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Structural Design of Flat Slabs to Eurocode 2

By
Ubani Obinna Uzodimma
-

Flat slabs can offer economical solutions to wider floor spans in a reinforced concrete building. Flat slabs are slabs that are supported directly by columns without floor beams. The columns may or may not have drops. There are many advantages of flat slab such as increased head room, easier flow of mechanical and electrical services, ease in construction of form work, faster construction etc.


Flat slabs can be analysed using the frame equivalent method, or by using the simplified analysis method which involves the use of coefficients. On the other hand, finite element analysis using computer programs can be used.  In the Eurocodes, the analysis  of flat slab is the same as that recommended in BS 8110. According to clause 3.7.2.7 of BS 8110, the simplified method can be used for flat slabs that the lateral stability is not dependent on the slab and columns provided that the following conditions are met;

(1) The slab is loaded with a single load case of all the panels loaded with maximum ultimate load. This means that the following requirements must be satisfied (clause 3.5.2.3);
(a) The area of each bay exceeds 30 m2
(b) The ratio of the characteristic imposed load to the dead load does not exceed 1.25
(c) The imposed load does not exceed 5 kN/m2

(2) There are at least three rows of approximately equal panels in the direction  being considered
(3)  Moments at supports can be reduced by 0.15Fhc

Table 2Bof 2Bcoefficients

In this post, we are going to analyse panel 6 of the flat slab general arrangement that is given above with the following data;

Thickness of slab = 230 mm (governed by deflection requirements)
fck = 30 Mpa
fyk = 500 Mpa
Imposed load = 4 kN/m2
Concrete cover = 25 mm

Load Analysis

Permanent Actions
Self weight = 0.23 × 25 = 5.75 kN/m2
Finishes = 1.2 kN/m2
Partition allowance = 1.5 kN/m2
Total gk = 8.45 kN/m2

Variable Actions
Imposed floor load qk = 4 kN/m2

Total load on the floor F = 1.35gk + 1.5qk = 17.41 kN/m2
F = 17.41 × 6.5 × 6.5 = 735.57 kN

A little consideration will show that the panel under consideration met the conditions for use of coefficients;

division 2Bof 2Bpanels 2Binto 2Bstrips

For reference on how the panel was divided into middle and column strips, see Figure 3.12 of BS 8110-1:1997.


Bending Moments (In the longitudinal direction)
(1) Centre of the interior span
M = 0.063 × 735.57 × 6.5 = 301.217 kNm

The width of the middle strip is = 6.5 – 1.6 – 1.6 = 3.3 m

Therefore, the middle strip positive moment = 0.45 × 301.217 = 135.547 kNm
The column strip positive moment is given by = 0.55 × 301.217 = 165.669 kNm

(2)  Interior support
M = -0.063 × 735.57 × 6.5 = 301.217 kNm

Therefore, the middle strip moment = 0.25 × 301.217 = 75.304 kNm
The column strip negative moment is given by = 0.75 × 301.217 = 225.912 kNm

Bending Moments (In the lateral direction)
(1) Centre of the interior span
M = 0.063 × 735.57 × 6.5 = 301.217 kNm

The width of the middle strip is = 6.5 – 1.6 – 1.6 = 3.3 m

Therefore, the middle strip positive moment = 0.45 × 301.217 = 135.547 kNm
The column strip positive moment is given by = 0.55 × 301.217 = 165.669 kNm

(2)  Interior support
M = -0.086 × 735.57 × 6.5 = 411.183 kNm

Therefore, the middle strip moment = 0.25 × 411.183 = 102.795 kNm
The column strip negative moment is given by = 0.75 × 411.183 = 308.387 kNm

The reinforcement design is done as usual…


Simplified Design of Welded Connections

By
Ubani Obinna Uzodimma
-

Welded connections are one of the most fundamental and versatile methods for joining steel members in structural applications. Their inherent strength, reliability, and cost-effectiveness make them a popular choice across various industries, from building construction and bridges to shipbuilding and machinery.

However, ensuring the safety and efficiency of these connections requires careful design and adherence to established codes of practice. In this article, we review the design of welded connections, exploring their key design principles, common types, and considerations, with Eurocode 3 (EN 1993) serving as our primary reference for best practices.

Fundamental Principles of Welded Connections

At the heart of welded connection design lies the understanding of stress distribution and material behaviour. Welds introduce inherent discontinuities within the structure, potentially creating stress concentrations. Therefore, selecting the appropriate weld type, size, and location becomes crucial. Eurocode 3 provides detailed guidance on various weld types, such as butt welds, fillet welds, and groove welds, each with its own load-carrying capacity and suitability for specific scenarios.

Material Properties

Understanding the base metal and filler metal properties is equally important. Eurocode 3 classifies different steel grades based on their yield strength, ultimate tensile strength, and ductility. The chosen weld metal must be compatible with the base metal, ensuring adequate strength and avoiding excessive brittleness.

Load Analysis and Design Methods

Accurately determining the loads acting on the connection is paramount. Eurocode 3 outlines various load combinations including dead loads, imposed loads, wind loads, and seismic loads. Once the loads are defined, different design methods can be employed, such as the ultimate limit state (ULS) and the serviceability limit state (SLS). The ULS ensures sufficient strength to prevent weld failure, while the SLS addresses factors like excessive deformations and fatigue resistance.

Common Types of Welded Connections

  • Butt Welds: Join two members along their full cross-section, often used for plates and pipes.
  • Fillet Welds: Connect two members at an angle, suitable for various member combinations.
  • Groove Welds: Similar to butt welds, but with a prepared groove in one or both members for greater weld penetration.
  • Tee and Corner Joints: Connect members forming an L or T shape, requiring careful consideration of stress concentrations.

Eurocode 3 Requirements

Eurocode 3 provides comprehensive design rules and guidance for welded connections. Key aspects include:

  • Weld throat thickness: Defines the effective depth of the weld, crucial for calculating its load-carrying capacity.
  • Electrode selection: Specifies suitable electrodes based on material properties and desired weld characteristics.
  • Joint geometry: Provides recommendations for joint preparation, weld size, and edge distance to ensure proper stress distribution.
  • Fatigue design: Addresses welds subjected to cyclic loading, requiring specific design considerations.
  • Fabrication and quality control: Outlines requirements for welding procedures, welder qualifications, and inspection methods.

Additional Considerations

  • Residual stresses: Welding introduces residual stresses within the structure, which can affect the overall behavior and need to be accounted for in the design.
  • Distortion and shrinkage: The heating and cooling process during welding can cause distortion and shrinkage of the members, requiring measures to control these effects.
  • Fatigue resistance: Connections experiencing repeated loading require careful design for fatigue resistance, potentially involving specific weld details and material selection.

Welded Connection Design Example

In our previous post, we were able to present an example of the design of fillet welds for truss members using two different approaches. In this article, we are going to simplify it further by presenting an example of how the strength of fillet welds can be easily verified, according to the requirements of Eurocode 3.

Solved Example
A 150 mm diameter hollow steel pipe is welded to a plate of thickness 25 mm of grade S275. If the welding side is 12 mm, determine the shear resistance of the welded connection (see the connection example with the picture below).

Typical welded connection construction
Typical welded connection construction

Solution
Steel grade = S275
Thickness of plate (t) = 25 mm < 40 mm, therefore fu = 430 Mpa (Table 3.1 BS EN 1993-1-1)

For steel grade S275, βw = 0.85 (Table 4.1, BS EN 1993-1-1)

The throat of the weld
a = (√2/2) × welding side
a = 0.7071t
a = 0.7071(12) = 8.485 mm

Design weld shear strength
fvwd = (fu/√3)/(βwγm2)
fvwd = (430/√3)/(0.85 × 1.25) = 233.65 Mpa

Weld resistance per length
Fw,Ed =   fvwda
Fw,Ed =  233.65 × 8.485 = 1982.58 N/mm = 1.982 kN/mm

Since the weld is done right round the perimeter of the pipe, the weld length L = πd = π × 150 = 471.24 mm

The total weld resistance = Fw,Ed × L = 1.982 × 471.24 = 934 kN

Therefore, the shear force at the base must not exceed 934 kN

Conclusion

Designing welded connections effectively requires a combination of engineering principles, material understanding, and adherence to established codes like Eurocode 3. This comprehensive guide serves as a starting point, highlighting key aspects and considerations. However, for real-world applications, seeking the expertise of qualified engineers and strictly adhering to relevant code provisions remain essential for ensuring safe and reliable structural performance.

Review on the Quantity of Cement Required to Lay 9 inches (230 mm) Blocks

By
Ubani Obinna Uzodimma
-

In our previous post, we were able to establish that we need 0.03 m3 of mortar to lay one square metre of 9 inches hollow block wall. If you missed the post, or wish to know how we arrived at that value, follow the link below.


How to Calculate the Quantity of Mortar for Laying Blocks

The post has some set backs in the fact that it was not simplified to the most basic level. In this post, I wish to clearly establish a value that can be appreciated on site for all construction purposes.

The typical mix ratio for mortar on construction sites in Nigeria is 1:5 (1 bag of 50 kg cement to 10 head pans of sand).

Therefore, the quantity of cement required per cubic metres is given by;
1/6 x 1440 = 240 kg/m3
For wet mortar, use 1.30 x 240 = 312 kg/m3

NB: We are using a factor of 1.30 to account for shrinkage in volume when water is added to the dry mix.

Therefore, the quantity of cement required per m3 of mortar = 312/50 = 6.24 bags of cement


Quantity of mortar required per 1 m2 (10 blocks) of wall = 0.03 m3
For 5 m2 (50 blocks) of wall = 0.15 m3

Therefore, if 6.24 bags if required in 1 m3 of mortar, then 0.936 bags of cement is required in 0.15 m3 of mortar.

By inference, we can say that we need 1 bag of cement to lay 50 pieces of hollow 9 inches block (i.e 5 m2 of wall).

Design of Waffle Slabs

By
Ubani Obinna
-

Waffle slabs are reinforced concrete slabs that are normally employed for large-span construction. They can be described as the equivalent of 2-way spanning solid slabs but with a ribbed slab system spanning in both directions. In waffle slabs, the grids run in two directions to provide support for the floor.

Waffle slabs have a flat top and a surface resembling a grid made of concrete joists on the bottom. After the concrete has set and cured, the waffle moulds are removed to create the grid. Due to the rigidity of the waffle slab, this form of structure is suggested for buildings like manufacturing plants and laboratories that require little vibration.

Additionally, waffle slabs are utilized in structures like theatres and train stations that call for expansive open areas. Waffle slabs may be more expensive than other types of slabs because they require intricate formwork, but depending on the project and the amount of concrete used, they may be less expensive to construct.

waffle slab design

The design concept of waffle slabs is similar to that of ribbed slabs, with the difference being that waffle slabs have ribs spanning in both directions, and the coefficients used for analysing the slab are similar to those used for a two-way restrained slab. Waffle slabs are supported on beams or columns, where the support zones are made to be uniformly thick.

Design Example of Waffle Slab

A church building has a square grid of 7.5m x 7.5m of waffle slab, and is to support an imposed load of 5 kN/m2, design an interior panel of the waffle slab, and a supporting beam. fck = 30 Mpa, fyk = 500 Mpa. The design will be done according to Eurocode 2 (EN 1992-1-1:2004).

GENERAL 2BARRANGEMENT 2BOF 2BWAFFLE 2BSLAB

For details of waffle mould, see details of technical data sheet.

Solution
From basic span/effective depth ratio of 26, the trial depth of slab to be adopted = 7500/26 = 288 mm

From the table, let us select the following mould details;
Mould height = 225 mm
Top layer thickness = 75 mm
Total height of slab = 300 mm
Average rib width = 176 mm

Load Analysis
Self weight of concrete = 5.2 kN/m2 (see technical data sheet)
Finishes = 1.2 kN/m2
Partition allowance = 1.5 kN/m2
Total permanent action = 7.9 kN/m2

Imposed load = 5 kN/m2

At ultimate limit state = 1.35gk + 1.5qk = 1.35(7.9) + 1.5(5) = 18.165 kN/m2

Load per rib = 0.9 × 18.165 = 16.35 kN/m2

Analysis of the slab
Treating as all sides fixed:
k = Ly/Lx = 1.0
Span coefficient = +0.024
Support coefficients = -0.032

Design of the span
Designing the span as a T-beam;
MEd =  0.024 × 16.35 × 7.52 = 22.07 kN.m

Effective depth (d) = h – Cnom – ϕ/2 – ϕlinks
Assuming ϕ12 mm bars will be employed for the main bars, and ϕ10 mm bars for the stirrups (links)
d = 300 – 25 – (12/2) -10 = 259 mm

k = MEd/(fckbd2) = (22.07 × 106)/(30 × 500 × 2592) = 0.0219
Since k < 0.167, no compression reinforcement required

z = d[0.5+ √(0.25 – 0.882k)]
k’ = 0.0219
z = d[0.5+ √((0.25 – 0.882(0.0219))] = 0.95d = 246.05 mm

Depth to neutral axis x = 2.5(d – z) = 2.5(259 – 246.05) = 32.375 mm < 1.25hf (93.75 mm)
Therefore, we design rib as a rectangular section

Area of tension reinforcement As1 = MEd / (0.87fyk z)
As1 = MEd / (0.87fyk z) = (22.07 × 106) / (0.87 × 500 × 0.95 × 259) = 206 mm2

Provide 2H12  Bot (ASprov = 226 mm2)

Check for deflection
ρ = As,req /bd = 206 / (900 × 259) = 0.0008837
ρ0 = reference reinforcement ratio = 10-3√(fck) = 10-3√(30) = 0.00547
Since  ρ ≤ ρ0;
L/d = k [11 + 1.5√(fck) ρ0/ρ + 3.2√(fck) (ρ0/ρ – 1)(3⁄2)]
k = 1.5 (continuous system)
L/d = 1.5 [11 + 1.5√(30) × (0.00547/0.0008837) + 3.2√(30) × [(0.00547 / 0.0008837) – 1](3⁄2)]
L/d = 1.5[11 + 50.855 + 207.227] = 269.082

βs = (500 Asprov)/(fyk Asreq) = (500 × 226) / (500 × 206) = 1.09

beff/bw = 900/176 = 5.11

Therefore multiply basic length/effective depth ratio by 0.8

Therefore limiting L/d = 1.09 × 0.8  × 269.082 = 234.639
Actual L/d = 7500/259 = 28.96

Since Actual L/d (28.96) < Limiting L/d (234.639), deflection is satisfactory.

Design of Supports;
MEd = 0.032 × 16.35 × 7.52 = 29.43 kN.m

k = MEd/(fckbd2) = (29.43 × 106)/(30 × 176 × 2592) = 0.083
Since k < 0.167, no compression reinforcement required

z = d[0.5+ √(0.25 – 0.882k)]
k = 0.083
z = d[0.5+ √((0.25 – 0.882(0.083))] = 0.92d

Area of tension reinforcement As1 = MEd / (0.87fyk z)
As1 = MEd / (0.87fyk z) = (29.43 × 106) / (0.87 × 500 × 0.92 × 259) = 283 mm2
Provide 3H12 TOP (ASprov = 339 mm2)

Shear Design 
Maximum shear force in the rib 
VEd = 0.33 × 16.35 × 7.5 = 40.466 kN/m

VRd,c = [CRd,c.k.(100ρ1 fck)(1/3) + k1.σcp]bw.d ≥ (Vmin + k1.σcp) bw.d

CRd,c = 0.18/γc = 0.18/1.5 = 0.12
k = 1 + √(200/d) = 1 + √(200/259) = 1.88 < 2.0, therefore, k = 1.88
Vmin = 0.035k(3/2) fck0.5
Vmin = 0.035 × (1.88)1.5 × 300.5 = 0.494 N/mm2
ρ1 = As/bd = 339/(176 × 259) = 0.00743 < 0.02; Therefore take 0.00743

VRd,c = [0.12 × 1.88 (100 × 0.00743 × 30 )(1/3)] × 176 × 259 = 28941.534 N = 228.94 kN

Since VRd,c (28.94 kN) < VEd (40.466 kN), shear reinforcement is required.
The compression capacity of the compression strut (VRd,max) assuming θ = 21.8° (cot θ = 2.5)

VRd,max = (bw.z.v1.fcd) / (cot⁡θ + tanθ)
V1 = 0.6(1 – fck/250) = 0.6(1 – 30/250) = 0.528
fcd = (αcc fck) / γc = (0.85 × 30) / 1.5 = 17 N/mm2
Let z = 0.9d

VRd,max = [(176 × 0.9 × 259 × 0.528 × 17) / (2.5 + 0.4)] × 10-3 = 126.981 kN

Since < VEd < VRd,c < VRd,max

Hence Asw / S = VEd / (0.87 Fyk z cot θ) = 40466 / (0.87 × 500 × 0.9 × 259 × 2.5 ) = 0.159

Minimum shear reinforcement;
Asw / S = ρw,min × bw × sinα (α = 90° for vertical links)
ρw,min = (0.08 × √(fck)) / fyk = (0.08 × √30) / 500 = 0.000876
Asw/Smin = 0.000876 × 176 × 1 = 0.154
Maximum spacing of shear links = 0.75d = 0.75 × 259 = 194.25 mm

Provide H8mm @ 175 mm c/c as shear links (Asw / S = 0.574)

Slab Topping
A142 BRC Mesh can be provided or H8 @ 250mm c/c

Watch out for design of supporting beams…


8 Top Civil Engineering Questions (Part 2)

By
Ubani Obinna Uzodimma
-

(1) For the structure shown in the figure below, which of the following is the most likely bending moment diagram considering linear first order elastic analysis?

FB IMG 1556340232949

(A)
(B)
(C)


(2) For the structure shown in the figure below, what is the most likely shear force diagram?

FB IMG 1556340286893

(A)
(B)
(C)

(3) A building is generally divided into how many parts?

(A) Four
(B) Three
(C) Two
(D) Six


(4) What is the recommended maximum spacing of shear links in a beam according to British and European codes?

(A) d
(B) 1.5d
(C) 0.75d
(D) Width of beam

(5) What is the recommended minimum compressive strength for water retaining structures according to the Eurocodes?

(A) C20/25
(B) C25/30
(C) C30/37
(D) C35/45

(6) The process through which water reacts with free cement after hydration reaction in a concrete structure to seal up cracks is known as?

(A) Post-hydration reaction
(B) Consolidation healing
(C) Efflorescence
(D) Autogenous healing

(7) How many percentage of 28 days strength is expected of concrete after 7 days?

(A) 50%
(B) 65%
(C) 30%
(D) 75%

(8) A beam of span L is subjected to a uniformly distributed load w with support conditions pinned and fixed at either ends. What is the bending moment at the fixed end?

(A) wL2/8
(B) wL2/12
(C) wL2/16
(C) wL2/24


Use of Polystyrene in Ribbed Slabs: Structural Design Example

By
Ubani Obinna Uzodimma
-

For long-span slabs in a building subjected to a low live load, ribbed slabs are more economical than reinforced concrete solid slabs. In ribbed slab construction, a reduction in the volume of concrete is achieved by removing some of the concrete below the neutral axis of the section, under the assumption that the tensile strength of concrete is negligible.

In effect, closely spaced reinforced concrete beams (more like joists) spanning in one direction are used to support thinner slab topping in ribbed slabs. The closely spaced beams are the major load-bearing elements of the floor and resist major bending and shear forces. The space between the rib beams may or may not be filled.

image 26
Figure 1: Ribbed slab

Therefore, the two major forms of construction of ribbed slabs are;

(1) Ribbed slab without permanent blocks
(2) Ribbed slab with permanent blocks

The major advantage of a ribbed slab with permanent blocks is the reduction in the quantity of formwork required, given the fact that the strength of the blocks is not usually assumed to contribute to the strength of the slab in design.

In Nigeria, clay hollow pots and sandcrete blocks have been successfully used, but on the rise is the use of polystyrene. Polystyrene offers the advantage of lightweight, improved fire resistance, and economy, at the expense of less rigidity compared to clay hollow pots or blocks.

image 27
Figure 2: Expanded Polystyrene (EPS)

In a generic sense, the use of polystyrene is to encourage savings in formwork costs, and may not be assumed to contribute to the strength of the floor in any way.

polystyrene in floor construction
Figure 3: Polystyrene in floor construction

In this article, we are going to consider the design of a floor (see Figure 4) subjected to the following loading conditions;

Structural 2Bscheme 2Bfor 2Bribbed 2Bslab
Figure 4: Preliminary Structural Arrangement

Finishes – 1.2 kN/m2
Partition allowance – 1.5 kN/m2
Imposed load – 2.5 kN/m2

fck = 30 Mpa, fyk = 500 Mpa, Concrete cover = 25 mm

The structural engineer decided the scheme the structural layout as shown in Figure 5. The slab is made solid close to the supports, to enable adequate transmission of shear forces.

general 2Barrangement 2Bof 2Bribbed 2Bslab 1
Figure 5: Layout of the rib beams

From a basic span/effective depth ratio of 26, let us choose a trial depth;
6000/26 = 231 mm

Let us try a total depth of 250 mm

SECTION 2BTHROUGH 2BRIBBED 2BSLAB 1
Figure 6: Transverse Section through the Slab
RIBBED 2BSLAB
Figure 7: Longitudinal Section through the slab

Load Analysis

For 500mm c/c rib spacing;

Permanent Actions
Weight of topping: 0.075 × 25 × 0.5 = 0.9375 kN/m
Weight of ribs: 0.15 × 0.175 × 25 = 0.656 kN/m
Weight of finishes: 1.2 × 0.5 = 0.6 kN/m
Partition allowance: 1.5 × 0.5 = 0.75 kN/m
Weight of heavy-duty EPS (16 kg/m3) = 0.156 × 0.175 × 0.5 = 0.01365 kN/m
Total dead load gk = 2.95 kN/m

Variables Action(s)
Occupancy load qk: 2.5 × 0.5 = 1.25 kN/m

At ultimate limit state; 1.35gk + 1.5qk = 1.35(2.95) + 1.5(1.25) = 5.8575 kN/m

Structural Analysis
The design load per span at the ultimate limit state (FEd) is given by = 5.8575 kN/m × 6m = 35.145 kN/m

From clause 3.5.2.3 of BS 8110-1:1997, we can use the simplified method for structural analysis.

At the middle of span 1-2; M = 0.075 × 35.145 × 6 = 15.815 kNm
At support 2, MEd = -0.086 × 35.145 × 6 = 18.134 kNm

Designing the span as a T-beam;
MEd = 15.815 kN.m

Effective depth (d) = h – Cnom – ϕ/2 – ϕlinks
Assuming ϕ12 mm bars will be employed for the main bars, and ϕ10 mm bars for the stirrups (links)
d = 250 – 25 – (12/2) -10 = 209 mm

k = MEd/(fckbd2) = (15.815 × 106)/(30 × 500 × 2092) = 0.024
Since k < 0.167, no compression reinforcement required

z = d[0.5+ √(0.25 – 0.882k)]
k’ = 0.024
z = d[0.5+ √((0.25 – 0.882(0.024))] = 0.95d = 198.55 mm

Depth to neutral axis x = 2.5(d – z) = 2.5(209 – 198.55) = 26.125 mm < 1.25hf (93.75mm)
Therefore, we design rib as a rectangular section

Area of tension reinforcement As1 = MEd / (0.87fyk z)
As1 = MEd / (0.87fyk z) = (15.815 × 106) / (0.87 × 500 × 0.95 × 209) = 183 mm2

Provide 2H12  Bot (ASprov = 226 mm2)

Check for deflection
ρ = As,req /bd = 183 / (500 × 209) = 0.00175
ρ0 = reference reinforcement ratio = 10-3√(fck) = 10-3√(30) = 0.00547
Since  ρ ≤ ρ0;
L/d = k [11 + 1.5√(fck) ρ0/ρ + 3.2√(fck) (ρ0 / ρ – 1)(3⁄2)]
k = 1.3
L/d = 1.3 [11 + 1.5√(30) × (0.00547/0.00175) + 3.2√(30) × [(0.00547 / 0.00175) – 1](3⁄2)]
L/d = 1.3[11 + 25.68 + 54.321] = 118.3

βs = (500 Asprov)/(fyk Asreq) = (500 × 226) / (500 × 183) = 1.23

beff/bw = 500/150 = 3.33

Therefore multiply basic length/effective depth ratio by 0.8

Therefore limiting L/d = 1.23 × 0.8  × 118.3 = 116.41
Actual L/d = 6000/209 = 28.708

Since Actual L/d (28.708) < Limiting L/d (116.41), deflection is satisfactory.

Design of Support 2;
MEd = 18.134 kN.m

Effective depth (d) = h – Cnom – ϕ/2 – ϕlinks
Assuming ϕ12 mm bars will be employed for the main bars, and ϕ10 mm bars for the stirrups (links)
d = 250 – 25 – (12/2) -10 = 209 mm

k = MEd/(fckbd2) = (18.134 × 106)/(30 × 150 × 2092) = 0.0922
Since k < 0.167, no compression reinforcement required

z = d[0.5+ √(0.25 – 0.882k)]
k = 0.092
z = d[0.5+ √((0.25 – 0.882(0.092))] = 0.91d

Area of tension reinforcement As1 = MEd / (0.87fyk z)
As1 = MEd / (0.87fyk z) = (18.134 × 106) / (0.87 × 500 × 0.91 × 209) = 219 mm2
Provide 2H12  Top (ASprov = 226 mm2)

Shear Design 
Maximum shear force in the rib VEd = 0.6F = 0.6 × 35.145 = 21.087 kN

VRd,c = [CRd,c.k.(100ρ1 fck)(1/3) + k1.σcp]bw.d ≥ (Vmin + k1.σcp) bw.d

CRd,c = 0.18/γc = 0.18/1.5 = 0.12
k = 1 + √(200/d) = 1 + √(200/209) = 1.97 < 2.0, therefore, k = 1.97
Vmin = 0.035k(3/2) fck0.5
Vmin = 0.035 × (1.97)1.5 × 300.5 = 0.53 N/mm2
ρ1 = As/bd = 226/(150 × 209) = 0.0072 < 0.02; Therefore take 0.0072

VRd,c = [0.12 × 1.97 (100 × 0.0072 × 30 )(1/3)] × 150 × 209 = 20639.67 N = 20.639 kN

Since VRd,c (20.639 kN) < VEd (21.087 kN), shear reinforcement is required.
The compression capacity of the compression strut (VRd,max) assuming θ = 21.8° (cot θ = 2.5)

VRd,max = (bw.z.v1.fcd) / (cot⁡θ + tanθ)
V1 = 0.6(1 – fck/250) = 0.6(1 – 30/250) = 0.528
fcd = (αcc fck) / γc = (0.85 × 30) / 1.5 = 17 N/mm2
Let z = 0.9d

VRd,max = [(150 × 0.9 × 209 × 0.528 × 17) / (2.5 + 0.4)] × 10-3 = 87.333 kN

Since < VEd < VRd,c < VRd,max

Hence Asw / S = VEd / (0.87 Fyk z cot θ) = 21087 / (0.87 × 500 × 0.9 × 209 × 2.5 ) = 0.103

Minimum shear reinforcement;
Asw / S = ρw,min × bw × sinα (α = 90° for vertical links)
ρw,min = (0.08 × √(fck)) / fyk = (0.08 × √30) / 500 = 0.000876
Asw/Smin = 0.000876 × 150 × 1 = 0.131
Maximum spacing of shear links = 0.75d = 0.75 × 209 = 156.75 mm

Provide H8mm @ 150 mm c/c as shear links.

Slab Topping
A142 BRC Mesh can be provided or H8 @ 250mm c/c

For more information on design and consultancy to accomplish the most challenging design brief, contact the author on info@structville.com. Thank you, and God bless you.

Advise on the Stability of this Structure

By
Ubani Obinna Uzodimma
-
Screenshot 20190421 181440 1 1 1

As a structural engineer, advise the client/architect on the stability of the structure, distinguishing between transient and permanent situation.

Height – 11.28 m
Length – 20.00 m


Screenshot 20190421 182059 1 1

Screenshot 20190421 182125 1

Note: This is an existing observatory structure in Tielt-Winge Belgium. Report has it that it was recently vandalised and is looking to be rebuilt…

Comparative Analysis of Cylindrical Water Tanks

By
Ubani Obinna Uzodimma
-

Water tanks are usually rectangular or cylindrical in shape. Cylindrical tanks are normally used for surface or elevated water tanks, while rectangular tanks can be used for underground, surface, or elevated water tank construction. The aim of this article is to compare the analysis of cylindrical water tanks subjected to hydrostatic load using the classical method and finite element analysis.

Some researchers have tried to compare the structural performance of cylindrical and rectangular water tanks under the same conditions such as volume, height above the ground, or aspect ratio. A study carried out in the year 2017 showed that circular water tanks require lesser materials than rectangular water tanks. However, some of the challenges in the construction of cylindrical tanks are increased cost of setting out, formwork preparation/installation, reinforcement installation, and general increased labour cost when compared with a rectangular tank.

Cylindrical tanks are subjected to radial pressure from the stored water, and/or from the retained earth when they are buried under the ground. Just like rectangular water tanks, the analysis of cylindrical water tanks can be easily done with the use of coefficients picked from tables (see Table 1). The values of the coefficients are usually based on the support condition of the wall relative to the base.

Analysis of cylindrical water tanks
Fig 1: Elevated circular water tank

The internal stresses normally analysed for in circular tanks are;

  • Circumferential tension (hoop tension)
  • Radial shears
  • Vertical moments

If the walls of the tank are supported on the base in such a way that no radial movement can occur, radial shear and vertical bending result, and the circumferential tension are zero at the bottom of the wall. This is usually referred to as the fixed joint condition and has been considered in this article.

Analysis of Cylindrical Water Tanks – Worked Example

Determine the maximum service values for circumferential (hoop) tension, vertical moment and radial shear in the wall of a cylindrical tank that is free at the top edge and fixed at the bottom. The wall is 400 mm thick, the tank is 4 m deep, the diameter is 8 m, and the water level is taken to the top of the wall.

Hydrostatic pressure at the base of the tank (n) = 10 kN/m3 × 4 m = 40 kN/m2
From Table 1;
lz2/Dh = 42/(8 × 0.4) = 5
Maximum hoop tension (t) = αtnr = 0.477 × 40 × 4 = 76.32 kN/m

Table 1: Coefficients for Analysis of Cylindrical Tanks Fixed at the Base (Reynolds and Steedman, 2008)

Table 2Bfor 2BAnalysis 2Bof 2BCylindrical 2BTanks

Maximum vertical negative moment (outside face) = αmnlz2 = 0.0059 × 40 × 42 = 3.776 kNm/m
Maximum vertical positive moment (inside face) = αmnlz2 = 0.0222 × 40 × 42 = 14.208 kNm/m
Radial shear V = αvnlz = 0.213 × 40 × 4 = 34.08 kN/m

When analysed on Staad Pro as a concrete cylindrical tank with a Poisson ratio of 0.2, fixed at the bottom, and subjected to a hydrostatic pressure of 40 kN/m2 the following results were obtained;

Modelling 2Bof 2Btank 2Bon 2BStaad 2BPro 2BFEM
Fig 2: Finite Element Meshing of Cylindrical Tank on Staad Pro
Mx
Fig 3: Horizontal Moment on the Tank Walls
My
Fig 4: Vertical Bending Moment on the Tank Walls

The summary of the finite element analysis results from Staad Pro is shown in Table .

Table 2: Summary of the finite element analysis results from Staad Pro

table 2Bof 2Bresults 1

Maximum vertical bending moment = 10.714 kNm/m
Maximum vertical shear stress = 0.073 N/mm2 = 29.2 kN/m
Maximum hoop stress (membrane) = 0.195 N/mm2 = 78 kN/m
Let us compare the results from Staad Pro with results from the use of coefficients (see Table 3).

Table 3: Comparison of analysis result using classical method and finite element analysis

Comparison 2BTable 1

Thank you for visiting Structville today and God bless you. If you have some time, you take a look at our QUIZ and evaluate your performance.

Solution to 25 Top Civil Engineering Questions

By
Ubani Obinna Uzodimma
-
Civil 1


Examiner: Structville Integrated Services

(1) Mechanics of deformable bodies is usually studied under?
(A) Statics
(B) Strength of Materials
(C) Dynamics

Answer is (B): Strength of Materials is deformable bodies mechanics, while statics is rigid body mechanics. Dynamics is the mechanics of bodies in motion.

(2) Which structural model is used for assessing shear resistance of RC structures in BS 8110-1:1997?
(A) Strut and Tie Method
(B) Theorem of three moments
(C)  Euler’s Theorem

Answer is (A): Shear resistance of concrete in BS 8110 is modelled based on Strut and Tie Method. Concrete acts as the compression chord, while main reinforcements act as tension chord. Shear links act as diagonals.

(3) The vertical deformation of soils per unit pressure is usually referred to as?
(A) Stiffness
(B) Modulus of subgrade reaction
(C) Bearing Capacity

Answer is (B). The unit for modulus of subgrade reaction is properly expressed as kN/m2/m.

(4) What is the basic unit of flow rate?
(A) m3/s
(B) m2/s
(C) m/s

Answer is (A).

(5) An S275 hot rolled I-Section has a flange thickness of 20.5mm. What is the design yield strength?
(A) 255 Mpa
(B) 265 Mpa
(C) 275 Mpa

Answer is (B). In both BS 5950 and Eurocode 3, once an S275 steel section has a thickness > 16mm but < 40mm, the yield strength to be used in design calculations is 265 Mpa

front
Click on the Image to Download Textbook


(6) The natural foundation of a road way is called?
(A) Base
(B) Sub-base
(C) Subgrade

Answer is (C). Foundation of the roadway is the subgrade. In design of highways, the CBR of the subgrade has effect on the thickness of the base and Sub-base.

(7) Why are return bars provided at the edges of RC slabs under simply supported assumptions?
(A) To resist torsion
(B) To resist negative moment
(C) None of the above
(D) All of the above

The answer is (D). In this case, apart from anchorage length requirements, return bars assist is resisting the minor fixity at edges, inclusive of possible twisting.

(8) A simply supported beam is pinned at both ends. Therefore the beam is?
(A) Statically determinate
(B) Unstable
(C) Statically indeterminate

Answer is (C). A beam that is pinned at both ends is statically indeterminate to the first order.

(9) How can we control deflection in a solid RC slab?
(A) By increasing the depth of the slab
(B) By introducing compression reinforcements
(C) By increasing the area of tension reinforcement
(D) All of the above

Answer is (D). Any of the above processes can help reduce deflection in slabs, even though introduction of compression bars is rarely done.


(10) In the design of highway bridges (BS 5400), HA loading comprises of ?
(A) Knife Edge Load and UDL
(B) Tandem Axle Loads
(C) UDL only

Answer is (A). HA load comprises of UDL which depends on the bridge length, and knife edge load of 120 kN.

(11) In concrete slabs on grade, the most critical loading condition occurs at?
(A) Mid span
(B) Edge
(C) 2.5d from the edge

Answer is (B). Critical loading stress for slab on grade occurs at the edges. Westagaard’s equation can be used for such evaluation.

(12) For a rectangular lamina immersed in a fluid at rest, the hydrostatic thrust passes through?
(A) Centre of pressure
(B) Centroid
(C) Meta Centre

Answer (A). Hydrostatic thrust passes through the centre of pressure, which may or may not coincide with the centroid of the body.

(13) At a construction site, the leveling staff reads +2.45m at point A and +2.76m at point B. Which of the following points is higher in elevation?
(A) Point A
(B) Point B
(C) Depends on the datum

Answer is (A). Point A is higher than point B, and the question of datum is out since both points are above the datum.

(14) A 0.8m long cantilever beam is subjected a concentrated load of 750 kN at the free edge. Which of the following poses the most serious problem?
(A) Bending
(B) Shear
(C) All of the above

Answer is (B). Both shear and bending is critical in this case, but shear is more critical.

(15) The continuous shear deformation of  a fluid is referred to as?
(A) Pressure
(B) Flow
(C) Turbulence

Answer is (B). Fluid does not have capacity to resist shear stress. Once continuous shear is applied to a fluid, it will flow continuously. 

(16) In the continuous support of a floor beam, how many percent of top tension reinforcement should be allowed into 0.15L of the adjacent spans?
(A) 50%
(B) 65%
(C) 100%

Answer is (C). At a continuous support in beams, 100% of top tension reinforcement must extend into 0.15L of the span.

(17) If the diameter of a pile foundation is 400mm, the recommended minimum thickness of the pile cap is?
(A) 900 mm
(B) 750 mm
(C) 400 mm

Answer is (A). For piles with diameter less than 550 mm, the recommended minimum thickness of the pile cap is 2d + 100mm (where d is the diameter of the pile in mm). Thickness of pile caps should satisfy shear and anchorage length requirements.

(18) Punching shear in pad footings is usually checked at?
(A) d from the column face
(B) 1.5 d from the column face
(C) at the column face

Answer is (B). Though we check for shear at all the locations listed. Critical diagonal shear is checked at d from column face, while we ensure that shear stress at column face does not exceed allowable shear stress.

(19) The complete loss of shear strength in soils is referred to as?
(A) Earthquake
(B) Settlement
(C) Liquefaction

Answer is (C).

(20) The deflection of a steel beam depends on?
(A) The elastic Modulus
(B) The length
(C) The support conditions
(D) All of the above

Answer is (D).

(21) Curtain walls are usually used in high rise building for which of the following reasons?
(A) Aerodynamics of winds
(B) To reduce dead load
(C) Reflection of night sky

Answer is (B). 

(22) In a single span generic beam, deflection is maximum at?
(A) Centre of the beam
(B) Point of maximum moment
(C) Point of zero slope

Answer is (C). Maximum deflection will not always occur at the point of maximum moment, but must occur at point of zero slope on the elastic curve.

(23) What is the recommended maximum water/cement ratio for production of concrete for water retaining structures?
(A) 0.4
(B) 0.5
(C) 0.6

Answer is (B). Research has shown that water tightness is compromised when water to cement ratio exceeds 0.5

(24) The difference between plastic limit and liquid limit in a soil is called?
(A) Activity
(B) Plasticity Index
(C) Cohesion
(D) Shrinkage Limit

Answer is (B). PI = PL – LL

(25) Adequate concrete cover to reinforcement does not enhance one of the following in RC structures?
(A) Fire Resistance
(B) Durability
(C) Cracking
(D) None of the above

Answer is (C). SLS crackwidth depends on concrete cover amongst other factors. The bigger the concrete cover, the bigger the crackwidth. But since the question says ‘adequate cover’, we are certain that fire resistance and durability requirements are good. It is only crackwidth that ‘adequate cover’ cannot guarantee.




How to Calculate Crackwidth Due to Bending According to EC2 (Download Excel Spreadsheet)

By
Ubani Obinna Uzodimma
-

Cracking is normal in reinforced concrete structures subjected to bending, shear, twisting, axial tension, and restraint from movement. This is mainly due to the low tensile strength of concrete. Cracking is usually a serviceability limit state problem, but apart from ruining the appearance of the concrete surface, it also posses durability issues, and leakage problem in water retaining structures.

Cracking is assumed to occur in a concrete section when the restraint strain exceeds the tensile strength capacity of the concrete. This means that for cracking to occur, some part or the whole of the concrete section must be in tension. Crackwidth in concrete is predicted by multiplying crack inducing strain (strain dissipated by the occurence of cracking) by the crack spacing.

The crack inducing strain due to flexure (bending) is given in expression 7.9 of EN 1992-1-1 as;

cracking 2B2

where;
εsm = mean strain in reinforcement.
εcm = mean strain in the concrete between cracks.
σs = stress in the reinforcement based on cracked section properties under quasi permanent load combination
αe = modular ratio, Es/Ecm (generally a value of 7 may be used.)
kt = 0.6 for short term loading and
0.4 for long term loading.
fct,eff = fctm at 3 days and/or 28 days.
ρp,eff = As/Ac,eff  (this is calculated for each face)

Where; As = area of reinforcement provided, mm2
Ac,eff = Area of concrete in tension whose depth is:
min[0.5h , 2.5(c + 0.5ϕ) , (h – x)/3] for each face of a wall

where;
h = thickness of wall
c = nominal cover
ϕ = bar diameter
x = depth to neutral axis
d = effective depth.
Es = elastic modulus for reinforcement = 200,000 MPa

Solved Example
Calculate the crackwidth due to externally applied load on a 200 mm thick slab with the following data;

Design serviceability bending moment = 14.96 kNm/m
Area of tension reinforcement provided at ultimate limit state As1 = H12@200 c/c (Asprov = 565 mm2/m)
Concrete cover = 25 mm
Effective depth d = 200 – 12/2 – 25 = 169 mm

Solution
STEP 1: Calculate the depth to the neutral axis of the section
The full procedure on how to calculate the depth of neutral axis of slabs and walls can be obtained by downloading our textbook on Design of Swimming Pool and Underground Water Tank, HERE

Otherwise, you can download our fully functional EXCEL Spreadsheet for a cheaper price (NGN 1000 only) HERE

Depth 2Bto 2Bneutral 2Baxis 1

From excel spreadsheet, x = 32.77 mm

STEP 2: Calculate the compressive stress in concrete

Compressive 2BStress 2Bin 2BConcrete 1

1000 × 32.77 × 0.5fc (169 – 32.77/3) = 14.96 × 106
On solving; fc = 5.78 Mpa

STEP 3: Calculate the tensile stress in steel

tensile 2Bstress 2Bin 2Bsteel 1

7 × 5.78 × (169 – 32.77)/32.77 = 168.19 Mpa


STEP 4: Calculate the area of concrete in tension
Ac,eff = Area of concrete in tension whose depth is:

min[0.5h , 2.5(c + 0.5ϕ) , (h – x)/3] = min[100, 77.5, 55.74]mm = 55.74 mm

ρp,eff  = 565/(55.74 × 1000) = 0.0101

STEP 5: Calculate the crack inducing strain in the section
(εsm – εcm) = [σs – kt(fct,eff/ρp,eff)(1 + αeρp,eff)]/Es

Note that fct,eff is the mean tensile strength of the concrete at 28 days = 2.21 Mpa for C20/25

(εsm – εcm) = [168.19 – 0.4(2.21/0.0101)(1 + 7 × 0.0101)]/200000
(εsm – εcm) = (168.19 – 93.712)/200000 = 372.386 × 10-6

372.386 × 10-6 < 0.6 σs/Es (504.57 × 10-6)

Therefore take (εsm – εcm) = 504.57 × 10-6

STEP 6: Calculate the maximum crack spacing
Sr,max = 3.4c + 0.425(k1 k2φ/ ρp,eff)
Take k1 = 0.8 (high bond bars, EN 1992)
k2 = 0.5 (for bending)

Sr,max = 3.4(25) + 0.425 × (0.8 × 0.5 × 12)/0.0101 = 286.98 mm

STEP 7: Calculate the crackwidth
Actual crackwidth wk = (εsm – εcm)  × Sr,max = 504.57 × 10-6 ×  286.98 mm = 0.14 mm

0.14 mm < 0.3 mm (Therefore crackwidth due to bending is very ok for the floor slab)

At Structville Integrated Services Limited, we have developed an EXCEL SPREADSHEET that can take care of the following calculations;

(1) Design of floor slabs
(2) Calculation of neutral axis (both for singly and doubly reinforced sections)
(3) Calculation of minimum area of reinforcement
(4) Calculation of early and long term thermal cracking
(5) Calculation of cracking due to loading

A screenshot from the excel spreadsheet is given below;

SPREADSHEET 1

To download the excel spreadsheet (restricted version), click HERE

To obtain the fully functional excel spreadsheet for NGN 1000 only, click HERE


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