Structville daily questions
From now henceforth, Structville will be publishing daily questions on different aspects of civil engineering. You are expected to enter your response in the comment section. At the end of every week, exceptional participants will stand a chance to win some gifts. This exercise is open to participants all over the world.
Today’s Question For the compound frame loaded as shown above, find the bending moment just to the left of point E and the horizontal support reaction at support B assuming linear elastic response.
Thank you for participating in exercise today, remember to enter your answer in the comment section. The main aim of this exercise to stimulate knowledge of structural analysis on the internet in a fun and exciting way. We are always happy to hear from you, so kindly let us know how you feel about Structville.
E-mail:info@structville.com WhatsApp: +2347053638996 You can also visit Structville Research for downloads of civil engineering materials.
STRUCTVILLE REINFORCED CONCRETE DESIGN MANUAL
We have this very affordable design manual available…
Do you want to preview the book, click PREVIEW To download full textbook, click HERE
Large span cantilevers are delightful to everyone, but very challenging to engineers during design. This is because the engineer is battling with a lot of design factors, and he must guaranty the stability and good performance of the structure.
The most prominent issues in the design of reinforced concrete cantilever structures are;
(1) Excessive deflection and
(2) Balance of moment for stability
In this post, I am going to comment on some common design solutions to large span cantilevers. The simplest solutions are discussed below;
(1) Controlling deflection
For relatively shorter spans (say less than 1.5m), increasing the depth of the section or increasing the quantity of steel reinforcement looks like an express solution without very serious consequences. However as the span of the cantilever increases, increasing the depth will increase the design load and add to the design challenges. You may have to increase the compressive strength of the concrete (employing high strength concrete), and heavily reinforce the tension and compression zones of the section to assist in resisting deflection. On the other hand, you can try prestressing, and if your architect permits, the use of trusses to support the cantilever will provide a very fast and simple solution.
Furthermore, forming a web kind of design (like introducing grids) will improve the load distribution on the members, offer higher resistance, and ultimately bring about shallower sections. This is one of the most efficient ways of dealing with large span cantilevers. For improved aesthetics, the web can be adequately covered with finishes to make the whole arrangement look like one single section. See the example below where the use of multiple members have been used to approach a relatively large span cantilever problem.
(2) Balance of moment for stability
For cantilever structures to stand, the moment generated at the fixed end must be balanced, otherwise equilibrium problem will ensue. If the cantilever structure has no backspan (discontinuous), then the foundation must be used to provide the balance needed, otherwise the structure will topple (EQU problem). This also means that the column of the cantilever must be designed to resist heavy moment (STR problem). The exercise below has been presented to give you a little idea of what is being discussed. You can attempt the problem and place your solution in the comment section of this post.
Exercise For the structure loaded as shown below, proportion the dimensions of base to counteract the moment from externally applied load. (Take width of base = 1.2 m, Unit weight of concrete = 24 kN/m3). Make all other necessary assumptions in your design.
On the other hand, if the cantilever has a backspan (continuous), then the bending moment is distributed to all members meeting at that node, and column moment is alleviated to some extent. So we really do not need the base to help provide equilibrium, but the backspan of the structure and interaction with other members provide the balance needed. So we have mainly STR problem. The picture below shows a cantilever with a backspan. However, in some cases, the base will be required to assist in maintaining equilibrium.
There is a structural design challenge that is currently ongoing, and some Nigerian civil engineering students are carrying out a design on a structure that has many cantilevers that are spanning as long as 3 meters. We are all looking forward to the solutions that they will come up with. So let us anticipate.
To check out the nature of the competition, click HERE.
Check out this beautiful cantilever structure below. Have you designed long span cantilever structures before? Kindly let us know how you approached it.
Structville daily questions From now henceforth, Structville will be publishing daily questions on different aspects of civil engineering. You are expected to enter your response in the comment section. At the end of every week, exceptional participants will stand a chance to win some gifts. This exercise is open to participants all over the world.
Today’s Question For the compound beam loaded as shown above, find the bending moment and support reaction at support C assuming linear elastic response.
Thank you for participating in exercise today, remember to enter your answer in the comment section. The main aim of this exercise to stimulate knowledge of structural analysis on the internet in a fun and exciting way. We are always happy to hear from you, so kindly let us know how you feel about Structville.
E-mail:info@structville.com WhatsApp: +2347053638996 You can also visit Structville Research for downloads of civil engineering materials.
STRUCTVILLE REINFORCED CONCRETE DESIGN MANUAL We have this very affordable design manual available…
Do you want to preview the book, click PREVIEW To download full textbook, click HERE
Structville daily questions From now henceforth, Structville will be publishing daily questions on different aspects of civil engineering. You are expected to enter your response in the comment section. At the end of every week, exceptional participants will stand a chance to win some gifts. This exercise is open to participants all over the world.
Today’s Question For the two beams loaded as shown above, which formular is common to each of them when evaluating the internal forces in the final loaded state? (assume linear elastic response)
Thank you for participating in exercise today, remember to enter your answer in the comment section. The main aim of this exercise to stimulate knowledge of structural analysis on the internet in a fun and exciting way. We are always happy to hear from you, so kindly let us know how you feel about Structville.
E-mail:info@structville.com WhatsApp: +2347053638996 You can also visit Structville Research for downloads of civil engineering materials.
STRUCTVILLE REINFORCED CONCRETE DESIGN MANUAL We have this very affordable design manual available…
Do you want to preview the book, click PREVIEW To download full textbook, click HERE
Formworks are structures used to support and hold fresh concrete in place in order to obtain the desired shape prior to setting, curing, and hardening. Formwork can be temporary (struck from concrete after curing) or permanent. Normally, formwork is designed to support load from fresh concrete (including hydrostatic pressure), their self-weight, live load from working personnel, and load from other equipment. A typical formwork configuration for floor slabs is shown in the figure below;
Before construction commences, it is very important that the formwork be well-designed. The design effort required will be determined by the form’s size, complexity, height above the ground, and materials (considering reuses). In all cases, the strength and serviceability of the formwork should be considered in the design. Furthermore, the entire system’s stability and member buckling should be checked.
Formwork for reinforced concrete structures should be designed to safely handle all vertical and lateral loads that may be imposed until the concrete structure can take such loads. The weight of reinforcing steel, fresh concrete, the weight of the forms themselves, and numerous live loads exerted during the construction are all loads on the forms.
Furthermore, wind load can cause lateral forces that must be resisted by the formwork to avoid lateral failure. The formwork system or in-place structure with suitable strength for that purpose should carry vertical and lateral loads to the ground.
Loads and Pressures on Formworks
Unsymmetrical concrete pouring/placement, impact from machine-delivered concrete, uplift, and concentrated loads caused by keeping supplies on the freshly constructed slab must all be considered while designing the forms. There will rarely be precise knowledge about the loads that will be applied to the forms, therefore the designer must make certain safe assumptions that will hold true in most situations. The sections that follow are intended to assist the designer in establishing the loading on which to base form design for typical structural concrete situations.
Lateral Pressure of fresh Concrete on Formwork
The gravity load on a horizontal slab or beam formwork differs from the loads imposed by fresh concrete against wall or column formwork. The freshly placed concrete behaves like a fluid for a short time until the concrete begins to set, causing hydrostatic pressure to act laterally on the vertical forms.
Concrete pressure on formwork is primarily determined by several or all of the following factors:
Rate of placing concrete in forms (R)
The temperature of concrete (T)
Weight or density of concrete (ρ)
Cement type or blend used in the concrete.
Method of consolidating the concrete.
Method of placement of the concrete.
Depth of placement.
Height of form.
Lateral Pressure equations
The American Concrete Institute has spent a lot of time and effort researching and studying form design and construction techniques. The complete hydrostatic lateral pressure, as defined by the following equations, is the maximum pressure on formwork, according to ACI Committee (347).
p = ρgh (kPa) ——— (1)
The set characteristics of a concrete mixture should be understood, and the level of fluid concrete can be calculated using the rate of placement. When more than one placement of concrete is to be made, (h) should be taken as the complete height of the form or the distance between horizontal construction joints for columns or other forms that can be filled quickly before stiffening of the concrete occurs.
When working with mixtures using newly introduced admixtures that increase set time or increase slump characteristics, such as self-consolidating concrete, Eq. (1) should be used until the effect on formwork pressure is understood by measurement.
For concrete having a slump of 175 mm or less and placed with normal internal vibration to a depth of 1.2 m or less, formwork can be designed for a lateral pressure as follows:
For columns: For determining pressure of concrete on formwork ACI 347 defines a column as a vertical structural member with no plan dimensions greater than 2m. For concrete with a slump (175 mm):
pmax = CwCc [7.2 + (785R/T + 17.8)] ——— (2)
With a minimum of 30 Cw kPa, but in no case greater than ρgh.
For walls: For determining pressure of concrete on formwork ACI 347 defines a wall as a vertical structural member with at least one plan dimension greater than 2m. Two equations are provided for wall form pressure:
(a) With a rate of placement of less than 2.1 m/hr and a placement height not exceeding 4.2 m : pmax = CwCc [7.2 + (785R/T + 17.8)] ——— (3)
With a minimum of 30Cw kPa, but in no case greater than ρgh.
(b) with a placement rate less than 2.1 m/hr where placement height exceeds 4.2 m, and for all walls with a placement rate of 2.1 to 4.5 m/hr: pmax = CwCc [7.2 + 1156/(T + 17.8) + 244/(T + 17.7)] ——— (4) With a minimum of 30Cw kPa, but in no case greater than ρgh.
Where; Cw = Unit weight coefficient which depends on the unit weight of the concrete Cc = Chemistry coefficient which depends on the type of cementitious materials
Alternatively, a method based on appropriate experimental data can be used to determine the lateral pressure used for form design.
The unit weight coefficient Cw can be calculated using the Table below;
Density of concrete
Unit weight coefficient (Cw)
Less than 2240 kg/m3
Cw = 0.5[1 + ρc/2320] but not less than 0.8
2240 – 2400 kg/m3
Cw = 1.0
More than 2400 kg/m3
Cw = ρc /2320
The value of the chemistry coefficient Cc can be picked from the Table below;
Type of cement
Chemistry Coefficient Cc
Types I, II, and III without retarders
1.0
Types I, II, and III with retarders
1.2
Other types or blends containing less than 70% slag or 40% fly ash without retarders
1.2
Other types or blends containing less than 70% slag or 40% fly ash with retarders
1.4
Other types or blends containing more than 70% slag or 40% fly ash
1.4
In the UK, the formula for calculating the pressure on formworks according to CIRIA Report 108 is given by the equation below, which must not be greater than the hydrostatic pressure.
Pmax = [C1√R + C2K √(H1 – C1√R)]γ
Where: Pmax = Maximum lateral pressure against formwork (kPa) R = Rate of placement (m/h) C1 = Coefficient for the size and shape of the formwork (1 for walls). C2 = Coefficient for the constituent materials of the concrete (0.3 – 0.6). γ = Specific weight of concrete (kN/m3). H1 = Vertical form height (m). K = Temperature coefficient K = (36/T + 16)2
Vertical loads on Formwork
In addition to lateral pressure, vertical loads are also imposed on formwork. Vertical loads consist of dead and live loads. The weight of formwork, the weight of the reinforcement and freshly placed concrete is dead load. The live load includes the weight of the workers, equipment, material storage, runways, and impact.
Vertical loads assumed for shoring and reshoring design for multistory construction should include all loads transmitted from the floors above as dictated by the proposed construction schedule.
The majority of all formwork involves concrete weighting 22 – 24 kN/m3. Minor variations in this weight are not significant, and for the majority of cases, 24 kN/m3 including the weight of reinforcing steel is commonly assumed for design. Formwork weights vary from as little as 0.15 to 0.7 kN/m2. When the formwork weight is small in relation to the weight of the concrete plus live load, it is frequently, neglected.
ACI committee 347 recommends that both vertical supports and horizontal framing components of formwork should be designed for a minimum live load of 2.4 kN/m2 of horizontal projection to provide for the weight of workmen, runways, screeds and other equipment. When motorized carts are used, the minimum should be 3.6 kN/m2. Regardless of slab thickness, the minimum design load for combined dead and live loads should not be less than 4.8 kN/m2 or 6.0 kN/m2 if motorized carts are used.
Horizontal loads on Formwork
Horizontal loads include the assumed value of load due to wind, dumping of concrete, inclined placement of concrete, cable tensions and equipment. The impact of wind increases with height. Horizontal loads should be not less than 1.5 kN/m of floor edge or 2% of total dead load on the form.
Bracing should be provided to withstand the side sway effects which occur when concrete is placed unsymmetrical on a slab form. Wall form bracing should be designed to meet the minimum wind load requirements of the local building code with adjustments for shorter recurrence intervals.
For wall forms exposed to the elements, the minimum wind design load should not be less than (0.72 kN/m2). Bracing for wall forms should be designed for a horizontal load of at least 1.5 kN/m of wall length, applied at the top. Wall forms of unusual height or exposure should be given special consideration.
Formwork Requirement for Suspended Slabs
In Nigeria, the standard given below works for normally proportioned reinforced concrete slabs.
Sheathing – 20 mm thick marine plywood Joists – 200mm deep wooden H-beams spaced at 600 mm c/c Stringers – 200mm deep wooden H-beams spaced at 1000 mm c/c Shores – Steel props spaced at 1000 mm c/c
For low-cost construction, the following can be used;
Sheathing – 25 mm thick wooden planks Joists – 2″ x 3″ wood (50 mm x 75 mm) spaced at 400mm c/c Stringers – 2″ x 4″ wood (50 mm x 100 mm) spaced at 600 mm c/c Shores – Bamboo wood spaced at 600 mm c/c
Now having known how to arrange formwork for floor slab, the next phase is to determine how to place your order of materials. We will use a typical example to show how it is done.
The first-floor plan of a building is shown below. We are to determine the formwork requirement of the floor slab. We will be neglecting the floor beams in this calculation.
On studying the drawing, Length of longer side of building = 16.120 m Length of shorter side of building = 12.530 m Therefore, gross area of building = 201.9836 m2
Area of openings (lift and staircase) = 18.21 m2
Therefore net area of slab = 201.9836 – 18.21 = 183.7736 m2
Marine Plywood Requirement Area of each marine plywood = 2.4m x 1.2m = 2.88 m2
Number of marine plywood required = 183.7736/2.88 = 63.8 Therefore, supply 64 pieces of (2.4m x 1.2m) marine plywood (no allowance for wastage)
If plank were to be used; Area of one plank = 3.6m x 0.3m = 1.08 m2 Therefore, supply 183.7736/1.08 = 171 pcs of (1″ x 12″ x 12′ plank)
Floor Joist Requirement Length of wooden H-beam = We normally have variety of 3.9 m or 2.9 m Considering the longer side of the building;
Number of H-beam required per line = 16.12/3.9 = 4.13 No We can therefore say, provide 4 No of 3.9m H-beam and 1 No of 2.9m H-beam per line (there will be projections though, and appropriate considerations should be made on site)
Spacing = 600mm
Therefore number of lines required = 12.53/0.6 + 1 = 22 lines
Hence provide; 3.9m H-beam = 4 x 22 = 88 pieces 2.9m H-beam = 1 x 22 = 22 pieces
If it were to be that 2″ x 3″ wood will be used; Supply length of 2″ x 3″ wood in Nigeria = 3.6m
Number of 2″ x 3″ wood required per line = 16.12/3.6 = 4.47 pcs At a spacing of 400 mm, we have 12.53/0.4 + 1 = 33 lines
Therefore number of 2″ x 3″ wood required for floor joists = 4.47 x 33 = 148 pieces
Stringer Requirement We will also be using wooden H-beams for stringers. This will run parallel to the shorter side of the building;
Number of H-beam required per line = 12.53/3.9 = 3.21 We can therefore say, provide 3 No of 3.9m H-beam and 1 No of 2.9m H-beam per line
Spacing = 1000mm
Therefore number of lines required = 16.12/1.0 + 1 = 17 lines
Summarily provide; 3.9m H-beam = 3 x 17 = 51 pieces 2.9m H-beam = 1 x 17 = 17 pieces
Shoring Requirement Steel acrow props will be used for the shoring; Spacing = 1m c/c
Number required on the longer side = 16.12/1 + 1 = 17 pcs Number required on the shorter side = 12.53/1 + 1 = 14 pcs
Acrow props required = 17 x 14 = 238 pieces
Number of props required in the lift and staircase area Length of lift area and staircase area = 5.055m (6 props) Width of lift and staircase area = 3.60m (5 props) Number of props that would have been in lift area = 5 x 6 = 30 pieces
Therefore, the total number of acrow props required = 238 – 30 = 208 pieces
Formwork summary for floor slab;
Marine ply wood = 64 pieces
3.9m H-beam = 88 + 51 = 139 pieces
2.9m H-beam = 17 + 22 = 39 pieces
Acrow Props = 208 pieces
Note that all these materials can be hired because they are very reusable. This is one of the advantages. For instance, marine plywood can be used 8 times before it gets damaged, and wooden H-beams (joists) are very durable for a long period of time provided they are well handled and protected from long exposure to moisture.
I hope you found this piece of information helpful. Thank you for visiting Structville today and God bless you. Kindly contact info@structville.com and find out how we can be of help to you.
Slabless staircases (also called sawtooth staircases) offer aesthetically pleasing alternatives in buildings, and are often a source of wonder to those who do not understand the design and detailing principles underlying their construction. The processes of determining the design moments in sawtooth staircases have been presented in this post.
According to Reynolds and Steedman (2005), Cusens (1966) showed that if axial shortening is neglected, and the strain energy due to bending only considered, the mid span moment in a slabless staircase is given by the general expression;
where;
k0 is the ratio of the stiffness of the tread to the stiffness of the tread j = number of treads
If j is odd;
If j is even;
Design Example
Let us obtain the design forces in a slabless staircase with the following properties;
Thickness of tread and riser = 100mm
Height of riser = 175mm
Width of tread = 300mm
Number of treads = 7
Width of staircase = 1500mm
At ultimate limit state;
n = 1.35gk + 1.5qk
n = 1.35(5) + 1.5(3) = 11.25 kN/m2
k0 = 175/300 = 0.583
Reading from chart, the support moment coefficient can be taken as -0.086
The support bending moment is therefore;
Ms = 0.086 × 11.25 × 2.12 = 4.267 kN.m
j = 7 (odd)
The free bending moment is therefore
Mf = 1/8 × 11.25 × 2.12 × [(72 + 1)/72] = 6.325 kNm
Therefore, the span moment = Mf – Ms = 6.325 – 4.267 = 2.058 kNm
Comparing this answer with finite element analysis from Staad Pro;
The 3D rendering of the staircase is as shown below;
The linear elastic analysis of the structure gave the following results;
Longitudinal Bending Moment
From the result above;
Maximum support moment = 4.11 kNm/m
Mid span moment = 2.84 kN/m
(Kindly compare this answer with the result from manual analysis)
Transverse Bending Moment
Twisting Moment
Detailing
The reinforcement detailing for slabless staircase is provided in the form of links. See a sample detailing image below;
That is how far we will go with this post. For more information, kindly contact info@structville.com
Few months ago, I made a publication on the design of pile foundation founded on sandy soils by using Lekki Pennisula Lagos, as case study. Engr. Maxwell Azu (Nigeria) has submitted modifications to the post, and his inputs will be published soonest. We are grateful for his contributions.
In this post, I want to repeat the same calculation by using Prokon software. The design parameters are repeated below;
Depth of pile = 20 m
Diameter of pile = 600 mm
First layer (3m thick) – Loose to Medium Silty Sand, Cu = 0, ϕ = 32°, γ = 18.5 kN/m2 Second layer (1.5m thick) – Soft silty clay, Cu = 24 kN/m2 , ϕ = 28°, γ = 15 kN/m2 Third layer (4.5m thick) – Very loose sand, Cu = 0, ϕ = 30°, γ = 18.5 kN/m2 Fourth layer (11m thick) – Medium dense sand, Cu = 0, ϕ = 33°, γ = 19.5 kN/m2
Analysis Using Prokon
(1) Input Pile Dimensions as given below
(2) Input soil parameters
Note that the angle of internal friction has been modified by 0.75ϕ in the table above.
(3) Input factor of safety
(4) Input other design parameters and design load
(5) On analysis, the output below can be seen;
The total allowable load on the pile = 1258.01 kN
This can be compared to the load from manual analysis which is obtained as 1048 kN using DA1 and 1248 kN using DA2 of Eurocode 7.
However, once the appropriate soil parameters have been entered, Prokon can be trusted to give reliable result for pile foundations.
Other useful results;
However, study the previous example on this topic, and notice the differences in assumptions made. Your comments and reactions are highly welcome.
Structville Integrated Services is calling on civil engineering students in various universities and polytechnics in Nigeria to participate in her first structural design competition. This is in line with our ambition of promoting civil engineering knowledge among students in Nigeria, and encouraging the hardworking ones to keep their efforts up. We also believe that this will help ignite the love of structural engineering among students. For this competition, serving NYSC members are also eligible to participate. NOTE THAT PARTICIPATION IN THIS COMPETITION IS BY CHOICE, AND IT IS ABSOLUTELY FREE.
The prize money to be won is as follows; 1st Position ~ =N= 10,000 2nd Position ~ =N= 3,000 3rd Position ~ =N= 2,000
Competition Details Structure: Reinforced Concrete Office Building Location: Lekki Free Trade Zone, Lagos Nigeria Concrete covers: Design and Select Fire rating of building: 1 hour Yield strength of reinforcement = Fy = 460 N/mm2 Design strength of concrete:Design and select The relevant drawings for the building are given below. You should however download the questions and soft copy of drawings for more clarity. This can be obtained by following the link below.
Design Questions (1) Design the floor slab of the conference room and do the detailing sketches (100 marks) (2) Design the beam on axis F with detailing sketches (100 marks) (3) Design beam on axis 3 and do the detailing sketches (100 marks) (4) What is the design axial force and bending moment on column G3 considering first floor loads only. Go ahead and provide reinforcement (100 marks) (5) Comment on the technicalities of the design, your assumptions, the challenges you encountered, how you approached it, why you made the decisions you made, and recommendations (if any) to the architect/client. (100 marks)
Instructions (1) This design may only be carried out using BS 8110-1:1997 or Eurocode 2 only. (2) All designs must be FULLY referenced to the code of practice used. (3) You are to provide the member sizes, except the thickness of the floor slab. (4) All assumptions made in the design must be explicitly stated, and all other references well cited. (5) All analysis and design must be manually done. Do not skip any step on calculation or you will lose marks. All works must be shown. (6)You can type set your calculations using MS Word (Font style: Comic Sans Size: 12). If you are using pen, the design must be neatly done on plain A4 paper, scanned and compiled to MS Word or PDF format before being sent. Documents sent as jpeg will not be assessed. (7) Detailing sketches can be done manually or using AUTOCAD. (8) Where necessary, use sketches to drive your point home. (9) Your name and application number must be clearly stated on your design sheet. (10) Neglect roof load in your analysis.
Scoring You design will be assessed based on the following;
(1) Adherence to instructions (10 marks) (2) Attention to details (10 marks) (3) Accuracy of calculations and adherence to the code of practice (20 marks) (4) Robustness and durability of design (20 marks) (5) Economy (20 marks) (6) Detailing sketches (20 marks) Total = 100 marks for each question
Application Procedure (1) Download application form here (2) To confirm that you are still a student, you are expected to forward either of the following; JAMB admission slip, School ID card, or Departmental ID card to info@structville.com. Corp Members must provide Corpers ID card. This submission must be done before final submission of answers. (3) You can download the question booklet below (4) You can also download the architectural drawing in .DWG (AutoCAD) format below. (5) If your application is approved, you will be issued an application number. However, you can start your design pending your approval. (6) Your well compiled and completed design and drawings should be forwarded to the following e-mail addresses;
Additional Benefits: Outstanding students/participants will have their designs published so that the whole world will see what they have done, and why they were selected as the best. The most outstanding student will also be featured on Structville, where his profile will be published, and his achievements celebrated. So by participating in this competition, you are consenting to this. The whole aim of this exercise is to teach, and to inspire.
In the design of civil engineering structures, retaining walls are normally used to retain soil (earth materials) and possible hydrostatic pressure, and they are usually found on embankments, highways, basements of buildings, etc. This publication presents an example of the design of cantilever retaining walls.
The fundamental requirement of retaining wall design is that the wall must be able to hold the retained material in place without causing excessive movement due to deflection, overturning, or sliding. Furthermore, the wall must be structurally sound to withstand internal stresses such as bending moment and shear due to the retained soil. This is ensured by providing adequate wall thickness and reinforcement.
Retaining wall design can be separated into three basic phases:
(1) Stability analysis – ultimate limit state (EQU and GEO) (2) Bearing pressure analysis – ultimate limit state (GEO), and (3) Member design and detailing – ultimate limit state (STR) and serviceability limit states.
Construction of a reinforced concrete retaining wall
Stability Analysis
A retaining wall must be stable under the action of the loads corresponding to the ultimate limit state (EQU) in terms of resistance to overturning. This is exemplified by the straightforward example of a gravity wall shown below.
Gravity retaining wall
When a maximal horizontal force interacts with a minimal vertical load, overturning becomes critical. It is common practice to apply conservative safety factors of safety to the pressures and weights in order to prevent failure by overturning. The partial factors of safety that are important to these calculations are listed in Table 10.1(c).
If the permanent load Gk has “favourable” effects, a partial factor of safety of γG = 0.9 is applied, and if the effects of the permanent earth pressure loading at the rear face of the wall have “unfavourable” effects, a partial factor of safety of γf = 1.1 is multiplied. The variable surcharge loading’s “unfavorable” consequences, if any, are compounded by a partial factor of safety of γf = 1.5.
The moment for overturning resistance is normally taken about the toe of the base, at point A on figure 1. Consequently, it is necessary that 0.9Gkx > γfHky.
Friction between the base’s bottom and the ground provides the resistance to sliding, which is why it is also influenced by the entire self-weight Gk. The base’s front face may experience some resistance from passive ground pressure, but as this material is frequently backfilled against the face, this resistance cannot be guaranteed and is typically disregarded.
Failure by sliding is taken into account while analyzing the loads that correspond to the GEO’s final limit condition. The variables that apply to these calculations are listed in the table above. Typically, it is observed that sliding resistance is usually more critical than overturning. If the sliding requirement is not satisfied, a heel beam may be employed, and the force from passive earth pressure across the heel’s face area may be added to the force resisting sliding.
Bearing Pressure Analysis
When evaluating the necessary foundation size, the bearing pressures behind retaining walls are evaluated using the geotechnical ultimate limit state (GEO). The analysis is comparable to what would happen to a foundation if an eccentric vertical load and an overturning moment were to act simultaneously.
Typical bearing pressure under a retaining wall
The distribution of bearing pressures will be as shown below, provided the effective eccentricity lies within the ‘middle third’ of the base, that is;
M/P ≤ B/6
Therefore;
pmax = P/B + 6M/B2 pmin = P/B – 6M/B2
In general, section 9 of EN 1997-1 applies to retaining structures supporting ground (i.e. soil, rock or backfill) and/or water and is sub-divided as follows:
§9.1. General §9.2. Limit states §9.3. Actions, geometrical data and design situations §9.4. Design and construction considerations §9.5. Determination of earth pressures §9.6. Water pressures §9.7. Ultimate limit state design §9.8. Serviceability limit state design
According to EN 1997-1, ultimate limit states GEO and STR must be verified using one of three Design Approaches (DAs). The United Kingdom proposed to adopt Design Approach 1, and this has been used in this design.
The design of gravity walls to Eurocode 7 involves checking that the ground beneath the wall has sufficient:
bearing resistance to withstand inclined, eccentric actions
sliding resistance to withstand horizontal and inclined actions
stability to avoid toppling
stiffness to prevent unacceptable settlement or tilt
Verification of ultimate limit states (ULSs) is demonstrated by satisfying the inequalities: Vd ≤ Rd Hd ≤ Rd MEd,dst ≤ MEd,stb
In the example downloadable in this post, the retaining wall shown below has been analysed for resistance against sliding, overturning, and bearing. All the calculations were carried out at ultimate limit state.
Member Design and Detailing
The design of bending and shear reinforcement, like that of foundations, is based on a consideration of the loads for the ultimate limit state (STR), along with the accompanying bearing pressures. Steel will rarely be needed in gravity walls, whereas the walls in counterfort and cantilever walls may be designed as slabs. Unless they are quite large, counterforts often have a cantilever beam-like design.
Construction of cantilever retaining wall
The stem of a retaining wall of the cantilever type is made to resist the moment brought on by the horizontal forces. The thickness of the wall can be calculated as 80mm per metre of backfill for the purposes of preliminary sizing. In most cases, the base’s thickness is the same with the stem’s. The heel and toe need to be made to resist the moments brought on by the downward weight of the base and soil and the upward earth-bearing pressures.
As necessary, reinforcement details of retaining walls must adhere to the general guidelines for slabs and beams. To prevent shrinkage and thermal cracking, special attention must be paid to the reinforcement’s detailing. Because they typically entail enormous concrete pours, gravity walls are particularly prone to damage. Minimal restrictions on thermal and shrinkage movement should be used. The need for strong soil-base friction, however, negates this in the design of bases, making it impossible to create a sliding layer. Therefore, the bases’ reinforcement needs to be sufficient to prevent cracking from being brought on by a lot of constraint.
Due to the loss of hydration heat during thermal movement, long retaining walls supported by inflexible bases are especially prone to cracking, thus detailing must make an effort to disperse the cracks to maintain appropriate widths. There must be full vertical movement joints available. Waterbars and sealants should be used, and these joints frequently include a shear key to stop differential movement of adjacent wall sections.
Groundwater hydrostatic forces are typically applied to the back sides of retaining walls. By including a drainage passage at the face of the wall, these might be decreased. A layer of earth or porous blocks with pipes to remove the water, frequently through the front of the wall, is the customary method for creating such a drain. Water is less likely to flow through the retaining wall, cause damage to the soil beneath the wall’s foundations, and cause hydrostatic pressure on the wall to decrease as a result.
Retaining Wall Analysis and Design Example
A cantilever retaining wall is loaded as shown below. Design the retaining wall in accordance with EN1997-1:2004 incorporating Corrigendum dated February 2009 and the UK National Annex incorporating Corrigendum No.1.
Retaining wall details Stem type; Cantilever Stem height; hstem = 3000 mm Stem thickness; tstem = 300 mm Angle to rear face of stem; α = 90 deg Stem density; γstem = 25 kN/m3 Toe length; ltoe = 500 mm Heel length; lheel = 1500 mm Base thickness; tbase = 350 mm Base density; γbase = 25 kN/m3 Height of retained soil; hret = 2500 mm Angle of soil surface; β = 0 deg Depth of cover; dcover = 500 mm Depth of excavation; dexc = 200 mm
The effect of wind on a building gets more significant as the height of the building increases. Due to the various flow scenarios that result from wind’s interaction with structures, wind load analysis on tall buildings is a very complex phenomenon. In a general stream of air flowing in relation to the earth’s surface, wind is made up of several eddies with different diameters and rotational properties. The wind is turbulent or gusty because of these eddies.
Strong winds’ lower atmosphere gustiness is mostly caused by contact with surface structures. While the gustiness of the wind tends to diminish with height, the average wind speed tends to increase over a period of time of the order of ten minutes or more. The size of the eddies affects the dynamic loading on a structure as a result of turbulence. Large eddies that are comparable in size to the structure create well-correlated pressures as they encircle it. On the other hand, minor eddies cause stresses on different regions of a structure that are essentially unrelated to the separation between them.
Some buildings, especially tall or slender ones, react to the influences of wind in a dynamic way. The causes of a structure’s dynamic response to wind are a variety of different occurrences. These include galloping, fluttering, buffeting, and vortex shedding. Due to turbulence buffeting, thin structures are likely to be sensitive to dynamic reactions in the direction of the wind.
Although turbulence buffeting can also cause a transverse or cross-wind reaction, vortex shedding or galloping is more likely to cause one. Flutter is a linked motion that frequently combines torsion and bending and can cause instability.
Wind load simulation on a highrise building
Design Wind Load Pressure
The geometry of the structure under consideration, the geometry and proximity of the structures upwind, and the characteristics of the approaching wind all influence the features of wind pressures on that structure. The wind is gusty, which contributes to the pressure fluctuations, but local vortex shedding around the borders of the buildings themselves is also a factor.
If a structure is dynamically wind-sensitive, the varying pressures can cause fatigue damage to it as well as dynamic excitation. Additionally, the pressures are not constant along the surface of the structure but rather change according to position.
When using a design document, it is important to keep in mind the complexity of wind loading. The maximum wind loads that a structure would sustain over the course of its lifetime may differ significantly from those estimated during design due to the numerous unknowns involved. As a result, the success or failure of a structure in a windstorm cannot always be interpreted as a sign of the conservatism or non-conservatism of the Wind Loading Standard.
Buildings and constructions with distinctive shapes or locations are exempt from the Standards’ application. Some types of constructions, such as tall skyscrapers and thin towers, are designed according to wind loading. It frequently becomes appealing to use experimental wind tunnel data instead of the Wind Loading Code coefficients for these buildings.
In this article, wind load analysis has been carried out on a 60m tall high-rise building using the method described in EN 1991-1-4:2005 (General actions – Wind action). The structure is assumed to be located in an area with a basic wind speed of 40 m/s.
Basic Data Height of building = 60m Width of a building = 30m Structure – Framed building The structure is located at terrain category II (see Table below)
Basic wind velocity The fundamental value of the basic wind velocity Vb,0 is the characteristic 10 minute mean wind velocity irrespective of wind direction and time of the year, at 10 m above ground level in open-country terrain with low vegetation such as grass, and with isolated obstacles with separations of at least 20 obstacle heights.
The basic wind velocity Vb,0 is calculated from; Vb = Cdir . Cseason .Vb,0
Where: Vb is the basic wind velocity defined as a function of wind direction and time of the year at 10m above the ground of terrain category II Vb,0 is the fundamental value of the basic wind velocity Cdir is the directional factor (defined in the National Annex, but recommended value is 1.0) Cseason is the season factor (defined in the National Annex, but recommended value is 1.0)
For the area and location of the building that we are considering; Basic wind velocity Vb,0 = 40 m/s Vb = Cdir. Cseason.Vb,0 = 1.0 × 1.0 × 40 = 40 m/s
Mean Wind The mean wind velocity Vm(z) at a height z above the terrain depends on the terrain roughness and orography, and on the basic wind velocity, Vb, and should be determined using the expression below;
Vm(z) = cr(z). co(z).Vb
Where; cr(z) is the roughness factor (defined below) co(z) is the orography factor often taken as 1.0
The terrain roughness factor accounts for the variability of the mean wind velocity at the site of the structure due to the height above the ground level and the ground roughness of the terrain upwind of the structure in the wind direction considered. Terrain categories and parameters are shown in Table 2.0.
cr(z) = kr. In (z/z0) for zmin ≤ z ≤ zmax cr(z) = cr.(zmin) for z ≤ zmin
Where: Z0 is the roughness length kr is the terrain factor depending on the roughness length Z0 calculated using;
kr = 0.19 (Z0/Z0,II)0.07
Where: Z0,II = 0.05m (terrain category II) Zmin is the minimum height = 2 m z = 60 m Zmax is to be taken as 200 m Kr = 0.19 (0.05/0.05)0.07 = 0.19 cr(60) = kr.In (z/z0) = 0.19 × In(60/0.05) = 1.347
Wind turbulence The turbulence intensity Iv(z) at height z is defined as the standard deviation of the turbulence divided by the mean wind velocity. The recommended rules for the determination of Iv(z) are given in the expressions below;
Iv(z) = σv/Vm = kl/(c0(z).In (z/z0)) for zmin ≤ z ≤ zmax Iv(z) = Iv.(zmin) for z ≤ zmin
Where: kl is the turbulence factor of which the value is provided in the National Annex but the recommended value is 1.0 Co is the orography factor described above Z0 is the roughness length described above.
For the building that we are considering, the wind turbulence factor at 60m above the ground level;
Peak Velocity Pressure The peak velocity pressure qp(z) at height z is given by the expression below;
qp(z) = [1 + 7.Iv(z)] 1/2.ρ.Vm2(z) = ce(z).qb
Where: ρ is the air density, which depends on the altitude, temperature, and barometric pressure tobe expected in the region during wind storms (recommended value is 1.25kg/m3)
ce(z) is the exposure factor given by; ce(z) = qp(z)/qb qb is the basic velocity pressure given by; qb = 0.5.ρ.Vb2
External Pressure Coefficients From Table 7.1 of EN 1991-1-4:2005 (E) For the building, taking the height to width ratio h/d = 2.0 Pressure coefficient for windward side = +0.8 Pressure coefficient for leeward side = –0.6 The net pressure coefficient Cpe10 = +0.8 – (–0.6) = 1.4 The net external surface pressure on the structure = qp(z) Cpe10 = 3.6057 × 1.4 = 5.05 kN/m2 Therefore, we = 5.05 kN/m2
Detailed Calculation
Wind loading of a high-rise building in accordance with EN1991-1-3:2005+A1:2010 and the UK national annex.
Building data Type of roof; Flat Length of building; L = 30000 mm Width of building; W = 30000 mm Height to eaves; H = 60000 mm Eaves type; Sharp Total height; h = 60000 mm
Basic values Location; Ibadan, Nigeria Wind speed velocity (FigureNA.1); vb,map = 40.0 m/s Distance to shore; Lshore = 250.00 km Altitude above sea level; Aalt = 100.0m Altitude factor; calt = Aalt/1m × 0.001 + 1 = 1.100 Fundamental basic wind velocity; vb,0 = vb,map × calt = 44.0 m/s Direction factor; cdir = 1.00 Season factor; cseason = 1.00 Shape parameter K; K = 0.2 Exponent n; n = 0.5 Air density; ρ = 1.226 kg/m3
Probability factor; cprob = [(1 – K × ln(-ln(1-p)))/(1 – K × ln(-ln(0.98)))]n = 1.00 Basic wind velocity (Exp. 4.1); vb = cdir × cseason × vb,0 × cprob =44.0 m/s Reference mean velocity pressure; qb = 0.5 × ρ × vb2 = 1.187 kN/m2
Orography Orography factor not significant; co = 1.0
The velocity pressure for the windward face of the building with a 0 degree wind is to be considered as 2 parts as the height h is greater than b but less than 2b (cl.7.2.2)
The velocity pressure for the windward face of the building with a 90 degree wind is to be considered as 2 parts as the height h is greater than b but less than 2b (cl.7.2.2)
