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Loading and Design of Box Culverts to the Eurocodes

A culvert is a drainage structure designed to convey stormwater or stream of limited flow across a roadway. Culverts can consist of single or multi-span construction, with a minimum interior width of 6 m when the measurement is made horizontally along the centreline of the roadway from face to face of side walls.

Technically, any such structure with a span over 6 m is not a culvert but can be treated as a bridge. Box culverts consist of two horizontal slabs at the top and bottom, and two or more vertical side walls which are built monolithically.

For proper performance of culverts in their design life, there must be a hydraulic design, which will give the geometric dimensions or openings that will convey the design flood.  It is typical for culverts to be designed for the peak flow rate of a design storm of acceptable return period.

The peak flow rate may be obtained from a unit hydrograph at the culvert site, or developed from a stream flow and rainfall records for a number of storm events. In the absence of hydraulic data, it is wise to make conservative assumptions based on visual inspection of the site, performance of existing culverts and other drainage infrastructures, or by asking locals questions.

Structural Design of Box Culverts

Structural design begins when the structural design units receive the culvert survey and hydraulic design report from the hydraulics unit. The report in conjunction with the roadway plans shall be used to compute the culvert length, design fill, and other items that lead to the completed culvert plans.

Box culverts are usually analysed as rigid frames, with all corner connections considered rigid and no consideration for side sway. The centreline of slabs, walls and floor are used for computing section properties and for dimensional analysis. Standard fillets which are not required for moment or shear or both shall not be considered in computing section properties.

Design Loads

The structural design of a reinforced concrete box culvert comprises the detailed analysis of a rigid frame for bending moments, shear forces, and axial forces due to various types of loading conditions outlined below:

(i) Permanent Loads

  • Dead loads
  • Superimposed dead loads
  • Horizontal earth pressure
  • Hydrostatic pressure and buoyancy
  • Differential settlement effects

(ii) Vertical Live Loads

  • HA or HB loads on the carriageway (Load Model 1 of Eurocode)
  • Footway and Cycle Track Loading
  • Accidental Wheel Loading
  • Construction Traffic

(iii) Horizontal Live Loads

  • Live Load Surcharge
  • Traction
  • Temperature Effects
  • Parapet Collision
  • Accidental Skidding
  • Centrifugal Load

I believe that these loads are very familiar to designers, otherwise, the reader should consult standard textbooks. However, I am going to point out some important considerations worthy of attention while assessing design loads on culverts.

Loading of Box Culverts to Eurocode 1 Part 2 (EN 1991-2)

The traffic loads to be applied on box culverts are very similar to those to be applied on bridges. The box culvert will have to be divided into notional lanes as given in Table 1;

Carriageway width (w)Number of notional lanes (n)Width of the notional laneWidth of the notional lane
w < 5.4 m13 mw – 3m
5.4 ≤ w < 6m20.5w0
6m ≤ wInt(w/3)3 mw – 3n
Table 1: Division of Carriageway into Notional Lanes

The loading of the notional lanes according to Load Model 1 (LM1) is as given in Figure 1;

load 2Bmodel 2B1
Fig 1: Application of Traffic Load on Notional Lanes

Concentrated loads

According to BD 31/01, no dispersal of load is necessary if the fill is less than 600 mm thick for HA loading. However, once the fill is thicker than 600 mm, 30 units of HB loads should be used with adequate dispersal of the load through the fill. This same concept can be adopted for LM1 of EN 1991-2.

Earth Pressure

Depending on the site conditions, at rest pressure coefficient ko = 1 – sin (∅) is usually used for analysing earth pressure.

Loading Example

A culvert on a roadway corridor has the parameters given below. The culvert was found at a location with no groundwater problem. Using any suitable means, obtain the design internal forces induced in the members of the culvert due to the anticipated loading conditions when the culvert is empty under the following site conditions:

(1) The top slab of the culvert is in direct contact with the traffic carriageway and overlaid with 75 mm thick asphalt
(2) There is a 1.2 m thick fill on the top of the culvert before the carriageway formation level.

The geometry of the culvert
Total length of culvert = 8 m
Width of culvert c/c of side walls = 2.5 m
Height of culvert c/c of top and bottom slabs = 2.0 m
Length of wing walls = 2.12 m
The thickness of all elements = 300 mm
The thickness of the asphalt layer = 75 mm

Materials property
The angle of internal friction of fill soil = 30°
Unit weight of water = 9.81 kN/m3
Unit weight of backfill soil = 19 kN/m3
Unit weight of concrete = 25 kN/m3
Unit weight of asphalt concrete = 22.5 kN/m3
fck = 30 Mpa
fyk = 500 Mpa
Concrete cover = 50 mm

front

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Load Analysis
Width of carriageway = 8 m
Number of notional lanes = 8/3 = 2 notional lanes
Width of the remaining area = 8 – (2 × 3) = 2 m

(1) Case 1: Box culvert with no earth fill
(a) Applying the recommended traffic actions on the notional lanes

Notional 2Blane 2Bon 2Bculvert
Fig 2: Division of the Culvert Carriageway into Notional Lanes
Eurocode 2Btraffic 0A 0A 2Bload 2Bon 2Bculvert 1
Fig 3: Loading of the Notional Lanes
traffic wheel load on box culvert
Fig. 4: Section of the Culvert across Notional Lane 1

(b) Permanent actions
The self-weight of the structure should be calculated by Staad Pro software, but let us show how we can easily compute and apply it to the structure;

(i) Self-weight of top slab
Thickness of top slab = 300 mm = 0.3 m
Self weight of the slab per unit length = 0.3 m × 25 kN/m3 = 7.5 kN/m2

(ii) Permanent action from the asphalt layer
Thickness of asphalt = 75 mm = 0.075 m
Self weight of the asphalt per unit length = 0.075 m × 22.5 kN/m3 = 1.69 kN/m2

For the purpose of simplicity, let us combine these two actions such that the permanent action is given by gk = 7.5 + 1.69 = 9.19 kN/m2

TOP 2BSLAB
Fig 5: Permanent Action on Top of the Box Culvert

(iii) Earth Pressure
At rest earth pressure coefficient ko = 1 – sin (∅) = 1 – sin (30) = 0.5
Maximum earth pressure on the side walls p = koρH = 0.5 × 19 kN/m3  × 2.3m = 21.85 kN/m2

earth 2Bpressure 2Bon 2Bculvert 2Bwall
Fig 6: Horizontal Earth Pressure on the Culvert Walls

(iv) Live load Surcharge
Consider a live load surcharge of q = 10 kN/m2
Therefore horizontal surcharge pressure = koq = 0.5 × 10 kN/m2  = 5.0 kN/m2

surcharge 2Bon 2Bculvert 2Bwall
Fig 7: Live Load Surcharge on the Walls of the Culvert

When the culvert is full, the hydrostatic pressure profile inside the culvert can also be easily obtained. However, this was not considered in this analysis.

(2) Case 2: Box culvert with 1.2 m thick earth fill
(a) Traffic Load on the Box Culvert
In this case, since the thickness of the fill is greater than 0.6 m, we are going to consider the wheel load of the traffic actions dispersed to the top slab of the culvert as a uniformly distributed load. The UDL of traffic action will not be considered.

For this case, let us use Load Model 1 of EN 1991-2 which is recommended by clause 4.9.1 of EN 1991-2. The tandem load can be considered to be dispersed through the earth fill and uniformly distributed on the top of the box culvert. The contact surface of the tyres of LM1 is 0.4m x 0.4m, which gives a contact pressure of about 0.9375 N/mm2 per wheel.

load 2Bmodel 2B1 2Btandem 2Bsystem 1
Fig 8: LM1 Tandem System

We are going to disperse the load through the earth fill to the box culvert by using the popular 2(vertical):1(horizontal) load increment method. This is the method recommended by BD 31/01, otherwise, Boussinesq’s method can also be used. However, clause 4.9.1 (Note 1) of EN 1991-2:2003 recommends a dispersal angle of 30° to the vertical for a well-compacted earth fill. A little consideration will show that this is not so far away from the 2:1 load increment method.

Load 2BDispersal 2Bon 0A 0A 2BBox 2BCulvert 1
Fig 9: Single Wheel Load Distribution Through Compacted Earth Fill

For the arrangement in Fig 9 above;
P1 = 150 kN
L1 = 0.4 m
L2 = 0.4 + D = 0.4 + 1.2 = 1.6 m

Therefore, the equivalent uniformly distributed load from each wheel to the culvert is;
qec = 150/(1.6 × 1.6) = 58.593 kN/m2

It is acknowledged that the pressure from each wheel in the axles can overlap when considering the tandem system as shown in the figure below. This is considered in the lateral and longitudinal directions.

tandem 2Bload 2Bon 0A 0A 2Bbox 2Bculvert
Fig 10: Overlapping Tandem Axle Load Dispersion Through Earth Fill

When considering the tandem system as shown in Figure 10;
∑Pi = 150 + 150 + 150 + 150 = 600 kN
L2 = 1.2 + 0.4m + 1.2m = 2.8 m (Spacing of wheels + contact length + depth of fill)
B2 = 2.0 m + 0.4m + 1.2m = 3.6 m (Spacing of wheels + contact length + depth of fill)
qec = 600/(2.8 × 3.6) = 59.523 kN/m2
As can be seen, the difference between considering the entire tandem system and one wheel alone is not much. But to proceed in this design, we will adopt the pressure from the tandem system.

Therefore the traffic variable load on the box culvert is given in Fig. 11 below;

traffic 2Bvariable 2Bload
Fig 11: Equivalent Traffic Load Distribution on Top of the Box Culvert
final 2Bfront 2Bcover

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(b) Earth load on top of the box culvert
At a depth of 1.2 m, the earth pressure on the box culvert is given by;
p = 1.2 × 19 kN/m= 22.8 kN/m2

Earth 2Bload
Fig 12: Earth Load on Buried Culvert

(c) Horizontal Earth Pressure on the Box Culvert
Since the box culvert is buried under the ground, the pressure distribution is as given in Figure 13.

The maximum pressure at the base of the culvert (at 2.3 m) is given by;
pmax =  koρH = 0.5 × 19 kN/m3  × 3.5 m = 33.25 kN/m2

The minimum pressure at the top of the culvert (at 1.2 m below the ground) is given by;
pmin =  koρH = 0.5 × 19 kN/m3  × 1.2 m = 11.40 kN/m2

horizontal 2Bearth 2Bpressure 1
Fig 13: Horizontal Earth Pressure on Buried Box Culvert

(d) Surcharge load
The horizontal surcharge load distribution on the buried box culvert will be the same as that of case A.

Thank you for visiting Structville today. We are going to present the actual analysis and design of box culverts using Staad Pro in our next post which will come shortly. Please stay tuned and God bless you.

Preliminary Plastic Analysis of Portal Frames

Steel portal frames provide economical solution for construction of industrial structures, warehouses, churches or other buildings with large enclosures. The loading scheme on portal frames usually includes dead load, wind load, imposed load, service loads, snow loads, imperfections, thermal loads, crane loads etc.

In the plastic analysis of portal frames, preliminary analysis is usually carried out to determine member sizes before detailed checks are carried out to determine the adequacy of the selected sections. Weller’s chart is usually used to quickly determine the plastic moments that can enable initial sizing of members. The simplicity of the charts depends on a series of three charts developed by Weller. The application of the method relies on the following assumptions;

(1) The rafter depth is approximately span / 55
(2) The hunch length is approximately span /10
(3) The rafter slope lies between 10° and 20°.
(4) The ratio of span to eaves height is between 2 and 5.
(5) The hinges in the mechanism are formed at the level of the underside of the haunch in the column and close to the apex.

Each chart requires a knowledge of the geometry of the frame and the design loading as input data in order to determine approximate sizes for the column and rafter member.

Application Example

roof 2Blayout 2Bgeneral 2Barrangement
Fig 1: General Arrangement of Steel Structure


portal 2Bframe 2Bsection
Fig 2: Portal Frame

Length of building = 40 m
Width of building = 30 m
Height to eaves = 7.0 m
Height to roof top = 11.0 m

Consider the portal frame on gridline 2. The gravity load on the portal frame is analysed as follows;

Load Analysis 
(1) Dead Load
Aluminium roof sheeting = 0.04 kN/m2
Insulation (assume gypsum boards per 10 mm thickness) = 0.077 kN/m2
Other permanent services = 0.15 kN/m2
Steel purlins (take) = 0.03 kN/m2
Self weight of rafter (take) = 0.15 kN/m2
Total = 0.447 kN/m2 on slope

Angle of inclination of roof = 15°
Load on plan = 0.447/cos 15° = 0.4627 kN/m2

Spacing of portal frame = 5 m centre to centre
Therefore dead load on roof gk = 0.4627 kN/m2 x 5m = 2.31 kN/m

(2) Imposed Load
According to BS 6399 Part 3, for roof with slope less than 30°, a minimum imposed load of 0.6 kN/m2 on plan should be provided as long as no access except for cleaning and maintenance is provided.

For this design, let us adopt an imposed load of 0.75 kN/m2

qk = 0.75 kN/m2

Imposed load on roof = 0.75 x 5m = 3.75 kN/m

At ultimate limit state;
p = 1.4gk + 1.6qk = 1.4(2.31) + 1.6(3.75) = 9.234 kN/m

portal 2Bframe 2Bloaded
Fig 3: Portal Frame with ULS Gravity Load


Step by Step Analysis

(a) Calculate the span/height to eaves ratio = L/h
Span/height to eaves ratio L/h = 30/7 = 4.285

(b) Calculate the rise/span ratio = r/L
Height of Rise / span ratio r/L = 4/30 = 0.133


(c) Calculate the total design load FL on the frame and then calculate FL2, where F is the load per unit length on plan of span L
Ultimate gravity load = 9.234 kN/m (see Fig. 3)
Total load on the frame (wL) = 9.234 x 30 = 277.02 kN
Parameter wL2 = 9.234 x 302 = 8310.6 kNm

(d) From Fig 4 (Chart 1) obtain the horizontal force ratio HFR at the base from r/L and L/h

Chart 2B1
Fig 4: Chart 1

H/wL = 0.27
Hence, horizontal force ratio HFR = 0.27

(e) Calculate the horizontal force at the base of span H = HFR wL
Hence, H = 0.27 x 277.02 = 74.795 kN

(f) From Fig 5 (Chart 2), obtain the rafter Mp ratio MpR from r/L and L/h.

Chart 2B2
Fig 5: Chart 2

Rafter MpR/wL2  ratio = 0.028

(g) Calculate the Mp required in the rafter from Mp(rafter) = MpR x wL2
Mp,Rafter = 0.028 x 8310.6 = 232.696 kNm

(h) From Fig 6 (Chart 3), obtain the column Mp ratio from r/L and r/h

Chart 2B3
Fig 6: Chart 3

Column MpC/wL2 ratio = 0.057

(i) Calculate the Mp required in the column from Mp,Column = MpC x wL2
Mp,Column  = 0.057 x 8310.6 = 473.70 KNm

(j) Determine the plastic moduli for the rafter Wpl,y,R and the column Wpl,y,C from
Wpl,y,RMp,Rafter /fy
Wpl,y,CMp,Column/fy

Where fy is the yield strength = 275 Mpa.

Using the plastic moduli, the rafter and column sections may be chosen from the range of plastic sections as so defined in the section books

Selection of Members
(a) For Rafters
Plastic Moment Mp,Rafter = 232.696 kNm
Plastic Modulus Required Sx
Sx = (232.696 x 103) / 275 = 846 cm3
Try UB 356 x 171 x 57 kg/m (Plastic modulus = 1010 cm3)

(b) For Columns
Plastic Moment Mp,Columns = 473.7 kNm
Plastic Modulus Required
Sx = (473.7 x 103) / 275 = 1722 cm3
Try UB 457 x 191 x 82 kg/m (Plastic modulus = 1830 cm3)

These trial sections are usually checked for the following in the detailed design;
(a) In-plane stability
(b) Column Stability
(c) Rafter Stability
(d) Haunch Stability
(e) Deflection etc.

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Structural Design of Flat Slabs

Flat slabs are slabs are two-way concrete slabs that are supported directly by columns without floor beams. Therefore, they offer economical solutions to larger floor spans in reinforced concrete buildings. The columns supporting the slab panel may or may not have drop panels.

Drop panels are particularly advantageous in situations where substantial negative moments occur at supports or where punching shear stresses are critical and the use of punching reinforcement is undesirable.

There are many advantages of flat slabs such as increased headroom, easier flow of mechanical and electrical services, ease in construction of formwork, faster construction etc. Unlike traditional beam and slab systems, flat slabs eliminate the use of floor beams, resulting in a cleaner, more open space.

Analysis of Flat Slabs

The following methods are usually employed in the analysis of flat slabs:

  • simplified coefficient approaches,
  • equivalent frame analysis,
  • yield line theory,
  • grillage analysis, and
  • finite element methods.

The application of simplified coefficients is subject to specific conditions outlined in Annex I of BS EN 1992-1-1. Yield line and finite element methods are particularly well-suited for computer-based analysis. Given this, this discussion will concentrate on the use of simplified coefficients.

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

Design Example of Flat Slab

FLAT 2BSLAB 2BCONFIGURATION

In this article, we are going to analyse gridline 2 (same as gridline 5) of the flat slab general arrangement that is given above with the following data;

The thickness of slab = 230 mm (governed by deflection requirements)
fck = 30 Mpa
fyk = 500 Mpa
Imposed load = 3 kN/m2
Concrete cover = 30 mm
Size of columns = 300 x 300 mm

Load Analysis

Permanent Actions

Self weight = 0.23 × 25 = 5.75 kN/m2
Finishes = 1.2 kN/m2
Partition allowance = 1.2 kN/m2
Total permanent actions gk = 8.15 kN/m2

Variable Actions
Imposed floor load qk = 3 kN/m2

Total load on the floor F = 1.35gk + 1.5qk = 1.35(8.15) + 1.5(3) = 15.5 kN/m2

Effective span of the slab = 6500 – 2(300/2) + 2(230/2) = 6430 mm
Width of bay = 6500 mm
Width of column strip = 2(lx/4)= 2(6500/4) = 3250 mm
Width of middle strip = 6500 – 1625 – 1625 = 3250 mm

F = 15.5 × 6.5 × 6.5 = 654.875 kN

A little consideration will show that the slab under consideration met the conditions for use of coefficients.

image

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)

Sagging – First Span
M = 0.086 × 654.87 × 6.43 = 362.13 kNm

Hogging – First interior support
M = -0.086 × 654.87 × 6.43 = -362.13 kNm

The apportionment of moments between column strips and middle strips is shown in Table 2:

HoggingSagging
Column strip70%50%
Middle strip30%50%

(Column Strip)
Hogging moment (Support B and C)
MEd = 0.7 × (362.13/3.25) = -77.99 kNm/m

Sagging moment:
MEd = 0.5 × (362.13/3.25) = 55.71 kNm/m

(Middle Strip)
Hogging moment (Support B and C)
MEd = 0.3 × (362.13/3.25) = -33.427 kNm/m

Sagging moment:
MEd = 0.5 × (362.13/3.25) = 55.71 kNm/m

Structural Design – Flexure: column strip and middle, sagging

MEd = 55.71 kNm/m
d = h – Cc – ϕ/2

Assuming ϕ16mm bars will be employed for the construction
d = 230 – 30 – 8 = 192 mm; b = 1000mm (designing per unit width)

k = MEd/(fckbd2) = (55.71 × 106)/(30 × 1000 × 1922) = 0.086

Since k < 0.167 No compression reinforcement required

z = d[0.5 + √(0.25 – 0.882k)]
z = d[0.5 + √(0.25 – 0.882 × 0.086)] = 0.917d = 0.917 × 192 = 176.12 mm

As1 = MEd/(0.87fykz) = (55.71 × 106)/(0.87 × 500 × 176.12) = 727 mm2/m

Provide H16mm @ 200 mm c/c BOT (ASprov = 1005 mm2/m)

Check the minimum area of steel required;
fctm = 0.3 × fck2⁄3 = 0.3 × 302⁄3 = 2.89 N/mm(Table 3.1 EC2)

As,min = 0.26 × fctm/fyk × b × d = 0.26 × 2.89/500 × 1000 × 192 = 289 mm2/m
Check if As,min < 0.0013 × b × d (250 mm2/m)
Since, As,min = 250 mm2, the provided reinforcement is adequate.

Check for deflection
We check for deflection of flat slabs at the long span of the slab

Allowable L/d = N × K × F1 × F2 × F3 ≥ Actual L/d
K = 1.2 for flat slabs
ρ = As,req/bd = 727/(1000 × 192) = 0.003786 > 10-3√fck (0.00547)

Since ρ > ρ0
l/d = [11 + 1.5√fck0/(ρ – ρ’)) + 0.0833√fck0/ρ)0.5] if ρ > ρ0
ρ0/ρ= 0.00547/0.003786 = 1.44

N = [11 + 1.5√(30) × 1.444 + 0.0833√30 × (1.444)0.5] = 11 + 11.86 + 0.548 = 23.4

F1 = 1.0 (beff/bw = 1.0)
F2 = 1.0 (no brittle partitions)
F3 = 310/σs ≤ 1.5

where:

σsu = (500/1.15) × (8.15 + 0.3 × 3)/15.5 = 253.856 MPa
σs = σsu × (As,req/As,prov) = 253.856 × (727/1005) = 183.6 N/mm2
βs = 310/183.6 = 1.688 > 1.5 (take 1.5)

Therefore allowable L/d = 23.4 × 1.2 × 1.0 × 1.0 × 1.5 = 42.12

Actual L/d = 6430/192 = 33.49

Therefore, deflection is okay.

Flexure: column strip, hogging
MEd = 77.99 kNm/m

k = MEd/(fckbd2) = (77.99 × 106)/(30 × 1000 × 1922) = 0.0705
Since k < 0.167 No compression reinforcement required

z = d[0.5 + √(0.25 – 0.882k)]
z = d[0.5 + √(0.25 – 0.882 × 0.0705)] = 0.933d = 0.933 × 192 = 179.2 mm

As1 = MEd/(0.87fykz) = (77.99 × 106)/(0.87 × 500 × 179.2) = 1000 mm2/m

Provide T16mm @ 175 mm c/c Top (ASprov = 1149 mm2/m)

(c) Flexure: Middle strip, hogging
MEd = 33.43 kNm/m

k = MEd/(fckbd2) = (33.43 × 106)/(30 × 1000 × 1922) = 0.030
Since k < 0.167 No compression reinforcement required

z = d[0.5 + √(0.25 – 0.882k)]
z = d[0.5 + √(0.25 – 0.882 × 0.030)] = 0.95d = 0.95 × 192 = 182.4 mm

As1 = MEd/(0.87fykz) = (33.43 × 106)/(0.87 × 500 × 182.4) = 421 mm2/m
Provide T12mm @ 200 mm c/c Top (ASprov = 565 mm2/m)

Perpendicular to edge of slab at edge column
Design transfer moment to column Mt = 0.17 bed2fck
be = cz + y = 300 + 300 = 600 mm

Mt = 0.17 × 600 × 1922 × 30 × 10−6 = 112.8 kNm
k = MEd/(fckbd2) = (112.8 × 106)/(30 × 600 × 1922) = 0.1699

z = d[0.5 + √(0.25 – 0.882k)]
z = d[0.5 + √(0.25 – 0.882 × 0.1699)] = 0.82d = 0.82 × 192 = 157.44 mm

As1 = MEd/(0.87fykz) = (112.8 × 106)/(0.87 × 500 × 157.44) = 1647 mm2/m
This reinforcement to be placed within cx + 2cy = 900 mm

Try H16T1 U-bars in pairs @ 100 (Asprov = 2010 mm2) local to column (max. 200 mm from column)

Punching Shear Check

See worked example on punching shear verification of flat slabs.

Simplified Design of Welded Connections

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

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

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 × 7.5= 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 × 7.5= 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) + k1cp]bw.d ≥ (Vmin + k1cp) 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)

(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

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) + k1cp]bw.d ≥ (Vmin + k1cp) 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

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

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

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