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)
Shrinkage in concrete refers to the volume reduction or contraction that occurs in concrete as it dries and ages. It is a natural and inevitable process caused by several factors, including the loss of water from the concrete matrix and the chemical reactions that take place during hydration.
The magnitude of shrinkage in concrete depends on various factors, including the water-to-cement ratio, cement type, aggregate properties, temperature, relative humidity, and curing conditions. Higher water-to-cement ratios generally lead to increased shrinkage, as there is more water available for evaporation and drying. Additionally, certain types of cement, such as high-early-strength (R) or low-alkali cement, can exhibit higher shrinkage characteristics.
For instance, grade 42.5R cement is expected to exhibit higher shrinkage characteristics than grade 42.5N cement.
Typical early shrinkage in concrete
Shrinkage in concrete can have several consequences and implications for construction. It can cause cracking, which not only affects the aesthetic appearance of the concrete but also compromises its structural integrity and durability. Cracks can provide pathways for the ingress of harmful substances, such as water, chlorides, and other aggressive chemicals, leading to the corrosion of reinforcing steel and the deterioration of the concrete.
There are different types of shrinkage in concrete such as drying shrinkage, plastic shrinkage, carbonation shrinkage, and autogenous shrinkage. However, shrinkage in concrete is usually the sum of autogenous shrinkage and drying shrinkage. In this article, we will be reviewing the difference between autogenous and drying shrinkage in concrete.
Autogenous shrinkage is caused by self-desiccation in young concrete as water is consumed during the hydration reaction. This occurs majorly within the early days of casting the concrete. However, drying shrinkage is the reduction in volume caused mainly by the loss of water during the drying process and this continues perhaps for years after the concrete is placed. This means that drying shrinkage commences after curing.
Autogenous Shrinkage
Autogenous shrinkage is a phenomenon that occurs in concrete without any external drying or thermal effects. It refers to the volume reduction or contraction of concrete due to the self-desiccation process, which is caused by the chemical reactions between cement and water during hydration.
When cement reacts with water, it forms a gel-like substance called C-S-H (calcium silicate hydrate), which gives concrete its strength. As the hydration process continues, this gel continues to absorb water, leading to a decrease in the water content within the concrete matrix. As a result, the volume of the gel decreases, causing autogenous shrinkage.
Autogenous shrinkage can be influenced by various factors, including the water-to-cement ratio, cement composition, temperature, relative humidity, and the presence of supplementary cementitious materials. Higher water-to-cement ratios generally lead to greater autogenous shrinkage because more water is available for hydration. Additionally, higher temperatures and lower relative humidity can accelerate the rate of shrinkage.
Autogenous shrinkage can have several consequences for concrete structures. First, it can lead to cracking if the concrete is restrained from freely contracting. These cracks can impair the durability and structural integrity of the concrete. Furthermore, shrinkage-induced cracks can provide pathways for the ingress of harmful substances such as water, chloride ions, and other aggressive agents, which can result in deterioration over time.
Mechanism of Autogenous Shrinkage (1) During the hardening process of concrete, water from the mix accumulates into the pores, voids, and capillaries of the mixture. (2) Water binds itself into the grains of the binder, as well as moving into the surroundings whose relative humidity is usually less than that of the concrete. (3) In order to move water from the tiny voids, considerable forces are required to overcome the surface tension, as well as the forces of adhesion which binds the water to the voids. (4) These forces act on the load-bearing structure of the concrete which at that time has low stiffness, can cause deformations in the concrete which can lead to cracking.
Autogenous shrinkage cracks can be largely avoided by keeping the surface of the concrete continuously wet; conventional curing by sealing the surface to prevent evaporation is not enough and water curing is essential. With wet curing, water is drawn into the capillaries and the shrinkage does not occur.
Practically, autogenous shrinkage happens in the interior of a concrete mass. According to BS EN 1992-1-1, autogenous shrinkage occurs in all concretes and has a linear relationship to the concrete strength.
According to EN 1992-1-1, autogenous shrinkage is given by Equation (1)
εca(t) = βas(t)εca(∞) ————————— (1)
Where; εca(∞) = 2.5(fck – 10) x 10-6 βas(t) = 1 – e(√0.2t) fck = characteristic strength of concrete (MPa) Where t is given in days.
Drying Shrinkage
As concrete matures, moisture is gradually lost from the concrete mass to the atmosphere. This moisture loss is accompanied by volume change which is referred to as drying shrinkage.
Drying shrinkage, therefore, depends on the relative humidity (for instance, indoor concrete will shrink more readily when compared with external concrete), the quantity and class of cement (rich concrete mixes will shrink more than leaner mixes), and the member size (thinner sections will shrink more quickly than thicker sections). According to EN 1992-1-1, drying shrinkage strain develops slowly, since it is a function of the migration of the water through the hardened concrete.
Fig 2: Drying shrinkage crack in concrete
Expression 3.9 in EN 1992-1-1 gives the development of drying shrinkage strain with time as given in Equation (2);
εcd(t) = βds(t, ts).kh.εcd,0 ———————- (2)
Where kh is a coefficient depending on the notional size h0 (see Table 1).
Table 1: Variation of ho with kh
βds(t, ts) = (t – ts)/[(t – ts) + 0.04√(h03)]
Where; t is the age of the concrete at the moment considered (in days) ts is the age of the concrete (in days) at the beginning of drying shrinkage (normally this is at the end of curing) h0 is the notional size (in mm) of the cross-section = 2Ac/u
Where; Ac is the concrete cross sectional area (for slabs, Ac = thickness x 1000 mm) U is the perimeter of that part of the cross-section exposed to drying (for slabs, take u = 2000 mm)
εcd,0 is the basic drying shrinkage strain. This is given in Annex B of EN 1992-1-1 as Equation (3);
εcd,0 = 0.85[(220 + 110.αds1).e(-αds2.0.1fcm)] x 10-6 x βRH ——— (3) βRH = 1.55[1- (RH/RH0)3] ————- (4)
Where; fcm is the mean compressive strength (Mpa) αds1 is a coefficient which depends on the type of cement = 3 for cement class S = 4 for cement class N = 6 for cement class R
αds2 is a coefficient which depends on the type of cement = 0.13 for cement class S = 0.12 for cement class N = 0.11 for cement class R
RH is the ambient relative humidity (%) RH0 = 100%
Water retaining structures, from dams to swimming pools and water tanks, demand concrete that adheres to strict performance criteria. Beyond structural integrity, the concrete must exhibit exceptional durability in an aqueous environment, resisting water penetration (leakage), chemical attack, and freeze-thaw cycles. Specifying appropriate concrete for these applications requires balancing multiple factors, ensuring both long-term functionality and cost-effectiveness.
In the year 2018, we made a successful publication on the Analysis and Design of Swimming Pools and Underground Water Tanks. In this article, we will briefly review some construction aspects of water retaining structures with emphasis on the specification of concrete to be used for the construction of water retaining structures. To download the textbook on the design of swimming pools, with Staad Pro video tutorials and an Excel spreadsheet for the calculation of crack widths, click HERE.
After carrying out the structural analysis and design of a water retaining structure, the next step is to ensure that the construction is properly executed such that the water tightness and strength of the element will not be compromised. A good design and a bad construction are as good as a failed project. The recommendations given in the following sections can be followed to ensure good results in the construction of water retaining structures.
Figure 1: Construction of Water Retaining Structure
Key Considerations in Concrete for Water Retaining Structures
Strength and Durability
Strength class: Minimum compressive strength is typically in the range of 30-50 MPa, depending on structure type and design loads. Consider future expansions or hydrostatic pressure when selecting a strength class. However, the characteristic compressive strength of concrete for water-retaining structures should not be less than 30 MPa (C30/37) after 28 days of curing.
Permeability: Low permeability minimizes water ingress, reducing the potential for leaching and reinforcement corrosion. Specify a maximum water-to-cement ratio (typically 0.50) and use dense, well-graded aggregates.
Air content: Adequate air content (4-7%) improves freeze-thaw resistance and reduces internal stresses.
Cracking: Minimize by proper joint design, reinforcement detailing, and shrinkage-reducing admixtures.
Material Selection
For the construction of water retaining/excluding structures, class N (normal hardening) cement is recommended and should be used ahead of Class R (rapid hardening) due to shrinkage issues. In general, the concrete should be specified according to BS EN 206 and BS 8500 Parts 1 and 2. For water tanks, all materials in contact with potable water will need to comply with specific regulations and should be non-toxic. This is why all admixtures that will be used must be approved.
Cement: Portland cement types N (normal) or NA (normal air-entrained) are common choices. For severe environments, consider sulfate-resistant cement or supplementary cementitious materials (SCMs) like fly ash or slag.
Aggregates: Dense, well-graded, and clean aggregates minimize voids and permeability. Specify maximum size based on wall thickness and reinforcement spacing.
Admixtures: Utilize admixtures judiciously. Air-entraining admixtures enhance freeze-thaw resistance, while water reducers improve workability without compromising strength. Ensure compatibility with other ingredients and potable water regulations if applicable.
Construction Practices
Mixing and placing: Employ proper mixing procedures and equipment to achieve uniform consistency. Place concrete promptly and avoid segregation.
Curing: Implement effective curing practices, such as water spraying or curing compounds, to ensure proper hydration and minimize shrinkage cracking.
Joints: Design and construct joints to accommodate movement and prevent water leakage. Utilize waterstops (water bars) where necessary.
Additional Considerations
Chemical exposure: For structures exposed to aggressive chemicals, specify cement and additives with appropriate resistance.
Water quality: If the structure holds potable water, ensure all materials comply with relevant regulations for safety and non-toxicity.
Sustainability: Explore options like SCMs or recycled materials to reduce environmental impact while maintaining performance.
Standards and Codes
Refer to relevant national or international standards like ACI 318 (Building Code Requirements for Structural Concrete) or BS EN 206 (Concrete – Specification, performance, production and conformity) for detailed guidance and specific requirements.
Figure 2: Olympic standard swimming pool
Permeability Considerations
Concrete for water-retaining structures must have low permeability. Water tightness refers to the ability of concrete to hold back or retain water without visible leakage (Kosmatka et al., 2003). It is common knowledge that the permeability of concrete is related to the water/cement ratio because the mix design factor is directly related to the permeability of the hardened cement paste. After the hydration reaction is completed, the remaining water leaves the concrete slowly, thereby leaving pores which reduce the strength of the concrete and the durability.
It is widely believed that a water/cement ratio of 0.2 (about 10 litres of water to 50 kg bag of cement) is needed to complete the hydration reaction, while the rest is to improve the workability of the concrete. However according to Mather et al. (2006), for a given volume of cement to hydrate completely, the quantity of original mixing water required is 1.2 times the solid volume of the cement.
The reason for this is that water should be available to fill up the 30% pore space that must be present after the hydration reaction. Ultimately, the authors opined that not all the cement will hydrate if the water/cement ratio is less than 0.4, even though only half of the water will go into a chemical combination. Research carried out by Kim et al. (2014) showed that porosity in concrete increased by 150% when the water/cement ratio was increased from 0.45 to 0.6.
Fig 2: Experimental Relationship Between Hydraulic Permeability and Water/Cement Ratio Under Different Curing Conditions (Whiting 1989, cited by Kosmatka et al, 2003)
According to Powers et al., (1954) cited by Kosmatka et al. (2003), the permeability of mature hardened cement paste kept continuously moist ranges from 0.1 x 10-12 to 120 x 10-12 cm/sec for water-cement ratios ranging from 0.3 to 0.7 (see Figure 2). Also, in a leakage test conducted by Portland Cement Association on cement mortar disks of 25 mm thickness subjected to a water pressure of 140 kN/m2; the disks made with a water/cement ratio of 0.5 or less showed no sign of leakage after being moist cured for seven days.
However, the disks made with a water/cement ratio of 0.8 showed leakage after being cured for the same period of time. Research has also shown that concrete cured in water is less permeable than concrete that hardened in air, therefore it is recommended that immediately after removing the formwork of tank walls, the tank should be flooded with water. This also helps in autogenous healing of cracks.
Furthermore, the permeability and gradation of the aggregates, the quality of the aggregate/cement paste interface, and the ratio between cement paste and aggregates affect the overall permeability of concrete. Workability is usually an issue when we try to keep the water/cement ratio low. The best way out is to use water-reducing admixtures to make the concrete workable.
This is a better option than using waterproofing admixtures and a high water/cement ratio. Waterproofing admixtures reduce absorption and water permeability by acting on the capillary structure of the cement paste. They will not significantly reduce water penetrating through cracks or through poorly compacted concrete which are two of the more common reasons for water leakage in concrete structures.
Summary
Table 1 gives the recommended concrete material requirements for water-retaining structures. A concrete specified and prepared as given in Table 1 should not give problems for water retaining structures provided it is well placed and compacted.
Table 1: Recommendation for concrete to be used water retaining structure
In effect, cracking can only be controlled by structural design based on the structural arrangement adopted: introducing movement joints, or by reinforcing the structure properly to limit the crack widths. Otherwise, Type A water protection should be specified. For water-retaining structures, the recommended minimum thickness for water tightness is 250 mm, unless the hydrostatic pressure in the tank walls is very low.
Specifying concrete for water retaining structures requires a comprehensive understanding of performance demands, material properties, and construction practices. By carefully considering strength, durability, material selection, and construction techniques, engineers can ensure long-lasting, leak-free structures that fulfil their intended purpose for decades to come.
References Kim Yun-Yong , Kwang-Myung Lee, Jin-Wook Bang, and Seung-Jun Kwon (2014): Effect of W/C Ratio on Durability and Porosity in Cement Mortar with Constant Cement Amount. Advances in Materials Science and Engineering Volume 2014, Article ID 273460, 11 pages http://dx.doi.org/10.1155/2014/273460
Kosmatka, S. H., Kerkhoff B., and Panarese, W. C. (2003): Design and Control of Concrete Mixtures, EB001, 14th edition, Portland Cement Association, Skokie, Illinois, USA
Mather, B. and Hime, W. G. (2002): Amount of Water Required For Complete Hydration of Portland Cement. American Concrete Institute (ACI) Volume: 2 Issue Number: 6 ISSN: 0162-4075 pp 56 -58 http://worldcat.org/oclc/4163061
Powers, T. C.; Copeland, L. E.; Hayes, J. C.; and Mann, H. M (1954): Permeability of Portland Cement Pastes, Research Department Bulletin RX053, Portland Cement Association, http://www.portcement.org/pdf_files/RX053.pdf
Whiting, D. (1989): “Permeability of Selected Concretes,” Permeability of Concrete, SP108, American Concrete Institute, Farmington Hills, Michigan, pp 195 -222.
When the most economical number of piles that will satisfy design requirements is three, a triangular-shaped pile cap can be adopted. In this case, the piles are arranged in a triangular manner, and equidistant from each other (usually spaced at 3 times the diameter of the pile).
Following the same arrangement, a similarly shaped pile cap can be used to transfer the loads to the piles. Triangular pile caps can be analysed and designed using the strut-and-tie method, beam analogy, or finite element analysis.
In our post on Structural Aspects of Pile Foundation Design, we showed how we can determine the number of piles required to support a column load, and we went ahead to design a triangular pile cap based on the strut-and-tie method (truss analogy).
In this article, we are going to analyse the same triangular pile cap using bending theory and finite element analysis and compare the results obtained.
The layout of the pile cap is shown in Figure 1 and we are going to assume that it is subjected to an axial load of 2500 kN from a 300mm x 300mm column (neglecting the orientation of the column in Figure 1).
Figure 1: Layout of triangular pile cap
Bending TheoryAnalysisof Triangular Pile Cap
The load on each pile is computed based on its distance from the centroid of the column
Load on the single pile N1 = N(Ay – Ay‘)/Ay Where: N = Axial load from column = 2500 kN Ay = Vertical distance between the centriod of piles in the y-direction = 515 + 1045 = 1560 mm Ay‘ = Vertical distance between the centroid of the column to the centroid of the single pile in the y-direction = 1045 mm
Bending moment in the x-direction Mx = N2 x (Ax – Cx)/2 Ax = Spacing of the twin piles = 1800 mm Cx = Width of the column parallel to x-direction = 300 mm
Mx = 837.3395 x (1.8 – 0.3)/2 = 628 kNm
Bending moment in the y-direction My = N1 x (Ay‘ – Cy/2) Ay‘ = Spacing of the twin piles = 1045 mm Cy = Width of the column parallel to y-direction = 300 mm My = 825. 321 x (1.045 – 0.3/2) = 738.662 kNm
Finite Element Analysisof Triangular Pile Cap
The pile cap was modelled as a 1300 mm thick concrete plate element on Staad Pro, while the piles were modelled as 600 mm diameter columns with a length of 2m supported on a fixed base. The short column was adopted to avoid wide variation of result due to second-order effects and to avoid large displacement which is not very practicable and also not considered in manual analysis.
Figure 2: 3D rendering of the pile cap model
The results obtained from Staad Pro are given in Figures 3 – 6 below;
Figure 3: Axial load on Single Pile
Figure 4: Axial load on twin pile
Figure 5: Bending moment in the x-direction
Figure 6: Bending moment in the y-direction
The summary of the results between manual analysis (bending theory) and Staad Pro (finite element analysis) is shown in Table 1.
Action effect
Bending theory
FEA (Staad Pro)
Percentage difference
Single pile load (kN)
825.32
825.77
0.054%
Twin pile load (kN)
837.32
837.12
0.024%
Mx (kNm)
628
756.07
18.507%
My (kNm)
738.66
762.904
3.22%
Table 1: Comparison of bending theory and finite element analysis results
From Table 1, it could be seen that the distribution of load to the piles was found to be approximately the same for both approaches, but a wide variation was observed for Mx bending moment between the two methods. The My bending moment was found to be close with a percentage difference of about 3.229%.
Therefore, Staad Pro can be used to model triangular pile caps and the result can be used for design purposes. The result obtained from this study is open to discussion, especially based on the difference obtained in bending moment values. We will be glad to hear from you!
To download the full structural analysis and design calculation sheet, click HERE
A site instruction is a formal written order given by the heads of a project to contractors or sub-contractors on specific projects issues such as delay in progress, defective work or materials, guidance on how to carry out a specific item of work, permission to proceed with an item of work, instruction on procurement and logistics, amendments to procedure, or instruction on general project challenges. Site instructions are formal documents that can be presented during disputes, claims, variations, and arbitration. As a result, they should be issued and treated with utmost care and caution by both parties.
Due to its formal nature, a site instruction must be specific, direct, and understandable. A good site instruction must contain the following information;
(1) The name of the firm and the individual issuing the instruction. (2) The name of the firm and the individual receiving the instruction. (3) The place, date, and time that the instruction was issued. (4) The team that issued the instruction and the head of the team (if applicable). (5) The current observation on site for the item of work for which instruction is being issued (6) The instruction on what is to be done (with sketches if applicable). (7) The signature of the person issuing instruction and the signature of the person receiving the instruction
It should be issued in a minimum of 3 duplicates where one copy is kept on site with the site manager, the consultant takes one copy, and another copy is filed in the project’s file at the company’s office.
Who issues site instruction? Site instruction can be issued by the project manager to the consultants or contractors. The consultants can issue instruction to the contractors, and contractors can issue instruction to the sub-contractors. Building regulatory agencies from the government can also issue site instruction to the contractors or consultants of a project.
For example, if a client decides to make HVAC changes, which will lead to need for new ductworks (say for chiller pipes), the project manager (usually the architects) will have to instruct the HVAC consultants to modify their design, and perhaps instruct the structural engineers to make provision for passage of pipes through the beams. Note that this should be a technically coordinated decision. In as much as the project has started already, such alterations cannot be instructed verbally, hence the need for written site instruction to that effect. The structural engineers may deem the changes not too critical, and may opt to issue site instruction (with sketches) to the main contractors (usually civil works) on how to make adjustments for the pipes to pass through.
Also, during site inspection, regulatory agencies can issue site instruction to contractors to increase set-backs if found contrary to the approved drawing. They can also order demolition of defective works, or rearrangement of wrongly placed bars etc.
When do you issue/request for site instruction? (1) A project head should issue site instruction when proposing a change to what is in the approved construction drawing.
(2) If an error or challenge is discovered in the working drawing, the consultant must either issue a new drawing or give a site instruction on how to proceed.
(3) A project head should issue site instruction when proposing a solution to an unforeseen problem or challenge on site which can impact on time, procedure, and cost.
(4) A project head can issue notice of delay in form of a site instruction.
(5) A site manager/contractor should request for an instruction from consultants/regulatory agencies to proceed after an item of work has been satisfactorily completed. For instance, after checking arrangement and placements of reinforcements and formwork, the consultant/regulatory agency should issue an instruction to the contractor to proceed with concreting.
(6) A site manager/contractor should request for a written site instruction when asked to do something he is not comfortable or very familiar with.
(7) A site manager/contractor should request for site instruction to proceed and use a material or procedure he has not tried before.
What do you do after receiving site instruction? Since site instructions are formal documents, the site manager should therefore take the following steps after receiving instruction:
(1) Contact the company’s head office using the approved means of communication and forward a duplicate copy of the instruction received for filing (an e-mail attachment or posting it to an online workplace platform can suffice).
(2) Discuss and analyse the impact of the instruction on delivery time, procedure, and cost of the project with the project team.
(3) The contractor should send a formal response to the instructor on the impact of the instruction if necessary. It is worthwhile to note that some instructions are very normal in construction and do not need deliberation for an experienced site manager. For instance, a site instruction from a consultant that formwork should not be removed until after 21 days should be seen as normal. However, if this will affect the project delivery time, the contractor can request for an instruction to remove the formwork at say 14 days. The project team can look at the situation technically and a new instruction can be issued to proceed or there can be an outright rebuttal by the consultant. However, it is now on record that there might be 7 days delay depending on the initially agreed programme of work.
Instructions to demolish, remove, or make good a major defective work must be approved from the general office with the input of the quantity surveyor of the project. Such decisions do not require a site manager’s unilateral action.
(4) A good site manager should work hard and implement all approved corrections according to the instruction.
(5) After all the corrections have been effected, the consultant or agency should be invited to check that the instruction has been carried out properly, after which they will issue a new instruction to proceed with the next item of work.
Do you need good project managers, drawings, and construction advice for your construction works?Send an e-mail to structville@gmail.com or info@structville.com. Alternatively, send a Whatsapp message to +2347053638996
In the design of a pile foundation, the geotechnical engineer is expected to hand over the soil investigation report to the structural engineer, who will proceed to provide the longitudinal reinforcements needed for the piles and also design the pile cap. The structural design of a pile cap is an important aspect of pile foundation design and the method of carrying out the design has been presented in this article.
The soil investigation report handed to the structural engineer for the design of pile foundation purpose should contain the embedment length of the piles, the recommended sizes of the pile, the safe working load of each pile size, and other information that may be necessary for the structural engineer to do his design properly. All aggressive materials in the soil should be stated as well so that adequate protection will be given to the pile material(s) for durability.
The first step in the structural design of a pile cap usually involves obtaining the number of piles required to support each column load. This is usually done using the service loads of the column, and relating them to the safe working load of the piles from the soil investigation report. In this article, we are going to show how the structural design of reinforced concrete pile foundations and pile caps can be done based on practical design and site experience.
Design Example The frame of a 5 storey building is shown in Figure 1, and it is expected to be supported on piles with an embedment length of 20 m. The allowable working loads of the bored pile (CFA) is shown in Table 1. fy = 460 Mpa, fcu = 30 Mpa
Table 1: Pile safe working load
Pile diameter (mm)
300
450
600
750
900
Safe working load (kN)
246.74
370.11
493.48
616.85
740.22
Fig 1: Frame of a 5-storey building
Fig 2: Column load layout
Design for Column A1 Service axial load on column = 647 kN Ultimate axial load on column = 885 kN Size of column = 450 x 230 mm
Try 2 No of piles Service load per pile = 647/2 = 323.5 kN Let us adopt 600 mm diameter piles for uniformity and to have fewer pile boring points
Safe working load of φ600 mm piles = 493.48 kN > 323.5 kN Okay
Centre to centre spacing of piles = 3φ = 3 x 600 = 1800 mm Overhang of pile cap edge from the pile = 150 mm
Total length of pile cap = 1800 + 600 + 2(150) = 2700 mm Width of pile cap = 600 + 150 + 150 = 900 mm Thickness of pile cap = 2φ + 100 = 2(600) + 100 = 1300 mm
The layout of the pile cap is therefore given as shown in Figure 3.
From Table 3.61 of Reynolds et al. (2008), the tensile force to be resisted within the pile cap is given by;
Ft = N/(12ld)[3l2 – a2]
Where; N = Column axial load at ultimate limit state l= Centre to centre spacing of the piles d = Effective depth of the pile cap a = dimension of the column side parallel to the length of the pile cap
Self weight of the pile cap (ULS) = 1.4 x Area x depth x 24 kN/m3 = 1.4 x 2.7m x 0.9m x 1.3m x 24 kN/m3 = 106.14 kN
N = 885 kN + 106.14 = 991.142 kN l = 1.8 m d = 1300 – 100 = 1200 mm = 1.2 m a = 0.45 m
Ft = [991.142/(12 x 1.8 x 1.2)] x [3 x 1.82 – 0.452] = 364 kN Ast = Ft/0.95fy = (364 x 1000)/(0.95 x 460) = 833 mm2 Asmin = 0.13bh/100 = 1690 mm2 Provide 6T20 @ 175 c/c (Asprov = 1974 mm2)
Check for shear Critical position for shear on vertical section across full width of pile-cap occurs at distance from face of column given by: av = 0.5(l – c) – 0.3φ = 0.5(1800 – 450) – (0.3 x 600) = 495 mm
The shear force carried by the piles V = 991.142/2 = 495.571 kN
The shear stress ν = V/bd = (495.571 x 1000)/(900 x 1200) = 0.458 MPa Concrete resistance shear stress vc = 0.632(100As/bd)1/3(400/d)1/4
vc = 0.632 x [(100 x 1974)/(900 x 1200)]1/3 x (400/1200)1/4 = 0.632 x 0.557 x 0.759 = 0.275 MPa For grade 30 concrete, vc = 0.275 x (30/25)1/3 = 0.292 Mpa vc(2d/av) = 0.292 x [(2 x 1200)/495] = 1.415 MPa > 0.458 MPa This is okay
Shear stress at column perimeter ν = V/ud = (885 x 1000)/[(2 x 225 + 2 x 450) x 1200] = 0.546 MPa This is less than 0.8√fcu = 4.38 Mpa. Therefore, this is okay.
Anti-burst bars of T12 @ 200 spacing should be provided Main bars should be returned at least 900 mm into the sides to satisfy anchorage length requirements. You can take anchorage length to be conservatively 50 x diameter of reinforcement = 50 x 20 = 1000 mm
Design of Pile Cap Type 2 Service axial load on column = 1077 kN Ultimate axial load on column = 1476 kN Size of column = 450 x 225 mm The number of φ600 mm piles required = 1077/493.48 = 2.184
Using 3 No of φ600 piles Service load per pile = 1077/3 = 359 kN
Safe working load of φ600 mm piles = 493.48 kN > 359 kN This is okay
Let us adopt a triangular pile cap arranged in such a way that the column load will be equally distributed to the piles. This arrangement can be found in Table 3.16 of Reynolds et al (2008), and it is shown in Figure 4.
Fig 4: Dimensions of triangular pile cap for equal load distribution (Reynolds et al, 2008)
hp = φ = diameter of pile = 600 mm Spacing of piles = 3φ = 3 x 600 = 1800 mm Overhang of pile cap edge from the pile = 150 mm (α + 1)φ + 300 = (3 + 1)600 + 300 = 2700 mm φ + 250 = 600 + 250 = 850 mm φ + 300 = 600 + 300 = 900 mm (6α/7 + 1)φ + 300 = 2442.857 mm (say = 2445 mm) (2α/7 + 0.5)φ + 150 = 964.285 mm (say = 965 mm) Thickness of pile cap = 2φ + 100 = 2(600) + 100 = 1300 mm
The layout of the pile cap is shown in Figure 5.
Fig 5: Structural arrangement of 3 pile caps
Self weight of the pile cap (ULS) = 1.4 x Area x depth x 24 kN/m3 = 1.4 x 5.166 m2 x 1.3m x 24 kN/m3 = 225.61 kN
Total load on pile cap at ULS = 1476 kN + 225.61 kN = 1701.61 kN l = 1.8 m a = 0.225 m b = 0.45 m
Tensile force to be resisted by the reinforcement in the direction parallel to X-X; Ft,x = N/(36ld)[4l2 + b2 – 3a2] Ft,x = [1701.61/(36 x 1.8 x 1.2)] x [4 x 1.82 + 0.452 – 3 x 0.2252] = 284 kN
Tensile force to be resisted by the reinforcement in the direction parallel to Y-Y; Ft,y = N/(18ld)[2l2 – b2] Ft,y = [1701.61/(18 x 1.8 x 1.2)] x [2 x 1.82 – 0.452] = 275 kN
Let us use the highest value for design in anticipation that we will provide the same reinforcement in both directions Ast = Ft/0.95fy = (284 x 1000)/(0.95 x 460) = 649 mm2 Asmin = 0.13bh/100 = 1690 mm2 Provide T20 @ 175 c/c in both directions (Asprov = 1974 mm2)
Shear resistance The shear force carried by the piles V = 1701.61/3 = 567.2 kN The shear stress ν = V/bd = (567.2 x 1000)/(1000 x 1200) = 0.472 MPa vc(2d/av) = 0.292 x [(2 x 1200)/495] = 1.415 MPa > 0.472 MPa This is okay Shear will obviously not be a problem.
Design of Pile Cap Type 3 Service axial load on column = 1825 kN Ultimate axial load on column = 2545 kN Size of column = 400 x 400 mm The number of φ600 mm piles required = 1825/493.48 = 3.69
Use 4 No of φ600 piles Service load per pile = 1825/4 = 456.25 kN
Safe working load of φ600 mm piles = 493.48 kN > 456.25 kN This is okay
Let us adopt a square pile cap arranged in such a way that the column load will be equally distributed to the piles. This arrangement can be found in Table 3.16 of Reynolds et al (2008), and it is shown in Figure 6.
Fig 6: Pile Cap Type 3
Self weight of the pile cap (ULS) = 1.4 x Area x depth x 24 kN/m3 = 1.4 x 7.29 m2 x 1.3m x 24 kN/m3 = 318.43 kN Total load on pile cap at ULS = 2545 kN + 318.43 kN = 2863.43 kN
Tensile force to be resisted by the reinforcement in both directions; Ft = N/(24ld)[3l2 – a2] Ft = [2863.43/(24 x 1.8 x 1.2)] x [3 x 1.82 – 0.402] = 528 kN
Ast = Ft/0.95fy = (528 x 1000)/(0.95 x 460) = 1208 mm2 Asmin = 0.13bh/100 = 1690 mm2 Provide T20 @ 175 c/c in both directions (Asprov = 1974 mm2)
Shear resistance The shear force carried by the piles V = 2863.43/4 = 715.9 kN The shear stress ν = V/bd = (715.9 x 1000)/(1000 x 1200) = 0.595 MPa vc(2d/av) = 0.292 x [(2 x 1200)/495] = 1.415 MPa > 0.472 MPa This is okay
The structural engineer is expected to provide the following drawings;
(1) Setting out drawing showing the piling points and layout with a known reference point (2) General column/pile cap layout/arrangement (3) Pile cap/ground beam/ground floor slab layout (3) Columns, piles and pile cap reinforcement drawings (detailing) (5) Ground beams and ground floor slab reinforcement detailing (6) Construction procedure sketches
Construction Considerations (1) Ground beams are usually provided to chain the pile caps together, and to provide the needed support for the ground floor slab. There are construction scenarios where the ground floor slab is placed directly on the pile caps, but note that this concept is quite different from piled raft foundation. The ground beams are usually embedded into the pile caps, or may be allowed to sit directly on the pile caps depending on the site level. A typical construction drawing showing this interaction is given in Figure 7.
(2) Contractor should maintain a minimum concrete cover of 75 mm
(3) There may be need to be cast the pile cap in two stages to achieve the configuration shown in Figure 7. The first casting will get to the bottom level of the ground beams (see Figure 8), then the ground beam reinforcements are laid (see Figure 9), before the final casting of the pile cap and ground beams to the required level (see Figure 10). Read about bonding of old and new concrete.
Figure 8: Typical casting of pile cap to ground beam level.
On getting to the stage shown in Figure 10, the bays are filled with sharp sand, and the ground floor slab is cast as appropriate.
Do you need help in design, consultancy, production of construction drawings, supervision, and project management, contact us today at Structville Integrated Services Limited. We are excellent at what we do, and we pride ourselves in professionalism and integrity. Send an e-mail now to info@structville.com copy structville@gmail.comor send a whatsapp message to +2347053638996.
References [1] Reynolds C.E., Steedman J.C., Threlfall A.J. (2008): Reynolds’ Reinforced Concrete Designer’s Handbook 11th Edition. Taylor and Francis, New York
Civil engineers are always faced with one challenge or another on-site, and one of the major challenges on-site has to do with the nature of the soil to support the structure to be constructed. Some soils exhibit excessive change in volume under constant load and changes in moisture content. Under field conditions, if the addition of water under constant load causes the void ratio to reduce drastically, the soil is said to have collapsed.
These soils are predominantly found in the arid regions and appear strong and stiff in their dry natural state, but lose strength and undergo high compression upon wetting. According to Behzad (2013), almost all natural deposits of collapsible soils are either debris flow deposits or wind deposited soils (loess).
Soils which exhibit this behaviour are usually unsaturated granular soils which in a loose state are maintained by apparent cohesion. This apparent cohesion may be due to the presence of clays at the intergranular contact areas or the accumulation of soluble salts as binders (Murthy, 2012). In South-Western Nigeria, collapsible soils have been observed and studied by Owolabi and Ola (2014).
Collapsible Soils
For a soil sample to collapse, the soil structure must have an open, partially unstable, and unsaturated fabric which must be held in place by apparent cohesion. On wetting and application of high enough stress, the soil fabric will be expected to collapse.
For compacted cohesive soils, the collapse behaviour will depend on the percentage of fines, the moulding water content, the dry density, and the compaction energy employed Murthy (2012). Any soil compacted at the dry side of the optimum moisture content and at low dry density will also be expected to develop a collapsible structure.
Calculation of Collapse Settlement
A method of calculating the collapse settlement of a soil sample was suggested by Jennings and Knight (1975) and reported by Murthy (2012). This involves carrying out double oedometer tests of identical undisturbed samples in the laboratory. Both specimens are kept under a constant pressure of 1 kPa for a period of 24 hours at their natural moisture content after which one specimen is flooded and the other allowed to remain in its natural state.
After flooding for 24 hours, the consolidation test is allowed to progress as usual after which the e-LogP curve is plotted for both samples. The collapse settlement is calculated in two parts according to equations (1) and (2).
Where; Δen = Change in void ratio due to load ΔP as per the e-logP plot without change in moisture content Δec = Change in void ratio due to load ΔP as per the e-logP plot with increase in moisture content (settlement due to the collapse of the soil structure). H = Thickness of the collapsible layer e0 = Initial void ratio
Collapse settlement (S1) = S1 + S2
Solved Example
A building was constructed on collapsible soil and the double oedometer test carried out on the undisturbed sample gave the results shown in Table 1. The thickness of the collapsible layer is 4 m and the average overburden pressure was 60 kPa. Calculate the collapse settlement for an increase in pressure of 40 kPa.
Applied Pressure (kPa)
10
20
40
100
200
400
800
Void ratio at natural moisture content
0.80
0.79
0.78
0.75
0.725
0.68
0.61
Void ratio at soaked condition
0.75
0.71
0.66
0.58
0.51
0.43
0.32
Table 1-
Solution
(1) The first step is to plot the e-log P curve for the two samples as shown below. This has been done using Microsoft Excel.
Figure 1: e-log P plot for double oedometer test
(2) The virgin compression curve is plotted by drawing a line tangential to the soaked sample curve.
(3) After drawing the virgin compression line, the overburden pressure is identified, and for this problem, it is given at 60 kPa. This is traced up to meet the virgin compression line, and this marks the initial void ratio (e0).
(4) From point e0, an adjusted moisture curve is plotted to be perfectly parrallel to the natural moisture content curve.
(5) The increment pressure p + ΔP = 60 + 40 = 100 kPa. A vertical line is drawn from the pressure of 100 kPa to meet the soaked sample curve and adjusted moisture curve.
Steps (2) to (5) are shown in Figure 2.
Fig 2: Full analytical plot for collapse settlement of the soil
(6) Trace down the points where the pressure increment makes contact with the soaked sample curve and adjusted moisture content curve and identify the associated void ratios e1 and e2.
For the problem at hand; e0 = 0.64 e1 = 0.62 e2 = 0.58
(3) In a laboratory study carried out by AlShaba et al.. (2018) at Egypt, iron powder was used to reduce collapse settlement of collapsible soil by about 46% at a mix ratio of 6% of the dry weight of the soil. When the same mixture was compacted, a reduction ratio of about 86.66% was obtained.
(4) Prewetting and compaction of the collapsible soil.
(5) Ayeldeen et al (2016) obtained promising results on improving collapsible soils using biopolymers.
References [1] Ayeldeen M., Nrgm A., El-Sawwaf M., Kitazume M. (2016): Enhancing the Behaviour of Collapsible Soils using Bioploymers. Journal of Rock Mechanics and Geotechnical Engineering. doi:10.1016/j/jrmge.2016.11.007 [2] AlShaba A.A., Abdelaziz T.M., Ragheb A.M. (2018): Treatment of Collapsible Soils by Mixing with Iron Powder. Elsevier – Alexandria Engineering Journal (57):3737-3745 [3] Behzad Kalantari (2013): Foundations of Collapsible Soils: A Review. Proceedings of the Institute of Civil Engineers Forensic Engineering 166(FE2):57-63 [4] Jennings J.E., and Knight K. (1975): An Additional Settlement of Foundation Due to Collapse Structure of Snady Soils on Wetting. Proceedings to thhe 4th International Conference on Soil Mechanics and Foundation Engineering, Vol 1. Paris [5] Murthy V.N.S (2012): Textbook of Soil Mechanics and Foundation Engineering (First Edition). CBS Publishers and Distributors Pvt Ltd, India [6] Owolabi F.A., Ola S.A. (2014): Geotechnical Properties of a Typical Collapsible Soil i South-Western Nigeria. Electronic Journal of Geotechnical Engineering (19):1721-1738
Gravity actions are normally used for the verification of portal frames for member resistances, lateral buckling, and torsional stability. Elastic design of portal frames is permitted by Eurocode 3, and the most significant load case (from experience) is the situation where the frame is subjected to gravity load from permanent actions and variable actions, taking into account the second-order effects and imperfection.
The load combination used for this is usually of the form shown in Equation (1);
p = 1.35gk + 1.5qk (1)
In the past, we have solved the problem of two hinged portal frames which are statically determinate due to the presence of internal hinge at the apex. However, it is not a very practical scenario as portal frames are required to be rigidly connected in order to achieve in-plane stability.
In this article, we are going to show how you can obtain the effects of actions (bending moment, shear force, and axial force) for a rigid portal frame subjected to an ultimate limit state load of 12 kN/m as shown in Figure 1. We will assume that the same universal beam section will be used for the column and the rafters (EI = Constant).
Fig 1: Portal frame
A little consideration will show that the frame shown in Figure 1 is statically indeterminate to the first order. Therefore, using the force method of analysis, we will need to reduce the structure to a basic system.
This is a process by which we remove the redundants (excess reactive forces) in the structure, to make it statically determinate. The basic system adopted must also be stable. This is shown in Figure 2.
Fig 2: Reduction of the frame to basic system
To proceed, we will now assign a unit force to the removed redundant horizontal reaction, and plot the bending moment diagram due to the unit force on the basic system as shown in Figure 3.
Fig 3: Bending moment diagram due to horizontal unit force at point A
Since there is no other redundant support, we can apply the external load on the basic system. Due to the symmetry of the structure and the loading, we can verify that the vertical support reactions at A and E are given as shown in Figure 4;
Ay = Ey = (12 x 18)/2 = 108 kN
Fig 4: External load and support reaction on the basic system
On observing Figure 4, we can see that there will be no bending moment in sections A-B and D-E, rather they will be subjected to compressive axial force of 108 kN. For member B-C, the bending moment will induced due to the reactive force and the externally applied uniformly distributed load. The free-body diagram of member B-C is shown in Figure 5.
We can verify that the geometrical properties of B-C are as follows; The angle of inclination β = tan-1(1.5/9) = 9.462º The length of the member z = √(1.52 + 92) = 9.124 m
Fig 5: Free body diagram of member of B-C
The equation for the bending moment of the section can be given by; Mz = [108 x cos(β) x z] – [12 x cos(β) x z2]/2 Mz = 106.53z – 5.918z2
At z = 9.124 m MCL = 106.53(9.124) – 5.918(9.124)2 = 479.32 kNm
The same thing is also happening at section C-D, and the bending moment diagram of the structure is shown in Figure 6.
Figure 6: Bending moment diagram of external load on basic system
The canonical equation for the structure is given by Equation (2);
δ11X1 + Δ1P = 0 ——– (2)
Where; δ11 = Deflection at point 1 due to unit force at point 1 X1 = The actual external reaction at point 1 in the direction of the deflection due to externally applied load Δ1P = Deflection at point 1 due to externally applied load
We are going to evaluate δ11 and Δ1P according to Vereshchagin’s rule which involves combining of the bending moment diagrams. For another example of how to apply Vereshchagin’s rule, click here.
Evaluation of δ11 In this case, the bending moment diagram of the unit force will combine with itself.
Fig 7: Load Case combining with itself
δ11 = 2 x [1/3 x 8 x 8 x 8] + 2 x 1/3 x [(8 x 8) + (9.5 x 8) + (9.5 x 9.5)] x 9.124 = 1741.867/EI
Evaluation of Δ1P Here, the bending moment diagram due to the unit force will combine with the bending moment diagram due to the externally applied load as shown in Figure 8.
Fig 8: Load case 1 combined with externally applied load
Δ1P = 2 x (1/12) x 479.32 x [(5 x 9.5) + (3 x 8)] x 9.124 = 52115.345
Substituting back into the canonical equation; X1 = 52115.345/1741.867 = 29.919 kN (This obviously represents the horizontal reactions Ax and Ex)
The final bending moment diagram can be obtained therefore as follows; Mdef = M1X1 + M0 MA = ME = 0 MB = MD = (29.919 x -8) + 0 = -239.352 kNm MC = (29.919 x -9.5) + 479.32 = 195.089 kNm
The bending moment diagram is shown in Figure 9, and the shear and axial forces can also be obtained accordingly.
Fig 9: Final bending moment diagram
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