Structville Question of the Day

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May 23, 2018

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.

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Design and Calculation of Formwork Requirement of Slabs

By

Ubani Obinna Uzodimma

-

May 22, 2018

Table of Contents

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):

p_max = C_wC_c[7.2 + (785R/T + 17.8)] ——— (2)

With a minimum of 30 C_w 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 :
p_max = C_wC_c[7.2 + (785R/T + 17.8)] ——— (3)

With a minimum of 30C_w 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:
p_max= C_wC_c [7.2 + 1156/(T + 17.8) + 244/(T + 17.7)] ——— (4)
With a minimum of 30C_w kPa, but in no case greater than ρgh.

Where;
C_w = Unit weight coefficient which depends on the unit weight of the concrete
C_c = 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 C_w can be calculated using the Table below;

Density of concrete	Unit weight coefficient (C_{_w})
Less than 2240 kg/m³	Cw = 0.5[1 + ρ_c/2320] but not less than 0.8
2240 – 2400 kg/m³	C_w = 1.0
More than 2400 kg/m³	C_w = ρ_c /2320

The value of the chemistry coefficient C_c can be picked from the Table below;

Type of cement	Chemistry Coefficient C_c
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.

P_max = [C₁√R + C₂K √(H₁ – C₁√R)]γ

Where:
P_max = 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.

Structural Design of Slabless (Sawtooth) Staircase

By

Ubani Obinna Uzodimma

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May 20, 2018

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

Load Analysis
Self weight of staircase;
{[(0.3 × 0.1) + (0.175 × 0.1)] / 0.3} × 24 = 3.8 kN/m²
Self weight of finishes = 1.2 kN/m²

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;
M_s = 0.086 × 11.25 × 2.1² = 4.267 kN.m

j = 7 (odd)
The free bending moment is therefore
M_f = 1/8 × 11.25 × 2.1² × [(7² + 1)/7²] = 6.325 kNm

Therefore, the span moment = M_f – M_s = 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

Thank you.

Design of Pile Foundation Using Prokon Software

By

Ubani Obinna Uzodimma

-

May 19, 2018

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.

You can read the original post by clicking HERE

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, C_u = 0, ϕ = 32°, γ = 18.5 kN/m²
Second layer (1.5m thick) – Soft silty clay, C_u = 24 kN/m² , ϕ = 28°, γ = 15 kN/m²
Third layer (4.5m thick) – Very loose sand, C_u = 0, ϕ = 30°, γ = 18.5 kN/m²
Fourth layer (11m thick) – Medium dense sand, C_u = 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 Design Competition for Students (Prizes to be won)

By

Ubani Obinna Uzodimma

-

May 14, 2018

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;

info@structville.com
Copy Cc;
rankiesubani@gmail.com
djkally2be@gmail.com
oprite.bobmanuel@gmail.com

(6) The deadline for submission is 12 mid night, 31st May, 2018.

DOWNLOAD ALL REQUIREMENTS HERE

* APPLICATION FORM
* QUESTION BOOKLET
* ARCHITECTURAL DRAWING

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.

Design of Retaining Walls

By

Ubani Obinna

-

May 9, 2018

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 — 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 G_k has “favourable” effects, a partial factor of safety of γ_G = 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 γ_f = 1.1 is multiplied. The variable surcharge loading’s “unfavorable” consequences, if any, are compounded by a partial factor of safety of γ_f = 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 0.9G_kx > γ_fH_ky.

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 G_k. 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.

bearing pressure distribution on a retaining wall — 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;

p_max = P/B + 6M/B²
p_min = 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:
V_d ≤ R_d
H_d ≤ R_d
M_Ed,dst ≤ M_Ed,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.

cantilever retaining wall construction — 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; h_stem = 3000 mm
Stem thickness; t_stem = 300 mm
Angle to rear face of stem; α = 90 deg
Stem density; γ_stem = 25 kN/m³
Toe length; l_toe = 500 mm
Heel length; l_heel = 1500 mm
Base thickness; t_base = 350 mm
Base density; γ_base = 25 kN/m³
Height of retained soil; h_ret = 2500 mm
Angle of soil surface; β = 0 deg
Depth of cover; d_cover = 500 mm
Depth of excavation; d_exc = 200 mm

Retained soil properties
Soil type; Medium-dense well graded sand
Moist density; γ_mr = 21 kN/m³
Saturated density; γ_sr = 23 kN/m³
Characteristic effective shear resistance angle; φ’_r.k = 30 deg
Characteristic wall friction angle; δ_r.k = 0 deg

Base soil properties
Soil type; Medium-dense well graded sand
Soil density; γ_b = 18 kN/m³
Characteristic cohesion; c’_b.k = 0 kN/m²
Characteristic effective shear resistance angle; φ’_b.k = 30 deg
Characteristic wall friction angle; δ_b.k = 15 deg
Characteristic base friction angle; δ_bb.k = 30 deg

Loading details
Variable surcharge load; Surcharge_Q = 10 kN/m²

Retaining wall geometry
Base length; l_base = l_toe + t_stem + l_heel = 2300 mm
Moist soil height; h_moist = h_soil = 3000 mm
Length of surcharge load; l_sur = l_heel = 1500 mm
– Distance to vertical component; x_{sur_v} = l_base – (l_heel / 2) = 1550 mm
Effective height of wall; h_eff = h_base + d_cover + h_ret = 3350 mm
– Distance to horizontal component; x_{sur_h} = h_eff / 2 = 1675 mm

Area of wall stem; A_stem = h_stem × t_stem = 0.9 m²
– Distance to vertical component; x_stem = l_toe + t_stem / 2 = 650 mm

Area of wall base; A_base = l_base × t_base = 0.805 m²
– Distance to vertical component; x_base = l_base / 2 = 1150 mm

Area of moist soil; A_moist = h_moist × l_heel = 4.5 m²
– Distance to vertical component; x_{moist_v} = l_base – (h_moist × l_heel² / 2) / A_moist = 1550 mm
– Distance to horizontal component; x_{moist_h} = h_eff / 3 = 1117 mm

Area of base soil; A_pass = d_cover × l_toe = 0.25 m²
– Distance to vertical component; x_{pass_v} = l_base – [d_cover × l_toe × (l_base – l_toe/2)] / A_pass = 250 mm
– Distance to horizontal component; x_{pass_h} = (d_cover + h_base) / 3 = 283 mm

Area of excavated base soil; A_exc = h_pass × l_toe = 0.15 m²
– Distance to vertical component; x_{exc_v} = l_base – [h_pass × l_toe × (l_base – l_toe/2)] / A_exc = 250 mm
– Distance to horizontal component; x_{exc_h} = (h_pass + h_base) / 3 = 217 mm

Design approach 1 – Combination 1

Partial factor set; A1
Permanent unfavourable action; γ_G = 1.35
Permanent favourable action; γ_Gf = 1.00
Variable unfavourable action; γ_Q = 1.50
Variable favourable action; γ_Qf = 0.00

Partial factors for soil parameters – Table A.4 – Combination 1
Soil parameter set; M1
Angle of shearing resistance; γ_{_φ‘} = 1.00
Effective cohesion; γ_c’ = 1.00
Weight density; γ_g = 1.00

Retained soil properties
Design moist density; γ_mr‘ = γ_mr / γ_g = 21 kN/m³
Design saturated density; γ_sr‘ = γ_sr / γ_g = 23 kN/m³
Design effective shear resistance angle; φ’_r.d = tan^-1(tan(φ’_r.k) / γ_φ’) = 30 deg
Design wall friction angle; δ_r.d = tan^-1(tan(δ_r.k) / γ_φ’) = 0 deg

Base soil properties
Design soil density; γ_b‘ = γ_b/γ_g = 18 kN/m³
Design effective shear resistance angle; φ’_b.d = tan^-1(tan(φ’_b.k) / γ_φ’) = 30 deg
Design wall friction angle; δ_b.d = tan^-1(tan(δ_b.k) / γ_φ’) = 15 deg
Design base friction angle; δ_bb.d = tan^-1(tan(δ_bb.k) / γ_φ’) = 30 deg
Design effective cohesion; c’_b.d = c’_b.k / γ_c’ = 0 kN/m²

Using Rankine theory
Active pressure coefficient; K_A = (1 – sin(φ’_r.d)) / (1 + sin(φ’_r.d)) = 0.333
Passive pressure coefficient; K_P = (1 + sin(φ’_b.d)) / (1 – sin(φ’_b.d)) = 3.000

Sliding check

Vertical forces on wall
Wall stem; F_stem = γ_Gf × A_stem × γ_stem = 22.5 kN/m
Wall base; F_base = γ_Gf × A_base × γ_base = 20.1 kN/m
Moist retained soil; F_{moist_v} = γ_Gf × A_moist × γ_mr‘ = 94.5 kN/m
Base soil; F_{exc_v} = γ_Gf × A_exc × γ_b‘ = 2.7 kN/m
Total; F_{total_v} = F_stem + F_base + F_{moist_v} + F_{exc_v} = 139.8 kN/m

Horizontal forces on wall
Surcharge load; F_{sur_h} = K_A × γ_Q × Surcharge_Q × h_eff = 16.8 kN/m
Moist retained soil;F_{moist_h} = γ_G × K_A × γ_mr‘ × h_eff² / 2 = 53 kN/m
Total; F_{total_h} = F_{sur_h} + F_{moist_h} = 69.8 kN/m

Check stability against sliding
Base soil resistance; F_{exc_h} = γ_GfK_Pγ_b‘ × (h_pass + h_base)² / 2 = 11.4 kN/m
Base friction; F_friction = F_{total_v} × tan(δ_bb.d) = 80.7 kN/m
Resistance to sliding; F_rest = F_{exc_h} + F_friction = 92.1 kN/m

Factor of safety; FoS_sl = F_rest / F_{total_h} = 1.32

PASS – Resistance to sliding is greater than sliding force

Overturning check

Vertical forces on wall
Wall stem; F_stem = γ_Gf × A_stem × γ_stem = 22.5 kN/m
Wall base; F_base = γ_Gf × A_base × γ_base = 20.1 kN/m
Moist retained soil; F_{moist_v} = γ_Gf × A_moist × γ_mr‘ = 94.5 kN/m
Base soil; F_{exc_v} = γ_Gf × A_exc × γ_b‘ = 2.7 kN/m
Total; F_{total_v} = F_stem + F_base + F_{moist_v} + F_{exc_v} = 139.8 kN/m

Horizontal forces on wall
Surcharge load; F_{sur_h} = K_A × γ_Q × Surcharge_Q × h_eff = 16.8 kN/m
Moist retained soil; F_{moist_h} = γ_G × K_A × γ_mr‘ × h_eff² / 2 = 53 kN/m
Base soil; F_{exc_h} = -γ_Gf × K_P × γ_b‘ × (h_pass + h_base)² / 2 = -11.4 kN/m
Total; F_{total_h} = F_{sur_h} + F_{moist_h} + F_{exc_h} = 58.4 kN/m

Overturning moments on wall
Surcharge load; M_{sur_OT} = F_{sur_h} × x_{sur_h} = 28.1 kNm/m
Moist retained soil; M_{moist_OT} = F_{moist_h} × x_{moist_h} = 59.2 kNm/m
Total; M_{total_OT} = M_{sur_OT} + M_{moist_OT} = 87.3 kNm/m

Restoring moments on wall
Wall stem; M_{stem_R} = F_stem × x_stem = 14.6 kNm/m
Wall base; M_{base_R} = F_base × x_base = 23.1 kNm/m
Moist retained soil; M_{moist_R} = F_{moist_v} × x_{moist_v} = 146.5 kNm/m
Base soil; M_{exc_R} = F_{exc_v} × x_{exc_v} – F_{exc_h} × x_{exc_h} = 3.1 kNm/m
Total; M_{total_R} = M_{stem_R} + M_{base_R} + M_{moist_R} + M_{exc_R} = 187.4 kNm/m

Check stability against overturning
Factor of safety; FoS_ot = M_{total_R} / M_{total_OT} = 2.147

PASS – Maximum restoring moment is greater than the overturning moment

Bearing pressure check

Vertical forces on wall
Wall stem; F_stem = γ_G × A_stem × γ_stem = 30.4 kN/m
Wall base; F_base = γ_G × A_base × γ_base = 27.2 kN/m
Surcharge load; F_{sur_v} = γ_Q × Surcharge_Q × l_heel = 22.5 kN/m
Moist retained soil; F_{moist_v} = γ_G × A_moist × γ_mr‘ = 127.6 kN/m
Base soil; F_{pass_v} = γ_G × A_pass × γ_b‘ = 6.1 kN/m
Total; F_{total_v} = F_stem + F_base + F_{sur_v} + F_{moist_v} + F_{pass_v} = 213.7 kN/m

Horizontal forces on wall
Surcharge load; F_{sur_h} = K_A × γ_Q × Surcharge_Q × h_eff = 16.8 kN/m
Moist retained soil; F_{moist_h} = γ_G × K_A × γ_mr‘ × h_eff² / 2 = 53 kN/m
Base soil; F_{pass_h} = -γ_Gf × K_P × γ_b‘ × (d_cover + h_base)² / 2 = -19.5 kN/m
Total; F_{total_h} = max(F_{sur_h} + F_{moist_h} + F_{pass_h} – F_{total_v} × tan(d_bb.d), 0 kN/m) = 0 kN/m

Moments on wall
Wall stem; M_stem = F_stem × x_stem = 19.7 kNm/m
Wall base; M_base = F_base × x_base = 31.2 kNm/m
Surcharge load; M_sur = F_{sur_v} × x_{sur_v} – F_{sur_h} × x_{sur_h} = 6.8 kNm/m
Moist retained soil; M_moist = F_{moist_v} × x_{moist_v} – F_{moist_h} × x_{moist_h} = 138.5 kNm/m
Base soil; M_pass = F_{pass_v} × x_{pass_v} – F_{pass_h} × x_{pass_h} = 7 kNm/m
Total; M_total = M_stem + M_base + M_sur + M_moist + M_pass = 203.4 kNm/m

Check bearing pressure
Distance to reaction; x’ = M_total / F_{total_v} = 952 mm
Eccentricity of reaction; e = x’ – l_base / 2 = -198 mm
Loaded length of base; l_load = 2x’ = 1903 mm
Bearing pressure at toe; q_toe = F_{total_v} / l_load = 112.3 kN/m²
Bearing pressure at heel; q_heel = 0 kN/m²

Effective overburden pressure; q = (t_base + d_cover)γ_b‘ = 15.3 kN/m²
Design effective overburden pressure; q’ = q / γ_g = 15.3 kN/m²

Bearing resistance factors;
N_q = Exp(π × tan(φ’_b.d)) × (tan(45 deg + φ’_b.d / 2))² = 18.401
N_c = (N_q – 1)cot(φ’_b.d) = 30.14
N_γ = 2(N_q – 1)tan(φ’_b.d) = 20.093

Foundation shape factors;
s_q = 1
s_γ = 1
s_c = 1

Load inclination factors;
H = F_{sur_h} + F_{moist_h} + F_{pass_h} = 50.3 kN/m
V = F_{total_v} = 213.7 kN/m
m = 2

i_q = [1 – H / (V + l_load × c’_b.d × cot(φ’_b.d))]^m = 0.585
i_γ = [1 – H / (V + l_load × c’_b.d × cot(φ’_b.d))]^{(m + 1)} = 0.447
i_c = i_q – (1 – i_q) / (N_c × tan(φ’_b.d)) = 0.561

Net ultimate bearing capacity
n_f = c’_b.dN_cs_ci_c + q’N_qs_qi_q + 0.5γ_b‘l_loadN_γs_γi_γ = 318.6 kN/m²

Factor of safety;
FoS_bp = n_f / max(q_toe, q_heel) = 2.838

PASS – Allowable bearing pressure exceeds maximum applied bearing pressure

Design approach 1 – Combination 2

Partial factor set; A2
Permanent unfavourable action; γ_G = 1.00
Permanent favourable action; γ_Gf = 1.00
Variable unfavourable action; γ_Q = 1.30
Variable favourable action; γ_Qf = 0.00

Soil parameter set; M2
Angle of shearing resistance; γ_f’ = 1.25
Effective cohesion; γ_c’ = 1.25
Weight density; γ_g = 1.00

Retained soil properties
Design moist density; γ_mr‘ = γ_mr/γ_g = 21 kN/m³
Design saturated density; γ_sr‘ = γ_sr/γ_g = 23 kN/m³
Design effective shear resistance angle; φ’_r.d = tan^-1(tan(φ’_r.k) / γ_φ’) = 24.8 deg
Design wall friction angle; δ_r.d = tan^-1(tan(δ_r.k) / γ_φ’) = 0 deg

Base soil properties
Design soil density; γ_b‘ = γ_b/γ_g = 18 kN/m³
Design effective shear resistance angle; φ’_b.d = tan^-1(tan(φ’_b.k) / γ_φ’) = 24.8 deg
Design wall friction angle; δ_b.d = tan^-1(tan(δ_b.k) / γ_φ’) = 12.1 deg
Design base friction angle; δ_bb.d = tan^-1(tan(δ_bb.k) / γ_φ’) = 24.8 deg
Design effective cohesion; c’_b.d = c’_b.k /γ_c’ = 0 kN/m²

Using Rankine theory
Active pressure coefficient;
K_A = (1 – sin(φ’_r.d)) / (1 + sin(φ’_r.d)) = 0.409

Passive pressure coefficient;
K_P = (1 + sin(φ’_b.d)) / (1 – sin(φ’_b.d)) = 2.444

Sliding check

Vertical forces on wall
Wall stem; F_stem = γ_Gf × A_stem × γ_stem = 22.5 kN/m
Wall base; F_base = γ_Gf × A_base × γ_base = 20.1 kN/m
Moist retained soil;F_{moist_v} = γ_Gf × A_moist × γ_mr‘ = 94.5 kN/m
Base soil;F_{exc_v} = γ_Gf × A_exc × γ_b‘ = 2.7 kN/m
Total; F_{total_v} = F_stem + F_base + F_{moist_v} + F_{exc_v} = 139.8 kN/m

Horizontal forces on wall
Surcharge load; F_{sur_h} = K_A × γ_Q × Surcharge_Q × h_eff = 17.8 kN/m
Moist retained soil; F_{moist_h} = γ_G × K_A × γ_mr‘ × h_eff² / 2 = 48.2 kN/m
Total; F_{total_h} = F_{sur_h} + F_{moist_h} = 66 kN/m

Check stability against sliding
Base soil resistance; F_{exc_h} = γ_Gf × K_P × γ_b‘ × (h_pass + h_base)² / 2 = 9.3 kN/m
Base friction; F_friction = F_{total_v} × tan(d_bb.d) = 64.6 kN/m
Resistance to sliding; F_rest = F_{exc_h} + F_friction = 73.9 kN/m

Factor of safety; FoS_sl = F_rest / F_{total_h} = 1.119

PASS – Resistance to sliding is greater than sliding force

Overturning check

Vertical forces on wall
Wall stem; F_stem = γ_Gf × A_stem × γ_stem = 22.5 kN/m
Wall base; F_base = γ_Gf × A_base × γ_base = 20.1 kN/m
Moist retained soil; F_{moist_v} = γ_Gf × A_moist × γ_mr‘ = 94.5 kN/m
Base soil; F_{exc_v} = γ_Gf × A_exc × γ_b‘ = 2.7 kN/m
Total; F_{total_v} = F_stem + F_base + F_{moist_v} + F_{exc_v} = 139.8 kN/m

Horizontal forces on wall
Surcharge load; F_{sur_h} = K_A × γ_Q × Surcharge_Q × h_eff = 17.8 kN/m
Moist retained soil; F_{moist_h} = γ_G × K_A × γ_mr‘ × h_eff² / 2 = 48.2 kN/m
Base soil; F_{exc_h} = -γ_Gf × K_P × γ_b‘ × (h_pass + h_base)² / 2 = -9.3 kN/m
Total; F_{total_h} = F_{sur_h} + F_{moist_h} + F_{exc_h} = 56.7 kN/m

Overturning moments on wall
Surcharge load; M_{sur_OT} = F_{sur_h} × x_{sur_h} = 29.8 kNm/m
Moist retained soil; M_{moist_OT} = F_{moist_h} × x_{moist_h} = 53.8 kNm/m
Total; M_{total_OT} = M_{sur_OT} + M_{moist_OT} = 83.7 kNm/m

Restoring moments on wall
Wall stem; M_{stem_R} = F_stem × x_stem = 14.6 kNm/m
Wall base; M_{base_R} = F_base × x_base = 23.1 kNm/m
Moist retained soil; M_{moist_R} = F_{moist_v} × x_{moist_v} = 146.5 kNm/m
Base soil; M_{exc_R} = F_{exc_v} × x_{exc_v} – F_{exc_h} × x_{exc_h} = 2.7 kNm/m
Total; M_{total_R} = M_{stem_R} + M_{base_R} + M_{moist_R} + M_{exc_R} = 186.9 kNm/m

Check stability against overturning
Factor of safety; FoS_ot = M_{total_R} / M_{total_OT} = 2.234

PASS – Maximum restoring moment is greater than overturning moment

Bearing pressure check

Vertical forces on wall
Wall stem; F_stem = γ_G × A_stem × γ_stem = 22.5 kN/m
Wall base; F_base = γ_G × A_base × γ_base = 20.1 kN/m
Surcharge load; F_{sur_v} = γ_Q × Surcharge_Q × l_heel = 19.5 kN/m
Moist retained soil; F_{moist_v} = γ_G × A_moist × γ_mr‘ = 94.5 kN/m
Base soil; F_{pass_v} = γ_G × A_pass × γ_b‘ = 4.5 kN/m
Total; F_{total_v} = F_stem + F_base + F_{sur_v} + F_{moist_v} + F_{pass_v} = 161.1 kN/m

Horizontal forces on wall
Surcharge load; F_{sur_h} = K_A × γ_Q × Surcharge_Q × h_eff = 17.8 kN/m
Moist retained soil; F_{moist_h} = γ_G × K_A × γ_mr‘ × h_eff² / 2 = 48.2 kN/m
Base soil; F_{pass_h} = -γ_Gf × K_P × γ_b‘ × (d_cover + h_base)² / 2 = -15.9 kN/m
Total; F_{total_h} = max(F_{sur_h} + F_{moist_h} + F_{pass_h} – F_{total_v} × tan(δ_bb.d), 0 kN/m) = 0 kN/m

Moments on wall
Wall stem; M_stem = F_stem × x_stem = 14.6 kNm/m
Wall base; M_base = F_base × x_base = 23.1 kNm/m
Surcharge load; M_sur = F_{sur_v} × x_{sur_v} – F_{sur_h} × x_{sur_h} = 0.4 kNm/m
Moist retained soil; M_moist = F_{moist_v} × x_{moist_v} – F_{moist_h} × x_{moist_h} = 92.6 kNm/m
Base soil; M_pass = F_{pass_v} × x_{pass_v} – F_{pass_h} × x_{pass_h} = 5.6 kNm/m
Total; M_total = M_stem + M_base + M_sur + M_moist + M_pass = 136.4 kNm/m

Check bearing pressure
Distance to reaction; x’ = M_total / F_{total_v} = 847 mm
Eccentricity of reaction; e = x’ – l_base / 2 = -303 mm
Loaded length of base; l_load = 2x’ = 1693 mm
Bearing pressure at toe; q_toe = F_{total_v} / l_load = 95.2 kN/m²
Bearing pressure at heel; q_heel = 0 kN/m²

Effective overburden pressure; q = (t_base + d_cover)γ_b‘ = 15.3 kN/m²

Design effective overburden pressure; q’ = q/γ_g = 15.3 kN/m²

Bearing resistance factors;
N_q = Exp[πtan(φ’_b.d)] × [tan(45 deg + φ’_b.d / 2)]² = 10.431
N_c = (N_q – 1) × cot(φ’_b.d) = 20.418
N_γ = 2(N_q – 1) × tan(φ’_b.d) = 8.712

Foundation shape factors;
s_q = 1
s_γ = 1
s_c = 1

Load inclination factors;
H = F_{sur_h} + F_{moist_h} + F_{pass_h} = 50.1 kN/m
V = F_{total_v} = 161.1 kN/m
m = 2

i_q = [1 – H / (V + l_load × c’_b.d × cot(φ’_b.d))]^m = 0.475
i_γ = [1 – H / (V + l_load × c’_b.d × cot(f’φ_b.d))]^{(m + 1)} = 0.327
i_c = i_q – (1 – i_q) / (N_c × tan(φ’_b.d)) = 0.419

Net ultimate bearing capacity
n_f = c’_b.dN_cs_ci_c + q’N_qs_qi_q + 0.5γ_b‘l_load N_γs_γi_γ = 119.1 kN/m²

Library item: Drained bearing output

Factor of safety;
FoS_bp = n_f / max(q_toe, q_heel) = 1.252

PASS – Allowable bearing pressure exceeds maximum applied bearing pressure

Analysis summary

Description	Unit	Capacity	Applied	F o S	Result
Sliding stability	kN/m	73.9	66	1.119	PASS
Overturning stability	kNm/m	187.4	87.3	2.147	PASS
Bearing pressure	kN/m²	119.1	95.2	1.252	PASS

Structural Design

In accordance with EN1992-1-1:2004 incorporating Corrigendum dated January 2008 and the UK National Annex incorporating National Amendment No.1

Concrete details

Concrete strength class; C20/25
Characteristic compressive cylinder strength; f_ck = 20 N/mm²
Characteristic compressive cube strength; f_ck,cube = 25 N/mm²
Mean value of compressive cylinder strength; f_cm = f_ck + 8 N/mm² = 28 N/mm²
Mean value of axial tensile strength; f_ctm = 0.3(f_ck)^2/3 = 2.2 N/mm²
5% fractile of axial tensile strength; f_ctk,0.05 = 0.7f_ctm = 1.5 N/mm²
Secant modulus of elasticity of concrete; E_cm = 22 kN/mm²(f_cm/10)^0.3 = 29962 N/mm²
Partial factor for concrete – Table 2.1N; γ_C = 1.50
Compressive strength coefficient – cl.3.1.6(1); a_cc = 0.85
Design compressive concrete strength – exp.3.15; f_cd = a_cc(f_ck/γ_C) = 11.3 N/mm²
Maximum aggregate size; h_agg = 20 mm
Ultimate strain – Table 3.1; ε_cu2 = 0.0035
Shortening strain – Table 3.1; ε_cu3 = 0.0035
Effective compression zone height factor; λ = 0.80

Effective strength factor; h = 1.00
Bending coefficient k₁; K₁ = 0.40
Bending coefficient k₂; K₂ = 1.00(0.6 + 0.0014/ε_cu2) = 1.00
Bending coefficient k₃; K₃ =0.40
Bending coefficient k₄; K₄ = 1.00(0.6 + 0.0014/ε_cu2) =1.00

Reinforcement details
Characteristic yield strength of reinforcement; f_yk = 500 N/mm²
Modulus of elasticity of reinforcement; E_s = 200000 N/mm²
Partial factor for reinforcing steel – Table 2.1N; γ_S = 1.15
Design yield strength of reinforcement; f_yd = f_yk / γ_S = 435 N/mm²

Cover to reinforcement
Front face of stem; c_sf = 40 mm
Rear face of stem; c_sr = 50 mm
Top face of base; c_bt = 50 mm
Bottom face of base; c_bb = 75 mm

Check stem design at base of stem

Depth of section; h = 300 mm

Flexural Design (Bending)

Design bending moment combination 1;
M = 65 kNm/m
d = h – c_sr – φ_sr / 2 = 244 mm
K = M / (d²f_ck) = 0.055
K’ = (2h a_cc/γ_C) × [1 – λ(d – K₁)/(2K₂)] × [λ(d – K₁)/(2K₂)]
K’ = 0.207

K’ > K – No compression reinforcement is required

Lever arm; z = min(0.5 + 0.5(1 – 2K / (h × a_cc / γ_C))^0.5, 0.95)d = 232 mm

Depth of neutral axis; x = 2.5(d – z) = 31 mm

Area of tension reinforcement required; A_sr.req = M / (f_ydz) = 646 mm²/m
Tension reinforcement provided; 12 dia.bars @ 150 c/c ( A_sr.prov = 754 mm²/m)
Minimum area of reinforcement – exp.9.1N; A_sr.min = max(0.26f_ctm / f_yk, 0.0013)d = 317 mm²/m
Maximum area of reinforcement – cl.9.2.1.1(3);A_sr.max = 0.04h = 12000 mm²/m
max(A_sr.req, A_sr.min) / A_sr.prov = 0.856

PASS – Area of reinforcement provided is greater than area of reinforcement required

Deflection control

Reference reinforcement ratio; ρ₀ = √(f_ck) / 1000 = 0.004
Required tension reinforcement ratio; ρ = A_sr.req / d = 0.003
Required compression reinforcement ratio; ρ’ = A_sr.2.req / d₂ = 0.000
Structural system factor – Table 7.4N; K_b = 0.4
Reinforcement factor – exp.7.17; K_s = min(500 / (f_ykA_sr.req / A_sr.prov), 1.5) = 1.168

Limiting span to depth ratio – exp.7.16.a;
min(K_sK_b[11 + 1.5 × √(f_ck) ρ₀/ρ + 3.2√(f_ck) × (ρ₀/ρ – 1)^3/2], 40K_b) = 14.3

Actual span to depth ratio; h_stem / d = 12.3

PASS – Span to depth ratio is less than deflection control limit

Rectangular section in shear

Design shear force; V = 57.5 kN/m
C_Rd,c = 0.18 / γ_C = 0.120
k = min(1 + √(200 mm / d), 2) = 1.905

Longitudinal reinforcement ratio; ρ_l = min(A_sr.prov / d, 0.02) = 0.003
v_min = 0.035k^3/2f_ck^0.5 = 0.412 N/mm²

Design shear resistance – exp.6.2a & 6.2b;
V_Rd.c = max(C_Rd.ck(100ρ_lf_ck)^1/3, v_min)d
V_Rd.c = 102.4 kN/m
V / V_Rd.c = 0.562

PASS – Design shear resistance exceeds design shear force

Horizontal reinforcement parallel to face of stem – Section 9.6

Minimum area of reinforcement – cl.9.6.3(1);
A_sx.req = max(0.25A_sr.prov, 0.001t_stem) = 300 mm²/m
Maximum spacing of reinforcement – cl.9.6.3(2); s_{sx_max} = 400 mm
Transverse reinforcement provided; H10 dia.bars @ 200 c/c (A_sx.prov = 393 mm²/m)

PASS – Area of reinforcement provided is greater than area of reinforcement required

Check base design at toe

Depth of section; h = 350 mm

Rectangular section in flexure

Design bending moment combination 1;
M = 13.8 kNm/m

d = h – c_bb – φ_bb / 2 = 269 mm
K = M / (d² f_ck) = 0.010
K’ = (2h × a_cc/γ_C) × (1 – λ(d – K₁)/(2K₂)) × (λ(d – K₁)/(2K₂))
K’ = 0.207

K’ > K – No compression reinforcement is required

Lever arm;
z = min(0.5 + 0.5(1 – 2K / (h × a_cc/γ_C))^0.5, 0.95)d = 256 mm
Depth of neutral axis; x = 2.5(d – z) = 34 mm

Area of tension reinforcement required; A_bb.req = M / (f_ydz) = 124 mm²/m
Tension reinforcement provided;
12 dia.bars @ 200 c/c ( A_bb.prov = 565 mm²/m)

Minimum area of reinforcement – exp.9.1N;
A_bb.min = max(0.26f_ctm/f_yk, 0.0013)d = 350 mm²/m

Maximum area of reinforcement – cl.9.2.1.1(3);
A_bb.max = 0.04h = 14000 mm²/m
max(A_bb.req, A_bb.min) / A_bb.prov = 0.618

PASS – Area of reinforcement provided is greater than area of reinforcement required

Rectangular section in shear

Design shear force;
V = 53.3 kN/m
C_Rd,c = 0.18/γ_C = 0.120
k = min(1 + √(200 mm / d), 2) = 1.862

Longitudinal reinforcement ratio; ρ_l = min(A_bb.prov / d, 0.02) = 0.002

v_min = 0.035 k^3/2f_ck^0.5 = 0.398 N/mm²

Design shear resistance – exp.6.2a & 6.2b;
V_Rd.c = max(C_Rd.ck(100ρ_lf_ck)^1/3, v_min) d
V_Rd.c = 107 kN/m
V / V_Rd.c = 0.498

PASS – Design shear resistance exceeds design shear force

Check base design at heel

Depth of section; h = 350 mm

Rectangular section in flexure

Design bending moment combination 1;
M = 51.9 kNm/m

d = h – c_bt – φ_bt / 2 = 294 mm
K = M / (d²f_ck) = 0.030
K’ = (2h × α_cc/γ_C) × (1 – λ(d – K₁)/(2K₂)) × (λ(d – K₁)/(2K₂))
K’ = 0.207

K’ > K – No compression reinforcement is required

Lever arm; z = min(0.5 + 0.5(1 – 2K/(h × α_cc/γ_C))^0.5, 0.95)d = 279 mm
Depth of neutral axis; x = 2.5(d – z) = 37 mm

Area of tension reinforcement required;
A_bt.req = M / (f_ydz) = 427 mm²/m

Tension reinforcement provided;
12 dia.bars @ 200 c/c (A_bt.prov = 565 mm²/m)

Minimum area of reinforcement – exp.9.1N;
A_bt.min = max(0.26f_ctm/f_yk, 0.0013)d = 382 mm²/m

Maximum area of reinforcement – cl.9.2.1.1(3);
A_bt.max = 0.04h = 14000 mm²/m
max(A_bt.req, A_bt.min) / A_bt.prov = 0.755

PASS – Area of reinforcement provided is greater than area of reinforcement required

Rectangular section in shear

Design shear force;
V = 53.5 kN/m
C_Rd,c = 0.18 / γ_C = 0.120
k = min(1 + √(200 mm / d), 2) = 1.825

Longitudinal reinforcement ratio;
ρ_l = min(A_bt.prov / d, 0.02) = 0.002
v_min = 0.035k^3/2f_ck^0.5 = 0.386 N/mm²

Design shear resistance – exp.6.2a & 6.2b;
V_Rd.c = max(C_Rd.ck (100 ρ_lf_ck)^1/3, v_min)d
V_Rd.c = 113.4 kN/m
V / V_Rd.c = 0.472

PASS – Design shear resistance exceeds design shear force

Secondary transverse reinforcement to base – Section 9.3
Minimum area of reinforcement – cl.9.3.1.1(2);
A_bx.req = 0.2A_bb.prov = 113 mm²/m

Maximum spacing of reinforcement – cl.9.3.1.1(3);
s_{bx_max} = 450 mm

Transverse reinforcement provided;
10 dia.bars @ 200 c/c (A_bx.prov = 393 mm²/m)

PASS – Area of reinforcement provided is greater than area of reinforcement required

Summary

Description	Unit	Provided	Required	Utilisation	Result
Stem ρ₀ rear face – Flexural reinforcement	mm²/m	754.0	645.7	0.86	PASS
Stem ρ₀ – Shear resistance	kN/m	102.4	57.5	0.56	PASS
Base top face – Flexural reinforcement	mm²/m	565.5	427.2	0.76	PASS
Base bottom face – Flexural reinforcement	mm²/m	565.5	349.7	0.62	PASS
Base – Shear resistance	kN/m	107.0	53.3	0.50	PASS
Transverse stem reinforcement	mm²/m	392.7	300.0	0.76	PASS
Transverse base reinforcement	mm²/m	392.7	113.1	0.29	PASS

How to Apply Wind Load on High-Rise Buildings

By

Ubani Obinna

-

May 6, 2018

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 V_b,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 V_b,0 is calculated from;
V_b = C_dir . C_season .V_b,0

Where:
V_b 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
V_b,0 is the fundamental value of the basic wind velocity
C_dir is the directional factor (defined in the National Annex, but recommended value is 1.0)
C_season 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 V_b,0 = 40 m/s
V_b = C_dir. C_season.V_b,0 = 1.0 × 1.0 × 40 = 40 m/s

Mean Wind
The mean wind velocity V_m(z) at a height z above the terrain depends on the terrain roughness and orography, and on the basic wind velocity, V_b, and should be determined using the expression below;

V_m(z) = c_r(z). c_o(z).V_b

Where;
c_r(z) is the roughness factor (defined below)
c_o(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.

c_r(z) = k_r. In (z/z₀) for z_min ≤ z ≤ z_max
c_r(z) = c_r.(z_min) for z ≤ z_min

Where:
Z₀ is the roughness length
k_r is the terrain factor depending on the roughness length Z₀ calculated using;

k_r = 0.19 (Z₀/Z_0,II)^0.07

Where:
Z_0,II = 0.05m (terrain category II)
Z_min is the minimum height = 2 m
z = 60 m
Z_max is to be taken as 200 m
K_r = 0.19 (0.05/0.05)^0.07 = 0.19
c_r(60) = k_r.In (z/z₀) = 0.19 × In(60/0.05) = 1.347

Therefore;
V_m(60) = c_r(z). c_o(z).V_b = 1.347 × 1.0 × 40 = 53.88 m/s

Wind turbulence
The turbulence intensity I_v(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 I_v(z) are given in the expressions below;

I_v(z) = σ_v/V_m = k_l/(c₀(z).In (z/z₀)) for z_min ≤ z ≤ z_max
I_v(z) = I_v.(z_min) for z ≤ z_min

Where:
k_l is the turbulence factor of which the value is provided in the National Annex but the recommended value is 1.0
C_o 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;

I_v(60) = σ_v/V_m = k₁/[c₀(z).In(z/z₀)] = 1/[1 × In(60/0.05)] = 0.141

Peak Velocity Pressure
The peak velocity pressure q_p(z) at height z is given by the expression below;

q_p(z) = [1 + 7.I_v(z)] 1/2.ρ.V_m²(z) = c_e(z).q_b

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)

c_e(z) is the exposure factor given by;
c_e(z) = q_p(z)/q_b
q_b is the basic velocity pressure given by; q_b = 0.5.ρ.V_b²

q_p(60m) = [1 + 7(0.141)] × 0.5 × 1.25 × 53.88² = 3605.23 N/m²

Therefore, q_p(60m) = 3.605 kN/m²

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 C_pe10 = +0.8 – (–0.6) = 1.4
The net external surface pressure on the structure = q_p(z) C_pe10 = 3.6057 × 1.4 = 5.05 kN/m²
Therefore, w_e = 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); v_b,map = 40.0 m/s
Distance to shore; L_shore = 250.00 km
Altitude above sea level; A_alt = 100.0m
Altitude factor; c_alt = A_alt/1m × 0.001 + 1 = 1.100
Fundamental basic wind velocity; v_b,0 = v_b,map × c_alt = 44.0 m/s
Direction factor; c_dir = 1.00
Season factor; c_season = 1.00
Shape parameter K; K = 0.2
Exponent n; n = 0.5
Air density; ρ = 1.226 kg/m³

Probability factor; c_prob = [(1 – K × ln(-ln(1-p)))/(1 – K × ln(-ln(0.98)))]ⁿ = 1.00
Basic wind velocity (Exp. 4.1); v_b = c_dir × c_season × v_b,0 × c_prob = 44.0 m/s
Reference mean velocity pressure; q_b = 0.5 × ρ × v_b² = 1.187 kN/m²

Orography
Orography factor not significant; c_o = 1.0

Terrain category; Country
Displacement height (sheltering effect excluded); h_dis = 0mm

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)

Peak velocity pressure – windward wall (lower part) – Wind 0 deg

Reference height (at which q is sought); z = 30000mm
Displacement height (sheltering effects excluded); h_dis = 0 mm
Exposure factor (Figure NA.7); c_e = 3.05
Library item: Peak velocity factors output
Peak velocity pressure; q_p = c_e × q_b = 3.62 kN/m²

Structural factor
Structural damping; d_s = 0.100
Height of element; h_part = 30000 mm
Size factor (Table NA.3); c_s = 0.895
Dynamic factor (Figure NA.9); c_d = 1.000
Structural factor; c_sCd = c_s × c_d = 0.895

Peak velocity pressure – windward wall (upper part) – Wind 0 deg and roof
Reference height (at which q is sought); z = 60000mm
Displacement height (sheltering effects excluded); h_dis = 0 mm
Exposure factor (Figure NA.7); c_e = 3.50
Peak velocity pressure; q_p = c_e × q_b = 4.16 kN/m²
Structural damping; d_s = 0.100
Height of element; h_part = 30000 mm
Size factor (Table NA.3); c_s = 0.907
Dynamic factor (Figure NA.9); c_d = 1.000
Structural factor; c_sCd = c_s × c_d = 0.907

Structural factor
Structural damping; d_s = 0.100
Height of element; h_part = 60000 mm
Size factor (Table NA.3); c_s = 0.888
Dynamic factor (Figure NA.9); c_d = 1.000
Structural factor; c_sCd = c_s × c_d = 0.888

Peak velocity pressure – windward wall (lower part) – Wind 90 deg

Reference height (at which q is sought); z = 30000mm
Displacement height (sheltering effects excluded); h_dis = 0 mm
Exposure factor (Figure NA.7); c_e = 3.05
Peak velocity pressure; q_p = c_e × q_b = 3.62 kN/m²

Structural factor
Structural damping; d_s = 0.100
Height of element; h_part = 30000 mm
Size factor (Table NA.3); c_s = 0.895
Dynamic factor (Figure NA.9); c_d = 1.000
Structural factor; c_sCd = c_s × c_d = 0.895

Peak velocity pressure – windward wall (upper part) – Wind 90 deg and roof
Reference height (at which q is sought); z = 60000mm
Displacement height (sheltering effects excluded); h_dis = 0 mm
Exposure factor (Figure NA.7); c_e = 3.50
Peak velocity pressure; q_p = c_e × q_b = 4.16 kN/m²
Structural damping; d_s = 0.100
Height of element; h_part = 30000 mm
Size factor (Table NA.3); c_s = 0.907
Dynamic factor (Figure NA.9); c_d = 1.000
Structural factor; c_sCd = c_s × c_d = 0.907

Structural factor
Structural damping; d_s = 0.100
Height of element; h_part = 60000 mm
Size factor (Table NA.3); c_s = 0.888
Dynamic factor (Figure NA.9); c_d = 1.000
Structural factor; c_sCd = c_s × c_d = 0.888

Peak velocity pressure for internal pressure
Peak velocity pressure – internal (as roof press.); q_p,i = 4.16 kN/m²

Pressures and forces
Net pressure; p = c_sCd × q_p × c_pe – q_p,i × c_pi;
Net force; F_w = p_w × A_ref;

Roof load case 1 – Wind 0, c_pi 0.20, -c_pe

Zone	Ext pressure coeff c_pe	Peak velocity pressure q_p (kN/m²)	Net pressure element, p_e (kN/m²)	Net pressure structure p_s (kN/m²)	Area A_ref (m²)	Net force element F_w,e (kN)	Net force structure F_w,s (kN)
F (-ve)	-2.00	4.16	-9.15	-8.22	45.00	-411.69	-369.93
G (-ve)	-1.40	4.16	-6.65	-6.00	45.00	-299.41	-270.18
H (-ve)	-0.70	4.16	-3.74	-3.42	360.00	-1347.33	-1230.43
I (-ve)	-0.20	4.16	-1.66	-1.57	450.00	-748.52	-706.77

Total vertical net force; F_w,v = -2577.31 kN
Total horizontal net force; F_w,h = 0.00 kN

Walls load case 1 – Wind 0, c_pi 0.20, -c_pe

Zone	Ext pressure coeff c_pe	Peak velocity pressure q_p (kN/m²)	Net pressure element, p_e (kN/m²)	Net pressure structure p_s (kN/m²)	Area A_ref (m²)	Net force element F_w,e (kN)	Net force structure F_w,s (kN)
A	-1.20	4.16	-5.82	-5.36	360.00	-2095.85	-1928.18
B	-0.80	4.16	-4.16	-3.85	1440.00	-5988.15	-5541.03
D_b	0.80	3.62	2.07	1.76	900.00	1859.33	1585.51
D_u	0.80	4.16	2.50	2.18	900.00	2245.55	1966.11
E	-0.55	4.16	-3.12	-2.91	1800.00	-5613.89	-5229.65

Overall loading

Equivalent leeward net force for upper section;
F_l = F_w,wEs / A_ref,wE × A_ref,wu = -2614.8 kN

Net windward force for upper section;
F_w = F_w,wus = 1966.1 kN

Lack of correlation (cl.7.2.2(3) – Note);
f_corr = 0.89; as h/W is 2.000

Overall loading upper section;
F_w,u = f_corr × (F_w – F_l + F_w,h) = 4065.6 kN

Equiv leeward net force for bottom section;
F_l = F_w,wEs / A_ref,wE × A_ref,wb = -2614.8 kN

Net windward force for bottom section;
F_w = F_w,wbs = 1585.5 kN

Lack of correlation (cl.7.2.2(3) – Note);
f_corr = 0.89; as h/W is 2.000

Overall loading bottom section;
F_w,b = f_corr × (F_w – F_l) = 3727.8 kN

Roof load case 2 – Wind 0, c_pi -0.3, +c_pe

Zone	Ext pressure coeff c_pe	Peak velocity pressure q_p (kN/m²)	Net pressure element, p_e (kN/m²)	Net pressure structure p_s (kN/m²)	Area A_ref (m²)	Net force element F_w,e (kN)	Net force structure F_w,s (kN)
F (+ve)	-2.00	4.16	-7.07	-6.14	45.00	-318.12	-276.37
G (+ve)	-1.40	4.16	-4.57	-3.92	45.00	-205.84	-176.62
H (+ve)	-0.70	4.16	-1.66	-1.34	360.00	-598.81	-481.91
I (+ve)	0.20	4.16	2.08	1.99	450.00	935.65	893.90

Total vertical net force; F_w,v = -41.00 kN
Total horizontal net force; F_w,h = 0.00 kN

Walls load case 2 – Wind 0, c_pi -0.3, +c_pe

Zone	Ext pressure coeff c_pe	Peak velocity pressure q_p (kN/m²)	Net pressure element, p_e (kN/m²)	Net pressure structure p_s (kN/m²)	Area A_ref (m²)	Net force element F_w,e (kN)	Net force structure F_w,s (kN)
A	-1.20	4.16	-3.74	-3.28	360.00	-1347.33	-1179.66
B	-0.80	4.16	-2.08	-1.77	1440.00	-2994.07	-2546.96
D_b	0.80	3.62	4.15	3.84	900.00	3730.63	3456.80
D_u	0.80	4.16	4.57	4.26	900.00	4116.85	3837.40
E	-0.55	4.16	-1.04	-0.83	1800.00	-1871.30	-1487.06

Overall loading

Equivalent leeward net force for upper section;
F_l = F_w,wEs / A_ref,wE × A_ref,wu = -743.5 kN

Net windward force for upper section;
F_w = F_w,wus = 3837.4 kN

Lack of correlation (cl.7.2.2(3) – Note);
f_corr = 0.89; as h/W is 2.000

Overall loading upper section;
F_w,u = f_corr × (F_w – F_l + F_w,h) = 4065.6 kN

Library item: Overall loading output

Equiv leeward net force for bottom section;
F_l = F_w,wEs / A_ref,wE × A_ref,wb = -743.5 kN

Net windward force for bottom section;
F_w = F_w,wbs = 3456.8 kN

Lack of correlation (cl.7.2.2(3) – Note);
f_corr = 0.89; as h/W is 2.000

Overall loading bottom section;
F_w,b = f_corr × (F_w – F_l) = 3727.8 kN

Roof load case 3 – Wind 90, c_pi 0.20, -c_pe

Zone	Ext pressure coeff c_pe	Peak velocity pressure q_p (kN/m²)	Net pressure element, p_e (kN/m²)	Net pressure structure p_s (kN/m²)	Area A_ref (m²)	Net force element F_w,e (kN)	Net force structure F_w,s (kN)
F (-ve)	-2.00	4.16	-9.15	-8.22	45.00	-411.69	-369.93
G (-ve)	-1.40	4.16	-6.65	-6.00	45.00	-299.41	-270.18
H (-ve)	-0.70	4.16	-3.74	-3.42	360.00	-1347.33	-1230.43
I (-ve)	-0.20	4.16	-1.66	-1.57	450.00	-748.52	-706.77

Total vertical net force; F_w,v = -2577.31 kN
Total horizontal net force; F_w,h = 0.00 kN

Walls load case 3 – Wind 90, c_pi 0.20, -c_pe

Zone	Ext pressure coeff c_pe	Peak velocity pressure q_p (kN/m²)	Net pressure element, p_e (kN/m²)	Net pressure structure p_s (kN/m²)	Area A_ref (m²)	Net force element F_w,e (kN)	Net force structure F_w,s (kN)
A	-1.20	4.16	-5.82	-5.36	360.00	-2095.85	-1928.18
B	-0.80	4.16	-4.16	-3.85	1440.00	-5988.15	-5541.03
D_b	0.80	3.62	2.07	1.76	900.00	1859.33	1585.51
D_u	0.80	4.16	2.50	2.18	900.00	2245.55	1966.11
E	-0.55	4.16	-3.12	-2.91	1800.00	-5613.89	-5229.65

Overall loading

Equiv leeward net force for upper section;
F_l = F_w,wEs / A_ref,wE × A_ref,wu = -2614.8 kN

Net windward force for upper section;
F_w = F_w,wus = 1966.1 kN

Lack of correlation (cl.7.2.2(3) – Note);
f_corr = 0.89; as h/L is 2.000

Overall loading upper section;
F_w,u = f_corr × (F_w – F_l + F_w,h) = 4065.6 kN

Library item: Overall loading output

Equiv leeward net force for bottom section;
F_l = F_w,wEs / A_ref,wE × A_ref,wb = -2614.8 kN

Net windward force for bottom section;
F_w = F_w,wbs = 1585.5 kN

Lack of correlation (cl.7.2.2(3) – Note);
f_corr = 0.89; as h/L is 2.000

Overall loading bottom section;
F_w,b = f_corr × (F_w – F_l) = 3727.8 kN

Roof load case 4 – Wind 90, c_pi -0.3, +c_pe

Zone	Ext pressure coeff c_pe	Peak velocity pressure q_p (kN/m²)	Net pressure element, p_e (kN/m²)	Net pressure structure p_s (kN/m²)	Area A_ref (m²)	Net force element F_w,e (kN)	Net force structure F_w,s (kN)
F (+ve)	-2.00	4.16	-7.07	-6.14	45.00	-318.12	-276.37
G (+ve)	-1.40	4.16	-4.57	-3.92	45.00	-205.84	-176.62
H (+ve)	-0.70	4.16	-1.66	-1.34	360.00	-598.81	-481.91
I (+ve)	0.20	4.16	2.08	1.99	450.00	935.65	893.90

Total vertical net force; F_w,v = -41.00 kN
Total horizontal net force; F_w,h = 0.00 kN

Walls load case 4 – Wind 90, c_pi -0.3, +c_pe

Zone	Ext pressure coeff c_pe	Peak velocity pressure q_p (kN/m²)	Net pressure element, p_e (kN/m²)	Net pressure structure p_s (kN/m²)	Area A_ref (m²)	Net force element F_w,e (kN)	Net force structure F_w,s (kN)
A	-1.20	4.16	-3.74	-3.28	360.00	-1347.33	-1179.66
B	-0.80	4.16	-2.08	-1.77	1440.00	-2994.07	-2546.96
D_b	0.80	3.62	4.15	3.84	900.00	3730.63	3456.80
D_u	0.80	4.16	4.57	4.26	900.00	4116.85	3837.40
E	-0.55	4.16	-1.04	-0.83	1800.00	-1871.30	-1487.06

Overall loading

Equiv leeward net force for upper section;
F_l = F_w,wEs / A_ref,wE × A_ref,wu = -743.5 kN

Net windward force for upper section;
F_w = F_w,wus = 3837.4 kN

Lack of correlation (cl.7.2.2(3) – Note);
f_corr = 0.89; as h/L is 2.000

Overall loading upper section;
F_w,u = f_corr × (F_w – F_l + F_w,h) = 4065.6 kN

Library item: Overall loading output

Equiv leeward net force for bottom section;
F_l = F_w,wEs / A_ref,wE × A_ref,wb = -743.5 kN

Net windward force for bottom section;
F_w = F_w,wbs = 3456.8 kN

Lack of correlation (cl.7.2.2(3) – Note);
f_corr = 0.89; as h/L is 2.000

Overall loading bottom section;
F_w,b = f_corr × (F_w – F_l) = 3727.8 kN

Simple Proofs Why Shorter Spans Are More Critical in Slab Design

By

Ubani Obinna Uzodimma

-

May 2, 2018

Table of Contents

A slab is a structural member whose depth is considerably smaller than the lateral dimensions. They provide useful surfaces for floors in buildings and support the occupancy loads. Reinforced concrete slabs may be supported by beams, columns, walls, or masonry. When a slab is supported on two opposite sides, they are referred to as one-way slabs because the load is transferred from the slab to the beam in the perpendicular direction.

one%2Bway%2Bslab — Typical one-way slab system

However, if a slab is supported by beams on the four sides, and the ratio of length to width is greater than 2, majority of the load is transferred to the short direction to the supporting beams, and one-way action is obtained in effect, even though supports are provided on all sides.

The structural action of one-way slabs may be visualized in terms of the deflected surface. This can be approximated as a cylindrical surface, and curvatures and bending moments are parallel to the short side L_x. The slab is normally analysed a beam of a unit strip, with the bending moment, shear forces, and reinforcement determined per unit strip.

56777 — Cylindrical one-way bending action of a thin pate

In many design circumstances, rectangular slabs have dimensions where the ratio of the longer side to the shorter side is less than 2 (and are also supported in such a way that two-way action results). When loaded, such slabs bend into a dished surface rather than a cylindrical one. This means that at any point the slab is curved in both principal directions, and since bending moments are proportional to curvatures, moments also exist in both directions.

dttttu — Two-way bending action of a thin plate

To resist these moments, the slab must be reinforced in both directions, by at least two layers of bars perpendicular, respectively, to two pairs of edges. The slab must be designed to take a proportionate share of the load in each direction.

Why are Short Span Critical in Slabs?

Whenever civil engineering students enter a structural design classroom for the first time, they are told that the shorter span is more critical in the design of slabs. This is usually source of wonder for first timers because by mere instincts, the longer span should be more critical.

This article shows very simple proofs why the short span is more critical in slabs.

Deflection of centre-strip approach

This approach offers an extremely simplified concept that shows that the load transmitted to the shorter span is greater than the load transmitted to the longer span.

Let us consider the two way action of the slab shown below;

Looking at the centre-strips highlighted in red, a little consideration will show that at the intersection point of the strips, the deflection is equal. Logical right?

Let the uniform load on the longer strip be q_y
Let the uniform load on the shorter strip be q_x

We can therefore say that;

Deflection at centre w = 5q_yL_y⁴/384EI = 5q_xL_x⁴/384EI
5q_yL_y⁴/384EI = 5q_xL_x⁴/384EI
q_yL_y⁴= q_xL_x⁴

We can verify from the above relationships that;
q_x/q_y= L_y⁴/ L_x⁴

Since L_y > L_x, we can accept that the load on the short span (q_x) is greater than the load on the long span (q_y) since L_y/L_x > 1.0
This is an oversimplification of the behaviour of slabs though.

Yield Line Approach

The yield line method was developed to determine the limit state of slabs by considering the yield lines that occur at the slab collapse mechanism. The yield lines are usually approximated to originate at the corners, forming at an angle of 45° until they intersect. These yield lines usually show trapezoidal loads going to the longer supporting beam of the slab, and triangular loads going to the shorter supporting beam.

Let us assume that the slab is subjected to a unit pressure load (1 kN/m²)

L_y = 6m
L_x = 5m

Area of trapezium = 8.6825 m²
Area of triangle = 6.316 m²

Therefore;
The load parallel to the short span (P_x) = 2 × 8.6825m² × 1 kN/m² = 17.365 kN
The load parallel to the long span (P_y) = 2 × 6.316m² × 1 kN/m² = 12.632 kN

Total load on slab = (5 × 6)m² × 1 kN/m² = 30 kN

Therefore, you can see why the shorter span is more heavily loaded based on the yield line pattern.

Finite Element Analysis

When we carry out finite element analysis on slabs, the result offers an insight;
Let us check out the slab investigated above using finite element analysis results from Staad Pro.

Let us assume that the slab is subjected to a unit pressure load (1 kN/m²)

L_y = 6m
L_x = 5m
Thickness of slab = 150 mm
v = 0.2
E = 21.7 kN/mm²

From the result, the bending moment parallel to the short span is given below;

Simple Proofs Why Shorter Spans Are More Critical in Slab Design 91

In the longer direction;

If you look at the results above, the moments parallel to the the short span is more critical than the moment parallel to the longer side.

Conclusion

The load transferred to supporting beams through the short span of the slab is heavier than the load transferred through the long span of the slab. For a one way slab, the slab is analysed as a beam of unit width , and the main reinforcement is provided parallel to the short span, while distribution bars are provided parallel to the long span. Minimum steel can be used as distribution bars.

But for two way slabs, the slab is designed for strength in the two principal directions, but the main bars are placed at the bottom, parallel to the short span, while the other is placed near to the bottom (on top of the bottom bar) parallel to the long span

An Introduction To Isogeometric Analysis (IGA)

By

Ubani Obinna Uzodimma

-

April 30, 2018

1.0 Introduction
Isogeometric Analysis (IGA) is a computational approach that integrates Finite Element Analysis (FEA) and Computer Aided Design (CAD). Isogeometric Analysis is developed for the purpose of utilizing the same data set in both design and analysis (Raknes, 2011). In today’s CAD and FEA packages one have to convert the data generated in design to a data set suitable for FEA. Converting the data is not trivial, as the computational geometric approach is different in CAD and FEA. IGA makes it possible to utilize the NURBS geometry, which is the most used basis in CAD packages, in FEA directly. Isogeometric analysis is thus a great tool for optimizing models, as one easily can make refinements and perform testing and analysis during design and development.

2.0 CAD and FEA
CAD systems are characterised by “exact” (small gaps within predefined tolerances are allowed) descriptions of the inner and outer shells of the object and 3D models are sufficiently accurate for production purposes (Ferreira, 2014). In FEA the object is represented with a description of composed structures of finite elements, where there is a cruder representation of the shape of inner and outer hulls. In addition FEA needs compact geometric descriptions, where unnecessary detail is removed to focus the analysis on essential properties of the objects, namely, the simulation is performed only in the regions of interest to study the physical problem.

Right now, numerical simulation often involves using the Finite Element Method (FEM) where the geometry model, derived from CAD, usually will suffer a reparameterization of the CAD geometry by piecewise low order polynomials (mesh), generally applying linear Lagrange polynomials to approximate the geometry. This information transfer between models suitable for design (CAD) and analysis (FEM) is considered being a bottleneck in industry (industrial applications) of today, because it introduces significant approximation errors, or makes the simulations computational costly in FEM models with fine mesh, and entails a huge amount of man-hours to generate a suitable finite element mesh (Ferreira, 2014).

Nowadays most CAD systems use spline basis functions and often Non-Uniform Rational B-Splines (NURBS) of a different polynomial order. For that reason, IGA is based on NURBS, where the main idea is to use the same mathematical description for the geometry in the design process within CAD and the numerical procedures used in FEA simulations. In other words, IGA is characterised by the use of a common spline basis for geometric modelling and Finite Element Analysis of a given object and can be considered a generalisation of FEM analysis (Ferreira, 2014). Much research in application of IGA have been done at various topics of FEA: linear and non-linear static and dynamic analysis of thin-walled structures, fluid mechanics, fluid structure interaction, shape and topology optimisation, vibration analysis, buckling and others.

Isogeometric analysis based on NURBS (Non-Uniform Rational B-splines) was introduced by Tom Hughes (2005) and further expanded by Cortrell et al, (2006). The objectives of IGA are to generalise and improve upon FEA in the following ways:

Get volumetric spline models having the expected behaviour according to the Partial Differential Equations system (PDEs) and according with boundary conditions and loads behind the physical phenomena in analysis;
Represent common engineering shapes (circles, cylinders, spheres, ellipsoids, etc.) and provide accurate modelling of complex geometries. This means that geometrical errors optimally should be eliminated;
Simplify mesh refinement of geometries by eliminating the need for communication with CAD geometry once the initial mesh is constructed;
Provide systematic refinements (h, p and k refinements) that exhibit improved accuracy and efficiency compared with classical FEA.

3.0 Differences Between Isogeometric Analysis and Finite Element Analysis

The concept of isogeometric analysis is that the basis functions that are used to model the exact geometry are also used as a basis for the solution field. In classical FEA it is the other way around; the basis functions we choose to approximate the unknown solution field is used to approximate the already known geometry (Raknes, 2011). NURBS based Galerkin finite element method is somewhat similar to classical FEA, only now, different basis functions are being used.

For example, in finite element analysis, a circle is an ideal achieved in the limit of mesh refinement (i.e., h-refinement) but never achieved in reality, whereas a circle is achieved exactly for the coarsest mesh in isogeometric analysis, and this exact geometry, and its parameterization, are maintained for all mesh refinements. It is interesting to note that, in the limit, the isogeometric model converges to a polynomial representation on each element, but not for any finite mesh. This is the obverse of finite element analysis in which polynomial approximations exist on all meshes, and the circle is the idealized limit (Cotrell, 2006). The table below shows a comparison between FEA and IGA.

Example

Analysis of a CCCC (all sides clamped) thin plate (Kirchoff’s theory)

Data

a = 6.0 m

b = 6.0 m
Thickness of plate (h) = 150 mm

Pressure Load (q) = 6.2 kN/m²

Poisson ratio (v) = 0.2
Modulus of elasticity (E) = 21.7 kN/mm²

By Isogeometric Analysis
(a) Meshing

(b) Enforcement of Boundary Conditions

ENFORCEMENT%2BOF%2BBOUNDARY%2BCONDITIONS

(c) Deflection profile of the plate (MATLAB)

Maximum deflection at mid span = – 0.001593681267073 m
w_max = 1.59368 mm

Comparing this result with other methods;
Classical solution (Timoshenko) = 1.593 mm
Finite element Analysis (Staad Pro) = 1.560 mm (rectangular mesh size division = 25 x 25)

References
[1] Ferreira J.P. (2014): Elasto-plastic analysis of structures using an Isogeometric Formulation. M.Eng thesis presented to the Department of Mechanical Engineering , University of Porto, Portugal

[2] Cottrell J, A. Reali, Y. Bazilevs, and T. Hughes (2006): “Isogeometric analysis of structural vibrations,” Computer Methods in Applied Mechanics and Engineering, vol. 195, no. 41– 43, pp. 5257 – 5296. John H. Argyris Memorial Issue. Part II.

[3] Raknes Siv Bente (2011): Isogeometric Analysis and Degenerated Mappings. An M.Sc thesis submitted to the Department of Mathematical Sciences, Norwegian University of Science and Technology.

[4] Timoshenko S. And S. Woinowsky-Krieger (1959): Theory of plates and shells. McGraw-Hill Book Company, Inc

Design of Bolted Beam Splice Connections | EN 1993

By

Ubani Obinna Uzodimma

-

April 29, 2018

In the construction of steel structures, there is always a need to join steel members for the purpose of continuity. It is not always possible to have the full lengths of members to span the desired length due to production, handling, and transportation issues. As a result, it is very common to bolt or weld two or more steel members, so as to achieve the desired length.

A bolted beam splice connection is a structural joint used in the construction of steel beams. It is designed to connect two or more beams end-to-end to form a longer beam, increasing the span length or overall structural capacity. The design of a bolted beam splice connection involves several considerations to ensure its strength, stability, and reliability.

The design of a bolted beam splice connection involves careful consideration of load transfer, bolted connection details, splice configuration, stiffness requirements, structural analysis, connection detailing, and material properties. By implication, all the anticipated design forces acting at the point of the beam splice must be taken into consideration during the design.

Therefore, the design involves performing structural analysis to determine the forces and moments acting on the connection. This analysis considers the applied loads, beam properties, and connection geometry. Finite element analysis or other advanced analysis methods may be employed to evaluate the connection’s behaviour under different loading conditions.

In this article, we are going to look at a design example of a beam splice connection (beam-to-beam connection using steel plates). This joint should be able to transmit bending, shear, and axial forces.

Installation of bolted beam splice connection on site

Design and functions of Bolted Splice Connections

The primary function of a beam splice connection is to transfer the loads between the beams effectively. The connection should be designed to transfer axial forces, shear forces, and bending moments from one beam to another without significant deformation or failure.

The connection typically consists of bolts, washers, and nuts. High-strength bolts are commonly used due to their ability to withstand heavy loads. The number, size, and grade of bolts are determined based on the design requirements and the applied loads.

The configuration of the splice connection depends on the structural requirements and the type of loads being transferred. Common types include full-depth end plate splices, extended end plate splices, and flange plate splices. The choice of splice configuration is influenced by factors such as the beam profile, connection stiffness, and fabrication constraints.

The stiffness of the splice connection plays a crucial role in maintaining the overall structural integrity and load distribution. Adequate stiffness helps to limit deflections and minimize differential movement between the spliced beams. The connection should be designed to ensure that it does not introduce excessive additional stiffness or flexibility to the overall beam system.

Design Example

Let us design a bolted beam splice connection for a UB 533 x 210 x 101 kg/m section, subjected to the following ultimate limit state loads;

M_Ed = 610 kNm
V_Ed = 215 kN
N_Ed = 55 kN

At serviceability limit state;
M_Ed,ser = 445 kNm
V_Ed,ser = 157 kN
N_Ed,ser = 41 kN

Properties of UB 533 x 210 x 101
h = 536.7 mm
b = 210 mm
d = 476.5 mm
t_w = 10.8 mm
t_f = 17.4 mm
r = 12.7 mm
A = 129 cm²
I_y = 61500 cm⁴
I_z = 2690 cm⁴
W_el,y = 2290 cm³
W_pl,y = 2610 cm³;
h_w = h – 2t_f = 476.5 mm

Cover plates
Let us assume 20 mm thick cover plates for the flanges and 15 mm thick plates for the web. The thickness and dimension to be confirmed later in the post.

Bolts
M24 preloaded class 8.8 bolts
Diameter of bolt shank d = 24mm
Diameter of hole d₀ = 26mm
Shear area A_s = 353 mm²

Materials Strength
Beam and cover plates;
f_y,b = f_y,wp = 275 N/mm²
f_u,b = f_u,wp = 410 N/mm²

Bolts
Nominal yield strength f_yb = 640 N/mm²
Nominal ultimate strength f_ub = 800 N/mm²

Partial Factors for Resistance (BS EN 1993-1- NA.2.15)
Structural Steel
γ_m0 = 1.0
γ_m1 = 1.0
γ_m2 = 1.1

Parts in connections
γ_m2 = 1.25 (bolts, welds, plates in bearing)
γ_m3 = 1.0 (slip resistance at ULS)
γ_m3,ser = 1.1 (slip resistance as SLS)

Step 1: Internal Forces at Splice
For a splice in flexural member, the parts subject to shear (the web cover plates) must carry, in addition to the shear force and moment due to the eccentricity of the centroids of the bolt groups on each side, the proportion of moment carried by the web, without any shedding to the flanges.

The second moment of area of the web is;
I_w = [(h – 2t_f)³t_w/12]
I_w = [(476.5³ × 10.8)/12] × 10^-4 = 9737 cm⁴

Therefore the web will carry 9737/61500 = 15.8% of the moment in the beam (assuming an elastic stress distribution), while the flange will carry the remaining 84.2%.

The area of the web is;
A_w = 476.5 × 10.8 × 10^-2 = 51.462 cm²
Therefore, the web will carry 51.462/129 = 39.8% of the axial force in the beam, while the flange will carry the remaining 60.1%.

Forces at ULS
The force in each flange due to bending is therefore given by;
F_f,M,Ed = 0.842 × [(M_Ed)/(h – t_f)]
F_f,M,Ed = 0.842 × [(610 × 10⁶)/(536.7 – 17.4)] × 10^-3 = 989 kN

And the force in each flange due to axial force is given by;
F_f,N,Ed = 0.601 × (55/2) = 16.53 kN

Therefore
F_tf,Ed = 989 – 16.53 = 972.47 kN
F_bf,Ed = 989 + 16.53 = 1005.53 kN

The moment in the web = 0.158 × 610 = 96.38 kNm
The shear force in the web = 215 kN
The axial force in the web = 0.398 × 55 = 21.89 kN

Forces at SLS
The force in each flange due to bending is therefore given by;
F_f,M,Ed = 0.842 × [(M_Ed,ser)/(h – t_f)]
F_f,M,Ed = 0.842 × [(445 × 10⁶)/(536.7 – 17.4)] × 10^-3 = 721.53 kN

And the force in each flange due to axial force is given by;
F_f,N,Ed = 0.601 × (41/2) = 12.3205 kN

Therefore
F_tf,Ed = 721.53 – 12.32 = 709.21 kN
F_bf,Ed = 721.53 + 12.32 = 733.85 kN

The moment in the web = 0.158 × 445 = 70.31 kNm
The shear force in the web = 157 kN
The axial force in the web = 0.398 × 41 = 16.318 kN

Choice of Bolt Number and Configuration
Resistances of Bolts
The shear resistance of bolts at ultimate limit state
For M24 bolts in single shear = 132 kN
For M24 bolts in double shear = 265 kN

Flange Splice
For the flanges, the force of 1005.53 kN at ULS can be provided by 8 M24 bolts in single shear.

The full bearing resistance of an M24 bolt in a 20 mm cover plate (without reduction for spacing and edge distance) is;

F_b,max,Rd = (2.5f_udt)/γ_m2
F_b,max,Rd = [(2.5 × 410 × 24 × 20)/1.25] × 10^-3 = 393.6 kN

This is much greater than the resistance of the bolt in single shear, and thus the spacings do not need to be such as to maximise the bearing distance. Four lines of 2 bolts at a convenient spacing may be used.

Web Splice
For the web splice, consider one or two lines of 4 bolts on either side of the centreline. The full bearing resistance on the 10.8 mm web is;
F_b,max,Rd = (2.5f_udt)/γ_m2
F_b,max,Rd = [(2.5 × 410 × 24 × 10.8)/1.25] × 10^-3 = 212.54 kN

This is less than the resistance in double shear, and will therefore determine the resistance at ULS. To achieve this value, the spacings will need to be;
e₁ ≥ 3d₀ = 3 × 26 = 78mm
p₁ ≥ 15d₀/4 = (15 × 26)/4 = 97.5 mm
e₂ ≥ 1.5d₀ = 1.5 × 26 = 39 mm
p₂ ≥ 3d₀ = 3 × 26 = 78mm

Initially, try 4 bolts at a vertical spacing of 120mm at a distance of 80mm from the centreline of the splice.

The additional moment due to the eccentricity of the bolt group is;
M_add = 215 × 0.08 = 17.2 kNm

Bolt forces at ULS
Force/bolt due to vertical shear = 215/4 = 53.75 kN
Force/bolt due to axial compression = 21.89/4 = 5.47 kN
Force/bolt due to moment (considering top and bottom bolts only) = (96.38 + 17.2)/0.36 = 315.5 kN
Thus, a single row is adequate.

Considering the second line of bolts at a horizontal pitch of 100mm
Force/bolt due to vertical shear = 215/4 = 53.75 kN
Force/bolt due to axial compression = 21.89/4 = 5.47 kN
Force/bolt due to moment (considering top and bottom bolts only) = (96.38 + 17.2)/0.36 = 315.5 kN

The additional moment due to eccentricity of this bolt group is;
M_add = 215 × (0.08 + 0.1/2) = 27.95 kN.m

The polar moment of inertia of the blot group is given by;
I_bolts = 6 × 120² + 8 × (100/2)² = 106400 mm²

The horizontal component of the force on each top and bottom bolt is;
F_M,horiz = {[(96.38 + 27.95) × 120]/106400} × 10³ = 140.22 kN

The vertical component of the force on each top and bottom bolt is;
F_M,vert = {[(96.38 + 27.95) × 50]/106400} × 10³ = 58.425 kN

Therefore, the resultant force on the most highly loaded bolt is;
F_v,Ed = sqrt[(53.75 + 58.425)² + (140.22 + 5.47)²] = 183.87 kN

This is less than the full bearing resistance, and is therefore satisfactory for such a bolt spacing.

Chosen Joint Configuration

Summary of cover plate dimensions and bolt spacing

Flange cover plates;
Thickness t_fp = 20 mm
Length h_fp = 660 mm
Width b_fp = 200 mm
End distance e_1,fp = 70 mm
Edge distance e_2,fp = 30 mm

Spacing
In the direction of the force p_1,f = 100 mm
Transverse to direction of force p_2,f = 120 mm
Across the joint in direction of force p_1,f,j = 120 mm

Web cover plates;
Thickness t_wp = 12 mm
Length h_wp = 460 mm
Width b_wp = 460 mm
End distance e_1,wp = 50 mm
Edge distance e_2,wp = 50 mm

Spacing
Vertically p_1,w = 120 mm
Horizontally p_2,w = 100 mm
Horizontally across the joint p_1,w,j = 160 mm

Resistance of flange splices
Resistance of net section
The resistance of a flange cover plate in tension (N_t,fp,Rd) is the lesser of N_pl,Rd and N_u,Rd.

Here;
N_u,Rd = (0.9A_net,tpf_u,fp)/γ_M2

Where;
A_net,fp = (b_fp – 2d₀)t_fp = [200 – (2 × 26)] × 20 = 2960 mm
Therefore;
N_u,Rd = [(0.9 × 2960 × 410 )/1.1] × 10^-3 = 992.945 kN

N_pl,Rd = (A_fpf_y,fp)/γ_M0 = [(200 × 20 × 275)/1.0] × 10^-3 = 1100 kN

Since 1100 kN > 992.945 KN
N_t,fp,Rd = 992.945 kN

For the flange in tension;
F_tf,Ed = 989 – 16.53 = 972.47 kN

Therefore, the tension resistance of a flange cover plate is adequate.

Block Tearing Resistance;
n_1,fp = 3 and n_2,fp = 2;

The resistance to block tearing (N_t,Rd,fp) is given by:
N_t,Rd,fp = {(f_u,fpA_nt,fp)/γ_m2} + {[A_nv,fp(f_y,fp/√3)]/γ_m0}

Where:
A_nt,fp = t_fp(2e_2,fp – d₀)
A_nt,fp = 20 × [(2 × 40) – 26] = 1080 mm²

A_nv,fp = 2t_fp[(n_1,fp – 1)p_1,fp + e_1,fp – (n_1,fp – 0.5)d₀]
A_nv,fp = 2 × 20[(3 – 1) × 100 + 70 – (3 – 0.5) × 26] = 8200 mm²

Therefore;
N_t,fp,Rd = {[(410 × 1080)/1.1] + [(8200 × 265/√3)/1.0]} × 10^-3 = 1657.127 kN

Resistance of cover plate to compression flange
Local buckling resistance between the bolts need not be considered if:
p₁/t ≤ 9ε
ε = sqrt(235/f_y,fp) = sqrt(235/265) = 0.94
9ε = 9 × 0.94 = 8.46

Here, the maximum spacing of bolt across the centreline of the splice p_1,f,j = 120 mm
p_1,f,j/t_fp = 120/20 = 6

Since 8.46 > 6, no local buckling verification required.

RESISTANCE OF WEB SPLICE
Resistance of web cover plate in shear;
The gross shear area is given by:
V_wp,g,Rd = (h_wpt_wpf_y,wp/1.27γ_m0√3)

For two web cover plates;
V_wp,g,Rd = 2{[(460 × 12)/1.27] × [275/√3]} × 10^-3 = 1380.185 kN
V_Ed = 215 kN. Therefore, the shear resistance is adequate.

The net shear area is given by:

V_wp,net,Rd = A_v,wp,net(f_u,wp/√3)/γ_m2
A_v,net = (h_wp – 3d₀)t_wp
= (460 – 3 × 26) × 12 = 4584 mm²

For two web cover plates;
V_n,Rd = {2 × [4584 × (410/√3)]/1.1} × 10^-3 = 1972.9 kN
Therefore, shear resistance is adequate.

Resistance to block tearing

V_b,Rd = (f_u,wpA_nt)/γ_m2 + (f_y,wpA_nv.A_nv)/γ_m0√3

For a single vertical line of bolts;
A_nt = t_wp(e₂ – d₀/2)
A_nt = 12 × (50 – 26/2) = 444 mm²
A_nv = t_wp – e₁ – (n₁ – 0,5)d₀)
A_nv = 12 × [460 – 50 – (3 – 0.5)26] = 4140 mm²

For two web plates
V_b,Rd = 2 × {[(410 × 460)/1.1] + [(275 × 4140)/(√3 × 1.0)]} × 10^-3 = 1657.53 kN

V_Rd = min{1657.53, 1972.9, 1380.185 }
V_Rd = 1380.185 kN

V_Ed/V_Rd = 215/1380.185 = 0.1557 < 1.0
This is ok.

Resistance of beam web
Resistance of net shear area
V_n,w,Rd = [A_v,net(f_u/√3)]/γ_m2

Where;
A_v,net = A_v – 3d₀t_w
A_v = A – 2bt_f + (t_w + 2r)t_f but not less than ηh_wt_w

A_v= 12900 – (2 × 210 × 17.4) + (10.8 + 2 × 12.7) × 17.4 = 6221.88 mm²
A_v= 6221.88 – (3 × 26 × 10.8) = 5379.48 mm²

Thus;
V_n,w,Rd = {[5379.48 × (410/√3)]/1.1} × 10^-3 = 1157.632 kN
This is ok.

Resistance of web cover plate to combined bending, shear, and axial force

V_wp,Rd = 1380.185 kN
Vsub>Ed = 215 kN < 1380.185 kN

Therefore, the effects of shear can be neglected
A_wp = 12 × 460 = 5520 mm²

Modulus of the cover plate = (12 × 460²)/6 = 423200 mm³

N_wp,Rd = 12 × 460 × 275 × 10^-3 = 1518 kN

Therefore, for two web cover plates
M_c,wp,Rd = [(2 × 423200 × 275)/1.0] × 10^-6 = 232.76 kNm

For two web cover plates;
N_pl,Rd = 2 × 1518 = 3036 kN

M_wp,Ed = 96.38 + 27.95 = 124.33 kNm
N_wp,Ed = 21.89 kN

M_wp,Ed/M_c,wp,Rd + N_wp,Ed/N_wp,Rd
124.33/232.76 + (21.89/3036) = 0.541 < 1.0

Therefore, the bending resistance of the web cover plates is adequate.

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Formwork Requirement for Suspended Slabs

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