and Geotechnical Engineering ISSN 2319-5347, Vol. 05, No. 04, October 2016 [3] Raison Chris (2017): Pile Design to BS EN 1997-1:2004 (EC7) and the National Annex. Raison Foster Associates, University of Birmingham, UK [4] NAVFAC DM 7.2 (1984): Foundation and Earth Structures, U.S. Department of the Navy. [5] EN 1997-1:2004:Geotechnical design – Part 1: General rules, European Committee for Standardization [6] Meyerhof, G.G. (1951): The ultimate bearing capacity of foundations, Géotechnique, 2, 301-332 [7] Warmate T. and Nwankwoala H. O. (2019): Geotechnical Indications and Shallow Bearing Capacity Analysis within Lekki Peninsula, Lagos using Direct Shear Analysis. Cur Trends Civil & Struct Eng. 1(4): 2019. http://dx.doi.org/10.33552/CTCSE.2019.01.000516
1.0 Introduction
In the design of structures, engineers are always faced with the task of carrying out interpolations as supported by various codes of practice. These interpolations are often encountered when we are carrying out wind load analysis, designing our columns, designing our two-way slabs, etc.
Some people may not be very familiar with the process of linear interpolation, and this is the problem this post attempts to address, showing how to use Gregory-Newton forward difference formular for carrying out interpolations. I am not going to go through the process of deriving the formular, but I will present it exactly the way it is applied, and use practical examples to consolidate it.
2.0 Gregory-Newton Forward Difference Formular
Let x0, x1, x2,…, xn be equally spaced values, so that xi = x0 + ih, for i = 1, 2,…, n.
If the values f0, f1, f2,…, fn are known, where fi = f(xi), for some function f.
The Gregory–Newton forward difference formula is a formula involving finite differences that gives an approximation for f(x), where x = x0 + ph, and f(x) ≈ f0 + pΔf0 gives the result of linear interpolation. For each pair of consecutive function values f(x0) and f(x1), the forward difference is obtained by subtracting f(x0) from f(x1). However, when the series is terminated after one more term provides an example of quadratic interpolation.
In this post, we are considering linear interpolation only, and cases where it can be applied.
Quickly, let us use the table below to show how this process is carried out. The relationship between the values is fairly linear.
For the table given below, we are required to calculate the value of f(3.5).
Note that the value we seek (3.5) is between 3 and 4
So; x0 = 3; x1 = 4; xp = 3.5
h = x1 – x0 = 4 – 3 = 1
p = (xp – x0)/h = (3.5 – 3)/1.0 = 0.5 Δf0 = f(x1) – f(x0) = 59 – 47 = 12
Therefore; f(3.5) = 47 + 0.5(12) = 53
This one is just as simple as that.
Now let us go into practical scenarios encountered in structural design.
3.0 Applications in Structural Design Example 1 Calculation of bending moment coefficient for two-way slabs
Table 3.14 of BS 8110-1:1997 gives coefficients that are used to obtain the bending moment coefficients of rectangular slabs subjected to uniform pressure under various support conditions. This same coefficient is also used in the Eurocodes, and an excerpt is given below.
Let us assume that we have a slab with one short edge discontinuous with an aspect ratio of 1.46. We wish to obtain the positive bending moment coefficient for the mid-span by using linear interpolation. x0 = 1.4; x1 = 1.5; xp = 1.46
h = x1 – x0 = 1.5 – 1.4 = 0.1
p = (xp – x0)/h = (1.46 – 1.4)/0.1 = 0.6 Δf0 = f(x1) – f(x0) = 0.043 – 0.041 = 0.002
Therefore; f(1.46) = 0.041 + 0.6(0.002) = 0.0422
I guess that was very simple and straightforward.
Example 2 External Wind Pressure coefficient on Buildings
Table 7.1 of EN 1991-1-4:2004 gives the recommended values of external pressure coefficients for buildings that are rectangular in plan. Design engineers usually encounter this table when carrying out wind load analysis. The code permits us to determine intermediate values by interpolation.
Let us assume that we are to determine Cpe,10 for a building in zone A, having a h/d ratio of 2.3 x0 = 1; x1 = 5; xp = 2.3
h = x1 – x0 = 5 – 1 = 4
p = (xp – x0)/h = (2.3 – 1)/4 = 0.325 Δf0 = f(x1) – f(x0) = -1.2 – (-1.2) = 0
Therefore; f(2.3) = -1.2 + 0.325(0) = -1.2
Example 3 Compressive Strength of Steel to BS 5950-1:2000
Table 24 of BS 5950-1:2000 gives values for compressive strength of steel members. We are also permitted to obtain intermediate values through interpolation.
The slenderness ratio of a steel stanchion is 17.5, the grade of the steel is S275, and the thickness of the section is less than 40mm. So we are to interpolate from the table above to obtain the compressive strength. x0 = 15; x1 = 20; xp = 17.5
h = x1 – x0 = 20 – 15 = 5
p = (xp – x0)/h = (17.5 – 15)/5 = 0.5 Δf0 = f(x1) – f(x0) = 271 – 275 = -4
Therefore; f(2.3) = 275 + 0.5(-4) = 273 N/mm2
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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.
