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Normally Consolidated and Overconsolidated Soils

By
Ubani Obinna
-

When a saturated clay is subjected to external pressure, the pressure is initially taken up by the water in the pores thereby leading to excess pore water pressure. If drainage is permitted in the system, a hydraulic gradient is developed and the excess water begins to flow out of the soil mass.

As water dissipates from the system, the pressure gradually gets transferred to the soil skeleton (soil grain) and they begin to rearrange and compress until the water is completely dissipated from the soil. This process continues until the entire pressure is transferred to the soil grains. This process is known as consolidation and occurs in cohesive soils only. Pore water dissipation in granular soil is immediate.

Consolidation can occur in a soil mass for a lot of reasons such as;

  • Application of external static load from structures
  • Lowering of the groundwater table
  • Desiccation
  • Self-weight of the soil (an example is recently placed fill)

Clay soil is said to be normally consolidated if the effective overburden pressure that it is currently experiencing is the maximum it has ever experienced in its history. On the other hand, it is said to be overconsolidated if the present overburden pressure is less than the effective overburden pressure it has experienced in the past.

The ratio of the maximum overburden pressure it has experienced in the past and the current overburden pressure is known as the ‘overconsolidation ratio’ (OCR). The overconsolidation ratio of a normally consolidated clay is unity, while the overconsolidation ratio of an overconsolidated soil is greater than unity.

Overconsolidation can occur in clays due to reasons such as;

  • Self-weight of the soil which has eroded
  • Weight of continental ice sheet which has melted
  • Desiccation of layers close to the surface

The nature of consolidation of a clay soil affects its behaviour in the field and when tested in the laboratory. For instance, the natural moisture content of a normally consolidated clay is usually close to the liquid limit, while the natural moisture content of an overconsolidated clay is usually close to the plastic limit.

Clay soils in their natural state have memory of the magnitude of the highest pressure they have experienced in the past. This memory is locked in the soil structure and can only be broken when the clay is remoulded or reconstituted at a moisture content that is equal to or greater than the liquid limit.

The stress-strain curve of an overconsolidated clay is likely to exhibit more elastic behaviour when compared with the stress-strain curve of a normally consolidated clay. Since a normally consolidated clay is experiencing the maximum pressure of its history, it is more likely to get compressed without recovery (plastic recovery) when subjected to additional external pressure in their natural state. This is not the same for overconsolidated soils which will exhibit elastic behaviour whenever an external pressure that is less than what they have experienced in the past is placed and removed.

Preconsolidation pressure can be determined from the field or in the laboratory from the e-log P curve plot (void ratio against log of pressure) after carrying out a consolidation test in the laboratory. The most widely used method was proposed by Casagrande in the year 1936 (see Figure below).

consolidation curve

The method involves locating the maximum point of curvature in the curve and drawing a tangent and horizontal line at that point. The angle between these two lines is then bisected. The abscissa of the point of intersection of this bisector with the upward extension of the inclined straight part corresponds to the pre-consolidation pressure pc.

Pre-consolidation pressure is the yield point that indicates the beginning of the plastic deformation during the compressive loading in a soil. The ratio of the preconsolidation pressure and the applied effective pressure is the overconsolidation ratio.

Design of Glass Swimming Pools

By
Ubani Obinna
-

The use of swimming pools is increasingly becoming an important part of the urban lifestyle. Apart from the public swimming pools in parks and other recreation areas, a private swimming pool in a home offers a lot of benefits to the homeowner, and also increases the value of the property.

Swimming pools can be constructed of different materials such as reinforced concrete, shotcrete, plastics, composites, glass, etc. In all cases, the material to be used in the construction of a swimming pool must be able to resist the water pressure, some degree of impact, and must be watertight.

Glass swimming pools have gained popularity over the years especially due to their aesthetic appeal. The use of glass is more popular in suspended or surface swimming pools for obvious reasons. For suspended pools, glass can be used as the side panels and/or the base, while for surface pools it can be used for the side panels only. Advances in glass technology have improved the engineering properties of glass to make them less brittle, with improved toughness and durability. The basic design requirements for glass swimming pools are;

  • Maximum transparency
  • Robustness
  • Water tightness, and
  • Strength
images
Suspended glass swimming pool

Structural glass for swimming pools is usually made of heat strengthened laminated glass, which offers shatterproof behaviour when subjected to an impact force. Multiple glass sheets laminated together are usually used in glass engineering for additional functionality in case one of the glass sheet fails.

Furthermore, different materials such as PVB (Polyvinyl butryl) or ionoplast can be used as the interlayer between different panes of glass to form laminated glass. Ionoplast interlayer, however, offers the best behaviour for glass subjected to adverse loading conditions. The safety of the people inside the pool and the people in the building should be given top priority in the design.

laminated glass
Laminated glass

Structural Design of glass swimming pools

In the structural design of glass swimming pools, the thickness of the glass is determined from the anticipated hydrostatic pressure. The glass can be modelled as plate elements with simply supported boundary conditions. Glass possesses some elastic properties which allows it to regain its shape after unloading, but has no plastic properties.

This implies that fracture occurs in the material before any permanent deformation can take place. As a result, linear elastic analysis using Kirchoff’s thin plate theory is sufficient for the analysis of glass swimming pools, and the material can be idealised as a homogenous and isotropic. Failure in a glass is usually in tension because glass has good compressive strength.

Laminated glass has a modulus of elasticity several thousand times larger than the modulus of elasticity of the weakest interlayer material, PVB or ionoplast. This great mismatch makes the behaviour complex as well as analysis and modelling of laminated glass. To account for this during modelling, full relaxation of the interlayer (no cooperation between the sheets) can be assumed. This is deemed a conservative approach. Therefore, the thickness of the glass to be used in finite element modelling can be taken as;

tm = tpl √n

Where tm is the thickness to be used in modelling, tpl is the nominal thickness of each individual sheet, and n is the number of glass layers.

It is a good practice to investigate different types of support stiffness depending on the actual support conditions of the glass. Furthermore, the pool can be analysed for other conditions such as when one or more sheets are broken, and the effect evaluated.

For bottom glass panels, the self-weight of the glass and the water pressure can be applied as UDL on the surface of the glass, while for the walls, the water pressure should be applied as a triangular hydrostatic load. For suspended swimming pools, wind pressure and other anticipated forces should also be checked. After the analysis, the deflection, tensile stresses, and support reactions of the glass panes should be checked. The occurring tensile stress from the applied loading should not exceed the tensile strength capacity of the glass.

Detailing
Since every joint in a water retaining structure is a potential source of leakage, the number of joints should be minimised. The openings in the concrete floor or wall supporting the glass should be closed with a glass made of one piece. The joint between the concrete and the glass should be filled with a selected adhesive which should not react with the interlayer material. The concrete notch for the glass should be treated with primer to bind properly with the adhesive.


Application of Effective Stress Concept in the Analysis of Cohesive Soils

By
Ubani Obinna
-

The application of load on a saturated soil increases the pore water pressure. This sets up a hydraulic gradient which causes the water to flow out of the soil. As water dissipates from the soil mass, the pore water pressures gradually move toward their long-term equilibrium value. This is referred to as consolidation and it is time-dependent. The time taken for consolidation increases with decreasing stiffness and decreasing permeability. In sands and gravel, it is almost instantaneous.

Effective stress condition is used to represent the state when all the excess pore water pressure within the soil mass have dissipated. i.e. the drained state. The reverse is usually referred to as ‘total stress state’. Granular (cohesionless) soils are free-draining, therefore excess pore water pressures created during construction will dissipate so quickly that “effective stress” conditions exist in both the short and long term. Hence the effective angle of internal friction (ø′) is used for any stress analysis involving granular materials.

Clays are naturally complicated materials, and most clay deposits in the world are usually overconsolidated. The analysis of clay soils are usually idealised using;

  • Total stress approach (purely cohesive behaviour), or,
  • Effective stress approach (purely frictional behaviour)

Frictional behaviour corresponds to conditions where pore water pressures can be defined or assumed and thus allow the strength of the soil to be characterised in terms of effective stress. Cohesive behaviour refers only to the immediate short term when pore water pressures cannot be conveniently defined or assumed. Then, the strength of the soil is represented by the original undrained shear strength, cu.

In cohesive soils, the change from total stress (undrained conditions) to effective stress (drained conditions) generally occurs over a much longer period of time. The exception being the presence/addition of fine silts/granular material which can greatly reduce the time in which effective stress conditions are reached. During this period, the strength parameters of the cohesive soil may change significantly due to pore water pressures changes induced following the construction of a retaining structure. The change in strength is caused by equalisation of negative pore water pressure in the soil and results in reduced values of cohesion c′ but increased values of angle of internal friction (ø′).

Mohr CoulombFailureCriterionintermsofeffectivestresses

Whilst all cohesive soils are subject to these changes, the effective stress condition is not usually critical when fine silts and naturally consolidated and slightly overconsolidated clays (those with cohesion values of less than about 40kN/m2), are involved, since the change from effective parameters gives an overall increase in soil strength. However, the reverse is true for over-consolidated clays, (those with undrained values in excess of about 40kN/m2). The overall strength will, in most cases, be reduced as the stress condition changes from total to effective because the loss of substantial cohesive strength is not compensated adequately by the increasing angle of internal friction. Hence it is advised that for cohesive soils both short and long-term stress analyses be carried out to determine the more onerous design case.

Typically, long-term calculations are carried out using effective stress parameters – drained analysis. Short-term calculations use total stress parameters – undrained analysis. Traditionally, many generalist engineers have used the cohesion model and total stress procedures. However, any tendency for the pore water to drain will lead to changes in voids ratio and hence the value of undrained cohesion. Voids in soil are not fixed in place; water can move and voids can collapse or expand. Therefore undrained cohesion values are difficult to predict with certainty. In the long term, clays behave as granular soils exhibiting friction and dilation. For this reason, calculations to predict long-term values of pressures are undertaken using the effective stress method.

When using the effective stress method the friction and cohesion can be characterised by the two parameters angle of shearing resistance, φ’ and cohesion intercept in terms of effective stress, c’ (rather than peak angle of shearing resistance, φ’max). In the absence of test data, BS 8002 recommends values of φ’ that can be used with c’ = 0, and these are given in the Table below.

Plasticity index (%)15305080
φ’ (degrees)30252015


Engineered Wood Products and their Applications in Structural Engineering

By
Ubani Obinna
-

A wide range of derived wood products are obtained by binding and fixing of wood particles, boards, veneers, fibres, or strands with adhesives or other mechanical means to form composites. The reconstituted wood products derived using such means are generally referred to as engineered wood products (EWPs). There are different types of engineered wood products such as plywood, particleboard, laminated timber, finger joints, cross-laminated timber, etc. In the field of civil engineering, EWPs have been used in construction for over 40 years. Glued laminated timber (glulam), connected plated trusses, finger joints, plywood, mechanically and adhesive bonded web beams, etc have all found useful applications in structural engineering for a long period of time.

finger jointed timber
Timber Finger Joint (Source: http://www.structuraltimber.co.uk/)

Engineered wood products are increasingly becoming important in the field of structural engineering. Over the last decade, high rise timber structures have been constructed in different parts of the world as a result of improved engineered wood products. Moreover, timber has been identified as the most environmentally friendly construction material when compared with steel and concrete. According to Structural Timber Engineering Bulletin, there have been significant developments in the range of EWPs for structural applications with materials such as laminated veneer lumber (LVL), parallel strand lumber (PSL), laminated strand lumber (LSL), prefabricated I-beams, metal web joists and ‘massive’ or cross-laminated timber (CLT) becoming more widely available.

open web joists
Open Web Joist (Source: http://www.structuraltimber.co.uk/)

According to Tupenaite et al (2019), the most popular engineered timber products used in high-rise timber buildings are produced based on laminating and gluing. The examples of such products are;

  • Glued laminated timber (glulam)
  • Cross laminated timber (CLT)
  • Laminated veneer lumber (LVL)

Glued Laminated Timber (glulam)
Glued laminated timber (glulam) is a structural member made by gluing together a number of graded timber laminations with their grain parallel to the longitudinal axis of the section. Glulam is the oldest glued structural product (over 100 years). It is generally composed of lumber layers (2×3 to 2×12), planed and pre-finger-jointed, and then bonded together with moisture- resistant structural adhesives longitudinally. Members can be straight or curved, horizontally or vertically laminated and can be used to create a variety of structural forms (see figure below).

glued laminated timber
Curved glulam structural members (Source: http://www.structuraltimber.co.uk/)


Laminations are typically 25mm or 45mm thick but smaller laminations may be necessary where tightly curved or vertically laminated sections are required. The requirements for the manufacture of glulam are contained in product standard BS EN 14080: 2013.

Glulam is used in large structural elements such as portal frames, bridges, beams, columns, trusses etc.

glulam in construction
Glulam in Construction

Cross Laminated Timber (CLT)
Cross laminated timber (CLT) is a structural timber product with a minimum of three cross-bonded layers of timber, of thickness 6mm to 45mm, strength graded to BS EN 14081-1:2005 and glued together in a press which applies pressure over the entire surface area of the panel. CLT was developed around 15 years ago in Central Europe and is made up of a solid engineered wood panel, made up of cross angled timber boards which are glued together.

cross laminated timber
Cross laminated timber (Source: http://www.structuraltimber.co.uk/)

CLT panels typically have an odd number of layers (3,5,7,9) which may be of differing thicknesses but which are arranged symmetrically around the middle layer with adjacent layers having their grain direction at right angles to one another .

The structural benefits of CLT over conventional softwood wall framing and joisted floor constructions, include:


• large axial and flexural load-bearing capacity when used as a wall or slab
• high in-plane shear strength when used as a shear wall
• fire resistance characteristics for exposed applications
• superior acoustic properties

CLT can be used in floor slabs, roofs, beams, columns, load bearing walls, and shear walls. Length up to 20m can be pproduced, with thickness ranging from 50 – 300 mm. Width of up to 4800 mm can be achieved with CLT.

clt wall in construction
CLT walls in building construction
clt beams and columns 1
CLT beams and columns in a building

Laminated veneer lumber (LVL)
Laminated veneer lumber (LVL) is a structural member manufactured by bonding together thin vertical softwood veneers with their grain parallel to the longitudinal axis of the section, under heat and pressure. In some cases cross grain veneers are incorporated to improve dimensional stability. LVL is a type of structural composite lumber. Due to its composite nature, it is much less likely than conventional lumber to warp, twist, bow, or shrink, and has higher allowable stress compared to glulam.

laminated veneer lumber
Laminated veneer lumber

LVL is often used for high load applications to resist either flexural or axial loads or a combination of both. It can provide both panels and beam/column elements. It can be used for beams, walls, other structures and forming of edges.

lvl frame
LVL framing for a building in New Zealand


The requirements for LVL are contained in product standard BS EN 14374:2004

High Rise Timber Building to be Constructed in Tokyo

By
Ubani Obinna
-

Mitsui Fudosan and Takenaka Corporation has commenced plans to develop a 70 metres high timber building in Tokyo’s Nihonbashi district. The 17 storey building when completed will be among the tallest timber buildings in Japan.

The interest in multi-storey timber buildings has increased around the world. Timber building materials cause considerably lower climate change impact compared to materials like steel and concrete. Moreover, modern engineered timber products provides opportunities to build high. In the last decade, 6 storeys timber buildings and higher have been constructed around the world, and engineers have begun to look at the possibility of building much taller with timber.

Construction of the timber high rise building in Tokyo is tentatively scheduled to start in the year 2023 with a possible completion date of 2025. The building’s floor area would be 26,000 square meters and will be constructed from 1,000 cubic meters of domestic lumber. The main structure would be a hybrid that incorporates Takenaka’s fire-resistant laminated wood with materials sourced from Mitsui-owned forestry in Hokkaido. Carbon dioxide emissions would be 20 percent less than those when construction a standard steel-frame office building of the same size and scale. 

Mitsui Fudosan Group owns approximately 5,000 hectares of forestry in Hokkaido, all of which has been certified by the Sustainable Green Ecosystem Council. 

Bricklaying Robot Constructs 3-Bedroom Apartment in the UK – Video

By
Ivy Ugorji
-

A block and bricklaying robot has been developed for use in the building industry by Construction Automation, a Yorkshire-based start up firm. The Automatic Bricklaying Robot (ABLR) has been put to test work on a 3 bedroom apartment project in the village of Everingham, Yorkshire.

According to the company, ABLR has been four years in development, and will be the first machine of its kind to build round corners without stopping. The robot has the capacity to lay mortar, centralise, and align the edges of blocks properly. It has already been deployed from factory testing to a test construction site.

Construction Automation was formed in May 2016 by entrepreneurs David Longbottom and Stuart Parkes.

“The house will contain around 10,000 bricks and will take the ABLR about two weeks to build”, said Longbottom. When completed, a farm manager will move into the three-bedroom single plot house in Everingham, Yorkshire.

According to Longbottom, “The ABLR comprises of the robot and a sophisticated software control system that reads digitised versions of architect’s plans.

“This instructs the robot exactly where to lay the blocks, bricks and mortar.”

The ABLR is controlled from a tablet, and further requires just two people to work on each house – a labourer to load bricks and mortar into the robot and a skilled person to install tie bars, damp courses, and lintels, and to do the pointing.

Sensors measure each individual brick and then to line it up, so it is precisely central on the wall. The sensors also align the edge of each brick to produce a perfect finish.

Although the ABLR is almost market-ready, the partners are already working on further innovations including another robot that has the ability to place tiles.

Evaluation of Surcharge Pressure of Pad Foundations on Retaining Walls

By
Ubani Obinna
-

Surcharge pressure on earth retaining walls can arise from different sources such as loads from adjacent buildings, fills, roadways and traffic actions, construction activities, and undulating/uneven ground surfaces, etc. Lateral pressures caused by surcharge loading may be calculated depending on the type of surcharge and the nature of load distribution. Elastic vertical stress calculated in the soil at the location is multiplied by the appropriate earth pressure coefficient to obtain the lateral pressure. In this article, we are going to evaluate the surcharge pressure exerted on a retaining wall from a nearby pad footing.

In geotechnical engineering, equations have been developed to compute stresses at any point in a soil mass based on the theory of elasticity. In non-elastic soil masses, the theory of elasticity still holds good as long as the stresses are relatively small, especially in overconsolidated soils. The real requirement in this case is the presence of constant ratio between the stresses and the strains, and not that soil is actually an elastic material. When a load is applied on a soil surface, it increases the vertical stress within the soil mass. The increased stress is greatest directly under the loaded area, but also extends indefinitely to other areas. Equations are available in geotechnical engineering textbooks for point loads, line loads, strips, and uniform pressure loads.

Standard textbooks provide solutions for vertical stress, σ’v,z, at depth z under a corner of a rectangular area carrying a uniform pressure q.

It is usually in the form;
σ’v,z = qK

where;
q = uniform pressure
K = coefficient. Values of K are provided for different aspect ratios of the loaded area to depth. Fadum’s chart (shown below) can be used to obtain the influence coefficients.

fadums chart
Fadum’s chart

The method of superposition can be used to determine the vertical stress under any point within or outside the loaded area. Generally, the effects of patch loads are derived by calculating pressures as if the load extends to the wall, and then subtracting the pressure for the patch load that does not exist. This is shown in the illustration below.

In the figure below, the surcharge from the pad footing is to be evaluated at different points on the retaining wall viz E,F, and G.

F1

At point E;

F2

Pressure at E = Pressure at the corner of rectangle EABH – Pressure at the corner of rectangle EDCH

At point F;

F3

Pressure at F = Pressure at the corner of rectangle FJAE + Pressure at the corner of rectangle FJBH – Pressure at the corner of rectangle FIDE – Pressure at the corner of rectangle FICH

At point G;

F4

Pressure at G = Pressure at corner of rectangle GLBH – Pressure at corner of rectangle GLAE – Pressure at corner of rectangle GKCH + Pressure at corner of rectangle GKDE

Solved Example
Calculate the lateral pressures caused by the existing pad foundation adjacent to the proposed basement shown below. The soffit of the pad is 2 m x 2 m in plan and bears at 175 kN/m2 at 1.5 m below the existing ground level. Assume the soil strata and soil properties as shown below.

Surcharge load from pad footing 1

Solution

footing plan

Earth Pressure coefficients
At rest earth pressure coefficient for fill = Ko = 1 – sinφ = 1 – sin(30) = 0.5
At rest earth pressure coefficient for sand = Ko = 1 – sinφ = 1 – sin(28) = 0.53
At rest earth pressure coefficient for clay (using effective stress approach) φ’ for clay of plasticity index of 15% = 30° = Ko,d = (1 – sinφ) x OCR0.5
Assuming overconsolidation ratio (OCR) of 2.0; = 1 – sin(28) x 20.5 = 0.707

Pressure calculation
Pressures are calculated at a corner of a rectangular area viz:
Pressure at F = 2 x [(pressure at F for a rectangle 4 m x 1 m) – (pressure at F for a rectangle 2 m x 1 m)]

At 1.5m below ground level; take z below footing base = 0
(For 4 x 1m rectangle)
m = L/Z = 4/0 = ∞
n = B/Z = 1/0 = ∞
Hence from Fadum’s chart, K = 0.25

(For 2 x 1m rectangle)
m = L/Z = 2/0 = ∞
n = B/Z = 1/0 = ∞
Hence from Fadum’s chart, K = 0.25

Pressure at P = 2[(175 x 0.25) – (175 x 0.25)] = 0

At 3.0 m below ground level; take z below footing base = 1.5 m

(For 4 x 1m rectangle)
m = L/Z = 4/1.5 = 2.67
n = B/Z = 1/1.5 = 0.667
From Fadum’s chart, K = 0.162

(For 2 x 1m rectangle)
m = L/Z = 2/1.5 = 1.333
n = B/Z = 1/1.5 = 0.667
Hence from Fadum’s chart, K = 0.16

Pressure at P = 2 x [(175 x 0.162) – (175 x 0.16)] = 0.7 kN/m2

At 4.0 m below ground level; take z below footing base = 2.5 m

(For 4 x 1m rectangle)
m = L/Z = 4/2.5 = 1.6
n = B/Z = 1/2.5 = 0.4
From Fadum’s chart, K = 0.113

(For 2 x 1m rectangle)
m = L/Z = 2/2.5 = 0.8
n = B/Z = 1/2.5 = 0.4
Hence from Fadum’s chart, K = 0.095

Pressure at P = 2 x [(175 x 0.113) – (175 x 0.095)] = 6.3 kN/m2

At 5.0 m below ground level; take z below footing base = 3.5 m

(For 4 x 1m rectangle)
m = L/Z = 4/3.5 = 1.142
n = B/Z = 1/3.5 = 0.285
From Fadum’s chart, K = 0.08

(For 2 x 1m rectangle)
m = L/Z = 2/3.5 = 0.571
n = B/Z = 1/3.5 = 0.285
Hence from Fadum’s chart, K = 0.062

Pressure at P = 2 x [(175 x 0.08) – (175 x 0.062)] = 6.3 kN/m2

At 6.0 m below ground level; take z below footing base = 4.5 m

(For 4 x 1m rectangle)
m = L/Z = 4/4.5 = 0.889
n = B/Z = 1/4.5 = 0.222
From Fadum’s chart, K = 0.056

(For 2 x 1m rectangle)
m = L/Z = 2/4.5 = 0.444
n = B/Z = 1/4.5 = 0.222
Hence from Fadum’s chart, K = 0.040

Pressure at P = 2 x [(175 x 0.056) – (175 x 0.04)] = 5.6 kN/m2

The at rest surcharge pressure on the retaining wall is therefore given by;

At 1.5m below the ground level; σ’kh = KoP = 0
At 3.0m below the ground level; σ’kh = KoP = 0.53 x 0.7 = 0.371 kN/m2
At 4.0m below the ground level; σ’kh = KoP = 0.53 x 6.3 = 3.339 kN/m2
At 4.0m below the ground level; σ’kh = KoP = 0.707 x 6.3 = 4.451 kN/m2
At 5.0m below the ground level; σ’kh = KoP = 0.707 x 6.3 = 4.451 kN/m2
At 6.0m below the ground level; σ’kh = KoP = 0.707 x 5.6 = 3.959 kN/m2

The characteristic surcharge pressure distribution on the basement wall from the pad footing is therefore given below;

surcharge earth pressure distribution on basement wall

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


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