Sandcrete Blocks: Production, Specifications, Uses, and Testing

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July 14, 2022

Table of Contents

Sandcrete blocks are precast composite masonry units made of cement, sand, and water and are moulded into various sizes. According to the British Standard (BS 6073: 1981 Part 1), a block is a heterogeneous building material with a unit that is larger in all dimensions than what is required for bricks, but no dimension should be larger than 650 mm or the height should be six times the thickness or greater than the length. When set in their normal aspect, sandcrete blocks are walling units whose dimensions exceed those of bricks (NIS 87: 2007).

However, it should be noted that in BS EN 771-3:2011 + A1:2005 (which replaced BS 6073:1981), the distinction between blocks and bricks has been removed, and replaced by ‘masonry units’. A masonry unit is defined as a preformed component intended for use in masonry construction. Furthermore, no standard dimensions have been provided. A manufacturer is expected to declare the dimensions of the masonry units in mm in terms of Length, Width, and Height. Therefore, the new standard serves as a performance standard, and not a recipe standard.

It has been reported by some researchers that sandcrete blocks are the major masonry units used in Nigeria’s construction industry, accounting for more than 90% of the country’s physical infrastructure. As a result, sandcrete blocks are important components in building construction. They are commonly utilised as load-bearing and non-load-bearing walling units in Nigeria, Ghana, and other African nations.

moulding of sandcrete blocks — Sandcrete blocks is used extensively in building construction

Sandcrete blocks are reasonably priced when compared to other building materials. They offer great damage resistance without the additional cost of protective equipment. Unlike other building materials, sandcrete bricks don’t rust, rot, or serve as a haven for pests that cause harm. Furthermore, they don’t contain any environmentally hazardous substances. According to the Nigeria Industrial Standard (NIS 87: 2007), sandcrete blocks must have a minimum compressive strength of 2.5 and 3.45 N/mm² for non-load bearing and load bearing walls, respectively.

Specifications for Sandcrete Blocks

The most popular sizes for sandcrete blocks are 450mm x 225mm x 225mm and 450mm x 150mm x 225mm. Sandcrete blocks can also be rectangular and solid or hollow. The Nigerian Industrial Standards (NIS 87: 2007) defined two types of blocks:

Type A load bearing blocks, and
Type B non-load bearing blocks

They both have the option of being solid or hollow.

The approved sizes for sandcrete blocks specified by the NIS are presented in Table 1;

Type	Work size (mm) Length x Height x Thickness	Web Thickness	Usage
Solid Block	450 x 225 x 100	–	For non-load bearing and partition walls
Hollow	450 x 225 x 113	25	For non-load bearing and partition walls
Hollow	450 x 225 x 150	37.5	For load bearing walls
Hollow	450 x 225 x 225	50	For load bearing walls

Table 1: The approved sizes for sandcrete blocks

Masonry units that have a core void area larger than 25% of the gross area are considered hollow blocks. Lightweight aggregate is used to make hollow sandcrete blocks, which can be utilised to build both load-bearing and non-load-bearing walls. Blocks with two cells are typically produced in Nigerian construction factories. Sandcrete hollow blocks have a void running through them from top to bottom that takes up around one-third of their volume, yet solid blocks are completely devoid of voids.

When hardened, sandcrete blocks often have significant compressive strengths, and these strengths typically increase with density. Sandcrete hollow blocks should have a minimum strength requirement of 2.5 N/mm² for 150 mm and 3.45 N/mm² for 225 mm, according to NIS 87:2007.

Mix Ratios for Sandcrete Blocks

Sandcrete blocks are frequently produced using cement-sand mixtures with a cement-to-sand ratio of 1:6, 1:7, 1:8, or 1:9 and coarse aggregates no larger than 10 mm (when required for concrete blocks). When properly cured, these combinations produce sandcrete blocks with a compressive strength that is significantly high enough to meet construction standards.

The typical compressive strength obtained for different mix ratios and water-cement ratio for 450 x 150 x 225 (6 inches block) are shown in Table 2;

W/C	1:10 (f_c at 28 days N/mm²)	1:8 (f_c at 28 days N/mm²)	1:6 (f_c at 28 days N/mm²)	1:4 (f_c at 28 days N/mm²)
0.3	2.40	4.08	5.40	6.10
0.4	3.00	4.39	5.58	6.23
0.5	3.80	4.47	6.85	7.60
0.6	3.60	4.24	6.41	7.00
0.7	3.20	4.21	5.81	6.54

Table 2 :Compressive Strength of sandcrete block at 28 days

Therefore, the optimum water-cement ratio for maximum compressive strength in sandcrete blocks is 0.5. The mix ratio recommended by the NIS for sandcrete blocks in Nigeria is 1:8. For a 225mm hollow sandcrete block produced with a mix ratio of 1:8, the 28 days compressive strength is expected to be a minimum of 3.5 N/mm² under laboratory controlled conditions. For 150 mm hollow block, a minimum compressive strength of 2.77 N/mm² should be expected after 28 days of curing.

The following recommendations can be adopted in the production of 225 mm (9 inches) sandcrete blocks;

Cement	Sand	Mix-Ratio	Expected Number of Blocks per bag (9 inches hollow)	Expected Minimum Compressive Strength (N/mm²) at 28 days using Manual Compaction
One bag	3 wheelbarrows (12 head pans)	1:6	15	5.6
One bag	3½ wheelbarrows (14 head pans)	1:7	17	4.2
One bag	4 wheelbarrows (16 head pans)	1:8	20	3.5
One bag	5 wheelbarrows (20 head pans)	1:9	25	3.3

Materials Used in the Production of Sandcrete Blocks

The following materials are used in the production of sandcrete blocks:

Cement

Cement is a binder material that is used to hold the constituent (sand, gravels, etc) aggregates together to form a composite matrix. It is a carefully controlled combination of lime, silica, alumina and iron oxide. However, compounds of lime are the main ingredients of cement.

Hydration reaction takes place whenever water is added to cement, which results in a significant heat release. When concrete hydrates, a gel is created that holds the aggregate particles together and gives concrete its strength and water tightness when it hardens. Ordinary Portland Cement (OPC) or Limestone Portland Cement are the most popular form of cement used in construction projects. When making sandcrete blocks, Portland cement must adhere to all the specifications given in EN 197-1:2011 and NIS 444-1:2003, respectively.

Fine Aggregates

Fine aggregates are granular materials obtained by processing natural materials which pass through a sieve with a mesh size of 9.35 mm, almost totally pass through a sieve with a mesh size of 4.75 mm, and are mostly retained on a sieve with a mesh size of 200 (75 μm). For the production of sandcrete blocks, four different types of sand may be employed. They are river sand, sea sand, crushed stone sand, and pit sand. River sand is the most common in Nigeria.

In terms of volume, sand makes up around 75% of the mixture. It serves as a filler and is a reliable predictor of the sandcrete block’s anticipated compressive strength. The fundamental water to cement ratio is exceeded when using much finer sand because more cement and water are needed to coat the particles. The result is the production of weaker, more porous blocks. Too small natural dust grains can replace cement paste, cover grain surfaces, and form thin films, which hinder cement paste from lubricating the aggregates.

Tests for Fine Aggregates

Sieve Analysis

Sieve analysis is a laboratory test that measures the particle size distribution of a soil by passing it through a series of sieves. Soil retained on it is termed as gravel fraction. A set of British standard (BS) sieves of sizes – 1.0mm, 0.85mm, 0.60mm, 0.50mm, 0.30, 0.25, 0.180 and pan and a weighing balance were used for the analysis. The sieves were arranged by keeping the largest aperture sieve at the top and smallest aperture at the bottom.

The grading of soil is best determined by direct observation of its particle size distribution curve. The equation below is usually adopted in calculating the Uniformity coefficient C_u.

C_u = D₆₀/D₁₀

Where C_u is the Uniformity coefficient, D₆₀ is the particle diameter corresponding to 60% finer on the cumulative particle-size distribution curve and D₁₀ is the particle diameter corresponding to 10% finer on the cumulative particle-size distribution curve. If C_u < 4.0 the soil is poorly graded; if > 4.0 the soil is well graded. Well graded sands should be used for the production of sandcrete blocks.

Specific Gravity

The weight of a particular volume of fine aggregate (sand) to the weight of an equivalent volume of water is known as its specific gravity. Sands have a specific gravity of about 2.65. Specific gravity is considered to be a measure of strength or quality of a material.

To carry out a specific gravity test, the following procedure can be adopted:
An empty density bottle will be cleaned, dried, weighed and designated (W₁). The bottle will be filled with one-third of the total volume of the sand sample, weighed and designated (W₂). The bottle is then filled with distilled water, weighed and designated (W₃). Then the content of the bottle is discarded and rinsed thoroughly. The bottle is then filled with distilled water to the meniscus, weighed and designated (W₄). The Specific gravity (G) is then calculated using the equation below;

G = (W₂ – W₁)/[(W₄ – W₁) – (W₃ – W₂)]

Water

Water reacts with cement to produce the hydration reaction. The amount of water utilised in the mixing process has a significant impact on the workability and strength of sandcrete. To make concrete or sandcrete, water must be devoid of suspended solids, inorganic salts, acids, and alkalis, as well as algae, oil contamination, and acids and alkalis. It is advised to use potable water that complies with NIS 554:2007 standard when making sandcrete blocks.

Mechanical Properties of Sandcrete Blocks

The mechanical properties that are frequently declared in sandcrete blocks are the bulk density, water absorption, and compressive strength.

Bulk Density

Density is the quantity of an element’s or material’s particles packed into a specific volume. The density of the substance increases with the degree of particle packing. Therefore, higher levels denote a similar level of compaction. Mathematically, this is the mass of the masonry unit divided by the dimensions volume:

Bulk Density = mass of block (kg)/dimensional volume of block (m³)

The density is masonry units is determined in accordance wit BS EN 772-13:2000. In Nigeria, a minimum density limit of 1920 kg/m³ is recommended for individual sandcrete blocks, and 2020 kg/m³ for an average of three or more blocks.

Water Absorption

This is the amount of water that a block unit will absorb when submerged for the specified amount of time in the water at room temperature. It is stated as a percentage of the dry unit of the block’s weight. The weight of water absorbed when the block unit is partially submerged in water for one minute is the absorption rate. Additionally, it is also defined as the amount of water that a brick absorbs in the first minute after coming into contact with water.

The procedure for obtaining the water absorption of masonry units is described in in EN 772-11:2011.

It is expressed mathematically as;
Water absorption = (mass of saturated block (kg) – mass of dry block (kg)) / volume of block (m³).

The water absorption rate is determined by measuring the decrease in mass of the saturated block and surface dry sample. To achieve this, block samples whose weights had been taken in the dry state and noted as (M₁), were fully immersed in water. The time taken for full immersion was noted, and a period of twenty-four (24) hours was allowed to elapse. After 24 hours, the weight of the wet block samples was recorded as (M₂). The difference between the dry and wet weights of each block was calculated by subtracting the dry weight from the wet weight. The percentage absorption was calculated using the Equation below.

Water absorption (%) = [(M₂ – M₁)/M₁] × 100

The ASTM C140 recommended maximum water absorption capacity of 240 kg/m³. The maximum water absorption specified by the Nigerian Standard is 12%.

Compressive Strength

A compressive strength test is used to assess the quality of a block unit and its response to curing. It is described as the unit’s capacity to sustain an axial load that is applied to either the block’s bed face or its edge or the ratio of the crushing load that a sample can sustain to its net area. The declared compressive strength of the block by the manufacturer shall be the characteristic 5% fractile f_c or the mean 50% fractile f_m.

The compressive strength of sandcrete blocks should be evaluated in accordance with EN 772-1:2000. When the anticipated compressive strength is less than 10 N/mm², the crushing machine should be loaded at a rate of 0.05 (N/mm²)/s.

sandcrete block crushing and testing 1 — Sandcrete block compressive strength test

It is expressed mathematically as;

Compressive strength = maximum crushing load (N) / minimum surface area (mm²)

Sandcrete hollow blocks have a minimum strength requirement of 2.5 N/mm² for 150 mm and 3.45 N/mm² for 225 mm, according to NIS 87:2007. The factors affecting this property are water-cement ratio, degree of compaction, fine aggregate (grade, texture and shape characteristics), cement type, the efficiency of curing, amount of mixing water, and mix proportion.

Production of Sandcrete Blocks

The advent of various quickly assembled machines and other manually operated frameworks for the manufacture of masonry units is a significant factor contributing to this rise in the number of such production facilities. The top three methods of sandcrete blocks production are:

Hand ramming compaction moulds.
Manual tamping compaction frame.
Motorised vibration machine.

All three methods employ both horizontal and vertical orientations in production. It is rare to see block manufacturing industries employing all three forms of production. Every block production industry typically uses only one orientation for the creation of block units for a specific type of compaction mechanism. During production, several manufacturers employ various compaction techniques with various orientations.

Hand Ramming Compaction Moulds

The equipment comprises of a steel mould box that has been prefabricated and moulded to the required size of the block. It is designed so that when the cement and sand mixture is rammed together to create a sandcrete block, the resulting shape precisely fits the mould and so adheres to the necessary specifications. The compressed wet unit can be removed after hand ramming thanks to a detachable steel plate sitting at the bottom. The mould box has two curved steel handles that make it easier to remove the compacted unit.

hand moulded block — Hand moulding of blocks

A wooden bat in the form of a chisel serves as the compaction tool. The cement-sand mixture is brought up against the flat end of the bat, which is then driven over it and inserted into the steel mould. The compacted unit is taken out by slamming the box upside down. In order to receive the block, the opening end is eventually supported by a wooden pallet. The unit is turned upside down, and a removable steel plate now rests on top of it.

This is the commonest approach of block moulding in the rural areas of Nigeria.

Manual Tamping Compaction Frame

This compaction technique uses a steel framework with a mould supported by four legs as its equipment. Although the exact finishing height varies depending on the manufacturer, it is often around 1.0 m high. The sole lever mechanism is attached to a base plate that accommodates the mould. The means of compaction is another plate (top plate) covering with an adjacent handle. When tamping, the top plate is always weighed down with a thick steel piece to apply pressure to the cement-sand mixture effectively.

hand operated block moulding machine — Sandcrete block production machine

Pulling down the tubular length by the frame moves the lever system. When compaction is complete, this action raises the completed product for collection. The adjacent handle is used to aid with compaction. Depending on the needed level of consolidation, the operation is repeated at least four times. For stability, the legs are braced, and the compacting pressure varies.

Motorised Vibration Machine

Unskilled labour can operate this equipment because it is designed to be simple to use. Because of the little maintenance it requires, the design is ideal for remote locations. Furthermore, the machine can be designed to run by diesel engines as well as electric motors. For stability and safe handling, robust frames are used in its design. It is controlled by three levers, and a constant hydraulic pressure is generated. The structure is 1.85 metres high.

It is powered by a motor that is tucked away beneath the wooden pallet on which the mould is set. The diesel variants use a roller that is turned by a fan belt that is fixed over the motor. The fixture on the motor, a metallic mass, collides with the underside of the wooden pallet, causing actual vibration. The longest lever, which is typically on the right side of the machine, is used to remove the moulded unit.

block moulding machine — Motorised Vibration Block Moulding Machine

Actual compaction is accomplished by pressing down on the cement-sand mixture in the mould below with the “presser.” The highest lever is used to achieve this. The vibration is turned off with a second lever. The final and longest lever on the right lifts the mould gradually to enable quick removal of the compacted unit. The amount of compacting pressure from the motorised vibration machine is fairly constant.

In research conducted in Nigeria, the NIS’s specified compressive strength for 150 mm (6 inches) sandcrete blocks at 28 days of curing was not met by either manually or mechanically operated methods (though a mix ratio of 1:9 was adopted in the study). Both manually and with the aid of a vibrating machine, nine (9) inches sandcrete blocks were produced, and they met all Nigerian Industrial Standards specifications.

However, blocks made using a vibrating machine had a larger compressive strength than those made by hand. The vibrating block moulding machine produces sandcrete blocks with the highest compressive strength of all the techniques used because it achieves appropriate compaction. Additionally, compared to the other two methods, the vibrating machine-produced sandcrete blocks absorbed more moisture. Another study from Ghana found that using compaction in the vertical orientation with motorised vibration satisfies the required standards.

Quality Control Tips for Block Production

Water, sand, and aggregates must be clean and devoid of organic contaminants.
Ideally, the sand to be used for block production should be dry. When wet, the moisture content and the water absorption should be determined to effectively control the water-cement ratio.
Cement and aggregate must be mixed until the colour is uniform.
Only add enough water to the mixture to make it workable.
The wheelbarrows used for measurement (65-litre builder’s wheelbarrows) should not be heaped.
All mixtures should never be retempered by adding more water because doing so weakens the final product and should be used up within two hours of initial mixing.
Complete compaction is required. Avoid too little or inadequate compaction since it reduces the strength of the block.
Blocks need to be cured after demoulding. Curing is the process of keeping the blocks at an ideal temperature and moisture level to promote the hydration of the cement and the development of maximum strength. The curing period for blocks should be at least 14 days.

Analysis of Partition Loads on Slabs | Wall Load on Slabs

By

Ubani Obinna

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July 11, 2022

The provision of partitions on suspended slabs of residential, commercial, or industrial buildings is widespread in the construction industry. Spaces in a building can be demarcated using a variety of partition materials such as sandcrete blocks, bricks, gypsum dry walls, timber stud walls, metal lath, etc. These partitions exert additional loads on a suspended slab, and should be accounted for in the design of the slab. This is necessary especially when there is no beam or wall directly under the slab supporting the partition.

It is important to note that wall and partition loads insist on suspended slabs as line loads instead of uniformly distributed loads. It is more complex to analyse line loads on plates than uniformly distributed loads. Therefore during designs, engineers usually attempt to represent line loads with equivalent uniformly distributed loads to make the computational effort easier.

drywall partition load — Typical dry wall partitioning in an office building

There are established guides in the building code for assessing all types of loads that a building might be subjected to, and partition loads is not an exception. Typically in the design of reinforced concrete solid slabs, a partition allowance of between 1.00 kN/m² to 1.5 kN/m² is usually made during the analysis of dead loads (permanent actions). This is usually sufficient to allow for all lightweight movable partitions that may be placed on the slab later.

According to clause 6.3.1.2 of EN 1991-1-1:2002, provided that a floor allows a lateral distribution of loads, the self-weight of movable partitions may be taken into account by a uniformly distributed load q_k which should be added to the imposed loads of floors obtained from Table 6.2. This defined uniformly distributed load is dependent on the self-weight of the partitions as follows:

for movable partitions with a self-weight < 1.0 kN/m wall length: q_k = 0.5 kN/m²
for movable partitions with a self-weight > 1 ≤ 2.0 kN/m wall length: q_k = 0.8 kN/m²
for movable partitions with a self-weight > 2 ≤ 3.0 kN/m wall length: q_k = 1.2 kN/m²

However, full design consideration should be taken for heavier partitions, accounting for the locations and directions of the partitions and the structural forms of the floors.

According to BS 6399 Part 1, when the position of the wall load is not known, the equivalent uniformly distributed load that is added to the slab load should be 0.33w_p (kN/m²), where w_p is the weight of the wall (kN/m).

However, when the direction of the partition is normal to the span of the slab, the equivalent uniformly distributed load is given by 2w_p/L for simply supported slabs and 3w_p/2L for continuous slabs (Where L is the span of the slab normal to the wall load).

Said et al (2012) used finite element analysis and multiple linear regression to derive a general relation between line loads acting on two-way slab system and the equivalent uniformly distributed loads. According to them;

W_UDL/W_Line= 0.32193 + 0.00473α – 0.10175(L₂/L₁) ——– (R² = 0.8327)

Where;
W_UDL/W_Line is the ratio of equivalent uniformly distributed load to actual line load,
α is the relative ratio of the stiffness of beam to slab,
L₂/L₁ is the aspect ratio of the slab.

Therefore, from the statistical relationship above, when the value of the line load is known, the equivalent UDL that will produce comparable bending moment values can be obtained using the aspect ratio of the slab and the ratio of the stiffness of the supporting beams to the slab.

Partitions such as sandcrete blocks exert a significant magnitude line load on reinforced concrete solid slabs. For a 225mm hollow block, the unit weight is about 2.87 kN/m². For a 12 mm thick plaster on both sides, the total weight of finishes is about 0.6 kN/m². This brings the total unit weight of the block to about 3.47 kN/m², which is usually approximated to 3.5 kN/m².

Therefore, for a wall height of 3m, the equivalent line load exerted on the supporting slab or beam is 3.5 kN/m² × 3m = 10.5 kN/m.

So many structural engineering software design packages have the option of applying line loads directly on slabs. For software like Staad Pro, it may not be possible to assign line loads directly on plates, however a dummy beam of negligible stiffness can be used to transfer the line load to the slab.

Example on Partition Load Modelling

To demonstrate the effects of line loads from block wall, let us consider a 150 mm thick 5m x 6m two-way slab that is simply supported at all edges by a 450 mm x 225mm beam. The slab is supporting a line load of 10.5 kN/m coming from a 225 mm thick block wall placed at the centre parallel to the short span as shown below.

We are going to consider several scenarios;

(a) When the slab is loaded directly with the line load (w = 10.5 kN/m)
(b) When the line load is represented with equivalent UDL given by 0.33w_p = 0.33 × 10.5 = 3.15 kN/m²
(c) When the line load is represented with an equivalent UDL given by 2w_p / L = (2 × 10.5)/6 = 3.5 kN/m²
(d) When the line load is represented with an equivalent UDL by Said et al (2012);
W_UDL/W_Line= 0.32193 + 0.00473α – 0.10175(L₂/L₁)

Stiffness of beam = (0.225 × 0.45³)/12 = 1.70859 × 10^-3m⁴
Stiffness of slab = (1 × 0.15³)/12 = 2.8125 × 10^-4 m⁴
α = (1.70859 × 10^-3)/(2.8125 × 10^-4) = 6.075
L₂/L₁ = 6/5 = 1.2
W_UDL/W_Line= 0.32193 + 0.00473(6.075) – 0.10175(1.2) = 0.2279
Therefore W_UDL = 0.2279W_Line = 0.2279 × 10.5 = 2.392 kN/m²

Analysis Results

(a) When the slab is loaded directly with the line load (w = 10.5 kN/m)

(b) When the line load is represented with equivalent UDL (w_e = 3.15 kN/m²)

Analysis of Partition Loads on Slabs | Wall Load on Slabs 24

Analysis of Partition Loads on Slabs | Wall Load on Slabs 26

(c) When the line load is represented with equivalent UDL (w_e = 3.5 kN/m²)

(d) When the line load is represented with equivalent UDL (w_e = 2.392 kN/m²)

Analysis of Partition Loads on Slabs | Wall Load on Slabs 30

Analysis of Partition Loads on Slabs | Wall Load on Slabs 32

Analysis Method	M_x (sagging)	M_x (hogging)	M_y (sagging)	M_x (hogging)	M_xy
Line Load	3.99 kNm	0.799 kNm	6.65 kNm	0.627 kNm	0.913 kNm
0.33w_p = 3.15 kN/m²	4.48 kNm	0.138 kNm	4.7 kNm	0.349 kNm	0.978 kNm
2w_p / L = 3.5 kN/m²	4.98 kNm	0.153 kNm	5.22 kNm	0.388 kNm	1.09 kNm
W_UDL = 0.2279W_Line = 2.392 kN/m²	3.4 kNm	0.104 kNm	3.57 kNm	0.265 kNm	0.742 kNm

From the analysis result, it can be seen that none of the proposed equations was able to capture the effect of the line load adequately. Therefore, when heavy wall loads are to be supported on suspended slabs, the line load should be properly modelled on the slab instead of being converted to equivalent uniformly distributed load.

References

Said A. A, Obed S. R., and Ayez S. M. (2012): Replacement of Line Loads acting on slabs to equivalent uniformly Distributed Loads. Journal of Engineering 11(18):1193 – 1200

Design of Cantilever Stairs

By

Ubani Obinna

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July 10, 2022

Table of Contents

Cantilever stairs are a unique type of staircase where one end of the tread is rigidly supported by a beam or reinforced concrete wall, and the other end free. By implication, one end of the tread of the cantilever staircase appears to be floating in the air without support.

The design and construction of a cantilever staircase are expected to maintain and/or enhance the aesthetic appeal, while at the same time, ensuring that the staircase satisfies ultimate and serviceability limit state requirements.

The design of cantilever stairs, therefore, involves the selection of the adequate size of the supporting beams/walls, treads, and other accessories to support the anticipated load on the staircase and to also ensure the good performance of the staircase while in service. For reinforced concrete spine beams, the dimensions and reinforcements provided must satisfy all design requirements, while for steel spine beams, the section selected must satisfy all the requirements.

Moreover, the thickness of the tread must be adequate such that it does not undergo excessive vibration, cracking, deflection, or failure. The tread can be made of reinforced/precast concrete, timber, steel, glass, or composite sections.

Structurally, there are two variations of cantilever staircases;

(a) Cantilever staircase supported by a central spine beam
(b) Cantilever staircase supported by a side spandrel beam or RC Wall

cantilever staircase with spine beam — (a) Cantilever staircase supported by a spine beam

cantilever staricase — (b) Cantilever staircase supported by a side spandrel beam or RC Wall

In the first case, the spine beam of the staircase is placed at the centre of the tread, with the two ends of the tread hanging free. In the second case, only one end of the tread is fixed to an adjacent wall where the spandrel beam is hidden to support the treads, and the other end is free. Alternatively, if the wall is a reinforced concrete wall, spandrel beams will no longer be required. Therefore, the latter has a longer moment arm than the former.

For the design of cantilever staircases, the use of uniformly distributed live loads should not be employed. Rather, the concentrated loads provided in Table 6.2 of EN 1991-1-1:2002 should be used. According to clause 6.3.1.2(5)P, the concentrated load shall be considered to act at any point on the floor, balcony or stairs over an area with a shape which is appropriate to the use and form of the floor. The shape may be assumed to be a square of 50 mm.

Furthermore, the possibility of upward loading on the cantilever staircase should also be considered.

Worked Example on the Design of Cantilever Stairs

Design the treads and spine beam of the cantilever staircase in a proposed residential dwelling with the following information;

Width of staircase = 1200 mm
Going (width of riser) = 250 mm
Riser = 150 mm
Thickness of riser = 100 mm
f_ck = 25 N/mm²
f_yk = 500 N/mm²

Load Analysis

For a staircase in a residential dwelling the uniformly distributed live load varies from 2.0 to 4.0 kN/m², while the concentrated load (which can be used for local verification) varies from 2.0 to 4.0 kN. The exact value can be decided by the National Annex of the country.

Permanent Actions
Self weight of thread (UDL) = 25 × 0.1 × 0.25 = 0.625 kN/m
Finishes (allow 1.2 kN/m²) = 1.2 × 0.25 = 0.3 kN/m
Total UDL g_k = 0.925 kN/m
Railings (allow) G_k = 0.5 kN (Concentrated load)

Variable Actions
Allow a variable concentrated load Q_k of 3 kN (at the free end)

Ultimate Limit State
Permanent load (g_k) = 1.35 × 0.925 = 1.25 kN/m
Permanent load (G_k) = 1.35 × 0.5 = 0.675 kN
Variable load (Q_k) = 1.5 × 3 = 4.5 kN

Maximum design moment (about the centreline of spine beam) = (1.25 × 0.6²)/2 + (5.175 × 0.6) = 3.33 kNm
Maximum shear (about the centreline of spine beam) = (1.25 × 0.6) + (5.175) = 5.925 kN

Flexural Design

Design bending moment; M = 3.3 kNm
Effective depth of tension reinforcement; d = 68 mm
Redistribution ratio; d = min(M_{neg_red_z3} / M_{neg_z3}, 1) = 1.000

K = M / (b × d² × f_ck) = 0.115
K’ = 0.207
K’ > K – No compression reinforcement is required
Lever arm; z = min(0.5 × d × [1 + (1 – 2 × K / (h × a_cc / γ_C))^0.5], 0.95 × d) = 60 mm
Depth of neutral axis; x = 2 × (d – z) / λ = 20 mm
Area of tension reinforcement required; A_s,req = M / (f_yd × z) = 127 mm²

Tension reinforcement provided; 3H12 A_s,prov = 339 mm²
Minimum area of reinforcement – exp.9.1N; A_s,min = max(0.26 × f_ctm / f_yk, 0.0013) × b × d = 23 mm²
Maximum area of reinforcement – cl.9.2.1.1(3); A_s,max = 0.04 × b × h = 1000 mm²
PASS – Area of reinforcement provided is greater than area of reinforcement required

Crack control

Maximum crack width; w_k = 0.3 mm
Design value modulus of elasticity reinf – 3.2.7(4); E_s = 200000 N/mm²
Mean value of concrete tensile strength; f_ct,eff = f_ctm = 2.6 N/mm²
Stress distribution coefficient; k_c = 0.4
Non-uniform self-equilibrating stress coefficient; k = min(max(1 + (300 mm – min(h, b)) × 0.35 / 500 mm, 0.65), 1) = 1.00

Actual tension bar spacing; s_bar = 93 mm

Maximum stress permitted – Table 7.3N; s_s = 326 N/mm²
Steel to concrete modulus of elast. ratio; a_cr = E_s / E_cm = 6.35
Distance of the Elastic NA from bottom of beam; y = (b × h² / 2 + A_s,prov × (a_cr – 1) × (h – d)) / (b × h + A_s,prov × (a_cr – 1)) = 49 mm
Area of concrete in the tensile zone; A_ct = b × y = 12195 mm²
Minimum area of reinforcement required – exp.7.1; A_sc,min = k_c × k × f_ct,eff × A_ct / s_s = 38 mm²
PASS – Area of tension reinforcement provided exceeds minimum required for crack control

Quasi-permanent moment; M_QP = 1.0 kNm
Permanent load ratio; R_PL = M_QP / M = 0.30
Service stress in reinforcement; s_sr = f_yd × A_s,req / A_s,prov × R_PL = 49 N/mm²
Maximum bar spacing – Tables 7.3N; s_bar,max = 300 mm
PASS – Maximum bar spacing exceeds actual bar spacing for crack control

Deflection control

Reference reinforcement ratio; ρ_m0 = (f_ck)^0.5 / 1000 = 0.00500
Required tension reinforcement ratio; ρ_m = A_s,req / (b × d) = 0.00748
Required compression reinforcement ratio; ρ’_m = A_s2,req / (b × d) = 0.00000
Structural system factor – Table 7.4N; K_b = 0.4

Basic allowable span to depth ratio ; span_to_depth_basic = K_b × [11 + 1.5 × (f_ck)^0.5 × ρ_m0 / (ρ_m – ρ’_m) + (f_ck)^0.5 × (ρ’_m / ρ_m0)^0.5 / 12] = 6.404

Reinforcement factor – exp.7.17; K_s = min(A_s,prov / A_s,req × 500/ f_yk, 1.5) = 1.500
Flange width factor; F1 = 1.000
Long span supporting brittle partition factor; F2 = 1.000

Allowable span to depth ratio; span_to_depth_allow = min(span_to_depth_basic × K_s × F1 × F2, 40 × K_b) = 9.606
Actual span to depth ratio; span_to_depth_actual = L_{m1_s1} / d = 8.824

Shear Design

Using the maximum shear force for all the spans
Support A; V_Ed (say) = 6 kN
V_Rd,c= [C_Rd,c.k. (100ρ₁ f_ck)^1/3 + k₁.σ_cp]b_w.d ≥ (V_min + k₁.σ_cp)b_w.d
C_Rd,c = 0.18/γ_c = 0.18/1.5 = 0.12
k = 1 + √(200/d) = 1 + √(200/68) = 2.714 < 2.0, therefore, k = 2.00
V_min= 0.035k^3/2f_ck^1/2
V_min = 0.035 × 2.00^3/2 × 25^1/2 = 0.494 N/mm²
ρ₁ = A_s/bd = 339/(250 × 68) = 0.0199 < 0.02;

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.) Take N_Ed= 0

V_Rd,c = [0.12 × 2 × (100 × 0.0199 × 25 )^1/3] × 250 × 68 = 15005.755 N = 15 kN

Since V_Rd,c (15 kN) < V_Ed (6 kN), No shear reinforcement is required.

However, nominal shear reinforcement can be provided as H8 @ 150 c/c.

Summary of the tread (going) design
Thickness = 100 mm
Reinforcement = 3H12 (Top and Bottom)
Links = H8 @ 150 c/c.

Design of the Spine Beam

To design the spine beam, it is very important to transfer the load from the treads to the top of the spine beam. In other words, the spine beam will be subjected to its self-weight and the load from the treads. It will be very important to also consider the effects of asymmetric loading (when the live load is acting only on side of the staircase). This can lead to the development of torsional stresses on the beam.

Ultimate limit state load transferred from treads to beam = 6 kN + 6 kN = 12 kN (concentrated loads)
Load transferred from the landing area (say) = 5 kN/m

Width of spine beam = 225 mm
Depth = 300 mm
Self weight of spine beam (drop) = 1.35 × 25 × 0.225 × 0.3 = 2.278 kN/m
Self weight of the stepped area = 1.35 × 0.5 × 0.15 × 0.225 = 0.02278 kN/m

Total uniformly distributed load on the flight = 2.3 kN/m
Total uniformly distributed load on the landing = 2.278 + 5 = 7.278 kN/m

The spine beam was loaded as shown below;

The analysis results are shown below;

Flexural Design

Design bending moment; M = 26.9 kNm
Effective depth of tension reinforcement; d = 264 mm
K = M / (b × d² × f_ck) = 0.086
K’ = 0.207

K’ > K – No compression reinforcement is required
Lever arm; z = min(0.5 × d × [1 + (1 – 2 × K / (h × a_cc / γ_C))^0.5], 0.95 × d) = 242 mm
Depth of neutral axis; x = 2 × (d – z) / λ = 54 mm

Area of tension reinforcement required; A_s,req = M / (f_yd × z) = 255 mm²
Tension reinforcement provided; 2H16 A_s,prov = 402 mm²
Minimum area of reinforcement – exp.9.1N; A_s,min = max(0.26 × f_ctm / f_yk, 0.0013) × b × d = 77 mm²
Maximum area of reinforcement – cl.9.2.1.1(3); A_s,max = 0.04 × b × h = 2700 mm²
PASS – Area of reinforcement provided is greater than area of reinforcement required

Shear Design

Angle of comp. shear strut for maximum shear; θ_max = 45 deg
Strength reduction factor – cl.6.2.3(3); v₁ = 0.6 × (1 – f_ck / 250) = 0.552
Compression chord coefficient – cl.6.2.3(3); α_cw = 1.00
Minimum area of shear reinforcement – exp.9.5N; A_sv,min = 0.08 N/mm² × b × (f_ck)^0.5 / f_yk = 161 mm²/m

Design shear force at support ; V_Ed,max = V_{Ed,max_s1} = 54 kN
Min lever arm in shear zone; z = 242 mm
Maximum design shear resistance – exp.6.9; V_Rd,max = α_cw × b × z × v₁ × f_cwd / (cot(θ_max) + tan(θ_max)) = 201 kN

PASS – Design shear force at support is less than maximum design shear resistance

Design shear force ; V_Ed = 54 kN
Design shear stress; v_Ed = V_Ed / (b × z) = 0.994 N/mm²
Angle of concrete compression strut – cl.6.2.3; θ = min[max(0.5 × sin^-1(min(2 × v_Ed / (a_cw × f_cwd × v₁),1)), 21.8 deg), 45deg] = 21.8 deg
Area of shear reinforcement required – exp.6.8; A_sv,des = v_Ed × b / (f_yd × cot(θ)) = 206 mm²/m
Area of shear reinforcement required; A_sv,req = max(A_sv,min, A_sv,des) = 206 mm²/m

Shear reinforcement provided; 2H8 @ 175 c/c
Area of shear reinforcement provided; A_sv,prov = 574 mm²/m
PASS – Area of shear reinforcement provided exceeds minimum required
Maximum longitudinal spacing – exp.9.6N; s_vl,max = 0.75 × d = 198 mm

EXTRA:
The structural analysis result revealed a high compressive axial load on the spline beam. As a result, it will be important to check the interaction of bending moment and axial force, as typically will be done in the design of reinforced concrete columns. This was checked and found satisfactory.

Furthermore, the possibility of a torsional moment on the spline beam when the live load is asymmetrically loaded should also be checked.

Wind Tunnel Testing for Bridges

By

Ubani Obinna

-

July 8, 2022

Table of Contents

It is common practise to test bridges with lengthy or unusually flexible decks in wind tunnels. In order to fully comprehend the structure’s aerodynamic behaviour, wind tunnel testing may occasionally be necessary, as described in BD 49/01 (Department of Transport, 2001a). The aerodynamics and wind-induced reaction of vehicles can be impacted by the wind field on a bridge deck. Studies of wind fields on bridge decks can be used as a guide when designing a bridge to guarantee the safety of moving vehicles.

For long span bridges, the use of wind tunnel testing is imperative. The term “long span bridges” typically refers to structures with a span length of between 1000 and 1500 metres. The first natural frequencies for these bridges are of the order of 0.1 Hz or lower, and the structure possesses very high flexibility. Bridges with span lengths over 1.5 km are categorised as “extremely long span bridges,” and their natural frequencies decrease in inverse proportion of the span length.

Objectives of Wind Tunnel Testing

There are three main objectives for the use of wind tunnel testing during bridge design.

First, wind tunnel testing is utilised to estimate structure-specific drag and lift coefficients if bridge decks or piers do not conform to the conventional cross-sections specified in design regulations. The mean lateral force, mean normal force, and pitching moment near the centre of the deck are used to measure static wind loads, which are then given as static wind load coefficients. Depending on the location of the bridge site, measurements are made in both smooth and turbulent flow conditions.

The second objective is to determine whether a structure is vulnerable to vortex shedding and divergent amplitude responses. These evaluations are typically carried out under conditions of uniform flow at various test wind incidence angles.

The third objective is to examine how nearby terrain or structures affect the structure’s dynamic reaction. The main impact of such obstructions is to change the wind’s angle of action on the structure. The wind field in built-up areas is frequently extremely complicated, and it is feasible for the wind to be directed around nearby structures, increasing the mean velocity on the bridge’s under-review zones.

Modelling for Wind Tunnel Testing

For prismatic (two-dimensional) constructions like long-span bridge decks, section model investigations are carried out. To simulate the attributes of the structure, a rigid model is fixed to a dynamic test rig made of a set of springs. The dynamic response caused by phenomena like vortex shedding and the wind speeds for the development of aerodynamic instabilities like galloping and flutter are measured.

The failure of the Tacoma Narrows Bridge demonstrated the catastrophic failure that flutter can cause, which motivated engineers to carefully consider the aerodynamic analysis inside bridge design procedures.

The section model for wind tunnel testing can be made from materials such balsa wood, carbon fibre sheet, and steel sections, and it is designed to correctly replicate the prototype’s scaled mass as well as its lowest torsional and bending frequencies. Handrails, parapets, guiding vanes, maintenance gantry rails, stay pipes, and any other non-structural accessories that may have an impact on the deck’s aerodynamic behaviour should be included in sectional models.

wind tunnel testing of a suspension bridge

Aerodynamically comparable representations may be employed when these elements are too small to be scale-modelled. Typically, dimensions must be accurate to ±0.05 mm. The target spectral density function must be closely matched where turbulent atmospheric conditions are required.

A two-dimensional grid of vertical and horizontal strips is usually placed at the entrance of the wind tunnel test section to create turbulent conditions in low-speed aeronautical wind tunnels. It is possible to recreate the necessary turbulence conditions by adjusting the grid spacing.

Hot-wire anemometry is used to measure the wind parameters immediately upstream of the test section’s installation, including mean wind speed, turbulence intensity, and wind spectra. These turbulence characteristics are measured at various wind speeds that reflect the anticipated experimental circumstances. The u– and w-component spectra, as well as the longitudinal and vertical turbulence intensities, are used to display the results of the turbulence simulation.

The produced by the Engineering Sciences Data Unit standards, as well as the measured and target spectra, should be compared (ESDU). Smoke visualisation can be used during tests to study the vortex-shedding reaction to see how the airflow over the structure is changing.

The model forces and moments observed in the wind tunnel are used to generate the static load coefficients in the wind axis system as follows:

Drag coefficient, C_D = D/0.5ρV_m²B
Lift coefficient, C_L = L/0.5ρV_m²B
Pitching coefficient, C_M = M/0.5ρV_m²B

where D, L and M are the wind axis along- and across-wind force and pitching moment. V_m is the mean wind speed, ρ is the density of air (in the UK taken as 1.225 kg/m³) and B is the reference dimension which is usually the deck width.

Force coefficients must however be adjusted to take into account blockage effects brought on by the wind tunnel’s constriction. There are several approaches for computing correction factors, hence it is advised to consult specialised literature like ESDU 80024 (ESDU, 1980) to determine which method is best for the specific blockage ratio being taken into account.

The Reynolds number (R_e) is likely to be sensitive to sections with round section members or curved surfaces, therefore adjustments based on full-scale data or theoretical considerations may be required.

Analysis of Wind Tunnel Test Models

A whole bridge’s aerodynamic model is also known as an aero-elastic model. For structures with considerable wind-induced motions that have an impact on the aerodynamic forces and subsequently the dynamic response, aero-elastic tests are carried out. These structures usually behave as three dimensional structures.

By simulating the structure’s mass and stiffness distribution, models are created to replicate the essential dynamic properties. Particularly, the basic modes that control the structure’s dynamic response must be accurately modelled at the model-scale frequency. Transducers that measure displacement or acceleration are used to directly measure modal responses.

The modal responses measured are then used to calculate dynamic loads, or in the case of vertical structures, they are directly measured. The models for long-span bridges may be quite large because, typically, models for aeroelastic tests are constructed at either 1:200 or 1:100 scales.

Boundary layer wind tunnels (BLWT) with wide sections are therefore necessary for the experiments. A 1:100 scale aeroelastic replica of the Japanese Akashi-Kaiko bridge was constructed, having an overall length of about 40 m. Figure 19 depicts an instance of an aeroelastic model in a BLWT.

In a BLWT, turbulent boundary conditions are produced by placing a configuration of roughness elements across the wind tunnel’s floor and placing a two-dimensional barrier with square vortex-generating posts at the test section’s entry. These elements’ size, form, and distribution are planned to produce the specific turbulence characteristic needed for the testing.

Calibration of the wind-tunnel-generated turbulence attributes should be carried out with reference to the target spectrum, just like the section model testing. At crucial locations on the structure, displacement transducers and accelerometers measure the time histories of acceleration and acceleration during the testing.

Mean, root mean square (RMS) background, and RMS resonant components are calculated from the displacement time histories. The RMS resonant components of the movement are determined using acceleration time histories. In order to provide enough coverage of the design wind speed range representative of the bridge site, time records are recorded at small wind speed increments.

The Fourier transformation is used to analyse each time history in order to identify the spectrum, allowing for the isolation of narrowband responses that are indicative of structural resonance. By removing the peaks that correspond to the resonant components of displacement, background components can be extracted from displacement spectra.

The elements of a typical measured wind spectrum are shown below. The mean displacements plus or minus a sum of the standard deviations of the background and resonant components of displacement due to each mode are then used to determine the peak displacements.

The following expression is commonly used for calculating the measured peak displacements:

D_peak = D_mean ± √(g₀D_sdev)² + ∑(g_kD_modek)²

where;
D_peak = peak displacement
D_mean = mean displacement
D_sdev = standard deviation of background displacement
D_mode= RMS inertial displacement due to response in mode k
g₀ = peak factor to apply to background standard deviation (typically 3.4 to 3.5)
g_k= peak factor to apply to narrow band displacement due to motion in mode k.

The required peak factors across the modal responses are determined for the modal or narrow band displacements using the Davenport gust factor, which is given as:

g_k = √[2In(f_kT₀) + 0.577/2In(f_kT₀)]

where;
f_k = natural frequency of kth mode (in hertz)
T₀ = storm duration (typically 3600 seconds).

Correction factors may also be required within the calculation of the peak displacements in order to account for scaling inaccuracies necessitated by virtue of the model scale, e.g. cable diameters.

Wind Tunnel Testing of Cable Stays

Similar to the static testing of deck sections, testing of stay cables in wind tunnels is done to determine the drag coefficient and check for any potential divergent instabilities. Stay cables are often tested at full scale to guarantee Reynolds number similarity between model and prototype.

The surface characteristics of the prototype, such as any helical fillets or dimple patterns, should be precisely reflected in the cable models. Using calibrated spring balances and turntables in the test section’s ceiling and floor, cable models are dynamically mounted. This makes it possible to test the cable model in a variety of wind directions and wind-inclination angles. Spray heads positioned in the tunnel ceiling in some specialised wind tunnels can imitate a variety of rain conditions, making it possible to analyse rain-induced cable vibrations.

Best Hydraulic Cross-section of Roadside Drains

By

Anuoluwapo Kolade

-

July 8, 2022

Table of Contents

Drains are one of the components of the drainage system of residential areas and/or public infrastructures. Generically, roadside drains are structures used for collecting and conveying surface water to their discharge point, and in many cases, are artificial. Furthermore, drains are essential in road construction as stormwater ponding on pavement surfaces can cause premature pavement failure especially when unevaporated and undrained water seeps into the pavements.

The main objective of road drainage is to keep the road surface and foundation as dry as possible to maintain its stability. Thus, a good drainage system is essential for efficient highway transportation with minimum maintenance costs. Similarly, proper planning and design of road drainage systems are very important to prevent water from in-filtering the road surface, removing it from driving lanes, and carrying it away from the roadway.

pavement with no roadside drains — Water ponding on highway due to poor drainage

Common Roadside Drain Cross-sections

The common cross-sections used for roadside drains construction are rectangular and trapezoidal sections. Trapezoidal drains have sloped sides and can be formed by excavating in-situ materials. The sloped sides and channel bottom may require paving for protection, depending on the stability of the sides and the resistance of the in-situ materials to erosion. However, trapezoidal drains are now formed with reinforced concrete, which may be precast units or cast-in-place.

Similarly, rectangular drains have vertical or near-vertical sides, formed with reinforced concrete retaining walls, I-walls, or U-frame structures. The bottom of the channel may be paved or unpaved depending on the resistance of the in situ material to erosion.

Picture3 102454 — Typical cross-sections for roadside drain

Cross-section	Flow Area (A)	Wetted Perimeter (P)	Top width (T)	Hydraulic Radius (R)
Rectangular	by	b + 2y	b	A/P
Trapezoidal	y(b + yz)	b + 2y√(1 + z²)	b + 2yz	A/P

Geometric dimensions of drain cross-sections

Best Hydraulic Cross-section

The best cross-section for a drainage channel provides adequate hydraulic capacity at the minimum cost. Economic considerations for selecting the channel section include design and construction costs, right-of-way, required relocations, maintenance and operation.

A trapezoidal channel is usually the most economical channel when right-of-way is available. In contrast, a rectangular channel may be required for channels located in urban areas where the right-of-way is severely restricted or available at a high cost. Furthermore, site developments, existing geophysical site conditions, and performance or service requirements affect the selection of channel type and the resulting construction costs.

For discharge to be maximum or for the best hydraulic cross-section in a rectangular channel;
b = 2y and R = y/2

For trapezoidal channel;
T + 2yz = 2y√(z² + 1)
R = y/2
A = 3^(1/2) × y²
b = 2y/3^(1/2)
Z = 1/3^(1/2) = 0.577

Freeboard for Drains

In open channels or drains, freeboard is of great importance. The provision of freeboard in open drains ensures enough room for wave action and flow surges not to overtopping or overflow the drain. Many uncontrolled causes may create wave action or water surface fluctuation, but mostly because of changes in specific energy of the flow in a channel.

Therefore, a freeboard is provided to ensure the drain is not filled with water. In addition, the calculated discharge rate does not consider deposited solids and lack of maintenance, which will usually reduce the system’s efficiency.

A table for minimum freeboard for ditches is provided below;

Picture2 102454 1 — Freeboards for roadside drains

Worked Example

It was estimated from hydrological analysis that the peak discharge from a catchment area is 1.351m³/s. Using Manning’s equation, determine the best hydraulic cross-section for a rectangular and trapezoidal roadside drain. Other assumptions for computation are provided below.

Q = VA
V= 1/n x R^2/3 × S^1/2

Where:
Q = Discharge (m³/s)
A = Flow area (m²⁾
V = Allowable velocity (m/s)
R = Hydraulic radius (m)

Manning’s roughness coefficient (n) = 0.014
Bed slope of channel (S) = 3% (0.03)

Solution

From
Q = VA ——– (1)
Q = 1/n × R^2/3 × S^1/2 × A ——– (2)

(a) For the best hydraulic section for the rectangular drain;
b = 2y and R = y/2

Cross-sectional area of drain (A) = by
Substituting b = 2y into the area formula;
Cross-sectional area of drain (A) = (2y)y = 2y²

Substituting Q, n, R, S and A into Equation (2);

1.351= 1/0.014 × (y/2)^2/3 × 0.03^1/2 × 2y²
1.351= 1/0.014 × (y/2)^2/3 × 0.1732 × 2y²
1.351= 1/0.014 × y^2/3/1.5874 × 0.1732 × 2 × y²

y^8/3 = 0.087m

y = 0.087^3/8 = 0.4 m

Cross-sectional area of drain (A) = 2y² = 2 × 0.4² = 0.32m²
Width of drain (b) = 2y = 2 × 0.4 = 0.8 m

It is necessary to provide freeboard of at least 1 ft (0.3 m)
Therefore, drain depth = 0.4 + 0.3 = 0.7 m

(b) For the best hydraulic section for the trapezoidal drain;

Area of drain (A) = 3^1/2 × y²
R = y/2
Bottom width (b) = 2y/3^1/2
Z = 1/3^1/2 = 0.577

Substituting Q, n, R, S and A into Equation (2)

1.351 = 1/0.014 × (y/2)^2/3 × 0.03^1/2 × (3^1/2 × y²)
1.351 = 1/0.014 × (y/2)^2/3 × 0.1732 × 1.7321 x y²
1.351 = 1/0.014 × y^2/3/1.5874 × 0.1732 × 1.7321 x y²

y^8/3 = 0.1 m
y = 0.087^3/8 = 0.422 m (approximated to 0.45 m)

Cross-sectional area of drain (A) = 3^1/2 × y² = 3^1/2 × 0.422² = 0.308 m²
Depth of drain = 0.422 + 0.3 = 0.722 m (approximated to 0.75 m)
Bottom width of drain (b) = 2y/3^1/2 = 2 x 0.422/3^1/2 = 0.487 m (approximated to 0.5m)
Top width (T) = b + 2yz = 0.487 + (2 x 0.722 x 0.577) = 1.32 m (approximated to 1.35m)

Conclusion

The best hydraulic section of a roadside drain is characterized by the provision of maximum discharge with a given cross-sectional area. Other advantages than the hydraulic performance, for instance, for a given discharge rate, the best hydraulic section could guarantee the least cross-sectional area of the channel. Substantial savings could be made by reducing excavation and using fewer channel linings such as reinforced concrete.

Lastly, any roadside drain section selected should be large enough to permit the required discharge, as deep as required to provide a satisfactory outlet for both surface and subsurface drainage needs of the area served, and of a width-depth ratio and side slopes which will result in a stable channel which can be maintained in a satisfactory condition at a reasonable cost.

Reference(s)

King, H. W. and Brater, E. F. (1963). Handbook of Hydraulics, Fifth Edition. McGraw-Hill Book Company, Inc,, New York.

Cost Comparison of Solid and Ribbed Slabs

By

Ubani Obinna

-

June 30, 2022

Table of Contents

Cost is a major controlling factor in civil engineering construction projects. Different types of floor systems are adopted in reinforced concrete slab designs such as solid slabs, waffle slabs, ribbed slabs, flat slabs, etc. Each floor system has its advantages, applications, and cost implications in construction. This article aims to evaluate the cost comparison of solid and ribbed slabs.

The idea behind the adoption of the ribbed slab system is the need to reduce the volume of concrete in the tension zone of a concrete slab. Theoretically, the tensile strength of concrete is assumed to be zero during the structural design of flexural structural elements such as beams and slabs. By implication, all the tensile stresses from bending are assumed to be resisted by the reinforcements (see the stress block of flexural concrete sections in Figure 1).

**Figure 1:** Eurocode 2 concrete stress block

If concrete is assumed to do no work in the tension zone, it then makes economical sense to reduce the volume of concrete in that zone (bottom of the slab). To achieve this, beams of relatively shallow depths (ribs) are spaced at intervals to resist the flexural stresses due to the bending of the floor, with a thin topping (of about 50 mm). When this is done, a thick volume of concrete is no longer uniformly provided in the tensile zone of the concrete.

As the span of a floor increases, the thickness of the concrete and the quantity of reinforcements required to satisfy ultimate and serviceability limit state requirements also increase. However, by introducing ribbed slabs, longer spans can be economically spanned.

To reduce the cost and labour of constructing ribbed slabs, clay hollow pots, sandcrete blocks, or polystyrene are usually provided as infills between the ribs. While clay hollow pots and sandcrete blocks will improve the stiffness of the floor, the same, however, cannot be confidently said about polystyrenes.

7F6AEF68 146A 483E 83FB D44806FDE5E0 — **Figure 2:** Use of polystyrene as infill in ribbed slab

There are significant cost implications of adopting either solid or ribbed slabs. In a study by Ajema and Abeyo (2018), they observed that frames with solid slabs are more economical than frames with a ribbed slab when subjected to seismic action. In another study by Nassar and Al-Qasem (2020) on the cost of different slab systems, they observed that the flat slab system reduces the total cost of construction by 7% compared to the solid slab system, 4 % compared to the one-way ribbed slab system, and 3.33% compared to the two-way ribbed slab system.

Mashri et al (2020) compared the cost of constructing solid slabs and hollow block ribbed slabs and concluded that ribbed slabs are cheaper than solid slabs. In a study on the assessment of the cost difference between solid and hollow floors, Dosumu and Adenuga (2013) observed that the cost of in-situ solid slabs are higher than that of hollow slab provided the hollow slab is a one-way hollow floor and not a waffle floor.

However, it is important to note that the scenario on the issue of cost can vary depending on the size and geometry of the slab. For short one-way slabs, solid slabs may be cheaper than ribbed slabs, while in large-span two-way slabs, ribbed slabs may be cheaper than solid slabs.

ribbed slab construction — **Figure 3:** Ribbed slab construction

With the aid of design examples and quantity estimation, let us compare the cost of constructing a two-way slab of dimensions 5.225m x 7.426m that is discontinuous at all edges using ribbed slab and solid slab. The slab is to support a live load of 2.5 kN/m².

Design of Ribbed Slab

Reinforcement details
Characteristic yield strength of reinforcement; f_yk = 410 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 = 357 N/mm²

Concrete details
Concrete strength class; C25/30
Aggregate type; Quartzite
Aggregate adjustment factor – cl.3.1.3(2); AAF = 1.0
Characteristic compressive cylinder strength; f_ck = 25 N/mm²
Mean value of compressive cylinder strength; f_cm = f_ck + 8 N/mm² = 33 N/mm²
Mean value of axial tensile strength; f_ctm = 0.3 N/mm² × (f_ck)^2/3 = 2.6 N/mm²
Secant modulus of elasticity of concrete; E_cm = 22 kN/mm² × (f_cm/10)^0.3 × AAF = 31476 N/mm²

Design Information
Spacing of ribs = 450 mm
Topping = 50 mm
Rib width = 150 mm
Span = 5.225 m (simply supported)
Total depth of slab = 250 mm
Design live load q_k = 2.5 kN/m²Weight of finishes = 1.2 kN/m²
Partition allowance = 1.5 kN/m²

Load Analysis

Dead Load
Self-weight of topping = 24 × 0.05 × 0.45 = 0.54 kN/m
Self-weight of ribs = 24 × 0.15 × 0.2 = 0.72 kN/m
Weight of finishes = 1.2 × 0.45 = 0.54 kN/m
Partition allowance = 1.5 × 0.45 = 0.675 kN/m
Self-weight of heavy duty EPS = (0.156 × 0.2 × 0.45) = 0.014 kN/m
Total dead load per rib g_k = 2.489 kN/m

Live Load
Characteristic live load = 2.5 kN/m²
Total live load per rib = 2.5 × 0.45 = 1.125 kN/m

At ultimate limit state;
P_Ed = 1.35g_k + 1.5q_k = 1.35(2.489) + 1.5(1.125) = 5.047 kN/m

Flexural Design
Design bending moment; M_Ed = 17.2 kNm
Effective flange width; b_eff = 2 × b_eff,1 + b = 450 mm
Effective depth of tension reinforcement; d = 209 mm
K = M / (b_eff × d² × f_ck) = 0.035
K’ = 0.207
Lever arm; z = 199 mm
Depth of neutral axis; x = 2 × (d – z) / l = 26 mm

lx <= h_f – Compression block wholly within the depth of flange
K’ > K – No compression reinforcement is required

Area of tension reinforcement required; A_s,req = M / (f_yd × z) = 243 mm²
Tension reinforcement provided; 2Y16 (A_s,prov = 402 mm²)
Minimum area of reinforcement – exp.9.1N; A_s,min = max(0.26 × f_ctm / f_yk, 0.0013) × b × d = 51 mm²
Maximum area of reinforcement – cl.9.2.1.1(3); A_s,max = 0.04 × b × h = 1500 mm²
PASS – Area of reinforcement provided is greater than area of reinforcement required

Deflection control
Reference reinforcement ratio; ρ_m0 = (f_ck )^0.5 / 1000 = 0.00500
Required tension reinforcement ratio; ρ_m = A_s,req / (b_eff × d) = 0.00259
Required compression reinforcement ratio; ρ’_m = A_s2,req / (b_eff × d) = 0.00000

Structural system factor – Table 7.4N; K_b = 1.0
Basic allowable span to depth ratio ; span_to_depth_basic = K_b × [11 + 1.5 × (f_ck)^0.5 × ρ_m0 / ρ_m + 3.2 × (f_ck)^0.5 × (ρ_m0 / ρ_m – 1)^1.5] = 39.900

Reinforcement factor – exp.7.17; K_s = min(A_s,prov / A_s,req × 500 N/mm² / f_yk, 1.5) = 1.500
Flange width factor; F1 = if(b_eff / b > 3, 0.8, 1) = 1.000
Long span supporting brittle partition factor; F2 = 1.000

Allowable span to depth ratio; span_to_depth_allow = min(span_to_depth_basic × K_s × F1 × F2, 40 × K_b) = 40.000
Actual span to depth ratio; span_to_depth_actual = L_{m1_s1} / d = 25.000
PASS – Actual span to depth ratio is within the allowable limit

Shear Design
Angle of comp. shear strut for maximum shear; θ_max = 45 deg
Strength reduction factor – cl.6.2.3(3); v₁ = 0.6 × (1 – f_ck / 250 N/mm²) = 0.540
Compression chord coefficient – cl.6.2.3(3); a_cw = 1.00
Minimum area of shear reinforcement – exp.9.5N; A_sv,min = 0.08 N/mm² × b × (f_ck )^0.5 / f_yk = 146 mm²/m

Design shear force at support ; V_Ed,max = 13 kN
Min lever arm in shear zone; z = 199 mm
Maximum design shear resistance – exp.6.9; V_Rd,max = a_cw × b × z × v₁ × f_cwd / (cot(θ_max) + tan(θ_max)) = 134 kN
PASS – Design shear force at support is less than maximum design shear resistance

Design shear force at 209 mm from support; V_Ed = 12 kN
Design shear stress; v_Ed = V_Ed / (b × z) = 0.407 N/mm²

Area of shear reinforcement required – exp.6.8; A_sv,des = v_Ed × b / (f_yd × cot(θ)) = 69 mm²/m
Area of shear reinforcement required; A_sv,req = max(A_sv,min, A_sv,des) = 146 mm²/m
Shear reinforcement provided; 2Y 8 legs @ 150 c/c
Area of shear reinforcement provided; A_sv,prov = 670 mm²/m
PASS – Area of shear reinforcement provided exceeds minimum required

Maximum longitudinal spacing – exp.9.6N; s_vl,max = 0.75 × d = 157 mm
PASS – Longitudinal spacing of shear reinforcement provided is less than maximum

However, calculations have shown that shear reinforcements are not required since V_Ed (13 kN) is less than V_Rdc (25.2 kN). According to clause 6.2.1(4) of EN 1992-1-1:2004, when, on the basis of the design shear calculation, no shear reinforcement is required, minimum shear reinforcement should nevertheless be provided according to clause 9.2.2. The minimum shear reinforcement may be omitted in members such as slabs (solid, ribbed or hollow core slabs) where transverse redistribution of loads is possible.

Therefore to save cost, let us provide a triangular pattern link spaced at 300mm c/c, and 1Y12 (A_sprov = 113 mm²) at the top of the rib.

RIB BEAM ELEVATION VIEW — **Figure 5:** Rib beam elevation view

RIB BEAM SECTION — **Figure 6**: Ribbed slab section

Design of Solid Slab

RC slab design

In accordance with EN1992-1-1:2004 incorporating corrigendum January 2008 and the UK national annex

solid slab — **Figure 7:** Solid slab panel

Slab definition
Type of slab; Two way spanning with restrained edges
Overall slab depth; h = 200 mm
Shorter effective span of panel; l_x = 5225 mm
Longer effective span of panel; l_y = 7426 mm
Support conditions; Four edges discontinuous

Loading
Characteristic permanent action; G_k = 6.5 kN/m²
Characteristic variable action; Q_k = 2.5 kN/m²
Partial factor for permanent action; γ_G = 1.35
Partial factor for variable action; γ_Q = 1.50
Quasi-permanent value of variable action; ψ₂ = 0.30
Design ultimate load; q = γ_G × G_k + γ_Q × Q_k = 12.5 kN/m²
Quasi-permanent load; q_SLS = 1.0 × G_k + ψ₂ × Q_k = 7.2 kN/m²

Concrete properties
Concrete strength class; C25/30
Characteristic cylinder strength; f_ck = 25 N/mm²
Partial factor (Table 2.1N); γ_C = 1.50
Compressive strength factor (cl. 3.1.6); a_cc = 0.85
Design compressive strength (cl. 3.1.6); f_cd = 14.2 N/mm²
Mean axial tensile strength (Table 3.1); f_ctm = 0.30 N/mm² × (f_ck )^2/3 = 2.6 N/mm²
Maximum aggregate size; d_g = 20 mm

Reinforcement properties
Characteristic yield strength; f_yk = 410 N/mm²
Partial factor (Table 2.1N); γ_S = 1.15
Design yield strength (fig. 3.8);f_yd = f_yk / γ_S = 356.5 N/mm²

Concrete cover to reinforcement
Nominal cover to outer bottom reinforcement; c_{nom_b} = 25 mm
Fire resistance period to bottom of slab; R_btm = 60 min
Axial distance to bottom reinft (Table 5.8); a_{fi_b} = 10 mm
Min. btm cover requirement with regard to bond; c_{min,b_b} = 12 mm
Reinforcement fabrication; Not subject to QA system
Cover allowance for deviation; Dc_dev = 10 mm
Min. required nominal cover to bottom reinft; c_{nom_b_min} = 22.0 mm
PASS – There is sufficient cover to the bottom reinforcement

Reinforcement design at midspan in short span direction (cl.6.1)

Bending moment coefficient; b_{sx_p} = 0.0881
Design bending moment; M_{x_p} = b_{sx_p} × q × l_x² = 29.9 kNm/m
Reinforcement provided; 12 mm dia. bars at 200 mm centres
Area provided; A_{sx_p} = 565 mm²/m
Effective depth to tension reinforcement; d_{x_p} = h – c_{nom_b} – f_{x_p} / 2 = 169.0 mm

K factor; K = M_{x_p} / (b × d_{x_p}² × f_ck) = 0.042
Redistribution ratio; δ = 1.0
K’ factor; K’ = 0.598 × d – 0.18 × d² – 0.21 = 0.208
K < K’ – Compression reinforcement is not required

Lever arm; z = min(0.95 × d_{x_p}, d_{x_p}/2 × (1 + √(1 – 3.53 × K))) = 160.5 mm
Area of reinforcement required for bending; A_{sx_p_m} = M_{x_p} / (f_yd × z) = 523 mm²/m
Minimum area of reinforcement required; A_{sx_p_min} = max(0.26 × (f_ctm/f_yk) × b × d_{x_p}, 0.0013 × b × d_{x_p}) = 275 mm²/m
Area of reinforcement required; A_{sx_p_req} = max(A_{sx_p_m}, A_{sx_p_min}) = 523 mm²/m

Check reinforcement spacing
Reinforcement service stress; s_{sx_p} = (f_yk / g_S) × min((A_{sx_p_m}/A_{sx_p}), 1.0) × q_SLS / q = 190.7 N/mm²
Maximum allowable spacing (Table 7.3N); s_{max_x_p} = 262 mm
Actual bar spacing; s_{x_p} = 200 mm
PASS – The reinforcement spacing is acceptable

Reinforcement design at midspan in long span direction (cl.6.1)
Bending moment coefficient; b_{sy_p} = 0.0560
Design bending moment; M_{y_p} = b_{sy_p} × q × l_x² = 19.0 kNm/m
Reinforcement provided; 12 mm dia. bars at 250 mm centres
Area provided; A_{sy_p} = 452 mm²/m
Effective depth to tension reinforcement; d_{y_p} = h – c_{nom_b} – f_{x_p} – f_{y_p} / 2 = 157.0 mm
K factor; K = M_{y_p} / (b × d_{y_p}² × f_ck) = 0.031
Redistribution ratio; d = 1.0
K’ factor; K’ = 0.598 × d – 0.18 × d² – 0.21 = 0.208
K < K’ – Compression reinforcement is not required

Lever arm; z = min(0.95 × d_{y_p}, d_{y_p}/2 × (1 + √(1 – 3.53 × K))) = 149.2 mm

Area of reinforcement required for bending; A_{sy_p_m} = M_{y_p} / (f_yd × z) = 358 mm²/m
Minimum area of reinforcement required; A_{sy_p_min} = max(0.26 × (f_ctm/f_yk) × b × d_{y_p}, 0.0013 × b × d_{y_p}) = 255 mm²/m
Area of reinforcement required; A_{sy_p_req} = max(A_{sy_p_m}, A_{sy_p_min}) = 358 mm²/m
PASS – Area of reinforcement provided exceeds area required

Check reinforcement spacing
Reinforcement service stress; s_{sy_p} = (f_yk / γ_S) × min((A_{sy_p_m}/A_{sy_p}), 1.0) × q_SLS / q = 163.1 N/mm²
Maximum allowable spacing (Table 7.3N); s_{max_y_p} = 296 mm
Actual bar spacing; s_{y_p} = 250 mm
PASS – The reinforcement spacing is acceptable

Basic span-to-depth deflection ratio check (cl. 7.4.2)
Reference reinforcement ratio; ρ₀ = (f_ck)^0.5 / 1000 = 0.0050
Required tension reinforcement ratio; ρ = max(0.0035, A_{sx_p_req} / (b × d_{x_p})) = 0.0035
Required compression reinforcement ratio; ρ’ = A_{scx_p_req} / (b × d_{x_p}) = 0.0000
Structural system factor (Table 7.4N); K_d = 1.0

Basic limit span-to-depth ratio (Exp. 7.16);
ratio_{lim_x_bas} = K_d × [11 +1.5 × (f_ck)^0.5 × ρ₀/ρ + 3.2 × (f_ck)^0.5 × (ρ₀/ρ -1)^1.5] = 26.20

Mod span-to-depth ratio limit;
ratio_{lim_x} = min(40 × K_d, min(1.5, (500 N/mm²/ f_yk) × (A_{sx_p} / A_{sx_p_m})) × ratio_{lim_x_bas}) = 34.54
Actual span-to-eff. depth ratio; ratio_{act_x} = l_x / d_{x_p} = 30.92
PASS – Actual span-to-effective depth ratio is acceptable

Reinforcement summary
Midspan in short span direction; 12 mm dia. bars at 200 mm centres B1
Midspan in long span direction; 12 mm dia. bars at 250 mm centres B2
Discontinuous support in short span direction; 12 mm dia. bars at 200 mm centres B1
Discontinuous support in long span direction; 12 mm dia. bars at 250 mm centres B2

solid slab reinforcement detailing — **Figure 8:** Solid slab reinforcement detailing (plan)

SOLID SLAB SECTION — **Figure 9:** Solid slab section

Cost Comparison of Solid and Ribbed Slabs

In this section, we are going to consider the cost of constructing ribbed slab and the cost of constructing solid slabs (considering the cost of materials only). In this case, the cost of labour is assumed to be directly proportional to the quantity of materials.

Cost analysis of solid slab

Concrete
Volume of concrete required for the slab = 5.45 × 7.65 × 0.2 = 8.3385 m³
Current unit cost of concrete materials = ₦55,000/m³
Cost of concrete materials = 8.3385 × 55000 = ₦ 458,618

Quantity of steel required
Bar mark 1 = 37 × 7.645 × 0.888 = 251.184 kg
Bar mark 2 = 21 × 10.485 × 0.888 = 195.524 kg
Bar mark 3 = 12 × 4.225× 0.617 = 31.28 kg
Bar mark 3 = 8 × 6.625 × 0.617 = 32.701kg
Total = 510.689 kg

Unit cost of reinforcement = ₦ 450/kg
Cost of reinforcement = 510.689 × 450 = ₦ 229,810

Formwork Required
Soffit of slab = 36 m² (treated as a constant)

Cost of constructing solid slab = ₦ 229,810 + ₦ 458,618 = ₦ 688,428

Cost Analysis of Ribbed Slab

Concrete
Volume of concrete required for the topping = 5.45 × 7.65 × 0.05 = 2.084 m³
Volume of concrete required for the ribs = 12 × 0.2 × 0.15 × 5 = 1.8 m³
Total volume of concrete = 3.884 m³
Cost of concrete materials = 3.884 × 55,000 = ₦ 213,620

Quantity of steel required
Bar mark 1 = 2 × 12 × 5.7 × 1.579 = 216 kg
Bar mark 2 = 1 × 12 × 5.8 × 0.888 = 61.8 kg
Bar mark 3 (triangular links) = 17 × 12 × 0.612 × 0.395 = 49.314 kg
Total = 327.114 kg
Cost of reinforcement = 327.114 × 450 = ₦ 147,205

BRC Mesh for Topping (A142) – 41.7 m²
Unit cost of BRC mesh = ₦ 1580/m²
Cost of BRC mesh = 41.7 × 1580 = ₦ 65,886

Total cost of reinforcement works = 147,205 + 65,886 = ₦ 213,620

Clay hollow pot/Sandcrete Blocks/Polystyrene
Number of block units required = 240 units
Unit price of hollow blocks = NGN 400 per unit
Cost of hollow blocks = 240 × 400 = ₦ 96,000

Formwork Required
Soffit of slab = 36 m² (treated as a constant)

Cost of constructing ribbed slab = ₦ 213,620 + ₦ 213,091 + ₦ 96,000 = ₦ 522,711

The Table for comparison is shown below;

Material	Solid Slab	Ribbed Slab	Percentage Reduction
Concrete	₦ 458,618	₦ 213,620	53.42%
Reinforcement	₦ 229,810	₦ 213,620	7.04%
Hollow Pots/Blocks	–	₦ 96,000	–
Total	₦ 688,428	₦ 522,711	24.07%

Conclusion

From the analysis of the two-way slab carried out, it can be seen that the volume of concrete required for a ribbed slab is 53.42% less than that required for a solid slab. Furthermore, the reinforcement required in the ribbed slab is 7.04% less than that required for a solid slab. If the same type of formwork is adopted (completely flat soffit supported with props), and if the cost of labour is directly related to the quantity of materials, then the adoption of the ribbed slab is expected to save cost by about 24.07% compared to solid slab.

References

[1] Ajema D. and Abeyo A. (2018): Cost Comparison between Frames with Solid Slab and Ribbed Slab using HCB under Seismic Loading. International Research Journal of Engineering and Technology 05(01):109-116
[2] Dosumu O. S. and Adenuga O. A.(2013): Assessment of Cost Variation in Solid and Hollow Floor Construction in Lagos State. Journal of Design and Built Environment 13(1):1-11
[3] Mashri M., Al-Ghosni K., Abdulrahman A., Ismaeil M., Abdussalam A. and Elbasir O. M. M. (2020):Design and cost comparison of the Solid Slabs and Hollow Block Slabs. GSJ 8(1):110-118 www.globalscientificjournal.com
[4] Reema R. Nassar 1, Imad A. Al-Qasem 2 (2020): Comparative Cost Study for A residential Building Using Different Types of Floor System. International Journal of Engineering Research and Technology 13(8): 1983-1991

Economic Analysis of Public Projects

By

Anuoluwapo Kolade

-

June 27, 2022

Table of Contents

The management of public projects includes making good decisions, and these decisions usually involve choosing between alternatives. Therefore, every decision-maker needs to answer questions like what public projects to implement, how to begin, how many project units are required, and where and when to execute these projects. Economic analysis helps the decision-maker to determine answers to these questions.

For every public project conceived, there must be very clear goals, benefits, and expectations from the project during and when it is completed. Complying with these objectives with minimum cost and maximum benefits is crucial because public projects are usually executed using public funds and tax revenues. Therefore, executing public projects involves many choices among physically feasible alternatives. Some examples of public projects are roads, bridges, hospitals, dams, leisure centres, etc

Furthermore, each choice among alternatives should be made on an economic basis. In addition, each alternative should be expressed in terms of money units before making the final choice. Money units can only measure the cost and benefits of different project alternatives.

Public parks are good examples of public projects

Public vs Private Projects

The economic analysis of public projects uses a different approach for evaluation compared to private projects. The focus is usually on benefits instead of profits, as in the case of private projects. In addition, the scope is the society; that is, it covers the interest of the owner and the society. Therefore, benefits becomes the performance criterion instead of profitability.

Economic Analysis of Public Projects

The best project is the one which gives the highest benefit-cost (B/C) ratios as it would give the maximum return on investment (ROI). On the contrary, public projects are usually executed to achieve maximum benefits and not maximum B/C ratios. However, projects should be economically viable and give some minimum return rate (RoR).

It has been observed in most public projects that benefit increase with the size of the project. However, project cost also increases with the increase in project size. Furthermore, a stage is reached beyond which an increase in project size may not yield minimum attractive returns. Therefore, the size of the project is fixed at this stage.

Key Test

It has always been easy to determine costs but determining full benefits remains challenging. Therefore, before economic analysis can proceed, efforts must be made to list possible benefits and estimated values. A decision-maker should accept projects as viable (economically acceptable) if benefits outweigh costs. In addition, the projects should be implemented, subject to the availability of funds.

Benefit-Cost Ratio

The benefit-cost ratio (B/C) compares the present value of all benefits to the present value of all costs. The ratio is merely used to see if a unit of costs (say, #1) will return at least the same unit (#1) in benefits. B/C ratios do not themselves provide enough information to make an economic choice. Therefore, additional calculations are necessary to use B/C ratios as a sound basis for project formulation.

Conditions for Viability

Accept projects as viable (economically acceptable) if benefits outweighs cost, that is, B/C > 1.0
Reject project if B/C < 1.0

Steps in Checking the Viability of Projects

Carryout incremental analysis on the projects, that is, arrange projects in increasing order of costs.
Put the projects in a portfolio from best to worst, noting cumulative cost values.
Stop when funding limit is reached.
Fund only the acceptable projects within limits.
If there is any extra fund, invest in some other venture or keep money in bank.

Worked Example

The information about 5 proposed public projects are given below.

Project Proposal	Annual Benefits (millions)	Annual Cost (Million)
A	12.00	10.30
B	16.80	12.40
C	22.00	16.53
D	25.80	22.70
E	27.20	26.83

a) Determine the projects that are viable.
b) Determine the best project.
c) If there is a budget of 40 million, determine the projects to adopt for execution.

Solution

a) All the projects are viable because their B/C > 1.0
b) The best project is B because it has the highest ∆AB/∆AC value of 2.286
c) With a budget of 40 million, projects A, B and C with overall cost of 39.23 million should be adopted for execution

Conclusion

The main purpose of economic analysis is to help select projects that contribute to the welfare of the people. Therefore, economic analysis is most useful when used early in the project cycle to catch bad projects and bad project components. However, if used at the end of the project cycle, economic analysis can only help decide whether or not to proceed with a project.

One of the most important steps in project evaluation is the consideration of alternatives throughout the project cycle, from identification through appraisal. Many important choices are made early when alternatives are rejected or retained for a more detailed study. Comparing mutually exclusive options is one of the principal reasons for applying economic analysis from the early stages of the project cycle.

Finally, an economical design is that project which gives the greatest excess benefits over cost. Moreover, a project should be evaluated in terms of RoR. The RoR (net gain or loss over a specified period) on investment may be estimated by calculating the cost and benefits of the project. Projects may also be ranked in the merit of RoR. Thus, the decision to go ahead can be made based on the minimum RoR and B/C ratio.

Reference

[1] Belli, P., Anderson, J., Barnum, H., Dixon, J. and Tan, J-P (1997), Handbook of Economic Analysis of Investment Operations, Operations Policy Department: Learning and Leadership Center. https://www.unisdr.org

Application of Genetic Algorithm to the Design of Pile Foundations

By

Ubani Obinna

-

June 22, 2022

Machine learning and deep learning techniques have been widely applied in the design of civil engineering structures in recent years. This is mainly due to the development of advanced computational capacity of computers in handling big data. Researchers from Ryerson University, Toronto, Canada have applied the concept of genetic algorithm to the design of pile foundations using Standard Penetration Test (SPT) data. The findings were published in the journal, Soils and Foundations.

Genetic Algorithm (GA) is a method of optimization based on Darwin’s theory of evolution. In nature, chromosomes give an organism its characteristics, and organisms can adapt and evolve to survive and thrive in their environments through reproduction. A GA represents the problem domain as a string or matrix termed a chromosome, then evolves the chromosome through reproduction mechanisms to find a solution.

One of the oldest geotechnical investigation procedures, the standard penetration test (SPT), is still widely utilised around the world. SPT is also used to empirically test the ultimate load-carrying capacity (Q_u) for the design of pile foundations, despite its significant shortcomings. There are two types of SPT-based design methods: direct and indirect.

Indirect approaches use SPT blow counts (N-values) to determine the shear strength of soils, such as the frictional angle (ϕ) or undrained shear strength (C_u), and then apply these parameters to the design. Unfortunately, correlating ϕ and C_u with SPT N-values is difficult. Some relationships between ϕ and N-values for sands have been proposed from various parts of the world.

According to the authors (Jesswein and Lui, 2022), it may be preferable to directly correlate SPT N-values to pile shaft (Q_s) and tip resistance (Q_p) given the uncertainties and additional inaccuracies caused by indirect correlations. However, many direct approaches may produce biased Q_u forecasts. This is mainly because they were based on simple linear regressions or trial-and-error procedures with the N-value as the only input, and they likely ignore or misrepresent several factors that influence pile-soil interaction, such as effective stress, excess pore pressure, repacking of soil grains during pile driving, and pile load-transfer response.

Procedure of Standard Penetration Test Q640 — **Fig. 2**. Standard penetration test

This inspired the authors to consider machine learning (ML) approach. ML algorithms are more efficient at regressing huge datasets and capturing nonlinear interactions between numerous variables than standard regression approaches. Furthermore, because the systems evaluate alternative solutions based on a set of criteria, which typically includes an objective or fitness function, they do not require prior knowledge of the problem domain. This eliminates the need to assume a relationship between variables while performing regression.

Due to its “blackbox” nature, the most popular ML method, artificial neural networks (ANN), does not allow the relationships to be stated in a practical formula. As a result, a genetic algorithm (GA) may be a better ML technique since it may provide useful, easy-to-understand functions that represent generic trends.

A total of 72 axial compression tests on driven steel piles were gathered from the literature and the Ministry of Transportation of Ontario (MTO) for the study. The 72 piles selected were divided into two groups based on their measurements.

Driven Steel Piles Nucor — **Fig. 3**. Typical driven steel piles

The load-transfer distribution data were used to extract both the unit shaft and tip resistances in the first group. The GA developed the design approach by correlating these unit resistances with soil data and pile parameters. The new design approach was validated and compared to three existing SPT-based design methods in the second group, where only total capacity measurements were provided. In the study both N₆₀ and (N₁)₆₀ are represented by N_cr.

The piles that were researched were chosen based on the following criteria:

Sufficient information was available on the subsurface ground conditions, particularly the soil classifications and SPT N-values, at the test site;
Non-organic soils were found along the pile length;
The pile types were either open-ended pipe (OEP), closed-ended pipe (CEP), and H piles;
The pile width or diameter and embedment length were available; and
The load–displacement curves were reported.

Based on a total of 72 full-scale static load tests reported in the study, the new SPT-based formula provides an unbiased prediction with a higher level of accuracy. The proposed design formulas are summarized below:

q_s= N_crσ’ / 2.55L ≤ 100 kPa (for cohesive soils) ——— (1)
(Training R² = 0.9; Testing R² = 0.87; All R² = 0.89)

q_s= 2(N_cr+ N_crσ’/L)/L ≤ 250 kPa (for cohesionless soils) ——— (2)
(Training R² = 0.87; Testing R² = 0.92; All R² = 0.89)

q_p = N_crp(σ_t‘/7.57 + 82) ≤ 1200 kPa (for all soils)——— (3)
(Training R² = 0.94; Testing R² = 0.89; All R² = 0.92)

Where;
L = Length of pile
σ’ = Effective overburden pressure at the depth considered
N_cr = Corrected SPT Number
q_s = shaft resistance
q_p = Tip resistance

According to the authors, Equations (1) and (2) can consider the changing ground conditions by using the characteristic values for Ncr and σ’ within a soil layer.

comparison 1 — **Fig. 4.** Comparison between Measured and Predicted Unit Side Resistance by Functions from the GA.(Jesswein and Lui, 2022)

Comparison 2 — **Fig. 5.** Comparison between Measured and Predicted Unit Tip Resistance
from the GA (Jesswein and Lui, 2022)

Although the proposed formulas provide better predictions than existing design methods, the authors insist that engineering judgement and knowledge of local ground conditions are critical components in correctly designing a pile. The quality of the produced model is dependent on the correctness of the input variables, particularly N_cr, because GA regressions are data driven. Because regressed models may not perform well under extrapolation, the range of the examined variables should be addressed while applying the provided formulas. The proposed approach can be used on a wide range of cohesive and cohesionless soils, but organic soils and soft clays are not recommended.

Article Source:
Jesswein M. and Liu J. (2022): Using a genetic algorithm to develop a pile design method. Soils and Foundations 62(2022)101175. https://doi.org/10.1016/j.sandf.2022.101175

This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/).

Fatigue Verification in RC Bridges (Eurocode 2)

By

Ubani Obinna

-

June 21, 2022

Deck slabs of bridges are likely to be among the elements that are influenced by fatigue verification calculations the most. This is due to the high live load to dead load ratio that these slabs have. Tests, on the other hand, have shown that the actual stress ranges in the reinforcement in these are significantly lower than what is suggested by the standard elastic calculations. As a result of this, the NA to BS EN 1992-2 identifies situations in which fatigue assessment is not necessary and gives regulations that are on the safe side.

The failure of a structure due to repeated loadings that are smaller than a single static force that exceeds the material’s strength is known as Fatigue. Fatigue occurs when a material fails due to direct tension or compression, torsion, bending, or a combination of these actions.

Since reinforced concrete is a composite material, it can fail due to fatigue in a variety of ways. Failure is frequently the result of a variety of reasons, and failure types can vary quite significantly. The concrete, the reinforcement, and the bond between the materials might all fail locally.

Compressive fatigue failure in reinforced concrete is referred to as ductile because cracks might appear in the concrete long before the structure falls. Because the crack propagation rate in the reinforcement at the end is rather fast, tensile fatigue failure in reinforced concrete has a more brittle behaviour.

According to clause 6.8.1 of BS EN 1992-2, a fatigue verification is generally not necessary for the following structures and structural elements:

Footbridges, with the exception of structural components very sensitive to wind action.
Buried arch and frame structures with a minimum earth cover of 1 m and 1.5 m for road and railway bridges.
Foundations.
Piers and columns which are not rigidly connected to superstructures.
Retaining walls of embankments for roads and railways.
Abutments of road and railway bridges which are not rigidly connected to superstructures, except the slabs of hollow abutments.
Prestressing and reinforcing steel, in regions where, under the frequent load combination of actions and P_k only compressive stresses occur at the extreme concrete fibres.

Furthermore, in the case of road bridges, fatigue verification is not required for the local effects of wheel loads given directly to a slab that spans across beams or webs, provided that the following conditions are met:

The slab does not contain welded reinforcement or reinforcement couplers.
The clear span to overall depth ratio of the slab does not exceed 18.
The slab acts compositely with its supporting beams or webs.
Either:
- the slab also acts compositely with transverse diaphragms; or
- the width of the slab perpendicular to its span exceeds three times its clear span.

In Eurocode there are two alternative methods by which fatigue verification can be calculated for bridges;

the λ-Coefficient Method and
the Cumulative Damage Method.

Both approaches take into account the loading during the course of a structure’s lifetime. The λ-Coefficient Method is a simplified method that uses a single load model amplified by several coefficients. The Cumulative Damage Method is a complicated model that takes into account the load history in greater detail. The λ-Coefficient Method simply evaluates if the building meets the code’s requirements, whereas the Cumulative Damage Method derives a fatigue damage factor that indicates the structure’s actual damage in relation to the design fatigue life.

Internal Forces and Stresses for Fatigue Verification

According to clause 6.8.2 of BS EN 1992-2, the calculation of stress for fatigue verification must be based on the assumption of cracked cross-sections, which must be done while ignoring the tensile strength of the concrete and satisfying compatibility of strains. The influence of the varied bond behaviour of prestressing steel and reinforcing steel must be taken into account by increasing the stress range in the reinforcing steel computed under the assumption of a perfect bond by the factor, η, which is given by:

η = (A_s + A_p)/[(A_s + A_p) × √ξ(ϕ_s/ϕ_p)]

where:
A_s= area of reinforcing steel
A_p = area of prestressing tendon or tendons
ϕ_s= largest diameter of reinforcement
ϕ_p = diameter or equivalent diameter of prestressing steel
= 1.6 √A_p for bundles
= 1.75 ϕ_wire for single 7-wire strands where ϕ_wire is the wire diameter
= 1.20 ϕ_wire for single 3-wire strands where ϕ_wireis the wire diameter
ξ = ratio of bond strength between bonded tendons and ribbed steel in concrete. The value is subject to the relevant European Technical Approval. In the absence of this the values given in Table 1 may be used.

Prestressing steel	ξ (pre-tensioned)	ξ (Bonded, post-tensioned) ≤ C50/60	ξ (Bonded, post-tensioned) ≥ C70/85
Smooth bars and wires	Not Applicable	0.3	0.15
Strands	0.6	0.5	0.25
Indented wires	0.7	0.6	0.30
Ribbed bars	0.8	0.7	0.50

Table 1: Ratio of bond strength, ξ, between tendons and reinforcing steel (Source: Table 6.2 BS EN 1992-1-1)

For intermediate values between C50/60 and C70/85, interpolation may be used.

Verification of Concrete under Compression or Shear

According to clause 7.6.2 of PD 6687-2, it is doubtful that National Authorities will have the S–N curves that are necessary to carry out a fatigue verification of concrete while it is being subjected to compression or shear. In the absence of such data, the simplified technique that is outlined in BS EN 1992-2, Annex NN may be utilised for railway bridges; however, there is no equivalent alternative available for highway bridges.

σ_c,max/f_cd,fat ≤ 0.5 + 0.45(σ_c,min/f_cd,fat)

where;
σ_c,max = maximum compressive stress at a fibre under the frequent load combination (compression measured positive)
σ_c,min= minimum compressive stress at the same fibre where σ_c,max occurs. If σ_c,min is a tensile stress, then σ_c,min should be taken as zero
f_cd,fat= design fatigue strength of concrete = 0.85 β_cc(t₀) f_cd (1 – f_ck/250)
f_cd = f_ck/ 1.5
β_cc(t₀) = coefficient for concrete strength at first load application = exp{s[1 – (28/t₀)^0.5]}

where;
t₀ = time of the start of the cyclic loading on concrete in days
s = 0.2 for cement of strength Classes CEM 42.5R, CEM 52.5N and CEM 52.5R (Class R)
= 0.25 for cement of strength Classes CEM 32.5R, CEM 42.5 (Class N)
= 0.38 for cement of strength Class CEM 32.5 (Class S)

The maximum value for the ratio σ_c,max / f_cd,fat is given in Table 2.

Concrete Strength	σ_c,max / f_cd,fat
f_ck ≤ 50 MPa	≤ 0.9
fck > 50 MPa	≤ 0.8

Table 2: Values for σ_c,max / f_cd,fat

S-N curve for fatigue verification — (a) Truss model for beams with shear reinforcement and (b) characteristic S-N curve for reinforcing steel

For members not requiring design shear reinforcement for the ultimate limit state it may be assumed that the concrete resists fatigue due to shear effects where the following apply:

Limiting Stress Range for Reinforcement under Tension

Adequate fatigue resistance may be assumed for reinforcing bars under tension if the stress range under the frequent cyclic load combined with the basic combination does not exceed 70 MPa for unwelded bars and 35 MPa for welded bars.

For UK highway bridges, the values in Tables 12.3 and 12.4 may be used for straight reinforcement. These are based on bars conforming to BS 4449. For bars not conforming to BS 4449, the rules for bars > 16 mm diameter should be used for all sizes unless the ranges for bars ≤ 16 mm diameter can be justified.

Span (m)	Adjacent spans loaded (Bars ≤ 16 mm)	Adjacent spans loaded (Bars > 16 mm)	Alternate spans loaded (Bars ≤ 16 mm)	Alternate spans loaded (Bars > 16 mm)
≤ 3.5	150	115	210	160
5	125	95	175	135
10	110	85	175	135
20	110	85	140	110
30 -50	90	70	110	85
100	115	90	135	105
≥ 200	190	145	200	155

Table 3: Limiting stress ranges – longitudinal bending for unwelded reinforcing bars in road bridges, MPa

For Tables 3 and 4, intermediate values can be interpolated.

Span (m)	(Bars ≤ 16 mm)	(Bars > 16 mm)
≤ 3.5	210	160
5	120	90
≥ 10	70	55

Table 4: Limiting stress ranges – transverse bending for unwelded reinforcing bars in road bridges, MPa

Full fatigue checks are to be carried out using the ‘damage equivalent stress range’ approach, following the instructions given in Annex NN of BS EN 1992-2, if the stress range limitations exceed the values specified in Tables 3 and 4 (e.g. for reinforcement over the pier). The stress ranges are derived using the ‘Fatigue Load Model 3’, which simulates a four-axle vehicle weighing 48 tonnes total. This weight is increased to 84 tonnes for intermediate supports and 67 tonnes for other locations in Annex NN.

Elastic Settlement of Pile Groups

By

Ubani Obinna

-

June 20, 2022

pile group settlement — Feature Image Source: Ata et al (2015)

The immediate or elastic settlement of pile groups, denoted by ρ_i and the long-term consolidation settlement, denoted by ρ_c, both contribute to the total settling of a pile group in clay. Equation (1) can be used as a general rule to calculate the elastic settlement for a flexible foundation at the level of the ground surface;

ρ_i = q_n × 2B × (1 – v²)/E_u× I_p ——– (1)

where ρ_i is the settlement at the centre of the flexible loaded area, q_n is the net foundation pressure, B is the width of an equivalent foundation flexible raft, v is the undrained Poisson’s ratio for clay (generally taken as equal to 0.5), I_p is an influence factor, and E_u is the deformation modulus for the undrained loading conditions.

The values of I_p are dependent on the ratio H/B of the depth of the compressible soil layer to the width of the pile group, as well as the ratio L/B of the length of the group to its width. The values of I_p and, consequently, the immediate settlement of a surface foundation can be obtained through the use of curves that were established by Steinbrenner and published by Terzaghi (1943).

These curves can be found in Figure 1 of this article. To calculate the elastic settlement of pile groups, the use of Fox’s correction curves (1948) is required in order to obtain the immediate settlement of a raft foundation that is equivalent to the pile group.

steibrenners influence factor chart — **Figure 1:** Values of Steinbrenner’s influence factor Ip (for v of 0.5)

To obtain the average immediate settlement of a foundation at a depth D below the surface where the deformation modulus is reasonably constant with depth, it will be found to be more convenient to use the influence factors of Christian and Carrier (1978). In general, this will be the case where the deformation modulus is reasonably constant with depth.

Average settlement ρ_i = μ_iμ₀q_nB/E_u ——– (2)

In the equation (2), it was assumed that Poisson’s ratio was equal to 0.5. Figure 2 shows the factors μ_i and μ₀, both of which are associated with the depth of the equivalent raft, the thickness of the compressible soil layer, and the length/width ratio of the equivalent raft foundation.

**Figure 2**: Influence factors for calculating immediate settlements of flexible foundations of width B at depth D below ground surface (after Christian and Carrier(5.5)).

The value of the deformation modulus E_u can be determined by analysing the stress-strain curve that is produced when compressive loading is applied to the soil while the conditions are undrained. Only when the stress level is relatively low (shown by the line AB in Figure 3) does a curve of this type show purely elastic behaviour, which requires the use of a modulus of elasticity.

It is possible that the immediate settlement will be underestimated due to the fact that (Young’s modulus) corresponds to the straight-line portion. The standard procedure is to draw a secant AC to the stress-strain curve that corresponds to a compressive stress that is equal to the net foundation pressure at the base of the equivalent block foundation.

The secant AD can be drawn at a compressive stress of 1.5 or any other suitable multiple of the foundation pressure for a more conservative approach. Following this, one can calculate the deformation modulus E_u, as shown in Figure 3.

As a result of sample disturbance, unreasonably low modulus values are obtained from stress-strain curves produced from conventional unconfined or triaxial compression tests in the laboratory. These results are not representative of the material’s true behaviour. Plate bearing tests carried out in boreholes or trial pits, as well as field testing carried out with a pressuremeter or Camkometer, provide the most reliable data for calculating modulus values.

Another possibility is that E_u is connected to the clay’s undrained shearing strength c_u. According to Butler (1975), the relationship E_u = 400c_u for London clay is a reasonable compromise between divergent data representing, on the one hand, the relationship established from laboratory testing, and on the other hand, the observations of the settlement of full-scale structures.

Recent years have seen the development of apparatus for obtaining piston-driven tube samples of stiff clays, as well as improvements in methods for coring weak rocks. Therefore, samples that have been subjected to a relatively minor disturbance can be provided for laboratory testing to obtain modulus values. There is apparatus available that can measure very small strains during uniaxial or triaxial compression tests, which can then be utilised to derive small strain modulus values.

In layered soils with different values of the deformation modulus E_u in each layer or in soils which show a progressively increasing modulus with increases in depth, the strata below the base of the equivalent raft are divided into a number of representative horizontal layers, and an average value of E_u is assigned to each layer.

This ensures that the values of the deformation modulus E_u are consistent across the entire profile of the soil. The values for the dimensions L and B in Figure 2 are derived from the hypothesis that the load is distributed across the surface of each layer at an angle of thirty degrees with respect to the edges of the equivalent raft (Figure 4). After that, the total settlement of the piled foundation can be calculated as the sum of the average settlements for each soil layer that were determined using Equation 2.

Elastic Settlement of Pile Groups — **Figure 4:** Load distribution beneath pile group in layered soil formation

The assumption that the deformation modulus remains the same with depth is the basis for the influence values presented in Figure 1. Calculations that are based on a constant modulus, on the other hand, give inflated estimates of the amount of settlement that will occur.

This is because the modulus increases with depth in the majority of natural soil and rock formations. Butler (1974) developed a method for settlement calculations for the conditions of a deformation modulus increasing linearly with depth within a layer of finite thickness. This method was used for determining the amount that a layer would settle. The following equation will give you the value of the modulus at a depth z below the level of the foundation:

E_u = E_f(1 + kz/B) ——– (3)

and

ρ_i = q_nBI’_p/E_u ——– (4)

where E_f represents the modulus of deformation at foundation level (the base of the equivalent raft), and ρ_i represents the settlement at the corner of the loaded area in the equation. In order to calculate the value of k, first plot the measured values of E_u against depth, and then draw a straight line through the points that are plotted. This will give you the values you need to substitute into equation 3.

In situations in which a plot of undrained shear strength versus depth has been obtained, the E_u vs depth line can be derived from the empirical relationships given earlier in this paragraph.

After you have determined k, you can use Butler’s curves, which are presented in Figure 5, to determine the appropriate factor for I’_p. These are for different ratios of L/B at the level of the equivalent raft, and they are applicable for a compressible layer thickness that is not greater than 9B. The curves have been constructed on the basis of the assumption that an undrained condition will have a Poisson’s ratio of 0.5; this is so that the load can be applied immediately.

Butler influence factor curves — **Figure 5:** Values of the influence factor for deformation modules increasing linearly with depth and Poisson’s ratio of 0.5 (after Butler (1974))

In situations in which the pile group covers a large area and is, as a result, relatively flexible, it is possible that it will be necessary to calculate the settlements at the corners in addition to those in the area’s centre.

Consolidation Settlement of Pile Groups

The results of oedometer tests carried out on clay samples in the laboratory are used as the basis for the calculation of the consolidation settlement ρ_c. The pressure-to-voids ratio curves that were obtained from these tests are what are used to calculate the volume compressibility coefficient, denoted by the symbol m_v.

It may be difficult to obtain satisfactory undisturbed samples for oedometer testing in hard glacial tills or in rocks that have been highly weathered and weakened by weathering. If the results of standard penetration tests are available, then the values of m_v (as well as c_u) can be obtained from the empirical relationships established by Stroud (1975) and shown in Figure 6.

CURVES — **Figure 6:** Relationship between mass shear strength, modulus of volume compressibility, plasticity index and standard penetration test N-values (after Stroud (1975)) (a) N-value vs. undrained shear strength (b) N-value vs. modulus of volume compressibility

Once a representative value of m_v has been obtained for each soil layer that is being stressed by the pile group, the oedometer settlement ρ_oed for this layer at the centre of the loaded area can be calculated using the equation;

ρ_oed = μ_dm_v× σ_z × H ——– (5)

where μ_d is a depth factor, σ_z is the average effective vertical stress imposed on the soil layer due to the net foundation pressure q_n at the base of the equivalent raft foundation and H is the thickness of the soil layer. The depth factor is obtained from Fox’s correction curves.

The oedometer settlement must now be corrected to obtain the field value of the consolidation settlement. The correction is made by applying a ‘geological factor’ μ_g to the oedometer settlement, where;

ρ_c = μ_g× ρ_oed ——– (6)

Published values of μ_g have been based on comparisons of the settlement of actual structures with computations made from laboratory oedometer tests. Values established by Skempton and Bjerrum (1975) are shown in Table 1.

Type of clay	μ_g value
Very sensitive clays (soft alluvial, estuarine and marine clays)	1.0 – 1.2
Normally-consolidated clays	0.7 – 1.0
Over-consolidated clays (London clay, Weald, Kimmeridge, Oxford and Lias clays)	0.5 – 0.7
Heavily over-consolidated clays (unweathered glacial till, Keuper Marl)	0.2 – 0.5

Table 1: Value of geological factor μ_g

After this, the total settlement of the pile group is determined by adding together the immediate and consolidation settlements that were calculated for each individual layer. A gradual decrease in compressibility with depth is a typical example of this phenomenon. When this occurs, the stressed zone beneath the pile group is segmented into a number of distinct horizontal layers.

The value of the modulus of elasticity (m_v) for each of these layers is obtained by plotting the m_v value against the depth, which is based on the results of the laboratory oedometer tests. At the level where the vertical stress has decreased to one tenth of q_n, the level at which the base of the lowest layer is determined to be is chosen.

When calculating the total consolidation settlements for each layer, the depth factor, denoted by d, is multiplied by that total. It does not apply to the immediate settlement if that settlement has already been computed based on the factors shown in Figure 5.

If a pile group is topped by a deep rigid cap or if it is used to support a rigid superstructure, then the pile group can be considered to be equivalent to a rigid block foundation that has a uniform settlement. This is because both of these conditions ensure that the pile group remains in its original position.

To calculate the value of the latter, a “rigidity factor” is applied to the consolidation settlement that was obtained from the equivalent flexible raft foundation. The immediate settlement can be calculated using Equation 2. This immediate settlement is the same as the average settlement that is given by a rigid foundation. The value of 0.8 for the rigidity factor is widely accepted as being appropriate.

Thus;
Settlement of rigid pile group/Settlement of flexible pile group = 0.8

It is possible, for the purposes of this condition, to consider a pile group that is composed of a number of small clusters or individual piles connected by ground beams or by a flexible ground floor slab to be equivalent to a flexible raft foundation at depth.

This is the case if the pile group is connected by ground beams. According to the calculations described above, the consolidation settlements that occurred at the corners of the piled area make up approximately one-half of the settlement that occurred in the centre of the group.

The use of Equation 1, with the substitution of a deformation modulus obtained for loading under drained conditions, is yet another method for estimating the total settlement of a structure that is resting on an over-consolidated clay.

This modulus has been given the name E_v‘, and it can be found at the bottom of Equation 1, where it stands in place of E_u. It roughly corresponds to the value of 1/m_v. When used to calculate consolidation settlements, the equation does not adhere to strict validity standards because it assumes a material that is both homogenous and elastic.

However, when applied to over-consolidated clays for which the settlements are relatively small, it has been observed through experience that the method gives predictions that are reasonably reliable. The success of utilising the method is dependent on the collection of sufficient data correlating the observed settlements of structures with the determinations of from plate loading tests and laboratory tests on good undisturbed samples of clay. This is necessary for a successful outcome.

In his analysis of the settlement of structures on over-consolidated clays, Butler (1974) related E_v‘ to the undrained cohesion c_u and determined that the relationship for London clay is E_v‘ = 130C_u.

In the settlement analysis of a group of piles, it is better to adopt an approach that is more rational, which is to consider immediate settlements and consolidation settlements separately. This appropriately takes into account the effects that time has had on the location as well as its geological history. The prediction of consolidation settlements based on oedometer tests conducted in the laboratory has been found to lead to reasonably accurate results, provided that a sufficient number of good undisturbed samples have been obtained at the site investigation stage.

The adoption of the method that is based on the total settlement deformation modulus is contingent upon the collection of adequate observational data, first regarding the relationship between the undrained shearing strength and the deformation modulus, and secondly regarding the actual settlement of structures from which the relationships can be checked.

The adoption of this method is dependent upon the collection of adequate observational data. It is highly unlikely that accurate results can be obtained from triaxial compression tests carried out in the lab, so any attempt to do so is likely to end in failure. The modulus can be determined most accurately using the E_u/c_u and relationship formulas, which need to be derived from plate bearing tests that have been competently carried out and field observations of settlement.

The reader is directed to a report written by Padfield and Sharrock (1983) for CIRIA that contains a general discussion on the subject of the settlement of foundations on clays. According to what they have found, the immediate settlement is approximately equal to 0.5 to 0.6 times the oedometer settlement, while the consolidation settlement, is approximately equal to 0.4 to 0.5 times the oedometer settlement. This is true for stiff overconsolidated clays. When it comes to normally consolidated soft clays, the immediate settlement is roughly equivalent to 0.1 times the oedometer settlement, and the consolidation settlement is roughly equivalent to the oedometer settlement.

The steps in making a settlement analysis of a pile group in, or transmitting stress to, a cohesive soil can be summarized as follows.

For the required length of pile, and form of pile bearing (i.e. friction pile or end-bearing pile), draw the equivalent flexible raft foundation represented by the group.
From the results of field or laboratory tests assign values to E_u and m_v for each soil layer significantly stressed by the equivalent raft.
Calculate the immediate settlement of ρ_i of each soil layer using equation 2, and assuming a spread of load of 30° from the vertical to obtain q_n at the surface of each layer (Figure 4). Alternatively calculate on the assumption of a linearly increasing modulus.
Calculate the consolidation settlement ρ_c for each soil layer from Equations 5 and 6, using relevant charts to obtain the vertical stress at the centre of each layer.
Apply a rigidity factor to obtain the average settlement for a rigid pile group.

The consolidation settlement calculated as described above is the final settlement after a period of some months or years after the completion of loading. It is rarely necessary to calculate the movement at intermediate times, i.e. to establish the time settlement curve, since in most cases the movement is virtually complete after a period of a very few years and it is only the final settlement which is of interest to the structural engineer. If time effects are of significance, however, the procedure for obtaining the time-settlement curve can be obtained from standard works of reference on soil mechanics.

References

TERZAGHI, K (1943). Theoretical Soil Mechanics, John Wiley, New York, p. 425.
FOX, E.N. (1948). The mean elastic settlement of a uniformly-loaded area at a depth below the ground surface, Proceedings of the 2nd International Conference, ISSMFE, Rotterdam, Vol. 1, pp. 129–32.
CHRISTIAN, J.T. and CARRIER, W.D. (1978). Janbu, Bjerrum and Kjaernsli’s chart reinterpreted, Canadian Geotechnical Journal, Vol. 15, pp. 123–8.
BUTLER, F.G. (1974). General report and state-of-the-art review, Session 3, Proceedings of the Conference on Settlement of Structures, Cambridge, Pentech Press, London, pp. 531–78.
STROUD, M.A. (1975) The standard penetration test in insensitive clays, Proceedings of the European Symposium on Penetration Testing, Stockholm, Vol. 2, pp. 367–75.
SKEMPTON, A.W. and BJERRUM, L. A (1957). contribution to the settlement analysis of foundations on clay , Geotechnique, Vol. 7, No. 4, pp. 168–78.
PADFIELD, C.J. and SHARROCK, M.J. (1983). Settlement of structures on clay soils, Construction Industry Research and Information Association (CIRIA), Special Publication 27, 1983.

Feature Image:
Ata A., Badrawi E., Nabil M. (2015): Numerical analysis of unconnected piled raft with cushion. Ain Shams Engineering Journal, 6(2):421-428 https://doi.org/10.1016/j.asej.2014.11.002.

Specifications for Sandcrete Blocks

Mix Ratios for Sandcrete Blocks

Materials Used in the Production of Sandcrete Blocks

Cement

Fine Aggregates

Tests for Fine Aggregates

Sieve Analysis

Specific Gravity

Water

Mechanical Properties of Sandcrete Blocks

Bulk Density

Water Absorption

Compressive Strength

Production of Sandcrete Blocks

Hand Ramming Compaction Moulds

Manual Tamping Compaction Frame

Motorised Vibration Machine

Quality Control Tips for Block Production

Example on Partition Load Modelling

Analysis Results

References

Worked Example on the Design of Cantilever Stairs

Load Analysis

Flexural Design

Crack control

Deflection control

Shear Design

Design of the Spine Beam

Flexural Design

Shear Design

Objectives of Wind Tunnel Testing

Modelling for Wind Tunnel Testing

Analysis of Wind Tunnel Test Models

Wind Tunnel Testing of Cable Stays

Common Roadside Drain Cross-sections

Best Hydraulic Cross-section

Freeboard for Drains

Worked Example

Conclusion

Reference(s)

Design of Ribbed Slab

Design of Solid Slab

Cost Comparison of Solid and Ribbed Slabs

Cost analysis of solid slab

Cost Analysis of Ribbed Slab

Conclusion

References

Public vs Private Projects

Economic Analysis of Public Projects

Key Test

Benefit-Cost Ratio

Conditions for Viability

Steps in Checking the Viability of Projects

Worked Example

Conclusion

Reference

Internal Forces and Stresses for Fatigue Verification

Verification of Concrete under Compression or Shear

Limiting Stress Range for Reinforcement under Tension

Consolidation Settlement of Pile Groups

References

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