8.7 C
New York
Wednesday, October 9, 2024
Home Blog Page 35

Design of Shear Walls

By
Ubani Obinna
-

Shear walls are structural members that provide additional lateral stiffness to a building by resisting shear, moment, and axial forces, which are produced due to gravity and lateral loads. In some design cases, the entire lateral loads coming to a building assumed to be resisted by shear walls alone, especially when it is the major stabilising/bracing component of the building. The design of shear walls involves providing adequate cross-section and reinforcements to resist bending, shear, axial, and twisting forces due to gravity and lateral loads.

For an element to be described as a reinforced concrete wall, the length to thickness ratio should be equal to or greater than 4 (Clause 9.6.1, Eurocode 2). A shear wall is said to be short when the height-to-width ratio is less than or equal to one, while it is said to be slender when the height-to-width ratio is greater than or equal to four. For slender walls, bending deformation is dominant, while in short walls, shear deformation is dominant. When the height-to-width ratio is in-between one and four, then the shear wall undergoes both shear and bending deformations.

shear walls and columns in a building
Shear walls and columns in a building

Forces Acting on Shear Walls

Shear walls are designed to resist bending moment, shear, axial, and uplift forces, especially when they are subjected to lateral actions. The lateral forces acting in the plane of a shear wall attempts to lift up one end of the wall and push the other end down. Uplift forces are greater on tall walls and less on low walls. Shear walls resist the shear force parallel to the plane of the wall by cantilever action.

Axial forces in a Shear wall
The axial load in a wall may be calculated assuming the beams and slabs transmitting the loads to it are simply supported.

Transverse moments
For continuous construction, transverse moments can be calculated using elastic analysis. The eccentricity is not to be less than h/30 or 20 mm where h is the wall thickness.

In-plane moments
Moments in the plane of a single shear wall can be calculated from statics. When several walls resist forces, the proportion allocated to each wall should be in proportion to its stiffness.

Location/Placement of Shear of Shear Walls

In a tall building where the shear wall is used for lateral stability, it should be located on each level of the structure. Preferably, shear walls of equal length should be placed symmetrically on all four exterior walls of the building to form an effective box structure. When the shear walls in the exterior frame could not provide sufficient strength and stiffness, the shear walls should be added to the interior frame as well.

If the shear wall is placed at the interior frame of a building, it attracts and resists higher internal forces but may not be too effective in reducing the maximum lateral deflection of the building. However, if a shear wall is placed at the ends of a building, the lateral deflection is reduced considerably when compared to the later.

Architectural disposition in a building may not always give the room for optimum placement of shear walls for good structural performance. In some areas such as lift areas or window/door openings, coupled shear walls may have to be used. However, the best position or arrangement of shear walls in a building is a symmetrical arrangement whereby the shear centre (centre of rotation) will coincide with the centre of gravity of the building in order to reduce or eliminate torsion (twisting) due to lateral loads.

pierced shear wall
Coupled shear wall in a building

Functions of Shear Walls

  • Shear walls must provide the necessary lateral strength to resist horizontal wind and earthquake forces.
  • They provide resistance against sliding through connections.
  • They also provide lateral stiffness to prevent the roof or floor above from excessive side sway.
  • When the shear walls are stiff enough, they prevent floor and roof framing members from moving off their supports.
  • Buildings that are sufficiently stiff usually suffer less structural damage due to the presence of shear walls.


Structural Action of a Shear Wall

In a shear wall, the primary mode of deformation is mainly due to flexure but not shear. The mode of deformation of shear walls is such that it has maximum slope at the top and least at the bottom which is the flexural mode shape. The structural action of a shear wall resembles that of a cantilever beam.

Displaced shape of a shear wall under lateral load
Typical deflected shape of a shear wall

Structural Design of Shear Walls

The amount of reinforcement needed for proper detailing of a reinforced concrete wall may be derived using strut-and-tie model. If a wall is however subjected predominantly to out-of-plane bending, the design rules and guidelines for slabs apply. In Eurocode 2, the requirements for the design of columns and shear walls are not so different except in the following areas;

■ The requirements for fire resistance
■ Bending will be critical about the weak axis
■ Rules for spacing and quantity of reinforcement

Reinforcement Detailing of Shear Walls

(a) Minimum and maximum area of vertical reinforcement
According to clause 9.6.2 of Eurocode 2, the minimum and maximum amounts of reinforcement required for a reinforced concrete wall are 0.002Ac and 0.04Ac outside lap locations respectively. It is further stated that where minimum reinforcement controls design, half of this area should be located on each face. The distance between two adjacent vertical bars should not exceed three times the wall thickness or 400 mm, whichever is lesser.

(b) Area of horizontal reinforcement
According to clause 9.6.3 of Eurocode 2, horizontal reinforcement should be provided at each face and should have a minimum area of 25% of the vertical reinforcement or 0.001Ac, whichever is greater. The spacing between two adjacent horizontal bars should not be greater than 400 mm.

(c) Provision of links
If the compression reinforcement in the wall exceeds 0.02Ac, links must be provided through the wall thickness in accordance with the rules for columns in clause 9.5.3 which are:

The diameter of the transverse reinforcement should not be less than 6 mm or one-quarter of the diameter of the largest longitudinal bar whichever is greater. The maximum spacing is to be Scl, max

Scl, max is the minimum of;

  • 20 times the diameter of the smallest longitudinal bar
  • The lesser dimension of the wall i.e. the thickness
  • 400 mm

The maximum spacing should be reduced by a factor of 0.6 in the following cases;

  • In sections within a distance equal to 4 × thickness of wall above or below a beam or slab.
  • Near lapped joints, if the diameter of the longitudinal bar is greater than 14 mm. A minimum of three bars evenly placed in the lap length is required.

Where the main reinforcement (i.e., vertical bars) is placed nearest to the wall faces, transverse reinforcement should be provided in the form of links with 4 per m2 of the wall area.

Design Example of Shear Walls

Design a 225 mm thick shear wall of 3.6 m height at the ground floor of the building. The shear wall is carrying a 200 mm thick slab on the first floor of the building. The action effects on the shear wall are as follows;

Vertical loads
Dead Load Gk = 300 kN/m
Live load Qk = 55 kN/m

Vertical load due to in-plane bending and wind Wk = ±650 kN/m
Vertical load due to in-plane bending and imperfections GkH = ±60 kN/m

Maximum moment out-of-plane, floor imposed load as leading action M = 40 kN/m @ ULS
Maximum moment out-of-plane, floor imposed load as accompanying action M = 35 kN/m @ ULS

fck = C25/30; fyk = 500 N/mm2

Check slenderness of wall at ground floor

Effective length, l0 = 0.75 × (3600 – 200) = 2550 mm (Table C16)
λ = 3.46 × l0/h = 3.46 × 2550/225 = 39.213 (Cl. 5.8.3.2(1))

Limiting slenderness, λlim = 20 ABC/n0.5 (Cl. 5.8.3.1(1))
where;
A = 0.7
B = 1.1
C = 1.7 – rm

where;
rm = M01/M02 = say = –0.25
C = 1.7 – (-0.25) = 1.95
n = NEd/Acfd

where;
NEd (assuming wind is the leading variable action) = 1.35Gk + 1.5Qk1 + 1.5Ψ0Qki
= 1.35(300 + 60) + 1.5(650) + (1.5 × 0.7 × 55) = 486 + 975 + 57.75 = 1518.75 kN/m
Acfd = (225 × 1000) × (0.85 × 30/1.5) = 3825 kN
Therefore, n = 1518.75/3825 = 0.397
λlim = 20 ×( 0.7 × 1.1 × 1.95)/0.3970.5 = 47.66

Therefore as λ < λlim wall is not slender and therefore no secondary moments.

Combinations of actions
(a) At ULS, for maximum axial load, Wk is leading variable action
NEd = 1.35Gk + 1.5Qk1 + 1.5Ψ0Qki
= 1.35(300 + 60) + 1.5(650) + (1.5 × 0.7 × 55) = 486 + 975 + 57.75 = 1518.75 kN/m

MEd = M + eiNEd ≥ e0NEd (Cl. 5.8.8.2(1), 6.1.4)

where
M = moment from 1st order analysis = 35 kNm/m
ei = l0/400 = 2550/400 = 6.375 mm (Cl. 5.2(7), 5.2(9))
e0 = h/30 ≥ 20 mm = 20 mm (Cl. 6.1.4)

MEd = 35 + (0.006375 × 1518.75) ≥ 0.020 × 1518.75
MEd = 35 + 9.68 ≥ 30.375 = 44.68 kNm/m

(b) At ULS, for minimum axial load, Wk is leading variable action
NEd = (1.0 × 300) – (1.35 × 60) – (1.5 × 650) + (0 × 55) = –756 kN/m (tension)

MEd = 35 + (0.006375 × 756) ≥ 0.020 × 756
= 35 + 4.82 ≥ 15.12 = 39.82 kNm/m

(c) At ULS, for maximum out of plane bending assuming Qk is leading variable action

NEd = 1.35(300 + 60) + (1.5 × 55) + (1.5 × 0.5 × 650) = 1056 kN/m

MEd = 40 + (0.006375 × 1056) ≥ 0.020 × 1056
MEd = 40 + 6.732 ≥ 21.12 = 46.732 kNm/m

or

NEd = (1.0 × 300) – (1.35 × 60) – (0 × 55) – (1.5 × 0.5 × 650) = -268.5 kN/m (tension)
MEd = 40 + (0.006375 × 268.5) = 41.71 kNm/m

Design load cases
Consolidate (c) into (a) and (b) to consider two load cases:

NEd = 1518.75 kN/m
MEd = 46.732 kNm/m (out of plane)

and

NEd = –756 kN/m (tension)
MEd = 46.732 kNm/m (out of plane)

Design: cover above ground
cnom = cmin + ∆cdev

where;
cmin = max[cmin,b; cmin,dur] (Exp. (4.1))

where
cmin,b = diameter of bar = 20 mm vertical or 10 mm lacers
cmin,dur = for XC1 = 15 mm
∆cdev = 10 mm
cnom = 15 + 10 = 25 mm to lacers (35 mm to vertical bars)

Design using charts
For compressive load:
d2/h = (25 + 10 + 16/2)/225 = 0.19

Interpolate between charts for d2/h = 0.15 and d2/h = 0.2 as shown below

Column interaction chart
Rectangular column interaction chart for d2/h = 0.15
column intreaction chart 2
Rectangular column interaction chart for d2/h = 0.2


NEd/bhfck = (1518.75 × 103)/(225 × 1000 × 30) = 0.225
MEd/bh2fck = (46.732 × 106)/(2252 × 1000 × 30) = 0.030

Gives:Asfyk/bhfck = 0
Therefore, minimum area of reinforcement required

As,min = 0.002Ac (Cl. 9.6.2 & NA)
As,min = 0.002 × 225 × 1000 = 450 mm2/m
As,min = 450 mm2/m = 225 mm2/m each face
maximum spacing = 400 mm c/c, minimum diameter of rebar = 12 mm diameter

Try H12 @ 300 c/c on each face

For tensile load and moment
Working from first principles, referring to the figure below and ignoring contribution from concrete in tension,

stresses and strain in a shear wall subjected to bending and tension
Stresses and strains in wall subject to tension and out of plane moment


NEd = (σst1 + σst2) × As/2
and MEd = (σst1 – σst2) × As/2 × (d – d2)

So σst1 + σst2 = 2NEd/As
and σst1 – σst2 = 2MEd/[(d – d2)As]

2σst1 = 2NEd/As + 2MEd/[(d – d2)As]
As = (NEd/σst1) + MEd/(d – d2)σst1

σst1 = fyk/γs = 500/1.15 = 434.8

As = (756 × 103/434.8) + (46.732 × 106)/[(182 – 43) × 434.8] = 1738.7 + 773 = 2512 mm2

σst2 = 2NEd/As – σst1 = 601.9 – 434.8 = 167 MPa

By inspection all concrete is in tension zone and may be ignored.
Use 7H16 @ 175 c/c on both sides for at least 1 m each end of wall (Asprov = 2814 mm2).

Horizontal reinforcement
As,hmin = 0.001Ac or 25% As,vert (Cl. 9.6.3(1) & NA)
= 225 mm2 or (0.25 × 2814) = 704 mm2/m
This therefore requires 352 mm2/m each side
Use H10 @ 200 (393 mm2/m) both sides

Links
Check 0.02Ac = 0.02 × 225 × 1000 = 4500 mm2 > 2814 mm2
Therefore, links not required.
.



Technical Guide: Detailing and Arrangement of Beam Reinforcements on Site

By
Mubarak Shittu
-

Beams are horizontal structural elements used for supporting lateral loads. In conventional reinforced concrete structures, beams usually receive load from the floor slab, but may also be subjected to other loads such as wall load, finishes, services installation, etc. The design and detailing of reinforced concrete beams involves the selection of the proper beam size and the quantity of longitudinal and shear reinforcement that will satisfy ultimate and serviceability limit state requirements. Afterward, it is very important that the beam reinforcements are placed and arranged properly on site, to avoid construction error.

This article presents guides and short notes on how to properly place and arrange beam reinforcements on site. This is very important for students (on industrial training) or fresh graduates that are new to construction site practices.

Detailing and arrangement of beam reinforcements
Typical detailing of a reinforced concrete beam

The technical guides to the detailing and arrangement of beam reinforcements are as follows;

(1) Confirm the formwork dimensions and stability
Beam reinforcement placement commences immediately after the carpenters complete the soffit formwork of the floor. At this point, it is important to verify that the formwork dimensions have been done according to the design specification. The checks should include measuring the beam width and the drop of the beam relative to the proposed surface of the slab. The sides of the beam drop should be checked for verticality to avoid having slanted beam sides, and the soffit should be checked for perfect horizontality. It is usually very difficult and depressing to make corrections after the reinforcement has been placed.

Poorly sized beam dimensions can also lead to compromised concrete cover, which can impact the bonding of concrete and reinforcement, the fire rating, and the durability of the building. It is typical to take a bent link (stirrup) and insert it in the beam drop to check if the desired concrete cover has been achieved. It can also be done to confirm that the beam size has been done according to specification. Furthermore, it is important to check the stability and bracing of the formwork to avoid collapse, bulging, or bursting.

(2) Confirm cut length and bending dimensions
Before the placement of the rebars commences, the site engineer should confirm that reinforcements have been cut and bent according to the design requirements. Bar bending schedule (BBS) can be used as a check document, provided it is very accurate and synchronised with what has been done on-site. All changes in dimensions or arrangements should be communicated ahead of time.

cut and bent reinforcement at a construction site awaiting installation
Cut and bent beam reinforcements at a construction site awaiting tying and placement

(3) Tying of Beam Reinforcements
Links and stirrups are used for resisting shear stresses and torsion in a reinforced concrete beam. They are also used for holding the top and bottom reinforcement of beams in place. The correct quantity required should be installed and spaced as recommended in the structural drawing.

tying of beam reinforcement on site
Tying and arrangement of beam reinforcement on site

(4) Arrangement of Primary and Secondary Beams
For an internal beam (secondary beam), which is supported by another beam on either or both sides, the top bars should rest on the top bars of the primary beam. Also, the bottom bars should rest on the bottom bars of the primary beam as shown in the image below.

beam reinforement

(5) Arrangement of Beam-Column Junction at Corners
For beams occurring at a column junction at the corner of a building, either beam reinforcements can rest on each other.

corner beam column junction
Corner beam-column junction

(6) Arrangement of Continuous Beam-Column Junction
At a continuous beam-column junction as shown below, the continuous beam with heavier reinforcement will have to be placed on the beam with lighter reinforcement or the beam terminating at that point.

Internal beam column junction
Internal beam-column Junction

(7) Reinforcement Arrangement of Overhang (Cantilever) Beams
For cantilever beams, the main bars are the top reinforcements. If a side beam is resting on it (when the cantilever beam is acting as a primary beam) as shown in the image below, the top bars of the secondary beams beam must rest on the top bars of the cantilever beams. In the same vein, the bottoms bars of the side beam (main rebar) must rest on the bottom bars of the cantilever beam.

typical cantilever beam
Well designed and constructed cantilever beam

The arrangement of beams at a site should be checked strictly to conform to the engineer’s design drawings. Failure to check construction details properly may lead to the failure of the member or lack of robustness. We believe that beams are arguably the most common structural members in a building. Hence, attention should be given to its arrangements for robustness and efficient load transfer.

New Insights on the Failure of Fukae Bridge (Kobe 1995)

By
Ubani Obinna
-

Researchers from the Department of Civil, Environmental, and Geomatic Engineering, ETH Zurich, Switzerland have offered new insights on the failure of Fukae Bridge during the Great Hanshin 1995 earthquake in Kobe, Japan. The magnitude 6.9 earthquake which left many damages behind its wake caused about $100 billion loss in properties, including the collapse of all 18 spans of the elevated Route No. 3 of Hanshin Expressway.

In the catastrophic seismic event, the deck of the bridge which was monolithically connected to 3.1m diameter piers failed and overturned dramatically. However, the massive 17–pile groups supporting the piers survived the earthquake and are still in use, supporting the new bridge. The piles were founded in alluvium sand and gravel formation.

seismicity of kobe
Figure 1: Japan, showing seismicity from 1961 to 1994, location of the 1995 Hanshin-Awaji
earthquake, and projected rupture areas of largest historical earthquakes to shake Kobe, which
were subduction-zone earthquakes in 1944 and 1946 (Source: https://www.geosociety.org/)

According to the authors, the lessons learned from the Kobe earthquake influenced substantially the seismic practices and codes not only in Japan but also worldwide. However, new insights have been offered on the failure of the bridge as the researchers carried out nonlinear finite element analysis of the bridge. The findings of the study were published in Elsevier -Soils and Foundations Journal.

Previous studies attributed the failure of the bridge deck to inadequate structural design regarding mainly the addition of a prematurely terminated third row of longitudinal reinforcement and insufficient shear capacity due to poor transverse reinforcement. Surprisingly, the failure did not occur at the bottom of the pier (location of maximum bending moment) but 2.5 m above the pilecap, where shear cracking initiated.

bridge pier
Figure 2: Reinforcement details and critical section where the 3rd row of longitudinal rebars was terminated and shear cracking initiated; lower section of
a collapsed pier exposed in the Hanshin Expressway earthquake museum (Sakellariadis et al, 2020)

The authors, therefore, re-examined the collapse of the bridge by comparatively assessing the performance of the original foundation, which survived the earthquake and is still in use (for the fully replaced bridge), to that of alternative design concepts, considering nonlinear soil-foundation interaction.

To achieve this, they carried out the static and dynamic response of a single segment of the Fukae bridge employing the FE method using ABAQUS software. Six different foundation configurations were explored, starting with the actual very stiff 17–pile foundation with its large cap and a highly nonlinear rocking shallow footing alternative consisting only of the pile cap. The soil profile was modelled with hexahedral (8-node) elements, adopting a thoroughly validated kinematic hardening model, with a modified pressure-dependent Von Mises failure criterion and associated plastic flow rule.

abaqus model
Key attributes of the FE model of a single segment of the Fukae bridge in Abaqus (Sakellariadis et al, 2020)

To gain deeper insights on the collapse mechanism, critical reinforced concrete (RC) structural members (pier and piles) were modelled and simulated with nonlinear solid elements, employing the Concrete Damaged Plasticity (CDP) model.

The numerical simulation successfully reproduced the shear-dominated failure mode at the longitudinal reinforcement cutoff region. The analysis also confirmed that, despite being highly overdesigned, the pile group foundation experienced limited but non-negligible swaying and rocking during shaking, as a result of which the piles were subjected to tension and combined shear-moment loading.

The resulting stiffness reduction of the cracked under tension piles leads to load redistribution towards the stiffer compressed piles, preventing plastic hinging of the weaker piles (under tension). These findings were found to be consistent with the post-earthquake in-situ testing.

Some of the conclusions inspired by the study are as follows;

  1. Reducing the number of pile rows in the critical direction of seismic loading is promising in improving the seismic performance, but not sufficient to prevent severe damage of the examined pier.
  2. The unconnected piled raft alternative reduces both the structural distress and the settlement.

References
L. Sakellariadis, I. Anastasopoulos and G. Gazetas, Fukae bridge collapse (Kobe 1995) revisited: New insights, Soils and Foundations, https://doi.org/10.1016/j.sandf.2020.09.005

Disclaimer
The original article cited above is an open-access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/) which permits sharing, copying, and redistribution of the material in any medium or format provided appropriate credit is given. It has been presented on www.structville.com in form of a news article.


Structural Design Challenge | Explore your creativity

By
Ubani Obinna
-

Building design concepts all over the world are evolving on a daily basis. Architects are coming up with different ways of making buildings appear sleeker, elegant, and more beautiful. While the aesthetic appeal offered by these contemporary buildings are commendable and decorate our environment, they can be a nightmare to structural engineers who have the responsibility of making the buildings safe and stable.

As a modern structural engineer, there is no good reason for not following up on the latest trends in design and construction. All the modern solutions available for analysis, design, and construction of complex-looking buildings should be handy to you. Furthermore, engineers should always follow up on the latest developments in construction materials and techniques by attending workshops, seminars, webinars, expos, conferences, etc that are relevant to their field of practice. That is one of the best ways of staying current and relevant in the industry.

A3

At Structville, we are interested in always driving up curiosity and creativity in the minds of civil engineers. It is not a good idea to ‘over-modify’ the concepts of an architect in the quest for structural stability. Therefore, you should not lack ideas on how to approach any problem that comes to you. That is the major sign of a quality engineer.

Ideas on how to support a complex architectural design may not always come immediately, but if you sleep and play over it, bearing in mind the ‘basic’ principles of statics of structures, and having a wide knowledge of different structural schemes, you will eventually find the solution. In practical design, you seldom need to think far or explore complex theories before you get an idea; the basic knowledge of statics of structures is usually sufficient.

I came across an interesting architectural design on Instagram. No credit was given for the design, but the handle that posted the pictures is @academiadeestruturas. I am posting the pictures of the design below for us to explore our ingenious ways of carrying out the structural design of the building.

It is my hope that by the time a lot of professionals have contributed their ideas on how to go about the design, a lot of knowledge must have been passed around. For this structural design challenge, you can post your ideas as a comment or you can support with relevant sketches if necessary. The requirements of the challenge are as follows;

(1) Describe the structural scheme that you will use to support the building
(2) Describe the possible load path of the structural scheme
(3) Comment on the stability of the selected structural scheme
(4) State the most likely material that will work for the design

In a general, let us assume that the floor should be able to withstand a live load of 3.0 kN/m2. No calculations whatsoever are needed.

The pictures of the architectural design are given below;

A1
A2
A10
A3
A4
A5
A6
A8
A9

You are also free to comment or advise on the structural scheme selected by others. However, keep all interactions and use of language civil, professional, and respectful. Every engineering solution is a solution, but some solutions are better than the others. Remember to share this challenge with your friends and colleagues. Thank you, and God bless you.

Photo credit: Instagram – @academiadeestruturas

Precast Lintels: A Cost and Time-Saving Solution in Construction

By
Mubarak Shittu
-

A lintel is a secondary structural member used to support a wall spanning above an opening in a masonry wall. They are usually categorised as elements of minor structural importance unless the span of the opening is very large. Lintels are usually horizontal members with a rectangular cross-section and are often made of reinforced concrete or steel. Other materials such as stone, brick, reinforced brick, and timber can also be used.

Lintels are important in a building, and the details can be included in the structural drawing in the form of lintel schedule to show the rebar specification based on the size of openings. Such a schedule can aid the quantity surveyor to capture the cost of the lintels in the bill of quantity.

In Nigeria, the most popular material used for lintel is reinforced concrete, which may be cast-in-situ or precast. Cast-in-situ lintels can be constructed to span around the entire length of walls in the building, and are often referred to chained lintels. On the other hand, they can be constructed to span across the opening concerned without extending far beyond the opening. Precast lintels are often used to span across the opening concerned only.

Building with chained lintel
Chaining of lintels in a building

Precast lintels always come in handy for residential buildings and are mostly employed by value engineers. The major benefits include cost and time saving for the project or contractor.

Precast lintels installed in a build
Precast concrete lintel

Essentially, the idea behind thin precast lintels is that the provision of top reinforcements and hanger bars can be neglected for lightly loaded one-span members in bending. Reinforcements are provided at the bottom of the lintel to take up all the flexural stresses. The conventional cast-in-situ lintel, however, has an edge over precast lintel if the building will be subjected to remodeling or if the lintels spans between reinforced concrete columns.

Advantages of Precast Lintels

The advantages of using precast lintels in construction are as follows;

  1. Significant saving in reinforcement is obtained by using offcuts from other elements
  2. It more sustainable because of the reduction in cement and aggregates requirements
  3. Quality control in the production of the lintel is enhanced
  4. It also saves and/or reduces the cost of formwork
  5. It reduces the cost of labour
  6. It is relatively easy to construct
  7. Saves time on the construction program since it can be produced and stored ahead of schedule.

Disadvantages of Precast Lintels

  1. A lifting mechanism may be required to put the lintel in place
  2. The robustness of the wall/element is reduced
  3. The bond strength of the wall and the lintel is weaker when compared with cast-in-situ

Specifications for Precast Lintels

(1) Sizing: The width of a lintel should be equal to the width of the wall they are supporting or sitting on, unless otherwise specified. The depth of a lintel should be a minimum of 100 mm, and the length depends on the span of opening. An end bearing length of at least 200 mm beyond the face of the opening should be provided. When the span of a lintel exceeds 2m, a more serious consideration should be given.

(2) Concrete: A minimum concrete grade of 15 MPa should be used for the construction of lintels, unless otherwise specified. The quality of the cement, aggregate, and water should be the minimum requirement used for the construction of major reinforced concrete elements.

(3) Reinforcements: A 2 or 3 number of high yield 12mm diameter bars (2Y12 or 3Y12) placed at the bottom of the lintel is usually sufficient. However, the minimum area of reinforcement required for the section provided should not be violated (0.13% of the cross-sectional area). Top reinforcements (hanger bars) with links (stirrups) may or may not be provided depending on the depth adopted. When links are provided, the minimum area and spacing required should be provided.

IMG 20201111 212104
Typical reinforcement cutting and fixing for a precast lintel

Technical Guide for Construction of a Precast Lintel

(1) Surface preparation: A flat surface is required for casting of a precast lintel. This can be achieved on a well-leveled floor or on a marine plywood placed on a hard surface. The marine board will cover all undulations that may come from the surface, and will give a smooth finished appearance.

(2) Ease of removal: A cement sack or membrane should be spread on the surface to receive the lintel concrete, and to aid in quick removal without sticking with the platform. Alternatively, the surface of the marine board can oiled using an approved material.

(3) Shuttering/Mould Construction: The mould for the precast concrete can be achieved using planks or marine boards joined together. Special moulds made of steel or plastic can also be used where available.

(4) Concrete cover: Provision of nominal concrete cover (to the links) not less than 15mm should be made. Normally, fire rating, bond, and durability (exposure conditions) of the lintel should inform the choice of concrete cover.

(5) Concreting: A concrete mix of strength not less than 15 MPa after 28 days should be carefully poured and consolidated in the form/shuttering prepared. A vibrator is usually not needed for precast lintels as consolidation can be achieved by tamping the concrete or tapping the sides of the form/shuttering. The top surface can be finished using a trowel.

(6) Curing: Remove the forms after 24 hours and cure using any suitable method for at least 7 days before placement.

(7) Handling and Storage: Precast concrete lintels are fragile elements that can get damaged when not properly handled. All forms of impact or dropping from height should be avoided. The elements should be stored on a flat surface and protected until they are due for placement.

IMG 20201111 212333
Arrangement of temporary forms on a flat surface for precast lintel
IMG 20201111 212132 1
A unit of precast lintel of 100mm x 230mm x 1300mm

Placement/Installation of Precast Lintels

  1. Place high-quality cement mortar around the end bearing of the masonry wall that will support the lintel
  2. Lift the precast lintel carefully and place it on the mortar taking note of alignment and level
  3. Apply temporary support to transfer the stresses efficiently.

Recommendation

  1. Architects should make an allowance of using precast lintels in their designs by making window or door opening spans on masonry walls. Though this may not always be feasible due to the presence of columns in important areas.
  2. Low-cost and mass housing schemes can employ precast lintels to save cost and improve on delivery time.
  3. Value Engineers, Contractors, and Consultants should adopt precast lintels in their constructions.

Conclusion

The use of precast lintel is a cost and time-saving alternative in construction. It is also a sustainable practice that can lead to the reuse of waste materials on-site like empty cement bag, old planks/formworks, reinforcement offcuts, and waste concrete materials.

PS: At Structville, we value sustainability and value engineering. Do you want a review and analysis of your drawings and bills to see where you can save cost and time without sacrificing quality? You can reach us at info@structville.com to have a deal with our team of value engineers.

Elastic Analysis of Circular Plates

By
Ubani Obinna
-

A plate is a flat structural element that has a thickness that is small compared with the lateral dimensions. Plates are bounded by two parallel planes, called faces, and a cylindrical surface called an edge or boundary. The thickness is usually constant but may be variable and is measured normal to the middle surface of the plate. Circular plates are common in many structures such as nozzle covers, end closures in pressure vessels, pump diaphragms, turbine disks, and bulkheads in submarines and airplanes, etc. They are also used as bases or covers of cylindrical water tanks and structures. When they are resting on the ground, they behave as plates on an elastic foundation.

circular concrete plate
Typical circular plate covering an undergound tank

In the elastic analysis of circular plates, it is convenient to express the governing differential equation in polar coordinates, even though rectangular (Cartesian) coordinates can be used (as applied in this article). This can be readily accomplished by a coordinate transformation and can be found in many pieces of literature.

If the coordinate transformation technique is used, the following geometrical relations between the Cartesian and polar coordinates are applicable;

elastic analysis of circular plate

x = a cosθ; y = a sinθ; x2 + y2 = a2 —- (1)
θ = tan-1(y/x)

Referring to the above;
∂r/∂x = x/r = cosθ; ∂r/∂y = y/r = sinθ
∂θ/∂x = -y/r2 = -sinθ/r; ∂θ/∂y = x/r2 = cosθ/r

The governing differential equation (biharmonic equation) for the elastic deflection of a plate in polar coordinates is given by equation (2);

∇r4w = (∂2/∂r2 + ∂/r∂r + ∂2/r2∂θ2)(∂2w/∂r2 + ∂w/r∂r + ∂2w/r2∂θ2) = q/D — (2)

This is equivalent to the moment-curvature relationship of a thin plate which is given by (3);

∂4w/∂x4 + 2∂4w/∂x2∂y2 + ∂4w/∂y4 = q/D — (3)

Where D is the flexural rigidity of the plate which is given by;
D = Eh3/12(1 – ν2) — (4)

h = thickness of the plate
E = Modulus of elasticity of the plate material
ν = Poisson ratio of the plate material

Circular Plate Clamped at the Edge and Subjected to Uniform Load

CLAMPED CIRCULAR CARRYING UNIFORM LOAD

Let us consider a circular plate clamped at all edges (fixed) and subjected to a uniform lateral load q. The boundary of the plate is given by y2 + x2 = a2

For a such a clamped circular plate, the slope and deflection at the edges are zero;
w = 0; ∂w/∂x = 0; ∂w/∂y = 0 along y2 + x2 = a2

The solution to the deflection equation of the plate is given by;

w = c(x2 + y2 – a2)2 which satisfies the condition given above.

Let f(x, y) = x2 + y2 – a2
The partial derivatives are given by;

∂w/∂x = 4cxf; ∂w/∂y = 4cyf
∂2w/∂x2 = 4c(2x2 + f); ∂2w/∂x∂y = 8cxy; ∂2w/∂y2 = 4c(2y2 + f)
∂3w/∂x3 = 24cx; ∂3w/∂x2∂y = 8cy; ∂3w/∂x∂y2 = 8cx; ∂3w/∂y3 = 24cy
∂4w/∂x4 = 24c; ∂4w/∂x2∂y2 = 8c; ∂4w/∂y4 = 24c

Substituting this into the biharmonic equation, we can verify that;

c = -q/64D

Hence the elastic deflection of a circular plate clamped at all edges and subjected to a uniform lateral load is given by;

w = -q/64D(x2 + y2 – a2)2

The maximum deflection occurs at the centre of the plate and it is therefore given by;

wmax = -qa4/64D

The moments occurring in the plate from the moment-curvature equations are;

Mx = -q/16[(3 + v)x2 + (3v + 1)y2 – (1 + v)a2]
My = -q/16[(3v + 1)x2 + (3 + v)y2 – (1 + v)a2]
Mxy = q/8(1 – v)xy

The out-of-plane shear forces can be given by;

Vx = -qx/2; Vy = -qy/2

At the centre of the plate;

Mmax = q/16 × (1 + v)a2; Vx = Vy = Mxy = 0
The maximum moment occurs at the edge of the plate and it is given by Mr = -qa2/8

Simply supported circular plate subjected to a uniform lateral load

SIMPLY SUPPORTED CIRCULAR PLATE CARRYING UDL

The boundary conditions are at the support, the deflection is zero, and the bending moment is zero.
w = 0; Mx = My = 0

The equation of the deflected surface is given by;

w = C3r2 + C4 + qr2/64D

The expression for the bending moment Mr can be given as;

Mr = -D[2C3(1 + v)+ qr2/16D × (3 + v)]

Introducing the equations for bending moment and deflection into the boundary conditions;

C3 = -qa2(3 + v)/32D(1 + v); C4 = qa4(5 + v)/64D(1 + v)

Therefore, the elastic deflection of a circular plate simply supported at the edges is given by;

w = [-q(a2 – r2)/64D × ((5 + v)/(1 + v) × (a2 – r2))]

The maximum deflection which occurs at x = y = 0 is therefore;

wmax = qa4/64D × (5 + v)/(1 + v)

The bending moments are as follows;

Mr = q/16(3 + v)(a2 – r2)
Mt = q/16[(3 + v)a2 – (1 + 3v)r2]
Note: r2 = x2 + y2

The maximum moment occurs at the centre of the plate (r = 0), and it is given by;

Mmax = qa2/16 × (3 + v)

Clamped circular plate with a concentrated force at the center

CLAMPED CIRCULAR CARRYING CONCENTRATED LOAD

Clamped circular plate with a concentrated force P at the center, the expression for the deflection is given by;

w = P/16πD× (2r2In(r/a) + a2 – r2)

The bending moments are given by;

Mr = P/4π [(1 + v)In(a/r) – 1]
Mt = P/4π [(1 + v)In(a/r) – v]

Simply supported circular plate with a concentrated force at the center

Simply supported circular plate with a concentrated force at the center

For a simply supported circular plate with a concentrated force P at the center, the expression for the deflection and bending moments are given by;

w = P/16πD× (2r2In(r/a) + (3 + v)/(1 + v) × a2 – r2)
Mr = P/4π [(1 + v)In(a/r)]
Mt = P/4π [(1 + v)In(a/r) + 1 – v]

Solved Examples on the Analysis of Circular Plates

A circular plate of thickness (150 mm) and diameter 10m is subjected to a uniform lateral load of 10 kN/m2. Calculate the deflection and bending moment at the centre of the plate assuming;
(a) Simply supported conditions
(b) Clamped edge conditions
(v = 0.2; E = 2.1 x 107 kN/m2)

Solution
The flexural rigidity of the plate is given by;
D = Eh3/12(1 – ν2) = (2.1 × 107 × 0.153)/12(1 – 0.22) = 6152.34 kNm

(a) Assuming simply supported edge
(i) Maximum deflection
wmax = qa4/64D × (5 + v)/(1 + v) = (-10 × 54)/(64 × 6152.34) × (5 + 0.2)/(1 + 0.2) = 0.06877 m = 68.78 mm

(ii) Maximum bending moment
Mmax = qa2/16 × (3 + v) =(-10 × 52)/16 × (3 + 0.2) = 50 kNm

When analysed using finite element analysis on Staad Pro;

deflected profile of circular plate

The maximum deflection at the centre of the plate was observed to be 68.749 mm as shown in the table below. This comparable with 68.78 mm obtained using classical theory.

finite element analysis result

The bending moment obtained using finite element analysis was observed to be 49.9 kNm/m as shown below. A value of 50 kNm was obtained using classical theory.

bending moment on circular plate


(b) Assuming clamped edge conditions;
(i) Maximum deflection
wmax = -qa4/64D = (-10 × 54)/(64 × 6152.34) = 0.0158 m = 15.87 mm

(ii) Bending moment at the center
Mmax = q/16 × (1 + v)a2 = 10/16 × (1 + 0.2) × 52 = 18.75 kNm

(iii) Bending moment at the edge
Mr = -qa2/8 = (-10 x 52)/8 = -31.25 kNm


When analysed on Staad Pro;

deflected profile of a clamped circular plate 1


The maximum deflection at the centre of the plate was observed to be 15.964 mm (compared with the value of 15.87 mm obtained using classical theory)

finite element analysis of result of a clamped circular plate

The bending moment values from the finite element analysis result are shown below;

bending moment of a clamped circular plate

Summary of Analysis Result

Type of plate/Action EffectStaad Pro (FEA)Classical MethodPercentage Difference
Simply supported
Deflection (mm)
68.75 mm
68.78 mm
-0.0436%
Span Moment (kNm)49.9 kNm50 kNm-0.2%
Clamped Plate
Deflection (mm)
15.964 mm
15.87 mm
0.563%
Span Moment (kNm)
Edge Moment (kNm)		18.71 kNm
-27.8 kNm		18.75 kNm
-31.25 kNm		-0.213%
-12.41%

Bar Bending Schedule: Preparation, Applications, and Standards

By
Mubarak Shittu
-

On a construction site, there are different quality control/quality assurance systems to ensure the efficient delivery of a project. One of these is the Bar Bending Schedule (BBS). A bar bending schedule is a document showing the list of structural members, bar mark, type of reinforcement, size of rebar, number of rebars for each member, cutting length, total length, shape, and location/spacing/position of all reinforcements in the working drawing.

Bar bending schedule ensures that reinforcement cutting, bending, and placement are carried out in the most efficient manner on site. It also guides against the excessive waste of reinforcement by minimising the number of useless offcuts. The document is prepared in a manner such that the reinforcement requirements and specification can be recognised and applied without confusion on site. It is issued for different structural elements such as reinforced concrete beams, columns, slabs, foundation, staircase, etc.

Typical bar bending schedule

In Nigeria, the table showing lists of standard shape codes given in BS8666 doesn’t necessarily conform to the style used in Nigerian construction sites. Some modifications are usually done to avoid confusion, mistakes, and to reduce the workload/effort of the fitters (iron benders). It is very typical for detailers to sketch the actual bending shape on the document instead of referring to the standard shape code.

Read Also…
Preparation of bar bending schedule for slabs
Structural design and detailing standards in Nigeria

The importance of bar bending schedule on a construction site cannot be overemphasized. Some of these includes:

  1. To serve as a control document for Clerk of work, Inspectors, Detailers, and Structural Engineers on reinforcement
  2. For quantification of materials by the quantity surveyor during pre-contract and post-contract operations
  3. To assist steel bender and fixer
  4. To fast track construction and supervision;
  5. To plan, detect inconsistencies, and save costs in the handling of reinforcements.

Preparation of Bar Bending Schedule

Preparation of bar bending schedule should not be a haphazard operation. It should begin with the appropriate study of the reinforcement detailing drawings. All inconsistencies or errors (such as repeated bar marks for different types of bars) discovered in the drawing should be reported to the detailer, because as practically as possible, the details of the structural drawing and the bar bending schedule should be synchronised.

The major check lists before the commencement of BBS includes but not limited to the following:

  1. Check the concrete cover to each face
  2. Check the lapping length for tensions and compression bars of each structural elements
  3. Check the direction of bend of each bar
  4. Group each structural unit and floor by floor. For example, all ground floor elements must be scheduled before proceeding to next floor and all footing must finish before proceeding to starter columns.
  5. Bar mark must start from 01 and increasing upward consecutively
  6. Good knowledge of detailing, and considerations affecting the practical construction.

Notations in reinforcement detailing

It is also very important to be familiar with the style used by the draughtsman and its corresponding meaning in the standard detailing practice. For example, in the detailing of a floor slab, the standard method and the corresponding preferred style in Nigeria is as follows:

British Standard
Bottom (face): B1 (outer layer) and B2(second layer)
Top (face): T1 (outer layer) and T2 (second layer)

Nigerian style
B1 = Bottom Bottom (BB) or Bottom (B)
B2 = Bottom Top (BT) or Near Bottom (N)
T1 = Top Top (TT) or Top (T)
T2 =Top Bottom (TB) or Near Top (NT)

It is however common for the notation used in the structural detailing to be given in the design notes.

SLAB%2BDET%2B2
SECTION

For example, a main bar of a slab is typically detailed as 12H12-01-250B1 according to standard. However, the common notation used in Nigeria is 12Y12-01-250 BB, which means 12 numbers of high yield deformed bars of 12mm nominal size, at a spacing of 250mm center to center, in the bottom-bottom layer. The bar mark is -01- (Note: BB is the same thing as bottom outer layer). There might not be so much difference in notation based on what is seen above, but it is very important for the detailer to familiarise himself with the loacal standards.

Sections of a Bar Bending Schedule

A typical BBS contains the heading and the schedule section. The heading usually contains information which includes the following:

COMPANY NAME: Provide the name of the company issuing the schedule e.g. Structville Integrated Services Ltd.
PROJECT: Provide the title of the project. E.g. Construction of UniAbuja Main Gate
JOB NO: This is the number assigned to the job by the company or the client.
PREPARED BY: This shows the person that prepared the schedule
DRAWING NO: This gives the particular page in the drawing which references the schedule
DATE: This shows the date the schedule was prepared with reference to any revised schedule
SHEET NO: This shows the page by page documentation/compilation of the schedule

The presentation style of the schedule section of a BBS varies. Different companies always present their schedule in a way that suits them and their workers. Some schedules include the length and number of offcuts expected from a standard length of reinforcement or structural element to guide them in the reuse of reinforcements. For instance, two variants of schedules are shown in the images below.

BBS 1
Bar%2BBending%2BSchedule%2BTable

However, a typical BBS schedule section contains 10 columns in the following order:

[1] MEMBER: This shows the particular member in consideration. E.g base type 1, base type 2, column type 1, column type 2 etc

[2] BAR MARK: This depicts different bar marks present in the drawing. E.g. 01, 02, 03 etc.

[3] TYPE and SIZE: Type relates to whether the specified reinforcement is of high or mild yield, while size depicts the diameter of the bar and typically includes 8, 10, 12, 16, 20, 25, 32, and 40mm bars.

[4] NO. OF MEMBER: This shows the number of times the particular structural elements occur. E.g In the foundation details, how many numbers of Base Type 1 are there?

[5] NO. OF REBAR IN EACH: This means the number of times that particular bar marks occur in that particular member.

[6] TOTAL NO. OF REBAR: This column is the multiplication of columns [5] and [6] to show the cumulative of that particular bar mark.

[7] CUT LENGTH: This column comes next after the previous column. However, in preparing the schedule, the column showing the rebar “shape drawing” has to be done first in order to arrive at the cutting length.

[8] TOTAL LENGTH: This column is the multiplication of columns [6] and [7] to give the overall length of a particular bar mark of a particular element in the entire drawing. The total length is what is used by a Quantity Surveyor to quantify and price. From there, each length or tonnage is gotten as an executive summary. It is also important for data to know how many labourers or man-hours that will be expended on the work.

[9] LOCATION OR POSITION OR SPACING: This is the 9th column on the schedule. Here, if it is a wall, slab or foundation, the layers, and spacing of the bar are shown. If it is a link (stirrup), the spacing is shown and if it is a beam, the position is shown. It is important information for steel fixers.

[10] SHAPE DRAWING: This is the 10th and final column on the schedule. It shows the drawing and the dimensions of all the bends, straights, and the direction each bar is assumed to face after placement. This approach is more detailed and straightforward compared to the main standard where you need a separate sheet of shape code list and separate columns for each dimensions.

For a contractor that uses a fast-track approach, this column is of utmost importance to prefabricate all the shapes and fix all elements ready for placement. E.g all the columns in a story building can be done immediately after mobilization to site. This saves time on the construction program.

Conclusion

The main standard has been domesticated and adapted to be peculiar to the design and construction industry in Nigeria. The Nigerian standard style conforms to BS8666:2005. It is satisfactory and praiseworthy to say that the Nigerian style is comprehensive and easy to understand by a layman. We believe that complex English, Mathematics, Schedule, or Drawing should not be used to confuse sense of good engineering.

Shear Wall-Frame Interaction in High-Rise Buildings

By
Ubani Obinna
-

Shear walls and frames (comprising of columns and beams) are distinct structural systems that can be used in resisting lateral actions in high-rise buildings. However, as a building goes higher, frames alone become inadequate for lateral stability, hence, the structure can be augmented by shear walls and/or cores. Shear wall-frame interaction for lateral load resistance is complex because shear walls deflect primarily in bending mode, while frames deflect in shear mode.

However, the interaction between shear walls and frames is beneficial for high-rise buildings, since the linkage and stiffness of the floor slab diaphragm and the stabilising elements give better lateral load resistance. Furthermore, since their mode of deflection varies, the frame tends to restrain the shear wall in upper storeys and the shear wall tends to restrain the frame in the lower storeys. This reduces the lateral deflection and improves the overall efficiency of the structural system.

Shear Wall-Frame Interaction in High-Rise Buildings
Frame-shear wall interaction

Shear walls are stiffer than columns, hence they take up most of the lateral load. This has sometimes led to the conservative approach of transferring the entire lateral load to the shear walls during design. When the building is very tall, the flexural deformation of the shear wall becomes very pronounced and hence, must be allowed for in the analysis. Shear wall-frame systems have been used successfully in buildings ranging from 10 storeys to 50 storeys.

The difference in behavior under lateral load, in combination with the in-plane rigidity of the floor slabs, causes nonuniform interacting forces to develop when walls and frames are present. This makes the analysis more difficult. If torsion is not considered in the analysis, two simplified manual methods of determining the interaction of frames and shear walls are:

  • Use of charts given by Khan and Sbarounis (1964) or PCA’S Advanced Engineering Bulletin No. 14, f7), and
  • Use of Equation (C) by PCA.

In order to use these charts, the structure must be reduced to a single frame and a single wall by the addition of the properties of the separate vertical units. In both references, the stiffnesses (Iw) of all the shear walls are summed to give an equivalent single wall.

In this article, we are going to investigate the effects of shear wall-frame interaction in the resistance of uniformly distributed lateral loads using finite element analysis. A 2D model of a 10-storey and 50-storey building will be used in the analysis.

Model 1A: Analysis of a 10-storey Frame Without Shear Wall

Height of building = 30 m
Inter-storey height of building = 3 m
Dimension of beams = 400 x 600 mm
Dimension of columns = 400 x 400 mm
Lateral load = 5 kN/m

10 STOREY FRAME
Model of a 10 storey frame

When analysed on Staad Pro software, the analysis results are as follows;

DELECTED SHAPE OF 10 STOREY BUILDING
Lateral deflection of a 10 storey frame
Bending moment diagram of a tall building 1
Bending moment diagram of the 10 storey frame

Summarily, the analysis result of the 10-storey frame without shear walls is as follows;

Deflection
Maximum deflection at roof level = 16.989 mm
Deflection at 1st floor = 2.393 mm

External Column (windward side)
Bending moment = 59.1 kNm
Shear force = 39 kN
Axial force = 146 kN

Internal Column (windward side)
Bending moment = 63.5 kNm
Shear force = 39.9 kN
Axial force = 72.8 kN

First-floor beams
Bending moment = 75 kNm
Shear force = 26.4 kN
Axial force = 7.68 kN

Model 1B: Analysis of a 10-storey frame with shear wall

Height of building = 30 m
Inter-storey height of building = 3 m
Dimension of beams = 400 x 600 mm
Dimension of columns = 400 x 400 mm
Length of shear wall = 2500 mm
Thickness of shear wall = 250 mm
Lateral load = 5 kN/m

SHEAR WALL FRAME INTERACTION
Model of a 10 storey frame with shear wall


Deflection
Maximum deflection at roof level = 7.120 mm
Deflection at 1st floor = 0.319 mm

V1
Lateral deflection of a shear wall-frame model

External Column (windward side)
Bending moment = 13 kNm
Shear force = 12.7 kN
Axial force = 105 kN

First-floor beams
Bending moment = 20.8 kNm
Shear force = 8.08 kN
Axial force = 18.9 kN

MAX ABSOLUTE
Maximum absolute stress on the shear wall

Shear Walls
Maximum absolute stress = 1913.65 kN/m2
Sx = 1884.42 kN/m2
Sy = 508 kN/m2

SHEAR WALL 55
Axial stress on the shear wall

Model 2A: Analysis of a 50-storey building without shear wall

Height of building = 150 m
Inter-storey height of building = 3 m
Dimension of beams = 400 x 600 mm
Dimension of columns = 400 x 400 mm
Lateral load = 5 kN/m

Summary of analysis results

Deflection
Maximum deflection at roof level = 1516.839 mm
Deflection at 1st floor = 13.739 mm

External Column (windward side)
Bending moment = 299 kNm
Shear force = 171 kN
Axial force = 4170 kN

Internal Column (windward side)
Bending moment = 340 kNm
Shear force = 208 kN
Axial force = 1142 kN

First-floor beams
Bending moment = 411 kNm
Shear force = 145 kN
Axial force = 7.94 kN

Read Also…
Aspects of shear wall modelling
Application of wind load to shear walls

Model 2B: Analysis of a 50-storey framed building with shear wall

Height of building = 150 m
Inter-storey height of building = 3 m
Dimension of beams = 400 x 600 mm
Dimension of columns = 400 x 400 mm
Length of shear wall = 2500 mm
Thickness of shear wall = 250 mm
Lateral load = 5 kN/m

Deflection
Maximum deflection at roof level = 1297.128 mm
Deflection at 1st floor = 2.522mm

External Column (windward side)
Bending moment = 66.5 kNm
Shear force = 38.6 kN
Axial force = 3887 kN

First-floor beams
Bending moment = 137 kNm
Shear force = 53.3 kN
Axial force = 46.5 kN

Shear Walls
Maximum absolute stress = 15734 kN/m2
Sx = 15604.6 kN/m2
Sy = 4547.11 kN/m2

Discussion of results

Model 1

Element (Action Effect)Without shear wallWith shear wallPercentage Difference (%)
Maximum deflection 16.989 mm7.120 mm58.09 %
Deflection at first storey2.393 mm0.319 mm86.67 %
Column bending moment59.1 kNm13 kNm78%
Column shear force39 kN12.7 kN67.43%
Column axial force146 kN105 kN28.08%
Beam bending moment75 kNm20.8 kNm72.26%
Beam shear force26.4 kN8.08 kN69.39%
Beam axial force7.78 kN18.9 kN-142.93%

Model 2

Element (Action Effect)Without shear wallWith shear wallPercentage Difference (%)
Maximum deflection 1516.8 mm1297.128 mm14.482 %
Deflection at first storey13.739 mm2.522 mm81.643 %
Column bending moment299 kNm66.5 kNm77.76 %
Column shear force171 kN38.6 kN77.42 %
Column axial force4170 kN3887 kN6.78 %
Beam bending moment411 kNm137 kNm66.67 %
Beam shear force145 kN53.3 kN63.24 %
Beam axial force7.94 kN46.5 kN-485.642 %

The presence of shear walls reduced the maximum lateral deflection of the 10 storey building by about 58% when compared with the 50 storey building where deflection reduced by just 14.482%. Therefore as a building gets higher, the arrangement of shear walls becomes more and more important. In a study carried out by Aginam et al (2015) at Nnamdi Azikiwe University, Awka, Nigeria, it was observed that shear wall positioning affects the lateral displacement of tall buildings. Shear walls placed externally reduced lateral deflection when compared with shear walls placed internally.

In both models, the column bending moment was reduced by about 78% when the shear wall was introduced in-between the frames. However, column axial force reduced by about 28% in the 10-storey building, while it reduced by just 6.78% in the 50-storey building. Bending moment and shear forces in beams of the framed structures reduced considerably when shear walls were introduced. On the other hand, the presence of shear walls increased the axial force in the horizontal beam members.

References

[1] Aginam C.H., Chidolue C.A., and Ubani O.U. (2015): Effect of Planar Solid shear wall-frame arrangement on the deformation behavior of multi-story frames. IOSR Journal of Mechanical and Civil Engineering 12(1):98-105

[2] Khan, F.R. and Sbarounis (1964): Interaction of Shear Walls and Frames Journal of the Structural Division, Proc.,ASCE 90(3):285-338

Bracing Systems in a High-Rise Building

By
Ubani Obinna
-

The bracing system (stabilising components) of a high-rise building is normally made up of different units: frameworks (beams and columns), shear walls, coupled shear walls, and cores. They all contribute to the overall lateral load resistance of the system, but their contributions can be very different both in weight and in nature, so it is essential for the designer to know their behaviour in order that the optimum bracing system can be produced [1].

Bracing components in a building are assumed to be fully fixed at the base and hinged at the top. Non-stabilising elements are assumed to be hinged at both ends. Since non-stabilising elements must be braced by stabilising elements, they have a negative contribution to load resistance [2].

The different types of stabilising components (bracing elements) in a building are;

  • Columns and frameworks
  • Shear walls
  • Towers and cores

Columns and frameworks

reinforced concrete beams
Typical columns and frameworks under construction

A linear structural member which takes vertical compressive loads can generally be called a column. Columns can be constructed of steel, wood, or reinforced concrete depending on the strength and/or the aesthetics required. Columns are found mainly in structures in order to provide support for beams or slabs.

When calculating stability in a structure with columns it is essential to ascertain if the column is stabilising or not [2]. This means that non-stabilising elements have to be held up by the stabilising elements so they have a certain negative effect on the overall stiffness of the system.

The behaviour of frameworks under lateral load is complex, mainly because they develop both bending and shear deformations when resisting lateral loads [1]. Due to the complexity of the problem, designers and researchers have made considerable efforts to develop approximate techniques and methods. Perhaps the best and most widespread method is the continuum method which is based on an equivalent medium that replaces the framework [1]. However, due to the availability of numerous commercial computer programs for finite element analysis, this computational challenge has largely been overcome.

Columns and frameworks alone are not suitable for resisting lateral loads once the building exceeds 10 storeys, and must be combined with shear walls and cores.

Shear walls

pierced shear wall
Typical shear wall under construction

Shear walls, made of reinforced concrete, are used in modern buildings because of their effectiveness in maintaining stability and for the freedom they offer the architect who is designing [3]. Shear walls are stiff in the direction of the wind and can be effectively utilised in resisting the internal stresses due to wind action. A shear wall’s position in a building is often initially decided by the architect.

An architect is trained to design for the building’s function and appearance and not for its stability. Therefore, when a structural engineer is not involved in the first phase of design, it may lead to the shear walls being situated in non-favourable positions [2].

Also, while choosing reinforced concrete walls as partition walls, the architect can be unintentionally gaining stabilising elements. Pierced shear walls or coupled shear walls are described as shear walls with holes. These holes can be windows or doors that are necessary for access or lighting for the building. The force from the horizontal wind load results in shear forces which act within the wall and tension and compression resulting at the ground.

Towers/Cores

lift core 2
Typical core of a building under construction

Towers are rigid cores situated inside tall buildings. Shear walls are often built together to create three-dimensional units. A good example is the U-shaped elevator core but many different shapes exist in building structures. The second moments of area of a reinforced concrete core are normally large and a small number of cores are often sufficient to provide the building with the necessary stiffness to resist lateral loading [1].

Usually, a core will exist with another core or combined with shear walls and/or with columns. The combined effect will give rise to greater resistance to torsion depending on how the units are situated in relation to each other. Ideally, they are situated as far apart as possible for creating a torsional resistance. A disadvantage with using a single core, on its own, is that it is susceptible to torsion and must therefore be heavily dimensioned in order to resist torque [2].

The use of cores is however favourable in that they can be used not only as stabilising units but also as elevator shafts or stairwells. Funnelling of ventilation shafts, water pipes and electric cables can also be hidden within the tower giving the architect more manoeuvrability and the client more effective use of the space provided. Towers which have open cross sections, for example U-shaped or H-shaped, have less resistance to torsion than closed sections and should, in general, be combined with other stabilising components.

References

[1] Zalka K. A. (2013): Structural Analysis of Regular Multi-storey Buildings. CRC Press – Taylor and Francis Group, USA
[2] Gustaffson D., and Hehir J. (2005): Stability of tall buildings. M.Sc thesis submitted to the Department of Civil and Environmental Engineering, Chalmers University of Technology, Sweden
[3] Aginam C.H., Chidolue C.A., and Ubani O.U. (2015): Effect of Planar Solid shear wall-frame arrangement on the deformation behaviour of multi-story frames. IOSR Journal of Mechanical and Civil Engineering 12(1):98-105

Cement and Types of Cement Used in Construction

By
Muyiwa Ajumobi
-

Generically, cement is an inorganic fine powder that when mixed with water undergoes hydration reaction, and forms a paste which becomes the binder that holds aggregates together in concrete. The raw materials from which cement is made are lime, silica, alumina, and iron oxide.

These constituents are crushed and blended in the correct proportions and burnt in a rotary kiln. The resulting product is called clinker. The cooled clinker can be mixed with gypsum and various additional constituents and ground to a fine powder in order to produce different types of cement. The main chemical compounds in cement are calcium silicates and aluminates.

Hydraulic cements set and harden by reacting chemically with water in a process known as hydration reaction. During this reaction, cement combines with water to form a stonelike mass, called paste. When the paste (cement and water) is added to aggregates (sand and gravel, crushed stone, or other granular material) it acts as an adhesive and binds the aggregates together to form concrete, the world’s most versatile and most widely used construction material.

Different types of cement are bagged and sold commercially
Cement powder in bag

The European standard for cements is BS EN 197-1:2011 Cement –Part 1: Composition, specifications, and conformity criteria for common cements. The Standard defines and gives the specifications of 27 distinct common cements, 7 sulfate resisting common cements, as well as 3 distinct low early strength blast furnace cements and 2 sulfate resisting low early strength blast furnace cements and their constituents.

Standard Designation of Cements

CEM cement designation includes the following information:

i. Cement type (CEM I – CEM V)
ii. Strength class (32.5-52.5)
iii. Indication of early strength
iv. Additional designation SR for sulfate resisting cement
v. Additional designation LH for low heat cement

Cement brands in Nigeria

Types of Cements

All types of cement are grouped into five major groups, all of which are mixtures of different proportions of clinker and another major constituent. The five groups are:

  1. CEM I – Portland cement: This comprises mainly ground clinker and up to 5% of minor additional constituents.
  2. CEM II – Portland composite cement: This comprises of seven types which contain clinker and up to 35% of another single constituent.

    i. Portland slag cement (CEM II/A-S and CEM II/B-S): This comprises of clinker and blast furnace slag which originates from the rapid cooling of slag obtained by smelting iron ore in a blast furnace. The percentage of the slag varies between 6 and 35%.

    ii. Portland silica fume cement (CEM II/A-D): This comprises of clinker and silica fume which originates from the reduction of high purity quartz with coal in an electric arc furnace in the production of silicon and ferrosilicon alloys.

    iii. Portland-Pozzolana cement (CEM II/A-P, CEM II/B-P, CEM II/A-Q, CEM II/B-Q): This comprises clinker and natural pozzolana such as volcanic ashes or sedimentary rocks with suitable chemical and mineralogical composition or Natural calcined pozzolana such as materials of volcanic origin, clays, shales or sedimentary rocks activated by thermal treatment.

    iv. Portland-fly ash cement (CEM II/A-V, CEM II/B-V, CEM II/A-W, CEM II/B-W): This mixture of clinker and fly ash dust-like particles precipitated from the flue gases from furnaces fired with pulverised coal.

    v. Portland burnt shale cement (CEM II/A-T, CEM II/B-T): This consists of clinker and burnt shale, specifically oil shale burnt in a special kiln at 800 °C.

    vi. Portland-limestone cement (CEM II/A-L, CEM II/B-L, CEM II/A-LL, CEM II/B-LL).

    vii. Portland-composite cement (CEM II/A-M, CEM II/B-M).
  3. CEM III – blast furnace cement (CEM III/A, CEM III/B, CEM III/C): This comprises clinker and a higher percentage (36-95%) of blast furnace slag than that in CEM II/A-S and CEM II/B-S.
  1. CEM IV – pozzolanic cement (CEM IV/A, CEM IV/B): This comprises of clinker and a mixture of silica fume, pozzolanas and fly ash.
  1. CEM V – composite cement (CEM V/A, CEM V/B): This comprises clinker and a higher percentage of blast furnace slag and pozzolana or fly ash.

The clinker contents in different types of cements are given in Table 1. The letters A, B and C designate respectively higher, medium and lower proportion of clinker in the final mixture. However, the percentage of clinker with the designations A, B, C can be different in different types of cement as shown in Table 1.

Cement TypeClinker Content – AClinker Content – BClinker Content – C
CEM II80-94%65-79%–
CEM III35-64%20-34%5-19%
CEM IV65-89%45-64%–
CEM V40-64%20-38%–

The second constituent in cement in addition to clinker is designated by the second letter as follows:

S = blast furnace slag
D = silica fume
P = natural pozzolana
Q = natural calcined pozzolana
V = siliceous fly ash
W = calcareous fly ash (e.g., high lime content fly ash)
L or LL = limestone
T = burnt shale
M = combination of two or more of the above components

Strength Class of Cement

The standard strength of a cement is the compressive strength determined in accordance with EN 196-1 at 28 days and shall conform to the requirements in Table 3 of EN197-1. The three standard classes of cement strength are;

  • class 32.5
  • class 42.5, and
  • class 52.5
Strength Class28 days Compressive strength (MPa)Initial setting time (mins)
32.5 N ≥ 32.5, ≤ 52.5≥ 75
32.5 R ≥ 32.5, ≤ 52.5≥ 75
42.5 N≥ 42.5, ≤ 62.5≥ 60
42.5 R≥ 42.5, ≤ 62.5≥ 60
52.5 N≥ 52.5≥ 45
52.5 R≥ 52.5≥ 45

Sulfate-Resisting Cement

Sulfate resisting cements are used particularly in foundations where the presence of sulfates in the soil which can attack ordinary cements. The sulfate resisting cements have the designation SR and they are produced by controlling the amount of tricalcium aluminate (C3A) in the clinker. The available types are:

  • i. Sulfate resisting Portland cements CEM I-SR0, CEM I-SR3, CEM I-SR5 which have the percentage of tricalcium aluminate in the clinker less than or equal to 0, 3 and 5% respectively
  • ii. Sulfate resisting blast furnace cements CEM III/B-SR, CEM III/C-SR (no need for control of C3A content in the clinker).
  • iii. Sulphate resisting pozzolanic cements CEM IV/A-SR, CEM IV/B-SR (C3A content in the clinker should be less than 9%).

Low Early Strength Cement

These are CEM III blast furnace cements. Three classes of early strength are available with the designations N, R and L respectively signifying normal, ordinary, high and low early strength.

Examples of Cement Designation

  1. CEM II/A-S 42.5 N
    This indicates Portland composite cement (indicated by CEM II), with high proportion of clinker (indicated by letter A) and the second constituent is slag (indicated by letter S) and the strength class is 42.5 MPa (indicating that the characteristic strength at 28 days is a minimum of 42.5 MPa) and it gains normal early strength (indicated by letter N).
  2. CEM III/B 32.5 N
    This indicates blast furnace cement (indicated by CEM III); with medium proportion of clinker (indicated by letter B) and the strength class is 32.5 MPa (indicating that the characteristic strength at 28 days is a minimum of 32.5 MPa) and it gains normal early strength (indicated by letter N).
  1. CEM I 42.5 R-SR3
    This indicates Portland cement (indicated by CEM I), the strength class is 42.5 MPa (indicating that the characteristic strength at 28 days is a minimum of 42.5 MPa) and it gains high early strength (indicated by letter R) and is sulfate resisting with C3A content in the clinker less than 3%.
  2. CEM III-C 32.5 L – LH/SR
    This indicates blast furnace cement (indicated by CEM III), the strength class is 32.5 MPa (indicating that the characteristic strength at 28 days is a minimum of 32.5 MPa) and it gains low early strength (indicated by letter L) and is sulfate resisting (indicated by letters SR) and is of low heat of hydration (indicated by LH).

Common Types of Cements

Of different types of cement, the most common ones are mainly six:

i. CEM I
ii. CEM II/B-S (containing 65-79% of clinker and 21-35% of blast furnace slag)
iii. CEM II/B-V (containing 65-79% of clinker and 21-35% of siliceous fly ash)
iv. CEM II/A-LL (containing 80-94% of clinker and 6-20% of limestone)
v. CEM III/A (containing 35-64% of clinker and 36-65% of other constituents)
vi. CEM III/B (containing 20-34% of clinker and 66-80% of other constituents)

1...343536...66Page 35 of 66

EDITOR PICKS

POPULAR POSTS

POPULAR CATEGORY

ABOUT US

Structville is a media channel dedicated to civil engineering designs, tutorials, research, and general development. At Structville, we stop at nothing in giving you new dimensions to the profession of civil engineering.

Contact us: info@structville.com

FOLLOW US

© (2016 - 2024) Structville Integrated Services Limited. All rights reserved