Effects of Diesel Contamination on the Engineering Properties of Natural Soils

September 27, 2020

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

Ubani Obinna

September 13, 2020

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).

f_yr = (1 – ρ)f_y —– (1)

Where;
ρ = (2V_Ed/V_pl,Rd – 1)² for (V_Ed > 0.5V_pl,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;

M_y,V,Rd= [(W_pl,y – ρA_w²/4t_w)f_y]/γ_m0 —– (2)

Where;
A_w = h_wt_w

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.

For the beam loaded as shown above;

M_Ed = PL/4 = (900 x 1.5)/4 = 337.5 kNm
V_Ed = P/2 = 900/2 = 450 kN

Since an advanced UK beam S275 is to be used for this design;
f_y = 275 N/mm²
γ_m0= 1.0 [Clause 6.1(1) NA 2.15 BS EN 1993-1- 1:2005]

Try section 406 x 178 x 74 UKB

Properties
h = 412.8 mm; b = 179.5 mm; t_w = 9.5 mm; t_f = 16.0 mm r = 10.2mm; A = 9450 mm²; W_pl,y = 1501000 cm³

h_w = h – 2t_f = 380.8 mm

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

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

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

c_f/t_f = 74.8/16.0 = 4.68

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

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

The limiting value for class 1 is c/t_w ≤ 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
M_Ed/M_c,Rd ≤ 1.0 (clause 6.2.5(1))

M_c,Rd = M_pl,Rd= (M_pl,y× F_y)/γ_m0

M_c,Rd = M_pl,Rd= [(1501 × 10³ × 275)/1.0] × 10^-6 = 412 kNm

M_Ed/M_c,Rd = 337.5/412 = 0.819 < 1.0 Ok

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

V_Ed/V_c,Rd ≤ 1.0

V_c,Rd = V_pl,Rd= A_v(F_y/ √3)/γ_m0(for class 1 sections)
For rolled I-section with shear parallel to the web, the shear area is;

A_v = A – 2bt_f + (t_w + 2r)t_f (for class 1 sections) but not less than ηh_wt_w

A_v = 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)
ηh_wt_w= (1.0 × 380.8 × 9.5) = 3618 mm²
4184 > 3618
Therefore, A_v = 4184 mm²

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

V_Ed/V_pl,Rd = 450/664.3 = 0.677 < 1.0 Ok

Since V_Ed (450 kN) > 0.5V_pl,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;

h_w/t_w≤ 72ε/η

h_w/t_w= 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 M_y,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.

M_y,V,Rd= [(W_pl,y – ρA_w²/4t_w)f_y]/γ_m0 but M_y,V,Rd ≤ M_c,Rd

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

A_w = h_wt_w = 380.8 x 9.5 = 3617.6 mm²

M_y,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

Ubani Obinna

September 13, 2020

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

Ubani Obinna

September 12, 2020

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;

∑F_x = 0
∑F_y = 0
∑M_i = 0

Here ∑F_x and ∑F_y 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 ∑M_i 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

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

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

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

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)

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)

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)

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)

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)

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)

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

Example

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

Ivy Ugorji

September 2, 2020

Table of Contents

Soil is a complex and highly variable product of nature, and the extent of this variation is not 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 the 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.

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 3m 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

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

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 2m to 3m 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

Ubani Obinna

September 2, 2020

Table of Contents

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;

Design	Code of Practice
Loading of buildings	BS 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 structures	BS 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 structures	BS 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 structures	BS 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 structures	BS 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.

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

Ivy Ugorji

August 8, 2020

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.

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.

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.

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.

d = 244.5 mm
t = 10.0 mm
A = 7370 mm²
W_el,y= 415000 mm³
W_pl,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 f_y for grade S355 steel is found from EN 10210-1 to be 355 N/mm².

E = 210000 N/mm²

Section classification
ε = √(235/f_y ) = √(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)
N_C,Rd = (A.f_y)/γ_mo = (7370 × 355) / 1.0 = 2616350 N = 2616 kN

N_Ed/N_C,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)

N_b,Rd = (χA.f_y)/γ_mo for class 1, 2, and 3 sections

χ = 1/[Φ + √(Φ² – λ²)] ≤ 1.0

Φ = 0.5 [1 + α(λ – 0.2) + λ²]

λ = √(Af_y/N_cr)

Where;

N_cr = π²EI/L_cr²

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

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

λ = √(Af_y/N_cr) = √(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 N_b,Rd = (χA.F_y)/γ_m1 = (0.842 × 7370 × 355) / (1.0 ) = 2202966.7 N = 2203 kN

N_Ed/N_b,Rd = 2000/2203 = 0.907 < 1

Therefore section is ok for buckling.

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

Effects of Wind on Billboard Structures

Ubani Obinna

August 7, 2020

Billboards are large outdoor advertising structures that are found along highways and city centres. Due to their large size and flexibility for creative output, billboards are usually eye-catching and offer a viable solution to advertisers for maximum exposure to vehicular traffic along highways. Billboard advertisements can be digital, painted, or printed. Most billboards on our highways are usually one-sided, two-sided, or three-sided, and supported on a single column. This type of cantilever structure can have a large windward area which can lead to high internal stresses in the members. Wind force is the most critical action on billboards, and failure due to wind and hurricane has often been reported.

Other effects such as imperfection and p-delta will also need to be checked for the optimal performance of such structures under wind load.
The types of failure that billboards can experience are;

Damage of the plate cladding
Failure of supporting structure
Overturning of the entire structure from the foundation

collapsed billboard — Failure of billboard from foundation

failure of billboard members — Failure of supporting structure

In this article, we are going to investigate the effect of wind on a three-sided highway billboard using Staad Pro software.

billboard model — 3D model of a 3-sided billboard

The members of the billboard are as follows

Circular column – CHS – 500mm diameter, 25 mm thick
Main cantilever beams (projecting from column) – 127 x 76 x 13 UK
Diagonals – UA 50 x 50 x 8
Member supporting billboard plates – UA 70 X 70 X 8

A rendered 3D view of the billboard is shown below.

ANOTHER VIEW OF BILLBOARD 3D MODEL — *3D rendering of a 3-sided billboard*

Wind Load Analysis

The wind load analysis on the billboard has been carried out in accordance with ASCE-7-2002 using Staad Pro v8i software. The wind load information is as follows;

The wind pressure on the structure is shown below.

wind load on billboard — *Wind load on a billboard*

Analysis Results

deflected shape of billboard — Deflected shape of the billboard structure

Maximum lateral deflection of billboard = 5.665 mm

Maximum bending moment on column = 31.552 kNm
Maximum axial force on column = 24.5 kN

A little consideration of other analysis results showed that the sections provided are a bit uneconomical for the structure. The design and modelling can be readjusted accordingly.

Design of Moment-Resisting Column Base Plates

Ubani Obinna

August 6, 2020

In the design of portal frames and other steel structures, moment-resisting bases can be provided if large lateral forces and bending moments are anticipated. This can lead to the economical design of the frame members, but the base will have to be carefully designed as a moment-resisting connection. Therefore such bases are expected to transfer bending moment and axial forces between the steel members and the concrete substructure. Depending on the direction of the moment and the magnitude, base plates can be symmetric or asymmetric.

For the design of moment-resisting base plates, the smallest diameter of bolts to be considered should be M20, even though M30 bolts are deemed the most appropriate. Larger bolts in smaller numbers are usually preferable. All holding-down bolts should be provided with an embedded anchor plate for the head of the bolt to bear against.

The sizes of anchor plates are chosen so as not to apply stress of more than 30 N/mm² at the concrete interface, assuming 50% of the plate is embedded in concrete. When the moments and forces are high, the holding-down system may need to be designed in conjunction with the reinforcement in the base.

Design Methodology

The design methodology for a moment-resisting base plate connection involves using an iterative approach/experience to select the best base dimensions and bolt configuration. This arrangement is then evaluated for its resistance against the anticipated actions from the superstructure.

In EN 1993-1-1:2005, the resistance of a base plate subjected to bending moment and axial force is assumed to be provided by two T-stubs, one in tension, and the other in compression. The resistance of the stub in tension is assumed to be provided by the holding down bolts, while the resistance of the stub in compression is assumed to be provided by a compression zone in the concrete, which is concentric with the column flange as shown below.

The design verification of a moment-resisting base plate, therefore, involves the following steps;

STEP 1
Evaluate the design forces in the equivalent T-stubs for both flanges. For a flange in compression, the force may be assumed to be concentric with the flange. For a flange in tension, the force is assumed to be along the line of the holding down bolts.

STEP 2
Evaluate the resistance of the equivalent T-stub in compression.

STEP 3
Evaluate the resistance of the equivalent T-stub in tension

STEP 4
Verify the adequacy of the shear resistance of the connection.

STEP 5
Verify the adequacy of the welds in the connection.

STEP 6
Verify the anchorage of the holding down bolts.

Design Example

Verify the capacity of the base plate arrangement in grade S275 steel shown below to resist the following actions;
M_Ed= 125 kNm
N_Ed = -870 kN (compression)
V_Ed= 100 kN

Column
From data tables for 254 x 254 x 132 UKC in S355
Depth h_c = 276.3 mm
Width b_c = 261.3 mm
Flange thickness t_f,c = 25.3 mm
Web thickness t_w,c = 15.3 mm
Root radius r_c = 12.7 mm
Elastic modulus (y-y axis) W_el,y,c = 1630000 mm³
Plastic modulus (y-y axis) W_pl,y,c = 1870000 mm³
Area of cross section A_c= 16800 mm²
Depth between flanges h_w,c = h_c – 2 t_f,c = 276.3 – (2 x 25.3) = 200.4 mm

Base plate
Steel grade S275
Depth h_bp = 500 mm
Gross width b_g,bp = 500 mm
Thickness t_bp = 50 mm

Concrete
The concrete grade used for the base is C30/37

Bolts
M24 8.8 bolts
Diameter of bolt shank d = 24 mm
Diameter of hole d₀ = 26 mm
Shear area (per bolt) A_s = 353 mm²
Number of bolts on either side n = 3

MATERIAL STRENGTHS

Column and base plate
The National Annex to BS EN 1993-1-1 refers to BS EN 10025-2 for values of nominal yield and ultimate strength. When ranges are given the lowest value should be adopted.

For S355 steel and 16 < t_f,c< 40 mm
Column yield strength f_y,c= R_eH = 345 N/mm²
Column ultimate strength f_u,c = R_m = 470 N/mm²

For S275 steel and t_bp > 40 mm
Base plate yield strength f_y,bp = R_eH = 255 N/mm²
Base plate ultimate strength f_u,bp = R_m= 410 N/mm²

Concrete
For concrete grade C30/37
Characteristic cylinder strength f_ck = 30 MPa = 30 N/mm²

The design compressive strength of the concrete is determined from:
f_cd = α_ccf_ck/γ_c
f_cd = (0.85 x 30)/1.5 = 17 N/mm²

For typical proportions of foundations, conservatively assume:
f_jd = f_cd = 17 N/mm²

Bolts
For 8.8 bolts
Nominal yield strength f_yb = 640 N/mm²
Nominal ultimate strength f_ub = 800 N/mm²

PARTIAL FACTORS FOR RESISTANCE

Structural steel
γ_M0 = 1.0
γ_M1 = 1.0
γ_M2 = 1.1

Parts in connections
γ_M2 = 1.25 (bolts, welds, plates in bearing)

DISTRIBUTION OF FORCES AT THE COLUMN BASE
The design moment resistance of the base plate depends on the resistances of two T-stubs, one for each flange of the column. Whether each T-stub is in tension or compression depends on the magnitudes of the axial force and bending moment. The design forces for each situation are therefore determined first.

Forces in column flanges
Forces at flange centroids, due to moment:
N_L,f = M_Ed/(h – t_f) = [125/(276.3 – 25.3)] x 10³ = 498 kN (tension)
N_R,f = -M_Ed/(h – t_f) = -[125/(276.3 – 25.3)] x 10³ = -498 kN (compression)

Forces due to axial force:
N_L,f = N_R,f = N_Ed/2 = -870/2 = -435 kN

Total force:
N_L,f = 498 – 435 = 63 kN (tension)
N_R,f = – 498 – 435 = -933 kN (compression)

In this case, the left side is in tension and the right side is in compression. The resistances of the two T-stubs will, therefore, be centred along the lines shown below:

Forces in T-stubs of the base plate
Assuming that tension is resisted on the line of the bolts and that compression is resisted concentrically under the flange in compression, the lever arms from the column centre can be calculated as follows:

z_t= 380/2 = 190 mm
z_c = (276.3 – 25.3)/2 = 125.5 mm
In this design situation, the left flange is in tension and the right in compression.
Therefore, z_L = z_t and z_R = z_c

N_LT= [M_Ed/(z_L + z_R)] + [(N_Ed x z_R)/(z_L + z_R)] = [125/(190 + 125.5)] x 10³ + [-(870 x 125.5)/(190 + 125.5)] = 50.127 kN
N_RT= -[M_Ed/(z_L + z_R)] + [(N_Ed x z_L)/(z_L + z_R)] = -[125/(190 + 125.5)] x 10³ + [-(870 x 125.5)/(190 + 125.5)] = -742.265 kN

RESISTANCE OF EQUIVALENT T-STUBS

Resistance of compression T-stub
The resistance of a T-stub in compression is the lesser of:

The resistance of concrete in compression under the flange (F_c,pl,Rd)
The resistance of the column flange and web in compression (F_c,fc,Rd)

Compressive resistance of concrete below the column flange. The effective bearing area of the joint depends on the additional bearing width, as shown below.

Determine the available additional bearing width (c), which depends on the plate thickness, plate strength and joint strength.

c = t_bp x [√f_ybp/(3f_jdγ_M0)] = 50 x [√255/(3 x 17 x 1.0)] = 112 mm

Thus the dimensions of the bearing area are,
b_eff = 2c + t_fc = (2 x 112) + 25.3 = 249.3 mm
l_eff = 2c + b_c = (2 x 112) + 261.3 = 485.3 mm

The area of bearing is,
A_eff= 249.3 x 485.3= 120985 mm²

Thus, the compression resistance of the foundation is,
F_c,pl,Rd= A_efff_jd
F_c,pl,Rd = 120985 x 17 x 10^-3 = 2057 kN, > N_R,T = 742.265 kN (maximum value).
This is therefore satisfactory

Resistance of the column flange and web in compression
The resistance of the column flange and web in compression is determined from:
F_c,fcRd = M_c,Rd/(h_c – t_fc)

M_c,Rd is the design bending resistance of the column cross-section = 645 kNm (see page 402 of P363)

If V_Ed > V_c,Rd/2, the effect of shear should be allowed for

V_c,Rd= 705 kN (see page 234 of P363)
V_Ed = 100 kN
By inspection:
V_Ed< V_c,Rd/2

Therefore, the effects of shear may be neglected and hence
M_c,Rd= 645 kNm

F_c,fcRd = [(645 x 10⁶)/(276.3 – 25.3)] x 10-3 = 2569.4 kN

As, F_c,pl,Rd < F_c,fc,Rd the compression resistance of the right hand T-stub is:
F_c,pl,Rd = 2057 kN
F_c,pl,Rd> N_R,T = 742.265 kN Satisfactory

RESISTANCE OF TENSION T-STUB
The resistance of the T-stub in tension is the lesser of:

The base plate in bending under the left column flange, and
The column flange/web in tension

Resistance of base plate in bending
The design resistance of the tension T-stub is given by:
F_t,pl,Rd = F_t,Rd= min{F_T,1-2,Rd; F_T,3,Rd}

Where F_T,1-2,Rd is the ‘Mode 1 / Mode 2’ resistance in the absence of prying and F_T,3,Rd is the Mode 3 resistance (bolt failure).

F_T,1-2,Rd = 2M_Pl,1,Rd/m

M_Pl,1,Rd = (0.25∑l_eff,1t_bp²f_ybp)/γ_M0

∑l_eff,1 = is the effective length of the T-stub, which is determined from the equations below;

b_p = 500 mm
p = 190 mm
e = 60 mm
e_x = 60 mm
m_x = 52 mm
n = 3 (number of rows of bolt)

Non-circular patterns

Single curvature; l_eff,nc = b_p/2 = 500/2 = 250 mm
Individual end yielding; l_eff,nc = 0.2n(4m_x + 1.25e_x) = 1.5(4 x 52 + 1.25 x 60) = 424.5 mm
Corner yielding of outer bolts, individual yielding between; l_eff = 2m_x + 0.625e_x + e + (n – 2)(2m_x + 0.625e_x) = 2(52) + 0.625(60) + 60 + 1 x (2 x 52 + 0.625 x 60) = 324.25 mm
Group end yielding; l_eff = 2m_x + 0.625e_x + 0.5(n – 1)p = 2(52) + 0.625(60) + (0.5 x 2 x 190) = 331.5 mm

Circular patterns

Individual circular yielding; l_eff,cp = nπm_x= 3 x π x 52 = 490 mm
Individual end yielding; l_eff,cp = 0.5n(πm_x + 2e) = 1.5(π x 52 + 2 x 60) = 425 mm

The minimum is l_eff,1 = 250 mm

M_Pl,1,Rd = (0.25∑l_eff,1t_bp²f_ybp)/γ_M0 = [(0.25 x 250 x 50² x 255)/1.0] x 10^-6 = 39.84 kNm
F_T,1-2,Rd = 2M_Pl,1,Rd/m = (2 x 39.84)/0.052 = 1532.3 kN

F_t,3,Rd = ∑F_t,Rd
For class 8.8. M24 bolts; F_t,Rd = 203 kN
F_t,3,Rd = 3 x 203 = 609 kN

F_t,pl,Rd = F_t,Rd= min{F_T,1-2,Rd; F_T,3,Rd} = min{1532.3; 609} = 609 kN

N_LT = 50.127 kN < F_t,pl,Rd = 609 kN Okay.

WELD DESIGN
Welds to the tension flange
The maximum tensile design force is significantly less than the resistance of the flange, so a full-strength weld is not required. The design force for the weld may be taken as that determined between column and base plate in STEP 1, i.e. 498 kN (N_L,f)

For a fillet weld with s = 12 mm, a = 8.4 mm
The design resistance due to transverse force is:

F_nw,Rd = (Kaf_u/√3)/β_wγ_M2

where K = 1.225, f_u = 410 N/mm² and β_w = 0.85 (using the properties of the material with the lower strength grade – the base plate)

F_nw,Rd = [(1.225 x 8.4 x 410) / √3]/(0.85 x 1.25) = 2.29 kN/mm

Length of weld, assuming a fillet weld all around the flange:
For simplicity, two weld runs will be assumed, along each face of the column flange. Conservatively, the thickness of the web will be deducted from the weld inside the flange. An allowance equal to the leg length will be deducted from each end of each weld run.

L = (261.3 – 2 x 12) + (261.3 – 12 – 4 x 12) = 438.6 mm
F_t,weld,Rd = 2.29 x 438.6 = 1004 kN, > 498 kN – Satisfactory

Welds to the compression flange
With a sawn end to the column, the compression force may be assumed to be transferred in bearing. There is no design situation with moment in the opposite direction, so there should be no tension in the right-hand flange. Only a nominal weld is required. Commonly, both flanges would have the same size weld.

Welds to the web
Although the web weld could be smaller and sufficient to resist the design shear, it would generally be convenient to continue the flange welds around the entire perimeter of the column.

Structural Design of Signposts and Billboards

Ubani Obinna

August 3, 2020

As the name implies, signposts are post bearing structures that offer information or guidance to people. They are a prominent feature of our highways, streets, city centers, villages, and areas of public gathering. Signposts are usually placed strategically away from obstruction as they are intended to show information like route direction, warnings, route assurance, traffic signs, commercial advertisements, etc.

EN 12899-1:2007 requires that signposts made of steel structures should conform with EN 1993-1-1:2005 (Eurocode 3). One of the major concerns in the design of billboards and signposts is the risk of failure under wind load, which has serious economic and safety consequences. A failed highway sign structure can cause injury to pedestrians, damage vehicles, and obstruct traffic. As a result, such structures that are exposed to the public must satisfy all needed safety considerations. Additional risks of vehicles colliding with sign structures should also be checked, with passive protection provided for such structures.

Actions on Sign Structures

Wind action on signposts and billboards can be evaluated according to EN 1991-1-4:2005 (Eurocode 1 Part 4). ASCE 7-10 code of practice can also be used for the evaluation of wind load on sign structures. The National Annex to BS EN 12899-1:2007 recommends suitable wind loads for the majority of signs in the UK. Whilst is it more conservative than performing a full analysis, it is simpler and quicker.

Other forces that may need to be taken into account when designing sign structures are point loads and dynamic snow load (not applicable in Nigeria). The UK National Annex recommends that signs should be able to withstand a force of 0.5 kN applied at any point. This represents the load that might be exerted by, for example, a glancing blow from a vehicle mirror, a falling branch or malicious interference with the sign. This point load is the critical factor only for very small signs, but for signs mounted on a single support, it causes torsional forces that need to be considered.

For large billboards, live loads and the weight of services should be accounted for the in the design.

Design example
Provide adequate sections for a sign structure with the configuration shown below. The sign post is located in an area that is 76 m above sea level with a wind speed of 35 m/s.

Mounting height above ground h_m = 2.0 m
Width of sign face l = 3.0 m
Height of sign face b = 2.0 m
Total height H = h_m + b = 4.0 m
Height to centroid of sign area z = h_m + b/2 = 3.0 m
Depth of post buried above foundation h_b = 200 mm

Basic wind velocity
Let the basic wind velocity from wind map = v_b,map = 35 m/s
The altitude of site above sea level A = 76 m
Altitude factor c_alt = 1 + 0.001A = 1 + (0.001 × 76) = 1.076
v_b,0 = v_b,map ⋅ c_alt= 35 × 1.076 = 37.66 m/s

Assess Terrain Orography
Site is not very exposed site on cliff/escarpment or in a site subject to local wind funnelling. Therefore c_o = 1.0

Determine Design Life Requirement

p = design annual probability of exceedence
p = 1/design life = 1/25 = 0.04 (for signs design life is 25 years)
K = Shape parameter = 0.2
n = exponent = 0.5

Basic Wind Velocity
v_b = c_dir ⋅ c_season ⋅ v_b,0
c_dir = directional factor = 1.0
c_season = season factor = 1.0
v_b = 1.0 × 1.0 × 37.66 = 37.66 m/s

10 minute mean wind velocity having probability P for an annual exceedence is determined by:

v_{b,25 years}= v_b ⋅ c_prob
v_{b,25 years}= 37.66 × 0.96 = 36.15 m/s

Mean Wind
The mean wind velocity V_m(z) at a height z above the terrain depends on the terrain roughness and orography, and on the basic wind velocity, V_b, and should be determined using the expression below;

V_m(z) = c_r(z). c_o(z).V_b

Where;
c_r(z) is the roughness factor (defined below)
c_o(z) is the orography factor often taken as 1.0

c_r(z) = k_r. In (z/z₀) for z_min ≤ z ≤ z_max
c_r(z) = c_r.(z_min) for z ≤ z_min

Where:
Z₀ is the roughness length
k_r is the terrain factor depending on the roughness length Z₀ calculated using;

k_r = 0.19 (Z₀/Z_0,II)^0.07

Where:
Z_0,II = 0.05m (terrain category II)
Z_min is the minimum height = 2 m
z = 3 m
Z_max is to be taken as 200 m
K_r = 0.19 (0.05/0.05)^0.07 = 0.19
c_r(3) = k_r.In (z/z₀) = 0.19 × In(3/0.05) = 0.78

Therefore;
V_m(3.0) = c_r(z). c_o(z).V_b = 0.78 × 1.0 × 36.15 = 28.197 m/s

Wind turbulence
The turbulence intensity I_v(z) at height z is defined as the standard deviation of the turbulence divided by the mean wind velocity. The recommended rules for the determination of I_v(z) are given in the expressions below;

I_v(z) = σ_v/V_m = k_l/(c₀(z).In (z/z₀)) for z_min ≤ z ≤ z_max
I_v(z) = I_v.(z_min) for z ≤ z_min

Where:
k_l is the turbulence factor of which the value is provided in the National Annex but the recommended value is 1.0
C_o is the orography factor described above
Z₀ is the roughness length described above.

For the structure that we are considering, the wind turbulence factor at 3 m above the ground level;

I_v(60) = σ_v/V_m = k₁/[c₀(z).In(z/z₀)] = 1/[1 × In(3/0.05)] = 0.244

Peak Velocity Pressure
The peak velocity pressure q_p(z) at height z is given by the expression below;

q_p(z) = [1 + 7.I_v(z)] 1/2.ρ.V_m²(z) = c_e(z).q_b

Where:
ρ is the air density, which depends on the altitude, temperature, and barometric pressure to be expected in the region during wind storms (recommended value is 1.25kg/m³)

c_e(z) is the exposure factor given by;
c_e(z) = q_p(z)/q_b
q_b is the basic velocity pressure given by; q_b = 0.5.ρ.V_b²

q_p(60m) = [1 + 7(0.244)] × 0.5 × 1.25 × 28.197² = 1345.66 N/m²

Therefore, q_p(3m) = 1.345 kN/m²

Determination of force coefficient (Table NA 2 BS EN 12899)
λ = effective slenderness ratio of sign or aspect ratio
λ = l/b = 3.0 / 2.0 = 1.5
Therefore c_f= 1.30

Calculation of the total wind force (Clause 5.3 of EN 1991-1-4)
F_w = c_sc_d ⋅ cf ⋅ q_p(z_e) ⋅ A_ref (A_ref = area of sign)
c_sc_d = 1.0 (for sign posts)

F_w = 1.0 × 1.30 × 1.345 × 3.0 × 2.0 = 10.5 kN

Partial Factor for Action γ_F (Table 6 EN 12899-1:2007 (E))
ULS (bending and shear) γ_F = 1.5
SLS (deflection) γ_F = 1.0
γ_f3 = 1.0

Design Wind Force on the sign
F_w,d = F_w ⋅ γ_F ⋅ γ_f3
F_w,d (ULS) = 10.5 × 1.5 × 1.0 = 15.75 kN
F_w,d (SLS) = 10.5 × 1.0 × 1.0 = 10.5 kN

Ultimate Action Effects
Ultimate design bending moment per post, M_Ed
M_Ed = Wind force × lever arm to foundation / number of posts
M_Ed = F_w,d (ULS) ⋅ (z + h_b) / n
M_Ed = 15.75 × (3.00 + 0.2) / 2 = 25.2 kNm

Ultimate design shear per post
V_Ed = Wind force / number of posts
V_Ed = F_w,d(ULS) /2 = 15.75 / 2 = 7.9 kN

The 0.5 kN point load on the sign is not critical, since it is less than the wind action and there are no torsional effects with 2 posts.

Try circular hollow section CHS 139.7 x 8 (S355)

A = 33.1 cm²; W_pl = 139 cm³; I_x = 720 cm⁴

Section classification
ε = √235/f_y = √(235/355) = 0.81
Tubular sections (Table 5.2, sheet 3 of EN 1993-1-1:2005):
d/t = 139.7/8 = 17.46
Limit for Class 1 section = 50ε² = 40.7
40.7 > 17.46; section is Class 1

Member resistance at ULS
According to Table 7 of EN 12899-1:2007 (E), the material factor of safety for steel is γ_m = 1.05
Moment Capacity M_Rd= f_y⋅W_pl/γ_m = [(355 x 10³ x 139 x 10^-6)/√3)]/1.05= 46.99 kNm
M_Ed/M_Rd = 25.2/46.99 = 0.536 < 1.0 Okay

Shear capacity V_Rd = A_v(f_y/√3)/γ_m
Av = 2A/π = (2 x 33.1)/π = 21.027 cm²
V_Rd = 21.027 × 10^-4× [(355 x 10³)/√3)]/1.05 = 430.96 kN
VEd/VRd = 7.9/430.96 = 0.018 < 1.0 Okay

Calculation of Temporary deflection
The wind velocity for calculating the temporary deflection (SLS) criterion is 75% of the reference wind velocity, as it is based upon a 1 year mean return period. The 0.96 factor below reverses the c_prob conversion from 50 to 25 year return period used above (Clause 5.4.1, note 1 EN 12899-1).

F_{wd(1 year)} = F_wd(SLS) x 0.75²/0.96² = 10.5 x 0.75²/0.96² = 6.41 kN

Uniformly distributed load along sign face = F_{w,d (1 year)} / b
F_{w,d (1 year)} / b = 6.41/2.0 = 3.2 kN/m
where b = height of the sign face

Maximum deflection at top of sign (bending), δ

E = 210000 N/mm²
I = 720 cm⁴
n = number of posts = 2
Other parameters are as defined above

δ = [3.2/(24 x 210000 x 720 x 10⁴ x 2)] x [3(4000 + 200)⁴– 4(2000 + 200)³ x (4000 + 200) + (2000 + 200)⁴] = 34.3 mm

Deflection per linear metre = δ’ = δ/(H + h_b) = 34.3/(4 + 0.2)= 8.16 mm/m
Maximum temporary deflection taken as class TDB4 = 25 mm/m
8.16 mm/m < 25 mm/m. Therefore deflection is okay.

References
The Institution of Highway Engineers (2010): SIGN STRUCTURES GUIDE Support design for permanent UK traffic signs to BS EN 12899-1:2007 and structural Eurocodes

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Norms in the design of slabs

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Effects of Wind on Billboard Structures

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Design Methodology

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