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Structural Aspects of Pile Foundation Design: A Practical Example

By
Ubani Obinna Uzodimma
-

In the design of a pile foundation, the geotechnical engineer is expected to hand over the soil investigation report to the structural engineer, who will proceed to provide the longitudinal reinforcements needed for the piles and also design the pile cap. The structural design of a pile cap is an important aspect of pile foundation design and the method of carrying out the design has been presented in this article.

The soil investigation report handed to the structural engineer for the design of pile foundation purpose should contain the embedment length of the piles, the recommended sizes of the pile, the safe working load of each pile size, and other information that may be necessary for the structural engineer to do his design properly. All aggressive materials in the soil should be stated as well so that adequate protection will be given to the pile material(s) for durability.

The first step in the structural design of a pile cap usually involves obtaining the number of piles required to support each column load. This is usually done using the service loads of the column, and relating them to the safe working load of the piles from the soil investigation report. In this article, we are going to show how the structural design of reinforced concrete pile foundations and pile caps can be done based on practical design and site experience.

Design Example
The frame of a 5 storey building is shown in Figure 1, and it is expected to be supported on piles with an embedment length of 20 m. The allowable working loads of the bored pile (CFA) is shown in Table 1. fy = 460 Mpa, fcu = 30 Mpa

Table 1: Pile safe working load

Pile diameter (mm)300450600750900
Safe working load (kN)246.74370.11493.48616.85740.22
Building frame modelled on Staad Pro
Fig 1: Frame of a 5-storey building
Column layout
Fig 2: Column load layout

Design for Column A1
Service axial load on column = 647 kN
Ultimate axial load on column = 885 kN
Size of column = 450 x 230 mm

Try 2 No of piles
Service load per pile = 647/2 = 323.5 kN
Let us adopt 600 mm diameter piles for uniformity and to have fewer pile boring points

Safe working load of φ600 mm piles = 493.48 kN > 323.5 kN Okay

Centre to centre spacing of piles = 3φ = 3 x 600 = 1800 mm
Overhang of pile cap edge from the pile = 150 mm

Total length of pile cap = 1800 + 600 + 2(150) = 2700 mm
Width of pile cap = 600 + 150 + 150 = 900 mm
Thickness of pile cap = 2φ + 100 = 2(600) + 100 = 1300 mm

The layout of the pile cap is therefore given as shown in Figure 3.

Structural arrangement of column supported by two piles
Fig 3: Pile cap type 1

Let us quickly carry out the structural design of pile cap Type 1 according to BS 8110-1:1997. You can also read Design of Pile Cap According to Eurocode 2.

From Table 3.61 of Reynolds et al. (2008), the tensile force to be resisted within the pile cap is given by;

Ft = N/(12ld)[3l2 – a2]

Where;
N = Column axial load at ultimate limit state
l = Centre to centre spacing of the piles
d = Effective depth of the pile cap 
a = dimension of the column side parallel to the length of the pile cap

Self weight of the pile cap (ULS) = 1.4 x Area x depth x 24 kN/m3 = 1.4 x 2.7m x 0.9m x 1.3m x 24 kN/m3 = 106.14 kN

N = 885 kN + 106.14 = 991.142 kN
l = 1.8 m
d = 1300 – 100 = 1200 mm = 1.2 m
a = 0.45 m

Ft = [991.142/(12 x 1.8 x 1.2)] x [3 x 1.82 – 0.452] = 364 kN
Ast = Ft/0.95fy = (364 x 1000)/(0.95 x 460) = 833 mm2
Asmin = 0.13bh/100 = 1690 mm2
Provide 6T20 @ 175 c/c (Asprov = 1974 mm2)

Check for shear
Critical position for shear on vertical section across full width of pile-cap occurs at distance from face of column given by:
av = 0.5(l – c) – 0.3φ = 0.5(1800 – 450) – (0.3 x 600) = 495 mm

The shear force carried by the piles V = 991.142/2 = 495.571 kN

The shear stress ν = V/bd = (495.571 x 1000)/(900 x 1200) = 0.458 MPa
Concrete resistance shear stress vc = 0.632(100As/bd)1/3(400/d)1/4

vc = 0.632 x [(100 x 1974)/(900 x 1200)]1/3 x (400/1200)1/4 = 0.632 x 0.557 x 0.759 = 0.275 MPa
For grade 30 concrete, vc = 0.275 x (30/25)1/3 = 0.292 Mpa
vc(2d/av) = 0.292 x [(2 x 1200)/495] = 1.415 MPa > 0.458 MPa  This is okay

Shear stress at column perimeter
ν = V/ud = (885 x 1000)/[(2 x 225 + 2 x 450) x 1200] = 0.546 MPa
This is less than 0.8√fcu = 4.38 Mpa. Therefore, this is okay.

Anti-burst bars of T12 @ 200 spacing should be provided
Main bars should be returned at least 900 mm into the sides to satisfy anchorage length requirements. You can take anchorage length to be conservatively 50 x diameter of reinforcement = 50 x 20 = 1000 mm

Design of Pile Cap Type 2
Service axial load on column = 1077 kN
Ultimate axial load on column = 1476 kN
Size of column = 450 x 225 mm
The number of φ600 mm piles required = 1077/493.48 = 2.184

Using 3 No of φ600 piles
Service load per pile = 1077/3 = 359 kN

Safe working load of φ600 mm piles = 493.48 kN > 359 kN This is okay

Let us adopt a triangular pile cap arranged in such a way that the column load will be equally distributed to the piles. This arrangement can be found in Table 3.16 of Reynolds et al (2008), and it is shown in Figure 4.

Layout of column supported by three piles for equal load distribution
Fig 4: Dimensions of triangular pile cap for equal load distribution (Reynolds et al, 2008)

hp = φ = diameter of pile = 600 mm
Spacing of piles = 3φ = 3 x 600 = 1800 mm
Overhang of pile cap edge from the pile = 150 mm
(α + 1)φ + 300 = (3 + 1)600 + 300 = 2700 mm
φ + 250 =  600 + 250 = 850 mm
φ + 300 =  600 + 300 = 900 mm
(6α/7 + 1)φ + 300 = 2442.857 mm (say = 2445 mm)
(2α/7 + 0.5)φ + 150 = 964.285 mm (say = 965 mm)
Thickness of pile cap = 2φ + 100 = 2(600) + 100 = 1300 mm

The layout of the pile cap is shown in Figure 5.

Structural layout of column supported by three piles
Fig 5: Structural arrangement of 3 pile caps

Self weight of the pile cap (ULS) = 1.4 x Area x depth x 24 kN/m3 = 1.4 x 5.166 m2 x 1.3m x 24 kN/m3 = 225.61 kN

Total load on pile cap at ULS = 1476 kN + 225.61 kN = 1701.61 kN
l = 1.8 m
a = 0.225 m
b = 0.45 m

Tensile force to be resisted by the reinforcement in the direction parallel to X-X;
Ft,x = N/(36ld)[4l2 + b2 – 3a2]
Ft,x = [1701.61/(36 x 1.8 x 1.2)] x [4 x 1.82 + 0.452 – 3 x 0.2252]  = 284 kN

Tensile force to be resisted by the reinforcement in the direction parallel to Y-Y;
Ft,y = N/(18ld)[2l2 – b2]
Ft,y = [1701.61/(18 x 1.8 x 1.2)] x [2 x 1.82 – 0.452]  = 275 kN

Let us use the highest value for design in anticipation that we will provide the same reinforcement in both directions
Ast = Ft/0.95fy = (284 x 1000)/(0.95 x 460) = 649 mm2
Asmin = 0.13bh/100 = 1690 mm2
Provide T20 @ 175 c/c in both directions (Asprov = 1974 mm2)

Shear resistance
The shear force carried by the piles V = 1701.61/3 = 567.2 kN
The shear stress ν = V/bd = (567.2 x 1000)/(1000 x 1200) = 0.472 MPa
vc(2d/av) = 0.292 x [(2 x 1200)/495] = 1.415 MPa > 0.472 MPa  This is okay
Shear will obviously not be a problem.

Design of Pile Cap Type 3
Service axial load on column = 1825 kN
Ultimate axial load on column = 2545 kN
Size of column = 400 x 400 mm
The number of φ600 mm piles required = 1825/493.48 = 3.69

Use 4 No of φ600 piles
Service load per pile = 1825/4 = 456.25 kN

Safe working load of φ600 mm piles = 493.48 kN > 456.25 kN This is okay

Let us adopt a square pile cap arranged in such a way that the column load will be equally distributed to the piles. This arrangement can be found in Table 3.16 of Reynolds et al (2008), and it is shown in Figure 6.

Structural layout of column supported by four piles
Fig 6: Pile Cap Type 3

Self weight of the pile cap (ULS) = 1.4 x Area x depth x 24 kN/m3 = 1.4 x 7.29 m2 x 1.3m x 24 kN/m3 = 318.43 kN
Total load on pile cap at ULS = 2545 kN + 318.43 kN = 2863.43 kN

Tensile force to be resisted by the reinforcement in both directions;
Ft = N/(24ld)[3l2 – a2]
Ft = [2863.43/(24 x 1.8 x 1.2)] x [3 x 1.82 – 0.402]  = 528 kN

Ast = Ft/0.95fy = (528 x 1000)/(0.95 x 460) = 1208 mm2
Asmin = 0.13bh/100 = 1690 mm2
Provide T20 @ 175 c/c in both directions (Asprov = 1974 mm2)

Shear resistance
The shear force carried by the piles V = 2863.43/4 = 715.9 kN
The shear stress ν = V/bd = (715.9 x 1000)/(1000 x 1200) = 0.595 MPa
vc(2d/av) = 0.292 x [(2 x 1200)/495] = 1.415 MPa > 0.472 MPa  This is okay

The structural engineer is expected to provide the following drawings;

(1) Setting out drawing showing the piling points and layout with a known reference point
(2) General column/pile cap layout/arrangement
(3) Pile cap/ground beam/ground floor slab layout
(3) Columns, piles and pile cap reinforcement drawings (detailing)
(5) Ground beams and ground floor slab reinforcement detailing
(6) Construction procedure sketches

Construction Considerations
(1) Ground beams are usually provided to chain the pile caps together, and to provide the needed support for the ground floor slab. There are construction scenarios where the ground floor slab is placed directly on the pile caps, but note that this concept is quite different from piled raft foundation. The ground beams are usually embedded into the pile caps, or may be allowed to sit directly on the pile caps depending on the site level. A typical construction drawing showing this interaction is given in Figure 7.

(2) Contractor should maintain a minimum concrete cover of 75 mm

(3) There may be need to be cast the pile cap in two stages to achieve the configuration shown in Figure 7.  The first casting will get to the bottom level of the ground beams (see Figure 8), then the ground beam reinforcements are laid (see Figure 9), before the final casting of the pile cap and ground beams to the required level (see Figure 10). Read about bonding of old and new concrete.

Pile cap and ground beam arrangement
Concreting of pile cap
Figure 8: Typical casting of pile cap to ground beam level.
Ground beam reinforcement
Fig 9: Typical ground beam reinforcement arrangement
Reinforced concrete beam and pile cap
Fig 10: Completed pile cap and ground beam

On getting to the stage shown in Figure 10, the bays are filled with sharp sand, and the ground floor slab is cast as appropriate.

Do you need help in design, consultancy, production of construction drawings, supervision, and project management, contact us today at Structville Integrated Services Limited. We are excellent at what we do, and we pride ourselves in professionalism and integrity. Send an e-mail now to info@structville.com copy structville@gmail.com or send a whatsapp message to +2347053638996.

References
[1] Reynolds C.E., Steedman J.C., Threlfall A.J. (2008): Reynolds’ Reinforced Concrete Designer’s Handbook 11th Edition. Taylor and Francis, New York

Calculation of Collapse Settlement of Collapsible Soils

By
Ubani Obinna Uzodimma
-

Civil engineers are always faced with one challenge or another on-site, and one of the major challenges on-site has to do with the nature of the soil to support the structure to be constructed. Some soils exhibit excessive change in volume under constant load and changes in moisture content. Under field conditions, if the addition of water under constant load causes the void ratio to reduce drastically, the soil is said to have collapsed.

These soils are predominantly found in the arid regions and appear strong and stiff in their dry natural state, but lose strength and undergo high compression upon wetting.  According to Behzad (2013), almost all natural deposits of collapsible soils are either debris flow deposits or wind deposited soils (loess).

Soils which exhibit this behaviour are usually unsaturated granular soils which in a loose state are maintained by apparent cohesion. This apparent cohesion may be due to the presence of clays at the intergranular contact areas or the accumulation of soluble salts as binders (Murthy, 2012). In South-Western Nigeria, collapsible soils have been observed and studied by Owolabi and Ola (2014).

Collapsible Soils

For a soil sample to collapse, the soil structure must have an open, partially unstable, and unsaturated fabric which must be held in place by apparent cohesion. On wetting and application of high enough stress, the soil fabric will be expected to collapse.

For compacted cohesive soils, the collapse behaviour will depend on the percentage of fines, the moulding water content, the dry density, and the compaction energy employed Murthy (2012). Any soil compacted at the dry side of the optimum moisture content and at low dry density will also be expected to develop a collapsible structure.

Calculation of Collapse Settlement

A method of calculating the collapse settlement of a soil sample was suggested by Jennings and Knight (1975) and reported by Murthy (2012). This involves carrying out double oedometer tests of identical undisturbed samples in the laboratory. Both specimens are kept under a constant pressure of 1 kPa for a period of 24 hours at their natural moisture content after which one specimen is flooded and the other allowed to remain in its natural state.

After flooding for 24 hours, the consolidation test is allowed to progress as usual after which the e-LogP curve is plotted for both samples. The collapse settlement is calculated in two parts according to equations (1) and (2).

S1 = ΔenH /(1 + e0)  ————————- (1)
S2 = ΔecH /(1 + e0)  ————————- (2)

Where;
Δen =  Change in void ratio due to load ΔP as per the e-logP plot without change in moisture content
Δec =  Change in void ratio due to load ΔP as per the e-logP plot with increase in moisture content (settlement due to the collapse of the soil structure).
H = Thickness of the collapsible layer
e0 = Initial void ratio

Collapse settlement (S1) = S1 + S2

Solved Example

A building was constructed on collapsible soil and the double oedometer test carried out on the undisturbed sample gave the results shown in Table 1. The thickness of the collapsible layer is 4 m and the average overburden pressure was 60 kPa. Calculate the collapse settlement for an increase in pressure of 40 kPa.

Applied Pressure (kPa)102040100200400800
Void ratio at natural moisture content0.800.790.780.750.7250.680.61
Void ratio at soaked condition0.750.710.660.580.510.430.32
Table 1-


Solution

(1) The first step is to plot the e-log P curve for the two samples as shown below. This has been done using Microsoft Excel. 

e-log P plot for double oedometer test
Figure 1: e-log P plot for double oedometer test

(2) The virgin compression curve is plotted by drawing a line tangential to the soaked sample curve.

(3) After drawing the virgin compression line, the overburden pressure is identified, and for this problem, it is given at 60 kPa. This is traced up to meet the virgin compression line, and this marks the initial void ratio (e0).

(4) From point e0, an adjusted moisture curve is plotted to be perfectly parrallel to the natural moisture content curve.

(5) The increment pressure p + ΔP = 60 + 40 = 100 kPa. A vertical line is drawn from the pressure of 100 kPa to meet the soaked sample curve and adjusted moisture curve.

Steps (2) to (5) are shown in Figure 2.

Graph for calculation of collapse settlement
Fig 2: Full analytical plot for collapse settlement of the soil

(6) Trace down the points where the pressure increment makes contact with the soaked sample curve and adjusted moisture content curve and identify the associated void ratios e1 and e2.

For the problem at hand;
e0 = 0.64
e1 = 0.62
e2 = 0.58

Therefore;
Δen = e0 – e1 = 0.64 – 0.62 = 0.02
Δec = e1 – e2 = 0.62 – 0.58 = 0.04

S1 = ΔenH /(1 + e0) = (0.02 x 4)/(1 + 0.64) = 0.0487 m
S2 = ΔecH /(1 + e0) = (0.04 x 4)/(1 + 0.64) = 0.0975 m

The total collapse settlement is therefore given by S1 + S2 = 0.0487 + 0.0975 = 0.146 m = 146.2 mm

Methods of Foundation and Construction on Collapsible Soils

(1) Utilising deep foundations such as piles to send the load beyond the collapsible layer.

(2) Chaining the foundation using continuous reinforced ground beams and slab to minimise differential settlement.

(3) In a laboratory study carried out by AlShaba et al.. (2018) at Egypt, iron powder was used to reduce collapse settlement of collapsible soil by about 46% at a mix ratio of 6% of the dry weight of the soil. When the same mixture was compacted, a reduction ratio of about 86.66% was obtained.

(4) Prewetting and compaction of the collapsible soil.

(5) Ayeldeen et al (2016) obtained promising results on improving collapsible soils using biopolymers.

References
[1] Ayeldeen M., Nrgm A., El-Sawwaf M., Kitazume M. (2016): Enhancing the Behaviour of Collapsible Soils using Bioploymers. Journal of Rock Mechanics and Geotechnical Engineering. doi:10.1016/j/jrmge.2016.11.007
[2] AlShaba A.A., Abdelaziz T.M., Ragheb A.M. (2018): Treatment of Collapsible Soils by Mixing with Iron Powder. Elsevier – Alexandria Engineering Journal (57):3737-3745
[3] Behzad Kalantari (2013): Foundations of Collapsible Soils: A Review. Proceedings of the Institute of Civil Engineers Forensic Engineering 166(FE2):57-63
[4] Jennings J.E., and Knight K. (1975): An Additional Settlement of Foundation Due to Collapse Structure of Snady Soils on Wetting. Proceedings to thhe 4th International Conference on Soil Mechanics and Foundation Engineering, Vol 1. Paris
[5] Murthy V.N.S (2012): Textbook of Soil Mechanics and Foundation Engineering (First Edition). CBS Publishers and Distributors Pvt Ltd, India
[6] Owolabi F.A., Ola S.A. (2014): Geotechnical Properties of a Typical Collapsible Soil i South-Western Nigeria. Electronic Journal of Geotechnical Engineering (19):1721-1738



Structural Analysis of Portal Frames Subjected to Gravity Load

By
Ubani Obinna Uzodimma
-

Gravity actions are normally used for the verification of portal frames for member resistances, lateral buckling, and torsional stability. Elastic design of portal frames is permitted by Eurocode 3, and the most significant load case (from experience) is the situation where the frame is subjected to gravity load from permanent actions and variable actions, taking into account the second-order effects and imperfection.

The load combination used for this is usually of the form shown in Equation (1);

p = 1.35gk + 1.5qk (1)


In the past, we have solved the problem of two hinged portal frames which are statically determinate due to the presence of internal hinge at the apex. However, it is not a very practical scenario as portal frames are required to be rigidly connected in order to achieve in-plane stability.

In this article, we are going to show how you can obtain the effects of actions (bending moment, shear force, and axial force) for a rigid portal frame subjected to an ultimate limit state load of 12 kN/m as shown in Figure 1. We will assume that the same universal beam section will be used for the column and the rafters (EI = Constant).

Portal Frame
Fig 1: Portal frame

A little consideration will show that the frame shown in Figure 1 is statically indeterminate to the first order. Therefore, using the force method of analysis, we will need to reduce the structure to a basic system.

This is a process by which we remove the redundants (excess reactive forces) in the structure, to make it statically determinate. The basic system adopted must also be stable. This is shown in Figure 2.

Basic System 1
Fig 2: Reduction of the frame to basic system

To proceed, we will now assign a unit force to the removed redundant horizontal reaction, and plot the bending moment diagram due to the unit force on the basic system as shown in Figure 3.

Bending Moment Diagram
Fig 3: Bending moment diagram due to horizontal unit force at point A

Since there is no other redundant support, we can apply the external load on the basic system. Due to the symmetry of the structure and the loading, we can verify that the vertical support reactions at A and E are given as shown in Figure 4;

Ay = Ey = (12 x 18)/2 = 108 kN

LOAD ON BASIC SYSTEM
Fig 4: External load and support reaction on the basic system

On observing Figure 4, we can see that there will be no bending moment in sections A-B and D-E, rather they will be subjected to compressive axial force of 108 kN. For member B-C, the bending moment will induced due to the reactive force and the externally applied uniformly distributed load. The free-body diagram of member B-C is shown in Figure 5.

We can verify that the geometrical properties of B-C are as follows;
The angle of inclination β = tan-1(1.5/9) = 9.462º
The length of the member z = √(1.52 + 92) = 9.124 m

section bc
Fig 5: Free body diagram of member of B-C

The equation for the bending moment of the section can be given by;
Mz = [108 x cos(β) x z] – [12 x cos(β) x z2]/2
Mz = 106.53z – 5.918z2

At z = 9.124 m
MCL = 106.53(9.124) – 5.918(9.124)2 = 479.32 kNm

The same thing is also happening at section C-D, and the bending moment diagram of the structure is shown in Figure 6.

BMD
Figure 6: Bending moment diagram of external load on basic system

The canonical equation for the structure is given by Equation (2);

δ11X1 + Δ1P = 0 ——– (2)

Where;
δ11 = Deflection at point 1 due to unit force at point 1
X1 = The actual external reaction at point 1 in the direction of the deflection due to externally applied load
Δ1P = Deflection at point 1 due to externally applied load

We are going to evaluate δ11 and Δ1P according to Vereshchagin’s rule which involves combining of the bending moment diagrams. For another example of how to apply Vereshchagin’s rule, click here.

Evaluation of δ11
In this case, the bending moment diagram of the unit force will combine with itself.

Combiner 1
Fig 7: Load Case combining with itself

δ11 = 2 x [1/3 x 8 x 8 x 8] + 2 x 1/3 x [(8 x 8) + (9.5 x 8) + (9.5 x 9.5)] x 9.124 = 1741.867/EI

Evaluation of Δ1P
Here, the bending moment diagram due to the unit force will combine with the bending moment diagram due to the externally applied load as shown in Figure 8.

Combiner 2
Fig 8: Load case 1 combined with externally applied load

Δ1P = 2 x (1/12) x 479.32 x [(5 x 9.5) + (3 x 8)] x 9.124 = 52115.345

Substituting back into the canonical equation;
X1 = 52115.345/1741.867 = 29.919 kN (This obviously represents the horizontal reactions Ax and Ex)

The final bending moment diagram can be obtained therefore as follows;
Mdef = M1X1 + M0
MA = ME = 0
MB = MD = (29.919 x -8) + 0 = -239.352 kNm
MC = (29.919 x -9.5) + 479.32 = 195.089 kNm

The bending moment diagram is shown in Figure 9, and the shear and axial forces can also be obtained accordingly.

Bending moment diagram of portal frame
Fig 9: Final bending moment diagram

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On the Consolidation Settlement of Pile Groups

By
Ubani Obinna Uzodimma
-

When a group of piles are embedded in saturated clays, the pile group undergoes time-dependent consolidation settlement. This can be calculated using the consolidation settlement equations and generally involves evaluating the increase in stress when the soil beneath a pile group is subjected to pressure due to load from the superstructure/column Qg.

The consolidation settlement of pile groups can be estimated by assuming an approximate distribution method that is commonly referred to as the 2:1 method.

The calculation procedure involves the following steps as described in Das (2008);

1. Obtain the total length of embedment of the pile L, and the total load (service load) transmitted from the column to the pile cap (Qg).
2. The load Qg is assumed to act on a fictitious footing depth of 2L/3 from the top of the pile. The load Qg spreads out along 2 vertical:1 horizontal lines from this depth.
3. Calculate the effective stress increase caused at the middle of each soil layer by the load Qg:

∆σi‘ = Qg/[(Bg + zi) (Lg + zi)]

Where;
∆σi‘ = effective stress increase at the middle of layer i
Bg, Lg = Width and length of the plan of pile group, respectively
zi = distance from z = 0 to the middle of the clay layer, i

4. Calculate the settlement of each layer caused by the increased stress:

∆Sc(i) = [∆ei / (1 + e0(i)] × Hi

where;
∆Sc(i) = consolidation settlement of layer i
∆ei = change of void ratio caused by the stress increase in layer i
e0(i) = initial void ratio of layer i (before construction)
Hi = Thickness of layer i

5. Calculate the total consolidation settlement of the pile group by;
∆Sc(g) = ∑∆Sc(i)

Solved Example

Calculate the consolidation settlement of a group of 4 piles founded on layers of clay as shown below. The piles are 600mm in diameter and form a square plan of 1800mm x 1800mm. The service load from the column on the pile is 1350 kN.

consolidation settlement of pile group

Solution
Depth of fictitious footing = 2L/3 = 2(20)/3 = 13.33 m
By implication, it can be seen that the influence depth went beyond the first layer of clay. Therefore, consolidation settlement calculation commences from the second clay layer of the soil profile. Let us therefore call this first layer of influence, layer 1.

Layer 1
∆σ1‘ = Qg/[(Bg + z1) (Lg + z1)] = 1350 / [(1.8 + 4.585) × (1.8 + 4.585)] = 33.114 kPa

σ0(1)‘ = (2 × 17) + 12(18.5 – 9.81) + 5.415(19 – 9.81) = 188.04 kPa

∆Sc(1) = [(Cc(1)H1)/(1 + e0)] × log[(σ0(i)‘ + ∆σi‘ )/σ0(i)‘]
∆Sc(1) = [(0.28 × 9.17)/(1 + 0.73)] × log[(188.04 + 33.114)/188.04] = 0.1045m = 104.473 mm

Layer 2
∆σ2‘ = Qg/[(Bg + z2) (Lg + z2)] = 1350 / [(1.8 + 12.67) × (1.8 + 12.67)] = 6.4475 kPa

σ0(2)‘ = (2 × 17) + 12(18.5 – 9.81) + 10(19 – 9.81) + 3.5(18.3 – 9.81) = 259.895 kPa

∆Sc(2) = [(Cc(2)H2)/(1 + e0)] × log[(σ0(2)‘ + ∆σ2‘ )/σ0(2)‘]
∆Sc(2) = [(0.25 × 7)/(1 + 0.77)] × log[(259.895 + 6.4475)/259.895] = 0.0104m = 10.48 mm

Therefore total settlement ∆Sc(g) = ∆Sc(1) + ∆Sc(2) = 104.473 + 10.48 = 114.953 mm

Thank you so much for visiting Structville today, and God bless you.

References
Das B.M. (2008): Fundamentals of Geotechnical Engineering (3rd Edition). Cengage Learning, USA

Analysis of Beams on Elastic Foundation

By
Ubani Obinna Uzodimma
-

When a beam lies on an elastic foundation under the action of externally applied loads, the reaction forces of the foundation are proportional at every point to the deflection of the beam. This assumption was introduced first by Winkler in 1867. There are cases in which beams are supported on foundations which develop essentially continuous reactions that are proportional at each position along the beam to the deflection of the beam at that position.

This is the reason for the name ‘elastic foundation’. There are many geotechnical engineering problems that can be idealized as beams on elastic foundations. This kind of modelling helps to understand the soil-structure interaction phenomenon and predict the contact pressure distribution and deformation within the medium (e.g. soil). The most common theory for a beam on elastic foundation modelling is the Winkler approach.

image 5
Figure 1: Schematic representation of beam on elastic foundation


For instance, the beam shown in Fig. 1 will deflect due to the externally applied load, and produce continuously distributed reaction forces in the supporting medium. The intensity of these reaction forces at any point is proportional to the deflection of the beam y(x) at that point via the constant ks:

R(x) = ks.y(x)  — Eq. (1)

Where ks is the soil’s modulus of subgrade reaction which is the pressure per unit settlement of the soil (unit in kN/m2/m). The reactions R(x) act vertically and oppose the deflection of the beam. Hence, where the deflection is acting downward there will be a compression in the supporting medium. Where the deflection happens to be upward in the supporting medium tension will be produced which is not possible (for soils).

If we assume that the beam under consideration has a constant cross-section with constant width b which is supported by the foundation. A unit deflection of this beam will cause a reaction equal to ks.b in the foundation, therefore the intensity of distributed reaction (per unit length of the beam) will be:

R(x) = b.ks.y(x) = k.y(x) —-Eq. (2)

where k = k0.b is the constant of the foundation, known as Winkler’s constant, which includes the effect of the width of the beam, and has unit of kN/m/m.

The general 4th order differential equation for beam on elastic foundation is given by equation (3);

EI(d4y)/dx4 + k.y = q  —- Eq. (3)

The homogenous equation is given by;

EI(d4y)/dx4 + k.y = 0

(d4y)/dx4 + 4β4y = 0

Where β = ∜(k/4EI) = (k/EI)(1/4)

The general solution for the equation is available, which is given by;
y = eβx (C1sinβx + C2cosβx) + e-βx(C3sinβx + C4cosβx)

Warren and Richard (2002) published tables containing equations for analysing beams on elastic foundation subjected to different loads. The method described in the book has been employed in this article to analyse a beam on an elastic foundation and the results compared with the results from Staad Pro software.

Solved Example
A 600mm x 400mm rectangular beam is resting on a homogenous soil of modulus of subgrade reaction ks = 10000 kN/m2/m. The beam is 10m long and carries a concentrated load of W = 300 kN at 3m from the left-hand side. Neglecting the self-weight of the beam, and assuming freely supported ends, obtain the bending moment at the point of the concentrated load. Take the modulus of elasticity of concrete Ec = 21.7 x 106 kN/m2.

BEAM 2BON 2BELASTIC 2BFOUNDATION
Example of beam resting on an elastic foundation
beam 2Bsection

Solution 
Second moment of area of concrete beam IB = (bh3)/12 = (0.4 × 0.63)/12 = 7.2 x 10-3 m4
Flexural rigidity of the beam EcIB = 21.7 × 106 × 7.2 × 10-3 = 156240 kN.m2
β = (bks/4EI)(1/4) = [(0.4 × 10000)/(4 × 156240)](1/4) = 0.2828
βl = 0.2828 × 10 = 2.828 m; β(l – a) = 0.282(10 – 3) = 1.979

Where a is the distance of the concentrated load from the left end of the beam.
Since βl < 6.0, we can use Table 8.5 of Roark’s Table for Stress and Strain (Warren and Richard, 2002).

For a beam with both ends free;
RA = 0; MA = 0

The equations for bending moment and shear force along the beam is as given below;

Mx = MAF1 + RA/2βF2 – yA2EIβ2F3 – θAEIβF4 – W/2βFa2
Vx = RAF1 – yA2EIβ3F2 – θAEIβ2F3 – MAβF4 – WFa1

Where;
θA (rotation at point A) = [W/(2EIβ2)] × [(C2Ca2 – 2C3Ca1)/C11]
yA (vertical deflection at point A) = [W/(2EIβ3)] × [(C4Ca1 – C3Ca2)/C11]

We can therefore compute the constants as follows;
C2 = coshβl.sinβl + sinhβl.cosβl = [cosh(2.828) × sin(2.828)] + [sinh(2.828) × cos(2.828)] = (8.485 × 0.3084) + (8.426 × –0.9512 ) = –5.398

C3 = sinhβl.sinβl = (8.426 × 0.3084) = 2.5985

C4 = coshβl.sinβl – sinhβl.cosβl = [cosh(2.828) × sin(2.828)] + [sinh(2.828) × cos(2.828)] = (8.485 × 0.3084) – (8.426 × –0.9512) = 10.631

Ca1 = coshβ(l – a).cosβ(l – a) = cosh(1.974) × cos(1.974) = 3.669 × -0.392 = -1.438

Ca2 = coshβ(l – a).sinβ(l – a) + sinhβ(l – a).cosβ(l – a) = [cosh(1.974) × sin(1.974)] + [sinh(1.974) × cos(1.974)] = (3.669 × 0.9198) + (3.530 × -0.392) = 1.9909

Ca3 = sinhβ(l – a).sinβ(l – a) = (3.530 × 0.9198) = 3.247

C11 = sinh2βl – sin2βl = 8.4262 – 0.30842 = 70.902

θA = W/(2EIβ2) × [(C2Ca2 – 2C3Ca1)/C11]
θA = [300/(2 × 156240 × 0.28282 ) × [(–5.398 × 1.9909) – (2 × 2.5985 × –1.438)/70.902] = (0.012 × –0.0462) = -0.0005544 radians

yA (vertical deformation at point A) = [W/(2EIβ3)] × [(C4Ca1 – C3Ca2)/C11]
y = [300/(2 × 156240 × 0.28283)] × [(10.631 × –1.438) –2.5985 × 1.9909)/70.902] = (0.0424 × –0.2885) = –0.01223 m = -12.23 mm

Bending moment at point C
Substituting the values of deflection and slope into the equation for bending moment (note that the first and second terms of the equation goes to zero since RA = MA = 0);

Mx = – yAEIβ2F3  – θAEIβF4 – W/2βFa2
Mx = – (–0.01223 × 2 × 156240 × 0.28282)F3 – (– 0.0005544 × 156240 × 0.2828)F4 – 300/(2 × 0.2828)Fa2

Mx = 305.638F3 + 24.495F4 – 530.410Fa2

Substituting F3, F4, and Fa2 (see Table 8.5, Warren and Richard, 2002) into the equation;

Mx = 305.638(sinhβx.sinβx) + 24.495(coshβx.sinβx – sinhβx.cosβx) – 530.410[coshβ(x –a).sin β(x –a) + sinh β(x –a).cos β(x –a)]

The bending moment under the concentrated load (point C);
x = 3m; (x – a) = 3 – 3 = 0

βx = 0.8484
Mx = 305.638[sinh(0.8484) × sin(0.8484)] + 24.495[cosh(0.8484) × sin(0.8484) – sinh(0.8484) × cos(0.8484)]

Mc = 305.638 (0.953 × 0.750) + 24.495(1.382 × 0.750 – 0.9539 × 0.6611) = 218.45 + 9.942 = 228.392 kN.m


Verification
This manual calculation has been verified using Staad Pro software. The steps adopted were as follows:

(1) Modelling
The 10m beam was modelled as a one-dimensional line element connected by nodes at 1m length interval. This was to represent/attach soil springs at 1m interval.

node 2Bconnection

(2) Section Properties

The beam was modelled using concrete of modulus of elasticity = 21.7 x 106 kN/m2, with dimensions of 600mm x 400mm.

(3) Support 

Using the elastic mat foundation option will not work since the support will not form a closed loop (analysing this way will give a ‘colinear support error’). Therefore, the soil spring was modelled using the ‘FIXED BUT’ support option. The support must be released for moment since this is the ideal scenario for the structure we are trying to model.

The soil modulus of the subgrade reaction was multiplied by the width of the beam thus;

kb = 10000 kN/m2/m x 0.4m = 4000 kN/m2

Now, given that the nodes are spaced at 1m interval, the vertical spring constant was taken as 4000 kN/m. The general form of the foundation is given below;

soil 2Bspring 2Bmodel
Modelling of beam on elastic foundation

(4) Analysis and Results

When analysed using the static check option, the following results were obtained;

(a) Soil deformation

settlement

The vertical deflection at point A (node 1) was observed to be 12.195mm, against 12.23mm obtained using manual analysis.

(b) Bending Moment

Bending 2Bmoment
Staad Pro result for beam on elastic foundation

The bending moment obtained at point C was 228.146 kNm which is comparable with 228.392 kNm obtained using manual calculations. Therefore, it can be seen that both methods give approximately the same result.

References
Warren C.Y., Richard B. Y. (2002): Roark’s Formula for Stress and Strain (7th edition). McGraw Hill, USA


Analysis and Design of Cantilever Retaining Walls on Staad Pro

By
Ubani Obinna Uzodimma
-
In this post, we are going to show how cantilever retaining walls can be analysed and designed on Staad Pro software, and also compare the answer obtained with classical solutions. We should know that retaining walls must satisfy geotechnical, equilibrium, structural, upheaval, seismic considerations, etc. As a result, the designer must ensure that by appropriate knowledge of materials, site conditions, etc, he/she will provide suitable dimensions of the retaining wall that will ensure resistance of the structure to overturning, sliding, bearing capacity failure, uplift, etc. After appropriate sizing of the retaining wall, the structural analysis and design will commence to determine the action effects (bending moments, shear forces, axial forces, deflection etc), and provision of proper reinforcements to resist the action effects.


In the past, Structville has published a 17 page document on geotechnical design of cantilever retaining walls subjected to earth load, pavement surcharge load, traffic load, etc. This loading situation can be found when retaining wall is used to support embankment carrying traffic road way.  It was interesting to see how Design Approach 1 (DA1) of Eurocode 7 was used to ensure the geotechnical stability of the wall. Just in case you missed it, kindly download the PDF from the link below;

Geotechnical Design of Cantilever Retaining Walls to Eurocode 7

In this post, let us consider the retaining wall sized and loaded as shown in Figure 2. This structure has been modelled on Staad Pro in order to determine the action effects due to the applied load.

Cantilever 2Bretaining 2Bwall
Fig 2: Cantilever retaining wall

The retaining wall is subjected to a 3m thick earthfill, and a variable surcharge pressure of 10 kPa. Given that the retained earth has an angle of internal friction of 30°, we can obtain the Rankine active earth pressure as follows;

ka = (1 – sin 30)/(1 + sin 30) = 0.333

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Therefore the actions on the retaining wall that will be input into Staad Pro as are as follows;

Vertical actions
(1) Self weight (to be calculated automatically by Staad)
(2) Weight of earthfill (19 kN/m3 × 3m) = 57 kN/m2
(3) Surcharge load = 10 kN/m2

Horizontal Actions
(4) Triangular earth pressure = (0.333 × 19 × 3) = 18.98 kN/m2
(5) Uniform surcharge pressure = (0.333 × 10) = 3.333 kN/m2

The wall has been modelled per metre run on Staad, and plate mat foundation was utilised with coefficient of subgrade modulus of 100000 kN/m2/m.

Steps to adopt
(1) Model the retaining wall utilising plate element meshing, assign thickness of 0.4m to the base, and 0.3m to the wall. Also assign plate mat foundation of subgrade modulus 100000 kN/m2/m to the base in the y-direction.

RET 2B1
Fig 3: Modelling and meshing of the retaining wall

(2) Assign the following loads to the structure

Load Case 1 (LC1)
(a) Self weight to the whole structure
(b) Weight of earth fill to the heel of the retaining wall (57 kN/m2)
(c) Assume that the base is buried 1m into the ground, hence apply vertical pressure load of (19 kN/m2) to the toe but neglect all passive pressures.

a1
Fig 4: Permanent vertical actions on the retaining wall base
final 2Bfront 2Bcover
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Load Case 2 (LC2)
(d) Triangular earth pressure to the wall (18.98 kN/m2)

a2
Fig 5: Horizontal active earth pressure on the wall

Load Case 3 (LC3)
(e) Uniform horizontal surcharge pressure to the walls (3.333 kN/m2)
(f) Uniform vertical surcharge pressure to the heel (10 kN/m2)

a3
Fig 6: Surcharge loads on the wall and on the base


Combination (Ultimate limit state)
pEd = 1.35LC1 + 1.35LC2 + 1.5LC3

A little consideration will show that the load cases 1 and 2 are treated as permanent actions, while load case 3 is treated as a variable action.

(3) Analyse the structure for the load cases
(N/B): You may need to increase the iteration limits for the load cases containing horizontal actions to converge

MX1
Fig 7: Main bending moment on the retaining wall
MXY
Fig 8: Twisting moment on the retaining wall
SQX
Fig 9: Shear stress on the retaining wall

On considering the bending moment diagram Mx (Figure 7), we can see that the maximum moment close to the base of the wall is 57 kNm/m. If we add the effect of torsion (Figure 8), the design moment can be taken as 57 + 2.99 = 59.99 kNm/m.

To carry out manual analysis, we will have to follow the steps given below to obtain the maximum moment at the base of the wall. The actions causing bending on the wall are the horizontal earth pressure and the horizontal surcharge pressure.

Moment from surcharge pressure = [(3.333 × 32)/2] = 14.998 kNm/m
Moment from horizontal earth pressure = [(18.98 × 3)/2] × (3/3) = 28.47 kNm/m

At ultimate limit state, MEd = 1.35(28.47) + 1.5(14.998) = 60.931 kNm/m

A little consideration will show that Staad Pro and Manual calculations gave almost the same value for wall bending moment. However, I expect the value of base moment from Staad Pro to be lower than the one from manual analysis. Kindly verify this at your private time.


Thank you for visiting Structville today and God bless you.

An Investigation on the Analysis of Beam and Raft Slab Using Staad Pro

By
Ubani Obinna Uzodimma
-

Modern codes of practice are increasingly recognising the computational power of structural analysis  softwares. Staad Pro is a renowned structural analysis and design software that is used all over the world, and this post is aimed at investigating the analysis of beam and raft foundation using Staad Pro V8i. Winkler’s model has been used as a basis for the analysis. This post will help engineers who use Staad Pro to make decisions on whether raft slab can be confidently analysed on Staad Pro V8i environment, or there will be need to move to Staad Foundation, or any other foundation analysis software.

Winklers 2BModel
Fig 1: Winkler’s Model

Winkler’s model assumes that soil possesses stiffness which is considered to be the ratio of the contact pressure of the soil, and the vertical deformation associated with it. This relationship is assumed to be linear, and can be given by what is called the coefficient of subgrade reaction (ks).

ks = q/S ——————- (1)

where;
q is the contact pressure at a point in the footing, and
S is the settlement at the same point.

Equation (1) assumes that for granular soils, the value of ks is independent of the magnitude of pressure, and is the same at all points on the surface of the footing. These two assumptions are however not very accurate for obvious reasons, especially for flexible foundations. There are lots of literature available on the topic of ‘coefficient of subgrade reaction’, and you are advised to read them up on Google. However, it is accepted that the method gives realistic contact pressures especially when considering very low order of settlement. Coefficient of subgrade reaction is obtained in the field using in-situ plate load test. Plate load tests are usually carried out using (300 x 300) mm plate, and there is usually need to correct the obtained value for the actual width of the foundation (for a quick check of procedure, see page 696 of Murthy, 2012). I will not actually dabble into the accuracy of application of coefficient of subgrade reaction for foundation design in this post, but the reader is advised to consult wide range of literature since opinion on this seems to vary.

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In Staad Pro software, the use of Winkler’s Model can easily be done by applying ‘foundation’ support option to the structure in question. Under this option, you can use ‘elastic mat’ or ‘plate mat’. Elastic mat is suitable for beam and plate combination (e.g beam and raft slab in a building) while plate mat is good when you have plate elements only forming the base (e.g base of an underground water tank). For training on how to use Staad Pro efficiently, contact the author on ubani@structville.com


In many parts of Nigeria, beam and raft slab is employed for low-medium rise buildings when the soil has low bearing capacity (usually between 100 kPa and 50kPa) or where providing separate bases will pose so many challenges. For manual analysis of such foundations, the rigid approach is usually used, which yields higher value of internal stresses, while on the other hand, finite element based foundation design softwares can be used, which usually yields lower internal stresses but higher settlement (flexible approach).

In the design of beam and raft slab foundation, it is usually assumed that the column load gets transferred/distributed to the ground beams, and tries to push them down into the soil. This action is resisted by the earth pressure intensity, which is mainly transferred to the ground floor slab. The schematic diagram of this action is shown below.

Schematic 2BDiagram 2Bof 2BBeam 2Band 2BRaft 2BFoundation
Fig 2: Load Path of Beam and Raft Foundation

Design Example
As an example, let us consider a two storey building with the foundation layout shown below.

GA
Fig 3: Typical Foundation Layout of Structure

For the beam and raft layout shown in Fig. 3, the necessary data are as follows;

Ground beam dimensions = 1200 x 250 mm
Thickness of ground floor slab = 150 mm
Dimension of all columns = 230 x 230 mm
First floor slab = 150 mm
First floor beams = 450 x 230 mm
Roof beams = 300 x 230 mm
Additional dead load on floors (finishes and partition allowance) = 2 kN/m2
Imposed load on building (variable action) = 2 kN/m2
Load at ultimate limit state = 1.35gk + 1.5qk
Coefficient of subgrade reaction ks =  of 20000 kN/m2/m (assumed for very soft clay, reference taken from CSC Orion Software)

The 3D view of the frame is shown in Figure 4 below;

3D 2BFRAME 2BOF 2BBUILDING
Fig 4: 3D Model of the Building

After analysis of the superstructure on Staad, the following ultimate column axial loads which are transferred to the foundation are shown below;

COLUMN 2BLOAD 2BON 2BRAFT 2BSLAB
Fig 5: Column Loads on the Foundation

From figure 5 above, I believe you can easily model the foundation in another software, or carry out manual analysis for checks of the procedure adopted here.

When the ground beams and elastic mat foundations are applied on Staad Pro, the results are as follows;

SUPPORTS 2BON 2BRAFT
Fig. 6: Typical View of the Structure with Elastic Mat Foundation




(1) Base Pressure

BASE 2BPRESSURE
Fig. 7: Base Pressure

A little consideration will show that we are obtaining a maximum base pressure of 73 kPa at the edges, which might be higher than the bearing capacity of the soil. The engineer is advised to review this properly. However, along grid lines 2 and 3, we have base pressure of 39 kPa which increases towards the edges. The minimum base pressure occurred at the mid-spans.

(2) Moment in the raft slab (x-direction)

MX
Fig 8: Mx Moment

The bending moment on the raft slab was discovered to be low when compared with the value that could be gotten using the rigid approach (the maximum moment can be seen to be within 10.9 and 13.4 kNm/m)

final 2Bfront 2Bcover
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(3) Bending moment in the Y-direction

MY
Fig 9: My Moment


(4) Twisting Moment

MXY
Fig 10: Mxy Moment

(5) Bending Moment in the ground beams

BENDING 2BMOMENT 2BIN 2BGROUND 2BBEAMS
Fig 11: Bending Moment Diagram of the Ground Beams

For simplicity, the maximum design forces have been presented in the table below;

Maximum 2BMoments

The bending moment on ground beam of grid line (2) is given below;

BM2
Fig 12: Bending Moment for Ground Beam Grid line 2

What are your thoughts on the values obtained? At your convenience, you can model the raft slab on the software you normally use, and compare the results obtained. You can present your observations in the comment section for review.

Thank you for visiting Structville, and God bless you.

Citation
Murthy V.N.S (2012): Textbook of Soil Mechanics and Foundation Engineering (Geotechnical Engineering Series) 1st Edition. CBS Publishers and Distributors Pvt Ltd, India ISBN: 81-239-162-1


Analysis and Design of Box Culvert Using Staad Pro

By
Ubani Obinna Uzodimma
-

In our last post, we were able to establish how we can load box culverts properly. If you missed the post, kindly follow the link below to read it;
Loading and Analysis of Box Culverts to Eurocode 2


In this post, we are going to describe how we can model, load, and analyse box culverts using Staad Pro software. Here is a quick a recap of the properties of the box culvert under consideration;

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
Thickness of all elements = 300 mm
Thickness of asphalt layer = 70 mm
Materials property
Angle of internal friction of fill soil = 30°
Unit weight of water = 9.81 kN/m3
Unit weight of back fill 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
box 2Bculvert 2Bdimensions
Fig 1: Section of the Box Culvert
The steps in modelling the structure on Staad Pro are as follows;

(1) Meshing
Here, the box culvert is idealised with dimensions based on centre to centre of the slabs and walls. This means that the width of the box culvert that will be input into Staad Pro is 2.5 m, while the depth will be 2.0 m. This can be started by forming the nodes in the global XY plane, and then copying and pasting for the length of 8 m in the Z global direction. The output of this operation is as given below;

initial 2Bmeshing
Fig 2: Initial Nodes for Commencement of Modelling

After forming this, the wing walls can also be formed, which is followed by meshing (rectangular or polygonal) to form the shell of the box culvert. The final output of the meshing operation is as shown below;

3D 2Bmeshing
Fig 3: Fully Meshed Box Culvert

The meshing process can be completed by adding plate thickness of 300 mm to all the elements.


(2) Assigning of support conditions/foundations
This is an important aspect of modelling structures. A purely rigid approach will involve using fixed supports, but note that employing 3D model for this purpose will not be very appropriate, but a simple 2D frame model will be better. There are many proposals on how culverts can be modelled as 2D frames, and the reader is advised to consult as many publications as possible. However, to incorporate the effects of soil-structure interaction (to a limited degree though) on our model, we can employ the use of ‘elastic plate‘ foundation option on our model.

If we assume that the supporting soil and the backfill are of the same material, then we can maintain the same angle of internal friction of 30°. Angle of internal friction of 30° can suggest a loose – medium dense sand in its undisturbed state, therefore we can take a modulus of subgrade reaction of 50,000 kN/m2/m for a well compacted sand. For a slightly compacted sand, you can take a value of 30,000 kN/m2/m.

So we can input this option into Staad Pro using ‘compression only’ option (see the dialog box below in Fig 4);

elastic 2Bmat 2Bfoundation 2Bon 2BStaad 2Bpro
Fig 4: Elastic Mat Foundation Dialog Box

When this is applied on the structure, we have the final model as shown in Fig 5;

3D 2BELASTIC 2BMAT 2BFOUNDATION
Fig 5: Application of Elastic Mat Foundation on the Model

What this model (Fig. 5)  means is that every node of the base slab is in contact with the soil, and the soil is represented by a spring of stiffness 50,000 kN/m2/m which is the subgrade modulus of the soil.

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(3) Loading
This is another important aspect of modelling structures because analysis results would go very wrong if the loads are not applied properly. If you review our previous post, we considered two construction cases;

(A) where the culvert is buried under the soil, and
(B) where is there no earth fill on top of the culvert.

When there is no earth fill on the culvert, the traffic load is directly on the top slab of the culvert as tandem loads and as UDL, but when there is earth fill, traffic load is dispersed in the ratio of 2:1 as UDL on the culvert. We are going to consider 5 load cases in our analysis in this model;

(1) Self weight and other superimposed actions
(2) Vertical earth load
(3) Traffic load
(4) Surcharge load
(5) Horizontal earth pressure load

These  load cases are considered independently at first, and then combined with appropriate partial factors of safety to determine the design actions. Please note that effects of ground water and the pressure in the shell of the culvert when it is filled with water is an important load case too but was not considered in this post. We have determined the magnitude of these loads in our previous post, and we are going to apply them on the box culvert for Case A and Case B.

In our previous post, we were able to analyse the loads on the culvert for Case A as follows;

(1) Self weight: To be calculated automatically by Staad + 1.69 kN/m2 (self weight of asphalt wearing course)
NB: In some cases, the partial factor of safety for self weight of concrete elements and other superimposed dead loads like asphalt wearing course might be different, so in that case, it is very advisable to treat each of them as a separate load case on Staad.
(2) Vertical earth load on the culvert = 22.80 kN/m2
(3) Traffic load dispersed as UDL = 59.523 kN/m2
(4) Horizontal surcharge load = 5.0 kN/m2
(5) Horizontal earth pressure load = trapezoidal distribution with minimum earth pressure of 11.40 kN/m2 at the top of the culvert and 33.25 kN/m2 at the bottom of the culvert

Load cases 1-4 are very easy to apply on Staad by utilising uniform pressure action on plates, however while load cases 1-3 are applied in the global Y direction (gravity loads), load cases 4-5 are applied in the global X direction (horizontally). I will only demonstrate how load case 5 can be applied, since it will involve using the hydrostatic load command on Staad. This is shown on Fig. 6.
HORIZONTAL 2BEARTH 2BPRESSURE 2BLOAD 2BON 2BSTAAD 2BPRO
Fig 6: Application of Horizontal Earth Pressure as Hydrostatic Load on Staad
The steps involved in applying hydrostatic load can be described as follows:
(i) Launch the hydrostatic plate pressure load on Staad
(ii) Select/highlight all the plates that will receive the load
(iii) Input the minimum and maximum pressure loads based on the notation given on Staad (in this case you can see that our maximum pressure load served as W1), and also input the sign conventions properly in order to identify the direction of pressure.
(iii) Put the interpolation direction (in this case we interpolated the load in the Y-direction)
(iv) Add the load case
After all said are done, the output should be as given below;
PRESSURE 2BLOAD
Fig 7: Earth Pressure on the Culvert
At this point, we are going to define the load combination for ultimate limit state as succinctly as possible. The reader is advised to consult the relevant code of practice and standard textbooks on how to group and combine loads involving traffic actions, earth load, etc. In this case, EN 1990, EN 1991-1, EN 1991-2, and EN 1997-1 can be consulted. Note that culverts are also analysed for other loads such as temperature effects, collision on head walls, centrifugal actions, braking actions, etc.
The load combination principle adopted here is based on expressions (6.10a and 6.10b) of EN 1990.
From Table A2.4(B) of EN 1990:2002 + A1:2005, we are going to adopt the following partial factors:
  • All permanent actions including superimposed dead load and vertical earth pressure  γG = 1.35
  • Leading/main traffic action γQ,1 = 1.35
  • Traffic surcharge γQ,2 = 1.5
  • Horizontal earth pressure and ground water γQ,2  = 1.5
We will now apply this load combination on Staad Pro as shown in Fig. 8.

ULS
Fig 8: ULS Load Combination Dialog Box


(4) Analysis
On analysing the structure, we obtain the following results at ultimate limit state;

MX
Fig 9: Transverse Bending Moment
final 2Bfront 2Bcover

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Do you wish to get trained in theory of structures, structural design, and use of Staad Pro for analysing and designing of complex structures? Then quickly contact us;
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MY
Fig 9: Longitudinal Bending Moment
BASE 2BPRESSURE
Fig 10: Base Pressure
SQX
Fig 11: Transverse Shear Stress

A little consideration will show that the top slab is subjected to an ultimate design moment MEd of 56.5 kNm/m and an axial pull of NEd of 91.5 kN/m (check for membrane stresses in your own Staad Model).

NB: Shear and membrane forces in plates are expressed in Mpa in Staad, so will you have to multiply them with the thickness of the element to get the values in kN/m.

For the top slab of the culvert, the M-N interaction chart is given in Fig. 11 below for obtaining the design reinforcement.

Column 2Bdesign 2Bchart
Fig 12: M-N Interaction Chart on the Box Culvert

When designed, the area of steel for the axial compression and bending is 1223 mm2/m. Therefore, provision of H16@200 c/c (Asprov = 2010 mm2) on each face will be adequate.

Design Case B: No Earth Fill on Box Culvert
(1) Traffic Loads
When there is no earth fill on the box culvert, all we have to do is to remove the vertical earth fill load, apply direct traffic load on the top slab, and edit the horizontal earth load from trapezoidal to rectangular. As a reminder, the nature of Load Model 1 (which can be used for global and local verification) on the culvert is given in Fig 14 below;

Eurocode 2Btraffic 2Bload 2Bon 2Bculvert
Fig 14: Load Model 1 on the Culvert

The ideal thing is to apply a moving load on Staad after vehicle definition, such that the worst effect can be obtained. Note that you cannot apply a moving load directly on plate elements on Staad Pro, but you will need to create dummy beam members of negligible stiffness so that the axles can sit on them. In this post, we are not going to bother ourselves with the process, but we are going to treat the wheel load as static.

Influence line has shown that the most onerous bending moment is obtained when the front axle is 0.26 m beyond the mid-span of the culvert. Therefore, we are going to apply static wheel load at that location. Remember that it is always recommended to apply the full tandem system of LM1 whenever applicable. The critical location of wheel load on the box culvert for maximum moment is given in Fig 15 below;

most 2Bcritical 2Bwheel 2Bload 2Blocation 2Bon 2Bbox 2Bculvert
Fig 15: Most Critical Location of Wheel Load on the Culvert
When the static traffic load is applied on Staad and viewed longitudinally on the box culvert, we can see it as given in Fig 16 (note that the unloaded areas represent the wing walls).
Traffic 2Bload
Fig 16: Application of LM1 on Staad Pro

(2) Non-traffic Loads
For Design Case A, it is observed that the self weight of the structure and asphalt layer remains the same, the surcharge load also remains the same, but the horizontal earth pressure changes to triangular distribution with a maximum pressure of 21.85 kN/m2 at the bottom of the culvert.


(3) Analysis
When the structure is analysed, the results at ultimate limit state are as shown below;

MX2
Fig 17: Transverse Bending Moment
MY2
Fig 18: Longitudinal Bending Moment
BASE 2BPRESSURE 2B2
Fig 19: Base Pressure

Our analysis results have shown that when there is earth fill, the bending moment at ultimate limit state on top of the culvert is about 56.5 kNm/m, but when traffic is directly on top of the culvert, the bending moment is about 62 kNm/m. This is about 8.8% difference.

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Loading and Design of Box Culverts to the Eurocodes

By
Ubani Obinna Uzodimma
-

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/m3 = 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

By
Ubani Obinna Uzodimma
-

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,R = Mp,Rafter /fy
Wpl,y,C = Mp,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.

Thank you for visiting Structville today and God bless you.


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