A free-standing sawtooth staircase is a type of slabless (without waist) staircase that is freely supported at the landing. By implication, this staircase comprises of the thread and risers only, which are usually produced using reinforced concrete.
Analytical and detailing solutions exist for reinforced concrete sawtooth and free-standing staircases, but when the two systems are combined, there may be a challenge with the detailing due to the well-known structural behaviour of cantilevered type structures.
Fig 1: Typical section of a free-standing sawtooth staircase
In cantilevers, the main reinforcements are provided at the top (the tension area), and furthermore, they must be properly anchored and/or extended into the back-span for good anchorage/development length, and to resist the hogging moment that exists at the back of the cantilever.
In reinforced concrete slabs, the cantilever reinforcement should extend into the back-span by at least 0.3 times the span of the back-span, or 1.5 times the length of the cantilever, whichever is greater. The same requirement of reinforcements extending cantilever reinforcements applies to reinforced concrete beams too.
It can therefore be seen that continuity of reinforcements is an important detailing requirement of cantilevered type structures. Without the introduction of haunches, this may be difficult to achieve in free-standing sawtooth staircases.
In sawtooth staircases, the main reinforcements are provided on the landing, which is connected to the risers using links. A typical detailing sketch of a sawtooth staircase is shown below;
Fig 2: Typical detailing sketch of sawtooth (slabless) staircase
Knowing full well that this type of staircase has been successfully designed and constructed (see Fig. 3). The pertinent questions to ask are, therefore;
(1) What is the detailing procedure of a free-standing sawtooth staircase without the use of haunches? (2) Is the continuity of reinforcements necessary at the landing-riser junction? (3) Are the links rigid enough to provide the needed continuity?
Fig 3: Well constructed free-standing sawtooth staircase
To provide an insight to the answers, let us consider a finite element model of a free-standing sawtooth staircase.
The properties of the staircase are as follows;
Width = 1000 mm Width of landing = 1000 mm Height of riser = 175 mm Width of thread = 275 mm Thickness of all elements = 150 mm
Loading (1) Self weight (2) Finishes of 1.2 kN/m2 (3) Imposed load of 3 kN/m2
Load Combination ULS = 1.35gk + 1.5qk
Fig 4: Finite element model of the free-standing sawtooth staircase
Some of the analysis results are given below;
Fig 5: Deflection profile of the staircase under uniformly distributed load
Fig 6: Transverse bending moment of the staircase (ULS)
Fig 7:Longitudinal bending moment of the staircase (ULS)
Fig 8:Twisting moment of the staircase (ULS)
From the nature of the internal stresses distribution, is it safe to say that the true cantilever behaviour of structures is not properly represented in free-standing sawtooth staircases? From the deflection profile, this is probably not the case. Can we get sketches of typical detailing guide from you?
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Precast concrete columns are reinforced concrete columns that are cast and cured on the ground before being hoisted up and installed in their desired positions. Just like in-situ columns, precast columns are capable of resisting shear, axial force, and bending moment, however, careful attention must be paid to their connection details. The design of precast concrete columns involves the provision of adequate member size, reinforcement, and connection details to satisfy internal stresses due to externally applied loads, second-order effects, and lifting.
Different connection conditions can be adopted by different manufacturers. The foundation connection of a precast column may be achieved by allowing reinforcement bars to project from the column which is then passed through established sleeves before being filled with concrete grout. Alternatively, a base plate can be connected to the column which is then installed in position on a concrete base using bolts and nuts.
Precast concrete columns with protruding reinforcements
Precast columns may have corbels or nibs for supporting the beams. Alternatively, precast beam-column connections can be made using dowels or mechanical couplers.
Precast concrete columns have the following advantages over in-situ concrete construction;
Increased speed in construction since production of precast elements can commence ahead of time
Greater flexibility in project management and site planning due to off-site production capacity
Improved and higher quality of concrete, dimensions, and surface finishes
Reduction in site labour
Reduction in formwork requirement
Less wastage of materials
The design of precast reinforced concrete columns is carried out by a structural engineer and involves the following steps;
Confirm all dimensions and tolerances of the column and other members.
Analyse the structure to obtain the design bending moments, axial, and shear force
Check for column slenderness
Obtain the final design moments taking into account imperfections and second-order effects (if applicable)
Provide reinforcements to satisfy bending and axial force
Check for biaxial bending
Check for shear
Check that reinforcement provided satisfies bending and shear due to factory lifting
Check that reinforcement provided satisfies bending and shear due to site pitching
Design the connections
Detail the column as appropriate
Worked Example on the Design of Precast Columns | EN 1992-1:2004
Check the capacity of a 4.5m high 450 x 250 mm precast column to resist the action effects given below. The column is reinforced with 6 numbers of H20 mm bars. fck = 35 N/mm2; fyk = 500 N/mm2; Concrete cover = 35 mm. The design has been executed using Tekla Tedds software.
Axial load and bending moments from frame analysis Design axial load; NEd = 1350.0 kN Moment about y-axis at top; Mtop,y = 55.0 kNm Moment about y-axis at bottom; Mbtm,y = 22.0 kNm Moment about z-axis at top; Mtop,z = 11.4 kNm Moment about z-axis at bottom; Mbtm,z = 5.5 kNm
Column geometry Overall depth (perpendicular to y-axis); h = 450 mm Overall breadth (perpendicular to z-axis); b = 250; mm Stability in the z-direction; Braced Stability in the y-direction; Braced
Concrete details Concrete strength class; C30/37 Partial safety factor for concrete (2.4.2.4(1)); γC = 1.50 Coefficient αcc (3.1.6(1)); αcc = 0.85 Maximum aggregate size; dg = 20 mm
Reinforcement details Nominal cover to links; cnom = 35 mm Longitudinal bar diameter; ϕ = 20 mm Link diameter; ϕv = 8 mm Total number of longitudinal bars; N = 6 No. of bars per face parallel to y-axis; Ny = 2 No. of bars per face parallel to z axis; Nz = ;3 Area of longitudinal reinforcement; As = N × π × ϕ2 / 4 = 1885 mm2 Characteristic yield strength; fyk = 500 N/mm2 Partial safety factor for reinft (2.4.2.4(1)); γS = 1.15 Es = 200000 MPa
Column effective lengths Effective length for buckling about y-axis; l0y = 3500 mm Effective length for buckling about z-axis; l0z =3900 mm
Effective depths of bars for bending about y-axis Area per bar; Abar = π × ϕ2/4 = 314 mm2 Spacing of bars in faces parallel to z-axis (centre to centre); sz = h – 2 × (cnom + ϕv) – ϕ)/ (Nz – 1) = 172 mm Layer 1 (in tension face); dy1 = h – cnom – ϕv – ϕ/2 = 397 mm Layer 2; dy2 = dy1 – sz = 225 mm Layer 3; dy3 = dy2 – sz = 53 mm
2nd moment of area of reinforcement about y axis; Isy = 2 × Abar × [Ny × (dy1 – h/2)2] = 3718 cm4 Radius of gyration of reinforcement about y-axis; isy = √(Isy/As) = 140 mm Effective depth about y axis (5.8.8.3(2)); dy = h/2 + isy = 365 mm
Effective depths of bars for bending about z-axis Area of per bar; Abar = π × ϕ2 / 4 = 314 mm2 Spacing of bars in faces parallel to y axis (c/c); sy = (b – 2 × (cnom + ϕv) – ϕ) / (Ny – 1) = 144 mm Layer 1 (in tension face); dz1 = b – cnom – ϕv – ϕ/2 = 197 mm Layer 2; dz2 = dz1 – sy = 53 mm Effective depth about z axis; dz = dz1 = 197 mm
Column slenderness about y-axis Radius of gyration; iy = h/√(12) = 13.0 cm Slenderness ratio (5.8.3.2(1)); ly = l0y / iy = 26.9
Column slenderness about z-axis Radius of gyration; iz = b/√(12) = 7.2 cm Slenderness ratio (5.8.3.2(1));lz = l0z / iz = 54.0
Design bending moments
Frame analysis moments about y axis combined with moments due to imperfections (cl. 5.2 & 6.1(4)) Eccentricity due to geometric imperfections (y axis); eiy = l0y /400 = 8.8 mm Min end moment about y-axis; M01y = min(|Mtopy|, |Mbtmy|) + eiyNEd = 33.8 kNm Max end moment about y-axis; M02y = max(|Mtopy|, |Mbtmy|) + eiyNEd = 66.8 kNm
Slenderness limit for buckling about y axis (cl. 5.8.3.1) A = 0.7 Mechanical reinforcement ratio; ω = As × fyd / (Ac × fcd) = 0.429 Factor B; B = √(1 + 2ω) = 1.363 Moment ratio; rmy = M01y / M02y = 0.506 Factor C; Cy = 1.7 – rmy = 1.194 Relative normal force; n = NEd / (Ac × fcd) = 0.706 Slenderness limit; llimy = 20 × A × B × Cy / √(n) = 27.1
ly < llimy – Second order effects may be ignored
Frame analysis moments about z-axis combined with moments due to imperfections (cl. 5.2 & 6.1(4)) Ecc. due to geometric imperfections (z axis); eiz = l0z /400 = 9.8 mm Min end moment about z axis; M01,z = min(|Mtopz|, |Mbtmz|) + eizNEd = 18.7 kNm Max end moment about z axis; M02,z = max(|Mtopz|, |Mbtmz|) + eizNEd = 24.5 kNm
Slenderness limit for buckling about y-axis (cl. 5.8.3.1) A = 0.7 Mechanical reinforcement ratio; w = As × fyd / (Ac × fcd) = 0.429 Factor B; B = √(1 + 2ω) = 1.363 Moment ratio; rmz = 1.000 Factor C; Cz = 1.7 – rmz = 0.700 Relative normal force; n = NEd / (Ac × fcd) = 0.706 Slenderness limit; llimz = 20 × A × B × Cz / √(n) = 15.9 lz > llimz – Second order effects must be considered
Design bending moments (cl. 6.1(4)) Design moment about y axis; MEdy = max(M02y, NEd × max(h/30, 20 mm)) = 66.8 kNm
Local second order bending moment about z-axis (cl. 5.8.8.2 & 5.8.8.3) Relative humidity of ambient environment; RH = 50 % Column perimeter in contact with atmosphere; u = 1400 mm Age of concrete at loading; t0 = 28 day Parameter nu; nu = 1 + w = 1.429 nbal = 0.4 Approx value of n at max moment of resistance; nbal = 0.4 Axial load correction factor; Kr = min(1.0 , (nu – n) / (nu – nbal)) = 0.703 Reinforcement design strain; εyd = fyd/Es = 0.00217
Moment capacity about y-axis with axial load (1350.0 kN) Moment of resistance of concrete By iteration: Position of neutral axis; y = 317.8 mm
Concrete compression force (3.1.7(3)); Fyc = h × fcd × min(lsb × y, h) × b = 1080.6 kN
Moment of resistance; MRdyc = Fyc × [h / 2 – (min(lsb × y, h)) / 2] = 105.8 kNm
Moment of resistance of reinforcement Strain in layer 1; εy1 = εcu3 × (1 – dy1/y) = -0.00087 Stress in layer 1; σy1 = max(-1 × fyd, Es × εy1) = -174.4 N/mm2 Force in layer 1; Fy1 = Ny × Abar × σy1 = -109.6 kN Moment of resistance of layer 1; MRdy1 = Fy1 × (h/2 – dy1) = 18.8 kNm
Strain in layer 2; εy2 = εcu3 × (1 – dy2 / y) = 0.00102 Stress in layer 2; σy2 = min(fyd, Es × εy2) – h × fcd = 187.4 N/mm2 Force in layer 2; Fy2 = 2 × Abar × σy2 = 117.8 kN Moment of resistance of layer 2; MRdy2 = Fy2 × (h/2 – dy2) = 0.0 kNm
Strain in layer 3; εy3 = εcu3 × (1 – dy3/y) = 0.00292 Stress in layer 3; σy3 = min(fyd, Es × εy3) – h × fcd = 417.8 N/mm2 Force in layer 3; Fy3 = Ny × Abar × σy3 = 262.5 kN Moment of resistance of layer 3; MRdy3 = Fy3 × (h/2 – dy3) = 45.2 kNm
Resultant concrete/steel force; Fy = 1351.2 kN PASS – This is within half of one percent of the applied axial load
Combined moment of resistance Moment of resistance about y axis; MRdy = 169.8 kNm PASS – The moment capacity about the y axis exceeds the design bending moment
Moment capacity about z-axis with axial load (1350.0 kN)
Moment of resistance of concrete By iteration, position of neutral axis; z = 171.9 mm Concrete compression force (3.1.7(3)); Fzc = h × fcd × min(lsb × z, b) × h = 1051.9 kN Moment of resistance; MRdzc = Fzc × [b / 2 – (min(lsb × z, b)) / 2] = 59.2 kNm
Moment of resistance of reinforcement Strain in layer 1; εz1 = εcu3 × (1 – dz1 / z) = -0.00051 Stress in layer 1; σz1 = max(-1 × fyd, Es × εz1) = -102.3 N/mm2 Force in layer 1; Fz1 = Nz × Abar × σz1 = -96.4 kN Moment of resistance of layer 1; MRdz1 = Fz1 × (b / 2 – dz1) = 6.9 kNm
Strain in layer 2; εz2 = εcu3 × (1 – dz2/z) = 0.00242 Stress in layer 2; σz2 = min(fyd, Es × εz2) – h × fcd = 417.8 N/mm2 Force in layer 2; Fz2 = Nz × Abar × σz2 = 393.8 kN Moment of resistance of layer 2; MRdz2 = Fz2 × (b/2 – dz2) = 28.4 kNm
Resultant concrete/steel force; Fz = 1349.2 kN PASS – This is within half of one percent of the applied axial load
Combined moment of resistance Moment of resistance about z-axis; MRdz = 94.5 kNm PASS – The moment capacity about the z-axis exceeds the design bending moment
Biaxial bending
Determine if a biaxial bending check is required (5.8.9(3))
Ratio of column slenderness ratios; ratiol = max(ly, lz) / min(ly, lz) = 2.01 Eccentricity in direction of y axis; ey = MEdz/NEd = 49.8 mm Eccentricity in direction of z axis; ez = MEdy/NEd = 49.5 mm
Equivalent depth; heq = iy × √(12) = 450 mm Equivalent width; beq = iz × √(12) = 250 mm
Relative eccentricity in direction of y-axis; erel_y = ey/beq = 0.199 Relative eccentricity in direction of z-axis; erel_z = ez/heq = 0.110
Ratio of relative eccentricities; ratioe = min(erel_y, erel_z)/max(erel_y, erel_z) = 0.552
ratiol > 2 and ratioe > 0.2 Therefore, biaxial bending check is required.
Biaxial bending (5.8.9(4)) Design axial resistance of section; NRd = (Ac × fcd) + (As × fyd) = 2732.0 kN Ratio of applied to resistance axial loads; ratioN = NEd / NRd = 0.494 Exponent a; a = 1.33
Effective depth of tension reinforcement; d = 397 mm Redistribution ratio; d = 1.000 K = M / (b × d2 × fck) = 0.002 No compression reinforcement is required
Lever arm; z = min(0.5 × d × (1 + (1 – 2 × K / (h × acc / γC))0.5), 0.95 × d) = 377 mm Depth of neutral axis; x = 2 × (d – z)/lsb = 50 mm
Area of tension reinforcement required; As,pos = M / (fyd × z) = 11 mm2 Tension reinforcement provided; 2H20 mm (As,prov = 628 mm2) Minimum area of reinforcement – exp.9.1N; As,min = max(0.26 × fctm/fyk, 0.0013) × b × d = 149 mm2 Maximum area of reinforcement – cl.9.2.1(3); As,max = 0.04 × b × h = 4500 mm2 Required area of reinforcement; As,req = 149 mm2 As,req / As,prov = 0.24 (okay)
Lifting check (negative moment) Design bending moment; M = γG × flifting × wself_precast × ((Lelement – Llift) / 2)2 / 2 = 3.3 kNm Effective depth of tension reinforcement; d = 397 mm Redistribution ratio; d = 1.000 K = M / (b × d2 × fck) = 0.003
Area of tension reinforcement required; As,neg = M / (fyd × z) = 20 mm2 Tension reinforcement provided; 2H20 mm (As,prov = 628 mm2) Minimum area of reinforcement – exp.9.1N; As,min = max(0.26 × fctm/fyk, 0.0013) × b × d = 149 mm2 Maximum area of reinforcement – cl.9.2.1(3); As,max = 0.04 × b × h = 4500 mm2 Required area of reinforcement; As,req = 149 mm2 As,req / As,prov = 0.24 (Okay)
Lifting check (Shear) Design shear force at critical shear plane; VEd = γG × flifting × wself_precast × max(Llift / 2, (Lelement – Llift) / 2) = 8.2 kN
CRd,c = 0.18/γC = 0.12
Tension reinforcement; Asl = Ny × π × ϕ2 / 4 = 628 mm2 Depth of tension reinforcement; dv = dy1 = 397 mm kshear = min(1 + (200 mm / dv)0.5, 2) = 1.710
Width of the cross section in tensile area; bw = b = 250 mm
Design shear resistance – exp. 6.2 a & b; VRd,c = max(CRd,c × kshear × (100 N2/mm4 × ρl × fck)1/3, vmin) × bw × dv VRd,c = 54.3 kN VEd / VRd,c = 0.15 (This is okay)
On-site Pitching Check
Precast element details Total length of column; Lelement = 4500 mm Distance to the pitching point; Lpitch = 1800 mm Distance from pitching point to end of column;Lend = 2700 mm Lifting load coefficient; fpitching = 1.25 Permanent load factor; gG = 1.35 Self weight of precast element; wself_precast = b × h × ρconc × gacc = 2.8 kN/m
Effective depth of tension reinforcement; d = 397 mm Redistribution ratio; d = 1.000 K = M / (b × d2 × fck) = 0.001
Area of tension reinforcement required; As,pos = M / (fyd × z) = 8 mm2 Tension reinforcement provided; 2H20 mm (As,prov = 628 mm2) Minimum area of reinforcement – exp.9.1N; As,min = max(0.26 × fctm/fyk, 0.0013) × b × d = 149 mm2 Maximum area of reinforcement – cl.9.2.1(3); As,max = 0.04 × b × h = 4500 mm2 Required area of reinforcement; As,req = 149 mm2 As,req / As,prov = 0.24 (Okay)
Lifting check (negative moment) Design bending moment; M = gG × fpitching × wself_precast × Lpitch2 / 2 = 7.5 kNm Effective depth of tension reinforcement; d = 397 mm Redistribution ratio; d = 1.000
K = M / (b × d2 × fck) = 0.006
Area of tension reinforcement required; As,neg = M / (fyd × z) = 46 mm2
Tension reinforcement provided; 2H20 mm (As,prov = 628 mm2) Minimum area of reinforcement – exp.9.1N; As,min = max(0.26 × fctm/fyk, 0.0013) × b × d = 149 mm2 Maximum area of reinforcement – cl.9.2.1(3); As,max = 0.04 × b × h = 4500 mm2 Required area of reinforcement; As,req = 149 mm2 As,req / As,prov = 0.24 (Okay)
Moving traffic is the major live load (variable action) on bridge decks. In the design of bridges, it is very important to consider the global and local effects of moving traffic loads on the bridge. While global effects can be used to distribute traffic wheel load to the girders (in the case of beam and slab bridges), local verification is very important in the design of deck slabs especially for bending moment and shear. In this article, we are going to show how to distribute single wheel load to bridge deck slabs.
In BS 5400, single HA load (with a value of 100 kN) is normally used for all types of local effects verification. This load is normally assumed to act on a square contact area of 300 mm x 300 mm to give a pressure of 1.11 N/mm2 on the surface where it is acting. For 45 units of HB load, a single wheel load of 112.5 kN can be considered where necessary.
In EN 1991-2 (Eurocode specification for bridges), the local wheel load is represented by Load Model 2 (LM2) which consists of a single axle load with a magnitude of 400 kN (inclusive of the dynamic amplification factor). The single axle consists of two wheels (200 kN each) spaced at a distance of 2m (centre to centre). This load model is intended to be used for local verification only and should be considered alone on the longitudinal axis of thebridge. Unless otherwise specificied, each wheel is assummed to act on a rectangular area of 600 x 350 mm.
To apply single wheel load on bridge decks for local verification, the wheel load is placed at the most adverse location on the bridge deck, and the contact pressure distributed to the neutral axis of the bridge deck. For all practical purposes, the neutral axis is normally taken at the mid-section of the bridge deck. For distribution of wheel load through concrete deck slabs, a ratio of 1:1 is normally adopted (dispersal angle of 45 degrees), while for distribution through asphalt surfacing, a dispersal ratio of 1:2 is normally adopted.
When the contact pressure has been obtained (together) with the dimensions of the contact area, the effects of the wheel load on the slab can be assessed using any suitable method. For manual analysis, Pigeaud’s curve can be used to obtain the design bending moments. For finite element analysis, the contact pressure can be applied as a patch load (partially distributed load) on a slab surface.
To show how this is done, let us consider a worked example.
Worked Example
Obtain the design moments in an interior panel of deck slab of the bridge system shown below due to the effect of a single HA wheel load on the bridge deck. The deck slab is overlain with a 75mm thick asphalt surfacing.
The dimensions of an interior panel of the bridge deck are as follows;
Length (ly) = 4.5 m Width (lx) = 1.8 Support condition: Continuous over all supports
Design wheel load = 100 kN Contact area = 300 x 300 mm
The contact pressure = 100 kN/(0.3 x 0.3)m = 1111.11 kN/m2
A 2(V):1(H) load distribution through the 75 mm thick asphalt to the surface of the deck slab will give a pressure of 711.11 kN/m2 acting on a square contact area of 375 x 375 mm.
A further 1:1 load distribution from the surface of the concrete deck slab to the neutral axis will give a pressure of 256 kN/m2 acting on a square contact area of 625 x 625 mm.
Therefore; ax = ay = 625 mm
To enable us use the Pigeaud’s curves; k = ly/lx = 4.5/1.8 = 2.5 ax/lx = 0.625/1.8 = 0.347 ay/ly = 0.625/4.5 =0.138
From Table 55 of Reynolds and Steedman (10th Edition), αx4 = 0.169 αy4 = 0.0964
Poisson’s ration v = 0.2
Transverse bending moment Mx = F(αx4 + vαy4) = 100 x (0.169 + 0.2 x 0.0964) = 18.828 kNm/m Longitudinal bending moment My = F(vαx4 + αy4) = 100 x (0.2 x 0.169 + 0.0964) = 13.02 kNm/m
Taking account of suggested allowances for continuity in the transverse direction; Mx1 = -0.25Mx = 0.25 x 18.828 = -4.707 kNm
These are therefore the bending moments due to the HA wheel load on the deck slab
Inclined or slanted columns are columns that are leaning at an angle away from perfect verticality (90 degrees to the horizontal). This is usually intentional and not due to imperfection from materials or construction. The degree of inclination can vary depending on the designer’s intentions. In this article, we are going to review the analysis and design of inclined columns using standard design codes.
Inclined columns can be introduced into a building to serve architectural or structural functions. The design of an inclined column is like the design of any other column but with special attention paid to the changes in stresses due to the eccentricity of the axial load on the column. In some cases, inclined columns can be more susceptible to second-order effects than perfectly vertical columns.
When a column in a structure is perfectly straight, first-order bending moment and other internal stresses are induced due to the externally applied loads. However, when a column is inclined at an angle to the beams or floor that it is supporting, changes in bending moments are observed due to the eccentricity of the axial load with respect to the longitudinal axis column. Such effects can be investigated for pinned supports or fixed supports.
In the design of inclined columns (say in reinforced concrete or steel), it is usually very sufficient to analyse the structure and obtain the design internal forces (bending moment, shear, and axial force) using first-order linear analysis. However, when required, the analysis can be extended to second-order non-linear analysis to account for secondary effects which may affect the stability of the column.
To verify the effect of inclination on the design of columns, let us investigate the following cases of a column with a fixed base in reinforced-concrete construction.
Dimensions of beam = 600 x 300 mm Dimensiond of column = 300 x 300 mm Ultimate load (factored load) on the beam = 70 kN/m
When the frames were analysed using first-order linear analysis, the following results were obtained;
CASE A
Maximum column design axial force = 220.219 kN Column design shear force = 22.880 kN Column design moment = 61.313 kNm
CASE B
Maximum column design axial force = 241.712 kN Column design shear force = 19.608 kN Column design moment = 56.624 kNm
CASE C
Maximum column design axial force = 226.991 kN Column design shear force = 20.892 kN Column design moment = 59.247 kNm
From the analysis results, it could be seen that case A gave the highest column moment but the least axial force, while case B gave the lowest column design moment but highest column axial force. The design of the column can then be carried out for each of the cases presented.
A flat slab is a type of reinforced concrete floor system where the slab is supported directly by columns, instead of beams. This system offers many advantages such as reduced headroom, flexibility in the distribution of services, reduced formwork complexity during construction, etc. Strengthening of flat slabs may become necessary especially when openings are introduced.
According to recent research (2021) from the Department of Civil Engineering, University of Wasit, Iraq, openings can be created in a slab after construction in order to pass the new services that satisfy the occupants’ desires, such as water or gas piping, ventilation, electricity, elevator, and staircase. The creation of openings in a flat slab has been found to significantly reduce the structural integrity of the floor.
Flat slabs can undergo brittle failure due to punching shear around the columns when the thickness of the slab is insufficient and/or when punching shear reinforcement is not provided. According to Hussein et al (2021), the possibility of punching shear failure increases by creating openings next to the columns due to removing a substantial amount of concrete from the critical section around columns, which is responsible for resisting the punching force. This is usually accompanied by the cutting of several flexural reinforcement bars.
This undesirable impact of openings gets more severe when they are not taken into account during the design, or they are installed after the construction of slabs. The opening influence could be minimized when they are positioned away from the columns, or the slab depth is increased.
From many research works carried out on flat slabs with openings, the following conclusions were reached. The full references and details of the research works can be found in Huessein et al., (2021).
For openings of the same size, the strength reduction was more significant as the number of openings around the column increased.
The influence of openings on the strength of slabs is greater when they are constructed in large sizes, especially when the size of the opening is greater than the size of the corresponding column.
The impact of the opening could be reduced by shifting them far away from the supporting columns. Two far openings were found to be less detrimental than ones adjacent to column’s faces.
Openings located at column corners reduced the strength of slabs more than openings placed in front of the column
For openings of the same number and size, the arrangement of openings as the letter (L) around the column caused a lesser drop in the slabs’ punching strength.
In general, the reported maximum reduction in the slabs’ strength due to openings was about 29% relative to the solid equivalents.
It, therefore, makes sense to strengthen flat slabs with cut-out opening, especially when the effect of the opening was not taken into account during the design. Therefore, Hussein et al (2021) decided to investigate the effects of strengthening flat slabs with cut-out openings using different materials. The strengthening techniques investigated in the study were Carbon Fiber Reinforced Polymer (CFRP), steel plates, steel bars, and near-surface mounted Engineered Cementations Composite (ECC) with steel mesh. The findings were published in Case Studies of Construction Materials (Elsevier).
Fig 1: Details of the specimens used in the investigation (Hussein et al, 2021)
In order to investigate how strengthening can retrieve the lost mechanical strength of slab with openings, six reinforced concrete slabs were prepared with dimensions of 1300 x 1300 x 120 mm. One of them was reference without opening, whereas the others contained a square edge opening of 350 mm side. For slabs with openings, one specimen was the control left without strengthening, and the remaining four were strengthened utilizing various methods stated above.
Fig 2: Details of the investigation techniques used in the research (Hussein et al, 2021)
The details of the preparation of the specimens can be found in Hussein et al, (2021). After the curing process completed, the specimens were tested under the influence of uniform loading. The samples were moved into a testing steel rig and rested at their corners on supporting steel plates, measuring 200 x 200 x 30 mm to simulate the real supporting columns. Nevertheless, the rotations around these plates were not restricted. The uniform loading was applied incrementally with the help of a hydraulic jack up to failure.
Based on the experiments by Hussein et al (2021) conducted on six reinforced concrete slabs, including edge openings and strengthened using four various methods, tested under uniformly distributed loads, the following points can be put forward as the essential outcomes;
1. Installation of an opening at the edge of a flat slab resting on four columns and subjected to uniformly distributed load did not change the failure modes. However, the losses in strength, ductility, and toughness were 20.6 %, 16.2 %, and 38 %, respectively when compared with the solid slab. 2. Among the four adopted methods of strengthening, the use of embedded steel bars and EEC altered the failure mode from brittle due to punching shear to more ductile combined flexural-shear modes. 3. The effectiveness of retrofitting techniques in restoring the lost mechanical properties of flat slabs due to openings was found to relate directly to the bond strength between the strengthening materials and the slab surface. Accordingly, the embedded steel bars technique was the most efficient, whereas the ECC was the least efficient method. 4. The strengthening materials should be extended as long as possible from the opening edge in order to achieve a sufficient development length. Hence, the retrofitting materials’ ultimate strength can arrive, and the debonding can be delayed. 5. None of the strengthening methods managed to restore the total missing strength of slabs due to openings. The use of embedded steel bars achieved the most significant upgrading in the slab strength, about 22.2 % over that of the slab with opening and 3 % below the solid slab’s strength.
References Hussein M. J., Jabir H. A., Al-Gasham T. S. (2021): Retrofitting of reinforced concrete flat slabs with cut-out edge opening. Case Studies in Construction Materials 14 (2021) e00537 https://doi.org/10.1016/j.cscm.2021.e00537
Disclaimer The contents of the above-cited research work have been presented on www.structville.com because it is an open-access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/)
In the UK, HB loads are used to represent abnormal traffic on a bridge deck. Even though the use of HA and HB loads as traffic actions has been replaced with Load Models 1 to 4 of EN 1991-2, the use of HB load is still applied in the design of bridges especially in Nigeria. It is very important, therefore, to know how to analyse bridge girders (in beam and slab bridges) for HB loads in order to obtain the design forces due to moving traffic.
There are many methods through which the effects of traffic wheel load can be obtained on the longitudinal components of a bridge deck. Some of these methods are;
The simplified methods of analysing bridge decks are known as the manual methods. While bridge design all over the world is majorly carried out using computers, the simplified methods (which are pretty quick to apply) can be used for quick checks or for preliminary and/or detailed design. The difference between the results from computer methods and simplified methods is often not very wide.
The basis of many simplified/manual methods is the distribution coefficient methods, e.g. Morice and Little (1956). These methods can be applied manually to obtain the values of various load effects at any reference point on a transverse section of the bridge.
For most of the load distribution coefficient methods, a right simply supported bridge is idealized as an orthotropic plate whose load distribution characteristics are governed by two dimensionless parameters α and θ. These parameters depend largely on the geometry of the bridge deck (which is usually known prior to the analysis).
In North America, the simplified method of bridge deck analysis is widely used. Unlike the distribution coefficient methods, the simplified methods of bridge analysis provide only the maximum action effect for the purposes of design. As a result, a lesser computational effort is required when using this method compared with the distribution coefficient method. These simplified methods are permitted by the current and past design codes, being the AASHTO Specifications (1998, 2010), the CSA Code (1988), the Ontario Highway Bridge Design Code (1992), and the Canadian Highway Bridge Design Code (2000, 2006).
The results from the North American simplified methods are reliable and can be obtained rather quickly. It is important to note that the analysis depends on the specification of the magnitude and placement of the design live loads, and accordingly are not always transportable between the various design codes.
The D-Method of Bridge Deck Analysis
In the simplified method of analysis, a longitudinal girder, or a strip of unit width in the case of slabs, is isolated from the rest of the structure and treated as a one-dimensional beam. This isolated beam is subjected to loads comprising one line of wheels of the design vehicle multiplied by a load fraction (S/D), where S is the girder spacing and D, having the units of length, has an assigned value for a given bridge type (Bakht and Mufti, 2015).
Fig 1: Transverse distribution of longitudinal moment intensity (Bakht and Mufti, 2015)
The concept of the factor D can be explained with reference to the figure above, which shows schematically the transverse distribution of live load longitudinal moment intensity in a slab-on-girder bridge at a cross-section due to one vehicle with two lines of wheels.
The intensity of longitudinal moment, having the units of kN.m/m, is obtained by idealizing the bridge as an orthotropic plate. A little consideration will show that the maximum girder moment, Mg, for the case under consideration occurs in the second girder from the left.
The moment in this girder is equal to the area of the shaded portion under the moment intensity curve. If the intensity of maximum moment is Mx(max) then this shaded area is approximately equal to SMx(max), so that:
Mg ≃ SMx(max) —— (1)
It is assumed that the unknown quantity Mx(max) is given by:
Mx(max) = M/D —— (2)
where M is equal to the total moment due to half a vehicle, i.e., due to one line of wheels. Substituting the value of Mx(max) from Eq. (2) into Eq. (1):
Mg = M(S/D) —— (3)
Thus if the value of D is known, the whole process of obtaining longitudinal moments in a girder is reduced to the analysis of a 1-dimensional beam in which the loads of one line of wheels are multiplied by the load fraction (S/D).
Simplified Method of Analysis for HB Load
A simplified method for two-lane slab-on-girder bridges in Hong Kong was presented by Chan et al. (1995). This was based on HB loads that are popularly used for bridge design in the UK.
The HB loading comprises four axles, with four wheels in each axle. The weights of the wheels are governed by the units of the HB loading. In Great Britain,we have 30 units, 37.5, and 45 units of HB. Each unit of HB is equal to 10 kN. For instance, for 45 units of HB load, each axle has a total load of 450 kN, while each wheel has a load of 112.5 kN. It should, however, be noted that the units of the design loading does not affect the simplified method presented by Chan et al. (1995).
Fig 2: HB wheel load arrangement
According to Chan et al (1995), the longitudinal flexural rigidity per unit width Dx, of slab-on-girder bridges in Hong Kong lies between two bounds defined by the following equations.
where the span of the bridge L is in metres and Dx in kN.m.
From a study of a large number of slab-on-girder highway bridges in Hong Kong, it was determined that the ranges of the various parameters which influence the transverse load distribution characteristics of a bridge are as follows.
(a) The deck slab thickness, t, varies between 150 and 230 mm, with the usual value being 200 mm. (b) The centre-to-centre spacing of girders, S, varies between 0.2 and 2.0 m, with the usual value being 1.0 m. (c) The lane width, We, varies between 3.2 and 3.8 m, with the usual value being 3.5 m. (d) The vehicle edge distance, VED, being the transverse distance between the centre of the outermost line of wheels of the HB loading and the nearest longitudinal free edge of the bridge, varies between 0.75 and 5.00 m, with the usual value being 1.00 m.
From the above observations, the following values of the various parameters were adopted for the developmental analyses conducted for developing the simplified method: t = 200 mm; S = 1.0 m; We = 3.5 m; and VED = 1.00 m. In addition, it was assumed that the deck slab overhang beyond the outer girders was 0.55 m. Bridges with spans of 10, 20, 30 and 40 m were selected for the developmental analyses.
Chan et al. (1995), having plotted the values of D from the above analyses against the span length L, found that these values of D are related to L according to the following equations with a reasonable degree of accuracy, provided that the design value of D, i.e. Dd, is corrected.
For internal girders having L < 25 m: D = 1.2 – 3.5/L —— (6)
For internal girders having L ≥ 25 m: D = 1.06 —— (7)
For external girders having L < 30 m: D = 0.95 + 2.1/L —— (8)
For external girders having L ≥ 30 m: D = 1.03 —— (9)
The correcting equation for obtaining Dd is as follows;
Dd = D(1 + µCw/100) —— (10)
Where; µ = (3.5 – We)/0.25 —— (11)
The values of cw for internal and external girders can be read from the chart below;
Fig 3: Values of cw (Chan et al, 1995)
The following conditions must be met for applying the simplified method.
The value of Dx lies between the upper and lower bound values given by Eqs. (4) and (5), respectively.
The bridge has two design lanes
The width is constant or nearly constant, and there are at least three girders in the bridge.
The skew parameter ε = (S tan ψ)/L does not exceed 1/18 where S is girder spacing, L is span and ψ is the angle of skew.
For bridges curved in plan, L2/bR does not exceed 1.0, where R is the radius of curvature; L is span length; 2b is the width of the bridge.
The total flexural rigidity of transverse cross-section remains substantially the same over at least the central 50% length of each span.
Girders are of equal flexural rigidity and equally spaced, or with variations from the mean of not more than 10% in each case.
The deck slab overhang does not exceed 0.6S, and is not more than 1.8 m.
The steps for applying the simplified method in the analysis of bending moment due to HB load on a bridge girder are;
Calculate the value of D from the relevant of Eqs. (6, 7, 8, and 9); for simply supported spans, L is the actual span length, and for continuous span bridges, the effective L for different spans should be obtained.
For the design lane width, We, obtain μ from Eq. (11) and Cw from the chart and thereafter obtain Dd from Eq. (10).
Isolate one girder and the associated portion of the deck slab, and analyse it by treating it as a one-dimensional beam under one line of wheels of the HB loading. The moment thus obtained at any transverse section of the beam is designated as M.
For any of the internal and internal girders, obtain the maximum moment at the transverse section under consideration by multiplying M with (S/Dd), where Dd is as obtained in Step (b) for the relevant internal or external girders.
Worked Example
Obtain the maximum sagging moment in an interior girder of a one-span bridge deck carrying 45 units of HB load. The configuration of the bridge deck is shown below;
Fig 4: Bridge deck section
Actual span lengths, L = 18.0 m Bridge width, 2b = 8.7 m Carriageway width = 7.2 m Number of notional lanes = 2 Design lane width (notional lane width) = We = 3.6 m Deck slab thickness = 0.20 m Girder spacing S = 1.80 m Deck slab overhang = 0.75 m Vehicle edge distance VED = 1.35 m
HB line beam analysis
Isolating a girder of the bridge and placing a line of HB wheel load on it to determine the maximum moment. The wheel load position that will produce the maximum moment is shown below;
Fig 5: Line beam analysis of an isolated girder under one line of HB load
When analysed;
M= 1203.75 kNm
From equations (6) and (8);
Internal girder (L < 25 m) D = 1.2 – 3.5/L = 1.2 – 3.5/18 = 1.0055
External girder (L < 25 m) D = 0.95 + 2.1/L = 0.95 + 2.1/18 = 1.067
Applying a correction factor to the value of D; Dd = D(1 + µCw/100)
Therefore the maximum moment in the internal girder = 1.8097 x 1203.75 = 2178 kNm Therefore the maximum moment in the external girder = 1.794 x 1203.75 = 2159 kNm
Kindly check for the accuracy of this answer using any method available to you and post your findings in the comment section.
Disclaimer: Most of the procedures presented here are as described by Bakht and Mufti (2015) with minor alterations after studying the original article cited (Chan et al, 2015). The copyright to those contents, therefore, belongs to the original owners. The worked example used to illustrate the procedure was however developed by Structville.com
References Bakht B., Mufti A. (2015): Bridge Analysis, Design, Structural Health Monitoring, and Rehabilitation. Springer Chan THT, Bakht B, Wong MY (1995): An introduction to simplified methods of bridge analysis for Hong Kong. HKIE Trans 2(1):1–8 Morice P. B. and Little G. (1956): The Analysis of Right BridgeDecks Subjected to Abnormal Loading. Cement and Concrete Association, London, Report Db 11
The Canadian Pacific Railway Company in the year 1913 constructed the Transcona grain elevators of about 36400 m3 capacity to provide relief for the Winnipeg Yards during the months of peak grain shipment. The structure consisted of a reinforced concrete work-house, and an adjoining bin-house, which contained five rows of 13 bins, each 28 m in height and 4.4 m in diameter.
The bins were based on a concrete structure containing belt conveyors supported by a reinforced concrete shallow raft foundation (Puzrin et al, 2010). The reinforced concrete raft foundation was 600 mm thick, with dimensions of 23.5 x 59.5 m.
Excavation for the construction of the elevator foundations started in 1911, and the first 1.5m depth of soil at the site was rather soft. Beyond the soft clay layer was a relatively stiff blue clay, typical for that area, and locally known as the “blue gumbo”.
According to literature cited by Puzrin et al (2010), no borings or extensive geotechnical investigations were carried out prior to the construction. This is not a surprise, given the level of technology and knowledge about soil engineering at that time. However, an in-situ bearing capacity test (test loading applied using a specially constructed wooden framework) was performed at a depth of 3.7 m.
According to the literature cited by Puzrin et al (2010), the plate load test result indicated that the soil was capable of bearing a uniformly distributed load of at least 400 kPa. The maximum foundation pressure from the bins at maximum load was not expected to exceed 300 kPa, therefore the tests appeared satisfactory to the engineers. Furthermore, they assumed that the “blue gumbo” at the site had similar characteristics and a depth to that on which similar raft foundations of many heavy structures had been founded in the vicinity of Winnipeg. This eventually turned out not to be so.
After the structure was completed, the filling was begun and grain was distributed uniformly between the bins. On October 18, 1913, after the elevator was loaded to 87.5% of its capacity, settlement of the bin-house was noted. Within an hour, the settlement had increased uniformly to about 300 mm following by a tilt towards the west, which continued for almost 24 hours until it reached an inclination of almost 27 degrees (Puzrin et al, 2010).
Fig 1: Transcona grain elevator (a) Before failure (b) After failure
Several wash-borings were made immediately after the failure, showing that the elevator was underlain by rather uniform deposits of clay. This finding was in agreement with the geological history of the area, according to which extensive fine-grained sediments were deposited in the waters of the glacial Lake Agassiz which came into being when the Wisconsin ice-sheet blocked the region’s northern outlet.
It was noted that no laboratory test was carried out on the samples collected during the wash borings, but classification was done based on visual observation. The wash-borings, therefore, confirmed the designers’ assumptions of uniform clay layer, and the failure of the Transcona Grain Elevator remained a mystery for another 40 years.
It was thought that, if the smaller-scale plate loading tests predicted a safety factor of more than 1.3 (400/300 = 1.33), and the soil profile is homogeneous, how could the foundation fail? The answer to this question was given by Peck and Bryant (1953) who, in 1951 (38 years later), made two additional borings, far enough from the zone of failure to be in material unaffected by the displacements (Puzrin et al, 2010).
They obtained undisturbed soil samples and performed unconfined compression strength tests (triaxial shear tests with zero confining stress), which produced some eye-opening results shown in Fig 2.
Fig 2: Soil profile below the elevator (after Peck and Bryant, 1953) (a) classification; (b) unconfined compression strength
On observation of the unconfined compressive strength (qu) result of the site, there are two easily distinguishable layers (Fig. bb). The upper one, a 7.5 m thick stiff clay layer with qu = 108 kPa (undrained shear strength cu = qu/2 = 54 kPa), appears to be resting on a softer clay layer with qu = 62 kPa (cu = 31 kPa). This finding suggests that the elevator failure was most likely caused by the insufficient bearing capacity of its foundation.
In a bearing capacity failure, a failure mechanism is formed below the foundation (Fig. 3b). The settlement takes place much faster and without decrease of the soil volume. Therefore, the displaced soil has to find itself an exit, causing a ground heave in the vicinity of the structure. This ground heave is a distinctive feature of the failure of the Transcona Grain Elevator (Puzrin et al, 2010).
Fig 3: Bearing capacity failure: (a) settlement; (b) failure; (c) the ground heave (Puzrin et al, 2010)
According to (Puzrin et al, 2010), the particular problem of the Transcona Grain Elevator was that the failure mechanisms of the plate loading tests were apparently confined to the upper stiffer clay layer, due to the relatively small size of the plates. The elevator foundation, however, developed a much deeper failure mechanism which entered the weaker clay layer, significantly reducing the bearing capacity.
This led the researchers to consider the two-layer model and other bearing capacity theories that are adequate to describe the situation on the site. Details of this can be found in Puzrin et al, (2010).
It was observed that the true failure contact pressure was 293 kPa. In the original design, it was assumed that the soil profile was homogeneous with the properties of the stiff upper layer uc1 = 54 kPa. In this case, the bearing capacity of the foundation was calculated as 386 kPa (applying the appropriate bearing capacity, shape, and depth correction factors). This was found to be close to the 400 kPa obtained from the plate load tests.
Note that if the soil was homogeneous but with the properties of the weaker lower layer (uc2 = 31 kPa), the resulting bearing capacity would be 251 kPa. If this value was available to designers, the result would actually not be that bad: not only the elevator would not fail, it would not even be too much overdesigned.
A more sophisticated analysis, based on a two-layer model, should produce more accurate predictions. First, the reseachers followed Peck and Bryant (1953) and used the approximate method based on the Prandtl solution using a weighted average of the undrained shear strength. This gave a bearing capacity of 321 kPa which is 10% larger than the failure pressure. While the Prandtl solution for a homogeneous soil is the exact solution, in the two-layer approximation, it was observed that it is not only inaccurate, but it is also not conservative and could lead to failure.
The scoop mechanism, in contrast, provides a remarkably good bearing capacity estimate of 297 kPa. Being an upper bound, this value, as expected, is higher than the true failure pressure of 293 kPa, but only marginally. It would provide an excellent estimate for the design of the elevator if only the Soil Mechanics was more mature in those days and the soil properties were properly determined.
Source: The information in this article was majorly obtained from: Puzrin A. M., Alonso E.E., Pinyol N. M. (2010): Geomechanics of Failures. Springer. DOI 10.1007/978-90-481-3531-8
We are pleased to announce the release of our publication on the ‘Design of Reinforced Concrete Bridges‘. This publication is part of the proceedings of our January 2021 Webinar on Bridge Design.
The publication essentially contains full calculation sheets on the analysis and design of different structural components of a T-beam bridge (beam and deck slab bridge) such as;
Both computer and manual methods were adopted in the analysis presented, and the design considers the effect of wind, temperature, and secondary traffic loads such as braking, traction, etc on the bridge. Other elements such as parapets, precast planks (filigree slab) for receiving the in-situ topping of the deck slab, etc were also considered in the design. The design was carried out according to the British Standards.
Trusses deflect when loaded. Under gravity loads, this is usually characterised by the sagging of the top and bottom chords, and the consequent movement of the web and diagonal members. In the design of trussed structures such as roofs and bridges, it is always important to keep the deflection of the trusses to a minimum in order to maintain the appearance and functionality of the structure. Deflection of trusses can be assessed manually using the virtual work method (unit load method), direct stiffness method, or finite element analysis.
The deflection of roof trusses is often limited to span/240. On the other hand, a span to depth ratio of 15 is often found adequate for all practical purposes. For bridges, AASHTO design code states that the maximum deflection of a truss bridge due to live load should not exceed span/800. The principle of virtual work can be used to compute the maximum deflection of the truss, which is then compared to the allowable deflection.
In this article, we are going to explore how to determine the deflection of trusses using the virtual work method (unit load method). In the virtual work method, the truss is analysed for the axial forces due to the externally applied load. Subsequently, the external forces are removed and replaced with a unit virtual load at the node where the deflection is to be obtained. The direction of the unit load should be in the same direction with where the deflection is sought. The truss is re-analysed for the axial forces in the members due to the virtual unit load.
The summation of the deflection of the individual members using the formula below gives the deflection of the roof truss at the point of interest.
δ = ∑niNiLi/EiAi
Where; n = axial force in member i due to virtual unit load N = axial load in member i due to externally applied load L = length of member E = Modulus of elasticity of member A = Area of member
Worked Example
A truss is loaded as shown below. Obtain the vertical deflection at point C due to the externally applied load (AE = constant).
Support Reactiondue to externally applied load Let ∑MG = 0; 9Ay – (6 × 6) – (4 × 3) + (10 × 3) = 0 Ay = 2 kN↑
Having obtained the internal forces in all the members due to the externally applied loads, let us remove all the loads and replace them with a unit vertical load at point C as shown below;
Support Reaction due to virtual load Let ∑MG = 0; 9Ay – (1 × 6) = 0 Ay = 6/9 = 0.667 ↑
Strap footings or cantilever footings are a special form of combined footings. They consist of two separate bases that are connected (balanced) by a strap beam. In the design of strap footing, it is assumed that the strap beam is rigid and does not transfer any load by bearing on the soil at its bottom contact surface.
Strap footing is necessary when the foundation of a column cannot be built directly under the column or when the column should not exert any pressure below. It is then necessary to balance it by a cantilever arm rotating about a fulcrum and balanced by an adjacent column (or a mass of concrete or by piles) in case of where footings cannot be built.
As was described above, balanced footings consist of two separate footings connected by a strap beam. In the design of balanced bases, uniform soil pressure can be assumed if we can make the centre of the areas of the system coincide with the centre of gravity of the loads. When these two centres do not coincide, we have the vertical load and moment acting on the system due to the eccentricity. As a result, the distribution of base pressure will not be uniform but can be assumed to be linearly varying.
The difference between balanced footings and cantilever footings can be described as follows. In a balanced footing, we make the centre of gravity of the loads and the centre of the areas coincide. Hence, the ground pressure will be uniform. In cantilever footings, in general, the two centres may not coincide, so we have a moment in addition to the vertical loads. Hence, the ground pressure will be varying.
Worked Example
Two columns with the following loading conditions are spaced 4 m apart. Due to site boundary constraints, design a strap footing for the columns, if the safe bearing capacity of the soil is 150 kN/m2; fck = 25 N/mm2; fyk = 500 N/mm2
Column
Size (mm)
Service Load (kN)
Ultimate Load (kN)
C1
300 x 300
450
617
C2
300 x 300
600
822
Solution We can either dimension each of the footings so that the CG of the areas and loads coincide, resulting in uniform pressure, or adopt suitable base dimensions and then check the resulting ground pressure taking the unit as a whole, which may not be uniform, and design for non-uniform pressure. We will adopt the first method.
Step 1: Preliminary sizing of the footing of the exterior column Base area (A1) needed for column C1 = Service load/Safe bearing capacity = 450/150 = 3m2 Try a rectangular footing of size 1.75m wide x 2 m long along the centre line (Area provided = 3.5 m2)
Hence, we fix the fulcrum at 2.0/2 = 1.0 m from the end near C1.
Therefore; Distance of R1 from C1 = L1 = 1.0 – 0.3 = 0.7 m from C1 Distance of C2 from R1 = L2 = 4 – 0.7 = 3.3 m
Taking moment about C2; 3.3R1 = 450 x 4 R1 = 545.45 kN
Let the summation of the vertical forces be equal to zero; R2 = 450 + 600 — 545.45 = 504.55 kN
Step 2: Check the factor of safety (FOS) against overturning using characteristic loads FOS = C2L2/C1L1 = (600 x 3.3)/(450 x 0.7) = 6.285 > 1.5 (Okay).
Step 3:Find the dimension of footing for R2 so that the CG of loads and areas coincides. This is given by; B = √(R2/SBC) = √(504.55/150) = 1.83 m Adopt a footing 1.85 m x 1.85 m
Step 4:Recalculate the necessary breadth of footing F1 so that CG of loads and areas of footings coincides. x1 = CG of loads = (C1 x 4)/(C1 + C2) = (450 x 4)/(450 + 600) = 1.714 m from C2
Let us find the CG of areas we have assumed. Area of F2 = A2 = 1.85 x 1.85 = 3.4225 m2. Find Ax required for CG to be same as that of loads.
x2 = (A1 x 3.3)/(A1 + A2) = 1.714 (for a balanced base) (A1 x 3.3)/(A1 + 3.4225) = 1.714 On solving; A1 = 3.698 m2
For a length of 2m, a width of 3.689/2 = 1.85 m is required. Therefore slightly increase the width of A1 to 1.85 m.
Hence; A1 = (1.85 x 2)m = 3.7 m2 A2 = (1.85 x 1.85)m = 3.4225 m2
You can verify independently; At ultimate limit state; R1 = (545.45 x 617)/450 = 747.87 kN Soil pressure under base 1 = 747.87/3.7 = 202.13 kN/m2
At umtimate limit state R2 = (504.55 x 822)/600 = 691.23 kN Soil pressure under base 2 = 691.23/3.4225 = 201.967 kN/m2
Step 6: Design of footing F1 Let the width of the strap beam be 0.3 m. The maximum moment will occur at the face of the strap beam (overhang of the strap beam).
Length of overhang = (1.85 – 0.3)/2 = 0.775 m on both sides of the beam
MEd = ql2/2 = (202.13 x 0.7752)/2 = 60.7 kNm
Assuming a footing depth h = 400 mm and concrete cover of 50 mm; Effective depth d = 400 – 50 – 8 = 342 mm.
Critical design moment at the face of the strap beam MEd = 125 kNm/m k = MEd/(bd2fck) = (60.7 x 106)/(1000 x 3422 x 25) = 0.0207 Lever arm = z = d[0.5 + √(0.25 – 0.882k)] = 0.95d ⇒ z = 0.95d = 0.95 x 342 = 325 mm ⇒ As = MEd/0.87fykz = (60.7 x 106)/(0.87 x 500 x 325) = 429 mm2/m Asmin = 520 mm2/m
Provide H12 @ 200 c/c (Asprov = 565 mm2/m)
Beam shear Check critical section d away from the face of the strap beam VEd = 202.13 x (0.775 – 0.342) = 87.522 kN/m vEd = 87.522/342 = 0.256 N/mm2
vRd, c = CRd, c × k × (100 × ρ1 × fck) 0.3333 CRd, c = 0.12 k = 1 + √ (200/d) = 1 + √ (200/342) = 1.76 ρ = 565/(342 × 1000) = 0.00165 vRd, c = 0.12 × 1.76 × (100 × 0.00165 × 25)0.333 = 0.338 N/mm2 => vEd (0.256 N/mm2) < vRd,c (0.338 N/mm2) beam shear ok
Step 7: Design of the strap beam Equivalent line load under column base 1 = 202.13 x 1.85 = 373.94 kN/m Equivalent line load under column base 2 = 202.13 x 1.85 = 373.94 kN/m
The shear force values at the critical points are; V1L = (0.15 x 373.94) = 56.09 kN V1R = 56.09 – 617 = -560.9 kN Vi = (373.94 x 2) – 617 = 130.88 kN V2L = 130.88 + (373.94 x 0.925) = 476.774 kN V2R = 476.774 – 822 = -345.226 kN Vj = -345.225 + (373.94 x 0.925) = 0
The maximum bending moment will occur at the point of zero shear which is 1.5 m from column A. This can be easily obtained by using similar triangles.
Mmax = (373.94 x 1.652)/2 – (617 x 1.5) = -416.474 kNm (hogging moment). The maximum sagging moment will occur under column C2; Mmax = (373.94 x 0.9252)/2 = 159.976 kNm
Assume a total depth of 600 mm for the beam; Effective depth = 600 – 50 – 10 – 10 = 530 mm k = MEd/(bd2fck) = (416.474 x 106)/(300 x 5302 x 25) = 0.1976
Since k < 0.167, compression is required. This means that the beam will have to be designed as a doubly reinforced beam. Area of compression reinforcement AS = (MEd – MRd) / (0.87fyk (d – d2))
Shear Design Using the maximum shear force for all the spans Support A; VEd = 560.9 kN VRd,c = [CRd,c.k. (100ρ1 fck)1/3 + k1.σcp]bw.d CRd,c = 0.18/γc = 0.18/1.5 = 0.12 k = 1 + √(200/d) = 1 + √(200/530) = 1.61 > 2.0, therefore, k = 1.61
Since VRd,c < VRd,max < VEd The beam is subjected to high shear load, we need to modify the strut angle. θ = 0.5sin-1[(VRd,max /bwd)/0.153fck(1 – fck/250)] (VRd,max /bwd) = 2.418 N/mm2 0.153fck(1 – fck/250) = 3.4425 θ = 0.5sin-1[(2.418/3.4425] = 22.31°
Since θ < 45°, section is OK for the applied shear stress
Hence Asw/S = VEd /(0.87fykzcotθ) = 560900/(0.87 × 500 × 0.9 × 530 × 2.437) = 1.109 Maximum spacing of shear links = 0.75d = 0.75 × 530 = 597.5 Provide 4 legs Y10 @ 250mm c/c as shear links (Asw/S = 1.256) Ok
