A box girder is formed when two web plates are joined by a common flange at both the top and the bottom. The closed cell which is formed has a much greater torsional stiffness and strength than an open section and this is the main reason why box girder configuration is usually adopted in long span bridges. The torsion and distortion rigidity of box beam is due to the closed section of the box beam.
It has been shown by reseachers that long span bridges with wider decks and eccentric loading on cross-section will suffer in curvature in longitudinal and transverse direction, thereby causing heavy distortion of cross-section. As a result, long span bridges with wide decks should have high torsional rigidity in order to reduce the distortion of cross-section deck to a minimum. This is one of the downsides of T-beam and deck system. Accordingly, box girders are more suitable for larger spans and wider decks, and any eccentric load that will cause high torsional stresses which will be counteracted by the box section (see Figure 1 for typical box girder bridge). However, despite being an efficient cross-section, we should know that that the analysis of such sections are more complicated due combination of flexure, shear, torsion, and distortion. In this post, we are going to evaluate the potentials of Staad Pro software in the analysis of box girder bridge subjected to Load Model 1 of Eurocode 1 Part 2.
Fig 1: Curved box girder bridge
The cross-section of the bridge deck is shown in Figure 2. The following data was used to model the bridge deck on Staad Pro.
Length of bridge = 30 m Support condition = Pinned to the piers Unit weight of concrete = 25 kN/m3 Thickness of sections = 250 mm (web and flanges) Unit weight of asphalt overlay = 22 kN/m3 Thickness of asphalt = 75 mm
Hence, the bridge deck will be subjected to the following actions;
(1) Self weight of girder (2) Weight of asphalt overlay (3) Load Model 1 uniformly distributed load (4) Load Model 1 tandem wheel load (modeled as moving load on Staad Pro) (5) Pedestrian load on the cantilever
In this analysis, self weight of girder and asphalt were modelled as one load case, Load Model 1 (UDL) and pedestrian action were modeled as a separate load case, while Load Model 1 (tandem wheel load) were modelled as another separate load case.
Fig 2: Cross-section of box girder bridge
The typical combined traffic action on the bridge deck is shown in Figure 3.
Fig 3: Box girder bridge subjected to Load Model 1 of Eurocode
When modelled on Staad Pro, the configuration in Figure 4 was obtained. The rendered 3D view is shown in Figure 5.
Fig 4: Model of box girder bridge on Staad Pro
Fig 5: 3D rendering of the box girder bridge
When the structure was analysed, the following results were obtained;
Fig 6: Transverse bending moment due to self weight of girder and weight of asphalt overlay
Fig 7: Longitudinal bending moment due to self weight of girder and weight of asphalt overlay
Fig 8: Transverse bending moment on girder due to traffic UDLFig 9: Flexural-distortional behaviour of the girder due to traffic UDLFigure 10: Transverse bending on the section (flexural – distortional) due to wheel load at the most critical location
Figure 11: Longitudinal bending on the section (flexural – distortional) due to wheel load at the most critical location
Fig 12: Torsional-distortional action effect due to wheel load at the most onerous section
From the results, it could be seen that uniform loads have barely no effect on the torsional-distortional behaviour of box girders, however, unbalanced wheel loads are critical in the behaviour of such structures. We will carry out more studies on this bridge deck behaviour using Staad Pro. Thank you for stopping by today.
Columns are vertical structural elements used in transmitting floor load to a lower level or to the substructure. As a result, their stability and structural resistance are of major importance in the integrity of a structure. In buildings, columns are usually classified as either slender, short, or intermediate.
Based on the loading they are subjected to, columns can be described as being axially loaded (subjected to axial load only), uniaxially loaded (subjected to axial load and bending moment in one direction), or biaxially loaded (subjected to axial load and bending in two directions).
While the procedure for the design of columns is well known by structural engineers, the process of constructing columns on site to achieve what is in the drawing is another different challenge altogether. The success and accuracy of any building have something to do with the setting out. In this article, we are going to briefly describe the procedures for setting out columns from the foundation and from the floor slab using locally available construction tools.
Let us briefly review this process.
Setting out columns from the foundation
Column starter bars will normally start from the foundation which can be a pad, raft, or pile cap. For local construction works, setting out works will involve establishing the building lines, and marking out column locations on the profile board which must be established all round the building.
Detailed information on how to set out buildings are abundant on the internet. These points marked out on the profile board are used as reference for excavation of the trenches for the foundation. After excavating to the required depth, the foundation should be blinded according to the drawing specifications.
Fig 1: Excavated trench for building foundation
After blinding the footing excavation, lines can be thrown as appropriate, and the proper location of each column can be identified. Therefore, the accuracy of column positioning at the foundation stage depends on the accuracy of the setting-out work. This is why proper setting out is important, especially during the foundation stage. The footing reinforcement and column reinforcement starters can be installed properly and held in place until the concreting of the foundation is done.
Fig 2: Setting out starter bars using lines
Setting out columns from the floor slab
If the foundation has been completed and there is a need to continue the structure from the ground floor or first-floor slab, then setting out will need to be done for the kickers of the columns. Setting out of kickers can be done by surveyors or the site engineer using simple tools such as tape, builder’s square, and lines.
It is usually erroneous to take a corner of a building (edge of a slab) as an absolute reference unless you are sure that the edges of your slab are perfectly straight and aligned in accordance with the drawing. Sometimes after casting the slab, you can experience an error of about ±20 mm with reference to the drawing. If you want to avoid cranking the column rebars, you should deal with such errors immediately.
Fig 3: Column reinforcements with kickers
The fastest way of dealing with this is to divide the floor of the building into two as shown in Figure 4 (red lines). The lines must not be strictly through the centre of the building, but from any convenient point that will enable you to take measurements with ease.
However, you should ensure that the intersection of the lines is at 90 degrees to each other. You can use a laser machine or the 3-4-5 method to ensure that the lines are square. Instead of using any edge of the building as a reference, you should the centrelines you established as a reference to set out the position of the columns. You can rarely go wrong using this method, and another advantage is that if there are errors in the dimensions of the floor slab, the error is shared equally.
Fig 4: Division of floor slab into parts for kicker setting out
Suggestions on improving this article are welcome and will be incorporated.
The characteristic compressive strength of concrete to be used in construction is usually specified in the working drawings, design calculation sheets, and contract documents. The grade of concrete specified is used by bidders/contractors in pricing the job. Therefore, it is expected that it must be achieved on site.
There are different ways of achieving grade 25 concrete on site, and it vary based on each contractor’s experience. This guide is from a personal site experience. If there is no special requirement from the consultants, contractors are usually free to use any acceptable way to achieve the specified grade of concrete.
Normally, trial mixes should be done before the commencement of any project, but experienced contractors are already familiar with the mixes that they have done over and over again unless there is a new consideration to be made. The aim of this article is to show you how you can achieve grade 25 concrete based on site experience only without any technical calculations.
The applicability of this procedure is for low scale – low budget construction where there is no batching plant and sophisticated machines/equipment. If you follow this simple procedure, your concrete cubes for grade 25 concrete will exceed 17 MPa and 25 MPa after 7 and 28 days crushing tests respectively.
To get started, you will need to have the following available:
(1) Trustworthy supervisors (2) Functional concrete mixer (3) Functional concrete vibrator/poker (4) Headpans, 25 litre buckets, and shovels (5) All materials needed for concrete production – gravel, sand, cement (grade 42.5R), and water (6) Concrete cube moulds
Fig 1: Concrete mixer
Fig 2: Concrete vibrator/poker
Steps for Achieving Grade 25 Concrete On-site
Having gotten all materials ready, you can follow the steps outlined below:
(1) Recommended mix ratio To achieve grade 25 concrete in local construction, I recommend a mix ratio of 1:2½:3½. What this means is that for 1 bag of cement, you add 5 headpans of sand, and 7 headpans of granite. The coarse aggregate that has been verified for this mix proportion is 3/4″ granite (19 mm size aggregate). The sand should be properly graded, and the cement should be sound with a strength grade of 42.5N or 42.5R.
Note: The popular 1:2:4 mix ratio sums up to 7 (that is 1 + 2 + 4 = 7), and the recommended 1:2½:3½ mix ratio also sums up to 7. It has been discovered on site that for 3/4″ granite in a 1:2:4 mix ratio, the quantity of granite usually appears excessive at about 0.4 – 0.6 water/cement ratio, thereby giving workmen difficulty in mixing, placing, and consolidating the concrete.
This leads to the addition of excess water which compromises the strength of the concrete. While this can be taken care of by the use of superplasticizer admixture, low-budget constructions do not normally factor admixtures into the cost. Therefore, by slightly adjusting the sand and gravel content from 2:4 to 2½:3½ respectively, a better workable mix is achieved without any serious consequences.
(2) Water Content For the mix ratio described above, the quantity of water to be added must not exceed 25 litres. If it exceeds 25 litres, you have deviated from the provisions of this guide for obtaining grade 25 concrete. On-site, you can achieve this by pouring one bucket of water only into the concrete mixer bunker.
There might be grumblings about the workability of the mix from the workmen, but stick to the plan as it will not affect your finished work. If you have a very tight reinforcement arrangement that you need very workable concrete, you might reconsider using this guide. The bucket size recommended for this is the 25 litres empty paint bucket shown in Figure 3. If you get the mix ratio right and add excess water, the test result will fail.
Fig 3: 25 Litres empty paint bucket
(3) Do not heap the headpans The granite and the sand must not be heaped if you are to achieve grade 25 concrete using this guide. The sand and granite must be made to flush with the top surface of the headpan as shown in Figure 4. If the sand and granite are heaped you have deviated from the provisions of this guide.
Fig 4: Unheaped headpan of sharp sand
(4) Supervision To achieve steps 1, 2, and 3 above, you need strict supervision and monitoring. First of all, you need a supervisor at the loading points, who must ensure that the granite and sand are not heaped. It is good to have different monitors for sand and granite.
To achieve this, after filling each head pan with sand/gravel, it must be levelled using a shovel or piece of wood before being carried to the bunker. Secondly, you need another supervisor at the bunker to ensure that the correct quantity of sand, granite, and water is being poured in for one bag of cement.
Fig 5: Supervision of concreting during local construction
(5) Placement and consolidation The concrete produced from the mixture should be transported without delay to the required point where it is poured and consolidated using a vibrator/poker. The curing process should also begin as soon as it sets.
Fig 6: Placement and Consolidation of concrete for a floor slab
(6) Collection of samples for testing The 150mm x 150mm steel moulds to be used for sample collection should be oiled, and the concrete poured in three layers. Each layer should be subjected to 25 tampings using at least 16 mm diameter reinforcement off-cut. You can do well by hitting the sides of the mould during compaction to ensure proper consolidation.
After filling the moulds and compacting the concrete properly, the surface is leveled and finished smoothly with a trowel. Remove the sample from under the sun (especially in tropics), and cover it with a polyethylene nylon sheet. You can label the sample after about one hour.
Demould the sample the following morning and cure in a clean water tank for the required number of days. 6 or 9 samples might be required for each batch of concreting. However, consultants usually require 7 days and 28 days test results, and an average of 3 cubes should be crushed for each test. In some cases, 14 days’ results might be required.
Figure 7: Preparation of concrete cube samples
If you have followed these procedures properly, you should get satisfactory results from the laboratory for grade 25 concrete. However, note that this does not substitute for proper concrete mix design where it is required. Find below some laboratory test results from the procedure described above. The target characteristic strength after 28 days was 25 MPa. In one of the cases (Figures 10 and 11), regulatory agencies came by themselves and collected samples from the point of casting, and the result was found satisfactory.
Fig 8: 14 days compressive test result of a floor slab (Average 14 days strength = 23.48 MPa)
Fig 9: 28 days test result of the same floor slab (Average 28 days strength = 26.93 MPa)
Fig 10: 7 days test result of a floor beam (Average 7 days strength = 18.46 MPa)
Fig 11: 28 days test result of the same floor beam (Average 28 days strength = 27.14 MPa)
A ribbed slab is a type of reinforced concrete slab in which some of the volume of concrete in the tension zone is removed and replaced with hollow blocks or left as voids. This reduction in the volume of concrete in the tension zone (below the neutral axis) is based on the assumption that the tensile strength of concrete is zero, hence all the tensile stress is borne by the reinforcements in the tension zone. The resulting construction is considerably lighter than a solid cross-section.
This design and construction concept is useful in long-span construction of floors (say spans greater than 5m), where the self-weight becomes excessive when compared to the applied dead and imposed loads, thereby resulting in an uneconomic method of construction. One method of overcoming this problem is to use ribbed slabs which are suitable for longer spans supporting light loading, as in residential or commercial buildings.
Design Example The layout of a floor slab is shown in Figure 1 below. Design the floor to satisfy ultimate and serviceability limit state requirements. (Concrete grade = 30 MPa, Yield strength of reinforcement = 500 MPa, variable action on floor = 2.5 kPa, Fire rating = 1 hour 30 minutes).
Fig 1: Ribbed slab layout
Fig 2: Section through the ribbed slab
Load Analysis For 550 mm c/c rib spacing;
Permanent Actions Weight of topping: 0.050 × 25 × 0.55 = 0.6875 kN/m Weight of ribs: 0.15 × 0.2 × 25 = 0.75 kN/m Weight of finishes: 1.2 × 0.55 = 0.66 kN/m Partition allowance: 1.5 × 0.55 = 0.825 kN/m Self weight of clay hollow pot = 0.65 kN/m Total dead load gk = 3.572 kN/m
Effective depth (d) = h – Cnom – ϕ/2 – ϕlinks Assuming ϕ12 mm bars will be employed for the main bars, and ϕ8 mm bars for the stirrups (links) d = 250 – 25 – (12/2) -8 = 211 mm
k = MEd/(fckbd2) = (21.56 × 106)/(30 × 550 × 2112) = 0.0293 Since k < 0.167, no compression reinforcement required
z = d[0.5+ √(0.25 – 0.882k)] k = 0.0709 z = d{0.5+ √[0.25 – (0.882 × 0.0293)]} = 0.95d = 200.45 mm
Depth to neutral axis x = 2.5(d – z) = 2.5(211 – 200.45) = 26.375 mm < 1.25hf (62.5 mm)
Therefore, we can design the rib as a rectangular section;
Area of tension reinforcement As1 = MEd/(0.87fyk z) As1 = MEd / (0.87fyk z) = (21.56 × 106) / (0.87 × 500 × 0.95 × 211) = 247.26 mm2
Since VRd,c (23.815 kN) > VEd (17.25 kN), no shear reinforcement is required.
According to clause 6.2.1(4), minimum shear reinforcements may be omitted in ribbed slabs where transverse distribution of loads is possible. But for this design, we will therefore provide minimum shear reinforcement.
Minimum shear reinforcement; Asw / S = ρw,min × bw × sinα (α = 90° for vertical links) ρw,min = (0.08 × √(fck)) / fyk = (0.08 × √30)/500 = 0.000876 Asw/Smin = 0.000876 × 150 × 1 = 0.131 Maximum spacing of shear links = 0.75d = 0.75 × 211 = 158.75 mm Provide H8mm @ 150 mm c/c as shear links.
Slab Topping A142 BRC Mesh can be provided or H8 @ 250mm c/c
For more information on design and consultancy to accomplish the most challenging design brief, contact the author on info@structville.com. Thank you, and God bless you.
Underground water retaining structures have wide applications in residential, commercial, and industrial buildings. According to Lagos State Development Laws (2005), a building with more than 5 floors must have an underground water tank with a minimum capacity of 20,000 litres.
Sometimes, the conventional method of constructing an underground water tank which involves excavating the area, preparing the bottom and slope of excavation, blinding, placing of reinforcements/formwork, bracing, etc before concreting and backfilling may not always be feasible or easy for the contractors. This may be a result of high water table, soft/unstable soils, lack of space, etc. In this case, a top-down construction approach is usually adopted, where the shell of the tank is cast on the ground surface and then sunk to the desired depth in a process similar to the construction of caissons or well foundations.
The advantages of this procedure are; (1) Improved quality assurance of the tank shells due to easy inspection. (2) The quality of concreting is assured since it is done on the surface. (3) The problem of dealing with groundwater is minimised. (4) The challenge of bracing and supporting the sides of excavation is completely obliterated. (5) There will be no need for backfilling the sides of the tank
The disadvantages of this procedure are; (1) The perfect verticality of the tank after sinking may not be guaranteed (2) The shell will need to cure and achieve enough strength before the sinking will begin, otherwise the sinking stresses might damage the shell (3) Adequate care will need to be taken during concreting of the base, since it may not be done in a perfectly dry condition. (4) Extra care is needed to guarantee the water tightness of the base. (5) Sinking gets difficult when an obstruction is encountered.
Let us now explain the process of constructing and sinking underground water tanks and septic tanks in high water table areas in full.
Step 1: Geometric design The sizing of the tank should be based on occupancy usage and projected water demand of the building. This can be carried out according to the process described in Ubani (2018). A typical tank dimension (plan and section) is shown in Figure 1.
Fig 1: Typical sizing of an underground tank (Ubani, 2018)
Step 2: Structural design The structural design of the tank should take into account the earth pressure, water pressure, imposed load on top of the tank (since the compound will serve as useful space), etc. The sizing of the tank members and the reinforcements provided should satisfy ultimate limit state requirements (for example bending, shear, etc) and serviceability limit state requirements (for example, cracking). The tank should also be able to resist uplift due to groundwater, and the structural engineer should detail the processes required for the water tightness of the structure. These processes are well described in Ubani (2018).
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Step 3: Setting out After the design has been properly done and approved, the contractor will need to set out the tank shell so that it will be cast at the exact intended position which is usually specified in the site layout drawing. The tank should be positioned in such a way that the supply and discharge pipes will have minimum obstruction. The contractor can set out the tank using the popular 3-4-5 method and ensure that the edges are square depending on the orientation of the tank. For cylindrical tanks, the appropriate setting out method should be used. Just like in a building, the four edges of the tank (centre to centre) should be established, and a little setback (say 600 mm) established to give the iron benders a good space to position their rebars (see Figure 2). A profile board should be established around the setback, and the exact width of the tank walls established on the profile board.
Fig. 2: Typical setting out sketch
Step 4: Leveling and blinding The ground surface (perimeter of the tank walls) should be levelled and made firm so that the base of the walls will be horizontal and level. After that, a mortar blinding should be done on the prepared surface to receive the reinforcements. The width of the blinding should be such that the carpenter will be able to nail or mark his kicker (or whichever method that is adopted) on the blinding. The blinding should be easily broken off as soon as sinking starts.
Step 5: Setting out of kickers From the established profile lines, the width of the tank walls should be transferred to the blinding on the ground (see Figure 3). The contractor can use laser, spirit levels, iron squares, or whatever instrument at his disposal to achieve this. Perfectly cut pieces of wood can be used as markers for the iron bender to position his rebars properly.
Fig 3: Typical kicker for wall
Step 6: Installation of reinforcements The iron bender should install the reinforcements according to the structural drawing (see example on Figure 4). The concrete cover to the surface facing the soil should typically be taken as 50 mm while the concrete cover to the water storage surface should be taken as 40 mm. The iron bender should ensure that the binding wires are neatly trimmed or pointed inwards after tying because roughly done binding wire can be a source of leakage in the tank. The reinforcement of the wall should be made to project into the base (in L-shape) before the formwork and concreting is done. The water bar (preferably stainless steel water bar) should be installed too as shown in Figure 5.
Fig. 4: Typical tank wall reinforcement details
Step 7: Formwork The formwork should be properly done and braced as suitable for water retaining structures. Lost ties with hydrophilic water bars wrapped round them with suitable plugs can be adopted to hold the panels in place. The panels should be made with suitable materials such as reusable plastics or marine plywoods. Acrow props should be used to support the bracings to ensure perfect verticality. A small toe of about 150 mm should be provided inwards the tank (see Figure 5). The carpenter or other contractors should also prepare a good scaffold/platform for casting of the concrete.
Step 8: Concreting A minimum of grade 30 concrete with water to cement ratio not exceeding 0.5 should be used. Read the specifications for concrete for water retaining structures here. Approved water proofing admixtures should be used also. The concreting should get to the level where the top slab should start. The reinforcement of the walls should be allowed to project into the top slab for proper shear connection. The toe and kickers could be cast first with water bars installed. This will aid the carpenters in achieving a perfect dimension. The first phase of casting is shown in Figure 5 while the full shell is shown in Figure 6. Contractors can adopt a different approach if they wish.
Fig 5: Typical kicker with toe showing all projecting reinforcements
Fig 6: Fully cast tank shell
Step 9: Curing and striking The concrete should be cured and the formwork removed after an approved period of time by the structural engineer.
Step 10: Sinking Labourers should climb into the tank and excavate the soil in order to sink the tank to the desired depth. This should be done as carefully as possible and two functional pumping machines (depending on the size of the tank) should be made available to enable the labourers excavate in the dry. The reinforcements projecting into the base should be bent upwards to enable the labourers do their excavation. The sinking is assisted by the weight of the tank which must overcome the frictional resistance from the soil.
Fig 7: Sinking of tank shell
Step 11: Concreting the base After sinking to the desired depth, the base surface should be prepared, cleaned, and properly blinded with concrete of high quality. The reinforcement should be straightened back, lapped properly, and the transverse reinforcements installed. All the water bars should be checked to ensure that they are perfect condition and the remaining portion of the base should be cast to the required thickness with waterproofing admixture. All through this process, there should be constant pumping out of the groundwater. The total weight of the tank at this point should be greater than the upthrust to avoid flotation.
Step 12: Check for water tightness Typically after casting, you should check the following day to ensure that the tank is dry (note that pumping of groundwater to stay below the base should continue while the concrete gains strength). If the tank is observed to be dry after setting of concrete and pumping of groundwater stops, it means that water exclusion was achieved. The tank can then be flooded with water to check drop in level and to also encourage the autogenous healing of any possible cracks.
References: (1) Lagos State Physical Planning and Development Regulations (2005): Lagos State Urban and Regional Planning Law. L.S.L.N. No 7 (2) Ubani O.U. (2018): Structural Design of Swimming Pools and Underground Water Tanks. Structville Integrated Services Limited, Lagos Nigeria. ISBN 9789-7869-5413
A multi-storey building can develop lateral vibrations in the two principal directions and torsional vibration around its vertical shear centre axis. The natural frequency of a structure is the frequency at its free or natural vibration. For a simple mass-spring system, the natural frequency is given by Equation (1);
f = 1/2π √(k/m) —– (1)
Where f is the natural frequency (Hz), and k and m are the stiffness and mass respectively.
Low-rise buildings have high natural frequencies, while tall flexible buildings are characterized by their low natural frequencies. Resonance occurs when the frequency of excitation is close to a system’s natural frequency. Therefore low-rise buildings are more susceptible to earthquake-induced load than to wind loads.
Although tall buildings are flexible and are therefore further away from the peak in the spectral density function of earthquakes, they are susceptible to the low-frequency range of seismic action. Wind poses a special hazard to tall buildings since their fundamental frequencies move towards the wind-spectrum peaks with higher flexibility. Therefore, it is important to calculate the natural frequency of multi-storey buildings.
Zalka (2013) suggested manual procedures for calculating the natural frequency of multi-storey frames. The equations used for calculating the natural frequency of a multi-storey frame is based on the whole bracing system, which can be characterised by the shear stiffness of the frameworks and coupled shear walls, the global bending stiffness of the frameworks and coupled shear walls, and the local bending stiffness of the individual columns/wall sections, shear walls, and cores.
By combining the individual bracing elements, linked by the floor slabs, to form a single cantilever, an equivalent system can be established with shear stiffness K, global bending stiffness EIg and local bending stiffness EI. The shear stiffness and the global bending stiffness are not independent of each other and can be incorporated into an effective shear stiffness, leading to a single equivalent column with effective shear stiffness Ke and bending stiffness EI.
Fig 1: Model for the lateral vibration analysis. a) bracing system consisting of frames, coupled shear walls, shear walls and cores, b) equivalent column (Zalka, 2013)
We will show how this is done with a solved example. Let us consider the frame of a tall building shown in Figure 2. The properties of the building are as follows;
Total height of building H = 90 m Number of storeys n = 30 Height of each storey h = 3 m Width of each bay L = 5 m Dimension of columns = 500 x 500 mm Dimension of beams = 600 x 400 mm Modulus of elasticity of concrete = 28 × 106 = kN/m2 Mass at each level = 2100 kg/m
Fig 2: Frame of a 30 storey building
Solution Second moment of area of beams IB = (bd3)/12 = (0.4 × 0.63)/12 = 0.0072 m4 Second moment of area of columns IC = (bd3)/12 = (0.5 × 0.53)/12 = 0.0052083 m4 Flexural rigidity of beams EIb = 28 × 106 × 0.0072 = 201600 kN.m2 Flexural rigidity of columns EIC = 28 × 106 × 0.0052083 = 145833.33 kN.m2
The part of shear stiffness associated with beams = KB = ∑(12EIb)/(Lih) = 2 × [(12 × 201600)/(5 × 3)] = 322560 kN
The part of the shear stiffness associated with the columns is; KC = ∑(12EIC)/h2 = 3 × [(12 × 145833.33)/32] = 583333.32 kN
From the above, the reduction factor r can be defined as; r = (KC)/(KB + KC) = (583333.32)/(322560 + 583333.32) = 0.6439
The original shear stiffness of the frame work can now be defined as; K = KB × r = 322560 × 0.6439 = 207696.384 kN
With the above shear stiffness, the square of the fundamental frequency (fs2) of the framework due to shear deformation can be calculated as;
fs’2= [1/(4H)2] × [rf2.K/m]
Where rf is the mass distribution factor given by: rf = √n/(n + 2.06) = √30/(30 + 2.06) = 0.967 (Note that n = number of storeys = 30)
For the local bending stiffness (EI = EIc r), the sum of the second moments of area of the columns should be produced (and multiplied by reduction factor r). As the bays of the framework are identical, the second moment of area of one column is simply multiplied by n and r (the reduction factor):
I = r∑Ic = 0.6439 x 3 x 0.0052083 = 0.010 m4
The fundamental frequency which is associated with the local bending stiffness is defined by;
fb2 = (0.313.rf2.EI)/H4m = (0.313 × 0.9672 × 28 × 106 × 0.010)/(904 x 2100) = 5.948 x 10-7 Hz2
Retaining walls are structures constructed for the purpose of retaining earth fill. There are different types of retaining walls such as cantilever retaining walls, gravity retaining walls, counterfort retaining walls, buttress walls, etc. Each of these has found wide applications in various civil engineering construction works such as embankments, highways, basements, etc.
Traditionally, cantilever retaining walls are designed as line elements where the width of the cantilever is analysed per unit strip (usually per metre length). The base of the cantilever is usually considered rigid, and as a result, a linear pressure distribution under the base is usually assumed.
However, when the dimensions of the retaining wall are such that a rigid behaviour cannot be obtained, or the soil is so soft or subjected to movement, then there will be significant soil-structure interaction. In this approach, the deformation of the soil and the structure itself contributes to the response of the structure to the externally applied load.
In this article, we are going to investigate the interaction of cantilever retaining walls with the soil supporting them in order to see how this affects the settlement, internal forces, and base pressure distribution. This will be carried out using Staad Pro software.
The factors usually checked out for in design of cantilever retaining walls are;
(1) Overturning
(2) Sliding
(3) Bearing capacity
(4) Uplift (rarely critical)
(5) Structural resistance
Methodology We are going to consider the analysis and design of the retaining wall with the dimensions and soil properties shown in Figure 1.
Figure 1: Cantilever retaining wall on sand
The retaining wall was modelled using concrete plate elements of mesh size 0.5m x 0.5m on Staad Pro software. The unit weight of concrete was set to be 24 kN/m3 for the computation of self-weight.
The soil elements were modelled using 8-noded solid element with the elastic properties indicated in Figure 1. Each soil element was modelled as a cuboid with a length and width of 0.5m, and thickness (depth) of 1m. The solid elements were stacked on each other to achieve the formation thickness of 7m. The rock layer was achieved by supporting the last nodes of the soil elements using pinned support. A width of 3m was adopted for modelling the retaining wall.
At ultimate limit state, the permanent actions (self-weight of concrete and earth fill) were factored with a partial factor of 1.35, while the variable action (surcharge) was multiplied by a partial factor of 1.5. The effect of groundwater was not considered in the analysis. The 3D rendering of the soil and retaining wall is shown in Figure 2. The concrete cantilever wall is shown in gold colour, while the soil element is shown in wine red colour (burgundy).
Figure 2: 3D Soil-Retaining Wall Modelling on Staad Pro.
Results (1) Settlement The settlement profile and deflection of the retaining wall at the serviceability limit state are shown in Figure 3.
Figure 3: Deflection and Settlement Profile of the soil
The maximum elastic settlement at the toe (SLS) = 19.647 mm Elastic settlement at the heel (SLS) = 14.856 mm Deflection at the free end of the cantilever = 35.493 mm
(2) Pressure distribution The pressure distribution under the retaining wall at the serviceability and ultimate limit states are shown in Figures 4 and 5 respectively. As expected, the maximum pressure occurred at the toe, while minimum pressure occurred at the heel. The pressure distribution was observed to be non-linear, but a little consideration will show that linear approximation can be adopted for design purposes. At SLS, a minimum pressure of 52.6 kPa was observed at the heel, and a maximum pressure of 184 kPa was found to be concentrated at the corners of the toe.
Figure 4: Base pressure distribution (SLS)
Figure 5: Base pressure distribution (ULS)
(3) Bending moment The bending moments on the wall at ULS are shown in Figures 6 and 7.
Figure 6: Vertical bending moment on the retaining wall
Figure 7: Horizontal bending moment on the retaining wall
(4) Stability Analysis (SLS)
(a) Sliding Summation of horizontal forces = 401.99 kN Summation of vertical forces = 1876.8 kN
Let the coefficient of friction of the base against sliding be tan(2/3 x 30o) = 0.363
Factor of safety against sliding = [1876.8 x 0.363] / 401.99 = 1.69 (Okay)
(b) Overturning Overturning moment = 2612.6 kNm Stabilising moment = 12356.42 kNm
Factor of safety against overturning = 12356.42/2612.6 = 4.729 (Okay)
The design for the reinforcements can be done accordingly. Kindly carry out the manual analysis of the same retaining wall and forward it to ubani@structville.com for consideration.
Every code of practice for bridge design has provision for modelling of traffic action as moving loads. This is similar to the idea of influence lines, where the wheel load action effect at the critical location of the influence line/surface is used in design combination. In this post, we are going to show how you can model bridges, and apply moving load on Staad Pro according to the principles of Eurocodes.
Let us consider a bridge deck with the configuration shown in Figure 1;
Fig 1: Bridge deck configuration
We can apply the tandem load of EN 1991-2 on the bridge deck in order to obtain the critical bending moment and shear forces. We have prepared a 21 page manual on how you can model a complete simply supported one-span bridge on Staad Pro. This can help you model a bridge of this nature by yourself on Staad Pro with full support of our technical team peradventure you need further help in getting it done. To obtain the publication for a nominal fee click HERE.
How to Apply Moving Load on Staad Pro
Step 1: Define the load Go to → Loads → Definitions → Moving Using the procedure described in Chapter 1, define the tandem wheel load of Eurocode on the bridge deck;
Fig 2: Vehicle definition on Staad Pro
Please note that the vehicle definition shown in Figure 2 is based on Eurocode Load Model 1 for tandem load on notional lane 1. The configuration of the wheel load is shown in Figure 3. You can define as many vehicle wheel load configurations as applicable.
Fig 3: Wheel load configuration for Load Model 1
Step 2: Generate the moving load on the bridge Go to → Load Case Details → Load Generation
When you click on Load Case details, click on Load Generation as circled in red pen in Figure 4, then click Add.
Fig 4: Load Case Generation on Staad Pro
Go over to the right hand side of the screen (Load and Definitions) and click on the load generated, and click Add in order to define the exact details such as the starting point, location, and step increment. This is shown in Figure 5.
Fig 5: Details of the generated load
For this example, one of the loads generated is shown in Figure 6.
Fig 6: Wheel load position on the bridge based on the definition given
Having gotten here, you can analyse the structure and see the variation of bending moment and shear forces as the vehicle travels through the bridge.
Fig 6: Bending moment on the girders due to moving load
The most basic test done on concrete is the compressive strength test. Sometimes, other properties of concrete such as tensile strength, modulus of elasticity, shrinkage values, etc are needed for design purposes.
Researchers and standards have come up with different relationships between the compressive strength of concrete and other properties. In this article, we are going to show the formulas which relate the compressive strength of concrete with other properties as applicable to the Eurocodes.
Characteristic strength of concrete (fck)
The characteristic strength is that strength below which 5% of results may be expected to fall during compressive strength test. Individual results below fck may be obtained but, in general, only need to be investigated if they fall more than 4 MPa below fck (BS EN 206-1, cl 8.2, table 14).
Design strength (fcd)
The compressive design strength of concrete is given by;
fcd = αcc fck/γc ——– (1)
where; fck = characteristic cylinder compressive strength of concrete at 28 days γc = partial (safety) factor for concrete (taken as 1.5 for UK National Annex) αcc = a coefficient taking account of long-term effects on the compressive strength (which is reduced under sustained load) and unfavourable effects resulting from the way the load is applied (conservatively taken as 0.85).
Target mean strength (fcm)
The target mean strength, fcm, is also the value used to establish the mix design and is intended to take account of the normal variability that will occur in concrete production. This margin of 8MPa for cylinders is consistent with a normal distribution with a standard deviation (SD) of about 5MPa:
fck = fcm – 1.64SD ——- (2)
Where 1.64SD = 8 Therefore SD = 8/1.64 ≈ 5MPa
N/B: For cubes, the margin is 10 MPa which gives a standard deviation of about of 6 MPa.
Development of compressive strength with time
While design is usually based on the 28-day strength, BS EN 1992-1-1, sub-clause 3.1.2(6) gives an expression for the development of the mean compressive strength of concrete with time at 20°C as follows:
fcm(t) = [βcc(t)]fcm ——— (3)
where; fcm(t) is the mean compressive strength at age t days. βcc(t) = exp{s[1 – (28/t)0.5]} ——— (3a)
where; s is a coefficient which depends on cement type = 0.20 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 42.5N (Class N) = 0.38 for cement of strength classes CEM 32.5N (Class S) (where Class R = high early strength; Class N = normal early strength; Class S = slow early strength).
Tensile strength
Tensile strength is commonly defined in one of three ways: direct tensile strength, tensile splitting strength, or flexural strength. For normal structural uses, the mean tensile strength, fctm, is related to the cylinder strength by the expressions:
BS EN 1992-1-1 provides expressions for calculating tensile strength at different maturities:
fctm(t) = [βcc(t)]α fctm ——— (7)
where: βcc(t) is as defined in Equation (3a) α = 1 for t < 28 days α = 2/3 for t ≥ 28 day
Modulus of elasticity
In design, the secant modulus, Ecm (in GPa), is derived from the mean compressive strength, fcm (in MPa), from the expression:
Ecm = 22 [fcm /10]0.3 GPa ——— (8)
Variation of Modulus of Elasticity with age
The variation of modulus of elasticity with time is estimated using the expression: Ecm(t) = [fcm(t)/fcm]0.3 Ecm ——— (9)
This formulas and relationships in this post are culled from: Bamforth P., Chisholm D., Gibbs J., Harrison T. (2008): Properties of concrete for use in Eurocode 2. The Concrete Centre, UK
Composite sections of concrete and steel have a lot of advantages, especially in the structural performance and fire resistance of a building. For columns and other compression members, they usually appear as steel-reinforced concrete columns (SRC) or as concrete-filled steel tubes. In this article, we are going to consider the structural design of concrete encased H-section subjected to concentric axial load using Eurocode 4 and BS 5950.
According to BS 5950, the steps to design composite steel columns are as follows;
Determine ultimate axial load Fc.
Select trial section and check if it is non-slender.
Determine rx, ry and Ag from steel tables.
Determine effective lengths, LEX and LEY
Calculate slenderness ratios, λEX (= LEX/rx) and λEY (= LEX/ry).
Select suitable strut curves from Table
Determine compressive strength, pc
Calculate compression resistance of member, Pc= Agpc.
Check Fc ≤ Pc. If unsatisfactory return to 2.
According to Eurocode 4, the simplified steps to design encased steel columns subjected to axial load are as follows;
Determine the ultimate axial load on the column NEd
Select a trial section and determine its properties
Obtain the buckling length of the column L
Obtain the effective flexural stiffness (EI)eff of the composite section
Calculate the plastic resistance to compression of the composite section Npl,Rk
Calculate the relative slenderness of the section (λ) using Euler’s critical load
Choose the appropriate buckling curve and calculate the corresponding reduction factor χ
Multiply the plastic resistance to compression with the reduction factor to obtain the buckling resistance of the section Nb,Rd
Check if NEd < Nb,Rd else return to step 2.
Solved Example Verify the capacity of UC 254 x 254 x 107 in grade S275 steel encased in a concrete section of 380 x 380 mm to resist a characteristic permanent axial force of 1900 kN and variable axial force of 800 kN using concrete grade C25/30. Column is 3m long and considered pinned at both ends (Area of reinforcement provided = 4H16 (804 mm2 , fyk = 500 N/ mm2).
Solution by BS 5950 The ultimate axial load on the column is given by; Fc = 1.4Gk + 1.6Qk = 1.4(1900) + 1.6(800) = 3940 kN
Properties of the UC section from Blue Book Area of UC section (Ag) = 13600 mm2 Radius of gyration (rx) = 113 mm Radius of gyration (ry) = 65.9 mm Design strength (py) = 265 N/mm2 (since thickness of flange T = 20.5 mm) Effective length (LE) = 3.0 m
Effective length Check that the effective length of column (L = 3000 mm) does not exceed the least of: (i) 40bc = 40 × 380 = 15200 mm (ii) 100bc2/dc = (100 × 3802)/380 = 38000 mm (iii) 250ry = 250 × 65.9 = 16475 mm OK
Radii of gyration for the cased section For the cased section rx is the same as for UC section = 113 mm For the cased section ry = 0.2bc = 0.2 × 380 = 76 mm but not greater than 0.2(B + 150) = 0.2(258.8 + 150) = 81.76 mm and not less than that for the uncased section (= 65.9 mm) Hence ry = 76 mm and rx = 113 mm
Compressive strength The relevant compressive strength values for buckling about the x–x axis are obtained from Table 24(b) of BS 5950 and from Table 24(c) of BS 5950 for bending about the y–y axis.
For λEX = 26.548 and py= 265 N/mm2, pc= 256.45 N/mm2 For λEY = 39.47 and py= 265 N/mm2, pc= 230.848 N/mm2
The compression resistance of the column is therefore given by;
Pc = (Ag + 0.45fcuAc/py)pc
Where: Ag = 13600 mm2 fcu = 30 N/mm2 Ac = bcdc = 380 x 380 = 144400 mm2 pc = 265 N/mm2 pc = 230.848 N/mm2
Pc = [13600 + (0.45 x 30 x 144400)/265] x 230.848 x 10-3 = 4837.703 kN
Check that Pc is not greater than the short strut capacity, Pcs , given by; Pcs = (Ag + 0.25fcuAc /py)py = [13600 + (0.25 x 30 x 144400)/265] x 265 x 10-3 = 4687 kN (this is less than Pc , therefore, take Pcs)
Fc /Pc = 3940 /4687 = 0.840 < 1.0 Okay
Design by Eurocode 4 At ultimate limit state; NEd = 1.35Gk + 1.5Qk = 1.35(1900) + 1.5(800) = 3765 kN
Effective length of the column L = 3000 mm Area of UC section (Aa) = 13600 mm2 Radius of gyration (iy) = 113 mm Radius of gyration (iz) = 65.9 mm Design strength (fy) = 265 N/mm2 (since thickness of flange T = 20.5 mm) Iy = 17500 cm4 Iz = 5930 cm4 E = 210000 N/mm2
The plastic resistance to compression Npl,Rk = Aa.fy + 0.85Acfck + Asfyk Aa = 13600 mm2 fy = 265 N/mm2 Ac = 380 x 380= 144400 mm2 fck = 25 N/mm2 As = 804 mm2 fyk = 500 N/mm2
Npl,Rk = [(13600 x 265) + (0.85 x 144400 x 25) + (804 x 500)] x 10-3 = 7074.5 kN The relative slenderness λi = ( Npl,Rk / Ncr )0.5 Ncr,i = π2(EI)eff,i /L2
(EI)eff,i = EaIa + 0.6EcmIc + EsIs
Ea = Es = Elastic modulus of the structural steel and reinforcement respectively = 210000 N/mm2 Ia = Moment of inertia of structural steel in the relevant axis Ecm = Modulus of elasticity of concrete = 22(fck/10)0.3 (GPa) = 28960 N/mm2 (see Table 3.1 of Eurocode 2) Ic = moment of inertia of the uncracked concrete section = bd3/12 = (380 x 3803)/12 = 17376.133 x 105 mm4 Is = moment of inertia of the reinforcement = πD4/64 = (π x 164)/64 = 3216.99 mm4 (for four bars = 4 x 3216.99) = 12867.96 mm4
Hence; (EI)eff,y = (210000 x 17500 x 104) + (0.6 x 28960 x 17376.133 x 105) + (210000 x 12876.96) = 6.69455 x 1013 N.mm2 (EI)eff,z = (210000 x 5930 x 104) + (0.6 x 28960 x 17376.133 x 105) + (210000 x 12876.96) = 4.26485 x 1013 N.mm2
Ncr,y= [(π2 x 6.69455 x 1013)/30002] x 10-3 = 73413.955 kN Ncr,z = [(π2 x 4.26485 x 1013)/30002] x 10-3 = 46769.313 kN
Check h/b ratio = 266.7/258.8 = 1.0305 < 1.2, and tf < 100 mm (Table 6.2 EN 1993-1-1:2005)
Therefore buckling curve b is appropriate for y-y axis, and buckling curve c for z-z axis. The imperfection factor for buckling curve b α = 0.34 and curve c = 0.49 (Table 6.1)
