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Structville Announces Webinar on Bridge Design | January 2021

Structville Integrated Services Limited is pleased to announce the first webinar of the year, 2021. The theme of the 2-day webinar will be focused on the “Design of Reinforced Concrete Bridges”. The details are as follows;

Date: 15th (Friday) and 16th (Saturday) January 2021
Time: 7:00 pm daily
Theme: Design of reinforced concrete bridges
Approach: Use of locally available software and manual approach
Platform: Zoom
Fee: NGN 7,000 only

Features

  • Considerations in the design of bridges
  • Components and parts of bridges
  • Selection of structural scheme for bridges
  • Loading/Actions on Bridges
  • Structural analysis of Bridges
  • Structural design of bridges
  • Questions and answers
  • Free design manual containing solved example on bridge design
bridge design

To register for this webinar, click HERE

To get the full structural drawings of a flyover bridge (a practical drawing for execution) in AUTOCAD format, kindly contact the author. Note that the drawing is for a reinforced concrete beam and slab deck system, and consists of both substructure and superstructure drawings. The drawing has been presented to be executed in Nigeria. This will cost you NGN 20,000 only, and it is separate from the webinar fee/requirements.

For further inquiries;
Whatsapp: +2347053638996
Call: +2348060307054
E-mail: ubani@structville.com


Akon’s Smart City Construction set to Begin in 2021 near Dakar, Senegal

Senegalese-American R&B singer Akon, is moving forward with plans to build a US$6 billion smart city in his country of birth, Senegal. Having finalized an agreement in January of this year, he claims to have raised at least one-third of the funding needed.

Dubbed Akon City and due for completion in about ten years, Akon’s 2,000-acre (809-hectare) project is being touted as an eco-friendly, mixed-use development open to all members of the African diaspora. Located two hours away from the capital Dakar and south of the West African nation’s relatively new Blaise Diagne International Airport, it may be the first LEED-certified project on the continent, if realized. Construction is slated to begin in 2021.

Akon is promoting his latest endeavor as a boon for both tourism and business in the region and has received considerable support from the Senegalese government, including through a partnership with SAPCO, the state’s tourism agency. While the vision is still low on specifics, Akon City’s official website markets the development as a venue for middle- and high-income residences, education and “training” services, and professional activities. It will run on Akoin, a new cryptocurrency that the singer suggests will be compatible with a variety of smartphones and other cellular devices, as well as with other cryptocurrencies like Bitcoin and Ethereum.

The project is being jointly developed by Los Angeles-based KE International and Dubai-based Bakri & Associates Development Consultants, with CEO Hussein Bakri as its lead architect. According to Business Insider, Bakri & Associates claims that Akon City will take advantage of traditional and newly-developed construction materials, including lighter-weight, more efficient glass-and-steel components. Buildings will be powered by clean energy, likely by Akon Lighting Africa, the singer’s solar energy project that aims to install solar-powered lights across Africa.

A series of dramatic renderings show an agglomeration of futuristic, ribbon-like structures towering into the sky. On the city’s official website, it is described as “an extension of the sea into the land with waves diving deep into the roots of each building.” Akon Tower, the plan’s architectural centerpiece, would far exceed Senegal’s current tallest structure, an 250-foot (76.2-meter) apartment building in Dakar.

As quixotic as it sounds, the plan is not without its skeptics. While certain government officials have praised Akon’s investment in Senegal at a time when the coronavirus pandemic has reduced tourism to a trickle, some local residents doubt the project’s efficacy. Senegal’s economy has experienced breakneck growth in recent decades, but poverty is still prevalent among its 15.4 million people. Akon himself told Business Insider that he sees more “elite local Senegalese” moving into the development first, including himself.

According to Reuters, Mayor Magueye Ndao of Ngueniene, the municipality where part of Akon City will be built, has expressed some cautious optimism about the proposal. Ndao hopes that the project will be realized and that many of its promised services, including youth and job training, will be brought to the area.

Others, though, are convinced that Akon City represents another smart city pipe dream—a project that will inevitably fizzle out due to lack of funding or government support. Xavier Ricou, an architect and the former director of Senegal’s APIX investment agency, told Reuters that Akon City will likely end up like most other city proposals seen in Senegal: just a proposal. If fears about investor commitments and feasibility prove valid, Ricou’s prediction could well be correct.

Construction of a Helical Staircase

A helical staircase is a staircase that is curved in plan and following the directrix of a full helix. In three-dimensional space, a helical surface is generated by moving a straight line such that the moving line is perpendicular to the helix.

A helix is a smooth space curve with tangent lines at a constant angle to a fixed axis. The construction of a helical staircase is a fairly challenging process that must be handled with care by the contractor. It is important to note that a helical staircase can be achieved using a variety of materials such as reinforced concrete, steel, or timber.

The aim of this article is to provide guidance on how to successfully construct a reinforced concrete helical staircase using locally available materials and tools.

Read Also…
Design of helical staircase

It is important to note that a reinforced concrete helical staircase can be in two forms;

  • Helical staircase with waist, and
  • Slabless (saw-tooth) helical staircase

A helical staircase with a waist has a reinforced concrete base slab of 100 – 200 mm thickness supporting the treads and the risers (which are usually unreinforced). The waist slab is the main structural component of the staircase where the rebars are placed. On the other hand, just like a typical sawtooth staircase, a slabless helical staircase does not have a waist slab and the structural component of the staircase are the treads and the risers.

helical staircase with waist slab
Curved staircase with waist slab
slabless curved staircase
Curved staircase without waist slab

Construction of a Helical Staircase

The construction of these two types of helical staircases present their own different challenges to a contractor, but generally, it is easier to construct a helical staircase with a waist than a slabless helical staircase. The material and construction effort demand is also greater in the latter than in the former. In this article, however, we are going to focus on the construction of a helical staircase with a waist slab.

For someone that is doing a helical staircase for the first time, it is very important to pay attention to the challenges it presents, especially when sophisticated tools are not available for the setting out. First and foremost, it is important to establish the layout of the staircase as given in the drawing.

The drawing should contain the exact number of treads and risers that will be required for the floor height. Other information such as the inner radius and the width of the staircase should also be provided. This is very important because without a proper setting out, it will be very challenging to get the staircase right.

From the drawing, locate the centre of the staircase and make a permanent mark there. Alternatively, you can get a straight and strong timber pole and attach it to a marine plywood (or 1″ x 12″ plank) base that is firmly attached to the floor. Make sure that the wood is plumb in all directions and properly placed at the centre of the curve.

Using the centre of the curve, establish the inner radius of the staircase and plot it on the floor of the building using any suitable tool/material of your choice. The outer radius of the curve can also be established, especially when the sides of the staircase are not attached to any wall.

setting out of a helical staircase
Typical setting out of a helical staircase

After establishing the curves in plan, you can use 1″ x 12″ or 1″ x 6″ planks (depending on the radius of the curve) that are standing upright to form the lines of the curves and to also establish a temporal wall for markings. The planks should be firmly nailed to the floor of the building using 2″ x 3″ wood and braced properly. The height of the planks should be determined with reference to how the staircase rises. This approach however consumes a lot of material.

planks for helical staircase
Installation of planks along the curve of a staircase

After joining the planks side by side, establish the markings of the staircase (tread and riser) on the planks which serves as a temporal wall. Make sure to establish the thickness of the waist on the planks too.

Construction of a helical staircase
Installation of planks along the curve of a staircase

After the markings, insert the stringers and support or brace the formwork with 2″ x 3″ wood. Bamboo or another strong material such as 2″ x 4″ wood can be used as props. The sheating of the staircase (soffit) can then be formed on the framing using suitable plywood. Depending on the configuration, the inner/outer curve of the staircase should be covered with another plywood with the equivalent height of the riser. Alternatively, the markings on the plank can be used if the height is sufficient but this might give a rough edge finish.

After the soffit of the formwork has been installed, the iron bender should neatly install the reinforcements according to the design specifications. The risers of the staircase can be installed by nailing perfectly cut planks to the edges of the inner and outer curves. To get this very well, you can tie a rope to the timber at the centre of the curve which you can easily adjust round it, and use it to trace the line of the plank that will form the riser which lies from the inner radius to the outer radius. Note that the tread is bigger in the outer radius because of a wider circumference/length.

Alternatively, you can prepare your framing by laying the 2”x 6” timber stringers from the central pole to your outer line of shores. The timber stringers should be laid flat on the floor and sawn accurately to dimension. The top elevation of the stringers will be determined by the thickness of the concrete slab and the thickness/arrangement of the formwork.

Once the stringers are complete and checked for position, level, and length, install 2” x 4” timber joists spaced at 400 mm centre to centre. After the stringers and joists are securely fixed in place, proceed with the plywood sheathing, and finally the edge forms. This approach requires knowing the dimensions as accurately as possible and cutting the timber forms properly.

formwork 2
Alternative framing of a helical staircase without planks

Once the formwork is complete, well secured, and inspected, the next step would be to install the reinforcement. Bars must be cut to the right length and placed neatly in size and spacing according to the construction drawings. Use chairs to maintain the proper clearance to the forms and observe the proper concrete coverage at the bottom of the reinforcing bars.

reinforcement of helical staircase
Typical reinforcement installation of a helical staircase

Following the inspection and approval of the reinforcing bar comes the careful preparation for pouring. We recommend a minimum concrete strength of 25 MPa at 28 days. The concrete mix should be properly designed, water content carefully controlled, vibrated, and compacted to its maximum density.

To request for this article or any other Structville article in PDF format, click HERE (premium but very affordable).

Question of the Day | 27-12-2020

The 2225 mm x 2225 mm pad footing shown above is meant to support a 230 x 230 mm column from the superstructure of a building. The characteristic dead and imposed loads on the column are 600 kN and 210 kN respectively, excluding the weight of the column stub. Making use of your personal engineering judgment, kindly answer the following questions using the information provided;

(1) Will you accept the size of the pad footing provided if the verified allowable bearing capacity of the soil is 175 kN/m2?

(2) What is your calculated value of the total service load for determining the footing dimensions?

(3) What informed your decision? Did you make any assumptions?

Design of Helical Staircase

Helical staircases provide alternatives for vertical circulation in buildings. Due to their curved and elegant nature, they inspire awe and admiration whenever they are properly designed and constructed. The analysis and design of a helical staircase can be complex due to an inherent interaction of vertical moments, horizontal moments, vertical shear, lateral shear, axial force, and torsion. The complexity of helical staircases is due to their geometry.

Geometrically, a helical surface is a three-dimensional structure in space consisting of a warped surface that is generated by moving a straight line touching a helix so that the moving line is always perpendicular to the axis of the helix. In an oblique helix, the generating line always maintains a fixed angle with the helix. Since a helical staircase is a space member, all the six internal actions come into play at any section, having varying directions and lines of actions.

internal stresses in a helical staircase
Fig 1: Internal stresses in a helical staircase

Despite the existence of the six internal stresses at any section in a helical staircase, they are usually designed for the horizontal bending moment only. Furthermore, helical staircases can sometimes be idealised as a fixed ended curved beam for the purposes of design due to their complexity.

Bergman (1956) proposed an approximate solution to helical staircases called the in-plane beam solution. The method reduces the problem to that of a horizontal bow girder, which fails to take into account the inherent structural strength of a helical beam.

Morgan (1960) and Scordelis (1960) presented the analysis of the longitudinal elastic axis of the helical beam as a three-dimensional structure indeterminate to the sixth degree. By selecting the redundant at mid-span and using the principle of symmetry, all but two of the redundants become zero. The two redundants at mid-span are the horizontal shear H and the vertical moment Mo.

Cusens and Trirojna (1964) carried out a test on half scaled fixed ended helical staircase and confirmed the equilibrium equations derived by Morgan. Santathadaporn and Cusens (1966) worked further on helicoidal stair slabs to simplify the design process. They proposed some simple equations for design purpose with the provision of using coefficients in the equations. They used computer program to solve those complicated equations and constructed a series of design charts to provide the values of the coefficients.

Reynolds and Steedman (1988) attempted further modifications of the charts provided by Santathadaporn and Cusens. The analytical approach they chose was the one made by Morgan and Scordelis. The proopsed equations and coefficients for determination of the internal stresses are as follows;

Geometry and internal stresses in helical staircases
Fig 2: Geometry and internal stresses in helical staircase

Horizontal moment Mn = Mo sinθsinϕ – HR2θtanϕcosθsinϕ – HR2sinθcosϕ + nR1sinϕ(R1sinθ – R2θ) —— (1)

Vertical moment My = Mocosθ + HR2θtanϕsinθ – nR12(1 – cosθ) —— (2)

Torsion T = (Mosinθ – HR2θcosθtanϕ + nR12sinθ – nR1R2θ)cosϕ + HR2sinθsinϕ —— (3)

Axial force N = -Hsinθcosϕ – nR1θsinϕ —— (4)

Vertical shear Vn = nR1θcosϕ – Hsinθsinϕ —— (5)

Radial horizontal shear Vh = Hcosθ —— (6)

Where;
Mo = Redundant moment acting tangentially at the midspan = k1nR22
H = Horizontal redundant force at midspan = k2nR2
Mvs = Vertical moment at supports= k3nR22
n = total design load on the staircase (kN/m2)
R1 = Radius of centreline of loading = 2(Ro3 – Ri3)/3(Ro2 – Ri2)
R2 = Radius of centreline of steps = 0.5(Ri + Ro)
θ = angle subtended in plan between point considered and mid-point of stair
ϕ = slope of tangent to helix centreline measured from horizontal
β = Total angle formed by the helical staircase in plan
k1 = GC/EI1
k2 = GC/EI2
I1, I2 = Second moment of area of stair section about horizontal axis and axis normal to slope respectively.

The 36 design charts presented by Santathadaporn and Cusens (1966) covered ranges of β of 60° to 720°, ϕ of 20° to 50°, b/h of 0.5 to 16 and R1/R2 of 1.0 to 1.1, based on a ratio of G/E of 3/7.

Four design charts are provided in Table 177 of (Reynolds and Steedman 10th Edition) and (Tables 2.90 and 2.91 of 11th Edition) for a ratio of G/E of 0.4 and by taking C to be one-half of the St. Venant value for plain concrete. These charts cover ranges of β of 30° to 360° and ϕ of 20° to 40°, with values of b/h of 5 and 10 and R1/R2 of 1.0 and 1.1, these being the ranges most frequently met in helical stair design. According to Reynolds and Steedman (1988), interpolation between the various curves and charts on the Tables will be sufficiently accurate for preliminary design purposes.

According to a 2015 research from the Department of Civil Engineering, Addis Ababa University, Ethiopia, the solution provided in Reynolds and Steedman is the best so far, even though it has its limitations. The charts presented in the book give good information about internal actions but only qualitatively, and indicate set of constants to put in his equations then solved. And, as Reynolds himself stated in his book, his results are only satisfactory for preliminary analysis only as lots of interpolation is involved.

Design Example

Design a helicoidal stair having an angle of inclination ϕ of 25° to the horizontal plane to support a uniform imposed load of 3 kN/m2. The stair is to have a width of 1.5 m and the minimum thickness of the waist is 150 mm, the radius to the inside of the stair Ri is 1000 mm, and the angle β turned through by the stair is 180°. (Weight of finishes = 1.2 kN/m2; weight of riser/thread = 1.875 kN/m2)

Solution
Permanent actions;
Self-weight of waist (150 mm thick) = 0.15 × 25 = 3.75 kN/m2
Weight of stepped area = 1.875 kN/m2
Weight of finishes = 1.2 kN/m2
Total dead load (characteristic permanent action) gk = 6.825 kN/m2

Variable action(s);
Characteristic variable action on staircase qk = 3 kN/m2

At ultimate limit state Ed = 1.35gk + 1.5qk = 1.35(6.825) + 1.5(3) = 13.71 kN/m2

Geometry
Inner radius Ri = 1000 mm = 1 m
Outer radius Ro = Ri + b = 1000 + 1500 = 2500 mm = 2.5 m

The radius of the centreline of the load R1 is;
R1 = 2(Ro3 – Ri3)/3(Ro2 – Ri2) = 2(2.53 – 1.03)/3(2.52 – 1.02) = 1.857 m
R2 = 0.5(Ri + Ro) = 0.5(1000 + 2500) = 1750 mm = 1.75 m

R1/R2 = 1.857/1.75 = 1.06
b/h = 1500/150 = 10
β = 180°; ϕ = 25°

Reading from chart (Tables 177 of Reynolds and Steedman, 2005);

(b/h = 10; R1/R2 = 1.0; β = 180°; ϕ = 25°) k1 = -0.02; k2 = +1.21; k3 = -0.1
(b/h = 10; R1/R2 = 1.1; β = 180°; ϕ = 25°) k1 = -0.09; k2 = +1.45; k3 = -0.19

Interpolating between R1/R2 = 1.0 and 1.1, we obtain;
k1 = -0.062; k2 = +1.354; k3 = -0.154

Structural Analysis
Mo = Redundant moment acting tangentially at the midspan = k1nR22 = -0.062 × 13.71 × 1.752 × 1.5 = -3.905 kNm
H = Horizontal redundant force at midspan = k2nR2 = 1.354 × 13.71 × 1.75 × 1.5 = 48.728 kN
Mvs = Vertical moment at supports = k3nR22 = -0.154 × 13.71 × 1.752 × 1.5 = -9.698 kNm

The values of Mo and H can now be substituted into equations (1) to (6) to obtain the internal stresses along the staircase.

For instance, when θ = 60°
n (as uniformly distributed load in kN/m) = 13.71 x 1.5 = 20.565 kN/m

Vertical moment My,Ed = Mocosθ + HR2θtanϕsinθ – nR12(1 – cosθ) = -3.905cos60 + (48.729 × 1.75 × (π60/180) × tan25 × sin60) – 20.565 × 1.8752 × (1 – cos60) = -1.9525 + 36.055 – 36.149 = -2.046 kNm

Horizontal moment Mn,Ed = Mo sinθsinϕ – HR2θtanϕcosθsinϕ – HR2sinθcosϕ + nR1sinϕ(R1sinθ – R2θ) = (-3.905 × sin60 × sin25) – [48.728 × 1.75 × (π60/180) × tan25 × cos60 × sin25] – (48.728 × 1.75 × sin60 × cos25) + [20.565 × 1.875sin25(1.875sin25 – 1.75(π60/180)] = -1.429 – 8.797 – 66.930 – 16.945 = -94.101 kNm

Torsion T = (Mosinθ – HR2θcosθtanϕ + nR12sinθ – nR1R2θ)cosϕ + HR2sinθsinϕ = [-3.905sin60 – (48.728 × 1.75 × (π60/180) × tan25 × cos60) + (20.565 × 1.8752 × sin 60) – (20.565 × 1.875 × 1.75 × (π60/180))]cos25 + (48.728 × 1.75 × sin60 × sin25) = -23.085 + 31.210 = 8.125 kNm


Axial force N = -Hsinθcosϕ – nR1θsinϕ = [-48.728 × sin60 × cos 25] – [20.565 × 1.857 × (π60/180) × sin25] = -38.245 – 16.898 = -55.143 kN

Vertical shear Vn = nR1θcosϕ – Hsinθsinϕ = [20.565 × 1.857 × (π60/180) × cos25] – [-48.728 × sin60 × sin 25] = 36.238 + 17.834 = 54.072 kN

Radial horizontal shear Vh = Hcosθ = -48.728 × cos60 = -24.364 kN

The calculation should be carried out for different angles, and the maximum values used for the structural design.

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References
[1] Bergman V.R. (1956): Helicoidal staircases of reinforced concrete, A.C.I. Journal Proceedings 53(4):403‐412
[2] Cusens A.R., Trirojna S. (1964): Helicoidal Staircase Study. ACI Journal, Proceedings, 61(1):85‐101.[2]
[3] Morgan V.A. (1960): Comparison of Analysis of Helical Staircases. Concrete and Construction Engineering, (London) 55(3):127‐132.
[4] Reynolds C.E., Steedman J.C. (1988): Reinforced Concrete Designers Handbook, Tenth Edition, E & FN Spon Ltd, London
[5] Santathadaporn, S. and Cusens, A R. (1966): Charts for the Design of Helical Stairs with Fixed Supports. Concrete and Construction Engineering, 61(2):46-54.
[6] Scordelis A. C. (1960): Closure to Discussion of Internal Forces in Uniformly Loaded Helicoidal Girder. ACI Journal Proceedings, 56(6):1491‐1502.




Merry Christmas & Season Greetings from all of us at Structville

As we celebrate the awesomeness of this special season, may your life, home, and career be filled with peace, love, joy, laughter, and happiness. Merry Christmas and Happy New Year in Advance.

christmas 2

From all of us at Structville

Question of the Day | 24-12-2020

The frame loaded as shown above is hinged at point D, simply supported at point F, and fixed at point C. Provide the following solutions based on the diagram.

(1) What is the vertical support reaction at point F?
(A) 2.5 kN
(B) 0.5 kN
(C) 0.75 kN
(D) 1.0 kN

(2) What is the bending moment just to the right of point B?
(A) 4 kNm
(B) -8 kNm
(C) -5 kNm
(D) 6 kNm

Types of Structural Systems for Tall Buildings

For a tall building to be successful, at a minimum, the structure should employ systems and materials appropriate to the building’s height and configuration. The structural system for a tall building must perform well and lend itself to efficient construction [1]. According to [1], a successful tall building should have the following features;

  1. Create a friendly and inviting image that has positive values to building owners, users, and observers.
  2. Fit the site, providing proper approaches with a congenial layout for people to live, work, and play.
  3. Be energy-efficient, providing interior space with a controllable climate.
  4. Allow flexibility in office layout with easily divisible space.
  5. Offer space oriented to provide the best views.
  6. Most of all, the building must make economic sense, without which none of the development would be a reality.

In the year 1969, Fazlur Khan classified structural systems for tall buildings relating to their heights with considerations for efficiency in the form of “Heights for Structural Systems” diagrams [2]. This marked the beginning of a new era of skyscraper revolution in terms of multiple structural systems [3].

Later, Khan upgraded these diagrams by way of modifications [4, 5]. He developed these schemes for both steel and concrete as can be seen from Figures 1 and 2 respectively [6]. Khan argued that the rigid frame that had dominated tall building design and construction so long was not the only system fitting for tall buildings.

Structural Systems for Tall Buildings
Fig 1: Classification of tall building structural systems for steel [1]
khan 2
Fig 2: Classification of tall building structural systems for concrete [1]

Bungale [1] also presented a table to show the appropriate structural system for reinforced concrete tall buildings. This is shown in Table 1.

Table 1: Appropriate structural systems for concrete tall building [1]

Appropriate structural systems for concrete tall building

We are going to give a brief description of the various and popular structural systems for tall buildings in the sub-sections below. The structural system to be adopted in any design should be able to carry different types of loads, such as gravity, lateral, temperature, blast and impact loads. The drift of the tower should be kept within limits, such as H/500.

The structural systems usually adopted for high rise buildings are as follows;

Column and slab systems

Concrete floors in tall buildings often consist of a two-way floor system such as a flat plate, flat slab, or a waffle system which can resist lateral loads (see Figure 3). In a flat plate system, the floor consists of a concrete slab of uniform thickness which frames directly into the columns. Two way flat slabs make use of either capitals in columns or drop panels in slab or both, requiring less than a flat plate because extra concrete is provided only at columns where the shears and moments are the greatest.

response of a flat slab to lateral load
Fig 3: Response of flat slab-frame to lateral load [1]

The waffle system is obtained using rows of joists at right angles to each other; the joists are commonly formed by square domes (see Figure 4). The domes are omitted around the columns to increase the moment and shear capacity of the slab [7]. Any of the three systems can be used to function as an integral part of the wind-resisting systems for buildings in the 10 to 20 storey range. The concept of an “effective width” is usually used in the analysis of such buildings subjected to lateral forces.

Typical floor systems for flat slab frames flat plate
Fig 4: Typical floor systems for flat slab-frames: (a) flat plate, (b) flat slab with drop panels, and (c) two-way waffle system [1]

Rigid Frames

A rigid frame is characterized by the flexure of beams and columns and rotation at the joints. Interior rigid frames for office buildings are generally inefficient because;

(1) The number of columns in any given frame is limited due to leasing considerations and
(2) The beam depths are often limited by the floor-to-floor height.

However, frames located at the building exterior do not necessarily have these limitations. An efficient frame action can thus be developed by providing closely spaced columns and deep spandrels at the building exterior. A rigid-frame high-rise structure typically comprises of parallel or orthogonally arranged bents consisting of columns and girders with moment-resistant joints [8].

The continuity of the frame also increases resistance to gravity loading by reducing the positive moments in the girders [1]. The advantages of a rigid frame are the simplicity and convenience of its rectangular form. Its unobstructed arrangement, clear of structural walls, allows freedom internally for the layout and externally for the fenestration.

Rigid frames are considered economical for buildings of up to about 25 stories, above which their drift resistance is costly to control. If, however, a rigid frame is combined with shear walls, the resulting structure is very much stiffer so that its height potential may extend up to 50 stories or more [1].

The horizontal stiffness of a rigid frame is governed mainly by the bending resistance of the girders, the columns, and their connections, and in a tall frame, also by the axial rigidity of the columns (see Figure 5). The accumulated horizontal shear above any story of a rigid frame is resisted by shear in the columns of that story.

Rigid frame Forces and deformations
Fig 5: Rigid frame – Forces and deformations [1]

Rigid frame systems are not efficient for buildings over 30 storeys in height because the shear racking component of deflection caused by the bending of columns and girders causes the building to sway excessively [3].

Braced frames are a type of moment-resisting frames that have single diagonal x-braces and k-braces. Lattice and knee bracing are also used. Concrete braced frames are often not used, since shear walls are superior for construction and lateral resistance. Lattice bracing is used in pre-cast panel construction.

Steel braced frames are used in interior cores, so connections could easily be made with wall panels. Composite braced frames may have steel bracings in concrete bracings in steel frames. Concrete encasement of columns and composite floor beams has also been used.

Shear wall-Frame Systems

In this system, resistance to horizontal loading is provided by a combination of shear walls and rigid frames [9]. The shear walls are often placed around elevator and service cores while the frames with relatively deep spandrels occur at the building perimeter. When a wall–frame structure is loaded laterally, the distinctly different deflected forms of the walls and the frames can be quite effective in reducing the lateral deflections to the extent that buildings of up to 50 stories or more are economical [1].

The potential advantages of a wall–frame structure depend on the intensity of horizontal interaction, which is governed by the relative stiffness of the walls and frames, and the height of the structure. The taller the building and the stiffer the frames, the greater the interaction [10].

The interaction of frame and shear walls has been understood for quite some time, the classical mode of the interaction between a prismatic shear wall and a moment frame is that the frame basically deflects in a so-called shear mode while the shear wall predominantly responds by bending as a cantilever (see Figure 6).

shear wall frame interaction
Fig 6: Shear wall-frame interaction [1]

Compatibility of horizontal deflection introduces an interaction between the two systems which tends to impose a reverse curvature in the deflection pattern of the system. The combined structural action, therefore, depends on the relative rigidities of different elements used in the makeup of the lateral-load-resisting system.

The distribution of total wind shear to the individual shear walls and frames as given by the simple interaction diagram is valid only if one of the following two conditions is satisfied.

1. Each shear wall and frame must have constant stiffness properties throughout height of the building.
2. If stiffness properties vary over the height, the relative stiffness of each wall and frame must remain unchanged throughout the height of the building.

Shear Truss-Outrigger Braced Systems

The structural arrangement for this system consists of a main concrete core connected to exterior columns by relatively stiff horizontal members such as one or two-story deep walls commonly referred to as outriggers. The core may be centrally located with outriggers extending on both sides or it may be located on one side of the building with outriggers extending to the building columns on one side [1].

The outrigger system is a development due to the desire to make inner cores and outer columns as one by linking them together at one or more levels with rigid arms – outriggers. It may be formed by any combination of steel, concrete or composite construction and reduce the structure’s internal overturning moment by up to 40 % compared to that of a free cantilever [10].

Multilevel outrigger systems can provide up to five times the moment resistance of a single outrigger system. Outrigger systems have been used for buildings up to 70 stories but the concept should hold for even higher buildings [11].

The basic structural response of the system is quite simple. When subjected to lateral loads, the column-restrained outriggers resist the rotation of the core, causing the lateral deflections and moments in the core to be smaller than if the freestanding core alone resisted the loading as shown in Figure 7 [3].

The external moment is resisted not by bending of the core alone, but also by the axial tension and compression of the exterior columns connected to the outriggers. As a result, the effective depth of the structure for resisting bending is increased when the core flexes as a vertical cantilever, by the development of tension in the windward columns, and by compression in the leeward columns.

Core supported outrigger structure
Fig 17: Core supported outrigger structure [3]

Framed-Tube System

In its simplest terms, a framed tube can be defined as a three-dimensional system that engages the entire building perimeter to resist lateral loads [1]. A necessary requirement to create a wall-like three-dimensional structure is to place columns on the building exterior relatively close to each other, joined by deep spandrel girders as shown in Figure 8.

Frame tube building
Fig. 8: Frame tube building. (a) Schematic plan and (b) isometric view [1]

The system works quite efficiently as a hollow vertical cantilever. However, lateral drift due to the axial displacement of the columns—commonly referred to as chord drift—and web drift, caused by shear and bending deformations of the spandrels and columns, may be quite large depending upon the tube geometry.

In framed tube systems, the “strong” bending direction of the columns is typically aligned along the face of the building, in contrast to a typical transverse rigid frame where it is aligned perpendicular to the face. The frames parallel to the lateral load act as webs of the perforated tube, while the frames normal to the load act as the flanges.

When subjected to bending, the columns on opposite sides of the neutral axis of the tube are subjected to tensile and compressive forces. In addition, the frames parallel to the direction of the lateral load are subjected to the in-plane bending and the shearing forces associated with an independent rigid frame action.

Trussed Tube Systems

A trussed tube system improves the efficiency of the framed tube by increasing its potential for use in taller buildings and allowing greater spacing between the columns. This is achieved by adding diagonal bracing at the faces of the tube as shown in Figure 9 to virtually eliminate the shear lag in both the flange and web frames [1].

hnn
Fig. 9: Trussed tube System [1]

The framed tube, as discussed previously, even with its close spacing of columns is somewhat flexible because the high axial stresses in the columns cannot be transferred effectively around the corners. For maximum efficiency, the tube should respond to lateral loads with the purity of a cantilever, with compression and tension forces spread uniformly across the windward and leeward faces.

The framed tube, however, behaves more like a thin-walled tube with openings. The axial forces tend to diminish as they travel around the corners, with the result that the columns in the middle of the windward and leeward faces may not sustain their fair share of compressive and tensile forces. An example of an application is the Onterie Centre building in Chicago (Figure 10).

onterie
Fig. 10: Onterie Center Building, Chicago (www.skyscrapercentre.com)

Bundled Tube Systems

The bundled tube structure consists of four parallel rigid frames in each orthogonal direction, interconnected to form nine bundled tubes (see Figure 11). The principle is the same as for the single tube structure where the frames in the horizontal load direction act as webs and the perpendicular frames acts as flanges.

bundled tube system
Fig. 11: Bundled tube structure [3]

By introducing the internal webs the shear lag is drastically reduced and as a result the stresses in the columns are more evenly distributed and their contribution to the lateral stiffness is more significant. This allows for the columns to be spaced further apart and to be less striking. In essence, the underlying principle to achieve a bundled tube response is to connect two or more individual tubes into a single bundle. The main purpose is to decrease shear lag effects.

References

[1] Bungale S. T. (2010): Reinforced Concrete Design of Tall Buildings. CRC Press, Taylor and Francis Group
[2] Khan, F.R. (1969): Recent structural systems in steel for high-rise buildings. In Proceedings of the British Constructional Steelwork Association Conference on Steel in Architecture. London: British Constructional Steelwork Association.
[3] Ali M.M., and Moon K.S. (2007): Structural developments in tall buildings: Current trends and future prospects. Architectural Science Review 50(3):205-223
[4] Khan, F.R. (1972): Influence of design criteria on selection of structural systems for tall buildings, In Proceedings of the Canadian Structural Engineering Conference. Toronto: Canadian Steel Industries Construction Council, 1-15.
[5] Khan, F.R. (1973): Evolution of structural systems for high-rise buildings in steel and concrete. In J. Kozak (Ed.), Tall Buildings in the Middle and East Europe: Proceedings of the 10th Regional Conference on Tall Buildings-Planning, Design and Construction. Bratislava: Czechoslovak Scientific and Technical Association
[6] Ali, M.M. (2001):  Art of the Skyscraper: The Genius of Fazlur Khan. New York: Rizzoli.
[7] Reddy S.V.B., and Eadukondalu M. (2018): Study of the lateral structural systems in tall buildings. International Journal of Applied Engineering Research 13(15):11738 – 11754
[8] Zalka K. A. (2013): Structural Analysis of Regular Multi-storey Buildings. CRC Press – Taylor and Francis Group, USA
[9] 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
[10] Sandelin C. and Bujadev E. (2013): The stabilization of high-rise buildings: An evaluation of the tubed mega frame concept. Dissertation submitted to the Department of Engineering Science, Applied Mechanics, Civil Engineering, Uppsala University
[11] Hallebrand E., and Jakobsson W. (2016): Structural design of high-rise buildings. M.Sc thesis presented to the Department of Construction Sciences (Division of structural mechanics), Lund University, Sweden


Out-of-Plane Two-Way Bending Analysis of Unreinforced Masonry Walls

Masonry is defined as an assemblage of masonry units (blocks, bricks, etc) laid in a specified pattern and joined together with mortar. They are usually used in components subjected to compressive loads such as walls, columns, arches, domes, vaults, etc. On other hand, masonry elements have limited capacity to support horizontal loads and bending moments.

Within the last decades, the efficiency of masonry units has increased due to higher allowable stresses, and refined possibilities in design. This, therefore, calls for more precision in the analysis, construction, and production of masonry to be used as structural members. This brings the design of masonry walls as a task of civil engineers.

Previous researches have shown that the out-of-plane (OOP) two-way bending failure of structural components can be one of the most predominant failure mechanisms in unreinforced masonry (URM) buildings. This is according to a submission from research carried out at the Faculty of Civil Engineering and Geosciences, Delft University of Technology, Netherlands. The study was published in the Elsevier – Structures journal in December 2020.

According to the researchers, extensive analytical formulations have been well developed for one-way vertically spanning walls, while analytical formulations for URM walls in OOP two-way bending require further improvement in accuracy and extension for the application range.

In order to bridge this knowledge gap, the researchers carried out an international testing campaign on a dataset of 46 testing specimens, in order to evaluate current analytical formulations. The current analytical formulations (codes of practice) used for the evaluation of the test specimens are;

The analytical formulations that have been developed in the past decades were incorporated into design standards to assess the wall capacity in engineering practice. The current analytical formulations are mainly based on the yield line method or on the virtual work method.

The underlying assumptions used in the development of the yield line method are:

  • masonry is simplified as a homogeneous material;
  • all cracks develop simultaneously;
  • the force capacity is calculated from the equilibrium between the applied forces and the reaction forces along cracking lines.

One drawback of the yield line method is that some crucial factors such as bonding patterns are neglected since masonry is considered as a homogenous material. This can affect the crack pattern therefore possibly resulting in misevaluation. Another drawback of the yield line method is that all cracks are assumed to develop concurrently, which can lead to inaccuracy for calculating the force capacity since contributions of all cracks are taken into account.

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Another category of analytical formulations originates from the virtual work method. The major underlying assumptions of the virtual work method are:

  • the contributions from horizontal cracks are neglected;
  • diagonal cracks start right from the wall corners;
  • the cracking pattern is assumed to follow the mortar joints and is determined by the aspect ratios of the units and of the wall; horizontal and diagonal bending moment capacities are calculated independently;
  • the virtual work done by external loads is equal to the strain energy along cracking lines in pre-assumed cracking patterns
yield line of masonry walls
Figure 1: Classic pre-assumed cracking patterns used in formulations based on the virtual work method [1]

Additionally, the virtual work method provides coefficients and formulas to consider the presence of openings, while the yield line method does not. It is important to note that the afore-mentioned formulations are all force-based methods.

The dataset of the study carried out consists of tests performed mostly on clay brick and calcium silicate (CS) brick masonry walls. 37 out of 46 testing specimens were subjected to quasi-static cyclic loading, while the others to dynamic loading.

The testing configuration of the specimens
Figure 2: The testing configuration of the testing specimens [1]

Eurocode 6 evaluates the force capacity w of a wall by following equation;

w = [(fx1 + σd)Z]/μα2Lw2 —— (1)

where the flexural strength ratio μ is defined as:

μ = (fx1 + σd)/fx2 —— (2)

Where;
fx1 and fx2 are the flexural strength of masonry obtained for planes of failure parallel to and perpendicular to the bed joints,
respectively;
σd is the vertical compressive stress at a specific height of the wall caused by self-weight and pre-compression σ;
Z is the section modulus of the wall;
α2 is the bending moment coefficient

The formulations AS3700, W2006 and G2019 assess the force capacity w of a wall by the following equation:

w = [2αf /Ld2] (k1Mh + k2Md) —— (3)

with the components of Eq. (3):

G = 2(hu + tj)/(lu + tj) —— (4)
α = GLd/Hd ——(5)

where;
Hd and Ld are the design height and design length of the wall, respectively
G is the assumed slope of the crack line;
α is the slope factor that identifies the expected cracking pattern including a vertical central crack in the case α < 1, or a horizontal central crack in the case α ≥ 1;
αf, k1 and k2 are coefficients determined by the presence of the openings, the slope factor α and the number of supported vertical edges
Mh and Md are the horizontal and the diagonal bending moment capacity of masonry, respectively.

The equations for Md and Mh are provided in the research article accordingly.

To compare the accuracy of the formulations, the tested force capacity from the dataset was predicted according to the equations of the analytical formulations. Lower and upper bounds for each testing specimen were calculated. The lower bound of the force capacity was estimated by considering the wall hinged on all sides in EC6 or assuming Rf = 0 in the other formulations; the upper bound of the force capacity was estimated by considering the wall clamped on all sides in EC6 or assuming Rf = 1 for the other formulations.

The comparison showed that EC6 has an incorrect prediction rate of 58.7% with the highest overestimation rate of 47.8%. W2006 and G2019 have incorrect prediction rates of 71.7% and 65.2%, respectively. Both these two formulations tend to underestimate the force capacity. AS3700 provides the lowest incorrect prediction rate of 56.6%. Also, the incorrect prediction rate on walls without openings of AS3700 is the lowest.

Table 1: Percentage of incorrect predictions for the considered dataset [1]

data set comparison table

Nevertheless, the accuracy of AS3700 requires further improvement considering 21.8% and 34.8% of testing specimens are overestimated and underestimated, respectively. The formulations based on the virtual work method provide close incorrect prediction rates for walls with and without openings.

In conclusion, the formulations based on the virtual work method returned the most accurate predictions for the testing specimens evaluated in the study, especially for partially clamped walls and walls with openings. Nevertheless, drawbacks and limitations were revealed when analytical formulations were applied to assess the influence of crucial factors on the force capacity such as precompression of the wall, bonding pattern, boundary conditions, material properties, area of openings, and eccentricity of load.

To improve the accuracy and application range of the analytical formulations, further study is suggested regarding the influence of above-mentioned crucial factors on the wall failure mechanisms and quantifying the relations between the force capacity and the crucial factors.

Disclaimer
The contents of this article are culled from [Lang-Zi Chang, Francesco Messali, Rita Esposito (2020): Capacity of unreinforced masonry walls in out-of-plane two-way bending: A review of analytical formulations. Structures 28 (2020) 2341-2477] and does not belong to www.structville.com. It has been presented here in accordance to the requirements of open access articles under the CC BY license (http://creativecommons.org/licenses/by/4.0/)

References
[1] Lang-Zi Chang, Francesco Messali, Rita Esposito (2020): Capacity of unreinforced masonry walls in out-of-plane two-way bending: A review of analytical formulations. Structures 28 (2020) 2341-2477 https://doi.org/10.1016/j.istruc.2020.10.060

Question of the Day | 20-12-2020

For the frame loaded as shown above, which of the following is the most likely bending moment diagram due to the externally applied load?

A
B
C
D