The provision of partitions on suspended slabs of residential, commercial, or industrial buildings is widespread in the construction industry. Spaces in a building can be demarcated using a variety of partition materials such as sandcrete blocks, bricks, gypsum dry walls, timber stud walls, metal lath, etc. These partitions exert additional loads on a suspended slab, and should be accounted for in the design of the slab. This is necessary especially when there is no beam or wall directly under the slab supporting the partition.
It is important to note that wall and partition loads insist on suspended slabs as line loads instead of uniformly distributed loads. It is more complex to analyse line loads on plates than uniformly distributed loads. Therefore during designs, engineers usually attempt to represent line loads with equivalent uniformly distributed loads to make the computational effort easier.
Typical dry wall partitioning in an office building
There are established guides in the building code for assessing all types of loads that a building might be subjected to, and partition loads is not an exception. Typically in the design of reinforced concrete solid slabs, a partition allowance of between 1.00 kN/m2 to 1.5 kN/m2 is usually made during the analysis of dead loads (permanent actions). This is usually sufficient to allow for all lightweight movable partitions that may be placed on the slab later.
According to clause 6.3.1.2 of EN 1991-1-1:2002, provided that a floor allows a lateral distribution of loads, the self-weight of movable partitions may be taken into account by a uniformly distributed load qk which should be added to the imposed loads of floors obtained from Table 6.2. This defined uniformly distributed load is dependent on the self-weight of the partitions as follows:
for movable partitions with a self-weight < 1.0 kN/m wall length: qk= 0.5 kN/m2
for movable partitions with a self-weight > 1 ≤ 2.0 kN/m wall length: qk = 0.8 kN/m2
for movable partitions with a self-weight > 2 ≤ 3.0 kN/m wall length: qk = 1.2 kN/m2
However, full design consideration should be taken for heavier partitions, accounting for the locations and directions of the partitions and the structural forms of the floors.
According to BS 6399 Part 1, when the position of the wall load is not known, the equivalent uniformly distributed load that is added to the slab load should be 0.33wp (kN/m2), where wp is the weight of the wall (kN/m).
However, when the direction of the partition is normal to the span of the slab, the equivalent uniformly distributed load is given by 2wp/L for simply supported slabs and 3wp/2L for continuous slabs (Where L is the span of the slab normal to the wall load).
Said et al (2012) used finite element analysis and multiple linear regression to derive a general relation between line loads acting on two-way slab system and the equivalent uniformly distributed loads. According to them;
Where; WUDL/WLine is the ratio of equivalent uniformly distributed load to actual line load, α is the relative ratio of the stiffness of beam to slab, L2/L1is the aspect ratio of the slab.
Therefore, from the statistical relationship above, when the value of the line load is known, the equivalent UDL that will produce comparable bending moment values can be obtained using the aspect ratio of the slab and the ratio of the stiffness of the supporting beams to the slab.
Partitions such as sandcrete blocks exert a significant magnitude line load on reinforced concrete solid slabs. For a 225mm hollow block, the unit weight is about 2.87 kN/m2. For a 12 mm thick plaster on both sides, the total weight of finishes is about 0.6 kN/m2. This brings the total unit weight of the block to about 3.47 kN/m2, which is usually approximated to 3.5 kN/m2.
Therefore, for a wall height of 3m, the equivalent line load exerted on the supporting slab or beam is 3.5 kN/m2 × 3m = 10.5 kN/m.
So many structural engineering software design packages have the option of applying line loads directly on slabs. For software like Staad Pro, it may not be possible to assign line loads directly on plates, however a dummy beam of negligible stiffness can be used to transfer the line load to the slab.
Example on Partition Load Modelling
To demonstrate the effects of line loads from block wall, let us consider a 150 mm thick 5m x 6m two-way slab that is simply supported at all edges by a 450 mm x 225mm beam. The slab is supporting a line load of 10.5 kN/m coming from a 225 mm thick block wall placed at the centre parallel to the short span as shown below.
We are going to consider several scenarios;
(a) When the slab is loaded directly with the line load (w = 10.5 kN/m) (b) When the line load is represented with equivalent UDL given by 0.33wp = 0.33 × 10.5 = 3.15 kN/m2 (c) When the line load is represented with an equivalent UDL given by 2wp / L = (2 × 10.5)/6 = 3.5 kN/m2 (d) When the line load is represented with an equivalent UDL by Said et al (2012); WUDL/WLine = 0.32193 + 0.00473α – 0.10175(L2/L1)
(a) When the slab is loaded directly with the line load (w = 10.5 kN/m)
(b) When the line load is represented with equivalent UDL (we = 3.15 kN/m2)
(c) When the line load is represented with equivalent UDL (we = 3.5 kN/m2)
(d) When the line load is represented with equivalent UDL (we = 2.392 kN/m2)
Analysis Method
Mx (sagging)
Mx (hogging)
My (sagging)
Mx (hogging)
Mxy
Line Load
3.99 kNm
0.799 kNm
6.65 kNm
0.627 kNm
0.913 kNm
0.33wp = 3.15 kN/m2
4.48 kNm
0.138 kNm
4.7 kNm
0.349 kNm
0.978 kNm
2wp / L = 3.5 kN/m2
4.98 kNm
0.153 kNm
5.22 kNm
0.388 kNm
1.09 kNm
WUDL = 0.2279WLine = 2.392 kN/m2
3.4 kNm
0.104 kNm
3.57 kNm
0.265 kNm
0.742 kNm
From the analysis result, it can be seen that none of the proposed equations was able to capture the effect of the line load adequately. Therefore, when heavy wall loads are to be supported on suspended slabs, the line load should be properly modelled on the slab instead of being converted to equivalent uniformly distributed load.
References
Said A. A, Obed S. R., and Ayez S. M. (2012): Replacement of Line Loads acting on slabs to equivalent uniformly Distributed Loads. Journal of Engineering 11(18):1193 – 1200
Cantilever stairs are a unique type of staircase where one end of the tread is rigidly supported by a beam or reinforced concrete wall, and the other end free. By implication, one end of the tread of the cantilever staircase appears to be floating in the air without support.
The design and construction of a cantilever staircase are expected to maintain and/or enhance the aesthetic appeal, while at the same time, ensuring that the staircase satisfies ultimate and serviceability limit state requirements.
The design of cantilever stairs, therefore, involves the selection of the adequate size of the supporting beams/walls, treads, and other accessories to support the anticipated load on the staircase and to also ensure the good performance of the staircase while in service. For reinforced concrete spine beams, the dimensions and reinforcements provided must satisfy all design requirements, while for steel spine beams, the section selected must satisfy all the requirements.
Moreover, the thickness of the tread must be adequate such that it does not undergo excessive vibration, cracking, deflection, or failure. The tread can be made of reinforced/precast concrete, timber, steel, glass, or composite sections.
Structurally, there are two variations of cantilever staircases;
(a) Cantilever staircase supported by a central spine beam (b) Cantilever staircase supported by a side spandrel beam or RC Wall
(a) Cantilever staircase supported by a spine beam
(b) Cantilever staircase supported by a side spandrel beam or RC Wall
In the first case, the spine beam of the staircase is placed at the centre of the tread, with the two ends of the tread hanging free. In the second case, only one end of the tread is fixed to an adjacent wall where the spandrel beam is hidden to support the treads, and the other end is free. Alternatively, if the wall is a reinforced concrete wall, spandrel beams will no longer be required. Therefore, the latter has a longer moment arm than the former.
For the design of cantilever staircases, the use of uniformly distributed live loads should not be employed. Rather, the concentrated loads provided in Table 6.2 of EN 1991-1-1:2002 should be used. According to clause 6.3.1.2(5)P, the concentrated load shall be considered to act at any point on the floor, balcony or stairs over an area with a shape which is appropriate to the use and form of the floor. The shape may be assumed to be a square of 50 mm.
Furthermore, the possibility of upward loading on the cantilever staircase should also be considered.
Worked Exampleon the Design of Cantilever Stairs
Design the treads and spine beam of the cantilever staircase in a proposed residential dwelling with the following information;
Width of staircase = 1200 mm Going (width of riser) = 250 mm Riser = 150 mm Thickness of riser = 100 mm fck = 25 N/mm2 fyk = 500 N/mm2
Load Analysis
For a staircase in a residential dwelling the uniformly distributed live load varies from 2.0 to 4.0 kN/m2, while the concentrated load (which can be used for local verification) varies from 2.0 to 4.0 kN. The exact value can be decided by the National Annex of the country.
Maximum design moment (about the centreline of spine beam) = (1.25 × 0.62)/2 + (5.175 × 0.6) = 3.33 kNm Maximum shear (about the centreline of spine beam) = (1.25 × 0.6) + (5.175) = 5.925 kN
Flexural Design
Design bending moment; M = 3.3 kNm Effective depth of tension reinforcement; d = 68 mm Redistribution ratio; d = min(Mneg_red_z3 / Mneg_z3, 1) = 1.000
K = M / (b × d2 × fck) = 0.115 K’ = 0.207 K’ > K – No compression reinforcement is required Lever arm; z = min(0.5 × d × [1 + (1 – 2 × K / (h × acc / γC))0.5], 0.95 × d) = 60 mm Depth of neutral axis; x = 2 × (d – z) / λ = 20 mm Area of tension reinforcement required; As,req = M / (fyd × z) = 127 mm2
Tension reinforcement provided; 3H12 As,prov = 339 mm2 Minimum area of reinforcement – exp.9.1N; As,min = max(0.26 × fctm / fyk, 0.0013) × b × d = 23 mm2 Maximum area of reinforcement – cl.9.2.1.1(3); As,max = 0.04 × b × h = 1000 mm2 PASS – Area of reinforcement provided is greater than area of reinforcement required
Crack control
Maximum crack width; wk = 0.3 mm Design value modulus of elasticity reinf – 3.2.7(4); Es = 200000 N/mm2 Mean value of concrete tensile strength; fct,eff = fctm = 2.6 N/mm2 Stress distribution coefficient; kc = 0.4 Non-uniform self-equilibrating stress coefficient; k = min(max(1 + (300 mm – min(h, b)) × 0.35 / 500 mm, 0.65), 1) = 1.00
Actual tension bar spacing; sbar = 93 mm
Maximum stress permitted – Table 7.3N; ss = 326 N/mm2 Steel to concrete modulus of elast. ratio; acr = Es / Ecm = 6.35 Distance of the Elastic NA from bottom of beam; y = (b × h2 / 2 + As,prov × (acr – 1) × (h – d)) / (b × h + As,prov × (acr – 1)) = 49 mm Area of concrete in the tensile zone; Act = b × y = 12195 mm2 Minimum area of reinforcement required – exp.7.1; Asc,min = kc × k × fct,eff × Act / ss = 38 mm2 PASS – Area of tension reinforcement provided exceeds minimum required for crack control
Quasi-permanent moment; MQP = 1.0 kNm Permanent load ratio; RPL = MQP / M = 0.30 Service stress in reinforcement; ssr = fyd × As,req / As,prov × RPL = 49 N/mm2 Maximum bar spacing – Tables 7.3N; sbar,max = 300 mm PASS – Maximum bar spacing exceeds actual bar spacing for crack control
Allowable span to depth ratio; span_to_depthallow = min(span_to_depthbasic × Ks × F1 × F2, 40 × Kb) = 9.606 Actual span to depth ratio; span_to_depthactual = Lm1_s1 / d = 8.824
Shear Design
Using the maximum shear force for all the spans Support A; VEd (say) = 6 kN VRd,c = [CRd,c.k. (100ρ1 fck)1/3 + k1.σcp]bw.d ≥ (Vmin + k1.σcp)bw.d CRd,c = 0.18/γc = 0.18/1.5 = 0.12 k = 1 + √(200/d) = 1 + √(200/68) = 2.714 < 2.0, therefore, k = 2.00 Vmin = 0.035k3/2fck1/2 Vmin = 0.035 × 2.003/2 × 251/2 = 0.494 N/mm2 ρ1 = As/bd = 339/(250 × 68) = 0.0199 < 0.02;
σcp = NEd/Ac < 0.2fcd (Where NEd is the axial force at the section, Ac = cross sectional area of the concrete), fcd = design compressive strength of the concrete.) Take NEd = 0
Since VRd,c (15 kN) < VEd (6 kN), No shear reinforcement is required.
However, nominal shear reinforcement can be provided as H8 @ 150 c/c.
Summary of the tread (going) design Thickness = 100 mm Reinforcement = 3H12 (Top and Bottom) Links = H8 @ 150 c/c.
Design of the Spine Beam
To design the spine beam, it is very important to transfer the load from the treads to the top of the spine beam. In other words, the spine beam will be subjected to its self-weight and the load from the treads. It will be very important to also consider the effects of asymmetric loading (when the live load is acting only on side of the staircase). This can lead to the development of torsional stresses on the beam.
Ultimate limit state load transferred from treads to beam = 6 kN + 6 kN = 12 kN (concentrated loads) Load transferred from the landing area (say) = 5 kN/m
Width of spine beam = 225 mm Depth = 300 mm Self weight of spine beam (drop) = 1.35 × 25 × 0.225 × 0.3 = 2.278 kN/m Self weight of the stepped area = 1.35 × 0.5 × 0.15 × 0.225 = 0.02278 kN/m
Total uniformly distributed load on the flight = 2.3 kN/m Total uniformly distributed load on the landing = 2.278 + 5 = 7.278 kN/m
The spine beam was loaded as shown below;
The analysis results are shown below;
Flexural Design
Design bending moment; M = 26.9 kNm Effective depth of tension reinforcement; d = 264 mm K = M / (b × d2 × fck) = 0.086 K’ = 0.207
K’ > K – No compression reinforcement is required Lever arm; z = min(0.5 × d × [1 + (1 – 2 × K / (h × acc / γC))0.5], 0.95 × d) = 242 mm Depth of neutral axis; x = 2 × (d – z) / λ = 54 mm
Area of tension reinforcement required; As,req = M / (fyd × z) = 255 mm2 Tension reinforcement provided; 2H16 As,prov = 402 mm2 Minimum area of reinforcement – exp.9.1N; As,min = max(0.26 × fctm / fyk, 0.0013) × b × d = 77 mm2 Maximum area of reinforcement – cl.9.2.1.1(3); As,max = 0.04 × b × h = 2700 mm2 PASS – Area of reinforcement provided is greater than area of reinforcement required
Shear Design
Angle of comp. shear strut for maximum shear; θmax = 45 deg Strength reduction factor – cl.6.2.3(3); v1 = 0.6 × (1 – fck / 250) = 0.552 Compression chord coefficient – cl.6.2.3(3); αcw = 1.00 Minimum area of shear reinforcement – exp.9.5N; Asv,min = 0.08 N/mm2 × b × (fck)0.5 / fyk = 161 mm2/m
Design shear force at support ; VEd,max = VEd,max_s1 = 54 kN Min lever arm in shear zone; z = 242 mm Maximum design shear resistance – exp.6.9; VRd,max = αcw × b × z × v1 × fcwd / (cot(θmax) + tan(θmax)) = 201 kN
PASS – Design shear force at support is less than maximum design shear resistance
Design shear force ; VEd = 54 kN Design shear stress; vEd = VEd / (b × z) = 0.994 N/mm2 Angle of concrete compression strut – cl.6.2.3; θ = min[max(0.5 × sin-1(min(2 × vEd / (acw × fcwd × v1),1)), 21.8 deg), 45deg] = 21.8 deg Area of shear reinforcement required – exp.6.8; Asv,des = vEd × b / (fyd × cot(θ)) = 206 mm2/m Area of shear reinforcement required; Asv,req = max(Asv,min, Asv,des) = 206 mm2/m
Shear reinforcement provided; 2H8 @ 175 c/c Area of shear reinforcement provided; Asv,prov = 574 mm2/m PASS – Area of shear reinforcement provided exceeds minimum required Maximum longitudinal spacing – exp.9.6N; svl,max = 0.75 × d = 198 mm
EXTRA: The structural analysis result revealed a high compressive axial load on the spline beam. As a result, it will be important to check the interaction of bending moment and axial force, as typically will be done in the design of reinforced concrete columns. This was checked and found satisfactory.
Furthermore, the possibility of a torsional moment on the spline beam when the live load is asymmetrically loaded should also be checked.
It is common practise to test bridges with lengthy or unusually flexible decks in wind tunnels. In order to fully comprehend the structure’s aerodynamic behaviour, wind tunnel testing may occasionally be necessary, as described in BD 49/01 (Department of Transport, 2001a). The aerodynamics and wind-induced reaction of vehicles can be impacted by the wind field on a bridge deck. Studies of wind fields on bridge decks can be used as a guide when designing a bridge to guarantee the safety of moving vehicles.
For long span bridges, the use of wind tunnel testing is imperative. The term “long span bridges” typically refers to structures with a span length of between 1000 and 1500 metres. The first natural frequencies for these bridges are of the order of 0.1 Hz or lower, and the structure possesses very high flexibility. Bridges with span lengths over 1.5 km are categorised as “extremely long span bridges,” and their natural frequencies decrease in inverse proportion of the span length.
Objectives of Wind Tunnel Testing
There are three main objectives for the use of wind tunnel testing during bridge design.
First, wind tunnel testing is utilised to estimate structure-specific drag and lift coefficients if bridge decks or piers do not conform to the conventional cross-sections specified in design regulations. The mean lateral force, mean normal force, and pitching moment near the centre of the deck are used to measure static wind loads, which are then given as static wind load coefficients. Depending on the location of the bridge site, measurements are made in both smooth and turbulent flow conditions.
The second objective is to determine whether a structure is vulnerable to vortex shedding and divergent amplitude responses. These evaluations are typically carried out under conditions of uniform flow at various test wind incidence angles.
The third objective is to examine how nearby terrain or structures affect the structure’s dynamic reaction. The main impact of such obstructions is to change the wind’s angle of action on the structure. The wind field in built-up areas is frequently extremely complicated, and it is feasible for the wind to be directed around nearby structures, increasing the mean velocity on the bridge’s under-review zones.
Modelling for Wind Tunnel Testing
For prismatic (two-dimensional) constructions like long-span bridge decks, section model investigations are carried out. To simulate the attributes of the structure, a rigid model is fixed to a dynamic test rig made of a set of springs. The dynamic response caused by phenomena like vortex shedding and the wind speeds for the development of aerodynamic instabilities like galloping and flutter are measured.
The failure of the Tacoma Narrows Bridge demonstrated the catastrophic failure that flutter can cause, which motivated engineers to carefully consider the aerodynamic analysis inside bridge design procedures.
The section model for wind tunnel testing can be made from materials such balsa wood, carbon fibre sheet, and steel sections, and it is designed to correctly replicate the prototype’s scaled mass as well as its lowest torsional and bending frequencies. Handrails, parapets, guiding vanes, maintenance gantry rails, stay pipes, and any other non-structural accessories that may have an impact on the deck’s aerodynamic behaviour should be included in sectional models.
Aerodynamically comparable representations may be employed when these elements are too small to be scale-modelled. Typically, dimensions must be accurate to ±0.05 mm. The target spectral density function must be closely matched where turbulent atmospheric conditions are required.
A two-dimensional grid of vertical and horizontal strips is usually placed at the entrance of the wind tunnel test section to create turbulent conditions in low-speed aeronautical wind tunnels. It is possible to recreate the necessary turbulence conditions by adjusting the grid spacing.
Hot-wire anemometry is used to measure the wind parameters immediately upstream of the test section’s installation, including mean wind speed, turbulence intensity, and wind spectra. These turbulence characteristics are measured at various wind speeds that reflect the anticipated experimental circumstances. The u– and w-component spectra, as well as the longitudinal and vertical turbulence intensities, are used to display the results of the turbulence simulation.
The produced by the Engineering Sciences Data Unit standards, as well as the measured and target spectra, should be compared (ESDU). Smoke visualisation can be used during tests to study the vortex-shedding reaction to see how the airflow over the structure is changing.
The model forces and moments observed in the wind tunnel are used to generate the static load coefficients in the wind axis system as follows:
Drag coefficient, CD = D/0.5ρVm2B Lift coefficient, CL = L/0.5ρVm2B Pitching coefficient, CM = M/0.5ρVm2B
where D, L and M are the wind axis along- and across-wind force and pitching moment. Vm is the mean wind speed, ρ is the density of air (in the UK taken as 1.225 kg/m3) and B is the reference dimension which is usually the deck width.
Force coefficients must however be adjusted to take into account blockage effects brought on by the wind tunnel’s constriction. There are several approaches for computing correction factors, hence it is advised to consult specialised literature like ESDU 80024 (ESDU, 1980) to determine which method is best for the specific blockage ratio being taken into account.
The Reynolds number (Re) is likely to be sensitive to sections with round section members or curved surfaces, therefore adjustments based on full-scale data or theoretical considerations may be required.
Analysis of Wind Tunnel Test Models
A whole bridge’s aerodynamic model is also known as an aero-elastic model. For structures with considerable wind-induced motions that have an impact on the aerodynamic forces and subsequently the dynamic response, aero-elastic tests are carried out. These structures usually behave as three dimensional structures.
By simulating the structure’s mass and stiffness distribution, models are created to replicate the essential dynamic properties. Particularly, the basic modes that control the structure’s dynamic response must be accurately modelled at the model-scale frequency. Transducers that measure displacement or acceleration are used to directly measure modal responses.
The modal responses measured are then used to calculate dynamic loads, or in the case of vertical structures, they are directly measured. The models for long-span bridges may be quite large because, typically, models for aeroelastic tests are constructed at either 1:200 or 1:100 scales.
Boundary layer wind tunnels (BLWT) with wide sections are therefore necessary for the experiments. A 1:100 scale aeroelastic replica of the Japanese Akashi-Kaiko bridge was constructed, having an overall length of about 40 m. Figure 19 depicts an instance of an aeroelastic model in a BLWT.
In a BLWT, turbulent boundary conditions are produced by placing a configuration of roughness elements across the wind tunnel’s floor and placing a two-dimensional barrier with square vortex-generating posts at the test section’s entry. These elements’ size, form, and distribution are planned to produce the specific turbulence characteristic needed for the testing.
Calibration of the wind-tunnel-generated turbulence attributes should be carried out with reference to the target spectrum, just like the section model testing. At crucial locations on the structure, displacement transducers and accelerometers measure the time histories of acceleration and acceleration during the testing.
Mean, root mean square (RMS) background, and RMS resonant components are calculated from the displacement time histories. The RMS resonant components of the movement are determined using acceleration time histories. In order to provide enough coverage of the design wind speed range representative of the bridge site, time records are recorded at small wind speed increments.
The Fourier transformation is used to analyse each time history in order to identify the spectrum, allowing for the isolation of narrowband responses that are indicative of structural resonance. By removing the peaks that correspond to the resonant components of displacement, background components can be extracted from displacement spectra.
The elements of a typical measured wind spectrum are shown below. The mean displacements plus or minus a sum of the standard deviations of the background and resonant components of displacement due to each mode are then used to determine the peak displacements.
Spectral density of a response to wind
The following expression is commonly used for calculating the measured peak displacements:
Dpeak = Dmean ± √(g0Dsdev)2 + ∑(gkDmodek)2
where; Dpeak = peak displacement Dmean = mean displacement Dsdev = standard deviation of background displacement Dmode = RMS inertial displacement due to response in mode k g0 = peak factor to apply to background standard deviation (typically 3.4 to 3.5) gk = peak factor to apply to narrow band displacement due to motion in mode k.
The required peak factors across the modal responses are determined for the modal or narrow band displacements using the Davenport gust factor, which is given as:
gk = √[2In(fkT0) + 0.577/2In(fkT0)]
where; fk = natural frequency of kth mode (in hertz) T0 = storm duration (typically 3600 seconds).
Correction factors may also be required within the calculation of the peak displacements in order to account for scaling inaccuracies necessitated by virtue of the model scale, e.g. cable diameters.
Wind Tunnel Testing of Cable Stays
Similar to the static testing of deck sections, testing of stay cables in wind tunnels is done to determine the drag coefficient and check for any potential divergent instabilities. Stay cables are often tested at full scale to guarantee Reynolds number similarity between model and prototype.
The surface characteristics of the prototype, such as any helical fillets or dimple patterns, should be precisely reflected in the cable models. Using calibrated spring balances and turntables in the test section’s ceiling and floor, cable models are dynamically mounted. This makes it possible to test the cable model in a variety of wind directions and wind-inclination angles. Spray heads positioned in the tunnel ceiling in some specialised wind tunnels can imitate a variety of rain conditions, making it possible to analyse rain-induced cable vibrations.
Drains are one of the components of the drainage system of residential areas and/or public infrastructures. Generically, roadside drains are structures used for collecting and conveying surface water to their discharge point, and in many cases, are artificial. Furthermore, drains are essential in road construction as stormwater ponding on pavement surfaces can cause premature pavement failure especially when unevaporated and undrained water seeps into the pavements.
The main objective of road drainage is to keep the road surface and foundation as dry as possible to maintain its stability. Thus, a good drainage system is essential for efficient highway transportation with minimum maintenance costs. Similarly, proper planning and design of road drainage systems are very important to prevent water from in-filtering the road surface, removing it from driving lanes, and carrying it away from the roadway.
Water ponding on highway due to poor drainage
Common Roadside Drain Cross-sections
The common cross-sections used for roadside drains construction are rectangular and trapezoidal sections. Trapezoidal drains have sloped sides and can be formed by excavating in-situ materials. The sloped sides and channel bottom may require paving for protection, depending on the stability of the sides and the resistance of the in-situ materials to erosion. However, trapezoidal drains are now formed with reinforced concrete, which may be precast units or cast-in-place.
Trapezoidal drain
Rectangular drain
Similarly, rectangular drains have vertical or near-vertical sides, formed with reinforced concrete retaining walls, I-walls, or U-frame structures. The bottom of the channel may be paved or unpaved depending on the resistance of the in situ material to erosion.
Typical cross-sections for roadside drain
Cross-section
Flow Area (A)
Wetted Perimeter (P)
Top width (T)
Hydraulic Radius (R)
Rectangular
by
b + 2y
b
A/P
Trapezoidal
y(b + yz)
b + 2y√(1 + z2)
b + 2yz
A/P
Geometric dimensions of drain cross-sections
Best Hydraulic Cross-section
The best cross-section for a drainage channel provides adequate hydraulic capacity at the minimum cost. Economic considerations for selecting the channel section include design and construction costs, right-of-way, required relocations, maintenance and operation.
A trapezoidal channel is usually the most economical channel when right-of-way is available. In contrast, a rectangular channel may be required for channels located in urban areas where the right-of-way is severely restricted or available at a high cost. Furthermore, site developments, existing geophysical site conditions, and performance or service requirements affect the selection of channel type and the resulting construction costs.
For discharge to be maximum or for the best hydraulic cross-section in a rectangular channel; b = 2y and R = y/2
For trapezoidal channel; T + 2yz = 2y√(z2 + 1) R = y/2 A = 3(1/2) × y2 b = 2y/3(1/2) Z = 1/3(1/2) = 0.577
Freeboard for Drains
In open channels or drains, freeboard is of great importance. The provision of freeboard in open drains ensures enough room for wave action and flow surges not to overtopping or overflow the drain. Many uncontrolled causes may create wave action or water surface fluctuation, but mostly because of changes in specific energy of the flow in a channel.
Therefore, a freeboard is provided to ensure the drain is not filled with water. In addition, the calculated discharge rate does not consider deposited solids and lack of maintenance, which will usually reduce the system’s efficiency.
A table for minimum freeboard for ditches is provided below;
Freeboards for roadside drains
Worked Example
It was estimated from hydrological analysis that the peak discharge from a catchment area is 1.351m3/s. Using Manning’s equation, determine the best hydraulic cross-section for a rectangular and trapezoidal roadside drain. Other assumptions for computation are provided below.
Q = VA V= 1/n x R2/3 × S1/2
Where: Q = Discharge (m3/s) A = Flow area (m2) V = Allowable velocity (m/s) R = Hydraulic radius (m)
Manning’s roughness coefficient (n) = 0.014 Bed slope of channel (S) = 3% (0.03)
Solution
From Q = VA ——– (1) Q = 1/n × R2/3 × S1/2 × A ——– (2)
(a) For the best hydraulic section for the rectangular drain; b = 2y and R = y/2
Cross-sectional area of drain (A) = by Substituting b = 2y into the area formula; Cross-sectional area of drain (A) = (2y)y = 2y2
y8/3 = 0.1 m y = 0.0873/8 = 0.422 m (approximated to 0.45 m)
Cross-sectional area of drain (A) = 31/2 × y2 = 31/2 × 0.4222 = 0.308 m2 Depth of drain = 0.422 + 0.3 = 0.722 m (approximated to 0.75 m) Bottom width of drain (b) = 2y/31/2 = 2 x 0.422/31/2 = 0.487 m (approximated to 0.5m) Top width (T) = b + 2yz = 0.487 + (2 x 0.722 x 0.577) = 1.32 m (approximated to 1.35m)
Conclusion
The best hydraulic section of a roadside drain is characterized by the provision of maximum discharge with a given cross-sectional area. Other advantages than the hydraulic performance, for instance, for a given discharge rate, the best hydraulic section could guarantee the least cross-sectional area of the channel. Substantial savings could be made by reducing excavation and using fewer channel linings such as reinforced concrete.
Lastly, any roadside drain section selected should be large enough to permit the required discharge, as deep as required to provide a satisfactory outlet for both surface and subsurface drainage needs of the area served, and of a width-depth ratio and side slopes which will result in a stable channel which can be maintained in a satisfactory condition at a reasonable cost.
Reference(s)
King, H. W. and Brater, E. F. (1963). Handbook of Hydraulics, Fifth Edition. McGraw-Hill Book Company, Inc,, New York.
Cost is a major controlling factor in civil engineering construction projects. Different types of floor systems are adopted in reinforced concrete slab designs such as solid slabs, waffle slabs, ribbed slabs, flat slabs, etc. Each floor system has its advantages, applications, and cost implications in construction. This article aims to evaluate the cost comparison of solid and ribbed slabs.
The idea behind the adoption of the ribbed slab system is the need to reduce the volume of concrete in the tension zone of a concrete slab. Theoretically, the tensile strength of concrete is assumed to be zero during the structural design of flexural structural elements such as beams and slabs. By implication, all the tensile stresses from bending are assumed to be resisted by the reinforcements (see the stress block of flexural concrete sections in Figure 1).
Figure 1: Eurocode 2 concrete stress block
If concrete is assumed to do no work in the tension zone, it then makes economical sense to reduce the volume of concrete in that zone (bottom of the slab). To achieve this, beams of relatively shallow depths (ribs) are spaced at intervals to resist the flexural stresses due to the bending of the floor, with a thin topping (of about 50 mm). When this is done, a thick volume of concrete is no longer uniformly provided in the tensile zone of the concrete.
As the span of a floor increases, the thickness of the concrete and the quantity of reinforcements required to satisfy ultimate and serviceability limit state requirements also increase. However, by introducing ribbed slabs, longer spans can be economically spanned.
To reduce the cost and labour of constructing ribbed slabs, clay hollow pots, sandcrete blocks, or polystyrene are usually provided as infills between the ribs. While clay hollow pots and sandcrete blocks will improve the stiffness of the floor, the same, however, cannot be confidently said about polystyrenes.
Figure 2: Use of polystyrene as infill in ribbed slab
There are significant cost implications of adopting either solid or ribbed slabs. In a study by Ajema and Abeyo (2018), they observed that frames with solid slabs are more economical than frames with a ribbed slab when subjected to seismic action. In another study by Nassar and Al-Qasem (2020) on the cost of different slab systems, they observed that the flat slab system reduces the total cost of construction by 7% compared to the solid slab system, 4 % compared to the one-way ribbed slab system, and 3.33% compared to the two-way ribbed slab system.
Mashri et al (2020) compared the cost of constructing solid slabs and hollow block ribbed slabs and concluded that ribbed slabs are cheaper than solid slabs. In a study on the assessment of the cost difference between solid and hollow floors, Dosumu and Adenuga (2013) observed that the cost of in-situ solid slabs are higher than that of hollow slab provided the hollow slab is a one-way hollow floor and not a waffle floor.
However, it is important to note that the scenario on the issue of cost can vary depending on the size and geometry of the slab. For short one-way slabs, solid slabs may be cheaper than ribbed slabs, while in large-span two-way slabs, ribbed slabs may be cheaper than solid slabs.
Figure 3: Ribbed slab construction
With the aid of design examples and quantity estimation, let us compare the cost of constructing a two-way slab of dimensions 5.225m x 7.426m that is discontinuous at all edges using ribbed slab and solid slab. The slab is to support a live load of 2.5 kN/m2.
Concrete details Concrete strength class; C25/30 Aggregate type; Quartzite Aggregate adjustment factor – cl.3.1.3(2); AAF = 1.0 Characteristic compressive cylinder strength; fck = 25 N/mm2 Mean value of compressive cylinder strength; fcm = fck + 8 N/mm2 = 33 N/mm2 Mean value of axial tensile strength; fctm = 0.3 N/mm2 × (fck)2/3 = 2.6 N/mm2 Secant modulus of elasticity of concrete; Ecm = 22 kN/mm2 × (fcm/10)0.3 × AAF = 31476 N/mm2
Figure 4: Typical ribbed slab panel
Design Information Spacing of ribs = 450 mm Topping = 50 mm Rib width = 150 mm Span = 5.225 m (simply supported) Total depth of slab = 250 mm Design live load qk = 2.5 kN/m2 Weight of finishes = 1.2 kN/m2 Partition allowance = 1.5 kN/m2
Load Analysis
Dead Load Self-weight of topping = 24 × 0.05 × 0.45 = 0.54 kN/m Self-weight of ribs = 24 × 0.15 × 0.2 = 0.72 kN/m Weight of finishes = 1.2 × 0.45 = 0.54 kN/m Partition allowance = 1.5 × 0.45 = 0.675 kN/m Self-weight of heavy duty EPS = (0.156 × 0.2 × 0.45) = 0.014 kN/m Total dead load per rib gk= 2.489 kN/m
Live Load Characteristic live load = 2.5 kN/m2 Total live load per rib = 2.5 × 0.45 = 1.125 kN/m
Flexural Design Design bending moment; MEd = 17.2 kNm Effective flange width; beff = 2 × beff,1 + b = 450 mm Effective depth of tension reinforcement; d = 209 mm K = M / (beff × d2 × fck) = 0.035 K’ = 0.207 Lever arm; z = 199 mm Depth of neutral axis; x = 2 × (d – z) / l = 26 mm
lx <= hf – Compression block wholly within the depth of flange K’ > K – No compression reinforcement is required
Area of tension reinforcement required; As,req = M / (fyd × z) = 243 mm2 Tension reinforcement provided; 2Y16 (As,prov = 402 mm2) Minimum area of reinforcement – exp.9.1N; As,min = max(0.26 × fctm / fyk, 0.0013) × b × d = 51 mm2 Maximum area of reinforcement – cl.9.2.1.1(3); As,max = 0.04 × b × h = 1500 mm2 PASS – Area of reinforcement provided is greater than area of reinforcement required
Allowable span to depth ratio; span_to_depthallow = min(span_to_depthbasic × Ks × F1 × F2, 40 × Kb) = 40.000 Actual span to depth ratio; span_to_depthactual = Lm1_s1 / d = 25.000 PASS – Actual span to depth ratio is within the allowable limit
Shear Design Angle of comp. shear strut for maximum shear; θmax = 45 deg Strength reduction factor – cl.6.2.3(3); v1 = 0.6 × (1 – fck / 250 N/mm2) = 0.540 Compression chord coefficient – cl.6.2.3(3); acw = 1.00 Minimum area of shear reinforcement – exp.9.5N; Asv,min = 0.08 N/mm2 × b × (fck )0.5 / fyk = 146 mm2/m
Design shear force at support ; VEd,max = 13 kN Min lever arm in shear zone; z = 199 mm Maximum design shear resistance – exp.6.9; VRd,max = acw × b × z × v1 × fcwd / (cot(θmax) + tan(θmax)) = 134 kN PASS – Design shear force at support is less than maximum design shear resistance
Design shear force at 209 mm from support; VEd = 12 kN Design shear stress; vEd = VEd / (b × z) = 0.407 N/mm2
Area of shear reinforcement required – exp.6.8; Asv,des = vEd × b / (fyd × cot(θ)) = 69 mm2/m Area of shear reinforcement required; Asv,req = max(Asv,min, Asv,des) = 146 mm2/m Shear reinforcement provided; 2Y 8 legs @ 150 c/c Area of shear reinforcement provided; Asv,prov = 670 mm2/m PASS – Area of shear reinforcement provided exceeds minimum required
Maximum longitudinal spacing – exp.9.6N; svl,max = 0.75 × d = 157 mm PASS – Longitudinal spacing of shear reinforcement provided is less than maximum
However, calculations have shown that shear reinforcements are not required since VEd (13 kN) is less than VRdc (25.2 kN). According to clause 6.2.1(4) of EN 1992-1-1:2004, when, on the basis of the design shear calculation, no shear reinforcement is required, minimum shear reinforcement should nevertheless be provided according to clause 9.2.2. The minimum shear reinforcement may be omitted in members such as slabs (solid, ribbed or hollow core slabs) where transverse redistribution of loads is possible.
Therefore to save cost, let us provide a triangular pattern link spaced at 300mm c/c, and 1Y12 (Asprov = 113 mm2) at the top of the rib.
Figure 5: Rib beam elevation view
Figure 6: Ribbed slab section
Design of Solid Slab
RC slab design
In accordance with EN1992-1-1:2004 incorporating corrigendum January 2008 and the UK national annex
Figure 7: Solid slab panel
Slab definition Type of slab; Two way spanning with restrained edges Overall slab depth; h = 200 mm Shorter effective span of panel; lx = 5225 mm Longer effective span of panel; ly = 7426 mm Support conditions; Four edges discontinuous
Concrete cover to reinforcement Nominal cover to outer bottom reinforcement; cnom_b = 25 mm Fire resistance period to bottom of slab; Rbtm = 60 min Axial distance to bottom reinft (Table 5.8); afi_b = 10 mm Min. btm cover requirement with regard to bond; cmin,b_b = 12 mm Reinforcement fabrication; Not subject to QA system Cover allowance for deviation; Dcdev = 10 mm Min. required nominal cover to bottom reinft; cnom_b_min = 22.0 mm PASS – There is sufficient cover to the bottom reinforcement
Reinforcement design at midspan in short span direction (cl.6.1)
Bending moment coefficient; bsx_p = 0.0881 Design bending moment; Mx_p = bsx_p × q × lx2 = 29.9 kNm/m Reinforcement provided; 12 mm dia. bars at 200 mm centres Area provided; Asx_p = 565 mm2/m Effective depth to tension reinforcement; dx_p = h – cnom_b – fx_p / 2 = 169.0 mm
K factor; K = Mx_p / (b × dx_p2 × fck) = 0.042 Redistribution ratio; δ = 1.0 K’ factor; K’ = 0.598 × d – 0.18 × d2 – 0.21 = 0.208 K < K’ – Compression reinforcement is not required
Lever arm; z = min(0.95 × dx_p, dx_p/2 × (1 + √(1 – 3.53 × K))) = 160.5 mm Area of reinforcement required for bending; Asx_p_m = Mx_p / (fyd × z) = 523 mm2/m Minimum area of reinforcement required; Asx_p_min = max(0.26 × (fctm/fyk) × b × dx_p, 0.0013 × b × dx_p) = 275 mm2/m Area of reinforcement required; Asx_p_req = max(Asx_p_m, Asx_p_min) = 523 mm2/m
Check reinforcement spacing Reinforcement service stress; ssx_p = (fyk / gS) × min((Asx_p_m/Asx_p), 1.0) × qSLS / q = 190.7 N/mm2 Maximum allowable spacing (Table 7.3N); smax_x_p = 262 mm Actual bar spacing; sx_p = 200 mm PASS – The reinforcement spacing is acceptable
Reinforcement design at midspan in long span direction (cl.6.1) Bending moment coefficient; bsy_p = 0.0560 Design bending moment; My_p = bsy_p × q × lx2 = 19.0 kNm/m Reinforcement provided; 12 mm dia. bars at 250 mm centres Area provided; Asy_p = 452 mm2/m Effective depth to tension reinforcement; dy_p = h – cnom_b – fx_p – fy_p / 2 = 157.0 mm K factor; K = My_p / (b × dy_p2 × fck) = 0.031 Redistribution ratio; d = 1.0 K’ factor; K’ = 0.598 × d – 0.18 × d2 – 0.21 = 0.208 K < K’ – Compression reinforcement is not required
Lever arm; z = min(0.95 × dy_p, dy_p/2 × (1 + √(1 – 3.53 × K))) = 149.2 mm
Area of reinforcement required for bending; Asy_p_m = My_p / (fyd × z) = 358 mm2/m Minimum area of reinforcement required; Asy_p_min = max(0.26 × (fctm/fyk) × b × dy_p, 0.0013 × b × dy_p) = 255 mm2/m Area of reinforcement required; Asy_p_req = max(Asy_p_m, Asy_p_min) = 358 mm2/m PASS – Area of reinforcement provided exceeds area required
Check reinforcement spacing Reinforcement service stress; ssy_p = (fyk / γS) × min((Asy_p_m/Asy_p), 1.0) × qSLS / q = 163.1 N/mm2 Maximum allowable spacing (Table 7.3N); smax_y_p = 296 mm Actual bar spacing; sy_p = 250 mm PASS – The reinforcement spacing is acceptable
Mod span-to-depth ratio limit; ratiolim_x = min(40 × Kd, min(1.5, (500 N/mm2/ fyk) × (Asx_p / Asx_p_m)) × ratiolim_x_bas) = 34.54 Actual span-to-eff. depth ratio; ratioact_x = lx / dx_p = 30.92 PASS – Actual span-to-effective depth ratio is acceptable
Reinforcement summary Midspan in short span direction; 12 mm dia. bars at 200 mm centres B1 Midspan in long span direction; 12 mm dia. bars at 250 mm centres B2 Discontinuous support in short span direction; 12 mm dia. bars at 200 mm centres B1 Discontinuous support in long span direction; 12 mm dia. bars at 250 mm centres B2
In this section, we are going to consider the cost of constructing ribbed slab and the cost of constructing solid slabs (considering the cost of materials only). In this case, the cost of labour is assumed to be directly proportional to the quantity of materials.
Cost analysis of solid slab
Concrete Volume of concrete required for the slab = 5.45 × 7.65 × 0.2 = 8.3385 m3 Current unit cost of concrete materials = ₦55,000/m3 Cost of concrete materials = 8.3385 × 55000 = ₦ 458,618
Quantity of steel required Bar mark 1 = 37 × 7.645 × 0.888 = 251.184 kg Bar mark 2 = 21 × 10.485 × 0.888 = 195.524 kg Bar mark 3 = 12 × 4.225× 0.617 = 31.28 kg Bar mark 3 = 8 × 6.625 × 0.617 = 32.701kg Total = 510.689 kg
Unit cost of reinforcement = ₦ 450/kg Cost of reinforcement = 510.689 × 450 = ₦ 229,810
Formwork Required Soffit of slab = 36 m2 (treated as a constant)
Concrete Volume of concrete required for the topping = 5.45 × 7.65 × 0.05 = 2.084 m3 Volume of concrete required for the ribs = 12 × 0.2 × 0.15 × 5 = 1.8 m3 Total volume of concrete = 3.884 m3 Cost of concrete materials = 3.884 × 55,000 = ₦ 213,620
Quantity of steel required Bar mark 1 = 2 × 12 × 5.7 × 1.579 = 216 kg Bar mark 2 = 1 × 12 × 5.8 × 0.888 = 61.8 kg Bar mark 3 (triangular links) = 17 × 12 × 0.612 × 0.395 = 49.314 kg Total = 327.114 kg Cost of reinforcement = 327.114 × 450 = ₦ 147,205
BRC Mesh for Topping (A142) – 41.7 m2 Unit cost of BRC mesh = ₦ 1580/m2 Cost of BRC mesh = 41.7 × 1580 = ₦ 65,886
Total cost of reinforcement works = 147,205 + 65,886 = ₦ 213,620
Clay hollow pot/Sandcrete Blocks/Polystyrene Number of block units required = 240 units Unit price of hollow blocks = NGN 400 per unit Cost of hollow blocks = 240 × 400 = ₦ 96,000
Formwork Required Soffit of slab = 36 m2 (treated as a constant)
From the analysis of the two-way slab carried out, it can be seen that the volume of concrete required for a ribbed slab is 53.42% less than that required for a solid slab. Furthermore, the reinforcement required in the ribbed slab is 7.04% less than that required for a solid slab. If the same type of formwork is adopted (completely flat soffit supported with props), and if the cost of labour is directly related to the quantity of materials, then the adoption of the ribbed slab is expected to save cost by about 24.07% compared to solid slab.
References
[1] Ajema D. and Abeyo A. (2018): Cost Comparison between Frames with Solid Slab and Ribbed Slab using HCB under Seismic Loading. International Research Journal of Engineering and Technology 05(01):109-116 [2] Dosumu O. S. and Adenuga O. A.(2013): Assessment of Cost Variation in Solid and Hollow Floor Construction in Lagos State. Journal of Design and Built Environment 13(1):1-11 [3] Mashri M., Al-Ghosni K., Abdulrahman A., Ismaeil M., Abdussalam A. and Elbasir O. M. M. (2020):Design and cost comparison of the Solid Slabs and Hollow Block Slabs. GSJ 8(1):110-118 www.globalscientificjournal.com [4] Reema R. Nassar 1, Imad A. Al-Qasem 2 (2020): Comparative Cost Study for A residential Building Using Different Types of Floor System. International Journal of Engineering Research and Technology 13(8): 1983-1991
The management of public projects includes making good decisions, and these decisions usually involve choosing between alternatives. Therefore, every decision-maker needs to answer questions like what public projects to implement, how to begin, how many project units are required, and where and when to execute these projects. Economic analysis helps the decision-maker to determine answers to these questions.
For every public project conceived, there must be very clear goals, benefits, and expectations from the project during and when it is completed. Complying with these objectives with minimum cost and maximum benefits is crucial because public projects are usually executed using public funds and tax revenues. Therefore, executing public projects involves many choices among physically feasible alternatives. Some examples of public projects are roads, bridges, hospitals, dams, leisure centres, etc
Furthermore, each choice among alternatives should be made on an economic basis. In addition, each alternative should be expressed in terms of money units before making the final choice. Money units can only measure the cost and benefits of different project alternatives.
Public parks are good examples of public projects
Public vs Private Projects
The economic analysis of public projects uses a different approach for evaluation compared to private projects. The focus is usually on benefits instead of profits, as in the case of private projects. In addition, the scope is the society; that is, it covers the interest of the owner and the society. Therefore, benefits becomes the performance criterion instead of profitability.
Economic Analysis of Public Projects
The best project is the one which gives the highest benefit-cost (B/C) ratios as it would give the maximum return on investment (ROI). On the contrary, public projects are usually executed to achieve maximum benefits and not maximum B/C ratios. However, projects should be economically viable and give some minimum return rate (RoR).
It has been observed in most public projects that benefit increase with the size of the project. However, project cost also increases with the increase in project size. Furthermore, a stage is reached beyond which an increase in project size may not yield minimum attractive returns. Therefore, the size of the project is fixed at this stage.
Key Test
It has always been easy to determine costs but determining full benefits remains challenging. Therefore, before economic analysis can proceed, efforts must be made to list possible benefits and estimated values. A decision-maker should accept projects as viable (economically acceptable) if benefits outweigh costs. In addition, the projects should be implemented, subject to the availability of funds.
Benefit-Cost Ratio
The benefit-cost ratio (B/C) compares the present value of all benefits to the present value of all costs. The ratio is merely used to see if a unit of costs (say, #1) will return at least the same unit (#1) in benefits. B/C ratios do not themselves provide enough information to make an economic choice. Therefore, additional calculations are necessary to use B/C ratios as a sound basis for project formulation.
Conditions for Viability
Accept projects as viable (economically acceptable) if benefits outweighs cost, that is, B/C > 1.0
Reject project if B/C < 1.0
Steps in Checking the Viability of Projects
Carryout incremental analysis on the projects, that is, arrange projects in increasing order of costs.
Put the projects in a portfolio from best to worst, noting cumulative cost values.
Stop when funding limit is reached.
Fund only the acceptable projects within limits.
If there is any extra fund, invest in some other venture or keep money in bank.
Worked Example
The information about 5 proposed public projects are given below.
Project Proposal
Annual Benefits (millions)
Annual Cost (Million)
A
12.00
10.30
B
16.80
12.40
C
22.00
16.53
D
25.80
22.70
E
27.20
26.83
a) Determine the projects that are viable.
b) Determine the best project.
c) If there is a budget of 40 million, determine the projects to adopt for execution.
Solution
a) All the projects are viable because their B/C > 1.0
b) The best project is B because it has the highest ∆AB/∆AC value of 2.286
c) With a budget of 40 million, projects A, B and C with overall cost of 39.23 million should be adopted for execution
Conclusion
The main purpose of economic analysis is to help select projects that contribute to the welfare of the people. Therefore, economic analysis is most useful when used early in the project cycle to catch bad projects and bad project components. However, if used at the end of the project cycle, economic analysis can only help decide whether or not to proceed with a project.
One of the most important steps in project evaluation is the consideration of alternatives throughout the project cycle, from identification through appraisal. Many important choices are made early when alternatives are rejected or retained for a more detailed study. Comparing mutually exclusive options is one of the principal reasons for applying economic analysis from the early stages of the project cycle.
Finally, an economical design is that project which gives the greatest excess benefits over cost. Moreover, a project should be evaluated in terms of RoR. The RoR (net gain or loss over a specified period) on investment may be estimated by calculating the cost and benefits of the project. Projects may also be ranked in the merit of RoR. Thus, the decision to go ahead can be made based on the minimum RoR and B/C ratio.
Reference
[1] Belli, P., Anderson, J., Barnum, H., Dixon, J. and Tan, J-P (1997), Handbook of Economic Analysis of Investment Operations, Operations Policy Department: Learning and Leadership Center. https://www.unisdr.org
Machine learning and deep learning techniques have been widely applied in the design of civil engineering structures in recent years. This is mainly due to the development of advanced computational capacity of computers in handling big data. Researchers from Ryerson University, Toronto, Canada have applied the concept of genetic algorithm to the design of pile foundations using Standard Penetration Test (SPT) data. The findings were published in the journal, Soils and Foundations.
Genetic Algorithm (GA) is a method of optimization based on Darwin’s theory of evolution. In nature, chromosomes give an organism its characteristics, and organisms can adapt and evolve to survive and thrive in their environments through reproduction. A GA represents the problem domain as a string or matrix termed a chromosome, then evolves the chromosome through reproduction mechanisms to find a solution.
Fig. 1. Procedures of Implementing the Developed Genetic Algorithm (Jesswein and Lui, 2022)
One of the oldest geotechnical investigation procedures, the standard penetration test (SPT), is still widely utilised around the world. SPT is also used to empirically test the ultimate load-carrying capacity (Qu) for the design of pile foundations, despite its significant shortcomings. There are two types of SPT-based design methods: direct and indirect.
Indirect approaches use SPT blow counts (N-values) to determine the shear strength of soils, such as the frictional angle (ϕ) or undrained shear strength (Cu), and then apply these parameters to the design. Unfortunately, correlating ϕ and Cu with SPT N-values is difficult. Some relationships between ϕ and N-values for sands have been proposed from various parts of the world.
According to the authors (Jesswein and Lui, 2022), it may be preferable to directly correlate SPT N-values to pile shaft (Qs) and tip resistance (Qp) given the uncertainties and additional inaccuracies caused by indirect correlations. However, many direct approaches may produce biased Qu forecasts. This is mainly because they were based on simple linear regressions or trial-and-error procedures with the N-value as the only input, and they likely ignore or misrepresent several factors that influence pile-soil interaction, such as effective stress, excess pore pressure, repacking of soil grains during pile driving, and pile load-transfer response.
Fig. 2. Standard penetration test
This inspired the authors to consider machine learning (ML) approach. ML algorithms are more efficient at regressing huge datasets and capturing nonlinear interactions between numerous variables than standard regression approaches. Furthermore, because the systems evaluate alternative solutions based on a set of criteria, which typically includes an objective or fitness function, they do not require prior knowledge of the problem domain. This eliminates the need to assume a relationship between variables while performing regression.
Due to its “blackbox” nature, the most popular ML method, artificial neural networks (ANN), does not allow the relationships to be stated in a practical formula. As a result, a genetic algorithm (GA) may be a better ML technique since it may provide useful, easy-to-understand functions that represent generic trends.
A total of 72 axial compression tests on driven steel piles were gathered from the literature and the Ministry of Transportation of Ontario (MTO) for the study. The 72 piles selected were divided into two groups based on their measurements.
Fig. 3. Typical driven steel piles
The load-transfer distribution data were used to extract both the unit shaft and tip resistances in the first group. The GA developed the design approach by correlating these unit resistances with soil data and pile parameters. The new design approach was validated and compared to three existing SPT-based design methods in the second group, where only total capacity measurements were provided. In the study both N60 and (N1)60 are represented by Ncr.
The piles that were researched were chosen based on the following criteria:
Sufficient information was available on the subsurface ground conditions, particularly the soil classifications and SPT N-values, at the test site;
Non-organic soils were found along the pile length;
The pile types were either open-ended pipe (OEP), closed-ended pipe (CEP), and H piles;
The pile width or diameter and embedment length were available; and
The load–displacement curves were reported.
Based on a total of 72 full-scale static load tests reported in the study, the new SPT-based formula provides an unbiased prediction with a higher level of accuracy. The proposed design formulas are summarized below:
Where; L = Length of pile σ’ = Effective overburden pressure at the depth considered Ncr = Corrected SPT Number qs = shaft resistance qp = Tip resistance
According to the authors, Equations (1) and (2) can consider the changing ground conditions by using the characteristic values for Ncr and σ’ within a soil layer.
Fig. 4. Comparison between Measured and Predicted Unit Side Resistance by Functions from the GA.(Jesswein and Lui, 2022)
Fig. 5. Comparison between Measured and Predicted Unit Tip Resistance from the GA (Jesswein and Lui, 2022)
Although the proposed formulas provide better predictions than existing design methods, the authors insist that engineering judgement and knowledge of local ground conditions are critical components in correctly designing a pile. The quality of the produced model is dependent on the correctness of the input variables, particularly Ncr, because GA regressions are data driven. Because regressed models may not perform well under extrapolation, the range of the examined variables should be addressed while applying the provided formulas. The proposed approach can be used on a wide range of cohesive and cohesionless soils, but organic soils and soft clays are not recommended.
Article Source: Jesswein M. and Liu J. (2022): Using a genetic algorithm to develop a pile design method. Soils and Foundations 62(2022)101175. https://doi.org/10.1016/j.sandf.2022.101175
Deck slabs of bridges are likely to be among the elements that are influenced by fatigue verification calculations the most. This is due to the high live load to dead load ratio that these slabs have. Tests, on the other hand, have shown that the actual stress ranges in the reinforcement in these are significantly lower than what is suggested by the standard elastic calculations. As a result of this, the NA to BS EN 1992-2 identifies situations in which fatigue assessment is not necessary and gives regulations that are on the safe side.
The failure of a structure due to repeated loadings that are smaller than a single static force that exceeds the material’s strength is known as Fatigue. Fatigue occurs when a material fails due to direct tension or compression, torsion, bending, or a combination of these actions.
Since reinforced concrete is a composite material, it can fail due to fatigue in a variety of ways. Failure is frequently the result of a variety of reasons, and failure types can vary quite significantly. The concrete, the reinforcement, and the bond between the materials might all fail locally.
Compressive fatigue failure in reinforced concrete is referred to as ductile because cracks might appear in the concrete long before the structure falls. Because the crack propagation rate in the reinforcement at the end is rather fast, tensile fatigue failure in reinforced concrete has a more brittle behaviour.
According to clause 6.8.1 of BS EN 1992-2, a fatigue verification is generally not necessary for the following structures and structural elements:
Footbridges, with the exception of structural components very sensitive to wind action.
Buried arch and frame structures with a minimum earth cover of 1 m and 1.5 m for road and railway bridges.
Foundations.
Piers and columns which are not rigidly connected to superstructures.
Retaining walls of embankments for roads and railways.
Abutments of road and railway bridges which are not rigidly connected to superstructures, except the slabs of hollow abutments.
Prestressing and reinforcing steel, in regions where, under the frequent load combination of actions and Pk only compressive stresses occur at the extreme concrete fibres.
Furthermore, in the case of road bridges, fatigue verification is not required for the local effects of wheel loads given directly to a slab that spans across beams or webs, provided that the following conditions are met:
The slab does not contain welded reinforcement or reinforcement couplers.
The clear span to overall depth ratio of the slab does not exceed 18.
The slab acts compositely with its supporting beams or webs.
Either:
the slab also acts compositely with transverse diaphragms; or
the width of the slab perpendicular to its span exceeds three times its clear span.
In Eurocode there are two alternative methods by which fatigue verification can be calculated for bridges;
the λ-Coefficient Method and
the Cumulative Damage Method.
Both approaches take into account the loading during the course of a structure’s lifetime. The λ-Coefficient Method is a simplified method that uses a single load model amplified by several coefficients. The Cumulative Damage Method is a complicated model that takes into account the load history in greater detail. The λ-Coefficient Method simply evaluates if the building meets the code’s requirements, whereas the Cumulative Damage Method derives a fatigue damage factor that indicates the structure’s actual damage in relation to the design fatigue life.
Internal Forces and Stresses for Fatigue Verification
According to clause 6.8.2 of BS EN 1992-2, the calculation of stress for fatigue verification must be based on the assumption of cracked cross-sections, which must be done while ignoring the tensile strength of the concrete and satisfying compatibility of strains. The influence of the varied bond behaviour of prestressing steel and reinforcing steel must be taken into account by increasing the stress range in the reinforcing steel computed under the assumption of a perfect bond by the factor, η, which is given by:
η = (As + Ap)/[(As + Ap) × √ξ(ϕs/ϕp)]
where: As = area of reinforcing steel Ap = area of prestressing tendon or tendons ϕs = largest diameter of reinforcement ϕp = diameter or equivalent diameter of prestressing steel = 1.6 √Ap for bundles = 1.75 ϕwire for single 7-wire strands where ϕwire is the wire diameter = 1.20 ϕwire for single 3-wire strands where ϕwire is the wire diameter ξ = ratio of bond strength between bonded tendons and ribbed steel in concrete. The value is subject to the relevant European Technical Approval. In the absence of this the values given in Table 1 may be used.
Prestressing steel
ξ (pre-tensioned)
ξ (Bonded, post-tensioned) ≤ C50/60
ξ (Bonded, post-tensioned) ≥ C70/85
Smooth bars and wires
Not Applicable
0.3
0.15
Strands
0.6
0.5
0.25
Indented wires
0.7
0.6
0.30
Ribbed bars
0.8
0.7
0.50
Table 1: Ratio of bond strength, ξ, between tendons and reinforcing steel (Source: Table 6.2 BS EN 1992-1-1)
For intermediate values between C50/60 and C70/85, interpolation may be used.
Verification of Concrete under Compression or Shear
According to clause 7.6.2 of PD 6687-2, it is doubtful that National Authorities will have the S–N curves that are necessary to carry out a fatigue verification of concrete while it is being subjected to compression or shear. In the absence of such data, the simplified technique that is outlined in BS EN 1992-2, Annex NN may be utilised for railway bridges; however, there is no equivalent alternative available for highway bridges.
σc,max/fcd,fat≤ 0.5 + 0.45(σc,min/fcd,fat)
where; σc,max = maximum compressive stress at a fibre under the frequent load combination (compression measured positive) σc,min = minimum compressive stress at the same fibre where σc,max occurs. If σc,min is a tensile stress, then σc,min should be taken as zero fcd,fat = design fatigue strength of concrete = 0.85 βcc(t0) fcd (1 – fck/250) fcd = fck / 1.5 βcc(t0) = coefficient for concrete strength at first load application = exp{s[1 – (28/t0)0.5]}
where; t0 = time of the start of the cyclic loading on concrete in days s = 0.2 for cement of strength Classes CEM 42.5R, CEM 52.5N and CEM 52.5R (Class R) = 0.25 for cement of strength Classes CEM 32.5R, CEM 42.5 (Class N) = 0.38 for cement of strength Class CEM 32.5 (Class S)
The maximum value for the ratio σc,max / fcd,fat is given in Table 2.
Concrete Strength
σc,max / fcd,fat
fck ≤ 50 MPa
≤ 0.9
fck > 50 MPa
≤ 0.8
Table 2: Values for σc,max / fcd,fat
(a) Truss model for beams with shear reinforcement and (b) characteristic S-N curve for reinforcing steel
For members not requiring design shear reinforcement for the ultimate limit state it may be assumed that the concrete resists fatigue due to shear effects where the following apply:
for VEd,min/VEd,max ≥ 0: |VEd,max|/|VRd,c| ≤ 0.5 + 0.45|VEd,min|/|VRd,c| ≤ 0.9 up to C50/60 ≤ 0.8 greater than C55/67
for VEd,min/VEd,max < 0: |VEd,max|/|VRd,c| ≤ 0.5 |VEd,min|/|VRd,c|
where; VEd,max = design value of the maximum applied shear force under frequent load combination VEd,min = is the design value of the minimum applied shear force under frequent load combination in the cross-section where VEd,max occurs VRd,c = design value for shear resistance
Limiting Stress Range for Reinforcement under Tension
Adequate fatigue resistance may be assumed for reinforcing bars under tension if the stress range under the frequent cyclic load combined with the basic combination does not exceed 70 MPa for unwelded bars and 35 MPa for welded bars.
For UK highway bridges, the values in Tables 12.3 and 12.4 may be used for straight reinforcement. These are based on bars conforming to BS 4449. For bars not conforming to BS 4449, the rules for bars > 16 mm diameter should be used for all sizes unless the ranges for bars ≤ 16 mm diameter can be justified.
Span (m)
Adjacent spans loaded (Bars ≤ 16 mm)
Adjacent spans loaded (Bars > 16 mm)
Alternate spans loaded (Bars ≤ 16 mm)
Alternate spans loaded (Bars > 16 mm)
≤ 3.5
150
115
210
160
5
125
95
175
135
10
110
85
175
135
20
110
85
140
110
30 -50
90
70
110
85
100
115
90
135
105
≥ 200
190
145
200
155
Table 3: Limiting stress ranges – longitudinal bending for unwelded reinforcing bars in road bridges, MPa
For Tables 3 and 4, intermediate values can be interpolated.
Span (m)
(Bars ≤ 16 mm)
(Bars > 16 mm)
≤ 3.5
210
160
5
120
90
≥ 10
70
55
Table 4: Limiting stress ranges – transverse bending for unwelded reinforcing bars in road bridges, MPa
Full fatigue checks are to be carried out using the ‘damage equivalent stress range’ approach, following the instructions given in Annex NN of BS EN 1992-2, if the stress range limitations exceed the values specified in Tables 3 and 4 (e.g. for reinforcement over the pier). The stress ranges are derived using the ‘Fatigue Load Model 3’, which simulates a four-axle vehicle weighing 48 tonnes total. This weight is increased to 84 tonnes for intermediate supports and 67 tonnes for other locations in Annex NN.
The immediate or elastic settlement of pile groups, denoted by ρiand the long-term consolidation settlement, denoted by ρc, both contribute to the total settling of a pile group in clay. Equation (1) can be used as a general rule to calculate the elastic settlement for a flexible foundation at the level of the ground surface;
ρi = qn × 2B × (1 – v2)/Eu × Ip ——– (1)
where ρiis the settlement at the centre of the flexible loaded area, qn is the net foundation pressure, B is the width of an equivalent foundation flexible raft, v is the undrained Poisson’s ratio for clay (generally taken as equal to 0.5), Ipis an influence factor, and Eu is the deformation modulus for the undrained loading conditions.
The values of Ip are dependent on the ratio H/B of the depth of the compressible soil layer to the width of the pile group, as well as the ratio L/B of the length of the group to its width. The values of Ip and, consequently, the immediate settlement of a surface foundation can be obtained through the use of curves that were established by Steinbrenner and published by Terzaghi (1943).
These curves can be found in Figure 1 of this article. To calculate the elastic settlement of pile groups, the use of Fox’s correction curves (1948) is required in order to obtain the immediate settlement of a raft foundation that is equivalent to the pile group.
Figure 1: Values of Steinbrenner’s influence factor Ip (for v of 0.5)
To obtain the average immediate settlement of a foundation at a depth D below the surface where the deformation modulus is reasonably constant with depth, it will be found to be more convenient to use the influence factors of Christian and Carrier (1978). In general, this will be the case where the deformation modulus is reasonably constant with depth.
Average settlement ρi = μiμ0qnB/Eu ——– (2)
In the equation (2), it was assumed that Poisson’s ratio was equal to 0.5. Figure 2 shows the factors μi and μ0, both of which are associated with the depth of the equivalent raft, the thickness of the compressible soil layer, and the length/width ratio of the equivalent raft foundation.
Figure 2: Influence factors for calculating immediate settlements of flexible foundations of width B at depth D below ground surface (after Christian and Carrier(5.5)).
The value of the deformation modulus Eucan be determined by analysing the stress-strain curve that is produced when compressive loading is applied to the soil while the conditions are undrained. Only when the stress level is relatively low (shown by the line AB in Figure 3) does a curve of this type show purely elastic behaviour, which requires the use of a modulus of elasticity.
It is possible that the immediate settlement will be underestimated due to the fact that (Young’s modulus) corresponds to the straight-line portion. The standard procedure is to draw a secant AC to the stress-strain curve that corresponds to a compressive stress that is equal to the net foundation pressure at the base of the equivalent block foundation.
The secant AD can be drawn at a compressive stress of 1.5 or any other suitable multiple of the foundation pressure for a more conservative approach. Following this, one can calculate the deformation modulus Eu, as shown in Figure 3.
Figure 3: Determining deformation modulus Eu from stress-strain curve
As a result of sample disturbance, unreasonably low modulus values are obtained from stress-strain curves produced from conventional unconfined or triaxial compression tests in the laboratory. These results are not representative of the material’s true behaviour. Plate bearing tests carried out in boreholes or trial pits, as well as field testing carried out with a pressuremeter or Camkometer, provide the most reliable data for calculating modulus values.
Another possibility is that Eu is connected to the clay’s undrained shearing strength cu. According to Butler (1975), the relationship Eu = 400cu for London clay is a reasonable compromise between divergent data representing, on the one hand, the relationship established from laboratory testing, and on the other hand, the observations of the settlement of full-scale structures.
Recent years have seen the development of apparatus for obtaining piston-driven tube samples of stiff clays, as well as improvements in methods for coring weak rocks. Therefore, samples that have been subjected to a relatively minor disturbance can be provided for laboratory testing to obtain modulus values. There is apparatus available that can measure very small strains during uniaxial or triaxial compression tests, which can then be utilised to derive small strain modulus values.
In layered soils with different values of the deformation modulus Euin each layer or in soils which show a progressively increasing modulus with increases in depth, the strata below the base of the equivalent raft are divided into a number of representative horizontal layers, and an average value of Eu is assigned to each layer.
This ensures that the values of the deformation modulus Eu are consistent across the entire profile of the soil. The values for the dimensions L and B in Figure 2 are derived from the hypothesis that the load is distributed across the surface of each layer at an angle of thirty degrees with respect to the edges of the equivalent raft (Figure 4). After that, the total settlement of the piled foundation can be calculated as the sum of the average settlements for each soil layer that were determined using Equation 2.
Figure 4: Load distribution beneath pile group in layered soil formation
The assumption that the deformation modulus remains the same with depth is the basis for the influence values presented in Figure 1. Calculations that are based on a constant modulus, on the other hand, give inflated estimates of the amount of settlement that will occur.
This is because the modulus increases with depth in the majority of natural soil and rock formations. Butler (1974) developed a method for settlement calculations for the conditions of a deformation modulus increasing linearly with depth within a layer of finite thickness. This method was used for determining the amount that a layer would settle. The following equation will give you the value of the modulus at a depth z below the level of the foundation:
Eu = Ef(1 + kz/B) ——– (3)
and
ρi = qnBI’p/Eu ——– (4)
where Efrepresents the modulus of deformation at foundation level (the base of the equivalent raft), and ρi represents the settlement at the corner of the loaded area in the equation. In order to calculate the value of k, first plot the measured values of Eu against depth, and then draw a straight line through the points that are plotted. This will give you the values you need to substitute into equation 3.
In situations in which a plot of undrained shear strength versus depth has been obtained, the Eu vs depth line can be derived from the empirical relationships given earlier in this paragraph.
After you have determined k, you can use Butler’s curves, which are presented in Figure 5, to determine the appropriate factor for I’p. These are for different ratios of L/B at the level of the equivalent raft, and they are applicable for a compressible layer thickness that is not greater than 9B. The curves have been constructed on the basis of the assumption that an undrained condition will have a Poisson’s ratio of 0.5; this is so that the load can be applied immediately.
Figure 5: Values of the influence factor for deformation modules increasing linearly with depth and Poisson’s ratio of 0.5 (after Butler (1974))
In situations in which the pile group covers a large area and is, as a result, relatively flexible, it is possible that it will be necessary to calculate the settlements at the corners in addition to those in the area’s centre.
Consolidation Settlement of Pile Groups
The results of oedometer tests carried out on clay samples in the laboratory are used as the basis for the calculation of the consolidation settlement ρc. The pressure-to-voids ratio curves that were obtained from these tests are what are used to calculate the volume compressibility coefficient, denoted by the symbol mv.
It may be difficult to obtain satisfactory undisturbed samples for oedometer testing in hard glacial tills or in rocks that have been highly weathered and weakened by weathering. If the results of standard penetration tests are available, then the values of mv (as well as cu) can be obtained from the empirical relationships established by Stroud (1975) and shown in Figure 6.
Figure 6: Relationship between mass shear strength, modulus of volume compressibility, plasticity index and standard penetration test N-values (after Stroud (1975)) (a) N-value vs. undrained shear strength (b) N-value vs. modulus of volume compressibility
Once a representative value of mv has been obtained for each soil layer that is being stressed by the pile group, the oedometer settlement ρoed for this layer at the centre of the loaded area can be calculated using the equation;
ρoed = μdmv × σz × H ——– (5)
where μd is a depth factor, σz is the average effective vertical stress imposed on the soil layer due to the net foundation pressure qn at the base of the equivalent raft foundation and H is the thickness of the soil layer. The depth factor is obtained from Fox’s correction curves.
The oedometer settlement must now be corrected to obtain the field value of the consolidation settlement. The correction is made by applying a ‘geological factor’ μg to the oedometer settlement, where;
ρc = μg × ρoed ——– (6)
Published values of μg have been based on comparisons of the settlement of actual structures with computations made from laboratory oedometer tests. Values established by Skempton and Bjerrum (1975) are shown in Table 1.
Type of clay
μg value
Very sensitive clays (soft alluvial, estuarine and marine clays)
1.0 – 1.2
Normally-consolidated clays
0.7 – 1.0
Over-consolidated clays (London clay, Weald, Kimmeridge, Oxford and Lias clays)
After this, the total settlement of the pile group is determined by adding together the immediate and consolidation settlements that were calculated for each individual layer. A gradual decrease in compressibility with depth is a typical example of this phenomenon. When this occurs, the stressed zone beneath the pile group is segmented into a number of distinct horizontal layers.
The value of the modulus of elasticity (mv) for each of these layers is obtained by plotting the mv value against the depth, which is based on the results of the laboratory oedometer tests. At the level where the vertical stress has decreased to one tenth of qn, the level at which the base of the lowest layer is determined to be is chosen.
When calculating the total consolidation settlements for each layer, the depth factor, denoted by d, is multiplied by that total. It does not apply to the immediate settlement if that settlement has already been computed based on the factors shown in Figure 5.
If a pile group is topped by a deep rigid cap or if it is used to support a rigid superstructure, then the pile group can be considered to be equivalent to a rigid block foundation that has a uniform settlement. This is because both of these conditions ensure that the pile group remains in its original position.
To calculate the value of the latter, a “rigidity factor” is applied to the consolidation settlement that was obtained from the equivalent flexible raft foundation. The immediate settlement can be calculated using Equation 2. This immediate settlement is the same as the average settlement that is given by a rigid foundation. The value of 0.8 for the rigidity factor is widely accepted as being appropriate.
Thus; Settlement of rigid pile group/Settlement of flexible pile group = 0.8
It is possible, for the purposes of this condition, to consider a pile group that is composed of a number of small clusters or individual piles connected by ground beams or by a flexible ground floor slab to be equivalent to a flexible raft foundation at depth.
This is the case if the pile group is connected by ground beams. According to the calculations described above, the consolidation settlements that occurred at the corners of the piled area make up approximately one-half of the settlement that occurred in the centre of the group.
The use of Equation 1, with the substitution of a deformation modulus obtained for loading under drained conditions, is yet another method for estimating the total settlement of a structure that is resting on an over-consolidated clay.
This modulus has been given the name Ev‘, and it can be found at the bottom of Equation 1, where it stands in place of Eu. It roughly corresponds to the value of 1/mv. When used to calculate consolidation settlements, the equation does not adhere to strict validity standards because it assumes a material that is both homogenous and elastic.
However, when applied to over-consolidated clays for which the settlements are relatively small, it has been observed through experience that the method gives predictions that are reasonably reliable. The success of utilising the method is dependent on the collection of sufficient data correlating the observed settlements of structures with the determinations of from plate loading tests and laboratory tests on good undisturbed samples of clay. This is necessary for a successful outcome.
In his analysis of the settlement of structures on over-consolidated clays, Butler (1974) related Ev‘ to the undrained cohesion cu and determined that the relationship for London clay is Ev‘ = 130Cu.
In the settlement analysis of a group of piles, it is better to adopt an approach that is more rational, which is to consider immediate settlements and consolidation settlements separately. This appropriately takes into account the effects that time has had on the location as well as its geological history. The prediction of consolidation settlements based on oedometer tests conducted in the laboratory has been found to lead to reasonably accurate results, provided that a sufficient number of good undisturbed samples have been obtained at the site investigation stage.
The adoption of the method that is based on the total settlement deformation modulus is contingent upon the collection of adequate observational data, first regarding the relationship between the undrained shearing strength and the deformation modulus, and secondly regarding the actual settlement of structures from which the relationships can be checked.
The adoption of this method is dependent upon the collection of adequate observational data. It is highly unlikely that accurate results can be obtained from triaxial compression tests carried out in the lab, so any attempt to do so is likely to end in failure. The modulus can be determined most accurately using the Eu/cu and relationship formulas, which need to be derived from plate bearing tests that have been competently carried out and field observations of settlement.
The reader is directed to a report written by Padfield and Sharrock (1983) for CIRIA that contains a general discussion on the subject of the settlement of foundations on clays. According to what they have found, the immediate settlement is approximately equal to 0.5 to 0.6 times the oedometer settlement, while the consolidation settlement, is approximately equal to 0.4 to 0.5 times the oedometer settlement. This is true for stiff overconsolidated clays. When it comes to normally consolidated soft clays, the immediate settlement is roughly equivalent to 0.1 times the oedometer settlement, and the consolidation settlement is roughly equivalent to the oedometer settlement.
The steps in making a settlement analysis of a pile group in, or transmitting stress to, a cohesive soil can be summarized as follows.
For the required length of pile, and form of pile bearing (i.e. friction pile or end-bearing pile), draw the equivalent flexible raft foundation represented by the group.
From the results of field or laboratory tests assign values to Eu and mv for each soil layer significantly stressed by the equivalent raft.
Calculate the immediate settlement of ρi of each soil layer using equation 2, and assuming a spread of load of 30° from the vertical to obtain qn at the surface of each layer (Figure 4). Alternatively calculate on the assumption of a linearly increasing modulus.
Calculate the consolidation settlement ρc for each soil layer from Equations 5 and 6, using relevant charts to obtain the vertical stress at the centre of each layer.
Apply a rigidity factor to obtain the average settlement for a rigid pile group.
The consolidation settlement calculated as described above is the final settlement after a period of some months or years after the completion of loading. It is rarely necessary to calculate the movement at intermediate times, i.e. to establish the time settlement curve, since in most cases the movement is virtually complete after a period of a very few years and it is only the final settlement which is of interest to the structural engineer. If time effects are of significance, however, the procedure for obtaining the time-settlement curve can be obtained from standard works of reference on soil mechanics.
References
TERZAGHI, K (1943). Theoretical Soil Mechanics, John Wiley, New York, p. 425.
FOX, E.N. (1948). The mean elastic settlement of a uniformly-loaded area at a depth below the ground surface, Proceedings of the 2nd International Conference, ISSMFE, Rotterdam, Vol. 1, pp. 129–32.
CHRISTIAN, J.T. and CARRIER, W.D. (1978). Janbu, Bjerrum and Kjaernsli’s chart reinterpreted, Canadian Geotechnical Journal, Vol. 15, pp. 123–8.
BUTLER, F.G. (1974). General report and state-of-the-art review, Session 3, Proceedings of the Conference on Settlement of Structures, Cambridge, Pentech Press, London, pp. 531–78.
STROUD, M.A. (1975) The standard penetration test in insensitive clays, Proceedings of the European Symposium on Penetration Testing, Stockholm, Vol. 2, pp. 367–75.
SKEMPTON, A.W. and BJERRUM, L. A (1957). contribution to the settlement analysis of foundations on clay , Geotechnique, Vol. 7, No. 4, pp. 168–78.
PADFIELD, C.J. and SHARROCK, M.J. (1983). Settlement of structures on clay soils, Construction Industry Research and Information Association (CIRIA), Special Publication 27, 1983.
Feature Image: Ata A., Badrawi E., Nabil M. (2015): Numerical analysis of unconnected piled raft with cushion. Ain Shams Engineering Journal, 6(2):421-428 https://doi.org/10.1016/j.asej.2014.11.002.
During the design and construction of pile foundations, there are instances of eccentric loading that may arise because column positions do not align with the centroid of the pile group. This eccentricity usually induces bending moments which affect the axial load on each pile in a group. Therefore, the evaluation of axial load distribution of piles under eccentric vertical loading is very important, so that the safe working load on each individual pile in not exceeded. In this article, we will discuss the effects of eccentric loading on a group’s axial load on piles.
Every deep foundation project is unique. Most clients frown at soil test reports recommending pile foundations for their projects due to the cost implication of deep foundations. Since there are a variety of pile types, professionals in the construction industry should always assure their clients of the most economical pile type for their projects.
Generally, pile foundations are used when suitable foundation conditions are not present at or near ground level, making the use of shallow traditional foundations uneconomical. Furthermore, it is often recommended to use more than one pile below a column, depending on the pile’s safe working load and the service axial load from the column. Piles in group are usually preferred, particularly if the piles are driven because driven piles in group have large load bearing capacity, a small settlement, and good stability.
Axial Load on Eccentrically Loaded Piles
When a pile group is subjected to eccentric loading, it raises questions about how safe and reliable the piles will become, especially if they were designed initially with zero loading eccentricity. Therefore, designers and construction engineers need to ascertain the eccentricity limit within which a design is still safe and applicable.
For a group of piles, the vertical load (P) is the sum of the column loads, the self-weight of the pile cap, any backfilling, and the surcharge on the pile cap. Using the figure above, the axial load (Rp) on any pile in a group can be calculated using the equation below.
Where; Rp = Axial load on any pile P = Vertical load on pile group n = Number of piles in pile group Mxx and Myy = Moment about x-x and y-y on pile group respectively x and y = Distance of any pile from x-x and y-y respectively ⅀x2 and ⅀y2 = Sum of the squares of the distances of all piles from x-x and y-y respectively
However, this method of calculating the axial load on any pile in a group is only valid for rigid pile caps, that is, very stiff pile caps. Therefore, large pile caps and rafts should be treated as flexible, and a rigorous grillage or finite element analysis should be undertaken to determine the axial load on any pile.
Solved Example
With reference to the pile layout in the diagram below, column position is on grid B2. Column load = 4500 kN. Assuming pile diameter = 400mm, pile cap depth = 900mm, and there is a surcharge load of 18.8 kN/m2
a) Determine the axial load on each pile in the group. b) If the column moves to a position x, that is, the right of gridline B by 0.3 m and below gridline 2 by 0.6 m simultaneously. What is the effect on the axial load on each pile?
Solution
Column load = 4500 kN Length and breadth of pile cap = 3100 mm Pile diameter = 400 mm Pile cap depth = 900mm Number of piles (n) = 9 Unit weight of concrete = 24 kN/m3 Surcharge load = 18.8 kN/m2
a) The axial load on each pile is the same because the vertical load is applied concentrically, and the piles are spaced at sufficient and equal distances apart. Axial load (Rp) = P/n= 4888.244/9 = 543.14 kN
b) Using equation (1) ⅀x2 = ⅀y2 = (1.2)2 + (1.2)2 + (1.2)2 + (1.2)2 + (1.2)2 + (1.2)2 + (0)2 + (0)2 + (0)2 = 8.64 m
From the solved example above, we can see that loading eccentricity significantly impacts the axial load on any pile in a pile group. Moreover, the effect is quite profound on PC3, with a 103.6% increment in axial load compared to when the vertical load (P) acts on the pile group concentrically.
Furthermore, piles are usually designed to transfer compression loads. As a result of the eccentric loading of the pile group, PA1 is now in tension. This means the pile is subjected to uplift forces that might otherwise cause it to be pulled out of the ground. Therefore, the pile group will fail if the piles are not designed to resist uplift forces.
Construction engineers must give more attention to the positioning of columns to ensure that column centerlines align with the centroid of pile groups. Indeed, the load capacity of piles is usually factored by 2 – 2.5, but this should not prevent construction engineers from constructing with due diligence.
Reference(s)
[1] Di Laora, R., de Sanctis, L. & Aversa, S. Bearing capacity of pile groups under vertical eccentric load. Acta Geotech. 14, 193–205 (2019). https://doi.org/10.1007/s11440-018-0646-5
