An orthotropic steel deck bridge is made up of a steel plate with welded stiffeners running in opposite directions. Longitudinal stiffeners are known as ribs, and transversal stiffeners are known as cross beams or floor beams. The entire deck is supported by main girders running longitudinally. The presence of two different members in the two orthogonal directions means that the stiffness of the deck is anisotropic (not the same in every direction). The term ‘orthotropic’ is coined from orthogonal-anisotropic.
Figure 1: Typical components on an orthotropic steel deck bridge
The major structural components of an orthotropic steel deck are;
The wearing surface
The deck plate
The transverse stiffeners
The longitudinal stiffeners (ribs), and
The main girders
Because the deck serves as a top flange for the longitudinal and transversal stiffeners, as well as the major girders, the orthotropic steel deck is a cost-effective and efficient system. This concept saves material, reduces self-weight, and increases the rigidity of the deck at the same time (Håkansson and Wallerman, 2015).
Structural Behaviour of Orthotropic Steel Deck
Almost every structure that exists is made up of several structural elements such as beams, columns, and slabs. In a sophisticated way, those elements contribute to the overall behavior of the structure, however, it is common for these members to be isolated and designed individually.
The elements of orthotropic steel deck bridges are linked in a more complicated way, and the same structural elements can perform several functions. The plate functions as a load distributer between the ribs as well as a top flange for ribs, crossbeams, and main girders, as previously stated. Due to this complex interaction, individual members should not be designed in isolation from each other. In other words, the structural elements cannot be treated individually if the true response of the bridge must be known (Håkansson and Wallerman, 2015).
The diagram below depicts the transfer of a concentrated load to the major girders. The load is applied to the deck plate, which distributes it among the ribs. The load is transferred from the ribs to the cross beams, which are then distributed between the main longitudinal girders.
Figure 2: Typical load path in an orthoptropic steel deck bridge (Karlsson and Wesley, 2015)
Subsystems for Analysis of Orthotropic Steel Deck Bridges
It has been proposed that the entire bridge deck be broken into subsystems in order to make hand computations and characterize the complex structural behavior of orthotropic steel decks. Because these subsystems are believed to work independently of one another, the impacts of the several subsystems can be combined using superposition (US Department of Transportation, 2012).
The proposed subsystems are described below;
Subsystem 1: Local Plate Deformation
In this subsystem, the deck plate should only transfer the imparted wheel load to the adjacent rib walls (US Department of Transportation, 2012). Deck plate bending is used to transfer the load. When a concentrated force is applied over a rib, the deck plate deforms as shown in Figure 3.
Figure 3: Local deformation of deck plate (Karlsson and Wesley, 2015)
Subsystem 2: Panel Deformation
Due to the fact that ribs share the same top flange, they are unable to operate independently, resulting in panel deformation. As explained in the preceding section, a concentrated force applied to the deck plate will be transferred to the neighboring ribs, but because of the shared top flange, even ribs that are not loaded will deflect. This action reduces the stresses in loaded ribs while increasing stresses in unloaded ribs (US Department of Transportation, 2012). Figure 4 shows how the panel deforms as a whole, with all ribs deflecting at the same time.
Figure 4:Panel deformation of deck plate (Karlsson and Wesley, 2015)
Subsystem 3: Longitudinal Flexure of the Ribs
Ribs are constructed in a continuous pattern over cross beams, and the fact that cross beams deflect under load must be taken into account. The ribs are treated as continuous across discrete flexible supports to account for this flexure. Cross-beams are simply supported between rigid main girders in this concept, and they deflect when loaded.
Subsystem 4: Cross Beam In-plane Bending
The ribs are built in a continuous pattern over the cross beams, as previously stated. This will result in cut-outs in the cross-section of the cross beam where the rib passes through. As a result, the geometry of the cross beam will change, making hand computations of in-plane stresses from bending and shear more difficult. According to the US Department of Transportation (2012), FE-analysis should be used to simulate the entire cross beam. The deformed shape of a cross beam subjected to in-plane forces is seen in Figure 5.
Figure 5: In-plane bending of cross beam (Karlsson and Wesley, 2015)
Subsystem 5: Cross beam distortion
Three separate effects affect the local stresses in the cross beam at the cross beam and rib intersection. Out-of-plane distortion from rib bending, distortion of rib walls due to shear stresses, and distortion of ribs due to unequal deflection are the local mechanisms at these intersections.
Subsystem 6: Rib Distortion
The rib will twist about its rotating centre if a concentrated force is applied in the mid-span between two cross beams and is eccentric about the axis of the rib (US Department of transportation, 2012). Because the rib-cross beam junction will be a fixed or partially fixed barrier, depending on how large cut-outs are employed, there will be substantial stress concentrations in the welds where twisting is inhibited. When the ribs are loaded, they distort as shown in Figure 6.
Figure 6: Distortion of ribs when loaded (Karlsson and Wesley, 2015)
Subsystem 7: Global Behaviour
When there is no consideration for local effects, the global system explains the displacement of the main girders as well as the overall behavior. It is possible to determine stresses and strains in the structure using traditional methods in this system (US Department of Transportation, 2012). Figure 7 depicts the orthotropic steel deck bridge’s global bending.
Figure 7: Global bending of the OSD (Karlsson and Wesley, 2015)
References
(1) Håkansson J. and Wallerman H. (2015): Finite Element Design of Orthotropic Steel Bridge Decks. Masters Thesis submitted toChalmers University of Technology, Göteborg, Sweden (2) Karlsson A. and Wesley C. (2015): Necessity of Advanced Fatigue Analysis for Orthotropic Steel Deck Bridges. Chalmers University of Technology, Göteborg, Sweden (3) US Department of Transportation (2012): Manual for Design, Construction, and Maintenance of Orthotropic Steel Deck Bridge. US Department of Transportation, Federal Highway Administration, Publication no. FHWA-IF-12-027. USA
In highway engineering, pavements should be constructed on subgrades with a known strength. The subgrade is the foundation or existing soil surface where the new highway will be built. Subgrade strength is normally described in terms of the CBR. The design idea for heavily trafficked roads is to improve the subgrade as needed and to establish a stable platform with the sub-base for the construction of the bound structural layer and surfacing above. In the long run, the capping and/or sub-base keep water out of the bound layers and offer a platform for compacting the bound layers.
The theory for both composite and flexible pavements is that the thickness of the sub-base/capping layer does not fluctuate with traffic, but rather with the strength of the subgrade. The thickness is the same in both types of roads.
Subgrade of a highway development
A pre-construction geotechnical site investigation shall be carried out for all sites prior to the start of any highway design/construction in order to assess a number of design issues, including the stiffness (CBR) of the material, its moisture sensitivity, and, if necessary, its suitability for earthworks and stabilisation to form a capping layer, sub-base, or road base material.
Despite problems in measurement, particularly on mixed fine and coarse graded soils and when moisture effects are considered, the California Bearing Ratio (CBR) remains the greatest predictor of soil strength. A method based on the Portable Dynamic Plate is available for measuring the stiffness of the sub-grade under dynamic loading. This is especially useful when a road is being widened such that the foundation is at or near equilibrium, while for new roads, additional approaches will be required to calculate equilibrium CBR values.
Regardless, any site investigation can only sample the soils at discrete areas inside the site. Variability is unavoidable, and it should be factored into the work’s design. A competent Geotechnical Engineer can provide guidance on the expected range of CBR values if necessary.
Typical laboratory CBR test
Aside from determining design CBR values for both short and long-term characterization of subgrade performance, several other elements might affect subgrade performance and must be considered during the design stage. The following are examples of typical issues to be addressed:
a) Water table depth/perched water tables b) Chemical contamination risk assessment c) Control of fine-grained soil piping d) The possibility of encountering loose Made Ground. a) The requirement for foundation soils to be improved on the ground (e.g. soft Alluvium, loose made ground etc.) f) The risk of collapse settlement of dry engineered fills g) The possibility of landslides. h) The possibility of subsurface caves, deneholes, and so forth. I The effect of nearby developments on locations with soft alluvium. j) The occurrence and handling of subgrade solution characteristics. k) Solution characteristics below drainage runs are treated. l) Treatment and frequency of other subgrade soft spots. m) Frost sensitivity of the subgrade. n) Risks of differential settlement/need for ground improvement. o) Subgrade soil chemistry if in-situ lime/cement stabilisation is considered. p) The ability of over-consolidated clays to shrink and expand. (Especially where trees have been destroyed) q) The possibility of open cracks in the underlying rock. r) The possibility of soft clay layers in granular soil.
The strength of a subgrade may be defined in terms of the soaked CBR. Subgrade strength classification in terms of CBR is given in the Table below;
Soaked CBR
Strength Classification
Comments
< 1%
Extremely weak
Geotextile reinforcement and separation layer with a working platform typically required.
1% – 2%
Very weak
Geotextile reinforcement and/or separation layer and/or a working platform typically required.
2% – 3%
Weak
Geotextile separation layer and/or a working platform typically required.
3%–10%
Medium
Geotextile separation layer and/or a working platform typically required.
10% – 30%
Strong
Good subgrade to Sub – base quality material
> 30%
Very Strong
Sub – base to base quality material
Selection of Method for Determination of CBR
The method for determining CBR should be chosen depending on the scale of the scheme, the precision required, and the likely soils encountered. Special precautions are required to provide adequate foundation support for CBR less than 2%.
The strength of most soils is highly dependent on moisture content. Some soils experience a rapid loss of strength as the moisture content increases. On such soils, structure protection as described in the Specification for Highway Works is very critical, as is a conservative approach when considering the effect of subgrade drainage.
For design purposes, two scenarios must be considered: the likely CBR at the time of construction and the long-term equilibrium value. Both of these are essential for design purposes. If the ‘as found’ CBR at the time of construction is less than that determined during the site investigation, a change in foundation layer thickness may be required.
The plasticity index should be computed for cohesive soils, and the description of the soil type from a grading examination of a bulk sample, as well as some knowledge of the probability of saturation in the future, should be examined for other soils. The table below can be used to approximate the CBR based on this information.
Methods of Determination of CBR
CBR’s performed upon pot samples
Advantages (i)It enables the evaluation of a soil’s CBR at various levels of saturation (CBRs, soaked/unsoaked)
Disadvantages (i) Considerable disturbance is created during the sample operation, which has a significant impact on the test results. (ii) Requires a large trial pit.
Plastic and Liquid Limits
Advantages (i) It enables a lower bound estimate of the CBR under recompacted circumstances for a wide range of effective stresses (i.e. construction conditions)
Disadvantages (i) Unless standard graphics are utilized, the analytical approach is costly. (ii) The typical graphs are based on the worst-case scenario. (iii) Can only be utilized on soils containing cohesive material (iv) Can be difficult to evaluate on granular and cohesive material mixtures
Soil Assessment Cone Penetrometer (MEXE Probe)
Advantages (i) It is simple and affordable to carry out.
Disadvantages (i) Correlation dependant (ii) Only offers the current CBR value (iii) Insensitive to the influence of the soil’s microstructure (iv) It is not suitable for usage in stony soils.
Measurement of Shear strength (hand vane/Triaxial tests)
Advantages (i) The hand vane is a quick and low-cost method of measuring undrained shear strength. (ii) Triaxial measurements of undrained shear strength take into consideration the soil microstructure (iii) For recompacted subgrades, remolded tests provide a lower bound.
Disadvantages (i) Only offers the existing CBR value when using the hand vane (ii) Triaxial measurements are somewhat expensive to perform, and sample and testing time might be lengthy (iii) It is dependent on correlation although this has a theoretical basis (iv) can only be applied for cohesive soils.
In-situ CBR
Advantages (i) Realistic CBR measurement (ii) Takes into account the macrostructure of the soil account (iii) can be used to determine the present chalk value in chalks
Disadvantages (i) Expensive to complete (ii) Only offers the present CBR value (iii) It is difficult to perform beneath the existing ground surface (iv) Not recommended for coarse granular soils.
Laboratory compaction test
Laboratory Compaction
Advantages (i) Calculates the CBR of remoulded soils at various moisture levels.
Disadvantages (i) Insensitive to soil macrostructure impacts (ii) Expensive to do (iii) Requires a big sample (iv) variable outcomes with coarse granular soils
Conclusion
CBR measurements are used to determine the current subgrade strength and to forecast the worst-case scenario for future service inside the pavement. In actuality, the current value may underestimate the subgrade’s strength at the time of construction because pessimistic conditions may not be present at the time of building, such as during summer construction. The current value, on the other hand, could have been taken in the summer and the plan built in the winter, when the subgrade may be weaker due to the moisture accumulating at that time of year.
As a result, it’s critical to think about the implications of the site investigation results in relation to those at the time of building. Unfortunately, there are no “hard and fast” guidelines, therefore field data may need to be interpreted by a professional. To develop an interpretative statement on the equilibrium CBR for design purposes, local knowledge of the effect of moisture on the relevant soil and the compaction/moisture content vs CBR connection should be combined with laboratory experiments. It will be necessary to make a request for this. Other elements impacting the performance of the subgrade, such as drainage and the possibility of foundation material settlement, must also be considered in the design.
Beams that have V-shapes (in the plan view) are commonly found in the corners of residential and commercial buildings. According to Hassoun and Al-Manaseer (2008), such beams can be analysed and designed using strain-energy principles especially when the beam is fixed at both ends.
In some scenarios, engineers may be tempted to design v-shaped beams as two cantilevers meeting at a point, but this is not strictly the case. There will be a need to assess other internal stresses such as torsion, and to understand the actual nature of the distribution of internal forces.
For a v-shaped beam subjected to a uniformly distributed load w (kN/m), the internal forces in the members can be obtained as follows;
(The bending moment at the centre of the beam Mc is given by; Mc = (wL2)/6 × [sin2θ/(sin2θ + λcos2θ)
Where; λ = EI/GJ L = half the total length of the beam AC θ = Half the angle between the two sides of the v-shape beam.
The torsional moment at the centreline of the section is given by; TC = (MC/sinθ) × cosθ = MCCotθ
At any section N along the length of the beam at a distance x from the centreline C,
A v-shaped beam at the corner of a building has a depth of 400mm and a width of 225 mm. The plan view of the beam is shown below. It is to support an ultimate uniformly distributed load of 30 kN/m inclusive of the factored self-weight. Design the beam according to the requirements of EC2. fck = 30 MPa, fyk = 500 MPa, Concrete cover = 35 mm
Solution
For fck = 30 MPa, Modulus of elasticity Ecm = 31476 MPa Shear modulus G = Ecm/2(1 + v) = 31476/2(1 + 0.2) = 13115 MPa Moment of inertia I = bd3/12 = (225 × 4003)/12 = 12 × 108 mm4 Polar moment of inertia J = 985033091.649413 mm4 λ = EI/GJ = (31476 × 12 × 108)/(13115 × 985033091.649413) = 2.9237
Mc = [wl2sin2θ/6(sin2θ + λcos2θ) = (30 × 2.52 × 0.75)/[6 × (0.75 + 2.9237 × 0.25)] = 140.625/8.88555 = 15.755 kNm MA = MB = MC – wl2/2 = 15.755 – (30 x 2.52)/2 = -77.995 kNm Torsional moment = TA = MCcotθ = 15.755 x 0.5773 = 9.095 kNm Shear at support = VA = VB = 30 × 2.5 = 75 kN
Let us check these answers using Staad Pro software;
By implication, one can confirm that the analysis method adopted is accurate. The beam can now be designed for torsion, bending, and shear using the guidelines provided in EN 1992-1-1:2004.
References Hassoun N. M. and Al-Manaseer A. (2008): Structural Concrete Theory and Design. Wiley and Sons Inc, New Jersey, USA
In the event of a fire, the general objectives of fire protection of structures are to limit hazards to individuals and society, adjacent property, and, where necessary, the environment or immediately exposed property (EN 1993-1-2:2005). The relevant time of fire exposure during which the associated fire resistance function of a structure is maintained despite fire actions is characterized as fire resistance in terms of time. Therefore, the fire resistance design of steel beams is concerned with maintaining the load-bearing function, integrity separating function, and thermal insulating separating function of a steel beam for a given period of time.
According to the Construction Products Directive 8911 06/EEC, the following are the essential requirement for the limitation of fire risks:
“The construction works must be designed and built in such a way, that in the event of an outbreak of fire;
the load bearing resistance of the construction can be assumed for a specified period of time
the generation and spread of fire and smoke within the works are limited
the spread of fire to neighbouring construction works is limited
the occupants can leave the works or can be rescued by other means
the safety of rescue teams is taken into consideration”.
According to the European standards, three major criteria are used to define the fire resistance of structures:
R – load-bearing function E – integrity separating function I – thermal insulating separation function
It is important to note that the above criteria may be required individually or in combination: • separating only: integrity (criterion E) and, when requested, insulation (criterion I) • load bearing only: mechanical resistance (criterion R) • separating and load-bearing: criteria R, E and, when requested I
Where mechanical resistance in the case of fire is required, steel structures shall be designed and constructed in such a way that they maintain their load-bearing function the relevant fire exposure. Criterion “R” is assumed to be satisfied in steel beams where the load-bearing function is maintained during the required time of fire exposure.
For a given load level, the temperature at which failure is expected to occur in a structural steel element for a uniform temperature distribution is called the critical temperature (θcrit). The thermal action on the steel member is a result of the heat flux transferred from the fire to the steel member. This is usually regarded as an indirect action.
The load bearing function of a structure is satisfied only if during the relevant duration of fire exposure t;
Efi,d,t ≤ Rfi,d,t
where Efi,d,t: design effect of actions (Eurocodes 0 and 1) Rfi,d,t: corresponding design resistance of the structure at instant t
It is however important to note that Eurocode permits the fire resistance of steel structures to be assessed in any of these three domains;
Time; tfi,d ≤ tfi,req (this usually feasible using advanced computational models) Load Resistance; Efi,d,t ≤ Rfi,d,t (This approach is feasible by hand calculation. Find reduced resistance at required resistance time) Temperature; θcr,d ≤ θd (The most simple approach, find the critical temperature for loading, compare with design temperature)
For ordinary structural fire structural design of steel beams, simple calculation models such as critical temperature can be used.
The thermal properties of structural steel at elevated temperature is shown below. At about 600°C, the elastic modulus of elasticity of steel is reduced by about 70% while the yield strength reduces by about 50%.
Under fire situation, the design loading is given by;
Efi,d,t = ∑Gk,j + ψ2,1Qk,1 + ∑ψ2,1Qk,I
More commonly, a reduction factor ηfi is applied to the loading under ambient conditions in order to represent the loading under fire conditions. This is given by;
Efi,d,t = ηfiEd
ηfi = Efi,d,t/Rd
In ambient temperature strength design; γG = 1.35 (Permanent loads) γQ.1 = 1.50 (Combination factor; variable loads)
In structural fire design γGA = 1.0 (Permanent loads; accidental design situations) γ2.1 = 0.3 (Combination factor; variable loads, offices)
Steps in the Fire Resitance Design of Steel Beams
The following are the steps to follow when assessing the fire resistance design of steel beams;
(1) Evaluate the actions on the structuture in fire situation Efi.d. This is commonly achieved by applying a reduction factor to the ambient temperature load. (2) Classify the member under class 1, class 2, class 3 or class 4 depending the rotation capacity using EC3 guidelines (3) Evaluate the resistance of the structure at ambient temperature (20°C) by fire rules Rfi.d.20. (4) Calculate the degree of utilisation μ0 (5) Calculate the critical temperature θcr (6) Calculate the section factor Am/V and the correction factor ksh (7) Carry out step by step calculation to verify if θfi,d= θcr at the time of fire exposure tfi,d (8) Check if tfi,dis greater than the required fire resistance period of the structure treq
Design Exampleof Fire Resistance of Steel Beams
Verify the fire resistance of an intermediate secondary beam (S275) in an office block supporting a slab of 175 mm thickness. The expected fire rating is 30 minutes. The slab is to support a variable action of 3 kN/m2, and finishes of 1.5 kN/m2.
Design Data
Imposed variable load qk = 3.0 kN/m2 Thickness of R.C. slab = 150 mm Finishes = 1.5 kN/m2
Analysis and Design at Normal Room Temperature Permanent loads
Unit weight of reinforced concrete = 25 kN/m3 Thickness of slab = 150 mm = 0.15 m
Self weight of slab = 25 × 0.15 = 3.75 kN/m2 Weight of finishes = 1.5 kN/m2 Total permanent actions (gk) = 5.25 kN/m2
Variable Actions Imposed variable load = 3.0 kN/m2 Total live load (qk) = 3.0 kN/m2
Normal TemperatureDesign of the Secondary Beam
The load transferred from the slab to the secondary beams considering a bay width of 2.667m
At ultimate limit state (neglecting reduction factors); PEd = 1.35gk + 1.5gk PEd = 1.35(5.25) + 1.5(3) = 11.6 kN/m2
Ultimate load transferred to every secondary beam; Fd = 11.6 kN/m2 × 2.667m = 30.937 kN/m Self weight of beam = 1.35 × 0.392 = 0.528 kN/m Total design load on the beam = 30.937 + 0.528 = 31.465 kN/m
An advanced UK beam S275 is to be used for this design. fy = 275 N/mm2 γm0 = 1.0 (Clause 6.1(1) NA 2.15 BS EN 1993-1- 1:2005)
The required section is supposed to have a plastic modulus about the y-y axis that is greater than; Wpl,y = My,Ed.γm0/fy Wpl,y = (150.093 × 103 × 1.0)/275 = 545.792 cm3
From steel tables, try section UB 305 x 165 x 40 Wpl,y = 623 cm3
Properties h = 303.4 mm; b = 165 mm; d = 265.2 mm; tw = 6.0 mm; tf = 10.2 mm; r = 8.9 mm; A = 51.3 cm4; Iy = 8500 cm4; Iz = 764 cm4; Wel,y = 560 cm3;
hw = h – 2tf = 265.2 mm E (Modulus of elasticity) = 210000 N/mm2 (Clause 3.2.6(1))
Classification of section ε = √(235/fy) = √(235/275) = 0.92 (Table 5.2 BS EN 1993-1- 1:2005)
Outstand flange: flange under uniform compression c = (b – tw – 2r)/2 = [165 – 6 – 2(8.9)]/2 = 70.6 mm c/tf = 70.6/10.2 = 6.921
The limiting value for class 1 is c/tf ≤ 9ε = 9 × 0.92 5.03 < 8.28 Therefore, outstand flange in compression is class 1
Internal Compression Part (Web under pure bending) c = d = 265.2 mm c/tw = 265.2/6 = 44.2 The limiting value for class 1 is c/tw ≤ 72ε = 72 × 0.92 = 66.24 44.2 < 66.24 Therefore, the web is plastic. Therefore, the entire section is class 1 plastic.
Member Resistance Verification
Moment Resistance
For the structure under consideration, the maximum bending moment occurs where the shear force is zeo. Therefore, the bending moment does not need to be reduced for the presence of shear force (clause 6.2.8(2)).
For rolled I-section with shear parallel to the web, the shear area is; Av = A – 2btf + (tw + 2r)tf (for class 1 sections) but not less than ηhwtw Av = (51.3 × 102 – (2 × 165 × 10.2) + [6 + 2(8.9)] × 10.2 = 2006.76 mm2 η = 1.0 (conservative) ηhwtw = (1.0 × 265.2 × 6) = 1591.2 mm2 2006.76 > 1591.2 Therefore, Av = 2006.76 mm2
The shear resistance is therefore; Vc,Rd = Vpl,Rd = [2006.76 × (275/ √3)/1.0] × 10-3 = 318.6159 kN VEd/Vc,Rd = 94.395/318.62 = 0.296 < 1.0 Ok
Shear Buckling
Shear buckling of the unstiffnened web will not need to be considered if; hw/tw ≤ 72ε/η hw/tw = 265.2/6 = 44.2 72ε/η = (72 × 0.92)/1.0 = 66 44.2 < 66 Therefore shear buckling need not be considered.
Serviceability limit state
Vertical deflections are computed based on unfactored variable loads. Permanent loads need not be considered (BS EN 1993-1-1 NA 2.23)
qk = 3.0 × 2.667 = 8 kN/m w = 5ql4/384EI w = (5 × 8 × 60004)/(384 × 210000 × 8500 × 104) = 7.563 mm Span/360 = 6000/360 = 16.667 (BS EN 1993-1-1 NA 2.23)
7.563 < 16.667 Therefore, deflection is satisfactory
Fire Design of the Secondary Beam
The loading due to fire design is given by;
Efi,d,t = ∑Gk,j + ψ2,1Qk,1 + ∑ψ2,1Qk,I
As a simplification to the loading above, the fire part of EC 3 permits us to apply a reduction factor ηfito the load of normal temperature design.
ηfi = (γGAGk + ψ2,1 Qk,1)/(γGGk + γQQk,1)
For the secondary beams; Gk = 5.25 × 2.667 = 14.0 kN/m Qk = 3.0 × 2.667 = 8.0 kN/m
According to equation 4.2 of EN 1993-1-2 for fire resistance design; ε = 0.85√(235/fy) = 0.786 for S275 grade of steel Outstand flange: flange under uniform compression c = (b – tw – 2r)/2 = [165 – 6 – 2(8.9)]/2 = 70.6 mm
c/tf = 70.6/10.2 = 6.921
The limiting value for class 1 is c/tf ≤ 9ε = 9 × 0.786 (c/tf)5.03 < (9ε) 7.074 Therefore, outstand flange in compression is class 1
Internal Compression Part (Web under pure bending) c = d = 265.2 mm c/tw = 265.2/6 = 44.2
The limiting value for class 1 is c/tw ≤ 72ε = 72 × 0.786 = 56.592 44.2 < 56.592 Therefore, the web is plastic. Therefore, the entire section is class 1 plastic under fire loading.
From our calculations for normal temperature design, it can be verified that; Mc,Rd = Mpl,Rd = 171.325 kNm Vc,Rd = Vpl,Rd = = 318.6159 KN
From relation 4.24 of EN 1993-1-2; We can calculate the utilisation ratio of the unprotected steel beam;
As the beam supports the concrete slab above, the impact of the kappa factors relative to the temperature gradient over its depth have to be taken into account. However, they have an impact on bending moment alone. No rule is provided for shear.
The modified degree of utilisation for bending moment; κ1 = 1.0 κ2 = 1.0
For the unprotected beam having all four sides exposed – clause 4.1(16) of EN 1993-1-2
For the protected section due to protection of the three sides exposed to fire; κ1 = 0.85 (three sides exposed; insulated) κ2 = 1.0 (for the unprotected beam having all four sides exposed – clause 4.1(16) of EN 1993-1-2
The modified degree of utilisation factor for vertical shear μ0,v,k = μ0,v = 0.166 (no adaptation factor) μ0 = max (μ0,m,k; μ0,v,k) μ0 = max (0.393; 0.166) = 0.393
Critical temperature of the protected beam From equation 4.22 of EN 1993-1-2; θcr = 39.19In[1/(0.9674μ03.833)- 1] + 482 θcr = 39.19In[1/(0.9674 × 0.3933.833 )- 1] + 482 = 622°C
From a more accurate linear interpolation from reduction table (Table 3.1); ky,θ (0.47) = 600°C ky,θ (0.393) = ?? ky,θ (0.23) = 700°C
Section factor is the ratio of the perimeter of the beam exposed to fire to the volume of the beam section (Am/V). In other words, it is the ratio between “perimeter through which heat is transferred to steel” and “steel volume”.
From steel table; Am/V = 240 m-1 (for UKB 305 x 165 x 40) (Am/V)b = 185 m-1 (boxed value)
The correction factor for shadow effect; ksh = 0.9(Am/V)b / (Am/V) ksh = (0.9 × 185)/240 = 0.6937
Heating of the unprotected beam The heating of the beam can be obtained from relation 4.25 of EN 1993-1-2.
Δθa.t = (ksh/(ca ρa) × Am/V ×hnet,d×Δt —– (1)
We will apply this relationship with the following assumptions; Δt = time interval increments = 5 seconds ρa = density of steel = 7850 kg/m3 ca = specific heat of steel = 600 J/kgK θa.t = 0.6937/(600 × 7850) × 240 × hnet,d× 5
Therefore; θa.t = (1.7664 × 10-4)hnet,d
hnet,d (heat flux) has two parts (the convective part hnet,c and the radiative part hnet,r) and varies with time as shown in the equation below. It can be easily calculated if the gas temperature θg is known.
Under standard fire situations; εres = εf × εm = 0.7 (εfis the emissivity of the fire usually taken as 1.0 while εm is the surface emissivity of the member usually taken as 0.7 for carbon steel) ϕ (view factor or configuration factor) = 1.0 αc = the coefficient of heat transfer by convection = 25 W/m2K
Therefore; hnet,r = 3.969 × 10-8[(θg + 273)4 – (θm + 273)4] hnet,c = 25(θg – θm) θg = 20 + 345log(8t + 1) (t in minutes) θm is the surface temperature of the steel member
The most relevant way to deal with hnet,d is to consider a mean value within the time interval ∆t (5 seconds in this case) between the instant ti and ti+1
When this is solved iteratively, the time at which the bare section reaches critical temperature can be obtained. The temperature development curve can also be plotted. By programming Eq. (1), it is easy to build tables or nomograms like the ones presented in Annex A ofFranssen and Real (2015)for unprotected steel profiles subjected to the ISO 834 fire curve. The use of these tables and nomograms avoids the need to solve Eq. (1). The nomograms from Annex A are reproduced in the Figure below.
kshAm/V = 166.5 m-1
For a critical temperature of 603°C (for the unprotected beam) and kshAm/V of 166.5 m-1, the time to achieve that temperature when exposed to fire is about 14 minutes. This is less than 30 minutes, therefore the beam cannot be designated as R30.
In the temperature domain; Table 4.6 gives the temperature after 30 minutes and 60 minutes of standard fire ISO 834 exposure, for different values of the modified section factor ksh[Am/V].
From Table 4.6, the critical temperature for 30 mins for ksh[Am/V] = 166.5 m-1 fire is 825.185°C. Therefore, the temperature in the section exposed on 4 sides after 30 min of standard fire ISO 834 exposure is 825.185°C. As this temperature is greater than the critical temperature, the member doesn’t fulfil the condition for temperature approach verification, because:
θd > θcr,d , at time tfi,requ
This further confirms that the member does not satisfy R30 requirements.
Maintaining an efficient HVAC system is crucial in delivering good indoor air quality in commercial establishments. Indoor Air Quality (IAQ) pertains to the air quality inside buildings and facilities. Understanding the risks involved in having air pollutants inside the facility is vital to appreciate the importance of maintaining the cleanliness of your HVAC unit.
In line with other significant building components, your HVAC system is primarily responsible for distributing heated or cooled air within the building. Having a properly functioning ventilation and air conditioning system not only promotes good IAQ but is also crucial in maintaining smooth operations for your business.
Due to its role in building operation, experts highly recommend regular inspection and cleaning of the commercial HVAC unit. With preventative maintenance, technicians can make repair and cleaning recommendations when necessary. A good inspection program is crucial in identifying minor leaks inside the ductwork, as minor issues quickly become severe and costly.
By scheduling regular maintenance, you can ensure the performance and longevity of your mechanical air conditioning system. While you can rely on technicians when it comes to assessment and repair, you can do your part by installing HVAC access doors and panels.
What is an HVAC Access Door?
A reliable commercial HVAC system needs to have its components functioning well to properly distribute conditioned air throughout the building. Its essential elements include the heat exchanger, blower motor, air ducts, combustion chamber, and thermostat. But how about the HVAC or duct access door? Can it be considered as part of the air conditioning unit?
The answer is a resounding yes. Access openings are without a doubt necessary in facilitating inspections, testings, repairs, and cleaning. These HVAC panels play a crucial role during maintenance services to ensure business operations — not only because they provide access but also because a good service ensures that the commercial AC unit is performing to expectations regarding safety and efficiency.
Ideally, contractors should install the HVAC panels near system components or either side of obstructions such as dampers and fans during installation. Purchasing a panel based on accurate measurements is also crucial for sufficient access to the air conditioning parts. Poorly constructed openings can harm the commercial HVAC unit in ways such as:
When improperly installed, the air ducts may compromise the system’s overall structural integrity.
Duct air leakage
Affect indoor air quality
Expose the mainframe to contamination and dust particles
No matter the panel used, it is vital to install the HVAC openings correctly and in a manner that facilitates proper closure. Therefore, it is highly ideal to hire professional contractors who have experience in physical installations.
Common Issues to Encounter if You Don’t Install HVAC Access Doors
There are many reasons why experienced contractors and technicians recommend using HVAC panels. The absence of a safe opening of the system’s mainframe and components presents many issues, particularly maintenance service providers. Without proper maintenance, your commercial air conditioner will inevitably affect its performance leading to poor air quality, among other things. Here are some possible issues you have to deal with if you choose not to install HVAC access panels.
Access Limitations – Technicians’ most common problem when undertaking service maintenance or repair is limited access to HVAC components. Access limitation doesn’t primarily refer to a lack of access, but it can also refer to insufficient space due to wrong measurements or improper door installation. The building owner needs to comply with standard access regulations and the request of the maintenance service provider to enable the required work to be safely accomplished.
Dust and Debris Accumulation – The absence of an entry door significantly promotes dust and debris accumulation inside the air ducts and the other components. Without proper cleaning, the dust particles can readily travel through the vents and into the interior, causing allergic reactions, dust build-ups, and poor air quality.
Animal Infestation – Without a good HVAC access point, your ductwork can quickly become a breeding ground for bacteria and mold growth. The molds can then attract small animals and insects, including rats, spiders, birds, and even snakes since your air passage has somehow become a thriving ecosystem for these animals. Having animals live in the ductwork can cause many issues, such as corrosion, destroyed wiring, blockage, and health issues due to animal wastes.
Lack of Protection – Keeping the components intact and in good condition is crucial for commercial HVAC systems to function appropriately. A duct access door offers additional protection from external elements or unauthorized access that may damage the system’s internal structure.
Reduced Aesthetic Appeal – There are residential housing properties with strict aesthetic requirements. The paint and materials used during construction must be similar on all floors, and an exposed HVAC unit can be an eyesore. Concealing the device is an excellent way of staying in line with the property’s overall appearance. In addition, there are now access door options that will allow the user to paint over the cover for the unit to blend seamlessly to the surface installation.
Delayed Repair and Maintenance – The primary purpose of an HVAC panel is to provide access. When it comes to commercial air conditioning systems, you cannot overestimate the importance of an efficient maintenance program. Technicians will undoubtedly have difficulties identifying and isolating the damage without sufficient access. Lack of repair and maintenance can cause a landslide of problems, eventually leading to either of the HVAC issues mentioned previously.
Takeaway
Maintaining good indoor air quality inside your building or facility is tremendously important, especially in residential units and office spaces. Turning a blind eye towards the significance of cleaning and maintaining your commercial air ducts could lead to a series of problems and, worse, the closure of your business due to negligence. Accepting that access doors have become an essential part of your commercial HVAC system is the key to a well-maintained ventilation system. Contact a licensed professional for more information, and remember to only purchase from a reputable store.
Trusses are important civil engineering structures that are used in the construction of roofs of buildings, bridges, towers, stadiums, etc. The main structural feature of trusses is that they are arranged in such a manner that they form triangulated systems that resist loads mainly by developing axial forces. The axial forces in the truss members may be tensile or compressive and will lead to the global deflection of the truss.
The configuration or arrangement of the members in a roof truss significantly affects the structural behaviour of the truss. This includes the distribution of the internal stresses and deflection behaviour. Some different types of trusses have been identified based on their shape and internal members’ arrangement. Examples of such are Howe truss, Fink truss, Pratt truss, K-Truss, Warren Truss, Warren truss with verticals, Baltimore Truss, etc. In this article, the effect of member configuration on the deflection behaviour of trusses was studied.
Theoretically, the factors that affect the deflection behaviour of trusses are;
Nature and magnitude of internal stresses in each member
However, in practice, it has been observed that factors such as connection details, size of holes for bolts, eccentricity of connection, etc can significantly affect the deflection value observed in roof trusses.
In order to achieve this, a 6m span truss was modelled on Staad Pro software and subjected to different member configurations. A constant nodal spacing of 1 m and section property of UA 90 x 90 x 10 were maintained for all the models in the study. All internal loads nodes were subjected to a concentrated load of 20 kN while the end nodes were subjected to a concentrated load of 10 kN. The typical truss configuration, loading, dimensions, and node numbering is shown below;
Figure 1: Typical loading and node numbering of the truss models
The analysis results are shown below;
Truss Type 1:Pratt Truss (Non-Parallel Diagonals)
Figure 2: Configuration, loading and deflection profile of Truss Type 1
Truss Type 1 is a typical Pratt truss that is balanced in symmetry and in loading, with non-parallel diagonals in the LHS and RHS of the structure. As expected, the maximum deflection occurred at the centre with a value of 2.239 mm as shown in Figure 2. The maximum compressive forces occurred at the middle of the top chord between members 3-4 and 4-5 as shown in Figure 3. Members 3-10 and 5-10 were observed to be zero-force members. The bottom chord members between nodes 9-13 were observed to be in tension, but the entire top chord members were observed to be in compression. The general compressive to tensile member ratio was observed to be 1.5, while the web compressive to tensile member ratio was observed to be 1.16.
Figure 3: Internal stresses distribution in Truss Type 1
Truss Type 2: Parallel Flat Truss
Figure 4: Configuration, loading and deflection profile of Truss Type 2
Truss Type 2 is a typical parallel flat truss comprising of verticals and diagonals that are all parallel to each other. By implication, the truss is not symmetrical in arrangement and the maximum deflection at the bottom chord occurred at the centre with a value of 2.172 mm (see Figure 4). The maximum compressive forces occurred at the top chord, while the maximum tensile forces occurred at the diagonals as shown in Figure 5. The general compressive to tensile member ratio was observed to be 2.0. In the web, the compressive to tensile member ratio was observed to be 1.6.
Figure 5: Internal stresses distribution in Truss Type 2
Truss Type 3: Pratt Truss (Parallel Diagonals)
Figure 5: Configuration, loading and deflection profile of Truss Type 3
Truss Type 3 is a typical perfect Pratt truss with parallel diagonals on the LHS and RHS of the structure. The value of the maximum deflection on the bottom chord was observed to be 2.7 mm as shown in Figure 6. All the diagonals are in tension, while the verticals are in compression (see Figure 7). Furthermore, the entire top chord is in compression while the mid-span of the bottom chord is in tension. The general compression to tensile member ratio is 1.6, while the web compressive to tensile member ratio is 1.16.
Figure 7: Internal stresses distribution in Truss Type 3
Truss Type 4: Howe Truss
Figure 8: Configuration, loading and deflection profile of Truss Type 4
Truss type 4 is a typical Howe truss with a non-parallel diagonal arrangement. The maximum deflection on the bottom chord was observed to be 1.782 mm as shown in Figure 8. Five (5) out of the twenty-five (25) members in the truss were observed to be zero-force members (see Figure 9). Furthermore, the general compressive to tensile member ratio was observed to be 4.0, while the web compressive to tensile member ratio was observed to be to be 4.0 also.
Figure 9: Internal stresses distribution in Truss Type 4
Truss Type 5: Long Truss
Figure 10: Configuration, loading and deflection profile of Truss Type 5
In Truss Type 5, additional internal members were introduced in the truss to form a typical Long Truss. A lower deflection value of 1.383 mm was observed at the centre of the roof truss as shown in Figure 9. This can be attributed to the additional constraints at the nodes of the trusses. The top chords are in compression (see Figure 11), and the general compressive to tensile member ratio was observed to be 1.5. In the web, a compressive to tensile member ratio of 1.11 was observed.
Figure 11: Internal stresses distribution in Truss Type 5
Truss Type 6: K-Truss
Figure 12: Configuration, loading and deflection profile of Truss Type 6
Truss type 6 is a typical K-truss. Despite having additional members, a deflection value of 2.178 mm was observed at the midspan as shown in Figure 12. Under gravity loading, a variation of axial stress distribution from compressive to tensile was observed in the vertical members (see Figure 13). A general compressive to tensile member ratio of 2.3 was observed in the truss, while the web’s compressive to tensile member ratio was 1.875.
Figure 13: Internal stresses distribution in Truss Type 6
Truss Type 7: Bailey Truss
Figure 14: Configuration, loading and deflection profile of Truss Type 7
Truss type 7 is a typical Bailey Truss. The deflection at the midspan was observed to be 1.904 mm (see Figure 14). A little consideration shows that a significant number of members can be removed without affecting the behavior of the truss as shown in Figure 15. A general compressive to tensile member ratio of 2.25 was observed in the truss, while a compressive to tensile member ratio of 3 was observed in the web.
Figure 15: Internal stresses distribution in Truss Type 7
Truss Type 8: Bailey Truss (with K at mid-span)
Figure 16: Configuration, loading and deflection profile of Truss Type 8
Truss Type 8 is a typical Bailey truss but with a K-configuration at the midspan. This configuration had a larger deflection than a pure K-truss or Bailey Truss arrangement. The maximum deflection observed in the truss was 2.371 mm as shown in Figure 16. A general compressive to tensile member ratio of 2.1 was observed in the truss, while a compressive to tensile member ratio of 2.83 was observed in the web (see Figure 17).
Figure 17: Internal stresses distribution in Truss Type 8
The summary of the findings is shown in the Table below;
Truss Type
Maximum deflection (mm)
C/T ratio (general)
C/T ratio (web)
Truss Type 1: Pratt Truss (Non-Parallel Diagonals)
2.239
1.5
1.16
Truss Type 2: Parallel Flat Truss
2.172
2.0
1.6
Truss Type 3: Pratt Truss (Parallel Diagonals)
2.700
1.6
1.16
Truss Type 4: Howe Truss
1.782
4.0
4.0
Truss Type 5: Long Truss
1.383
1.5
1.11
Truss Type 6: K-Truss
2.178
2.3
1.875
Truss Type 7: Bailey Truss
1.904
2.25
3.0
Truss Type 8: Bailey Truss (with K at mid-span)
2.371
2.1
2.83
The variation of vertical deflection with type of truss is shown in Figure 18.
Figure 18: Variation of deflection (mm) with Truss Type
From the result, the lowest deflection was observed in the Long Truss (Truss Type 5) due to the additional constraints that reduced the degree of movement at the nodes. The lowest deflection value occurred at the expense of additional members and increased fabrication cost. The most economical configuration in terms of deflection can be deemed to be the Howe Truss, which offered the lowest deflection with minimal members. The ratio of the number of compressive to tensile members did not have any recognisable effect on the deflection behaviour of trusses.
Reinforced concrete overhead tanks are water retaining structures that are placed at a height above the natural ground level. The major reason for placing the tanks at a height is to enable the water to flow under gravity to the point where they are needed. Since the tank is to be supported at a height, there is a need for a tank stand/support system, which is usually constructed of reinforced concrete, and cast monolithically with the tank shell.
The design of reinforced concrete overhead tank involves the determination of the dimensions of the tank shell to hold the desired volume of water, selection of an adequate concrete section and steel reinforcements to satisfy ultimate and serviceability limit state requirements of the tank shells, and selection of adequate column and beam sizes with the proper reinforcements to serve as the support system (tank stand).
Figure 1: Circular overhead reinforced concrete water tank
The design of overhead tanks follows the same basic principles as other reinforced concrete structures, with the exception of the special attention paid to the crack width of the tank shell in order to ensure the water-tightness of the structure. In principle, the elements usually designed for in reinforced concrete overhead tanks are;
The tank shell
The beams
The columns
Other bracing systems (if any),
Other ancillary components like stairs and rails, and
The foundation
Design of the Tank Shell
The shell of the tank is the major water-holding element of the structure, which usually comprises the walls, the base, and the cover. The volume of the tank shell can be estimated by considering the water demand of the end-users of the tank, and the anticipated frequency of pumping. Furthermore, space constraints and construction challenges can also influence the final volume and shape of the tank. Typically, it is more common to calculate the volume of the tank, and then select the desired height and plan area.
The cross-section of the tank can be rectangular or circular. Circular tanks usually give more economical sections and reinforcements than rectangular tanks, even though they are more difficult to construct. Other shapes can be designed for as long the engineer can guarantee the construction feasibility and economical implication of such decisions.
Since the tank shells are liquid containment elements, the walls and base are subjected to hydrostatic pressure from the stored water. The walls are also subjected to direct tension (for rectangular walls), and hoop tension (for circular walls). Therefore, the tank shell should have an adequate cross-section that will be able to withstand the pressure from the stored water. The design pressure is normally used to obtain the design moment, which can be used to provide the thickness of the sections and the reinforcements. The restraint condition of the tank walls influences the moment distribution (or the nature of the bending moment diagrams) which can ultimately influence the final arrangement of reinforcements.
Figure 2: Typical hydrostatic pressure on the walls of the tank
Figure 3: Typical pressure in the walls of the tank due to water pressure
In order to guarantee the water-tightness of the tank shell from a structural point of view, the crack width of the tank shell should not exceed 0.2 mm. This encompasses long and short-term cracking from flexure, shrinkage, and restraints within the concrete system. However, experience has shown that this serviceability limit state requirement usually governs the design of water retaining structures since the reinforcement required to achieve the minimum crack width is usually greater than that required to control flexure and shear.
Design of the Beamsin Reinforced Concrete Overhead Tanks
Depending on the structural scheme adopted, the load from the weight of the tank and the stored water may be transferred to the supporting beams before being transferred to the columns (see Figure 4). In some cases, a flat slab concept may be adopted where the base of the tank is supported directly by columns as shown in Figure 5.
Figure 4: Overhead circular tank supported on beams and columns
Figure 5: Overhead water tank supported on columns
However, the design concept of the beams is similar to the design concept of normal reinforced concrete beams. For the design of circular beams, it will be important to consider the effects of torsion. The procedure for the design of circular beams is readily available in literature.
Design of the columns
The design of the supporting columns in an reinforced concrete overhead tank is similar to the design of normal reinforced concrete columns. The columns will be subjected to axial force, bending, and shear from the self weight of the materials, the stored water, and wind. The size and arrangement of the columns will depend on the anticipated load from the water tank.
Furthermore, special attention must be paid to the to the concrete cover to the reinforcements in order to guarantee the durability of the structure. For the design of the columns, it will be very important to consider to consider second-order effects (p-delta) since such structures could be susceptible to such.
Design of the bracing system
More often than not, reinforced concrete overhead tanks are not provided with additional bracing systems apart from the normal beams and columns. In other words, the structure relies on the frame elements for structural stability. As a result of this, the columns should be designed as unbraced columns since they resist the horizontal loads coming to the structure.
Apart from introducing shear walls or cores, introducing diagonal or cross-bracings will likely pose significant construction and aesthetic challenges.
Design of the ancillarycomponents
Ancillary components such as stairs and rails are provided in reinforced concrete overhead tanks for ease of movement and safety up and down the tank. Such features are important for regular inspection and maintenance of the tank after it is constructed. The stair can be made up of steel straight ladders attached to the frame, or constructed as a regular flight staircase to the top. The decision on the type of staircase will depend on the size and complexity of the water tank.
For circular water tanks, a spiral or helical staircase can be provided to wrap around the columns as shown in Figure 6.
Figure 6: Different types of stairs in overhead water tanks
The stairs should be firm, durable, and be able to carry the anticipated design live load. Furthermore, the rails should be able to resist direct human actions such as pulling, leaning, and mild impacts.
Design of the Foundation
The type of foundation adopted for a reinforced concrete overhead tank should depend on the height of the tank, the anticipated load, and the type of soil. On very weak and marginal soils, deep foundations such as piles may be required to send to the superstructure load to a firmer stratum.
Furthermore, for lightly loaded water tanks on a firm soil, isolated pad foundations may be provided. However, it will be in the best of the structure to ‘chain’ the entire columns together such that the foundation will behave as one unit. This will guide against the detrimental effects of differential settlement which might compromise the integrity of the superstructure. It is important to guide against secondary/indirect stresses that may cause cracks in the tank shell.
Design Example
It is desired to provide a reinforced concrete overhead water tank (not to exceed 15m high) to serve an estate of 50 households. Design the structure using relevant design standards and by making reasonable assumptions. There are no space constraints. Allowable bearing capacity of the soil = 100 kN/m2
Solution
Let us assume that each household has an average of 5 persons with a water demand of 120 litres per capita per day.
Volume of water required per day = 50 × 5 × 120 = 30000 litres = 30 m3
Making allowance for other miscellaneous uses such as the washing of cars, emergency fire fighting, etc, let us increase the volume to 35 m3.
Let the height of the tank shell (in-to-in) be 2.5 m.
Therefore the plan area of the tank = 35/2.5 = 14 m2 Let us adopt a square tank of dimensions 3.75 m × 3.75 m (Area provided = 14.06 m2)
Total volume of water anticipated (L x B x H) = 3.75 m × 3.75m × 2.5m = 35.16 m3
Initial Dimensions
Thickness of the tank shells =250 mm Dimension of the tank shell supporting beams = 450 x 250 mm Dimension of the bracing beams = 250 x 250 mm Dimension of the columns = 250 x 250 mm
Therefore the centre-to-centre spacing of the square tank shell will be 4 m (such that the in-to-in dimensions will be 3.75 m).
Total weight of water = 10 kN/m3 × 35.16 m3 = 351.6 kN
The load combinations considered in the design of the reinforced concrete overhead tank are;
When analysed on Staad Pro, the support reactions obtained were as shown below. The maximum column axial force obtained at ultimate limit state was 315.328 kN, while the maximum service load was 244.688 kN. Therefore, with a soil bearing capacity of 100 kN/m2, a square pad footing of 1650 mm x 1650 mm should be adequate. A plinth beam beam can be introduced to chain the separate bases together.
The summary of the maximum internal stresses obtained from the analysis is shown below;
(a) Deflection of the tank stand due to wind load
(b) Bending moment of the tank stand due to the effect of wind
(c) Bending moment on the tank stand due to gravity actions
The summary of the column design results is shown below. The results show that 4-T16 as the main bars and T8@250 c/c spacing as links is sufficient for the columns.
A snippet of one of the beam design results is shown below;
The slab analysis result is shown below;
For the bottom slab of the tank shell, T12@160 c/c bottom reinforcements and T12@285 c/c Top reinforcements satisfied flexural and crackwidth requirements. See the snippet below.
Materials that are added to a concrete mix aside from the basic ingredients (cement, water, and aggregates) are generally referred to as admixtures. The addition of these substances which are usually in liquid or powder form helps achieve some special effects in the concrete either in the fresh or hardened state.
In concrete, admixtures are usually categorised under ‘chemical‘ and ‘mineral‘ admixtures.
Chemical admixtures are inorganic substances that are added in concrete just before or during mixing, in addition to cement, water, and aggregate. They usually fall under categories such as superplasticizers, air-entraining admixtures, accelerating admixtures, etc.
Mineral admixtures are generally pozzolans or cement replacement materials that are usually obtained from agricultural or industrial wastes. Some examples of mineral admixtures are fly ash, rice husk ash, silica fume, etc. Mineral admixtures are usually very important in the production of green concrete, self-compacting concrete, and also help in the management of municipal wastes.
Admixtures are largely used by producers to minimize the cost of concrete construction, adjust the properties of hardened concrete, ensure the quality of concrete during mixing, transportation, placement, and curing, and overcome certain problems during concrete operations. In the United States of America, the specification for chemical admixtures is covered by ASTM C494/C494M-08. In Europe, BS EN 934-2 covers the definitions, requirements, conformity, marking, and labelling of admixtures for concrete.
The type and amount of cement used, as well as the water content, slump, mixing time, air temperature, and concrete temperatures, all influence the efficiency of an additive. Furthermore, when the water-cement ratio is reduced, more cement is added, a new type of cement is used, or the aggregate and aggregate gradation are changed, the effects are sometimes comparable to those achieved by adding admixtures.
The most common types of concrete admixtures are listed below;
(a) Admixtures that improve the performance of fresh concrete mixes
Plasticizers (Ordinary water-reducing admixtures)
Superplasticizer (High-range water reducing admixures)
Air entrainers
Pumping admixtures
(b)Admixtures thatmodify concrete setting time and hardening performance
Set retarder
Early Strength Agent
Early strength water reducing agent
Set accelerators
Pumping agent
Pozzolanic admixtures
(c) Admixtures thatimprove concrete durability
Gas forming admixtures
Air entrainers
Water-repellent admixture
Alkali-silica reactivity inhibitors
(d) Admixtures thatadjust the air content of concrete
Gas forming admixtures
Air entrainers
Defoamer
(e) Admixtures thatprovide concrete special properties
Shrinkage reducing admixture
Expanding agent
Anti-freezing admixture
Curing agent
Coloring admixture
Underwater concrete anti-dispersant
(f)Admixtures that performother special functions
Mold release agent
Damp proofing admixture
Concrete bacteriostatic agent
Anti-corrosion admixture
Adhesive bonding admixture
Plasticizers (Ordinary Water Reducing Admixtures)
When ordinary water reducing agent is added to concrete, it can achieve the desired slump even if the water-cement ratio is lower than normal. It normally reduces the amount of water used in the concrete mixture by around 5% to 10%. As a result, when compared to untreated concrete, concrete containing normal water-reducing admixture uses less water to create the desired consistency.
Furthermore, the water ash content of the treated concrete is low. This usually means that higher-strength concrete may be made without adding more cement to the mix. A wood grade salt (calcium or sodium) of the series is a commonly used ordinary water-reducing agent.
Superplasticizers(High-Range Water Reducing Admixtures)
Superplasticizer is a high-efficiency water-reducing admixture, that is usually classified under high-range water-reducing agent (HRWR). This admixture can lower water content by 12-30% and can be used to manufacture high slump concrete from low water-cement ratio concrete mix. The use of superplasticizers serves two purposes. One is to make high-strength concrete with a slump of 175 to 225 mm, which is used for heavily reinforced concrete structures where sufficient consolidation cannot be achieved readily by vibration; the other is to make high-strength concrete with a water/cement ratio of 0.3 to 0.4.
Water-reducing admixtures have been shown in studies to increase the workability of concrete for most types of cement. However, slump loss is a significant issue when utilizing superplasticizers in concrete. The superplasticizer effect lasts 30 to 60 minutes, depending on the brand and dosing rate, before rapidly deteriorating function. Because of the slump loss, a superplasticizer is frequently added to the concrete on the job site.
Naphthalenesulfonic acid formaldehyde condensate, melamine sulfonate formaldehyde condensate, polycyclic aromatic hydrocarbon sulfonate formaldehyde condensate, sulfamate series, aliphatic series, and polycarboxylic acid series are the most common forms of superplasticizers.
Gas forming admixtures
To alter the air content and apparent density of concrete, gas-forming admixtures can be used to produce a high number of confined bubbles through a chemical reaction in the concrete. This procedure can also be adopted in the production of light-weight concrete. Gas forming admixtures can improve the performance of new concrete while reducing bleeding and segregation.
Furthermore, aerated concrete provides greater freeze-thaw endurance because entrained air bubbles function as a physical buffer to prevent cracking produced by stress generated by an increase in water volume at freezing temperatures. They also improve the resistance of concrete to severe frost or freeze-thaw cycles. Almost all concrete admixtures are compatible with gas-forming admixtures.
Air-Entraining Admixtures
Air-entraining admixtures froth the mixing water, resulting in the incorporation of billions of tightly spaced air bubbles into the concrete. Water moves into the air bubbles in concrete when it freezes, releasing the pressure in the concrete. When the concrete thaws, the water in the bubbles can escape, resulting in less cracking than if air entrainment hadn’t been used.
The bubble diameter ranges from 0.02 to 1.0 mm, with the majority being less than 0.2 mm. The process is that the air-entraining agent can reduce the surface tension of the concrete by acting on the air-liquid interface, resulting in stable and fine closed pores. Small air bubbles encased in each other in the mixture can act as balls, reducing friction between the particles and improving the fluidity of the concrete. Water consumption can be lowered if the fluidity remains constant. In general, every 1% increase in air content reduces water demand by 6% to 10%.
On the other hand, the water absorption rate is reduced as a result of the reduced connecting capillaries, and the internal water pressure created by the freezing of water can be buffered, considerably enhancing frost resistance. Rosin thermopolymer resin and polyether alkyl sulfonates are the most often used air entrainers.
Pumping Admixtures
Pumping admixture is normally added to a concrete mixture that is to be pumped through a delivery conduit that is smooth, has no obstruction, has no performance segregation, and is made of good plastic material. A fluidizing agent is a type of pumping agent. It can not only considerably improve the fluidity of the combination, but it can also keep it fluid for 60 to 180 minutes, with the remaining slump not falling below the original 55 percent. It’s also not a retarder, and the retarding time shouldn’t be more than 120 minutes (except for special circumstances).
Because the pumping agent is mostly made up of water-reducing agents, it is typically classified by the type of water-reducing agent used, such as lignosulfonate, naphthalene, melamine, aliphatic, sulfamate, and poly acid-based.
Set retarder
A set retarder is an additive that can extend the time it takes for concrete to set in the initial and final stages. High temperatures cause the pace of hardening to accelerate, making placement and finishing more challenging. The retarder makes the concrete workable during installation and delays the concrete’s first setting. The additive is used to slow down the chemical reaction that happens while the concrete hardens. These concrete admixtures can generally delay the rate of concrete setting and counteract the effects of high temperatures on the setting of concrete.
The use of curing retarding admixtures in concrete pavement construction allows for more time for repairs, lowers the expense of constructing a new concrete mixing plant on the job site, and aids in the elimination of cold joints in the concrete. Phosphoric acid, phosphate, metaphosphate, zinc salt, borax, silicofluoride, sodium sulfite, ferrous sulfate, iron, copper, zinc, cadmium sulfate, and some oxides are examples of inorganic retarders.
Set Accelerators
Concrete accelerator is an admixture that allows concrete to set and harden quickly without compromising its long-term strength. Shotcrete and fast plugging materials require it as an additive. Tunnels, basements, slopes, and other construction auxiliary structural measures such as structural reinforcement, anti-cracking, and anti-seepage engineering have all made extensive use of it.
Powder and liquid accelerators are the two types of accelerators. Accelerators with a high alkali content have a stronger detrimental impact on the concrete’s later strength, and alkali damage to human skin is also a severe concern. Currently, world-renowned businesses such as Sika, Grace, and BASF produce liquid accelerators with rather constant performance.
Water Proofing Admixture
The impermeability and waterproof function of concrete can be considerably improved by a concrete waterproofing admixture, and the impermeability level can approach P25 or higher. At the same time, it has retarding, early strength, water reduction, and fracture resistance properties, as well as the ability to improve the workability of fresh mortar: it can be used in place of lime paste. It is particularly well suited to concrete for concrete roofs, water tanks and swimming pools, large-volume waterproof concrete, hydraulic concrete, waterproof mortar, and other similar applications. It is capable of totally resolving roof leaks, wall wetness, ground water seepage, and other issues.
Colouring Admixtures
Concrete products can be colored artificially. Concrete with toner is a method of making concrete that is non-toxic, odorless, and harmless to the human body, as well as weathering and UV resistance, color stability. Red, green, yellow, black, and blue are basic colors, as well as other unique color palettes. Other colors necessary for self-preparation by the agent are simple to apply and have no effect on the mechanical qualities of concrete products.
Expanding Admixtures
An expansion agent is a type of admixture that allows concrete to expand in volume. The goal of adding an expansion agent is to compensate for the concrete’s drying shrinkage, autogenous shrinkage, and temperature deformation. This can prevent cracking and increase the concrete’s compactness and waterproof performance. Expansion agents are now being used in a growing number of building projects, including basement floor and side wall concrete, steel tube concrete, super long structure concrete, waterproof concrete projects, and so on.
Water tanks are containment structures that are used for the storage of water for household, industrial, or commercial usage. Water tanks can be constructed in a variety of shapes such as circular (cylindrical), rectangular, oval, octagonal, etc. Furthermore, different materials such as reinforced concrete, steel, high-density plastics, reinforced polymers, etc have been successfully used in the construction of water tanks. The material to be used in the construction of water tanks must be durable, watertight, non-toxic, and strong enough to resist the hydrostatic pressure exerted by the stored water.
Circular water tanks have been shown by numerous studies to be more economical in terms of materials than square or rectangular tanks of comparable volume. However, there are more significant challenges and technicalities during the construction of circular tanks than rectangular tanks.
Water tanks can be elevated above ground level, placed on the ground surface, or buried under the ground. The design of underground circular water tanks involves the determination of the volume that will satisfy water demand requirements, and the selection of the section and reinforcements that will satisfy structural serviceability and ultimate limit state requirements.
Circular water tanks that are buried under the ground can be subjected to a variety of forces such as;
hydrostatic lateral pressure from the stored water
gravity vertical load due to the stored water
lateral earth pressure from the retained earth
gravity forces from the self-weight of the tank and the cover
In order to adequately design underground circular water tanks, the effects of actions on the tank should be evaluated when the tank is empty, and when it is filled. The most critical service case should then be used in the structural design. For instance, when the tank is filled, the walls of the tank will be subjected to hydrostatic pressure from the stored water and lateral earth pressure from the retained earth. To a large extent, these pressures may neutralise each other due to their reverse directions and relatively close values, but it is often common to ignore earth pressure when assessing water tanks for ‘filled’ service conditions.
This makes sense because, during the testing or inspection of water tanks for leakage or sometime after the construction, it may be possible that the backfill earth may not be in place or removed. However, when the tank is empty, the walls will likely be subjected to earth pressure only. Therefore, it is logical to reinforce underground water retaining structures in the near face and the far face, just in case the loads are reversed.
Another important aspect in the design of reinforced concrete circular water tanks is ensuring that the section of the tank remains water-tight. This can be achieved using several methods but it is important that the crackwidth of the section remains less than 0.2 mm. Cracking is normal in reinforced concrete structures and can be caused due to direct tension or flexural stresses on the section. Early age cracking due to temperature and shrinkage of concrete can also be a problem.
The reinforcements provided, the thickness of the section, concrete cover, type of cement, type of formwork, and the temperature during casting are the main design factors that can influence the crack width of water retaining structures.
Stresses in Circular Water Tanks
Circular water tanks are designed to resist ring tensions due to the horizontal pressures of the contained liquid. The nature and distribution of internal stresses in circular water tanks depend mainly on the support conditions. According to Reynolds et al (2008), if the wall is free at the top and free-to-slide at the bottom then, when the tank is full, no vertical bending or radial shear exists in the walls of the tank. The ring tension at depth z is given by;
n = γzr
where γ is the unit weight of liquid, and r is the internal radius of the tank.
However, if the walls of the tank are connected to the base of the tank (say fixed) in such a way that no radial movement occurs, then the ring tension will be zero at the bottom of the wall. This affects the ring tensions throughout the lower part of the wall, and significant vertical bending and radial shear occur. Charts are available for the evaluation of these stresses under different types of loading and boundary conditions (see Reynolds et al, 2008; Table 2.75 – 2.77). The results gotten from the charts and finite element analysis using Staad Pro have been compared. To achieve a fixed support condition, the base of the tank will have to be expanded until the pressure distribution becomes uniform.
Finite element model of a circular water tank on Staad Pro
Alternatively, the base of the tank can be modelled as a plate on an elastic foundation. This can easily be achieved on Staad Pro software using the ‘plate mat’ or ‘elastic mat’ option as foundation/support. In this case, the property of the soil is represented using the modulus of subgrade reaction only.
According to Reynolds et al (2008), when slabs are supported on an elastic foundation, the influence on the behaviour of the tank will depend on the r/rk ratio. Where r is the radius of the base of the circular water tank and rk is the radius of the relative stiffness. The concept of the radius of relative stiffness rk was developed by Westergaard and is given by the Equation below;
rk = [Ech3/12(1 – v2)ks]0.25
where Ec, is the short-term modulus of elasticity of the concrete, h is the slab thickness, ks is the modulus of subgrade reaction, and v is the Poisson’s ratio.
Worked Example(s)
A circular water tank 3.8 m deep with a mean radius of 2 m is open at the top and fixed at the bottom. Determine, due to internal hydrostatic loading and lateral earth pressure, the maximum service values for circumferential tension, vertical moment, and radial shear in the walls of the tank. The walls of the tank are 300 mm thick, and the water level is taken to the top of the wall.
Solution
From Table 2.75 of Reynolds et al (2008); lz2/2rh = 3.82/(2 × 2 × 0.3) = 12.03
At z/lz = 0.7; αn1 = 0.633 αm1 = 0.0026 and -0.0104 αv1 = 0.145
Hoop tension n = αn1γlzr = 0.633 × 10 × 3.8 × 2 = 48.108 kN/m Vertical moment (span) m = αm1γlz3 = 0.0026 × 10 × 3.83 = 1.427 kNm/m Vertical moment (edge/support) m = αm1γlz3 = -0.0104 × 10 × 3.83 = -5.706 kNm/m Radial shear v = αv1γlz2 = 0.145 × 10 × 3.82 = 20.983 kN/m
Load Case 2 In the same way, if we assume that the tank is buried under the ground and backfilled with sand of angle of shearing resistance 30° and unit weight of 19 kN/m2. Then the coefficient of earth pressure at rest using Rankine’s theory is given by;
Ko = (1 – sin 30°)= 0.5
The maximum lateral earth pressure at the base of the wall will be given by; p = koγz = 0.5 × 19 × 3.8 = 36.1 kN/m2
However, for the purpose of the formula to be used in the analysis, the unit weight of the material to be plugged into the equation can be represented as koγ = 9.5 kN/m3
In the same vein, we calculate;
Hoop compression n = αn1γlzr = 0.633 × 9.5 × 3.8 × 2 = 45.7 kN/m Vertical moment (span) m = αm1γlz3 = 0.0026 × 9.5 × 3.83 = 1.355 kNm/m Vertical moment (edge/support) m = αm1γlz3 = -0.0104 × 9.5 × 3.83 = -5.421 kNm/m Radial shear v = αv1γlz2 = 0.145 × 9.5 × 3.82 = 19.891 kN/m
In the design of the walls of the tank, the hoop tension must be resisted entirely by the ring reinforcements. When considering the hoop compression from the retained earth, the contribution of the concrete and steel can be considered (just like in the design of columns). It is typical to design the section for hoop tension and check for its adequacy in hoop compression.
The vertical reinforcements in each face of the wall must be able to resist the vertical moments due to the water pressure and the retained earth. This applies also to the span bending moments. Typically, the vertical reinforcements can be reduced accordingly as the wall goes higher. The quantity of reinforcement to be provided is also heavily influenced by cracking requirements.
In summary, the design of circular tanks is an interesting endeavour that can be carried out using hand calculations or computer software. Feel free to contribute to knowledge here.
