Design of Pad Foundation

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December 2, 2020

Table of Contents

Pad foundations are isolated rectangular, square, or circular slabs that are provided under reinforced concrete columns or column stubs to safely transmit the column load to the ground. They are a type of shallow foundation that is widely used all over the world especially in areas where the soil possesses good bearing capacity. They are also referred to as isolated bases or spread footings. The design of pad foundations involves sizing the base slab to satisfy geotechnical requirements and providing adequate thickness and reinforcements to satisfy structural requirements.

The dimensions of a pad foundation should not be too small so as to cause excessive settlement or bearing capacity failure of the soil. As a matter of fact, allowable bearing capacity is normally used to control settlement during the design of a pad foundation, hence it is treated as a serviceability limit state parameter. The width of a pad foundation is expected not to be less than 1000 mm, and the thickness not less than 150 mm.

Geotechnical Design of Pad Foundation

The geotechnical design of a pad foundation can be carried out according to the requirements of EN 1997-1:2004 (Eurocode 7). Eurocode 7 gives three approaches for the geotechnical design of foundations and they are as follows;

Design Approach 1 (DA1): In this approach, partial factors are applied to actions and to ground strength parameters.
Design Approach 2 (DA2): In this approach, partial factors are applied to actions or to the effects of actions and to ground resistances.
Design Approach 3 (DA3): In this approach, partial factors are applied to actions or to the effects of actions from the structure and to ground strength parameters.

The three approaches can give very different results when applied in design. However, the UK National Annex to Eurocode 7 permits only Design Approach 1 (DA1). For the design of pad foundation using Design Approach 1, three limit states with their appropriate load combination shall be satisfied for the structure. These limit states are;

EQU: Loss of equilibrium of the structure
STR: Internal failure or excessive deformation of the structure itself
GEO: Failure due to excessive deformation of the ground supporting the structure
UPL: Failure due to uplift of the foundation due to water pressure
HYD: Failure due to hydraulic gradient

In the design of pad foundation using DA1, there are two sets of limit state combinations for STR and GEO limit states. Combination 1 is normally used for the structural design of the foundation, while combination 2 is normally used for sizing the foundation. The partial factors for the limit states are given in the table below;

The partial factors for EQU, UPL, and HYD are given in the Table below. They can also be used in the uplift verification of all kind of buried structures.

The partial factors for soil properties is given in the table below;

It should be noted that pad foundations fall under category 2 structures which means that they are conventional structures that are founded on non-difficult grounds. They offer no exceptional geotechnical risk. As a result, routine procedures for field and laboratory testing for design and execution may be used. Geotechnical design of pad foundations can be done by geotechnical or structural engineers. However, the geotechnical design of category 3 structures with abnormal risk can be done by geotechnical engineers only.

The design of pad foundations can be done using any of the following methods;

(a) Analytical (Direct) method
(b) Semi-empirical (Indirect) method
(c) Prescriptive method using presumed bearing resistance (BS 8004)

Using the analytical (direct) method, all limit states should be verified. The ultimate bearing capacity q_ult of a pad foundation should be verified using the expression below;

q_ult = c’N_cs_cd_ci_cg_cb_c+ q’N_qs_qd_qi_qg_qb_q + γ’BN_γs_γd_γi_γg_γb_γ/2

where;
c = cohesion
q = overburden
γ = body-weight
N_i = bearing capacity factors
s_i = shape factors
d_i = depth factors
i_i = inclination factors
g_i = ground inclination factors
b_i = base inclination factors

Example on the calculation of the bearing capacity of a pad foundation using Design Approach 1 (DA1)

Calculate the bearing capacity of a 1m x 1m pad foundation founded at 0.9 m below a lateritic soil deposit. The characteristic angle of shearing resistance φ_k of the soil is 21° while the effective cohesion c’ is 10 kN/m². The water table is 8m below the ground surface, and the unit weight of the soil is 18 kN/m³.

Solution

As the footing rests on a cohesive-frictional soil, the relevant material property is the angle of shearing resistance, φ and the effective cohesion c’.

Design values of angle of shearing resistance
Characteristic value φ_k = 21°. Note that the safety factor γ_φ is applied to tan φ_k not to φ_k.
Combination 1: γ_φ = 1.0, tan φ_d = tan φ_k/γ_φ= tan 21° = 0.383, φ_d = 21°
Combination 2: γ_φ = 1.25, tan φ_d = tan φ_k/γ_φ = tan 21/1.25 = 0.307, φ_d= 17°

Design values of cohesion
Characteristic value of effective cohesion c‘ = 10 kN/m²
Combination 1: γ_c’ = 1.0, c’_d = c’/γ_c’ = 10 kN/m²
Combination 2: γ_c’ = 1.25, c’_d = c’/γ_c’ = 10/1.25 = 8 kN/m²

Step 2: Calculate design bearing capacity factors. Use the equations in section D.4 of Annex D of Eurocode 7.

i. Overburden factor N_q
N_q = e^{(π × tanφ_d)} × tan²(45 + φ_d/2)
Combination 1: φ_d = 21°, N_q = 7.07
Combination 2: φ_d = 17°, N_q = 4.77

ii. Cohesion factor, N_c
N_c= (N_q − 1) cot φ_d
Combination 1: φ_d = 21°, N_c = 15.84
Combination 2: φ_d = 17°, N_c = 12.28

iii. Body weight factor N_γ
N_γ = 2(N_q − 1) tanφ_d
Combination 1: φ_d = 21°, N_γ = 4.66
Combination 2: φ_d = 17°, N_γ = 2.305

Step 3: Calculate the design shape factors. Use the equations in Annex D of Eurocode 7.

Square base
i. s_q = 1 + sin φ_d
Combination 1: φ_d = 21°, s_q = 1.358
Combination 2: φ_d = 17°, s_q = 1.292

ii. s_c = (s_qN_q − 1)/(N_q − 1)
Combination 1: N_q = 7.07, s_q = 1.358, s_c = 1.419
Combination 2: N_q = 4.77, s_q = 1.292, s_c = 1.369

iii. s_γ = 0.7 (for a square shape)

Step 4: Calculate the overburden pressure, q.
The unit weight of soil is 18 kN/m³ and the safety factor γ_γ = 1
q = 18 × depth of footing = γ_γ× 18 × 0.9 = 16.2 kN/m²

Step 5: Calculate the allowable q_ult:
q_ult = c’N_cs_cd_ci_cg_cb_c+ q’N_qs_qd_qi_qg_qb_q + γ’BN_γs_γd_γi_γg_γb_γ/2

For this calculation, all inclination factors have been ignored.

q_ult = c’N_cs_c + q’N_qs_q + 0.5γ’BN_γs_γ

Combination 1: q_ult = (10 × 15.84 × 1.419) + (16.2 × 7.07 × 1.358) + (0.5 × 18.0 × 4.66 × 1.0 × 0.7) = 409.66 kN/m²
Combination 2: q_ult = (8 × 12.28 × 1.369) + (16.2 × 4.77 × 1.292) + (0.5 × 18.0 × 2.305 × 1.0 × 0.7) = 248 kN/m²

For sizing of the foundation, Combination 2 allowable bearing capacity should be used (no other factor of safety is to be applied).

For the semi-empirical (indirect) method, a commonly recognized semi-empirical method such as bearing resistance estimation using pressuremeter test should be used. The use of experience and testing to determine SLS parameters that will also satisfy ULS requirements is generally done. An example can be found in Annex E of EN 1997-1:2004.

When using the prescriptive method, a presumed bearing resistance from BS 8004 should be used. When such a method is applied, the design result should be evaluated on the basis of comparable experience.

Example on the Structural Design of Pad Foundation

In the structural design of pad foundations, the reaction under an axially loaded column base may be assumed to be uniformly distributed if the load is concentric without any bending moment. Otherwise, the pressure distribution may be assumed to be varying linearly across the base as shown below.

pressure distribution of pad foundations

(a) Design ultimate bearing pressure
For a concentrically loaded pad foundation, the design earth pressure is given by;

q = P/A_prov

Where;
P = Design axial force of the column = 1.35G_k + 1.5Q_k (kN)
A_prov = Base area provided for the footing (m²)

(b)

(b) Bending
The critical section for bending is at the face of the column on a pad footing or the wall in a strip footing. The moment is taken on a section passing completely across a pad footing and is due to the ultimate loads on one side of the section. No redistribution of moments should be made.

(c) Beam Shear
The vertical shear force is the sum of the loads acting outside the section considered. Shear stress is checked at a distance d from the face of the column. It is normal practice to make the base sufficiently deep so that shear reinforcement is not required. The depth of the base is often controlled by the design for shear.

punching shear perimeter of pad foundations

(d) Punching Shear
Rules for checking for punching shear resistance are given in section 6.4 of EN 1992-1-1:2004. The punching shear force is the sum of the loads outside the periphery of the critical section. Two punching shear checks should be carried out – at the column perimeter and at between d – 2d from the face of the column.

typical reinforcement works of a pad foundation

Structural Design Example of Pad Foundation

Design a square pad footing for a 250 × 250 mm column carrying a characteristic permanent load G_k of 800 kN and characteristic variable load Q_k of 425 kN. The presumed allowable bearing pressure of the non-aggressive soil is 225 kN/m². f_ck = 30 N/mm²; f_yk = 500 N/mm²; Concrete cover = 50 mm

Pad foundation is a category 2 structure, and this design is to be done using prescriptive methods:

Let 10% of the service load account for the self-weight of the pad foundation.
Base area A = 1.1(800 + 425)/225 = 5.99 m²
Minimum dimensions of footing = √5.99 = 2.447m

Adopt a square foundation of 2500 mm x 2500m x (600 mm) trial depth (Area provided A_prov = 6.25 m²)

Loading at ultimate limit state N_Ed = (1.35 x 800) + (1.5 x 425) = 1717.5 kN

ULS bearing pressure N_Ed/A_prov = 1717.5/6.25 = 274.8 kN/m²

Critical design moment at the face of the column
M_Ed = (274.8 x 1.125²)/2 = 173.89 kNm/m

Effective depth d = 600 – 50 – 16 = 534 mm
k = M_Ed/(bd²f_ck) = (173.89 x 10⁶)/(100 x 534² x 30) = 0.0203
⇒ z = 0.95d = 0.95 x 534 = 507.3 mm
⇒ A_s = M_Ed/0.87f_ykz = (173.89 x 10⁶)/(0.87 x 500 x 507.3) = 788 mm²/m

Provide H16 @ 225 c/c both ways (A_sprov = 893 mm²/m)

Beam shear
Check critical section d away from column face
V_Ed = 274.8 x (1.125 – 0.534) = 162.4 kN/m
v_Ed = 162.4/534 = 0.304 N/mm²

v_{Rd, c} = C_{Rd, c} × k × (100 × ρ₁ × f_ck) ^0.3333
C_{Rd, c} = 0.12
k = 1 + √ (200/d) = 1 + √ (200/534) = 1.611
ρ = 893/(534 × 1000) = 0.00167
v_{Rd, c} = 0.12 × 1.611 × (100 × 0.00167 × 30)^0.333 = 0.33 N/mm²
=> v_Ed (0.304 N/mm²) < v_Rd,c (0.33 N/mm²) beam shear ok

Punching Shear
Punching shear: Basic control perimeter at 2d from face of column
v_Ed = βV_Ed/u_id < v_Rd,c

β = 1,
u_i = (250 x 4 + 534 x 2 x 2 x π) = 7710 mm

V_Ed = load minus net upward force within the area of the control perimeter)
V_Ed = 1717.5 – 274.8 x (0.25² + π x 1.068² + 1.068 x 0.25 x 4) = 422 kN
v_Ed = (422 x 10³)/(7710 x 534) = 0.102 N/mm²
Punching shear is therefore okay

Detailing sketches of pad foundation

What is the likely cause of this failure?

By

Ubani Obinna

-

December 2, 2020

Using the only data available from this picture (which is the crack pattern), what is likely caused failure of this structure?

The structure is a cantilever and we expect the maximum bending moment and shear force to occur at the support (usually taken as the centreline of the column). Predictions can be made on the cause of the failure based on the nature of the crack…

Kindly let us know your view in the comment section. Thank you and God bless you.

Design of Gabion Walls

By

Ubani Obinna

-

December 1, 2020

Table of Contents

Gabion walls are made up of row upon row of orthogonal cages or baskets (gabions) which are filled with rock fragments/cobbles and tied together. This arrangement forms a block of gravity structure that is able to withstand lateral pressure on it. Furthermore, the permeability of the rock fragments and the flexibility of gabion cages make them particularly suitable for use at sites which are liable to become saturated and where the foundation is composed of relatively compressible materials.

Gabion walls are relatively simple to construct. Where suitable rock is readily available, the use of gabion walls is particularly attractive for reasons of economy and speed of construction. In more recent times, gabion walls are generally used as decorative walls in compounds or gardens and are aesthetically pleasing. However, gabion walls can be designed as retaining structures.

decorative gabion retaining walls — Gabions being used as fence in a home

decorative gabion walls 2 — Gabion wall decoration of a garden

gabion acting as retaining wall — Gabions acting as a retaining wall

When gabions are to be used as retaining structures, they should be considered as gravity retaining walls. There is currently no universally accepted method for designing the individual gabion units. The individual units are placed on each other to form a stable gabion wall. The basic shape of a gabion wall is trapezoidal, but the front and rear faces may be straight or stepped.

It is recommended that back batter be provided for walls higher than about 3 m to improve stability and even out ground bearing pressures. A variety of cage sizes can be produced using suitable materials to suit the terrain. Gabion units are normally in modules of 2 m x 1 m x 1 m.

However, the design method for verification and appropriate functionality of a gabion wall can be carried out at the ultimate limit state analysis (ULS), verifying the following issues;

Soil Bearing: The base pressure applied by the wall must not exceed the ultimate bearing capacity of the supporting soil.
Sliding Resistance: There must be an adequate factor of safety for sliding between the base of the wall and the underlying soil due to the lateral earth pressure. Active earth pressure conditions should be assumed.
Overturning stability: The overturning of the wall due to horizontal earth pressure forces when the retained soil mass become unstable (active failure) should satisfy the required conditions.
Internal stability verification: For each layer of gabion a bearing capacity and sliding resistance verification should be made
Global Stability

Design Considerations of Gabion Walls

The maximum pressure transferred from the gabion wall to the ground should be compared with the safe bearing capacity of the soil. This issue depends on the type of gabion foundation also, which for this case it is considered a non-stiff footing (the gabion is not placed on top of concrete).

Since gabion retaining walls consist of several layers, the internal stability verification should verify that these layers are not collapsing, neither sliding nor the bearing capacity of lower layers lesser than required. These conditions must be satisfied for each layer starting from the top layer to the bottom layer.

The overturning stability of the gabion retaining wall ensures that the soil retained behind the gabion wall is not causing lateral collapse of the wall. For this analysis, it is assumed the gabion wall is rigid body, with no sliding occurrence or partial breakdown between layers. The stabilising force is the weight of the gabion wall, while the destabilising force is the lateral pressure from the retained material or surcharge.

Limit state checks should be carried out at selected planes through the gabion wall, ignoring the resistance contributed by the cage material and the connections between the cages. For stepped walls, stability checks should be carried out at each major change in section shape.

In order to limit deformation, gabion walls should be proportioned in such a way that the resultant force acts within the middle third of the wall’s cross-section. The mobilised angle of wall friction, δ, used in the design should not exceed ϕ’/2, where ϕ’ is the angle of shearing resistance of the compacted backfill. In order for the assumption of δ = ϕ’/2 to hold, the gabion infill must be placed in such manner to achieve a dense mass which will not settle relative to the backfill after construction. Otherwise δ should be assumed to be zero. For a wall to be founded on relatively compressible materials, δ should also be assumed to be zero.

Materials for construction of gabion retaining walls

(1) Gabion Baskets
Gabion baskets can be made from a range of materials. Nylon, polypropylene and polyethylene grids have been used. They have the advantage of being lightweight and corrosion-resistant. However, these materials are susceptible to attack by fire and ultraviolet light. A material widely used in the commercial production of gabions is steel wire-mesh, of which there are two types, hexagonal woven and square welded.

The wires used for the wire-mesh in gabion baskets should be mild steel wire with a minimum tensile strength of 350 N/mm². For permanent applications, the wires should be at least 2.7 mm in diameter and galvanized. Where it is suspected that the infill or retained materials or groundwater are aggressive, the wire mesh should be protected with PVC coating.

(2) Infill Material
Rock used for filling gabions should be sound, clean and well-graded. The maximum size of the rock should not exceed two-thirds the depth of the gabion to be filled or 300 mm, whichever is less. The preferred size is 150 mm to 300 mm. The smallest dimension of the rock should at least be twice the largest dimension of the mesh aperture.

(3) Backfill, Filter, and Drainage Materials
For a partially-submerged gabion wall, a free-draining granular backfill should be provided so that water pressure will not build up behind the wall when the water level in front of the wall is lowered.

Drainage Considerations for Gabion Walls

A geotextile filter should be provided behind the rear face of the gabion wall to prevent the migration of fines from the backfill into the coarse rock infill. Drainage layers at the rear face are normally not warranted. However, a drainage layer of adequate permeability should be provided at the base of the wall to guard against erosion of the foundation material.

The high permeability of the gabion units will permit direct infiltration through the body of the wall at times of heavy rainfall. In order to minimize the possibility of saturation and erosion of the foundation material under a non-submerged gabion wall, it is good practice to provide a blinding layer with adequate drainage provisions at the level of the foundation. For submerged gabion walls, appropriate measures should be incorporated to prevent scouring and erosion of the foundation.

Design Example of Gabion Walls

In the example below, a 3 m high gabion retaining wall is to be designed according to the requirements of Eurocode 7 (EN 1997-1).

analysis and design of gabion retaining walls

The design parameters and loadings are as follows;

Wall geometry
Width of gabion 1; w₁ = 2000 mm
Height of gabion 1; h₁ = 1000 mm
Width of gabion 2; w₂ = 1700 mm
Height of gabion 2; h₂ = 1000 mm
Step to front face between courses 1 and 2; s₂ = 150 mm
Width of gabion 3; w₃ = 1400 mm
Height of gabion 3; h₃ = 1000 mm
Step to front face between courses 2 and 3; s₃= 150 mm
Wall inclination; ε = 0 deg

Gabion properties
Unit weight of fill; γ_d= 16.0 kN/m³
Friction between gabions; δ_bg.k = 35.0 deg

Loading
Permanent surcharge; ρ_o,G = 5 kN/m²

Soil properties
Slope of retained soil; β = 0.0 deg
Characteristic effective shearing resistance angle; ϕ’_r.k = 30.0 deg
Characteristic saturated density of retained soil; γ_sr = 21.0 kN/m³
Coefficient for wall friction; k_membrane = 0.75
Wall friction angle; δ_r.k= 22.5 deg
Characteristic base friction angle; δ_bb.k= 30.0 deg
Soil bearing capacity; q = 100 kN/m²

The calculation sheet has been made available in PDF format and can be downloaded below;

Gabion-retaining-wall-analysis-design-EN1997-2Download

Thank you for visiting Structville today. God bless you.

ASDIP Announces the Release of STEEL-5 Software

By

Ubani Obinna

-

November 30, 2020

ASDIP Structural Engineering Software company has announced the release of STEEL-5, the latest version of ASDIP STEEL software module. The Florida, USA, based structural engineering software company has been operational since the year 1992 and offers different modules for civil engineering designs such as reinforced concrete, steel, foundation, and retaining walls. The STEEL-5 is the latest version of the steel package which offers advances and improvements from the previous versions.

ASDIP STEEL software is utilised by professional engineers for the design of steel members and connections. It works with any operating system, and assists structural engineers to transparently design, analyze, check and optimize structural design work. According to the company’s official website, the following are some of the most important features and benefits included in version 5 of ASDIP STEEL;

The new version includes the following five modules:

Base Plate / Anchorage Design – Any combination of axial vertical and horizontal loads, and biaxial moments. It includes uplift and partial bearing analysis. Design of anchor rods and shear lugs. Graphical generation of the breakout areas.
Steel Column Design – Either sway or non-sway columns. Second-order moments analysis to account for slenderness. Multiple load types and load cases. Results sorted by load combination. Graphical generation of the interaction diagram.
Steel / Composite Beam Design – Up to five spans and two overhangs. Multiple load types and load cases. Separate Construction and Final loads analysis. Graphical generation of the shear, and moment diagrams.
Shear Connection Design – Single angle, double angle, Shear Plate, and Tee connections. Multiple options to specify different conditions of the connection elements. Check of the limit states. Graphical generation of the connection in different views.
Moment Connection Design – Flange-Plated, and Welded Flange moment connections. Any combination of vertical loads and moments. Check of the limit states for moment and shear. Graphical generation of the connection in different views.

A Linkedin Post by the Founder of ASDIP, Javier Encinas PE, invited structural engineers across the world to download the 15-day trial of the new release in order to check the power and capabilities of the STEEL-5. According to him,

“For some weeks we have announced in advance the coming release of ASDIP STEEL 5, our software for structural steel design. As promised, ASDIP STEEL 5 has been released. In addition to an upgraded layout design and literally dozens of improvements, it includes great new features, such as:

Moment connections (NEW module)
AISC 360-16 and ACI 318-19 compliance
Custom load combinations
Biaxial base plates and anchorage
Continuous steel / composite beam

You are invited to download the 15-day free trial and check the software by yourself, hands-on. I have prepared a blog post and a short video with a brief overview of this new version of ASDIP STEEL. Your comments and suggestions are always welcome.”

Detailed information is available about this structural engineering software by visiting ASDIP STEEL. You are also invited to download the Free 15-day Software Trial, or go ahead and Place your Order.

Design of Reinforced Concrete (R.C.) Slabs

By

Ubani Obinna

-

November 30, 2020

Table of Contents

Reinforced concrete (R.C.) slabs are plate elements used to form the floors of buildings. In a typical reinforced concrete building, reinforcement bars arranged as mats are incorporated into a concrete plate of minimum thickness 125 mm to form a reinforced concrete solid slab. The provision of adequate reinforcement, slab thickness, and proper detailing to satisfy ultimate and serviceability limit state requirements forms the basis of the design of reinforced concrete (R.C.) slab. Satisfying other requirements such as durability, fire resistance, etc are also necessary.

Floor slabs are usually subjected to uniformly distributed loads, partially distributed loads, line loads, or concentrated loads in the transverse direction. A beam is similar to a slab in so many ways but there are fundamental differences in behaviour and stress distribution of the two elements. While a beam is generically a one-dimensional element, a plate is a two-dimensional element. Due to the two-dimensional nature of a slab, it is subjected not only to bending moments M_xx and M_yy and shear forces V_xand V_y but also to twisting moments M_xy on all the four faces.

Beam and plate element — **Fig 1**: Beam element and slab element

Types of Reinforced Concrete Slabs

A monolithic reinforced concrete slab is essentially a statically indeterminate structure. For a slab of a given shape and support conditions, the distribution of shear forces, bending, and twisting moments in the slab due to externally applied loads cannot be determined easily. The column layout in a monolithic reinforced concrete structure often forms a rectangular grid. Continuous beams may be provided in one direction or two orthogonal directions, to support slabs that may be solid or ribbed in cross-section. Alternatively, slabs can be supported directly on columns to form flat slabs.

Slabs may be simply supported or continuous over one or more supports and are classified according to the method of support as follows:

One-way spanning slab between beams or walls
Two-way spanning slab between the support beams or walls
Flat slabs carried on columns and edge beams or walls with no interior beams
Ribbed slab transferring slab load to beams and walls in one direction
Waffle slab transferring slab load to beams and slabs in two direction

One-way Spanning Solid Slabs

Clause 5.3.1(5) of EN 1992-1-1:2004 suggests that a slab subjected dominantly to uniformly distributed loads may be considered as one-way spanning if either:

It possesses two free (unsupported) and sensibly parallel edges.
It is the central part of a sensibly rectangular slab supported on four edges with a ratio of longer (L_y) to shorter span (L_x) greater than 2.

One way slab arrangement — **Fig 2**: Typical general arrangement of a one-way slab system

In other words, when beams are provided in one parallel direction only, the slab is a one-way slab. Furthermore, if the longer side of a slab panel exceeds twice the shorter side, the slab is generally designed as a one-way slab, but that does not mean that the slab is transmitting load in one direction only.

One-way slabs may be simply supported or continuous. For one-way slabs supported on two opposite sides, the bending moments are calculated in the same way as for beams. Continuity across a beam is treated as fixed support. In detailing, if a slab is assumed to be simply supported at an end support, it is advisable to provide reinforcement for a probable negative bending moment due to the monolithic construction of beams and slabs (Reynolds and Steedman, 2005).

Typical detailing of a one-way simply supported reinforced concrete slab — **Fig 3**: Typical reinforcement detailing of a simply supported one-way slab

The effective span for one-way slabs is the same as that of beams. If l_n is the clear span (distance between faces of supports), the effective span l_effis given by;

l_eff= l_n + a₁ + a₂

effective length of one way slabs 1 — **Fig 4**: Effective span of a simply supported one-way slab

One-way slabs should be designed to resist the most unfavourable arrangement of loads. In clause 5.1.3 of Eurocode 2, the following two loading arrangements are recommended for buildings.

Alternate spans carrying (γ_GG_k+ γ_QQ_k) other spans carrying only γ_GG_k.
Any two adjacent spans carrying (γ_GG_k + γ_QQ_k). All other spans carrying only γ_GG_k

Steps in the design of a one-way slab

The steps in the design of a slab are as follows;

Determine the design life of the structure
Choose a slab thickness determined using deflection requirements, experience, or otherwise
Establish the durability requirements, fire resistance, and adequate concrete cover
Calculate and apply the loads on the slab comprising of the dead and imposed loads
Apply the appropriate load combination
Idealise each slab element and analyse to determine the critical design moments M_Ed and shear forces V_Ed
Carry out the flexural design
Check the deflection
Check the shear capacity
Check bar spacing and cracking

For the flexural design of slabs, determine k from;

k = M_Ed/(f_ckbd²)
If k < 0.167, no compression reinforcement is required, and you can calculate the lever arm;
z = d[0.5 + √(0.25 – 0.882k)]

The area of reinforcement required is given by;
A_s1 = M_Ed/(0.87f_ykz)

(a) Concrete cover
The nominal concrete cover in slabs is expected to satisfy the requirement;

C_nom = C_min + ∆_c,dev

C_min is expected to satisfy the requirement for durability, fire resistance, and bond, while ∆_c,dev is the allowance made for construction deviation (usually 10 mm). The minimum cover for bond should not be less than the bar diameter. For a one-way slab to have a fire rating of one hour (REI 60), the minimum thickness should be 80 mm and the concrete cover (from the surface to the centre of bar) should be minimum of 20 mm. A minimum cover of 15mm + ∆_c,dev is adequate for the durability of slabs under exposure class XC1. Therefore under normal circumstances, a concrete cover of 25 mm is usually adequate for floor slabs.

(b) Minimum tension steel
The main moment steel spans between supports and over the interior supports of a continuous slab. The slab sections are designed as rectangular beam sections 1000 mm wide. The minimum area of main reinforcement has to satisfy clause 9.2.1.1(1) of Eurocode 2;

A_s,min = (0.26f_ctm/f_yk)b_td but not less than 0.0013bd

where b_t = width (for slab design 1000 mm), d = effective depth

The maximum area of steel provided should not exceed 0.04A_c.

(c) Distribution steel
The distribution, transverse, or secondary steel runs at right angles to the main moment steel and serves the purpose of tying the slab together and distributing non-uniform loads through the slab. Clause 9.3.1.1(2) states that in the case of one-way slabs, secondary reinforcement of not less than 20 percent of principal reinforcement should be provided. Note that distribution steel is required at the top parallel to the supports of continuous slabs. The main steel is placed nearest to the surface to give the greatest effective depth.

detailing of continuous one way slab — **Fig 5**: Typical reinforcement detailing of a continuous one-way slab

(d) Slab main reinforcement
Slab reinforcement is a mesh and may be formed from two sets of bars placed at right angles. The table below gives bar spacing data in the form of areas of steel per metre width for various bar diameters and spacings. Reinforcement in slabs consists of a large number bars both ways which need to be tied together to form a mat. This is actually an expensive operation (see cost of fixing reinforcement in Nigeria).

It is necessary to point out that the critical span in the analysis of solid slabs is the short span. Therefore, the main reinforcements will lie parallel to the short span, and that is where you check your deflections (see why short spans are critical in floor slabs).

Table 1: Area of reinforcement per metre width of spacing

Spacing	ϕ8 mm	ϕ10 mm	ϕ12 mm	ϕ16 mm	ϕ20 mm
100	502	785	1130	2010	3141
125	402	628	904	1608	2513
150	335	523	753	1340	2094
175	287	448	646	1148	1795
200	251	392	565	1005	1570
225	223	349	502	893	1396
250	201	314	452	804	1256
275	182	285	411	731	1142
300	167	261	376	670	1047

For instance, if the area of reinforcement required (A_s,req) from calculation is 523 mm²/m, you can provide H12@200 c/c (Area of steel provided A_s,prov= 565 mm²/m).

(e) Crack Control
According to Clause 9.3.1.1(3) of Eurocode 2, if h is the total depth of the slab, then the maximum spacing of reinforcements is normally restricted to;

3h ≤ 400 mm for principal reinforcement
3.5 h ≤ 450 for secondary reinforcement

However, in areas of maximum moment, maximum spacing is restricted to;

2h ≤ 250 mm for principal reinforcement
3 h ≤ 400 mm for secondary reinforcement

For slabs 200 mm thick or greater the bar size and spacing should be limited to control the crack width and reference should be made to section 7.3.3 of the Eurocode 2.

(f) Curtailment of bars in slabs
Curtailment of bars is done according to the moment envelope. However, clause 9.3.1.2(1) requires that half the calculated span reinforcement must continue up to support. It is further stated that in monolithic construction, where partial fixity occurs along an edge of a slab but is not taken into account, the top reinforcement should be capable of resisting at least 25 percent of the maximum moment in the adjacent span and this reinforcement should extend at least 0.2 times the length of the adjacent span measured from the face of the support.

simplified curtailment rules of floor slabs — **Fig 6:** Typical curtailment rules of solid slabs

The above situation occurs in the case of simply supported slabs or the end support of a continuous slab cast integral with an L-beam which has been taken as a simple support for analysis but the end of the slab might not be permitted to rotate freely as assumed. Hence negative moments may arise and cause cracking.

(g) Shear capacity of slabs
Under normal loads, shear stresses are not critical and shear reinforcement is not required in floor slabs. Shear reinforcement is provided in heavily loaded thick slabs but should not be used in slabs less than 200 mm thick (clause 9.3.2 (1)). To check the shear capacity of slabs, the shear stress (v_Ed = V_Ed/bd) must be checked against the shear capacity of an unreinforced section (V_Rc,d). This is given by;

V_Rd,c = [C_Rd,c k(100ρ₁ f_ck )^1/3 + k₁.σ_cp] ≥ (V_min + k₁.σ_cp)

Where;
C_Rd,c = 0.18/γ_c
k = 1 + √(200/d) < 0.02 (d in mm);
ρ₁ = A_s1/bd < 0.02 (In which A_s1 is the area of tensile reinforcement which extends ≥ (l_bd + d) beyond the section considered)
V_min= 0.035k^(3/2)f_ck^0.5
K₁ = 0.15; σ_cp = N_Ed/A_c < 0.2f_cd
(Where N_Ed is the axial force at the section, A_c = cross sectional area of the concrete), f_cd = design compressive strength of the concrete).

In a one-way slab, the design shear force is calculated from the support reactions or end-shears at the support, while in a two-way slab, they can be obtained from the coefficients in Table 3.15 of BS 8110-1:1997.

(h) Check for deflection
The check for deflection is a very important consideration in slab design and usually controls the slab depth. In normal cases, a strip of slab 1 m wide is checked against span-to-effective depth ratios. A slab should not deflect excessively under service load. Excessive deflection of slabs can cause cracking to partitions and finishes.

For deemed to satisfy basic span/effective depth (limiting to depth/250);

Actual L/d of the slab must be ≤ Limiting L/d × β_s

The limiting basic span/ effective depth ratio is given by;

L/d = K [11 + 1.5√(f_ck)ρ₀/ρ + 3.2√(f_ck) (ρ₀/ρ – 1)^1.5] if ρ ≤ ρ₀

L/d = K [11 + 1.5√(f_ck) ρ₀/(ρ – ρ’) + 1/12 √(f_ck) (ρ₀/ρ)^0.5 ] if ρ > ρ₀

Where;
L/d is the limiting span/depth ratio
K = Factor to take into account different structural systems
ρ₀ = reference reinforcement ratio = 10^-3 √(f_ck)
ρ = Tension reinforcement ratio to resist moment due to design load
ρ’ = Compression reinforcement ratio

The value of K depends on the structural configuration of the member and relates the basic span/depth ratio of reinforced concrete members. This is given in the table 2;

Table 2: Basic span/effective depth ratio of different structural systems

Structural System	K	Highly stressed ρ = 1.5%	Lightly stressed ρ = 0.5%
Simply supported slabs	1.0	14	20
End span of interior slabs	1.3	18	26
Interior span of continuous slabs	1.5	20	30
Flat slab	1.2	17	24
Cantilever slabs	0.4	6	8

β_s= (500A_s,prov)/(f_ykA_s,req)

Two-way Spanning Slabs

Two-way action occurs when a slab is supported on all four sides. If the two dimensions and support conditions are the same, then the load is distributed to all supporting beams equally. In design, a slab is considered to be two-way if the ratio of the longer side to the shorter side is less than two.

For two way slabs, the precise amount and distribution of the load taken by each support, and consequently the magnitude of the bending moments on the slab, are not easily calculated if assumptions resembling real conditions are made. Therefore, approximate analyses are generally used. The method applicable in any particular case depends on the shape of the slab panel, the conditions of restraint at the supports and the type of load.

Two basic methods are commonly used to analyse slabs that span in two directions. They are;

The theory of plates, which is based on elastic analysis, is particularly appropriate to the behaviour under service loads
Yield-line theory, which considers the behaviour of the slab as a collapse condition approaches.

Generally, for rectangular slabs with standard edge conditions and subject to uniformly distributed loads, normally the bending moments are obtained using tabulated coefficients. These coefficients are based on elastic analysis from thin plate theory. The loads used in the analysis are factored to represent the ultimate limit state condition. This is the approach used in BS 8110 for slabs with corners that are not held down (no consideration for torsion). The analysis must take into account the support conditions which can be idealised as fixed, hinged, or free.

For slabs with irregular plan shapes and slabs subject to a combination of point loads and distributed loads, Johansen’s yield line analysis and the Hillerborg strip method provide powerful methods for strength calculations.

Simply supported two-way slabs
Where the corners of slabs are free to lift and no provision is made to resist forces at the corners, the maximum moments per unit width are given by the following expressions:

M_sx = bending moment in strips with span l_x = M_sx = α_sxql_x²
M_sy = bending moment in strips with span l_y = M_sy = α_syql_x²

where l_x is the shorter span of the panel, l_y is the longer span of the panel and q is the design ultimate load per unit area. Values of α_sx and α_sy are given in Table 3.13 of BS 8110-1:1997 for different ratios of l_y and l_x, where l_y is the longer span.

Rectangular panels with restrained edges
Where corners of a two-way slab are prevented from lifting and reinforced to resist torsion, the maximum bending moments per unit width are given by the following expressions:

M_sx= β_sxql_x²
M_sy = β_syql_x²

where;
M_sxis the maximum design moment either over supports or at midspan on strips with span l_x
M_sy is the maximum design moment either over supports or at midspan on strips with span l_y
q is the design ultimate load per unit area,
l_x is the shorter span, and
l_y is the longer span

The coefficients can be obtained from Table 3.14 of BS 8110-1:1997.

Design Example of a two-way slab

The general arrangement of the floor plan of a building is shown below. Design and detail the panel 1 of the building using the following data;

Thickness of floor slab = 150 mm
Concrete cover = 25 mm
Characteristic variable load = 1.5 kN/m²
f_ck = 25 N/mm²
f_yk= 460 N/mm²

General arrangement of building — **Fig 7**: General arrangement of a floor slab

Load Analysis of Panel 1

Permanent Loads
Self weight of slab = 25 kN/m³ × 0.15m = 3.75 kN/m²
Weight of finishes = 1.2 kN/m²
Partition allowance = 1.5 kN/m²
Total dead load (g_k) = 6.45 kN/m²

Variable Load on slab
Leading variable action (Imposed load) q_k1 = 1.5 kN/m²

Total load on slab (ULS) = 1.35g_k + 1.5q_k= 1.35(6.45) + 1.5(1.5) = 10.9575 kN/m²

The floor slab (PANEL 1) is spanning in two directions, since the ratio (k) of the longer side (l_y) to the shorter side (l_x) is less than 2. Hence, k = Ly/Lx = 3.825/3.625 = 1.055 (say 1.1, however it is more appropriate to interpolate from the table).

Moment coefficient for two adjacent edges discontinuous

Short Span
Mid-span = 0.042
Continuous edge = 0.056

Long Span
Mid-span = 0.034
Continuous edge = 0.045

Design of short span
Mid span
M_Ed = αnl_x²= 0.042 × 10.9575 × 3.625² = 6.0475 kNm/m
d = h – C_c – ϕ/2

Assuming ϕ12mm bars will be employed for the construction
d = 150 – 25 – 6 = 119mm; b = 1000mm (designing per unit width)

k = M_Ed/(f_ckbd²) = (6.0475 × 10⁶)/(25 × 1000 × 119²) = 0.0171

Since k < 0.167 No compression reinforcement required

z = d[0.5 + √(0.25 – 0.882k)]
z = d[0.5 + √(0.25 – 0.882 × 0.0273)] = 0.95d

A_s1 = M_Ed/(0.87f_ykz) = (6.0475 × 10⁶)/(0.87 × 460 × 0.95 × 119) = 133.668 mm²/m

Provide Y12mm @ 250mm c/c BOT (A_Sprov= 452 mm²/m)

To calculate the minimum area of steel required;
f_ctm = 0.3 × f_ck^2⁄3 = 0.3 × 25^2⁄3= 2.5649 N/mm²(Table 3.1 EC2)

A_s,min = 0.26 × f_ctm/f_yk × b × d = 0.26 × 2.5649/460 × 1000 × 144 = 208.76 mm²/m
Check if A_s,min < 0.0013 × b × d (187.2 mm²/m)
Since, A_s,min = 187.2 mm², the provided reinforcement is adequate.

Check for deflection
We check for deflection at the short span of slabs
k = 1.3 for slab simply supported at one end and continuous at the other end

ρ = A_s/bd = 134/(1000 × 119) = 0.001126 < 10^-3√f_ck (0.005)

Since ρ < ρ₀
L/d = K [11 + 1.5√(f_ck ) ρ₀/ρ + 3.2√(f_ck) (ρ₀/ρ – 1)^3⁄2]

L/d = 1.3 [11 + 1.5√(25) × 0.005/0.001126 + 3.2√25 (0.005/0.001126 – 1)^3⁄2] = 1.3(44.3 + 102.106) = 190.327

Modification factor β_s = 310/σ_s
σ_s = (310f_ykA_s,req)/(500A_s,prov) = (310 × 460 × 133.688)/(500 × 452) = 84.341 N/mm²
β_s = 310/84.342 = 3.67 > 2.0 (take 2.0)

Taking the distance between supports as the effective span, L = 3625mm
The allowable span/depth ratio = β_s× 30.838 = 2.0 × 190.327 = 280.645
Actual deflection L/d = 3625/119 = 30.462
Since 280.645< 30.462 Therefore, deflection is ok.

Shear Verification
Maximum shear force coefficient for slab(β) = 0.44
V_Ed = βnl_x

Ultimate shear force V_Ed = 0.44 × 10.9575 × 3.625 = 17.477 kN/m

V_Rd,c = [C_Rd,c.k. (100ρ₁ f_ck)^1/3

C_Rd,c = 0.18/γ_c = 0.18/1.5 = 0.12

k = 1 + √(200/d) = 1 + √(200/119) = 2.296 > 2.0, therefore, k = 2

ρ₁ = A_s/bd = 452/(1000 × 119) = 0.003798 < 0.02; K1 = 0.15

V_Rd,c = [0.12 × 2 × (100 × 0.003798 × 25)^1/3] × 1000 × 119 = 60476.947 N = 60.476 kN

Since V_Rd,c (60.476 kN) > V_Ed (17.477 kN), no shear reinforcement is required.

Note that shear reinforcements are rarely required in solid slabs supported by beams. Hence no further checks for shear will be done on the slab.

Design of short span

Continuous edge
M_Ed = αnl_x² = 0.056 × 10.9575 × 3.6252 = 8.0633 kNm/m
Using the procedure described above;
k = 0.0228; La/d = 0.95; A_s1 = 178.2236 mm²; A_s,min= 172.5217 mm²
Provide Y12 @ 250mm c/c Top (A_s,prov = 452 mm²)

Design of long span
Mid span
M_Ed = αnl_x² = 0.034 × 10.9575 × 3.6252 = 4.895 kNm/m
k = 0.0138; La/d = 0.95; A_s1 = 108.1945 mm²; A_s,min = 172.5217 mm²;
Provide Y12 @ 250mm c/c N/Bottom (A_s,prov= 452 mm²)

Continuous edge
M_Ed = αnl_x² = 0.045 × 10.9575 × 3.6252 = 6.479 kNm/m
k = 0.0183; La/d = 0.95; A_s1 = 143.2057 mm²; A_s,min = 172.5217 mm²;
Provide Y12 @ 250mm c/c Top (A_s,prov= 452 mm²)

Reinforcement Detailing

detailing of floor slab — **Fig. 8:** Detailing of Panel 1

Cost of Reinforcement Works in Nigeria

By

Ubani Obinna

-

November 21, 2020

Installation of reinforcements is an important item of work in the construction of reinforced concrete structures. The cost of reinforcement works therefore significantly impacts the total contract sum of any construction project. In the preparation of a bill of quantity (BOQ) or bill of engineering measurement and evaluation (BEME), concrete and reinforcement works are listed as separate items of work.

Therefore, a contractor bidding for such a job is expected to provide his rate of buying, cutting, bending, and installing the reinforcements according to the construction drawing in one sum. While concrete is priced per cubic meters (m³), reinforcement is usually priced per kilogram (kg).

The price to be quoted per kg of reinforcement is expected to cover the cost of purchase and supply of the materials to the site, cost of cutting, bending, and installation according to the design specification. It is also expected to cover the contractor’s profit and overhead. The use of bar bending schedule (BBS) can enhance the process of reinforcement handling and evaluation.

Read Also…
How to quote/price for concrete in Nigeria

Therefore, a contractor bidding for a job in reinforced concrete construction is expected to know the basic cost of reinforcement in the market, the cost of transportation, the cost of labour, and the cost of machinery that will be needed to execute the job. Initial market research may be required before the tender is computed and submitted to the client.

In Nigeria, there are different categories of reinforcements that are used for construction works, and each of them is priced differently in the market. As found all over the world, the use of mild steel (f_y = 250 N/mm²) is no longer common, and it is rarely utilised in construction in Nigeria. However, some people still use R6mm (commonly called the ‘quarter rod’) and R8mm as links for columns and beams, but it is rarely specified in construction drawings these days. The categories of reinforcements in the Nigerian markets are;

Local reinforcements (manufactured locally)
TMT reinforcements (manufactured locally)
Foreign reinforcements (imported into the country)

Local reinforcements are manufactured locally by the steel industries in Nigeria and are the cheapest class of reinforcements in the market. They have a reputation of not always meeting the minimum yield strength requirements, and are not usually employed in large-scale or serious projects.

TMT reinforcements are high-quality reinforcements produced from recycled steel through the process of thermomechanical treatment. They usually satisfy the yield strength requirements and are widely used for all kinds of projects in the country, including bridges. They are more expensive than local reinforcements. As at July 2022, while one tonne of reinforcement is sold at about ₦415,000.

Foreign reinforcements are usually imported from Germany, Ukraine, or Russia, and are slightly more expensive than TMT reinforcements. However, the presence of TMT reinforcements is making foreign reinforcements less attractive.

Therefore, careful attention to design details and specifications is needed during the pricing of reinforcements. We are going to show below the process of pricing for reinforcement works in Nigeria. The information that will be needed for the evaluation of the cost of reinforcement works in Nigeria are as follows;

Cost of reinforcement in the market (based on the size and type of reinforcement)
Cost of binding wire
Cost of delivery to site
Cost of labour for cutting, bending, and installation
Cost of machinery (cutting machine and bending machine)

Calculation of the Cost of Reinforcement Works in Nigeria

For example, let us calculate the cost of installing 3650 kg of Y12 reinforcements in the floor slab of a building.

As a rule of thumb, about 12 kg of binding wire is required to tie 1 tonne of reinforcement (this also means that about 0.012 kg of binding wire is required to tie 1 kg of reinforcement). Therefore about 44 kg of binding wire is required for this project.

Cost calculation
Basic cost of reinforcement (including delivery to site) = say ₦430,000 per tonne = ₦430 per kg
Basic cost of binding wire (including delivery to site) = ₦17,000 per 20 kg = ₦ 850 per kg (= ₦10.2 per kg of reinforcement)
Basic cost of labour (cutting, bending, and installation) = ₦25,000 per tonne = ₦25 per kg
Assuming that the use of machinery is not applicable in the project or factored into the cost of labour, total basic cost = ₦430 + ₦10.2 + ₦25 = ₦465.2 per kg

Make 10% allowance for waste and laps = 1.1 x 465.2 = ₦512
Allow 20% for contractor’s profit and overhead = 1.2 x ₦465.2 = ₦ 559 per kg

Therefore the current cost of reinforcement works in Nigeria is ₦559 per kg of reinforcement.

Therefore, the cost of fixing the floor slab reinforcement = ₦559 x 3650 = ₦2,040,350:00 (Two million, forty-six thousand and three hundred and fifty Naira)

For your construction and building design works, contact;

Structville Integrated Services Limited (RC 1491828)
Whatsapp/Call: +2347053638996
E-mail: info@structville.com

Design of Shear Walls

By

Ubani Obinna

-

November 21, 2020

Table of Contents

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

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

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

Forces Acting on Shear Walls

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

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

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

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

Location/Placement of Shear of Shear Walls

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

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

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

pierced shear wall — Coupled shear wall in a building

Functions of Shear Walls

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

Structural Action of a Shear Wall

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

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

Structural Design of Shear Walls

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

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

Reinforcement Detailing of Shear Walls

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

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

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

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

S_{cl, max} is the minimum of;

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

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

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

Where the main reinforcement (i.e., vertical bars) is placed nearest to the wall faces, transverse reinforcement should be provided in the form of links with 4 per m²of the wall area.

Design Example of Shear Walls

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

Vertical loads
Dead Load G_k = 300 kN/m
Live load Q_k = 55 kN/m

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

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

f_ck = C25/30; f_yk = 500 N/mm²

Check slenderness of wall at ground floor

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

Limiting slenderness, λ_lim = 20 ABC/n^0.5 (Cl. 5.8.3.1(1))
where;
A = 0.7
B = 1.1
C = 1.7 – r_m

where;
r_m = M₀₁/M₀₂ = say = –0.25
C = 1.7 – (-0.25) = 1.95
n = N_Ed/A_cf_d

where;
N_Ed (assuming wind is the leading variable action) = 1.35G_k + 1.5Q_k1 + 1.5Ψ₀Q_ki
= 1.35(300 + 60) + 1.5(650) + (1.5 × 0.7 × 55) = 486 + 975 + 57.75 = 1518.75 kN/m
A_cf_d = (225 × 1000) × (0.85 × 30/1.5) = 3825 kN
Therefore, n = 1518.75/3825 = 0.397
λ_lim = 20 ×( 0.7 × 1.1 × 1.95)/0.397^0.5 = 47.66

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

Combinations of actions
(a) At ULS, for maximum axial load, W_k is leading variable action
N_Ed = 1.35G_k + 1.5Q_k1 + 1.5Ψ₀Q_ki
= 1.35(300 + 60) + 1.5(650) + (1.5 × 0.7 × 55) = 486 + 975 + 57.75 = 1518.75 kN/m

M_Ed = M + e_iN_Ed ≥ e₀N_Ed (Cl. 5.8.8.2(1), 6.1.4)

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

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

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

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

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

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

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

or

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

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

N_Ed = 1518.75 kN/m
M_Ed= 46.732 kNm/m (out of plane)

and

N_Ed = –756 kN/m (tension)
M_Ed = 46.732 kNm/m (out of plane)

Design: cover above ground
c_nom = c_min + ∆c_dev

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

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

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

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

Column interaction chart — Rectangular column interaction chart for d₂/h = 0.15

column intreaction chart 2 — *Rectangular column interaction chart for d*₂*/h = 0.2*

N_Ed/bhf_ck = (1518.75 × 10³)/(225 × 1000 × 30) = 0.225
M_Ed/bh²f_ck = (46.732 × 10⁶)/(225² × 1000 × 30) = 0.030

Gives:A_sf_yk/bhf_ck = 0
Therefore, minimum area of reinforcement required

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

Try H12 @ 300 c/c on each face

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

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

N_Ed= (σ_st1 + σ_st2) × A_s/2
and M_Ed = (σ_st1 – σ_st2) × A_s/2 × (d – d₂)

So σ_st1 + σ_st2 = 2N_Ed/A_s
and σ_st1 – σ_st2 = 2M_Ed/[(d – d₂)A_s]

2σ_st1 = 2N_Ed/A_s + 2M_Ed/[(d – d₂)A_s]
As = (N_Ed/σ_st1) + M_Ed/(d – d₂)σ_st1

σ_st1 = f_yk/γ_s = 500/1.15 = 434.8

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

σ_st2 = 2N_Ed/A_s – σ_st1 = 601.9 – 434.8 = 167 MPa

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

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

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

Technical Guide: Detailing and Arrangement of Beam Reinforcements on Site

By

Mubarak Shittu

-

November 20, 2020

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

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

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

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

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

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

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

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

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

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

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

corner beam column junction — Corner beam-column junction

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

Internal beam column junction — Internal beam-column Junction

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

typical cantilever beam — Well designed and constructed cantilever beam

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

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

By

Ubani Obinna

-

November 16, 2020

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Structural Design Challenge | Explore your creativity

By

Ubani Obinna

-

November 12, 2020

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

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

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

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

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

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

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

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

The pictures of the architectural design are given below;

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

Photo credit: Instagram – @academiadeestruturas

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