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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


Effects of Diesel Contamination on the Engineering Properties of Natural Soils

By
Ubani Obinna
-

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

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

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

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

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

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

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

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

The original article from which this post is developed 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. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.

Design of Steel Beams for Combined Bending and Shear

By
Ubani Obinna
-

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

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

fyr = (1 – ρ)fy —– (1)

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

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

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

Where;
Aw = hwtw

Solved Example

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

Steel beam with high shear load

For the beam loaded as shown above;

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

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

fa

Try section 406 x 178 x 74 UKB

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

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

hw = h – 2tf = 380.8 mm

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

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

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

cf/tf = 74.8/16.0 = 4.68

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

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

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

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

Mc,Rd = Mpl,Rd = (Mpl,y × Fy)/γm0

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

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

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

VEd/Vc,Rd ≤  1.0

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

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

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

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

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

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

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

hw/tw ≤  72ε/η

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

40.1 < 66 Therefore shear buckling need not be considered.

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

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

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

Aw = hwtw = 380.8 x 9.5 = 3617.6 mm2

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

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

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

By
Ubani Obinna
-

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

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

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

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

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

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

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

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

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

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