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Design of Timber Joists to EN 1995-1-1:2004

By
Ubani Obinna
-

Timber joists are flexural horizontal timber members that are used for framing an open space in a building in order to support a floor or sheathing. They are usually closely spaced (usually between 400 – 800 mm) with the plane of maximum strength positioned vertically.

The spans of the joists are usually supported or intercepted by blockings which may be linear or skewed (staggered). Timber joists transfer the load of the floor to the vertical compression members (such as timber columns) and can be made of solid timber, glulam, or other engineered wood products.

As flexural structural members, the design of timber joists is expected to satisfy the following requirements;

  1. Bending
  2. Shear
  3. Bearing
  4. Lateral buckling
  5. Deflection, and
  6. Vibration

Design Example of Timber Joists

In this article, let us design the timber floor joists for a domestic dwelling using timber of strength class C24 to support a medium-term permanent action of 0.75 kN/m2 and a variable (live) load of 1.5 kN/m2 given that the:

Loading on the floor joist of a building

a) floor width, b, is 3.6 m and floor span, l, is 3.0 m
b) joists are spaced at 600 mm centres
c) the bearing length is 100 mm

continuous timber floor joist loading

Self weight of timber = 3.4335 kN/m3 x 0.15m x 0.05m = 0.0257 kN/m
Chaaracteristic permaent action on each joist = 0.75 kN/m2 x 0.6m = 0.45 kN/m
Total characteristic permanent action gk = 0.476 kN/m

Total characteristic variable action qk = 1.5 kN/m2 x 0.6 m = 0.9 kN/m

At ultimate state; pEd = 1.35gk + 1.5qk = 1.9926 kN/m
At serviceability limit state pEd = 1.0gk + 1.0qk = 1.376 kN/m

Structural Analysis

design shear and bending moment
deflection

Structural Design
Member – Span 1
The partial factor for material properties and resistances
Partial factor for material properties (Table 2.3 of EN 1995-1-1:2004); γM = 1.300

Member details
Load duration (cl.2.3.1.2 of EC5); Medium-term
Service class – (cl.2.3.1.3 of EC5); 2

Timber section details
Number of timber sections in member; N = 1
Breadth of sections; b = 50 mm
Depth of sections; h = 150 mm
Timber strength class – EN 338:2016 Table 1; C24

timber section

Properties of 50 x 150 mm timber section
Cross-sectional area, A = 7500 mm2
Section modulus, Wy = 187500 mm3
Section modulus, Wz = 62500 mm3
Second moment of area, Iy = 14062500 mm4
Second moment of area, Iz = 1562500 mm4
Radius of gyration, iy = 43.3 mm
Radius of gyration, iz = 14.4 mm

Timber strength class C24
Characteristic bending strength, fm.k = 24 N/mm2
Characteristic shear strength, fv.k = 4 N/mm2
Characteristic compression strength parallel to grain, fc.0.k = 21 N/mm2
Characteristic compression strength perpendicular to grain, fc.90.k = 2.5 N/mm2
Characteristic tension strength parallel to grain, ft.0.k = 14.5 N/mm2
Mean modulus of elasticity, E0.mean = 11000 N/mm2
Fifth percentile modulus of elasticity, E0.05 = 7400 N/mm2
Shear modulus of elasticity, Gmean = 690 N/mm2
Characteristic density, rk = 350 kg/m3
Mean density, rmean = 420 kg/m3

Consider Combination 1 – 1.35Gk + 1.5Qk (Strength)

Typical arrangement and design of timber joists

Modification factors
Duration of load and moisture content – Table 3.1; kmod = 0.8
Deformation factor – Table 3.2; kdef = 0.8
Bending stress re-distribution factor – cl.6.1.6(2); km = 0.7
Crack factor for shear resistance – cl.6.1.7(2); kcr = 0.67
Load configuration factor – cl.6.1.5(4); kc,90 = 1.5
System strength factor – cl.6.6;  ksys = 1.1

At the start of span

Check compression perpendicular to the grain (cl.6.1.5 of EC5)
Design perpendicular compression – major axis; Fc,y,90,d = 2.394 kN
Effective contact length; Lb,ef = Lb = 100 mm

Design perpendicular compressive stress – exp.6.4; sc,y,90,d = Fc,y,90,d / (b × Lb,ef) = 0.479 N/mm2
Design perpendicular compressive strength; fc,y,90,d = kmod × ksys × fc.90.k / γM = 1.692 N/mm2
sc,y,90,d / (kc,90 × fc,y,90,d) = 0.189

PASS – Design perpendicular compression strength exceeds design perpendicular compression stress

Check shear force (Section 6.1.7 of EC5)
Design shear force;  Fy,d = 2.394 kN

Design shear stress – exp.6.60; ty,d = 1.5 × Fy,d / (kcr × b × h) = 0.714 N/mm2
Design shear strength;  fv,y,d = kmod × ksys × fv.k / γM = 2.708 N/mm2

ty,d / fv,y,d = 0.264
PASS – Design shear strength exceeds design shear stress

At the end of span

Check compression perpendicular to the grain – cl.6.1.5
Design perpendicular compression – major axis; Fc,y,90,d = 6.572 kN
Effective contact length; Lb,ef = Lb = 100 mm

Design perpendicular compressive stress – exp.6.4;
sc,y,90,d = Fc,y,90,d / (b × Lb,ef) = 1.314 N/mm2
Design perpendicular compressive strength;
fc,y,90,d = kmod × ksys × fc.90.k / γM = 1.692 N/mm2

sc,y,90,d / (kc,90 fc,y,90,d) = 0.518
PASS – Design perpendicular compression strength exceeds design perpendicular compression stress

Check shear force – Section 6.1.7
Design shear force; Fy,d = 3.583 kN
Design shear stress – exp.6.60; ty,d = 1.5 × Fy,d / (kcr × b × h) = 1.070 N/mm2
Design shear strength; fv,y,d = kmod × ksys × fv.k / γM = 2.708 N/mm2
ty,d / fv,y,d = 0.395
PASS – Design shear strength exceeds design shear stress

Check bending moment – Section 6.1.6
Design bending moment; My,d = 1.784 kNm
Design bending stress; sm,y,d = My,d / Wy = 9.517 N/mm2
Design bending strength; fm,y,d = kmod × ksys × fm.k / γM = 16.246 N/mm2
sm,y,d / fm,y,d = 0.586
PASS – Design bending strength exceeds design bending stress

Serviceability Limit State

Consider Combination 2 – 1.0Gk + 1.0Qk (Service)
Check y-y axis deflection – Section 7.2
Instantaneous deflection;  dy = 5.3 mm

Quasi-permanent variable load factor; y2 = 0.3
Final deflection with creep; dy,Final = 0.5 × dy × (1 + kdef) + 0.5 × dy × (1 + y2 × kdef) = 8.1 mm

Allowable deflection; dy,Allowable = L / 250 = 12 mm                
dy,Final / dy,Allowable = 0.676

Therefore, the final deflection is acceptable.

It is also typical to check the floor for vibration, but this was not considered in this design.

Question of the day | 23-01-2021

By
Ubani Obinna
-

For the structure loaded as shown above, answer the following questions given below. Note that the use of calculators is discouraged. The aim of this question is to improve our ability to predict the expected behaviour of a structure by just glancing at them. Good luck to you.

[1] What is the maximum bending moment in member DE?
(A) 0.5 kNm
(B) 1.0 kNm
(C) 2.0 kNm
(D) 4.0 kNm

[2] What is the value of the shear force at point G, just to the right?
(A) -1.5 kN
(B) +3 kN
(C) -3 kN
(D) +1.5 kNm

[3] What is the value of the bending moment at point C, just to the right?
(A) -3 kNm
(B) 1.5 kNm
(C) +3 kNm
(D) -1 kNm

[4] What is the value of the bending moment, at point F?
(A) -1 kNm
(B) -2 kNm
(c) -3 kNm
(D) -4 kNm

Worked Example | Design of RC beams for Torsion (EN 1992-1:2004)

By
Ubani Obinna
-

A full torsional design covering the ultimate and serviceability limit states is required when the equilibrium of a structure is dependent on the torsional resistance of the member. Reinforced concrete (RC) beams are subjected to torsion when the point of application of loads does not coincide with the shear centre of the beams. This can be due to the arrangement of the beams or the loading pattern as can be found in circular or canopy beams.

Typical examples of beams subjected to torsion
Fig 1: Typical examples of beams subjected to torsion

According to clause 6.3.1(3) of EN 1992-1:2004, the torsional resistance of a section may be calculated on the basis of a thin-walled closed section, in which equilibrium is satisfied by a closed shear flow. Solid sections may be modelled by equivalent thin-walled sections. Complex shapes, such as T-sections, may be divided into a series of sub-sections, each of which is modelled as an equivalent thin-walled section, and the total torsional resistance taken as the sum of the capacities of the individual elements.

NOTATTIONS AND DEFINITIONS FOR TORSION IN EUROCODE 2
Fig 2: Notations and definitions for torsion in Eurocode 2 [Source: Figure 6.11, EN 1992-1-1:2004]

In Eurocode 2, the shear stress in a wall of a section subject to a pure torsional moment may be calculated from:

τt,itef,i = TEd/2Ak

The shear force VEd,i in a wall i due to torsion is given by:

VEd,i = τt,itef,izi

where;
TEd is the applied design torsion
Ak is the area enclosed by the centre-lines of the connecting walls, including inner hollow areas.
τt,i is the torsional shear stress in wall i
tef,i is the effective wall thickness. It may be taken as A/u, but should not be taken as less than twice the distance between edge and centre of the longitudinal reinforcement. For hollow sections the real thickness is an upper limit
A is the total area of the cross-section within the outer circumference, including inner hollow areas
u is the outer circumference of the cross-section
zi is the side length of wall i defined by the distance between the intersection points with the adjacent walls

The required cross-sectional area of the longitudinal reinforcement for torsion ΣAsl may be calculated from Expression (6.28) of EC2:

ΣAsl/fyd = TEdcotθ/2Ak

where;
uk is the perimeter of the area Ak
fyd is the design yield stress of the longitudinal reinforcement Asl
θ is the angle of compression struts

According to clause 6.3.2(4) of EC2, the maximum resistance of a member subjected to torsion and shear is limited by the capacity of the concrete struts. In order not to exceed this resistance the following condition should be satisfied:

TEd/TRd,max + VEd/VRd,max ≤ 1.0 (Equation 6.29, EC2)

where:
TEd is the design torsional moment
VEd is the design transverse force
TRd,max is the design torsional resistance moment = TRd,max = 2ναcwfcdAktef,isinθcosθ
VRd,max is the maximum design shear resistance according to Expressions (6.9) of EC2

Worked Example

Carry out a full torsional design of a rectangular reinforced concrete (RC) beam subjected to an ultimate torsional moment of 55 kNm, and shear force of 225 kN. The section is 600 x 400 mm, and reinforcement of 4H25 (As,prov = 1964 mm2) has been provided to resist the bending moment. (fck = 28 N/mm2; fyk = 500 N/mm2)

Solution

Concrete strength class; C28/35 ×γα
Characteristic compressive cylinder strength;fck = 28 N/mm2
Partial factor for concrete -Table 2.1N; γC = 1.50
Compression chord stress coefficient – cl.6.2.3(3); αcw = 1.00
Compressive strength coefficient – cl.3.1.6(1); αcc = 0.85
Design compressive concrete strength – exp.3.15;   fcd = αcc × fck/γC = 15.9 N/mm2
Compressive strength coefficient – cl.3.1.6(1); αccw = 1.00
Design compressive concrete strength – exp.3.15;   fcwd = αccw × fck/γC = 18.7 N/mm2
Tensile strength coefficient – cl.3.1.6(2); αct = 1.00
Mean value of axial strength of conc. – cl.3.1.6(2);    fctm = 0.3 N/mm2 × (fck)2/3 = 2.77 N/mm2

Characteristic axial strength of conc. (5% factile);    fctk,0.05 = 0.7 × fctm = 1.94 N/mm2
Design axial strength of concrete; fct,d = act × fctk,0.05/γC = 1.29 N/mm2

Reinforcement details
Characteristic yield strength of reinforcement; fyk = 500 N/mm2
Partial factor for reinforcing steel – Table 2.1N; γS = 1.15
Design yield strength of reinforcement;  fyd = fyk/γS = 435 N/mm2

Beam dimensions
Section width; b = 400 mm
Section depth; h = 600 mm

Design forces
Maximum design torsional moment; TEd,max = 55.0 kNm
Maximum design shear force; VEd,max = 225.0 kN
Area of design longitudinal reinforcement; Asl = 1964 mm2
Effective depth to outer layer; d = 450 mm

equivalent thin walled section for torsion
Fig 3: Equivalent thin-walled section

Torsional resistance (Section 6.3, EC2)

Effective thickness of walls – cl.6.3.2(1); tef,i = (b × h) / (2 × (b + h)) = 120 mm
Area enclosed by centre lines of walls – cl.6.3.2(1);  Ak = (b – tef,i) × (h – tef,i) = 134400 mm2
Perimeter – cl.6.3.2(3); Uk = 2 × ((b – tef,i) + (h – tef,i)) = 1520 mm

Strength reduction factor;  v1 = 0.6 × (1 – fck / 250 N/mm2) = 0.533

Design shear stress; vt,Ed = VEd,max / (b × d) = 1.250 N/mm2

Torsional shear stress in wall – exp.6.26;  tt,Ed = TEd,max / (2 × Ak × tef,i) = 1.705 N/mm2

Concrete strut angle;qt = min[45°, max(0.5 × Asin(min(2 × (vt,Ed / 0.9 + tt,Ed) / (αcw × fcwd × v1), 1)), 21.8°)] = 21.8°

Max design value of torsional resist. mnt – exp 6.30; TRd,max = 2 × v1 × αcw × fcd × Ak × tef,i × sin(qt) × cos(qt) = 94.0 kNm

Max design shear force – exp.6.9; VRdt,max = acw × b × 0.9 × d × v1 × fcwd / (cot(qt) + tan(qt)) = 555.6 kN

Interaction formulae – exp.6.29; TEd,max / TRd,max + VEd,max / VRdt,max = 0.990

PASS – concrete section is adequate

Torsional and shear resistance of the concrete alone

Maximum torsional resist moment with no shear reinf. – cl.6.3.2(5);  TRd,c = 2 × Ak × fct,d × tef,i = 41.6 kNm
Shear resistance constant – cl.6.2.2; CRd,c = 0.18/γC = 0.120
Reinforcement ratio – cl.6.2.2; rl = min(Asl/(d × b), 0.02) = 0.011
Effective depth factor – cl.6.2.2; kv = min(1 + √(200mm/d), 2) = 1.667

Minimum shear stress; vmin = 0.035 N/mm2 × kv3/2 × (fck / 1N/mm2)0.5 = 0.4 N/mm2

Design value for shear resistance – exp.6.2.a;   VRd,c = max(CRd,c × kv × 1N/mm2 × (100 × rl × fck/1N/mm2)1/3 × b × d, vmin × b × d) = 112.5 kN

Interaction formulae – exp.6.31; TEd,max / TRd,c + VEd,max / VRd,c = 3.320
Therefore, additional reinforcement required

Required torsional reinforcement
Required area of add. long. reinf. for torsion (6.28); Asl,req = TEd,max × Uk × cot(qt) / (2 × Ak × fyd) = 1788 mm2
Provide 10Y16 side bars (5 on each face) for torsion (Asprov = 2010 mm2)

The longitudinal bars should be arranged so there is at least one bar at each corner with the other spaced around the periphery of the links at a spacing of 350mm or less (cl.9.2.3(4))

Required shear reinforcement for torsion (one leg); Asw,req = TEd,max / (2 × Ak × fyd × cot(qt)) = 188 mm2/m
Maximum spacing for torsion shear reinforcement;  sw,max = min(Uk/8, b, h) = 190 mm

Provide 2 legs H10@175 c/c as torsion/shear reinforcement

[Featured Image Credit] Chai H.K., Majeed A.A., Allawi A.A (2015): Torsional Analysis of Multicell Concrete Box Girders Strengthened with CFRP Using a Modified Softened Truss Model. ASCE Journal of Bridge Engineering, 20(8) https://doi.org/10.1061/(ASCE)BE.1943-5592.0000621

To download this article in PDF format, click HERE

Worked Example | Analysis and Design of Steel Sheet Pile Wall (EN 1997-1)

By
Ubani Obinna
-

This article contains a solved example of the analysis and design of steel sheet pile walls in accordance with BS EN1997-1:2004 – Code of Practice for Geotechnical design and the UK National Annex.

sheet pile wall

Geometry
Total length of sheet pile provided Hpile = 14500 mm
Number of different types of soil Ns = 2
Retained height of soil dret = 3500 mm
Depth of unplanned excavation dex = 500 mm
Total retained height ds = 4000 mm
Angle of retained slope β = 0.0 deg
Depth from ground level to top of water table retained side dw = 1500 mm
Depth from ground level to top of water table retaining side;  dwp = 4000 mm

Loading
Variable surcharge po,Q = 10.0 kN/m2

Soil layer 1
Characteristic shearing resistance angle ϕ’k,s1 = 30.0 deg
Characteristic wall friction angle  δk,s1 = 20.0 deg
Moist density of soil γm,s1 = 15.0 kN/m3
Characteristic saturated density of retained soil γs,s1 = 17.0 kN/m3
Height of soil 1  h1 = 8500 mm

Soil layer 2
Characteristic shearing resistance angle f’k,s2 = 27.0 deg
Characteristic wall friction angle dk,s2 = 16.0 deg
Moist density of soil γm,s2 = 16.0 kN/m3
Characteristic saturated density of retained soil γs,s2 = 19.0 kN/m3
Height of soil 2 h2 = 7000 mm

Partial factors on actions – Section A.3.1 – Combination 1
Permanent unfavourable action γG = 1.35
Permanent favourable action γG,f = 1.00
Variable unfavourable action γQ = 1.50
Angle of shearing resistance γϕ’ = 1.00
Weight density γg = 1.00

Design soil properties – soil 1
Design effective shearing resistance angle ϕ’d = tan-1[tan(ϕ’k)/γϕ’] = 30.0 deg
Design wall friction angle  δd = tan-1[tan(ϕk)/γϕ’] = 20.0 deg
Design moist density of retained soil γm.d1 = γm/γγ = 15.0 kN/m3
Design saturated density of retained soil γs.d1 = γs/γγ = 17.0 kN/m3
Design buoyant density of retained soil γd.d1 = γs.d1 – γw = 7.2 kN/m3

Active pressure using Coulomb theory Ka1 = sin(α + ϕ’d)2 / (sin(α)2 × sin(α – δd) × (1 + √(sin(ϕ’d + δd) × sin(ϕ’d – β)/(sin(α – δd) × sin(α + β))))2) = 0.297

Passive pressure using Coulomb theory Kp1 = sin(90 – ϕ’d)2 / (sin(90 + δd) × [1 – √[sin(ϕ’d + δd) × sin(ϕ’d) / (sin(90 + δd))]]2) = 6.105

Design soil properties – soil 2
Design effective shearing resistance angle ϕ’d = tan-1(tan(ϕ’k) /γϕ’) = 27.0 deg
Design wall friction angle  δd = tan-1(tan(δk)/γϕ’) = 16.0 deg
Design moist density of retained soil γm.d2 = γm/γγ = 16.0 kN/m3
Design saturated density of retained soil γs.d2 = γs /γγ = 19.0 kN/m3
Design buoyant density of retained soil γd.d2 = γs.d2 – γw = 9.2 kN/m3

Active pressure using Coulomb theory Ka2 = sin(α + ϕ’d)2 / (sin(α)2 × sin(α – δd) × (1 + √(sin(ϕ’d + δd) × sin(ϕ’d – β) / (sin(α – δd) × sin(α + β))))2) = 0.336

Passive pressure using Coulomb theory Kp2 = sin(90 – ϕ’d)2 / (sin(90 + δd) × [1 – √[sin(ϕ’d + δd) × sin(ϕ’d) / (sin(90 + δd))]]2) = 4.416

balanced pressure diagram of sheet pile wall

Overburden on the active side
Overburden at 0 mm below GL in soil 1; OB’a11 = po,Q × γQ = 15.0 kN/m2
Overburden at 1500 mm below GL in soil 1; OB’a21 = γG × γm.d1 × ha1 + OB’a11 = 45.4 kN/m2
Overburden at 4000 mm below GL in soil 1; OB’a31 = γG × γd.d1 × ha2 + OB’a21 = 69.6 kN/m2
Overburden at 8500 mm below GL in soil 1; OB’a41 = γG × γd.d1 × ha3 + OB’a31 = 113.3 kN/m2
Overburden at 8500 mm below GL in soil 2; OB’a42 = γG × γd.d1 × ha3 + OB’a31 = 113.3 kN/m2
Overburden at 11544 mm below GL in soil 2; OB’a51 = γG × γd.d2 × ha4 + OB’a42 = 151.1 kN/m2

Overburden on the passive side
Overburden at 4000 mm below GL in soil 1; OB’p31 = 0 kN/m2 = 0.0 kN/m2
Overburden at 8500 mm below GL in soil 1; OB’p41 = γG,f × γd.d1 × hp3 + OB’p31 = 32.4 kN/m2
Overburden at 8500 mm below GL in soil 2; OB’p42 = γG,f × γd.d1 × hp3 + OB’p31 = 32.4 kN/m2
Overburden at 11544 mm below GL in soil 2; OB’p51 = γG,f × γd.d2 × hp4 + OB’p42 = 60.3 kN/m2

Pressure on the active side
Active at 0 mm below GL in soil 1; p’a11 = Ka1 × OB’a11 = 4.5 kN/m2
Active at 1500 mm below GL in soil 1; p’a21 = Ka1 × OB’a21 = 13.5 kN/m2
Active at 4000 mm below GL in soil 1; p’a31 = Ka1 × OB’a31 + γγ × γw × (dL3 – dw) = 53.8 kN/m2
Active at 8500 mm below GL in soil 1; p’a41 = Ka1 × OB’a41 + γγ × γw × (dL4 – dw) = 126.4 kN/m2
Active at 8500 mm below GL in soil 2; p’a42 = Ka2 × OB’a42 + γγ × γw × (dL4 – dw) = 130.8 kN/m2
Active at 11544 mm below GL in soil 2; p’a51 = Ka2 × OB’a51 + γγ × γw × (dL5 – dw) = 183.8 kN/m2

Pressure on the passive side
Passive at 4000 mm below GL in soil 1; p’p31 = Kp1 × OB’p31 + γG,f × γw × (dL3 – max(ds, dw)) = 0.0 kN/m2
Passive at 8500 mm below GL in soil 1; p’p41 = Kp1 × OB’p41 + γG,f × γw × (dL4 – max(ds, dw)) = 241.7 kN/m2
Passive at 8500 mm below GL in soil 2; p’p42 = Kp2 × OB’p42 + γG,f × γw × (dL4 – max(ds, dw)) = 187.0 kN/m2
Passive at 11544 mm below GL in soil 2; p’p51 = Kp2 × OB’p51 + γG,f × γw × (dL5 – max(ds, dw)) = 340.4 kN/m2

By iteration the depth at which the active moments equal the passive moments has been determined as 11544 mm as follows:-

Active moment about 11544 mm

Moment level 1;Ma11 = 0.5 × p’a11 × ha1 × ((H – dL2) + 2/3 × ha1) = 36.9 kNm/m
Moment level 1; Ma12 = 0.5 × p’a21 × ha1 × ((H – dL2) + 1/3 × ha1) = 106.7 kNm/m
Moment level 2; Ma21 = 0.5 × p’a21 × ha2 × ((H – dL3) + 2/3 × ha2) = 155.3 kNm/m
Moment level 2; Ma22 = 0.5 × p’a31 × ha2 × ((H – dL3) + 1/3 × ha2) = 563.5 kNm/m
Moment level 3; Ma31 = 0.5 × p’a31 × ha3 × ((H – dL4) + 2/3 × ha3) = 731.8 kNm/m
Moment level 3; Ma32 = 0.5 × p’a41 × ha3 × ((H – dL4) + 1/3 × ha3) = 1292.3 kNm/m
Moment level 4; Ma41 = 0.5 × p’a42 × ha4 × ((H – dL5) + 2/3 × ha4) = 404.0 kNm/m
Moment level 4; Ma42 = 0.5 × p’a51 × ha4 × ((H – dL5) + 1/3 × ha4) = 283.8 kNm/m

Passive moment about 11544 mm

Moment level 3; Mp31 = 0.5 × p’p31 × hp3 × ((H – dL4) + 2/3 × hp3) = 0.0 kNm/m
Moment level 3; Mp32 = 0.5 × p’p41 × hp3 × ((H – dL4) + 1/3 × hp3) = 2471.0 kNm/m
Moment level 4; Mp41 = 0.5 × p’p42 × hp4 × ((H – dL5) + 2/3 × hp4) = 577.6 kNm/m
Moment level 4; Mp42 = 0.5 × p’p51 × hp4 × ((H – dL5) + 1/3 × hp4) = 525.7 kNm/m

Total moments about 11544 mm

Total active moment; SMa = 3574.5 kNm/m
Total passive moment; SMp = 3574.5 kNm/m

Required pile length
Length of pile required to balance moments; H = 11544 mm

Depth of equal pressure; dcontra = 5432 mm
Add 20% below this point; de_add = 1.2 × (H – dcontra) = 7334 mm

Minimum required pile length; Htotal = dcontra + de_add = 12766 mm

Pass – Provided length of sheet pile greater than the minimum required length of the pile

Pile capacity (EN1993-5)
Maximum moment in pile (from analysis); Mpile = max(abs(Mmin), abs(Mmax)) / 1m = 547.0 kNm/m
Maximum shear force in pile (from analysis); Vpile = 364.7 kN/m
Nominal yield strength of pile; fy_pile = 355 N/mm2
Name of sheet pile;  Arcelor PU(18)
Classification of pile; 2
Plastic modulus of pile; Wpl.y = 2134 cm3/m

steel sheet pile wall section and tables

Shear buckling of web (cl.5.2.2(6))
Width of section; c = h / sin(αpile) = 510 mm
Thickness of web;  tw = s = 9.0 mm
ε = √(235/fy_pile) = 0.814
c/tw = 56.6 = 69.6ε < 72ε
PASS – Shear buckling of web within limits

Bending 2
Interlock reduction factor (cl.5.2.2); βB = 1
Design bending resistance (eqn.5.2);                         
Mc,Rd = Wpl.y × fy_pile × βB / γM0 = 757.6 kNm/m
PASS – Moment capacity exceeds moment in pile

Shear
Projected shear area of web (eqn.5.6); Av = s × (h – t) = 3769 mm2
Design shear resistance (eqn.5.5); Vpl,Rd = Av × fy_pile / (√(3) × γM0) / b = 1287.6 kN/m
PASS – Shear capacity exceeds shear in pile

Partial factors on actions – Section A.3.1 – Combination 2

Permanent unfavourable action;  γG = 1.00
Permanent favourable action; γG,f = 1.00
Variable unfavourable action; γQ = 1.30
Angle of shearing resistance; γϕ’ = 1.25
Weight density; γγ = 1.00

Design soil properties – soil 1
Design effective shearing resistance angle; ϕ’d = tan-1(tan(ϕ’k)/γϕ’) = 24.8 deg
Design wall friction angle;  δd = tan-1(tan(δk)/γϕ’) = 16.2 deg
Design moist density of retained soil; γm.d1 = γm/γγ = 15.0 kN/m3
Design saturated density of retained soil; γs.d1 = γs/γγ = 17.0 kN/m3
Design buoyant density of retained soil; γd.d1 = γs.d1 – γw = 7.2 kN/m3

Active pressure using Coulomb theory; Ka1 = sin(α + ϕ’d)2 / (sin(α)2 × sin(α – δd) × (1 + √(sin(ϕ’d + δd) × sin(ϕ’d – β)/(sin(α – δd) ´ sin(α + β))))2) = 0.364

Passive pressure using Coulomb theory; Kp1 = sin(90 – ϕ’d)2 / (sin(90 + δd) × [1 – √[sin(ϕ’d + δd) × sin(ϕ’d) / (sin(90 + δd))]]2) = 3.977

Design soil properties – soil 2

Design effective shearing resistance angle; ϕ’d2 = tan-1(tan(ϕ’k)/γϕ’) = 22.2 deg
Design wall friction angle; δd2 = tan-1(tan(δk)/γϕ’) = 12.9 deg
Design moist density of retained soil; γm.d2 = γm/γγ = 16.0 kN/m3
Design saturated density of retained soil; γs.d2 = γs/γγ = 19.0 kN/m3
Design buoyant density of retained soil; γd.d2 = γs.d2 – γw = 9.2 kN/m3

Active pressure using Coulomb theory; Ka2 = sin(α + ϕ’d)2 / (sin(α)2 × sin(α – δd) × (1 + √(sin(ϕ’d + δd) × sin(ϕ’d – β) / (sin(α – δd) × sin(α + β))))2) = 0.406

Passive pressure using Coulomb theory; Kp2 = sin(90 – ϕ’d)2 / (sin(90 + δd) × [1 – √[sin(ϕ’d + δd) × sin(f’d) / (sin(90 + δd))]]2) = 3.154

pressure diagram of sheet pile wall

Overburden on the active side
Overburden at 0 mm below GL in soil 1; OB’a11 = po,Q × γQ = 13.0 kN/m2
Overburden at 1500 mm below GL in soil 1;  OB’a21 = γG × γm.d1 × ha1 + OB’a11 = 35.5 kN/m2
Overburden at 4000 mm below GL in soil 1; OB’a31 = γG × γd.d1 × ha2 + OB’a21 = 53.5 kN/m2
Overburden at 8500 mm below GL in soil 1; OB’a41 = γG × γd.d1 × ha3 + OB’a31 = 85.8 kN/m2
Overburden at 8500 mm below GL in soil 2; OB’a42 = γG × γd.d1 × ha3 + OB’a31 = 85.8 kN/m2
Overburden at 12532 mm below GL in soil 2;OB’a51 = γG × γd.d2 × ha4 + OB’a42 = 122.9 kN/m2

Overburden on the passive side
Overburden at 4000 mm below GL in soil 1; OB’p31 = 0 kN/m2 = 0.0 kN/m2
Overburden at 8500 mm below GL in soil 1; OB’p41 = γG,f × γd.d1 × hp3 + OB’p31 = 32.4 kN/m2
Overburden at 8500 mm below GL in soil 2; OB’p42 = γG,f × γd.d1 × hp3 + OB’p31 = 32.4 kN/m2
Overburden at 12532 mm below GL in soil 2;OB’p51 = γG,f × γd.d2 × hp4 + OB’p42 = 69.4 kN/m2

Pressure on the active side

Active at 0 mm below GL in soil 1; p’a11 = Ka1 × OB’a11 = 4.7 kN/m2
Active at 1500 mm below GL in soil 1; p’a21 = Ka1 × OB’a21 = 12.9 kN/m2
Active at 4000 mm below GL in soil 1; p’a31 = Ka1 × OB’a31 + γG × γw × (dL3 – dw) = 44.0 kN/m2
Active at 8500 mm below GL in soil 1; p’a41 = Ka1 × OB’a41 + γG × γw × (dL4 – dw) = 99.9 kN/m2
Active at 8500 mm below GL in soil 2; p’a42 = Ka2 × OB’a42 + γG × γw × (dL4 – dw) = 103.5 kN/m2
Active at 12532 mm below GL in soil 2; p’a51 = Ka2 × OB’a51 + γG × γw × (dL5 – dw) = 158.1 kN/m2

Pressure on the passive side

Passive at 4000 mm below GL in soil 1; p’p31 = Kp1 × OB’p31 + γG,f × γw × (dL3 – max(ds, dw)) = 0.0 kN/m2
Passive at 8500 mm below GL in soil 1; p’p41 = Kp1 × OB’p41 + γG,f × γw × (dL4 – max(ds, dw)) = 172.8 kN/m2
Passive at 8500 mm below GL in soil 2;p’p42 = Kp2 × OB’p42 + γG,f × γw × (dL4 – max(ds, dw)) = 146.2 kN/m2
Passive at 12532 mm below GL in soil 2; p’p51 = Kp2 × OB’p51 + γG,f × γw × (dL5 – max(ds, dw)) = 302.7 kN/m2

By iteration the depth at which the active moments equal the passive moments has been determined as 12533 mm as follows:-

Active moment about 12533 mm

Moment level 1; Ma11 = 0.5 × p’a11 × ha1 × ((H – dL2) + 2/3 × ha1) = 42.7 kNm/m
Moment level 1; Ma12 = 0.5 × p’a21 × ha1 × ((H – dL2) + 1/3 × ha1) = 111.8 kNm/m
Moment level 2; Ma21 = 0.5 × p’a21 × ha2 × ((H – dL3) + 2/3 × ha2) = 164.8 kNm/m
Moment level 2; Ma22 = 0.5 × p’a31 × ha2 × ((H – dL3) + 1/3 × ha2) = 515.1 kNm/m
Moment level 3; Ma31 = 0.5 × p’a31 × ha3 × ((H – dL4) + 2/3 × ha3) = 696.2 kNm/m
Moment level 3; Ma32 = 0.5 × p’a41 × ha3 × ((H – dL4) + 1/3 × ha3) = 1244.0 kNm/m
Moment level 4; Ma41 = 0.5 × p’a42 × ha4 × ((H – dL5) + 2/3 × ha4) = 561.3 kNm/m
Moment level 4; Ma42 = 0.5 × p’a51 × ha4 × ((H – dL5) + 1/3 × ha4) = 428.7 kNm/m

Passive moment about 12533 mm

Moment level 3; Mp31 = 0.5 × p’p31 × hp3 × ((H – dL4) + 2/3 × hp3) = 0.0 kNm/m
Moment level 3; Mp32 = 0.5 × p’p41 × hp3 × ((H – dL4) + 1/3 × hp3) = 2151.5 kNm/m
Moment level 4; Mp41 = 0.5 × p’p42 × hp4 × ((H – dL5) + 2/3 × hp4) = 792.7 kNm/m
Moment level 4; Mp42 = 0.5 × p’p51 × hp4 × ((H – dL5) + 1/3 × hp4) = 820.5 kNm/m

Total moments about 12533 mm

Total active moment; SMa = 3763.9 kNm/m
Total passive moment; SMp = 3763.7 kNm/m

Required pile length
Length of pile required to balance moments; H = 12533 mm

Depth of equal pressure; dcontra = 5694 mm
Add 20% below this point; de_add = 1.2 × (H – dcontra) = 8207 mm
Minimum required pile length; Htotal = dcontra + de_add = 13901 mm

PASS – Provided length of sheet pile greater than the minimum required length of pile

Pile capacity (EN1993-5)

Maximum moment in pile (from analysis); Mpile = max(abs(Mmin), abs(Mmax)) / 1m = 549.1 kNm/m
Maximum shear force in pile (from analysis); Vpile = 358.1 kN/m
Nominal yield strength of pile; fy_pile = 355 N/mm2
Name of pile;  Arcelor PU(18)
Classification of pile; 2
Plastic modulus of pile; Wpl.y = 2134 cm3/m

Shear buckling of web (cl.5.2.2(6))

Width of section; c = h / sin(apile) = 510 mm
Thickness of web; tw = s = 9.0 mm
ε = √(235/fy_pile)= 0.814
c / tw = 56.6 = 69.6ε < 72ε

PASS – Shear buckling of web within limits

Bending
Interlock reduction factor (cl.5.2.2); βB = 1
Design bending resistance (eqn.5.2);Mc,Rd = Wpl.y × fy_pile × βB / γM0 = 757.6 kNm/m

PASS – Moment capacity exceeds moment in pile

Shear
Projected shear area of web (eqn.5.6); Av = s × (h – t) = 3769 mm2
Design shear resistance (eqn.5.5); Vpl,Rd = Av × fy_pile / (√(3) × γM0) / b = 1287.6 kN/m

PASS – Shear capacity exceeds shear in the pile

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Cement Replacement Materials (CRM) in Concrete

By
Ubani Obinna
-

Cement replacement materials (CRM) are materials that can be used for substituting cement in the production of concrete or other cementitious products. For a material to be used as a cement replacement material, it must possess pozzolanic properties. In the recent wake of the need to produce sustainable concrete, conserve the environment, reduce greenhouse effects in construction, and modify the properties of concrete, the use of industrial and agricultural wastes have found wide applications as cement replacement materials.

Some basic examples of cement replacement materials are;

  1. Fly ash (Pulverised Fuel Ash)
  2. Ground Granulated Blast-furnace Slag (GGBS)
  3. Silica fume
  4. Limestone fines
  5. Rice husk ash
  6. Palm oil fuel ash
  7. Sugarcane bagasse ash

It is important to note that while fly ash, GGBS, silica fume, and limestone fines are industrial wastes, rice husk ash, palm oil fuel ash, and sugarcane bagasse ash are agricultural wastes. The use of so many other agro-wastes has also been investigated by researchers. The utilisation of these wastes in the production of concrete can reduce the carbon footprint associated with the construction industry, and at the same time help to solve the problems of waste disposal.

To varying degrees, these substitutes can be used in the partial replacement of cement due to the fact they can hydrate and cure like portland cement. Furthermore, they are “pozzolans,” which provide silica that reacts with hydrated lime, an unwanted by-product of concrete curing. While industrial byproducts are directly used as substitutes of cement, the agricultural wastes are usually burned in a controlled environment/temperature and ground finely before the ash can be used as a pozzolan.

If the silica, aluminum, and iron oxides obtained after burning the agricultural wastes are not up to 70%, the material will not function properly as a cement substitute. These are some of the drawbacks of using agricultural wastes. Furthermore, it is usually challenging to obtain a large quantity of agricultural wastes that can be used to execute a large-scale project when compared with industrials wastes.

Having said that, let us look at the properties of these cement replacement materials and how they influence the properties of concrete.

Fly Ash

Fly ash is a fine powder derived by burning pulverised coal with or without other combustion materials. It contains pozzolanic properties and consists mainly of aluminum oxide (Al203) and silicon oxide (SiO2). Typically, it is obtained as a byproduct of the combustion of coal from power stations. The ‘ash’ is recovered from the gases and used, amongst other functions, as a cement substitute. Fly ash can only be used as a partial replacement for cement because it relies on the water and lime from the cement to hydrate as part of the overall chemical reaction.

fly ash as a cement replacement materials
Fig 1: Fly ash

The use of fly ash offers beneficial properties to fresh and hardened concrete. It improves the workability of fresh concrete, the strength, and durability of hardened concrete. Furthermore, it is cost-effective and reduces the quantity of cement required for construction.  Typically, 15 percent to 30 percent of the portland cement is replaced with fly ash, with even higher percentages used for mass concrete placements. To achieve this, an equivalent or greater weight of fly ash is substituted for the cement removed.

In a 2017 study published in India, the following are the results of partial replacement of cement with fly ash;

S/No% of fly ashfck (7 days)fck (14 days)fck – 28 days
1026.736.740.2
21027.438.2541.9
32028.339.543.23
43030.2541.445.28
54027.7537.7442.00
65025.537.0339.15

As can be seen from the research, the replacement of cement with 20 – 30% fly ash gave better 28-days compressive strength than normal concrete. According to a 2010 study published in Poland, the concrete specimens containing 20 % of fly ash, related to cement mass, gained a compressive strength about 25% higher than normal concrete after 180 days of curing.

Ground Granulated Blast-furnace Slag (GGBS)

Ground-granulated blast-furnace slag (GGBS) is obtained by quenching molten iron slag (a by-product of iron and steel-making) from a blast furnace in water or steam, to produce a glassy, granular product that is then dried and ground into a fine powder. GGBS has off-white or near-white color and it exhibits excellent cementitious property when finely ground and combined with Portland cement.

GGBS
Fig 2: Ground-granulated blast-furnace slag (GGBS)

Essentially, GGBS comprises silicates and alumina silicates of calcium and other bases that are manufactured in a blast furnace under molten conditions simultaneously with the iron. The chemical composition of oxides in GGBS is similar to that of Portland cement, but the proportion varies.

The fineness of GGBS is a very important parameter when it is to be used in concrete production. This is measured by the specific surface area, and it controls the reactivity of GGBS with cement. Generally, increased fineness of GGBS results in better strength development, but in practice, fineness is limited by economic and performance considerations and factors such as setting times and shrinkage.

In a study carried out in the UK in the year 2017 on the strength development of concrete produced with GGBS, three environmental curing conditions were simulated in the laboratory;

C1 – Summer curing environment
C2 – Winter curing environment
C3 -Normal water curing environment

From the study, the 28-day compressive strength of concrete produced with GGBS are given below;

S/No% of GGBS28-day fck for C128-day fck for C2 28-day fck C3
1056.055.057.5
23056.549.058.5
34057.048.058.0
45053.047.554.0


As can be seen from the research result, the maximum 28-day compressive strength was obtained at 30% partial replacement of cement with GGBS under normal temperature curing conditions. In another study carried out by Vinayak and Nagendra (2014), the maximum compressive strength of grade 20 concrete was achieved at 30% of GGBS replacement of cement.

However, the negative effect of GGBS replacement on mechanical strength has been observed very noticeably for 60 % and 80 % replacement. For all days, the compressive strength of cement mortar was observed to decrease severely at early ages with the increased replacement levels of GGBS in cement.

Silica Fume

Silica fume is an ultrafine powder collected as a by-product of silicon and ferrosilicon alloy production. The raw materials are coal, quartz, and woodchips. It is basically an amorphous (non-crystalline) polymorph of silicon dioxide, silica.

Silica fumes consist of spherical particles with an average particle diameter of 150 nm. The surface area of silica fume particles is about six times that of cement because they are finer than cement. As a pozzolan and filler material, it can improve the strength and durability of hardened concrete. Due to the high surface area and high content of amorphous silica in silica fume, this highly active pozzolana reacts more quickly than ordinary pozzolans.

silica fume powder
Fig 3: Silica fume

Many experiments have shown that the addition of silica fume to concrete increases the compressive strength of concrete by between 30% and 100% depending on the type of cement, type of mix, use of plasticizers, amount of silica fume, aggregates type, and curing regimes.

In a 2012 study carried out in India, the effects of silica fume on the 7-days and 28-days compressive strength of concrete are shown in the table below;

S/No% of silica fumefck – 7 daysfck – 28 days
1025.2138.30
2529.3341.29
31034.1246.76
41538.3047.30
52035.9044.27

As can be seen from the study, the optimum partial replacement of cement with silica fume for maximum compressive strength of concrete occurred at 15%. Sakr (2006) reported that at 15% silica fume content, the compressive strength of concrete increased by 23.33% at 7 days, 21.34% at 28 days, 16.50% at 56 days, and 18.00% at 90 days.

Due to economic considerations, the use of silica fume is generally limited to high strength concretes or concretes in aggressive environmental conditions. The most commonly used proportion of silica fume in the UK – produced combinations is 10% by mass of total cementitious content.

Limestone Fines

Limestone fines is a powder that is obtained from the processing of limestone in quarries. According to concrete.org.uk, there is uncertainty over whether limestone fines should be classified as a Type I (nearly inert as a filler aggregate) or Type II addition (with pozzolanic or latent hydraulic properties, for example, materials like fly ash and GGBS).

Limestone is less reactive than either fly ash or GGBS, but research shows that it can have slight reactivity as well as beneficial physical effects conferred by virtue of its fine particle size.

BS 7979:2016 gives guidance on the use of limestone fines with Portland Cement.

According to Lomboy et al (2016), there are two methods by which limestone fines are incorporated into cementitious systems. The first is by addition, whereby limestone fines are to replace a percentage of cementitious materials or as filler, which is added during the mixing process. The other method is by co-grinding with Portland cement clinker, making the limestone a component of Portland cement.

In the study of Lomboy et al (2016), the effect of limestone fines replacement of cement is shown below.

Effect of limestone fines on compressive strength of concrete
Fig 4: Type I cement with limestone fines mortar compressive strength (Lomboy et al, 2016)
type 1p cement
Fig 5: Type IP cement with limestone fines mortar compressive strength (Lomboy et al, 2016)

Limestone fines used to replace Type IP (with 25% fly ash) cement slightly improved (e.g., 5% to 10%) concrete compressive strength. However, the replacement for Type I (normal portland) cement decreased the compressive strength. This is likely due to the size of the limestone fines being complementary to the combination of cement and fly ash to improve packing on Type IP. In addition, there may be a chemical interaction between the fly ash in Type IP and limestone fines, which also facilitates concrete strength gain.

Rice Husk Ash

Rice husk ash (RHA) is one of the promising pozzolanic materials that can be blended with Portland cement for the production of durable concrete. Under controlled burning, and if sufficiently ground, the ash that is produced can be used as a cement replacement material in concrete.

In a 2009 study carried out in Nigeria, the effects of partial replacement of concrete with RHA is shown below;

S/No% of RHAfck – 7 daysfck – 28 days
1028.032.3
21022.128.5
32018.624.3
43016.322.4
54014.418.2
6509.211.5

It can be seen from the table that the addition of RHA reduced the compressive strength of concrete. It always recommended that the optimum dosage of rice husk ash in concrete should be 10%, unless other effects apart from compressive strength are desired.

Palm Oil Fuel Ash

Palm oil fuel ash (POFA) is a by-product obtained during the burning of waste materials such as palm kernel shell, palm oil fiber, and palm oil husk. They are usually generated from biomass power plants, where palm oil residues such as fibers, shells, and empty fruit bunches are burned to generate electricity. The ash can be utilized to partially replace cement in a concrete mix due to its pozzolanic properties.

In a 2017 study carried out in Thailand, the effects of POFA on the compressive strength of concrete using water/binder ratio of 0.4 is shown the table below;

S/No% of POFAfck – 28 daysfck – 90 days
1050.255.9
21546.252.8
32543.350.9
43540.045.8

From the study, it can be seen that the replacement of POFA in OPC at 15 and 35% by weight of binder had strengths of 94–80% of the control concrete at 28 and 90 days. In addition, the use of ground POFA in the concrete required slightly higher amounts of superplasticizer than those required by the control concrete.

Sugarcane Bagasse Ash

Sugarcane bagasse is an agricultural waste that can be transformed to a pozzolan by burning it in a controlled environment and grinding it finely into a cement replacement material for various cementing purposes.

Ina 2017 study carried out in Malaysia, the effect of sugarcane bagasse ash on the compressive strength of concrete is given in the table below;

S/No% of SCBAfck – 7 daysfck – 14 daysfck – 28 days
1017.1620.3325.525
2520.9225.1128.5
31019.22521.0926.4

From the research result it can be seen that the optimum replace of cement with sugarcane bagasse ash in concrete is 5%.

Unlinked References
[1] Vinayak A., and Nagendra M. V (2014): Analysis of Strength Characteristics of GGBS Concrete, Int J Adv Engg Tech, Vol. 5, No. 2, pp. 82–84,
[2] Sakr K. (2006): Effects of Silica Fume and Rice Husk Ash on the Properties of Heavy Weight Concrete. ASCE Journal of Materials in Civil Engineering Vol. 18, Issue 3 https://doi.org/10.1061/(ASCE)0899-1561(2006)18:3(367)

To download this article in PDF format, click HERE.

The Problems of Using Timber in Building Construction

By
Ubani Obinna
-

In the late 19th and early 20th century, timber was used extensively in building construction, till the development of concrete and steel, which saw their use reduced. However, improvements in engineered wood products have resuscitated the use of timber in building construction.

Furthermore, environmental concerns, coupled with modern manufacturing technologies and prefabrication in timber that are able to produce building components that far exceed their old-growth ancestors, have enabled engineers to design and build larger and taller timber buildings than ever before [1].

However, the use of timber in building construction is not without problems and challenges. This is basically because timber is a natural and very complex material. Researchers from the Institute of Structural Engineering (IBK), Swiss Federal Institute of Technology (ETH) have highlighted some of the challenges of using timber in building construction. This was published in volume 227 of Elsevier- Engineering Structures journal.

According to the researchers, despite the numerous beneficial properties and benefits of timber buildings, there are some important challenges that must be highlighted, particularly for tall timber buildings. These are;

(1) Sensitivity to moisture: Timber is a natural grown, hygroscopic material which degrades significantly when it remains wet for a long time. From light swelling to complete loss of structural strength due to fungi attack, it is very important to design timber to remain protected from high moisture in its structural lifetime, particularly for highly loaded components.

decaying timber

(2) Light weight: The timber used in construction is approximately 5 times less dense than reinforced concrete and 15 times less dense than structural steel. The direct advantage of a lighter building with smaller foundations has the pitfall of being much more sensitive to critical lateral loads as the height of the building increases.

(3) Orthotropic: As a natural grown material, the properties of timber are not the same in every direction. Timber is strong along the fibres, but very weak across them. Failing to address this can have catastrophic consequences.

(4) Low stiffness: The timber used in construction is approximately 3 times less stiff than reinforced concrete and 20 times less stiff than structural steel. At increasing heights this can have a severe impact on deflections, accelerations, and occupant comfort. Furthermore, and combined with the orthotropic behaviour, it becomes challenging to construct stiff, moment resisting connections.

(5) Brittleness: Any timber member under tension, bending and shear is brittle in failure, although with careful design ductility can be achieved, in particular in connections with steel fasteners. Brittle behaviour is particularly unwelcome when a structure is called to redistribute loads, for example in the case of an accidental event requiring robustness.

(6) System effects: Although this applies to all materials, the way loads are distributed in a system is less clear, particularly for heterogeneous timber components. In the case of large timber structures subject to abnormal loads and potential damage, better knowledge of system behaviour is important.

(7) Size effects: A significant overall strength reduction is possible in the case of a large structural element. This applies to all materials, however the size effects in very large timber elements are still rather unknown, with preliminary indications that they can be significant.

(8) Time effects: Timber creeps with time, which can be critical in heavily loaded structures like tall timber buildings. Differential settlement in hybrid buildings including loadbearing timber elements is even more challenging.

The above properties make the design of a timber building everything but straightforward, particularly at larger scales.

[Source] Voulpiotis K., Jochen K., Jockwer R., Frangi A. (2021): A holistic framework for designing for structural robustness in tall timber buildings, Engineering Structures, Volume 227 https://doi.org/10.1016/j.engstruct.2020.111432 (http://www.sciencedirect.com/science/article/pii/S0141029620340335)

Question of the day | 07-01-2020

By
Ubani Obinna
-

For the frame loaded as shown above, answer the following question;

(1) What is the horizontal reaction at point D?
[A] 0.5 kN
[B] 1.0 kN
[C] 2.0 kN
[D] 4.0 kN

(2) What is vertical support reaction at point A?
[A] -0.5 kN
[B] 0.5 kN
[C] -1.0 kN
[D] 1.0 kN

(3) What is the most likely bending moment diagram of the structure?

A
B
C

Have you registered for Structville’s first webinar of the year on bridge design? See the flyer below for more.

bridge design webinar

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bridge design 1

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Structville Announces Webinar on Bridge Design | January 2021

By
Ubani Obinna
-

Structville Integrated Services Limited is pleased to announce the first webinar of the year, 2021. The theme of the 2-day webinar will be focused on the “Design of Reinforced Concrete Bridges”. The details are as follows;

Date: 15th (Friday) and 16th (Saturday) January 2021
Time: 7:00 pm daily
Theme: Design of reinforced concrete bridges
Approach: Use of locally available software and manual approach
Platform: Zoom
Fee: NGN 7,000 only

Features

  • Considerations in the design of bridges
  • Components and parts of bridges
  • Selection of structural scheme for bridges
  • Loading/Actions on Bridges
  • Structural analysis of Bridges
  • Structural design of bridges
  • Questions and answers
  • Free design manual containing solved example on bridge design
bridge design

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To get the full structural drawings of a flyover bridge (a practical drawing for execution) in AUTOCAD format, kindly contact the author. Note that the drawing is for a reinforced concrete beam and slab deck system, and consists of both substructure and superstructure drawings. The drawing has been presented to be executed in Nigeria. This will cost you NGN 20,000 only, and it is separate from the webinar fee/requirements.

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Akon’s Smart City Construction set to Begin in 2021 near Dakar, Senegal

By
Ivy Ugorji
-

Senegalese-American R&B singer Akon, is moving forward with plans to build a US$6 billion smart city in his country of birth, Senegal. Having finalized an agreement in January of this year, he claims to have raised at least one-third of the funding needed.

Dubbed Akon City and due for completion in about ten years, Akon’s 2,000-acre (809-hectare) project is being touted as an eco-friendly, mixed-use development open to all members of the African diaspora. Located two hours away from the capital Dakar and south of the West African nation’s relatively new Blaise Diagne International Airport, it may be the first LEED-certified project on the continent, if realized. Construction is slated to begin in 2021.

Akon is promoting his latest endeavor as a boon for both tourism and business in the region and has received considerable support from the Senegalese government, including through a partnership with SAPCO, the state’s tourism agency. While the vision is still low on specifics, Akon City’s official website markets the development as a venue for middle- and high-income residences, education and “training” services, and professional activities. It will run on Akoin, a new cryptocurrency that the singer suggests will be compatible with a variety of smartphones and other cellular devices, as well as with other cryptocurrencies like Bitcoin and Ethereum.

The project is being jointly developed by Los Angeles-based KE International and Dubai-based Bakri & Associates Development Consultants, with CEO Hussein Bakri as its lead architect. According to Business Insider, Bakri & Associates claims that Akon City will take advantage of traditional and newly-developed construction materials, including lighter-weight, more efficient glass-and-steel components. Buildings will be powered by clean energy, likely by Akon Lighting Africa, the singer’s solar energy project that aims to install solar-powered lights across Africa.

A series of dramatic renderings show an agglomeration of futuristic, ribbon-like structures towering into the sky. On the city’s official website, it is described as “an extension of the sea into the land with waves diving deep into the roots of each building.” Akon Tower, the plan’s architectural centerpiece, would far exceed Senegal’s current tallest structure, an 250-foot (76.2-meter) apartment building in Dakar.

As quixotic as it sounds, the plan is not without its skeptics. While certain government officials have praised Akon’s investment in Senegal at a time when the coronavirus pandemic has reduced tourism to a trickle, some local residents doubt the project’s efficacy. Senegal’s economy has experienced breakneck growth in recent decades, but poverty is still prevalent among its 15.4 million people. Akon himself told Business Insider that he sees more “elite local Senegalese” moving into the development first, including himself.

According to Reuters, Mayor Magueye Ndao of Ngueniene, the municipality where part of Akon City will be built, has expressed some cautious optimism about the proposal. Ndao hopes that the project will be realized and that many of its promised services, including youth and job training, will be brought to the area.

Others, though, are convinced that Akon City represents another smart city pipe dream—a project that will inevitably fizzle out due to lack of funding or government support. Xavier Ricou, an architect and the former director of Senegal’s APIX investment agency, told Reuters that Akon City will likely end up like most other city proposals seen in Senegal: just a proposal. If fears about investor commitments and feasibility prove valid, Ricou’s prediction could well be correct.

Construction of a Helical Staircase

By
Ubani Obinna
-

A helical staircase is a staircase that is curved in plan and following the directrix of a full helix. In three-dimensional space, a helical surface is generated by moving a straight line such that the moving line is perpendicular to the helix.

A helix is a smooth space curve with tangent lines at a constant angle to a fixed axis. The construction of a helical staircase is a fairly challenging process that must be handled with care by the contractor. It is important to note that a helical staircase can be achieved using a variety of materials such as reinforced concrete, steel, or timber.

The aim of this article is to provide guidance on how to successfully construct a reinforced concrete helical staircase using locally available materials and tools.

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Design of helical staircase

It is important to note that a reinforced concrete helical staircase can be in two forms;

  • Helical staircase with waist, and
  • Slabless (saw-tooth) helical staircase

A helical staircase with a waist has a reinforced concrete base slab of 100 – 200 mm thickness supporting the treads and the risers (which are usually unreinforced). The waist slab is the main structural component of the staircase where the rebars are placed. On the other hand, just like a typical sawtooth staircase, a slabless helical staircase does not have a waist slab and the structural component of the staircase are the treads and the risers.

helical staircase with waist slab
Curved staircase with waist slab
slabless curved staircase
Curved staircase without waist slab

Construction of a Helical Staircase

The construction of these two types of helical staircases present their own different challenges to a contractor, but generally, it is easier to construct a helical staircase with a waist than a slabless helical staircase. The material and construction effort demand is also greater in the latter than in the former. In this article, however, we are going to focus on the construction of a helical staircase with a waist slab.

For someone that is doing a helical staircase for the first time, it is very important to pay attention to the challenges it presents, especially when sophisticated tools are not available for the setting out. First and foremost, it is important to establish the layout of the staircase as given in the drawing.

The drawing should contain the exact number of treads and risers that will be required for the floor height. Other information such as the inner radius and the width of the staircase should also be provided. This is very important because without a proper setting out, it will be very challenging to get the staircase right.

From the drawing, locate the centre of the staircase and make a permanent mark there. Alternatively, you can get a straight and strong timber pole and attach it to a marine plywood (or 1″ x 12″ plank) base that is firmly attached to the floor. Make sure that the wood is plumb in all directions and properly placed at the centre of the curve.

Using the centre of the curve, establish the inner radius of the staircase and plot it on the floor of the building using any suitable tool/material of your choice. The outer radius of the curve can also be established, especially when the sides of the staircase are not attached to any wall.

setting out of a helical staircase
Typical setting out of a helical staircase

After establishing the curves in plan, you can use 1″ x 12″ or 1″ x 6″ planks (depending on the radius of the curve) that are standing upright to form the lines of the curves and to also establish a temporal wall for markings. The planks should be firmly nailed to the floor of the building using 2″ x 3″ wood and braced properly. The height of the planks should be determined with reference to how the staircase rises. This approach however consumes a lot of material.

planks for helical staircase
Installation of planks along the curve of a staircase

After joining the planks side by side, establish the markings of the staircase (tread and riser) on the planks which serves as a temporal wall. Make sure to establish the thickness of the waist on the planks too.

Construction of a helical staircase
Installation of planks along the curve of a staircase

After the markings, insert the stringers and support or brace the formwork with 2″ x 3″ wood. Bamboo or another strong material such as 2″ x 4″ wood can be used as props. The sheating of the staircase (soffit) can then be formed on the framing using suitable plywood. Depending on the configuration, the inner/outer curve of the staircase should be covered with another plywood with the equivalent height of the riser. Alternatively, the markings on the plank can be used if the height is sufficient but this might give a rough edge finish.

After the soffit of the formwork has been installed, the iron bender should neatly install the reinforcements according to the design specifications. The risers of the staircase can be installed by nailing perfectly cut planks to the edges of the inner and outer curves. To get this very well, you can tie a rope to the timber at the centre of the curve which you can easily adjust round it, and use it to trace the line of the plank that will form the riser which lies from the inner radius to the outer radius. Note that the tread is bigger in the outer radius because of a wider circumference/length.

Alternatively, you can prepare your framing by laying the 2”x 6” timber stringers from the central pole to your outer line of shores. The timber stringers should be laid flat on the floor and sawn accurately to dimension. The top elevation of the stringers will be determined by the thickness of the concrete slab and the thickness/arrangement of the formwork.

Once the stringers are complete and checked for position, level, and length, install 2” x 4” timber joists spaced at 400 mm centre to centre. After the stringers and joists are securely fixed in place, proceed with the plywood sheathing, and finally the edge forms. This approach requires knowing the dimensions as accurately as possible and cutting the timber forms properly.

formwork 2
Alternative framing of a helical staircase without planks

Once the formwork is complete, well secured, and inspected, the next step would be to install the reinforcement. Bars must be cut to the right length and placed neatly in size and spacing according to the construction drawings. Use chairs to maintain the proper clearance to the forms and observe the proper concrete coverage at the bottom of the reinforcing bars.

reinforcement of helical staircase
Typical reinforcement installation of a helical staircase

Following the inspection and approval of the reinforcing bar comes the careful preparation for pouring. We recommend a minimum concrete strength of 25 MPa at 28 days. The concrete mix should be properly designed, water content carefully controlled, vibrated, and compacted to its maximum density.

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