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Sheet Pile Walls and their Uses

By
Ivy Ugorji
-

Sheet pile walls are thin retaining walls constructed to retain earth, water, or any other fill material. They are typically thinner than masonry or reinforced concrete retaining walls such as cantilever retaining walls and can be constructed using materials such as steel, concrete, or timber. Timber sheet pile walls are used for resisting light loads and for temporary works such as braced sheeting in cuts and must be joined using tongue and groove joint. Reinforced concrete sheet piles are precast concrete members with tongue and groove joints. The piles are relatively and displace a huge volume of soil during driving.

The most common type of sheet pile walls is steel sheet piles. They have the advantage of being resistant to high driving stresses developed in hard or rocky formations. They are lighter in weight and can be reused several times in any type of condition. Steel sheet piling is used in many types of temporary works and permanent structures. The selected section must be designed to provide the maximum strength and durability at the lowest possible weight and good driving qualities.

Uses of sheet pile walls

(1) River control structures and flood defense
Steel sheet piling has traditionally been used for the support and protection of river banks, lock and sluice construction, and flood protection. Ease of use, length of life, and the ability to be driven through water make piles the obvious choice.

sheet pile in water protection
Sheet pile wall in river control / shore protection

(2) Ports and harbours
Steel sheet piling is a tried and tested material to construct quay walls speedily and economically. Steel sheet piles can be designed to cater for heavy vertical loads and large bending moments.

sheet pile 4
Sheet pile wall in port/harbour construction

(3) Pumping stations
Historically used as temporary support for the construction of pumping stations, sheet piling can be easily designed as the permanent structure with substantial savings in time and cost. Although pumping stations tend to be rectangular, circular construction should be considered as advantages can be gained from the resulting open structure.

sheet pile in construction of pump station
Sheet pile wall in pump station construction

(4) Bridge abutments
Abutments formed from sheet piles are most cost-effective in situations when a piled foundation is required to support the bridge or where the speed of construction is critical. Sheet piling can act as both foundation and abutment and can be driven in a single operation, requiring a minimum of space and time for construction.

sheet pile wall in bridge abutment
Sheet pile wall in bridge abutment

(5) Road widening retaining walls
Key requirements in road widening include minimised land take and speed of construction – particularly in lane rental situations. Steel sheet piling provides these and eliminates the need for soil excavation and disposal.

sheet pile road retaining
Sheet pile wall retaining earth embankment

(6) Basements
Sheet piling is an ideal material for constructing basement walls as it requires minimal construction width. Its properties are fully utilised in both the temporary and permanent cases and it offers significant cost and programme savings. Sheet piles can also support vertical loads from the structure above.

sheet pile being used in basement
Sheet pile wall in basement

(7) Underground car parks
One specific form of basement where steel sheet piling has been found to be particularly effective is for the creation of underground car parks. The fact that steel sheet piles can be driven tight against the boundaries of the site and the wall itself has minimum thickness means that the area available for cars is maximised and the cost per bay is minimised.

sheet pile in car parks
Sheet pile wall in underground car park

(8) Containment barriers
Sealed sheet piling is an effective means for the containment of contaminated land. A range of proprietary sealants is available to suit particular conditions where extremely low permeability is required.

contamination containment
Sheet pile wall used in ground contamination containment

(9) Load-bearing foundations
Steel sheet piling can be combined with special corner profiles to form small diameter closed boxes which are ideally suited for the construction of load-bearing foundations. Developed for use as a support system for motorway sign gantries, the concept has also been used to create foundation piles for bridges.

(10) Temporary works
For construction projects where a supported excavation is required, steel sheet piling should be the first choice. The fundamental properties of strength and ease of use – which steel offers – are fully utilised in temporary works. The ability to extract and re-use sheet piles makes them an effective design solution. However, significant cost reductions and programme savings can be achieved by designing the temporary sheet pile structure as the permanent works.

sheet piles in temporary construction
Sheet pile wall in temporary works

Lateral-Torsional Buckling of Steel Beams

By
Ubani Obinna
-

When a laterally unrestrained beam is subjected to bending about the major axis, there is a need to check for lateral-torsional buckling. Lateral-torsional buckling is a type of buckling that involves a combination of lateral deflection of beams and twisting, and typically occurs in open cross-sections.

The phenomenon occurs on the compression flange of the member and depends on factors such as the loading conditions, lateral restraint conditions, and geometry of the compression flange. Steel beams with sufficient lateral restraint to the compression flange may not need to be checked for lateral-torsional buckling. Cross-sections such as circular hollow sections or square box sections are also not susceptible to lateral-torsional buckling.

Lateral restraint to a steel beam in a building may be provided by;

  • Concrete floor slab on beams (inclusive of composite metal deck)
  • Sheeting or metal decking on roofs (spanning perpendicular to the beam)
  • Secondary beams to primary beams
  • Purlins on rafters
  • Bracings, etc
Decking spanning perpendicular to beam
Decking spanning perpendicular to beam (Gardner, 2011)
Beams supporting concrete slabs 1
Beams supporting concrete slabs (Garner, 2011)

In general, the bracing system assumed to provide effective lateral restraint must be capable of resisting an equivalent stabilising force qd (defined in clause 5.3.3(2) of EC3), the value of which depends on the flexibility of the bracing system.

Design for Lateral-Torsional Buckling

The design bending moment is denoted by MEd (bending moment design effect), and the lateral-torsional buckling resistance by Mb,Rd (design buckling resistance moment). The design requirement is that MEd must be shown to be less than Mb,Rd, and checks should be carried out on all unrestrained segments of beams (between the points where lateral restraint exists).

The design buckling resistance of a laterally unrestrained beam (or segment of beam) should be taken as;

Mb,Rd = χLTWyfy/γm0

where Wy is the section modulus appropriate for the classification of the cross-section, as given below. In determining Wy, no account need to be taken for fastener holes at the beam ends.


Wy = Wpl,y for Class 1 or 2 cross-sections
Wy = Wel,y for Class 3 cross-sections
Wy = Weff,y for Class 4 cross-sections
χLT is the reduction factor for lateral torsional buckling.

Solved Example

A simply supported primary beam is required to span 7m and to support two secondary beams as shown in the figure below. The secondary beams are connected through fin plates to the web of the primary beam, and full lateral restraint may be assumed at these points. Check the suitability of UKB UB 533 x 210 x 92 for the primary beam assuming grade S275 steel.

loaded laterally unrestrained beam

Let ∑MB = 0;
7VA – (350 × 5.7) – (375 × 1.3) = 0
VA = 354.64 kN

Let ∑MA = 0;
7VB – (350 × 1.3) – (375 × 5.7) = 0
VB = 370.36 kN

MB = 354.64 × 1.3 = 461.032 kNm
MC = (354.64 × 5.7) – (350 × 4.4) = 481.619 kNm

Bending moment and shear

UB 533 x 210 x 92

E = 210000 N/mm2
G = 81000 N/mm2

Properties of UB 533 x 210 x 92
h = 533.1 mm
b = 209.3 mm
tw = 10.1 mm
tf = 15.6 mm
r = 12.7 mm
A = 11700 mm2
Iy = 55200 cm4
Iz = 2390 cm4
IT = 7.57 x 106 mm4
IW = 1.6 x 1012 mm6
Wel,y = 2070 cm3
Wel,z = 228 cm3
Wpl,y = 2360 cm3
Wpl,z = 356 cm3

Section Classification

ε = √235/fy = √235/275 = 0.92

Web
cw = d = h – 2tf – 2r = 476.5 mm
cw/tw = 47.18
The limit for class 1 is 72ε = 66.24
cw/tw = 47.18 < 66.24
Therefore the web is class 1 Plastic

Flange
c = [(b – tw – 2r)]/2 = [209.3 – 10.1 – (2 × 12.7)]/2 = 86.9 mm
cf/tf = 5.57
The limit for class 1 is 9ε = 9 × 0.92 = 8.28
5.57 < 8.28
Therefore the flange is Class 1 (plastic)
Therefore the beam section is class 1

Bending Resistance (Clause 6.2.5 BS EN 1993-1-1)
Mpl,y,Rd = (Wplfy)/γm0 = (2360 × 103 × 275)/1.0 × 10-6 = 649 kNm

Maximum moment on the beam My,Ed = 481.619 kNm
481.619 < 649 kNm Ok

Shear Resistance (Clause 6.2.6 EN 1993-1-1)
Shear area Av = A – 2btf + (tw + 2r)tf but not less than ηhwtw

Av = 11700 – (2 × 209.3 × 15.6) + (10.1 + 2 × 12.7) × 15.6 = 5723.64 mm2
ηhwtw = 1.0 × 501.9 × 10.1 = 5069.19 mm2

Therefore take Av = 5723.64 mm2
Vpl,Rd = [Av(fy⁄√3)]/γm0 = [5723.64 (275⁄√3)]/1.0 × 10-3 = 908.749 kN
VEd = 370.36 kN < 908.749 kN Ok

Bending and Shear Interaction (clause 6.2.8 BS EN 1993-1-1)
When shear force and bending moment act simultaneously on a cross-section, the effect of the shear force can be ignored if it is smaller than 50% of the plastic shear resistance.

0.5Vpl,Rd = 0.5 × 908.749 = 454.374 kN
370.36 kN < 454.374 kN, therefore the effect of shear on the moment resistance can be ignored.

Lateral torsional buckling (segment B – C)
Lcr,T = 4.4 m
h/b = 533.1/209.3 = 2.54 > 2.0

Therefore select buckling curve : c = 0.49 (Table 6.5 EC3)

Moment diagram of the point between restraints

ratio of end moment

Ratio of end moments ψ = 461.032/481.619 = 0.957

C1 = 1.88 – 1.40ψ + 0.52ψ2 = 1.01 < 2.7 Okay

Mcr = C1 × (π2EIz)/(kL2 ) × [Iw/Iz + (kL2GIT)/(π2EIz )]0.5

Mcr = 1.01 x [(π2 × 210000 × 2390 × 104)/44002] × [(1.6 × 1012)/(2390 × 104) + (44002 × 81000 × 75.7 × 104)/(π2 × 210000 × 2390 × 104)]0.5 x 10-6 = 779.182 kNm

Non-dimensional lateral torsional slenderness λLT

λLT = √[(Wpl,yfy)/Mcr ] = √[(2360000 × 275)/(779.182 × 106] = 0.912

λLT,0 = 0.4, and β = 0.75
ϕLT = 0.5[1+ αLT (λLT – λLT,0) + βλLT2]
ϕLT = 0.5[1 + 0.49(0.912 – 0.4) + 0.75 × 0.9122] = 0.937

χLT = 1/[ϕLT + √(ϕLT2 – βλLT2)] but χ ≤ 1.0

χLT = 1/([0.937 + √(0.9372 – 0.75 × 0.9122)] = 0.6938

Mb,Rd = χLTWyfy/γm0 = (0.6938 × 2360 × 103 × 275)/1.0 × 10-6 = 450.33 kNm

MEd/Mb,Rd = 481.619/450.33 = 1.06 > 1.0

Therefore the section is not okay to resist lateral torsional buckling on the primary beam.

References
Gardner L. (2011): Stability of Steel Beams and Columns (In Accordance with the Eurocodes and UK National Annex). SCI – Steel Construction Institute, Berkshire UK.



Deflection of Structures According to Eurocode 2

By
Ubani Obinna
-

Checking for deflection is an important serviceability limit state (SLS) requirement in the design of structures. This check ensures that the structure does not deflect excessively in a manner that will impair the appearance, cause cracking to partitions and finishes, or affect the functionality or stability of the structure. In Eurocode 2, the deflection of a structure may be assessed using the span-to-effective depth ratio approach, which is the widely used method. It is also allowed to carry out rigorous calculations in order to determine the deflection of a reinforced concrete structure, which is then compared with a limiting value.

According to clause 7.4.1(4) of EN 1992-1-1:2004, the appearance of a structure (beam, slab, or cantilever) may be impaired when the calculated sag exceeds span/250 under quasi-permanent loads. However, span/500 is considered an appropriate limit for good performance.

Using the span-to-effective depth approach, the deflection of a structure must satisfy the requirement below;

Allowable l/d = N × K × F1 × F2 × F3 ≥ Actual l/d

Where;
N is the basic span-to-effective depth ratio which depends on the reinforcement ratio, characteristic strength of the concrete, and the type of structural system. The expressions for calculating the limiting value of l/d are found in exp(7.16) of EN 1992-1-1:2004. The expressions are given as follows;

l/d = K[11 + 1.5√fck(ρ0/ρ) + 3.2√fck(ρ0/ρ – 1)1.5] if ρ ≤ ρ0

l/d = K[11 + 1.5√fck(ρ0/(ρ – ρ’)) + 0.0833√fck(ρ0/ρ)0.5] if ρ > ρ0

Where:
l/d is the limit span/depth ratio
K is the factor to take into account the different structural systems
ρ0 is the reference reinforcement ratio = √fck /1000
ρ is the required tension reinforcement ratio at midspan to resist the moment due to the design loads (at supports for cantilevers)
ρ’ is the required compression reinforcement ratio at midspan to resist the moment due to the design loads (at supports for cantilevers)
fck is the characteristic compressive strength of the concrete in N/mm2

The values of K for different structural systems are given in Table 1;

Table 1: Values of K for different structural systems

Structural SystemK
Simply supported beam, one or two way spanning simply supported slab1.0
End span of continuous beam or one-way continuous slab or two-way spanning slab continuous over one long side1.3
Interior span of beam or one way or two-way spanning slab1.5
Slab supported on columns without beams (flat slab)1.2
Cantilever0.4

Some design aids are available for the evaluation of the limiting span/effective ratio. This is shown in Table 2 below (culled from Goodchild, 2009) and has been derived for K = 1.0 and ρ’ = 0.

Table 2: Basic ratios of span-to-effective-depth for members without axial compression (Goodchild, 2009)

basic span effective depth ratio table EC2


For the table above, ρ = As/bd (note that As is the area of steel required and not the area of steel provided). For T beams, ρ is the area of reinforcement divided by the area of concrete above the centroid of the tension reinforcement.

F1 = factor to account for flanged sections.
When beff/bw = 1.0, F1 = 1.0
When beff/bw is greater than 3.0, F1 = 0.8.
Intermediate values of beff/bw can be interpolated between 1.0 and 3.0

F2 = factor to account for brittle partition in long spans.
In flat slab where the longer span is greater than 8.5m, F2 = 8.5/leff
In beams and slabs with span in excess of 7.0m, F2 = 7.0/leff

F3 = factor to account for service stress in tensile reinforcement = 310/σs ≤ 1.5
Conservatively, if a service stress of 310 MPa is assumed for the designed reinforcement As,req, then F3 = As,prov/As,req ≤ 1.5

More accurately, the serviceability stress in the reinforcement may be stimated as follows;

σs = σsu[As,req/As,prov](1/δ)

Where;
σsu is the unmodified SLS steel stress taking account γM for reinforcement and of going from ultimate actions to serviceability actions.
σsu = fyk/γs(Gk + ψ2Qk)/(1.25Gk + 1.5Qk)
As,req/As,prov = Area of steel required divided by the area of steel provided
1/δ = factor to un-redistribute ULS moments

References
(1) EN 1992-1-1:2004 – Eurocode 2: Design of concrete structures – Part 1-1: General rules and rules for buildings. European Committee For Standardization
(2) Goodchild C.H. (2009): Worked Examples to Eurocode 2: Volume 1 (For the design of in-situ concrete elements in framed buildings to BS EN 1992-1-1:2004 and its UK National Annex 2005). MPA – The concrete Centre

Cover image credit: Sharooz et al (2014): Flexural Members with High-Strength Reinforcement: Behavior and Code Implications. ASCE Journal of Bridge Engineering Volume 19 Issue 5


Effects of Diesel Contamination on the Engineering Properties of Natural Soils

By
Ubani Obinna
-

Soil contamination is a serious problem and major environmental problem worldwide. Oil contamination of soils can occur as a result of oil drilling and exploration operations, leakage of storage tanks and wells, tanker accidents, spillage during transportation, etc. Research studies have shown that the presence of hydro carbons can affect the engineering properties of soils. The degree of alteration has been found to depend on the type of soil, the type of contaminant, and the concentration of the contaminant.

In an article published in the International Journal of Geo-Engineering from the Department of Civil Engineering, Federal University of Sao Carlos, Brazil, researchers demonstrated the capacity of lime treatment in improving the engineering properties of diesel contaminated soils. The research was geared towards demonstrating ways to reuse non-hazardous petroleum-contaminated soils in civil engineering applications, such as asphalt concrete, cold-mix asphalt, construction material, roadway sub-bases and alternative daily cover materials for landfills.

To carry out the study, the researchers collected an uncontaminated coarse grained soil sample from Sao Paulo, Brazil, and subjected it to laboratory tests such as specific gravity test, Atterberg limit test, particle size distribution test, compaction test, CBR, unconfined compression test, and pH test. Based on the Unified Soil Classification System, the coarse grain soil is classified as SC soil and mineralogical characteristics include quartz, feldspars, iron oxides and Kaolinite Hematite and Chlorite as secondary minerals. The diesel oil used in the study as the organic contaminant was a commercially available diesel oil named S500. Diesel properties include relative density of 0.834 at 20°C, kinematic viscosity of 2.0–5.0 cm2 s−1 at 40°C, pH of 6.0 and biodiesel up to 7.0% v/v. The diesel oil was directly mixed with dry soil to prepare oil-contaminated soil samples. The different ratios of diesel to dry soil were set to 4, 8, 12 and 16% to simulate different levels of contamination in the field. 

From the study, it was observed that the presence of diesel affected consistency limits of natural soil, while pH values were not affected. The liquid limit and plastic limit of the soil increased with increase in diesel content, thereby leading to increase in the plasticity index. High liquid limit and plasticity index is an indication of weak soil. No significant changes were observed in ΔpH values, remaining negative for all diesel contents, thus not altering soil electric charges with diesel content increase.

From the compaction test result, it was observed that the compaction properties were altered due to the presence of diesel contamination. Significant decrease in maximum dry unit weight and an increase in optimum moisture contents with diesel addition were observed. In the unconfined compression test, about 70% reduction in strength was observed as the diesel content increased.

compaction curve of lime contaminated soil
Compaction analysis: a aspect of 8% diesel contaminated soil; b curves of natural and diesel-contaminated soils (Correia et al, 2020)

However, when the contaminated soil was treated with lime, the pH and ΔpH values increased, while the consistency limits reduced. After lime addition, calcite minerals were predominate in natural and oil-contaminated soils. The addition of lime in the diesel-contaminated soil resulted in a flocculated structure soil similar to that of lime-treated natural soil. Compaction parameters of diesel-contaminated soils were less altered by the presence of lime. Lime allowed 8% diesel-contaminated to sufficiently increase in 300% the UCS soil and recover some mechanical properties of natural soil, evidencing possible carbonation reactions in the mixture.

Reference
Correia N., Portelinha F.N.M, Mendes I.S., Batista da Silva J. W. (2020): Lime treatment of a diesel‑contaminated coarse‑grained soil for reuse in geotechnical applications. International Journal of Geo-Engineering (2020) 11(8). https://doi.org/10.1186/s40703-020-00115-2

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Design of Steel Beams for Combined Bending and Shear

By
Ubani Obinna
-

It is very common to see bending moment and shear force act simultaneously in structural members. However, when the shear force under consideration is low, the effect of shear on moment resistance of the section can be ignored. Clause 6.2.8(2) of Eurocode 3 (EN 1993-1-1:2005) states that provided the applied shear force is less than half the plastic shear resistance of the cross-section its effect on the moment resistance may be neglected. The exception to this is where shear buckling reduces the resistance of the cross-section.

When the applied shear force is greater than half the plastic shear resistance of the cross-section, the moment resistance should be calculated using a reduced design strength for the shear area using equation (1).

fyr = (1 – ρ)fy —– (1)

Where;
ρ = (2VEd/Vpl,Rd – 1)2 for (VEd > 0.5Vpl,Rd)

An alternative to the reduced design strength for the shear area, defined by equation (1), which involves somewhat tedious calculations, is equation (2). Equation (2) may be applied to the common situation of an I section (with equal flanges) subjected to bending about the major axis. In this case, the reduced design plastic resistance moment allowing for shear is given by;

My,V,Rd = [(Wpl,y – ρAw2/4tw)fy]/γm0 —– (2)

Where;
Aw = hwtw

Solved Example

A short-span (1.5 m), simply supported, laterally restrained beam is to be designed to carry a central point load of 900 kN, as shown below. Assess the suitability of 406 x 178 x 74 UKB in grade S275 steel to carry the load.

Steel beam with high shear load

For the beam loaded as shown above;

MEd = PL/4 = (900 x 1.5)/4 = 337.5 kNm
VEd = P/2 = 900/2 = 450 kN

Since an advanced UK beam S275 is to be used for this design;
fy = 275 N/mm2
γm0 = 1.0 [Clause 6.1(1) NA 2.15 BS EN 1993-1- 1:2005]

fa

Try section 406 x 178 x 74 UKB

Read Also….
Design of Steel Beams to BS 5950 – 1: 2000
Structural Analysis of Compound Arch-Frame Structure

Properties
h = 412.8 mm; b = 179.5 mm; tw = 9.5 mm; tf = 16.0 mm r = 10.2mm; A = 9450 mm2; Wpl,y = 1501000 cm3

hw = h – 2tf = 380.8 mm

E (Modulus of elasticity) = 210000 N/mm2  [Clause 3.2.6(1)]

Classification of section
ε = √(235/fy) = √(235/275) = 0.92 (Table 5.2 BS EN 1993-1- 1:2005)

Outstand flange: flange under uniform compression cf = (b – tw – 2r)/2 = [179.5 – 9.5 – 2(10.2)]/2 = 74.8 mm

cf/tf = 74.8/16.0 = 4.68

The limiting value for class 1 is c/tf  ≤ 9ε = 9 × 0.92
4.68 < 8.28
Therefore, outstand flange in compression is class 1

Internal Compression Part (Web under pure bending)
cw = d = h – 2tf – 2r = 412.8 – 2(16) – 2(10.2) = 360.4 mm
cw/tw = 360.4/9.5 = 37.94

The limiting value for class 1 is c/tw ≤ 72ε = 72 × 0.92 = 66.24
37.94 < 66.24
Therefore, the web is plastic. Therefore, the entire section is class 1 plastic.

Bending Moment Resistance
MEd/Mc,Rd ≤  1.0 (clause 6.2.5(1))

Mc,Rd = Mpl,Rd = (Mpl,y × Fy)/γm0

Mc,Rd = Mpl,Rd = [(1501 × 103 × 275)/1.0] × 10-6 = 412 kNm

MEd/Mc,Rd = 337.5/412 = 0.819 < 1.0 Ok

Shear Resistance (clause 6.6.2)
The basic design requirement is;

VEd/Vc,Rd ≤  1.0

Vc,Rd = Vpl,Rd = Av(Fy / √3)/γm0 (for class 1 sections)
For rolled I-section with shear parallel to the web, the shear area is;

Av = A – 2btf + (tw + 2r)tf (for class 1 sections) but not less than ηhwtw

Av = 9450 – (2 × 179.5 × 16) + [9.5 + 2(10.2)] × 16 = 4184 mm2
η = 1.0 (clause NA.2.4 of the UK National Annex)
ηhwtw = (1.0 × 380.8 × 9.5) = 3618 mm2
4184 > 3618
Therefore, Av = 4184 mm2

The shear resistance is therefore;
Vc,Rd = Vpl,Rd = [4184 × (275/ √3)/1.0]  × 10-3 = 664.3 kN

VEd/Vpl,Rd = 450/664.3 = 0.677 < 1.0 Ok

Since VEd (450 kN) > 0.5Vpl,Rd (332.15 kN), the reduced moment resistance due to shear needs to be calculated.

Shear Buckling
Shear buckling of the unstiffnened web will not need to be considered if;

hw/tw ≤  72ε/η

hw/tw = 380.8/9.5 = 40.1
72ε/η  = (72 ×  0.92)/1.0  = 66.6

40.1 < 66 Therefore shear buckling need not be considered.

Resistance of cross-section to combined bending and shear
The applied shear force is greater than half the plastic shear resistance of the cross-section, therefore a reduced moment resistance My,V,Rd must be calculated. For an I section (with equal flanges) and bending about the major axis, clause 6.2.8(5) and equation (6.30) may be utilised.

My,V,Rd = [(Wpl,y – ρAw2/4tw)fy]/γm0 but My,V,Rd ≤ Mc,Rd

ρ = (2VEd/Vpl,Rd – 1)2 = {[(2 x 450)/664.3] – 1}2 = 0.126

Aw = hwtw = 380.8 x 9.5 = 3617.6 mm2

My,V,Rd = [(1501000 – 0.126 x 3617.62 /4 x 9.5) x 275]/1.0 x 10-6 = 400.84 kNm

400.84 kNm > 337.5 kNm Therefore section is adequate to resist combined bending and shear force on it.

Effect of Impact Load on Lap Length of Reinforcements in RC Structures

By
Ubani Obinna
-

Lap length and anchorage length of reinforcements is very important for continuity, strength, and ease of construction in reinforced concrete structures. When reinforcements are spliced (lapped), forces are transferred from one bar to another through the bond between the reinforcements and concrete. There are provisions for calculation of lap length in various codes of practice which are based on static loads, but researchers from Hunan University, Changsha, China have presented a new study on the effect of impact load on splice (lap) length. The study was published in the International Journal of Concrete Structures and Materials (Springer).

Impact load can occur in structures such as bridges when there is a vehicle or ship collision. Other instances are terror attacks on buildings, a mass of rock falling on retaining structures, etc. The impact resistance of RC structures depends on the material properties of concrete and reinforcing bars under high-strain rate (Hwang et al, 2020). Previous studies have shown that impact loads with short duration can increase the material properties of concrete and reinforcing bars in different ratio, which change ductile behavior into brittle behavior in RC structures. As a result, the load transfer from reinforcements to the concrete is important for the structural integrity and ductility of a structure under impact load. The research was therefore aimed at studying the bond between concrete and reinforcement under impact load.

To carry out the study, the researchers performed drop hammer test on 24 specimens of reinforced concrete beams with the reinforcements lapped at the mid-span. However, the lap lengths used in the study were smaller than the requirements of the ACI 318–19 code of practice within the range of 31 – 69%. The test parameters of the study were the drop height, hammer mass, bar diameter, and splice length, while the structural performances evaluated were the impact force, deflection, failure mode, and strain of reinforcing bar. The specimen set up (reinforced concrete beams) is shown in Figure 1 while the experimental set up (drop hammer test) is shown in Figure 2;

specimen setup
Figure 1: Test specimen set up (Hwang et al, 2020)
experimental set up
Figure 2: Experimental set up (Hwang et al, 2020)

The results of the study showed that the peak impact force, maximum mid-span defection, residual mid-span defection, maximum strain of reinforcing bars, and residual strain of reinforcing bars increased as the impact energy (i.e., impact velocity) increased. It was observed that bond failure occurred in the specimens under impact load since the lap lengths were smaller than that required for static loads. Furthermore, the tensile strength of bar splices was observed to be greater than that of static load due to strain rate effect under low impact energy. However, under high impact energy, the tensile strength of bar splices was higher than the dynamic yield energy but bond failure occurred due to the destruction of the concrete cover.

In the case of the same bar anchorage length, the larger diameter bars showed the larger impact resistance, but the bond strength was observed to be identical regardless of the bar diameter. In the case of the same bar diameter, the longer anchorage length showed the larger tensile strength of bar splices.

Ultimately, the authors proposed modifcation factors for bar stress prediction in existing methods to address the effect of effective impact energy on the bond strength. The proposed method was observed to predicted the test results very well.

Reference(s)
Hwang H., Yang F., Zang L., Baek J., Ma G. (2020): Effect of Impact Load on Splice Length of Reinforcing Bars. International Journal of Concrete Structures and Materials (2020) 14:40. https://doi.org/10.1186/s40069-020-00414-z

© The Author(s) 2020. The original version of this article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.

Static and Kinematic Determinacy of Structures

By
Ubani Obinna
-

A structure is a system of connected parts designed to resist load. Kinematical analysis is very important in the evaluation of the capacity of a structure to resist external loads. This concept is concerned with the rigid (unchangeable) part of a structure which may be made up of straight members, curved members, etc, and how they are connected to each other. The connection of members in a structure could be by the means of links, hinges, or fixed joints. Hence, the arrangement of structural members and their connection conditions define whether a structure is geometrically unchangeable (stable) or geometrically changeable (unstable).

In a geometrically unchangeable structure, any distortion of the structure occurs only with the deformation of its members. It means that this type of structure with absolutely rigid members cannot change its form. The simplest geometrically unchangeable structure is a triangle, which contains the pin-joined members. However, in a geometrically changeable structure, any finite distortion of the structure occurs without the deformation of its members. The simplest geometrically changeable system is formed as hinged four-bar linkage.

SIMPLE TRUSS
A simple truss is a geometrically unchangeable structure
4 HINGED BAR
A hinged four-bar linkage is a geometrically changeable structure

It may be recalled from statics that a structure or one of its members is in equilibrium when it maintains a balance of force and moment. In general, this requires that the force and moment equations of equilibrium be satisfied in a coplanar system of forces as follows;

∑Fx = 0
∑Fy = 0
∑Mi = 0

Here ∑Fx and ∑Fy represent, respectively, the algebraic sums of the x and y components of all the forces acting on the structure or one of its members, and ∑Mi represents the algebraic sum of the moments of these force components about an axis perpendicular to the x–y plane (the z-axis) and passing through point i. When all the forces in a structure can be determined strictly from the three equations of equilibrium, the structure is referred to as statically determinate. Structures having more unknown forces than available equilibrium equations are called statically indeterminate.

The required constraint of a structure is such a constraint which when removed changes the kinematical characteristic of the structure. It means that the entire unchangeable structure transforms into changeable. Note that constraint assumes not only the supports but the elements of a structure as well. For example, the elimination of any member of a simple truss can transforms it from unchangeable to changeable. A redundant constraint of a structure is such a constraint which when removed do not change a kinematical characteristic of the structure. It means that the entire unchangeable structure remains the unchangeable, and changeable structure remains a changeable one. Therefore, the presence of a redundant constraint is a characteristic of a statically indeterminate structure.

Generally, for a coplanar structure there are at most three equilibrium equations for each part, so that if there is a total of n parts and r force and moment reaction components, we have;

r = 3n (statically determinate structures)
r > 3n (statically indeterminate structures)
r < 3n (unstable)

Trusses

There are some formulas available for determining the degree of static determinacy of trusses.

D = s + m – 2n (plane trusses)
D = s + m – 3n (space trusses)

Where;
D = Degree of static indeterminacy
s = Number of support reactions
m = number of members
n = number of nodes

Examples

static determinacy of a truss

s = 3
m = 15
n = 9
D = 3 + 15 – 2(9) = 0 (statically determinate)

truss example 2

s = 3
m = 11
n = 7
D = 3 + 11 – 2(7) = 0 (statically determinate)

truss 3

s = 3
m = 26
n = 14
D = 3 + 26 – 2(14) = 1 (statically indeterminate to the first order)

truss 4

s = 2
m = 10
n = 7
D = 2 + 10 – 2(7) = -2 (unstable)

Beams and frames

For beams and framed structures, the formula below can be used to check the degree of static indeterminacy of the structure.

D = s + i + 3m – 3p

Where;
D = Degree of static indeterminacy
s = Number of support reactions
i = number of internal forces in hinges (usually 2 per internal hinge)
m = number of closed loops without hinges
p = number of parts

Examples

(1)

Beam 1

s = 6
i = 4 (2 internal hinges)
m = 0 (no closed loop)
p = 3 (3 parts)
D = 6 + 4 + 0 – 3(3) = 1 (statically indeterminate to the first order)

(2)

Beam 2 1

s = 6
i = 6 (3 internal hinges)
m = 0 (no closed loop)
p = 4 (4 parts)
D = 6 + 6 + 0 – 3(4) = 0 (statically determinate)

(3)

Beam 4

s = 5
i = 4 (2 internal hinges)
m = 0 (no closed loop)
p = 3 (3 parts)
D = 5 + 4 + 0 – 3(3) = 0 (statically determinate)

(4)

frame 1

s = 4
i = 2 (1 internal hinge)
m = 0 (no closed loop)
p = 2 (2 parts)
D = 4 + 2 + 0 – 3(2) = 0 (statically determinate)

(5)

frame 2

s = 6
i = 0 (no internal hinge)
m = 0 (no closed loop)
p = 1 (1 part)
D = 6 + 0 + 0 – 3(1) = 3 (statically indeterminate to the 3rd order)

(6)

frame 3

s = 6
i = 0 (no internal hinge)
m = 3 (3 closed loops)
p = 1 (1 part)
D = 6 + 0 + 3(3) – 3(1) = 12 (statically indeterminate to the 12th order)

3D Frames

For 3D structures, the formula below can be used to check the degree of static indeterminacy of the structure.

D = s + i + 6m – 6p

Where;
D = Degree of static indeterminacy
s = Number of support reactions
i = number of internal forces in hinges (usually 2 per internal hinge)
m = number of closed loops without hinges
p = number of parts

Example

3D FRAME

s = 11
i = 0 (no internal hinge)
m = 1 (1 closed loop)
p = 1 (1 part)
D = 11 + 0 + 6(1) – 6(1) = 11 (statically indeterminate to the 11th order)

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Recommended Depth of Boring for Site Investigation

By
Ivy Ugorji
-

Soil is a complex and highly variable product of nature, and the extent of this variation is known in advance. As a result, site investigation is very necessary before the commencement of any construction project. According to BS 5930:1999+A2:2010, the depth of boring and extent of the ground investigation is determined by the character and variability of the ground and groundwater, the type of project and the amount and quality of existing information.

It is usually important to know the depth that soil boring exploration should go, in order to determine all the parameters that will affect the behaviour of the soil/structure when the construction is completed. This will also guide the planning of the site investigation.

According to clause 12.7.1 of BS 5930:1999+A2:2010, the depth of boring and soil exploration depends on how new building work significantly affects the ground and groundwater, or is affected by them. Therefore, the boring should be undertaken below all soil deposits that may be unsuitable for foundations purposes, e.g. made ground and weak compressible soils, including weak strata overlain by a layer of higher bearing capacity.

Depth of Boring

The depth of boring and exploration should go through compressible cohesive soils likely to contribute significantly to the settlement of the proposed works, normally to a depth where stress increases cease to be significant, or deeper. If rock is found, a penetration of at least 3 m in more than one borehole may be required to establish whether bedrock or a boulder has been encountered, unless prior knowledge of the local geology obviates this.

The recommended depth of soil exploration for different construction purposes are therefore given below;

Foundations for Buildings

For the foundation of buildings and other structures, the depth of boring should be at least 1.5 times the width of the loaded area, unless the imposed stress change becomes insignificant when compared with the strength and stiffness of the ground at a lesser depth, e.g. in strong rock.

For foundations near the surface such as pad foundations, the loaded area is considered as either:

(i) the area of an individual footing; or
(ii) the plan area of the structures, where the spacing of foundation footings is less than about 3 times the breadth, or where the floor loading is significant; or
(iii) the area of a foundation raft.

In each case, the depth should be measured below the base of the footing or raft.

For pile foundations, a preliminary analysis of the possible lengths of different types and sizes of pile should be made using the desk study information. In addition, the following factors should be taken into account when planning the depth of boring.

(i) The depth of boring should be sufficient to identify any weak strata beneath the pile points, which might affect the bearing capacity of the pile groups.
(ii) Fills and weak compressible soils rarely contribute to the shaft resistance of a pile and may add down drag to the load on it.
The whole pile load, possibly with the addition of down drag, has to be borne by the stronger strata lying below the weak materials
(iii) In the case of end-bearing piles in strong rock, boreholes should be of sufficient depth to establish conclusively the presence of rock head.
(iv) Pile-supported rafts on clays are often used solely to reduce settlement. In these cases, the depth of boring is governed by the need to examine all strata that could contribute significantly to the settlement.
(v) In chalk and other weak rocks, the exploration should be taken to the base of the weathered materials and to a sufficient depth into the unweathered rock to prove its continuity.

Embankments

embankment
Highway embankment

For embankments, the depth of the boring should be sufficient to check possible shear failure through the foundation strata and to assess the likely amount of any settlement due to compressible strata. In the case of water-retaining embankments, investigation should explore all strata through which piping could be initiated or significant seepage occur.

Cuttings, Quarries and Opencast mines

For cuttings, quarries and opencast mines, the depth of boring should be sufficient to permit assessment of the stability of the future slopes; this may require proving the full depth of any relatively weak strata. Groundwater conditions, including the possibility of artesian water, should also be determined.

Roads, Highways, and Airfields

road construction
Road construction

For roads and airfields, the depth of boring should be sufficient to determine the strength and frost susceptibility of possible subgrades and the drainage conditions. Exploration to a depth 2 m to 3 m below the proposed formation level is probably sufficient in most cases. A greater depth may be needed where a road is to pass over a shrinkable subgrade through a heavily vegetated area.

Pipelines

For small, shallow pipelines, it is frequently sufficient to take exploration to 1 m below the invert level. For deeper pipelines the depth of boring should be sufficient to enable dewatering and to allow any likely difficulties in excavating trenches and supporting the pipelines to be investigated. This means that a depth of at least 1 m to 2 m below the invert level may be advisable. Large pipelines, especially those in ground of low bearing capacity, require special consideration.

It is not always necessary that every exploration should be taken to the depths recommended. In many instances, it is adequate
if one or more boreholes are taken to those depths in the early stages of the field work to establish the general ground profile, and then the remainder sunk a little higher to explore more thoroughly the zone near the surface which the initial exploration had shown to be most relevant to the problem in hand.

Structural Design and Detailing Standards in Nigeria

By
Ubani Obinna
-

The body mandated by law to regulate all engineering activities in Nigeria is the Council for the Regulation of Engineering in Nigeria (COREN). For any engineering design to be accepted as valid, the seal of a registered engineer in the field of interest must be on the drawings and the calculation sheets. The supervision and regulation of COREN, therefore, cover all aspects of structural design and detailing in Nigeria. Only COREN registered engineers are qualified to approve designs for the purpose of construction.

Structural design and detailing in Nigeria is usually carried out according to the requirements of the British code. However, the use of Eurocode is beginning to gain widespread acceptance. Therefore, any structural design and detailing conforming to the requirements of the British and Eurocodes will have no problem getting approved in Nigeria. All designs and drawings are expected to be presented in SI units. Apart from a few specialized industries like the oil and gas, drawings and calculation sheets presented in US customary units may not be approved by regulatory agencies since engineers and technologists in the country are not trained by such unit of measurement.

The foremost structural engineering body in Nigeria is the Nigerian Institution of Structural Engineers (NIStructE). The criteria for being called a structural engineer in Nigeria is to pass the NIStructE Part 3 exams and become a corporate member, or to possess a COREN seal with ‘Structural Engineer’ designated on it. Another important professional body for engineering consultancy in Nigeria is the Association for Consulting Engineering in Nigeria (ACEN).

The primary codes of practice that are widely accepted for different designs in Nigeria are shown in the Table below;

DesignCode of Practice
Loading of buildingsBS 6399: Part 1: 1996 – Loading for buildings – Code of practice for dead and imposed loads
BS 6399: Part 2: 1997 – Loading for buildings – Code of practice for wind loads
BS 6399: Part 3: 1988 – Loading for buildings – Code of practice for imposed roof loads
EN 1991: Part 1-1: Densities, self-weight, imposed loads for buildings
EN 1991: Part 1-4: General actions – Wind actions
Design of reinforced concrete structuresBS 8110: Part 1: 1997 – Structural use of concrete – Code of practice for design and construction
BS 8110: Part 2: 1985 – Structural use of concrete – Code of practice for special circumstances
EN 1992: Part 1-1: Design of concrete structures – General rules, and rules for buildings
Design of steel structuresBS 5950: Part 1: 2000 – Structural use of steelwork in building – Code of practice for design. Rolled and welded sections
EN 1993: Part 1-1: Design of steel structures – General rules and rules for buildings
Design of composite structuresBS 5950: Part 3: 1990 – Structural use of steelwork in building – Design in composite construction – Code of practice for design of simple and continuous composite beams
BS 5950: Part 4: 1994 – Structural use of steelwork in building – Code of practice for design of composite slabs with profiled steel sheeting
BS 5400: Part 5: 2005 – Steel, concrete and composite bridges. Code of practice for design of composite bridges
EN 1994: Part 1-1: Design of composite steel and concrete structures – General rules and rules for buildings
Design of timber structuresBS 5268: Part 2: 2002 – Structural use of timber – Code of practice for permissible stress design, materials, and workmanship
EN 1995: Part 1-1: Design of timber structures – General — Common rules and rules for buildings

The other parts of the codes mentioned above for other special designs such as bridges, fire resistance design, accidental actions, water tightness, etc are also applicable. Therefore structural designs and drawings conforming to the above mentioned are generally acceptable.

Engineers in Nigeria are allowed to be flexible and ingenious in their design, but their assumptions and models must be backed with sound engineering judgment, theory, and adequate experience. Of all things, safety during the execution and use of their design is paramount. A sound structural design revolves around balancing cost and stability/functionality. Therefore, the knowledge of different structural forms and static systems is an important skill a structural engineer should possess.

However, a few norms are generally found in structures designed and executed in Nigeria. Experience has shown that most engineers carry out design bearing in mind the challenges that are usually faced during execution/construction. For most construction works without expert contractors, engineers normally try to keep the structural layout simple and ‘familiar’. Regulatory and town planning authorities such as the Lagos State Building Control Agency (LASBCA) also encourage engineers to stick to some minimum standards that are deemed generally acceptable, due to uncertainty in quality control and assurance in materials and during execution. Unfortunately, testing of materials is done in a few projects in Nigeria unless the construction is serious or under serious professional supervision.

Talking about uncertainty in construction materials, the diameter of reinforcements bars supplied to a site may not be up to the specified size, and the yield strength may be less than the expected minimum. In some construction sites, the concrete mix ratio can be altered in a very bad manner, and the formwork may not be accurate enough to represent the design member sizes. As a result, some agencies in some states may not accept an ‘economical design’.

Norms in the design of slabs

In the design of reinforced concrete structures in Nigeria, the thickness of slabs is usually between 125 mm to 250 mm, with 150 mm being the most common for regular buildings with moderate spans. When the thickness of a slab in a regular building is less than 150 mm, there may be challenges with approval, even when calculations are showing that a lesser thickness is adequate. The thickness of solid slabs on beams can be increased from 150 mm to 175 mm (or 200 mm) when deflection is a concern. However, when the thickness exceeds 200 mm, the engineer is usually advised to consider another structural system such as ribbed, waffle, of flat slab.

Concrete placement
Concrete casting of reinforced concrete slab

The diameter of reinforcement bars recommended for slabs is 12 mm. When 10 mm or 8 mm bars are provided as the main reinforcements, approving agencies may reject your drawings, even when calculations show that they are adequate. Therefore for the design of slabs for a residential, industrial, or commercial building in Nigeria, it is advised to adopt a minimum slab thickness of 150 mm and 12 mm high yield bars as main reinforcements. The recommended maximum spacing of the main bars should be 250 mm. However, when it is very clear that lesser sizes will work due to very short spans or insignificant loading, there shouldn’t be problems with that. A 20 mm minimum cover is recommended in slabs irrespective of the exposure or fire resistance requirements.

Norms in the design of beams

In beams, the recommended size of reinforcement is 16 mm bars. Some regulatory agencies frown at seeing 12 mm bars in beams whether they are satisfactory or not. However, when 12 mm bars are used as hanger bars in beams, there shouldn’t be any problem but it is very common to see 16 mm bars go all the way. The use of mild steel as links is getting less popular, and 8mm and 10mm high yield bars are now more common as stirrups. Experience has shown that 8 mm bars are the most economical for links.

reinforced concrete beams
Reinforced concrete beam and columns in construction

The most popular width for floor beams in Nigeria is 225 mm (9 inches), and it corresponds with the width of sandcrete blocks that are widely available for construction of buildings. Using a beam width of 225 mm ensures uniform surfaces during the finishing of the building. However, when design requirements suggest otherwise, the engineer should look at it critically and evaluate how it will affect the final appearance of the building. On the other hand, the depth of beams for buildings with regular spans is commonly 450 mm and can be adjusted accordingly to suit design and architectural requirements. However, site experience has shown that achieving a complete 450 mm depth for beams using local 1″ x 12″ planks can be challenging or uneconomical unless the formwork is carefully sawn and selected. If this cannot be guaranteed, then a depth of 400 mm should be considered for design of beams. This can be ignored when marine boards are to be used for formwork.

Depending on the layout of the building, it is usually advised to keep the spacing of columns supporting beams to less than 4 m in low-cost low-expertise construction, unless there are serious restrictions. Due to uncertainty in quality control, some engineers recommend the use of at least three number of 16 mm bars in beams once the span exceeds 3 m. This does not make sense economically though, but can be given serious thought during design. However, if you have modelled, loaded, analysed, and designed the building properly, I suggest that you let your utilisation ratio guide you on whether an extra bar is considerable or not.

Norms in the design of columns

In normal circumstances, the most popular column size for duplexes and other low rise buildings is 225 x 225 mm square columns. The dimensions of columns are expected to increase as the number of floors increase. When larger columns are required in low-rise buildings, engineers usually adopt 225 mm x 300 mm or 225 x 450 mm columns aligned with the walls/beams instead of say 250 x 250 mm or 300 mm x 300 mm columns. This is to ensure that there will be no unnecessary projections during the finishes of the building.

concrete frame

16 mm high yield bars are more common in columns and should be considered the minimum bar size unless the construction seriously suggests otherwise. For bungalows and last floors of buildings, 12 mm bars can be used for columns if there are no serious roof loads. Using 12 mm bars in columns outside these conditions may have a problem with approval even when calculations show that they are adequate. The lapping of bars in columns should be kept to a minimum of 45 x diameter of the reinforcement, and the detailing of the column should follow standard detailing guidelines. For square and rectangular columns, a minimum of 4 bars should be provided while for circular columns, a minimum of 6 bars should be provided.

Norms in the design of foundations

The common shallow foundations in Nigeria are pad, strip, combined/continuous footing, and raft foundations. For low rise buildings on good soil, pad foundation is normally used. The minimum thickness of pad foundation starts at 225 mm or 300 mm and increases based on shear considerations. Any pad foundation with thickness less than 225 mm will likely not get a nod of approval. The minimum size of reinforcement used in pad foundations is 12 mm and the spacing should not exceed 250 mm.

Raft foundations are normally adopted when the soil bearing capacity is low. They are usually combined with ground beams that primarily carry the shear load and bending moment due to load from the superstructure. The depth of the ground beam is influenced by shear considerations and site levels. The requirements of the raft slab are similar to the requirements of a suspended slab with more careful attention to the exposure condition and layout of reinforcements.

Conclusion

When a design is carried out according to the recommended standards bearing in mind the peculiarities of Nigeria’s construction sector, there should not be issues with approval after other requirements are met.

Design of Circular Hollow Section (CHS) Steel columns

By
Ivy Ugorji
-

Circular Hollow Sections (CHS) are round tubular steel sections that are used for a variety of purposes in civil engineering. They are usually available in the market as hot-rolled or cold-formed sections. Hot-rolled circular hollow sections are usually employed for structural purposes such as columns, struts, ties, etc.

Cold-formed sections are usually used as purlins and as framing for lightweight building construction. However, EN 1993-1-1 permits the use of both cold-formed and hot-rolled sections for structural purposes, even though the weldability of the former is usually a concern to structural engineers.

CHS SECTIONS

In offshore structures, CHS are preferred due to their capacity to minimise wave forces and for their structural efficiency. The structural efficiency of CHS members in bending, axial, and torsion is due to the uniform distribution of section materials about the polar axis. CHS is also known to offer a better strength-to-weight ratio in buildings and can lead to economical and lightweight construction.

circular hollow steel column

The hollow in the member also offers the advantage of making provision for building services such as fire protection, ventilation, heating, etc. In terms of corrosion, CHS sections perform better than other open sections due to their round edges and smaller surface area. The aesthetic appeal of CHS members also makes them a preferred alternative to architects.

CHS MEMBER IN COMPRESSION

In this article, let us verify the axial load-carrying capacity of CHS members according to the provisions of EC3.

Solved Example

Assess the suitability of a hot-rolled 244.5 x 10 CHS in grade S355 steel as an internal column in a multi-storey building to resist an ultimate axial force of 2000 kN. The column has pinned boundary conditions at each end, and the inter-storey height is 4.5 m.

CH

d = 244.5 mm
t = 10.0 mm
A = 7370 mm2
Wel,y = 415000 mm3
Wpl,y = 550000 mm3
I = 50730000 mm4

For a nominal material thickness (t = 10.0 mm) of less than or equal to 16mm the nominal value of yield strength fy for grade S355 steel is found from EN 10210-1 to be 355 N/mm2.

E = 210000 N/mm2

Section classification
ε = √(235/fy ) = √(235/355) = 0.81

Tubular sections (Table 5.2, sheet 3 of EN 1993-1-1):
d/t = 244.5/10 = 24.45

Limit for Class 1 section = 50ε2 = 32.8

24.45 < 32.8. Therefore section is Class 1

Resistance of the member to uniform compression (clause 6.2.4, EN 1993-1-1)
NC,Rd = (A.fy)/γmo = (7370 × 355) / 1.0 = 2616350 N = 2616 kN

NEd/NC,Rd = 2000/2616 = 0.764 < 1

Therefore section is okay for uniform compression.

Buckling resistance of member to compression (clause 6.3.1, EN 1993-1-1)

Nb,Rd = (χA.fy)/γmo for class 1, 2, and 3 sections

χ = 1/[Φ + √(Φ2 – λ2)] ≤ 1.0

Φ = 0.5 [1 + α(λ – 0.2) + λ2]

λ = √(Afy/Ncr)

Where;

Ncr = π2EI/Lcr2

Since the member is pinned at both ends, critical buckling length is the same for all axis Lcr = 4500mm

Ncr = (π2 x 210000 x 50730000)/45002 = 5192289.213 N = 5192 kN

λ = √(Afy/Ncr) = √(7370 x 355)/(5192 x 103) = 0.71

Selection of buckling curve and imperfection factor α
For a hot-rolled CHS, use buckling curve a (Table 6.2 of EN 1993-1-1).
For buckling curve a, α = 0.21 (Table 6.1 of EN 1993-1-1).

Buckling curves
Φ = 0.5 [1 + α(λ – 0.2) + λ2]
Φ = 0.5 [1 + 0.21(0.71 – 0.2) + 0.712 ] = 0.8056

χ = 1/[Φ + √(Φ2 – λ2)]
χ = 1/[0.8056 + √(0.80562 – 0.712)] = 0.842 < 1 ok

Therefore Nb,Rd = (χA.Fy)/γm1 = (0.842 × 7370 × 355) / (1.0 ) = 2202966.7 N = 2203 kN

NEd/Nb,Rd = 2000/2203 = 0.907 < 1

Therefore section is ok for buckling.

Summarily, the section is okay to resist axial load on it.

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