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Design of Timber Roof Truss

The use of timber as trussed rafters for roofs of buildings is a very popular alternative all over the world. Timber roof trusses are made up of a series of interconnected timber members that are arranged in a triangular pattern. This pattern allows the truss to distribute the weight of the roof evenly across its supports, making it a strong and efficient structural system.

roof truss design
Figure 1: Members of a timber roof truss

The aim of this article is to show the design example of a timber roof truss (trussed rafter). As a direct product of nature, timber has so many variable properties that are more complex than that of concrete, steel, bricks, or aluminium. Some of the characteristics which influence the structural behaviour of timber are;

  • moisture content
  • the direction of applied load (perpendicular or parallel to the grain)
  • duration of loading
  • strength grading of the timber
commercial timber log
Figure 2: Unprocessed timber logs

Steps in the Design of Timber Roof Trusses

Here’s a step-by-step overview of the design process for timber roof trusses:

1. Preliminary Design: In this step, the designer is to determine the roof span, pitch, and overall architectural design of the building. It is important to choose a truss configuration based on structural performance, aesthetic preferences, functional requirements, and available space.

2. Load Analysis: In this case, it is important to identify and analyze all relevant loads, including dead loads (roofing materials, insulation), live loads (snow, occupants), and potential wind or seismic loads. Calculate the total design load that the truss system must support.

3. Material Selection: Choose an appropriate timber species based on its mechanical properties, availability, and durability. It is important to consider factors such as bending strength, compression and tension properties, and resistance to decay.

4. Truss Geometry and Configuration: Based on the chosen truss configuration (e.g., king post, queen post, Howe truss), determine the angles, lengths, and dimensions of each timber member. It is important to ensure that the truss geometry is consistent with both structural and architectural requirements.

5. Member Sizing and Design: Calculate the axial forces, bending moments, and shear forces in each timber member using static equilibrium equations. Size each timber member to withstand the calculated forces while considering factors such as serviceability and durability.

6. Connection Design: Design secure and efficient connections between timber members using appropriate fasteners (nails, screws, bolts) and connectors (metal plates, brackets). Ensure that connections provide load transfer, prevent excessive movement, and account for potential wood shrinkage.

7. Lateral Stability and Bracing: Incorporate diagonal bracing or other lateral stability measures to prevent buckling or twisting of the truss members under lateral loads (wind or seismic forces).

8. Serviceability and Deflection Control: Assess the truss deflection under the designed loads to ensure that it meets acceptable serviceability criteria. Incorporate additional members or adjust member sizes if necessary to control deflection.

9. Fire Resistance and Protective Treatments: Consider fire resistance requirements by choosing fire-rated timber or applying fire-retardant treatments if needed. Apply appropriate protective coatings or treatments to enhance the truss’s durability and resistance to decay.

10. Detailed Drawings and Documentation: Create detailed construction drawings that specify the geometry, dimensions, connections, and material details of the truss members. Prepare design calculations and documentation that outline the design methodology, load assumptions, and member sizing.

Analysis of Timber Roof Trusses

Ideal trusses are theoretical structures in which the members meet at points called nodes. These nodes are idealized as hinges or pins, which means that they cannot transmit bending moments. Loads are applied to ideal trusses only at the nodes. This keeps the truss members free from shear and bending stresses and makes the analysis of the truss much simpler.

However, practical construction does not allow roof trusses to behave exactly as ideal trusses. The members of real-world trusses are not pinned at the nodes, and loads are often applied along the length of the chords. This means that practical trusses must resist bending moments and shear in addition to axial stress.

As a result, the classical methods of analyzing trusses are only valid for ideal trusses. These methods do not account for the bending moments and shear in practical trusses. As a result, the results of classical methods of analysis can be inaccurate for real-world trusses, even though they are usually employed.

Quickly in this post, I am going to carry out a very simple design example of timber roof truss using BS 5268. A lot of information regarding timber as a structural material can be obtained from specialist textbooks. It is worth knowing that the most current design code for timber structures is Eurocode 5.

Timber%2Broof%2Btruss%2Bdesign

Note:
BS 5268 is based on permissible stress design. When using permissible stress design, the margin of safety is introduced by considering structural behaviour under working/service load conditions and comparing the stresses thereby induced with permissible values. The permissible values are obtained by dividing the failure stresses by an appropriate factor of safety. The applied stresses are determined using elastic analysis techniques, i.e.

Stress induced by working loads ≤  (failure stress/factor of safety)

Since BS 5268 is a permissible stress design code, mathematical modelling of the behaviour of timber elements and structures is based on assumed elastic behaviour.

Solved Example

Let us design the roof truss of a building subjected to the following medium-term loads. The configuration of the roof truss is as shown above.

Data
Span of roof truss = 4.8m
Spacing of the truss = 2.0m
Nodal spacing of the trusses = 1.2m
Service class of roof truss: Service class 2

Load Analysis

(i) Dead Loads

On rafter (top chord)
Self-weight of long-span aluminium roofing sheet (0.55mm gauge thickness) = 0.019 kN/m2
Weight of purlin (assume 50mm x 50mm African Mahogany hardwood timber)
Density of African Mahogany = 530 kg/m3 = 0.013  kN/m = (0.013 × 2m)/(2m × 1.2m) = 0.0108 kN/m2
Self-weight of rafter (assume) = 0.05 kN/m2
Total = 0.0885 kN/m2
Weight on plan = 0.0885 × cos 17.35 = 0.08 kN/m2

On Ceiling Tie Member (bottom chord)
Weight of ceiling (10mm insulation fibre board) = 0.077 kN/m2
Weight of services = 0.1 kN/m2
Self weight of ceiling tie = 0.05 kN/m2
Total = 0.227 kN/m2
Therefore the nodal permanent load on rafter (Gk) = 0.08 kN/m2 × 2m × 1.2m = 0.192 kN
Therefore the nodal permanent load on ceiling tie (Gk) = 0.227 kN/m2 × 2m × 1.2m = 0.5448 kN

(ii) Live Load
Imposed load on top and bottom chord (qk) = 0.75 KN/m(treat as medium-term load on plan)
Therefore the nodal permanent load on rafter (Gk) = 0.75 kN/m2 × 2m × 1.2m = 1.8 kN

Analysis of the rafter (top chord)
Span Length = 1.257m
Load = (0.0885 + 0.75) × 2m = 1.667 kN/m

Timber%2BTruss%2BRafter

Results
Analysis of the structure for the loads gave the following results;

on%2Bstaad


All load values are medium-term loads;

Medium-term load is defined in this case by:
Dead load + temporary imposed load

Top Chord Result
Axial force = 10.1 kN (Compression)
Bending Moment = 0.2 kNm
Length of member = 1.26 m

Design of the Top Chord
Let us try 38mm x 100mm timber
Strength class C18

Compression parallel to grain (σc,g,||) = 7.1 N/mm2
σc,adm,|| = σc,g,|| ×  k× k3 × k8 × k12

Bending parallel to grain (σm,g,||) = 5.8 N/mm2
σm,adm,|| = σm,g,|| ×  k× k3 × k6 × k× k8

k= wet exposure (does not apply in this case)
k= duration factor = 1.25 (medium-term loading)
k= shape factor = 1.0 (rectangular section)
k= Depth of section 72mm < h < 300mm
k= (300/h)0.11 = (300/100)0.11 = 1.128
k= Load sharing factor (does not apply since the spacing of the rafters exceed 610 mm).

Section Properties
Area = 3.8 × 10mm2
Zxx = 63.3 × 10mm3
Zyy = 24.1 × 10mm3
Ixx = 3.17 × 10mm4
Iyy = 0.457 × 10mm4
rxx = 28.9 mm
ryy = 11 mm

Applied bending stress
σm,a,|| = M/Z = (0.2 × 106)/(63.6 × 103) = 3.144 N/mm2

Axial compressive stress
σc,a,|| = P/A = (10.1 × 103)/(3.8 × 103) = 2.657 N/mm2

Check for slenderness
Effective length (Le) = 1260 mm (assuming pin end connection)

λ = Le/r = 1260/28.9 = 43.598 < 52 Ok  (clause 2.11.4)

Medium-term load
Compression parallel to grain (σc,g,||) = 7.1 N/mm2

Emin = 6000 N/mm2

k3 = 1.25 (Table 17)

σc,|| = 7.1 × 1.25 = 8.88 N/mm2

E/σc,|| = 6000/8.88 = 675.67
Slenderness λ = 43.598

We can obtain the value of k12 by interpolating from Table 22 of the code
We are interpolating for E/σc,|| = 675.67 and  λ = 43.598

E/σc,||           40         50
600           0.774       0.692
700           0.784       0.711

On interpolating (bivariate interpolation);
k12 = 0.7545

σc,adm,|| = σc,g,|| ×  k× k3 × k8 × k12
σc,adm,|| = 7.1 ×  1.0× 1.25 × 1.0 × 0.7545 = 6.699 N/mm2

σm,adm,|| = σm,g,|| ×  k× k3 × k6 × k× k8
σm,adm,|| = 5.8 ×  1.0× 1.25 × 1.0 × 1.128 × 1.0 = 8.178 N/mm2

Euler critical stress σ=  π2Emin2

σ=  π2(6000)/(43.598)= 31.154 N/mm2

For combined bending and compression

For%2Bcombined%2Bbending%2Band%2Bcompression

σm,a,|| =  3.144 N/mm2
σc,a,|| = 2.657 N/mm2
σc,adm,|| =  6.699 N/mm2
σm,adm,|| =  8.178 N/mm2
σ=  31.154 N/mm2

[3.144/(6.699 × 0.9034)] + [2.657/6.699] = 0.919 < 1.0
Therefore, 38mm x 100mm GS C18 Timber is adequate for the rafter

Consider portion over nodes (at supports)
Bending moment = 0.28 kN.m
Axial load (taking the average at that joint) = (10.81 + 8.43)/2 = 9.62 kN

Applied bending stress
σm,a,|| = M/Z = (0.28 × 106)/(63.6 × 103) = 4.40 N/mm2

Axial compressive stress
σc,a,|| = P/A = (9.62 × 103)/(3.8 × 103) = 2.531 N/mm2

At node point, λ < 5.0, and the rafter is designed as a short column at this point;

σc,adm,|| = σc,g,|| ×  k× k3 × k8
σc,adm,|| = 7.1 ×  1.0× 1.25 × 1.0  = 8.875 N/mm2

The interaction formula for this scenario is given below;

m,a,|| / σm,adm,||] + [σc,ma,|| / σc,adm,||] ≤ 1.0

[4.40 / 8.178] + [2.531 / 8.875] = 0.8232 < 1.0

This shows that the section is satisfactory for rafter.

Analysis of Tie Element
Span Length = 1.2m
Load = (0.227 + 0.75) × 2m = 1.954 kN/m

Results
Axial force = 9.74 kN (tension)
Bending Moment = 0.22 kNm
Length of member = 1.2m

Design of the Bottom Chord (ceiling tie)
Let us still try 38mm x 100mm timber
Strength class C18

Tension parallel to grain (σt,g,||) = 3.5 N/mm2
σt,adm,|| = σt,g,|| ×  k× k3 × k8 × k14
(width of section) k14 = (300/h)0.11 = (300/100)0.11 = 1.128
σt,adm,|| = 3.5 × 1.0× 1.25× 1.0 × 1.128 = 4.935 N/mm2

Bending parallel to grain (σm,g,||) = 5.8 N/mm2
σm,adm,|| = 5.8 ×  1.0× 1.25 × 1.0 × 1.128 × 1.0 = 8.178 N/mm2

Applied bending stress
σm,a,|| = M/Z = (0.22 × 106)/(63.6 × 103) = 3.459 N/mm2

Axial tensile stress
σc,a,|| = P/Effective Area = (9.74 × 103)/(3.8 × 103) = 2.563 N/mm2

Note: When the ceiling tie is connected to rafter by the means of a bolt, the projected area of the bolt hole must be subtracted from the gross area of the section.

Combined tension and bending
m,a,|| / σm,adm,||] + [σt,ma,|| / σt,adm,||] ≤ 1.0

[3.459 / 8.178] + [2.563 / 4.935] = 0.9422 < 1.0

This is ok.

Consider portion over nodes (at supports)
Bending moment = 0.3 kN.m
Axial load (taking the average at that joint) = (9.5 + 9.74)/2 = 9.62 kN

Applied bending stress
σm,a,|| = M/Z = (0.3 × 106)/(63.6 × 103) = 4.7169 N/mm2

Axial tensile stress
σc,a,|| = P/Effective Area = (9.62 × 103)/(3.8 × 103) = 2.531 N/mm2

Combined tension and bending
m,a,|| / σm,adm,||] + [σt,ma,|| / σt,adm,||] ≤ 1.0

[4.7169/ 8.178] + [2.531 / 4.935] = 1.089

In this case, the tie element may be increased to 38mm x 175mm or the grade of the timber could be changed to accommodate the combined flexural and axial stress in the member.

Check for deflection
Deflection of trussed rafter under full load = 6.095 mm (calculated on Staad)
Permissible deflection = 14 mm

Deflection is ok.

That is it for now. Thank you so much for visiting Structville today and God bless you. Remember to share with your folks.

Member Design of a 22m Span Steel Roof Truss

Introduction

We all need a roof over our head. Nowadays, architects frequently use roof pattern and design to enhance the aesthetics and functionality of a building, and it is very important that engineers follow up and ensure that the structural integrity of such roof systems are guaranteed. On the 10th of December 2016, the roof a church building collapsed at Uyo in Akwa Ibom state Nigeria, leaving about 60 people dead, and many other injured. This is why such designs are to be taken seriously, especially when the span is large.

truss

Large span construction of roof systems is usually found in public or commercial buildings. This comes in handy especially when there is need to have a large uninterrupted floor area. However, the efficiency of trusses in dealing with long span structures has been widely recognised in the structural engineering community, but there is no doubt that the longer the span, the more complex the design.



Some basic considerations in the design of trusses
(1)The connection design is as important as the member design. The connections could be welded or bolted or both.
(2) 2D idealisation of the structure is usually very sufficient for analysis and design.
(3) Joints are hardly pinned in reality, but the use of pinned joint is usually encouraged for the purpose of analysis and design.
(4) Trusses can be analysed as having continuous top and bottom chords, with the internal members pinned. In this case, loads can be applied at locations other than the joints with the resulting bending moment appropriately accounted for in the design.
(5)  It is usually important to consider out of plane buckling for compression members in trusses. However, purlins usually take care of the top chord, while bracing can be used to take care of the bottom chord.

Read Also…
Design of Roof Purlins

(6) For efficient structural performance, it is recommended that the truss span to depth ratio be kept between 10 to 15.
(7) In the layout of truss systems, it is more preferable (in terms of economy and efficiency) to have the shorter members in compression and the longer members in tension.
(8) Welded connections offer more advantages in terms of  deflection behaviour of trusses. Slack displacements are possible in bolted connections especially when non-preloaded bolts are used.
(9) In the design of compression members, buckling is the most critical.
(10) When member lines do not intersect at a node, it is important that the moment that arises due to the eccentricity be included in the design.


Design Example
The structural layout of a roof system is given below;

roof%2Blayout

We are required to carry out the member design of the trusses for this 23m span open hall. The design data is given below;

Data
Design code: BS EN 1993-1-1:2005
Design wind pressure: 1.5 kN/m2
Variable load = 0.75 kN/m2

To make this post simpler and shorter, a 29 page fully referenced design paper for all the trusses in the roof system has been attached.

Click HERE to download the calculation sheet in PDF format (premium).

Thank you for visiting Structville today, and remember to tell your colleagues about us.

truss33

Read Also...
Practical Analysis and Design of Roof Trusses

Structural Analysis of Free-Standing Staircase

Free-standing staircases offer a very pleasing solution for vertical circulation in residential and commercial buildings. They are usually constructed in such a way that the landing is freely supported, while the supports at the flight are fully fixed. It is possible to have the flights supported in other ways, but that makes the analysis more complicated. Freestanding staircase structures are complex in analysis and design, but with finite element analysis packages, simple solutions can be easily obtained as shown in this post.

The most widely available manual solution for the analysis of free-standing staircases was developed by Cusens and Kuang using the strain-energy principles. Expressions were developed relating the horizontal restraint force H and the bending moment Mo at the mid-point of the free-standing stair. By solving both equations simultaneously, the values of Mo and H can be substituted into the general expressions to obtain the bending moments and forces at any point in the structure.

In this article, we are going to compare the results obtained with Staad Pro software with results from manual analysis using the method proposed in Table 175, Reynolds and Steedman, 2005.

Dimensions

FREE%2BSTANDING%2BPLAN
SECTION


Solved Example
The geometry of a free-standing staircase is given below. We are expected to analyse the staircase for the ultimate moment using the formula given in Reynolds and Steedman (2005) and compare the answer with the result from Staad Pro.

QUESTION

Data
Thickness of waist of flight and landing = 250 mm
Depth of riser = 150mm
Unit weight of concrete = 25 kN/m3

Actions on the stairs
Concrete self weight (waist area) = 0.25 × 25 = 6.25 kN/m2 (normal to the inclination)
Stepped area = 1⁄2 × 0.15 × 25= 1.875 KN/m2 (global vertical direction)
Finishes (say) = 1.2 kN/m2

We intend to apply all gravity loads purely in the global y-direction, therefore we convert the load at the waist of the stair from local to global direction by considering the angle of inclination of the flight area to the horizontal;

γ = tan-1⁡(1.5/3) = 26.565°

Therefore UDL from waist of the stair in the global direction is given by = (6.25 × cos 26.565) = 5.59 kN/m2

Total permanent action on flight area (gk) = 5.59 + 1.875 + 1.2 = 8.665 kN/m2
Total permanent action on landing; (gk) = 6.25 + 1.2 = 7.45 kN/m2

Variable load on staircase (qk) = 4 kN/m2

The load on the flight area at ultimate limit state = 1.35gk + 1.5qk
nf = 1.35(8.665) + 1.5(4) = 17.67 kN/m2

The load on the landing at ultimate limit state = 1.35gk + 1.5qk
nl = 1.35(7.45) + 1.5(4) = 16.06 kN/m2

From Table 175, Reynolds and Steedman (2005), the approximate formula for calculating the critical design moments for free-standing stairs with the flights fully fixed is given below;

H
MO
K

From the given question;
Load on flight nf = 17.67 kN/m2
Load on landing nl  = 16.06 kN/m2
Thickness of flight hf = 250 mm
Thickness of landing hl = 250 mm
a = 3.35m
b = 1.4m
b1 = 2.0m
γ = 26.565°

Plugging these values into the equations above;
K = 0.746
H = 222.637 kN
M0 = 70.541 kNm

Comparing the above answer with ultimate limit state answer from Staad Pro;

Longitudinal Moment

MY

You can see that M0 from Staad Pro is 65.1 kNm. This about 8.3% less than the anser gotten from manual analysis, and further reinforces the fact that finite element analysis approach to this problem yields a more economical result.

Transverse Moment
The moment in the x direction due to ultimate load is given below;

The maximum moment in the x-direction can be found to be 45.5 kNm. Cusens and Kuang (1966) recommends that the transverse reinforcement be concentrated in the vicinity of the flight and the landing. This results offers a good insight.

Torsion
The twisting moment on the staircase due to the load is given below;

Torsion

A little consideration of the above result will show that considerable twisting is occurring at the mid-span section of the flights. This completely agrees with the conclusions made by Cusens and Kuang (1966). In their own words,

“Large torsional moments are present in the flights of free-standing stairs and a proper thickness of concrete must be chosen to resist these moments, due to the difficulty of reinforcing shallow-wide sections against torsion.”

Cusens and Kuang (1966)


Longitudinal Shearing Stresses

SY

As you can see, the maximum longitudinal stress is occurring at point O with a value of 1.64 N/mm2.

We are undertaking further studies on the dynamic behaviour of free standing stairs. We will update in due time. Thank you for visiting, and God bless you.

References
Cusens A.R., Jing Gwo Kuang (1966): Experimental Investigation of Free Standing Stairs. Journal of the American Concrete Institute, Proceedings V. 63, No. 5, May 1966.

Reynolds C.E., Steedman J.C (2005): Reinforced Concrete Designers Handbook. Spon Press, Taylor and Francis Group, London ISBN 0-419-14540-3

Introduction to Structville Research

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Structville Research is a unique page dedicated to download of research articles, textbooks, design/analysis calculations, spreadsheets, drawings, and other useful educational resources and materials. Currently, we have about 10 articles waiting for your free download! However, some contents are premium, and purchases can be made easily on the spot through Paystack (see www.paystack.com).
Structville Research will be updated on regular basis, and we encourage you to always visit the page for the latest article. From Structville’s home page, you can easily find Structville Research at the menu bar and at the side bar.
Furthermore, if there is any article, drawing, or spreadsheet that you have the copyright permission to distribute, and you will like to have it featured on Structville Research, kindly let us know by sending an e-mail to info@structville.com.
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Structural Design of Pile Caps Using Strut and Tie Model (EC 2)

Pile caps are concrete mats that rests on piles with adequate rigidity to transfer the column loads to the piles. Piles are provided as alternatives to shallow foundations when a firm and stable soil to carry column load is too deep below the surface, or when high lateral load is anticipated. More often than not, pile caps are usually so rigid that they make the entire group of piles to behave like one unit.

EN 1992-1-1:2004 permits us to use strut and tie models to analyse structures where non-linear strain distribution exists (e.g. pile caps, deep beams and corbels).

In strut and tie models, trusses are used with the following components:
• Struts (concrete)
• Ties (reinforcement)
• Nodes (intersections of struts and ties)

Eurocode 2 gives guidance for each of these.


Design Example
A 500mm x 500mm column is carrying an ultimate limit state load of 2140 kN. We are to design the pile cap using the following data;

Grade of concrete fck = 30 N/mm2
Fyk = 500 N/mm2
Concrete cover = 75mm
Spacing of pile = 1800mm
Diameter of piles = 600mm

Pile%2Bcap%2Bdesign%2BStrut%2Band%2BTie%2BMethod
Solution

The strut and tie model is as given below;

pile%2Bcap

Angle of inclination of strut γ = tan-1 (1100/900) = 50.710°
cos  γ = 0.633
sin γ = 0.7739

Force in strut Fs = 1070/sinγ = 1382.6075 kN

Force in tie Ft = 1382.6075 cos γ = 875.15 KN

Area of tension steel required As = Ft/0.87Fyk = (875.15 × 1000) / (0.87 × 500) = 2011.8 mm2
Provide 7H20 @ 150 c/c (Asprov = 2189 mm2 )
Asmin = 0.13bh/100 = 0.0013 × 900 × 1200 = 1404 mm2

Since pile spacing is less than three times pile diameter, the bars may be spread uniformly across the cap.

Check for Shear
Consider the critical section for shear to be located at 20% of the pile diameter inside the pile cap.

Distance of this section from the column face;
av = 0.5(Spacing between piles – width of column) – 0.3(pile diameter)
av = 0.5(1800 – 500) – 0.3(600) = 470 mm

Length of corresponding perimeter for punching shear
u = 2(900 + 1440) = 4680mm

Perimeter of pile cap = 2(900 + 2700) = 7200mm

Since the perimeter of the pile cap is less than 2u, normal shear extending across the full width of the pile cap is more critical than punching shear.

The contribution of the column load to the shear force may be reduced by applying a factor β = av/2d, where 0.5d ≤ a≤ 2d

But a little consideration will show that av(470 mm) < 0.5d(550 mm), therefore, take av as 0.5d (550)
Therefore β = 550 / 2(1100) = 0.25

v = βV/bd

V = 1070 KN + (Self weight of pile cap/2)
Self weight of pile cap = 1.35(25 × 2.7 × 0.9 × 1.2) = 98.415 KN

v = (0.25 × 1119.2 × 1000)/(2700 × 1100) = 0.0942 N/mm2

VRd,c = [CRd,c.k.(100ρ1 fck)(1/3) + k1cp] ≥ (Vmin + k1cp)


CRd,c = 0.18/γc = 0.18/1.5 = 0.12
k = 1 + √(200/d) = 1 + √(200/1100) = 1.426 > 2.0, therefore, k = 1.426
Vmin = 0.035k(3/2) fck0.5
Vmin = 0.035 × (1.426)1.5 × 300.5 = 0.326 N/mm2
ρ1 = As/bd = 2185/(2700 × 1100) = 0.0007356 > 0.02; Therefore take 0.02

VRd,c = [0.12 × 1.426 (100 × 0.0007356 × 30)(1/3)]  = 0.2227  N/mm2

Shear is ok.

Punching Shear Along Column Perimeter
V = 2140 KN

v = V/ud
v = (2140 × 1000)/(2000 × 1100) = 0.972 N/mm2

VRd,max = 0.2(1 –  fck/250)fck
VRd,max = 0.2 (1 – 30/250)30 = 5.28 N/mm2
This shows that the punching shear around column perimeter is ok.

Extra
This pile cap has been modelled on Staad Pro and the results obtained are given below.The set back in the model is the piles were modelled as columns that could sway under the load. What do you think?

PILE%2BCAP%2BSTAAD%2BPRO
pile%2Bcap%2Bbending
Thank you for visiting Structville today and God bless. Kindly share with your folks.

Structural Design of Steel Fin Plate Connection

1.0 Introduction
The fin plate connection is very popular in the construction industry due to its ease of erection, and the absence of shared bolts in two sided connections. It consists of a length of plate welded to the supporting member (eg column or primary beam), and the supported member bolted to the fin plate. Technically, fin plates derive their rotational capacity from shear deformation of the bolts, and hole distortion in the fin plate and/or the beam web.

Fin%2BPlate%2BConnection

2.0 Practical Considerations and Recommended Geometry
With 10mm thick fin plate in S275 steel, 8mm fillet weld to the supporting member will guard against any possibility of weld failure (CSI, 2011). Fin plates may be classified as short or long as follows;

Short tp/zp ≥ 0.15
Long tp/zp < 0.15

Where zp is the distance between the face of the support and the first line of bolts. For short fin plates, erection on site is usually more difficult, but with long fin plates, care must be taken against lateral torsional buckling, especially if the beam is laterally unrestrained.

According to SCI (2011), when detailing the joint, the following recommendations should be followed;
(1) Full strength fillet welds are provided
(2) The fin plate is positioned close to the top flange in order to provide positional restraint
(3) The depth of the fin plate is at least 0.6 times the supported beam depth in order to provide the beam with adequate torsional restraint
(4) The thickness of the fin plate or the beam web is ≤ 0.50d (for S275 steel)
(5) Property class 8.8 bolts non-preloaded are used
(6) All end and edge distances on the plate and the beam web are at least 2d

fin%2Bplate
3.0 Solved Example
A secondary beam UKB 305 x 165 x 40  is supported on a primary beam of  UKB 533 x 210 x 101 with an ultimate shear force of 100.25 kN. We are to design a fin plate connection for this joint using grade 8.8 non preloaded bolts.
 
Initial Considerations

Since the depth of beam is less than 610mm, let us adopt fin plate size of 100 x 10mm
Gap gh = 10mm
Bolt rows = 3
Length of plate = 0.6 × 303.4 = 182.04mm (provide plate length = 220mm)

Fin%2Bplate%2Bjoint%2Bconnection

(i) Supported beam Bolt Shear Check
Basic requirement VEd ≤ VRd

VRd = nFv,Rd/sqrt[(1 + αn)2 + (βn)2]

Fv,Rd = (αvfubA)/γm2

For M20 8.8 bolts;
Fv,Rd = (0.6 × 800 × 245)/1.25 = 94080 N = 94.08 kN

For a single vertical line of bolt (n2 = 1, and n = n1); α = 0

β = 6z/[n1(n1 + 1)p1]
β = (6 × 30)/[3 × (3 + 1) × 70] = 0.214

Therefore;
VRd = (3 × 94.08)/sqrt[(1 + 0 × 3)2 + (0.214 × 3)2)] = 237.506 kN

VEd = 100.25 kN < 237.506 kN
Therefore bolt group is ok for shear


(ii) Fin plate in bearing
Basic requirement VEd ≤ VRd

VRd = n/sqrt{[(1 + αn)/(Fb,ver,Rd )]2 + [βn/(Fb,hor,Rd]2}

α = 0; β = 0.214

The vertical bearing resistance of a single bolt
Fb,ver,Rd = (k1αbfupdtp)/γm2

k= min[2.8(e2/d0) – 1.7; 2.5) =  min[2.8(50/22) – 1.7; 2.5)
= min[4.66;2.5] = 2.5

α= min[e1/3d0; (p1/3d0 – 1/4); fub/fup; 1.0]
= min[40/(3 × 22); (70/(3 × 22) – 0.25); 800/410; 1.0]
= min[0.61; 0.81; 1.95; 1.0] = 0.61

Fb,ver,Rd = (2.5 × 0.61 × 410 × 20 ×10)/1.25 = 100040 N = 100.04 kN

The horizontal bearing resistance of a single bolt
Fb,hor,Rd = (k1αbfupdtp)/γm2

k= min[2.8(e1/d0) – 1.7; 1.4(p1/d0) – 1.7;  2.5)
=  min[2.8(40/22) – 1.7; 1.4(70/22) – 1.7;2.5)
= min[3.39; 2.75; 2.5] = 2.5

α= min[e2/3d0; fub/fup; 1.0]
= min[50/(3 × 22); 800/410; 1.0]
= min[0.76; 1.95; 1.0] = 0.76

Fb,hor,Rd = (2.5 × 0.76 × 410 × 20 ×10)/1.25 = 124640 N = 124.64 kN

VRd = (3)/sqrt{[(1 + 0 × 3)/100]2 + [(0.214 × 3)/124.64]2) = 288.218 kN

VEd = 100.25 kN < 288.218 kN Ok

(iii) Supported Beam – Fin Plate

Shear
Basic requirement VEd ≤ VRd,min

VRd,min = min(VRd,g;VRd,n;VRd,n)

Gross section
VRd,g = (hp.tp.ffy,p)/[1.27 × sqrt(3) × γm0]

VRd,g = [(220 × 10 × 275)/(1.27 × √3 × 1.0)] × 10-3 = 275.034 kN

Net Section
VRd,n = Av,net × [fu,p / (sqrt(3) × γm2)]

Net Area = Av,net =  tp.(h– n1.d0)
Av,net =  10(220 – 3 × 22) = 1540 mm2

VRd,n = 1540  × [410 / (√3 × 1.1)] × 10-3 = 331.399 kN

Block Tearing
VRd,b = [(0.5fu,p Antm2] +  [fy,p Anv / (sqrt(3) × γm0)]

Net area subject to tension = Ant = tp[e2 –  0.5d0]
Ant = 10[e2 –  0.5d0]
Ant = 10[50  –  0.5 × 22] = 390 mm2

Net area subject to shear = Anv = tp[h– e1 –  (n– 0.5)d0]
Anv = 10[20 – 40 –  (3 – 0.5) × 22] = 1250 mm2

VRd,b = [(0.5 × 410 × 390)1.1] +  [(275 × 1250)/ (sqrt(3) × 1.0)] = 72681.818 + 198464.155 = 271145.973 = 271.145 kN

VRd,min = min(275.034; 331.399; 271.145 ) = 271.145 kN

VRd,min = 100.25 kN < 271.145 kN Ok

Lateral Torsional Buckling of fin plates
Basic requirement VEd ≤ VRd
tp/zp = 220/50 = 4.4 > 0.15

Therefore fin plate is short;
VRd = [(Wel,p/z) × (fy,p / γm0)]

Wel,p = (tphp2)/6 = (10 × 2202)/6 = 73333.333 mm3

VRd = [(73333.333/50) × (275 / 1.0)] = 403333.333 N = 403.333 kN
VEd = 100.25 kN < 403.33 kN Ok

Supported Beam in Shear
Gross Section
VRd,g = [Av,web × (fy,b1  / (sqrt(3) × γm0))]

Gross Area Av  = ATee – btf,b1 + (tw,b1 + 2rb1× 0.5tf,b1

ATee =  (265.2 – 10.2) × 6 + (165 × 10.2) = 1530 + 1683 = 3213 mm2
Av  = 3213 – (165 × 10.2) +  (6 + 2 × 8.9) × 0.5 × 10.5 = 3213 – 1683 + 124.95 = 1654.95 mm2

VRd,g = 1654.95 × 275 / (sqrt(3) × 1.0) = 262758.603 N = 262.758 kN


Net Section
VRd,n = [Av,net × (fu,b1  / (sqrt(3) × γm2))]

Net area; Av,net = A– n1d0tw,b1 1654.95 – (3 × 22 × 6) = 1258.95 mm2

VRd,n = 1258.95 × 410 / (sqrt(3) × 1.1) = 270918.727 N = 270.918 kN

Block Tearing
VRd,b = [(0.5fu,b1 Antm2] +  [fy,b1 Anv / (sqrt(3) × γm0)]

Net area subject to tension = Ant = tw,b1[e2,b –  0.5d0]
Ant = 6[40  –  0.5 × 22] = 174 mm2

Net area subject to shear = Anv = tw,b1[e1,b + (n– 1)p– (n– 0.5)d0]
Anv = 6[40 + (3 – 1)70 –  (3 – 0.5)22] = 750 mm2

VRd,b = [(0.5 × 410 × 175)1.1] +  [(275 × 750)/ (sqrt(3) × 1.0)] = 35873.9 + 119081.986 = 154955.886 = 154.955 kN

VRd,min = min(262.758; 270.918; 154.955) = 154.955 kN

VRd,min = 100.25 kN < 154.955 kN Ok

Check for welding (supporting beam)
For a beam in S275 steel;

Basic requirement; a ≥ 0.5tp
0.5tp = 0.5 × 10 = 5mm
a = 5.7mm > 0.5tp

Thank you for visiting Structville today. God bless you.

Thickness Design of Column Base Plate Using Prokon Software

Last year, I made a submission here on how to carry out the thickness design of column base plates.

See post below;
Thickness Design of Column Base Plates According to EC3

Column%2BBase%2BPlate

Right now, we wish to check the result obtained from the manual analysis above with result from Prokon software. This present design has been carried out according to BS 5950:2000, and in this post, we will highlight all the input variables that are necessary to carry out the design, as well as the results.

Axial Load on Column 
Characteristic force from permanent action Gk = 620 kN
Characteristic force from leading variable action  Qk = 132 kN

Partial Factor for Loads 
Permanent action γG = 1.4
Variable action γG = 1.6

At ultimate limit state;
N = 1.4Gk + 1.6Qk
N = 1.4(620) + 1.6(132) = 1079.2 kN

Column Dimensions and Properties (UC 203 x 203 x 60)
Depth (d) = 209.6mm
Width (b) = 205.8mm
Thickness of web (tw) = 9.4mm
Thickness of flange = (tf) = 14.2mm
Root radius (r) = 10.2mm
Perimeter of section = 1206.4mm
Area of section = 76.4 cm2

Base Plate Bedding Details
Strength of concrete/grout = 25 N/mm2

Design Using Prokon


Step 1: Getting stared
Launch the Prokon Software, go to ‘Connections Module’, and select ‘Base Plate’. The window below comes up.

Window

This is a simple interactive interface that allows you to carry out your design in a very simple manner. As you input the design variables, the graphical representation of what you are doing comes up on the window, so that you will remain properly guided.


Step 2: Verify Design Code
At the upper left hand side of the screen, go to ‘File’, and make sure that the design code is set to BS 5950

Step 3: Select the Section Size

column

At the column dialog box at the upper left hand side of the screen, check the universal column symbol (see screenshot above). Now, go to the drop down list and select the appropriate column section. In this example, we are dealing with UC  203x203x60.

Step 4: Input Dimensions
As a guide, the area of base plate required should be equal to or greater than;
N/0.6fcu = (1079.2 x 1000)/(0.6 x 25) = 71946.667 mm2

It is always a good practice to allow at least 100mm from the edge of the column to the base plate.

Let us adopt base plate 350mm x 350mm (Aprov = 122500 mm2).

Now input the base plate dimensions as given below;

a1

You will discover that the drawing below automatically appears on the window;

a2

What this requires is for you to input the offset, in order to position the column on your desired location on the base plate. However, we want our column to centralised, and Prokon gives us that option by the click of a button. Go to the area where you input loads, and by the side of the table, click on ‘Centralise Column’. This brings the column to the centre of the base plate, and the offset value is calculated automatically for us. See the new image below;

a4

Now, by imputing the rest of the variables, we can position the bolts to suit the design appropriately. We are offsetting the bolts 35mm from the edge of the plate. See the input and the result below;

a5
a6


Step 5: Define materials properties (general parameters)
From the dialog box below, define the properties of the materials to be used in the construction.

a7

We are saying no to the use of studs because we want the column load to be transferred directly to the concrete bedding and not to the bolts. The grade of concrete and yield stress of steel selected is as given above.

Step 6: Define the loads
Since we have already factored our loads, we can input the ultimate load directly and set the load factor as unity (1.0). Alternatively, we can input the dead load and live load separately with their appropriate factor of safety, but note that they must be presented as the same load case. See below;

a8
Step 7: Design
If you have gotten to this point without any errors, we can then click on ‘Design’ function, and the window below comes up with the design completely done;

unstiffened

When the plate is unstiffened, the design result above holds good, with base plate thickness of 18mm.

When the plate is stiffened, the design result below holds good, which also includes the thickness of the stiffeners, and weld designs. The plate thickness in this case reduced to 10mm.

stiffe


Step 8: Detail Drawing
Click on drawing, and you will be able to see the shop drawing for the design. The images below gives the detailing for the unstiffened and stiffened plates respectively.

det1
det%2B2

Conclusion
I think the result above needs no interpretation. The details are quite crystal clear and I must say that it is very impressive. It is very interesting to note that when this same base plate was designed manually using EC3 (unstiffened), we obtained a thickness of 18.447mm, and then provided a base plate of 20mm. This shows reliability of both methods for all practical purposes. You can read the manual design below just in case you missed it.

Thickness Design of Column Base Plates According to EC3

Thank you for visiting Structville today, and God bless.

E-mail your thoughts to the author: ubani@structville.com

Analysis of Industrial Gable Frames

1.0 Introduction
Industrialization is one of the major keys to development and sustainable economy. Industrial structures are usually very easy to identify because of their unique  features that are quite different from residential or commercial buildings. Engineers are often tasked with analysing and designing industrial frames, and to simplify the analysis for manual calculations, they are usually idealised as 2D plane frames.


2.0 Solved Example
In this post, a gable industrial frame structure is subjected to a load regime as shown below. The frame is hinged at point D, and it is desired to obtain the internal forces (bending moment, axial forces, and shear forces) due to the externally applied load.

N/B: In practical construction, it is not advisable to hinge the structure at the apex due to the problem of excessive deflection and stability of the frame.

Industrial%2BFrame%2BAnalysis

Solution

(a) Support Reactions

support%2Breactions


Geometrical Properties
Angle of inclination of the rafter
⁡θ = tan-1(2.5/6) = 22.619°
cos ⁡θ = 0.923
sin ⁡θ = 0.385

Length of rafter (z) = sqrt(6+ 2.52) = 6.5m


Notations:
Example: NC= Axial Load at point C, just to the Right

Let ∑MG = 0
12Ay – [(10 × 122)/2] – (25 × 11) + (7 × 4) – (16 × 1) = 0
12Ay = 983
Ay = 81.917 kN

Let ∑MDL = 0
6Ay – 8Ax –  (25 × 5) – (7 × 4) – [(10cos22.619° × 6.52)/2]  = 0
6(81.917) – 8Ax –  (25 × 5) – (7 × 4) – [(9.23 × 6.52)/2]  = 0
8Ax = 143.518 kN
Ax = 17.939 kN

Let ∑MA = 0
12Gy – [(10 × 122)/2] – (25 × 1) – (7 × 4) – (16 × 11) = 0
12Gy = 949
Gy = 79.083 kN

Let ∑MDR = 0
6Gy – 8Gx –  (16 × 5) –  [(10cos22.619° × 6.52)/2]  = 0
6(79.083) – 8Gx –  (16 × 5) – [(9.23 × 6.52)/2]  = 0
8Gx = 199.515 kN
Gx = 24.939 kN

Equilibrium Check
All downward vertical forces = (10 × 12) + 25 + 16 = 161 kN
Upward reactive forces = 81.917 + 79.083 = 161 kN

All rightward horizontal forces = Ax + 7kN = 17.939 + 7 = 24.939 kN
All leftward horizontal forces = Gx = 24.939 kN

Therefore, equilibrium is ok. If you are having confusions, the post below might help.

Read Also….
Understanding Sign Conventions in Structural Analysis

(b) Internal Stresses
Section A – BB (0 ≤ y ≤ 4.0m)
(i) Bending moment
My = -Ax.y = -17.939y
At y = 0; MA = 0
At y = 4.0m; MBB = (-17.939 × 4) = -71.756 kNm

(ii) Shear Force
Qy = ∂My/∂y = -17.939 kN

(iii) Axial Force
Ny + Ay = 0
Ny = -Ay = -81.917 kN (compression)

Section 1 – BR (0 ≤ x ≤ 1.0m)
(i) Bending moment
Mx = -25.x
At x = 0; M1 = 0
At x = 1.0m; MBR = (-25 × 1) = -25 kNm

(ii) Shear force
Qx = ∂Mx/∂x = -25 kN

(iii) Axial Force
N– 7kN = 0
Nx = 7kN  (tension)

Section BUP – CB (4 ≤ y ≤ 5.5m)
(i) Bending moment
My = -Ax.y + (25 × 1) – 7(y – 4)
At y = 4m; MBUP = -(17.939 × 4) + 25 = -46.756 kNm
At y = 5.5m; MCB = -(17.939 × 5.5) + 25 – 7(1.5) = -84.1645 kNm

(ii) Shear force
Qx + Ax + 7kN = 0
Qx = -17.939 – 7 =  -24.939 kN

(iii) Axial force
Ny + Ay – 25kN = 0
Ny = -81.917 + 25 =  -56.917 kN (compression)


Section CR – DL (0 ≤  z ≤ 6.5m) (Pay careful attention here)
(i) Bending moment
The bending moment transferred transferred to the rafter at node C is -84.1645 kNm
Summation of vertical force transferred ∑V = Ay – 25 = 81.917 – 25 = 56.917 kN
Summation of horizontal force transferred ∑H = Ax + 7 = 17.939  + 7 = 24.939 kN

Mz =  (∑V.cos22.619-9°.z) – (∑H.sin22.619°.z) – [(10cos22.619° × z2)/2] – 84.1645
Mz =  52.539z – 9.591z – 4.6154z2 – 84.1645
Mz = -4.6154z42.9425– 84.1645

At z = 0; MCR = -84.1645 kNm
At z = 6.5m; MCB = -4.6154(6.5)42.948(6.5) – 84.1645 = -195 + 279.126 – 84.1645 = 0

Maximum span moment
Mz = -4.6154z42.9425– 84.1645

Maximum moment occurs at the point of zero shear
∂Mz/∂z = -9.23z + 42.9425 = 0
z = 42.9425/9.23 = 4.652m

Mmax  = -4.6154(4.652)42.9425(4.652) – 84.1645 = -99.882 + 199.768 – 84.1645 = 15.7215 kNm

(ii) Shear force
∂Mz/∂z = (∑V.cos22.619) – (∑H.sin22.619)

At z = 0; QCR = (56.917 × 0.923) – (24.939 × 0.385) = 42.933 kN
At z = 6.5m; QDL = [(56.917 – (10 × 6))   × 0.923] – (24.939 × 0.385) = -2.8456 – 9.6015 = -12.447 kN

(iii) Axial force

Nz = -(∑V.sin22.619) – (∑H.cos22.619)

At z = 0; NCR = -(56.917 × 0.385) – (24.939 × 0.923) = -21.913 – 23.018 =   -44.931 kN
At z = 6.5m; NDL = [-(56.917 – (10 × 6))  × 0.385] – (24.939 × 0.923) = 1.187 – 23.018 = -21.831  kN

 

Coming from the right
Section G – FB (0 ≤ y ≤ 4.0m)
(i) Bending moment
My = -Gx.y = -24.939y
At y = 0; MG = 0
At y = 4.0m; MFB = (-24.939 × 4) = -99.756 kNm

(ii) Shear Force
Qy = ∂My/∂y = +24.939 kN (note that we are coming from right to left)

(iii) Axial Force
Ny + Gy = 0

Ny = -Gy = -79.083 kN (compression)

Section 2 – FR (0 ≤  x ≤ 1.0m)
(i) Bending moment
Mx = -16.x
At x = 0; M2 = 0
At x = 1.0m; MFR = (-16 × 1) = -16 kNm

(ii) Shear force
Qx = ∂Mx/∂x = -16 kN

(iii) Axial Force
N– 0 = 0

Nx = 0 (no force)

Section FUP – EB (4 ≤ y ≤ 5.5m)
(i) Bending moment
My = -Gx.y + (16 × 1)
At y = 4m; MFUP = -(24.939 × 4) + 16 = -83.756 kNm
At y = 5.5m; MEB = -(24.939 × 5.5) + 16 = -121.1645 kNm

(ii) Shear force
Qx – Gx = 0
Qx =  24.939 kN

(iii) Axial force
Ny + Ay – 16kN = 0
Ny = -79.083 + 16 =  -63.083 kN (compression)

Section EUP – DR (0 ≤  z ≤ 6.5m)
(i) Bending moment
The bending moment transferred transferred to the rafter at node C is -121.1645 kNm
Summation of vertical force transferred ∑V = Gy – 16 = 79.083 – 16 = 63.083 kN
Summation of horizontal force transferred ∑H = Gx = 24.939 kN

Mz =  (∑V.cos22.619-9°.z) – (∑H.sin22.619°.z) – [(10cos22.619° × z2)/2] – 121.1645
Mz =  58.225z – 9.5915z – 4.6154z2 – 121.1645
Mz = -4.6154z48.6335– 121.1645

At z = 0; MER = -121.1645 kNm
At z = 6.5m; MCB = -4.6154(6.5)48.6335(6.5) – 121.1645 = -195 + 316.1175 – 121.1645 = 0

Maximum span moment
Mz = -4.6154z48.6335– 121.1645

Maximum moment occurs at the point of zero shear
∂Mz/∂z = -9.23z + 48.6335 = 0
z = 48.6335/9.23 = 5.269m

Mmax  = -4.6154(5.269)48.6335(5.269) – 121.1645 = -128.134 + 256.2499 – 121.1645 = 6.9514 kNm



(ii) Shear force
∂Mz/∂z = -(∑V.cos22.619) + (∑H.sin22.619)

At z = 0; QCR = (63.083 × 0.923) – (24.939 × 0.385) = -58.2256 + 9.6015 =  -48.624 kN
At z = 6.5m; QDL = -[(63.083 – (10 × 6)) × 0.923] + (24.939 × 0.385) = -2.8456 + 9.6015 = 6.756kN

(iii) Axial force

Nz = -(∑V.sin22.619) – (∑H.cos22.619)

At z = 0; NCR = -(63.083 × 0.385) – (24.939 × 0.923) = -24.287 – 23.018 =  -47.305kN
At z = 6.5m; NDL = [-(63.083 – (10 × 6))  × 0.385] – (24.939 × 0.923) = -1.187 – 23.018 = -24.205 kN

(b) Internal Stresses Diagram
(a) Bending Moment Diagram

BMD
(b) Shear Force Diagram
SFD


(c) Axial Force Diagram

AXIAL


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Design of Piles in Sand: Case Study of Lekki Pennisula, Lagos Nigeria


Construction experts in Lagos Island, Nigeria are no strangers to pile foundation. The poor soil profile in the area means that foundations are more economically founded at depths way beyond the weak upper soil strata. Faseki et al (2016) carried out assessment of sub-soil properties of some parts of Lekki Peninsula, Lagos.

The result of their findings form the basis of this post, which aims to explore the  design of pile foundations on sand, using Eurocode 7, and Lekki, Lagos State Nigeria, as case study. The summary of their findings regarding geotechnical properties of the soil in the area is given in the Table below;

Engineering%2BProperty%2Bof%2BLekki%2BSoil

The reports from soil investigation in most parts of Lekki indicate poor bearing capacity for shallow foundations. In a research work by Warmate and Nwankwoala (2019), an average bearing capacity of 55 kN/m2 was reported within 1m depth of the soil using direct shear analysis and SPT. However, in an independent soil investigation report carried out within Abijo Village, Ibeju Lekki, fibrous peat was observed within 0 – 5m depth of the area, necessitating the adoption of pile foundation in such areas.

Theoretical Background of Pile Foundation Design

When the static formulae is applied in the geotechnical design of piles, the ultimate load carrying capacity is calculated from the soil properties obtained from site investigation. This tends to be a more natural approach since it is the soil that actually carries the load, but more often than not, pile load tests are employed for validation of results. The ultimate capacity of piles is assumed to be reached when;

Qu = Qb + Qs + W——- (1)

Where:
Qu = Ultimate capacity of pile
Qb = Base resistance = quAb
Qs = Shaft resistance = fuAs
Ws = Weight of displaced soil

Quite often, the weight of the pile and the weight of the soil removed is assumed to be equal, such that the last term of the equation is eliminated.

qu = ultimate value of the resistance per unit area of the base due to shear strength of the soil
fu = ultimate value of the tangential force per unit area of the shaft due to adhesion and skin friction
As = surface area of the pile shaft
Ab = base area of the pile
L = Length of pile below ground.

Meyerhof (1951) expressed the base resistance per unit area as;

qu = cNc + ksPo‘Nq + 0.5BγNγ   ———(2)

In the case of a pile of normal proportions, L/B is about 30 or more, the term containing B in the above equation is so small that is usually ignored. The above equation now reduces to;

qu = cNc + ksPo‘Nq  ———-(3)

Where;
ks = the coefficient of earth pressure on the shaft within the failure zone
Nc, Nq = Bearing capacity factors dependent on the embedment ratio and angle of internal friction

In a soil where both adhesion and friction can be mobilised on the pile shaft, Meyerhof (1951) expressed the tangential force per unit area as;

fu = Ca + ksPo‘tanδ ——— (4)

Where;
Ca = adhesion per unit area
δ = angle of friction between the soil and the pile material
For a purely cohesive soil such as clay, δ is zero and for non-cohesive soil such as sand, Ca is zero.

Pile Foundation in Sand

Knowing full well that in purely cohesionless soil (sand), fu = ks‘Po‘tanδ. What this equation means is that the skin friction continues to increase linearly with increasing depth. But for piles founded on sand, it has been determined that the overburden pressure of the soil adjacent to the pile does not increase without limits. When a certain depth of penetration is reached, the overburden pressure remains more or less constant, and this is called the critical depth Dc.

The critical depth depends on the field condition and the dimensions of the pile. It is generally agreed that at a penetration depth between 10 and 20 pile diameters, a peak value of skin friction is reached which cannot be exceeded at greater penetration depth (see figure below). Therefore, the equation fu = ks‘Po‘tanδ gives increasingly unsafe values as the penetration depth increases and exceeds about 20 pile diameters. The assumptions usually made for piles as reported by Wrana (2015) are;

  • Pile in loose sand (10D)
  • Pile in medium dense sand (15D)
  • Pile in dense sand (20D)
Dile%2Bin%2BDense%2Bsand
Figure 1: Critical Depth For Pile Founded on Sand (Picked from Wrana, 2015)

Therefore, for piles in a homogenous dense sand for example;
Qu = Qb + Qs = qu.Ab + fu.As

Total skin friction = fu.As = ks‘tanδ × Area of PV diagram × Circumference of pile
Area of PV diagram = 0.5γ.Dc2
But Dc = 20B
qu = Pv. Nq = γ.Dc

Therefore;
fu.As = ks‘tanδ × 0.5γ.Dc2 × πD
qu.Ab = (γ.Dc)(Nq)(πD2/4)

The working load or design load for all pile types is equal to the sum of the base resistance and the shaft friction divided by a suitable factor of safety.

Pile Design to Eurocode 7 (EN 1997-1:2004)

EN 1997-1 clause 7.4(1)P states that the design of piles shall be based on one of the following approaches:

(1) The results of static load tests, which have been demonstrated by means of calculations or otherwise, to be consistent with other relevant experience,
(2) Empirical or analytical calculation methods whose validity has been demonstrated by static load tests in comparable situations,
(3) The results of dynamic load tests whose validity has been demonstrated by static load tests in comparable situations,
(4) The observed performance of a comparable piles foundation, provided that this approach is supported by the results of site investigation and ground testing.

Pile Resistance from ground parameters 

According to clause 7.6.2.3(8) pf EC7, the characteristic base and shaft resistances may also be determined directly from the ground parameters using the following equations;

Rb;k = Ab qb;k
Rs;k = ∑As;i qs;i;k

Where;
qb;k  is the characteristic unit base resistance
qs;i;k  is the characteristic unit shaft resistance

The design compressive resistance of a pile Rc,d may be obtained either by treating the pile resistance as a total resistance;

Rc,d = Rc;k / γt  (where Rc;k = Rb;k +  Rs;k)

or by separating it into base and shaft components Rb;k and Rs;k using the relevant partial factors, γb and γs

Rc,d = [Rb;k / γb  + Rs;k / γs]

The partial resistance factors in the UK National Annex have been modified to take account of the type of pile and whether the serviceability behaviour is to be determined either by load test or a rigorous and reliable calculation (Raison 2017).

The table below gives the partial factors for bored piles (culled from Raison, 2017)

Partial%2BResistance%2BFactors%2Bfor%2BBored%2BPiles%2BEC7

The equilibrium equation to be satisfied in the ultimate limit state design of axially loaded piles in compression is Fc,d  ≤ Rc,d .

Example of Pile Foundation Design for Lekki Peninsula

A structure to be built at Lekki Peninsula, Lagos has a group of piles to be subjected to a characteristic permanent load (Gk) of 665 kN and characteristic variable load Qk of 275 kN each. It has been decided to use piles of diameter 600mm, and soil investigation report from Faseki et al (2016) is to be used for the design. The design involves determining the length of embedment of the piles.

Solution

For details of this calculation and approach, go to clause 7.6.2.3 of EN 1997-1:2004.

Let us make an initial trial of 20m depth.

For this particular design, we will ignore the effect of shaft resistance on the first and second layers. Our interest is to determine the appropriate length of penetration into the fourth layer or possibly fifth layer. We are assuming that the water table lies at 1m below the ground.

For simplicity, let me summarise the properties of the layers below as reported by Faseki et al (2016);

First layer (3m thick) – Loose to Medium Silty Sand, Cu = 0, ϕ = 32°, γ = 18.5 kN/m2
Second layer (1.5m thick) – Soft silty clay, Cu = 24 kN/m2 , ϕ = 28°, γ = 15 kN/m2
Third layer (4.5m thick) – Very loose sand,  Cu = 0, ϕ = 30°, γ = 18.5 kN/m2
Fourth layer (11m thick) – Medium dense sand,  Cu = 0, ϕ = 33°, γ = 19.5 kN/m2

The decision to ignore the shaft (friction) resistance of the first two layers is conservative. I am not very confident about the behaviour of the 1st layer, and the clay layer (2nd layer) sandwiched between the upper and lower sandy formations. But a little consideration shows that the 3rd and 4th layer gives more confidence for design purposes.

(1) Calculation of the Base Resistance
Base resistance Rb;k = qb:k.Ab = (Po‘)(Nq)(πD2/4)

From Faseki’s report, the angle of internal friction at layer of interest φ = 33°

N/B: Table A.4 of EC7 recommends that we apply material factor of safety to some soil properties, but UK Annex to EC7 does not recommend application of partial safety factors except in the case of negative skin friction.

Values of bearing capacity factor Nq according to NAVFAC DM 7.2(1984), is given below;

Nq%2Bvs%2Bphi

From the table above; Nq at the receiving layer = 17 (pile will be bored)
Effective pressure at the base of the pile = (18.5 × 3) + (15.5 × 1.5) + (18.5 × 4.5) + (19.5 × 11) = 376.5 kN/m2
Pore water pressure = 10kN/m3  × 19m = 190 kN/m2
Effective Stress at foundation depth = 376.5 – 190 = 186.5 kN/m2

Base resistance = qb:k × Ab = (186.5 × 17) × (π × 0.62/4) = 896.553 kN

(2) Shaft Friction
We will ignore the shaft resistance of the first two layers as earlier stated, but typical calculation values are shown.

fu.As = ks‘ tanδ × 0.5Po‘ × πDL

Typical values for δ and ks‘ as suggested by Broms (1966) is given in the table below.

Broms

For the first layer (loose sand);
δ = 0.75ϕ = 0.75 × 32 = 24°
(tan 24°) = 0.445
Effective stress at the layer = (18.5 × 3) – (10 × 2) = 35.5 kN/m2
fu1.As1 = 1.0 × 0.445 × (0.5 × 35.5) × (π × 0.6 × 3) = 44.672 kN

For the second layer (silty clay);
Cohesion of soil Cu = 24.0 kN/m2
α = 1.0 since Cu < 25 kN/m2 (Wrana, 2015)
qs.A= 1.0 × 25 × (π × 0.6 × 1.5) = 70.695 kN

But we are neglecting the effects of these on the pile

For the third layer (loose sand);
δ = 0.75ϕ = 0.75 × 30 = 22.5°
(tan 22.5°) = 0.4142
Effective stress at the layer = (18.5 × 3) + (15.5 × 1.5) + (18.5 × 4.5) – (10 × 8) = 82.5 kN/m2
fu3.As3 = 1.0 × 0.4142 × (0.5 × 82.5) × (π × 0.6 × 4.5) = 144.945 kN

For the fourth layer (medium dense sand);
δ = 0.75ϕ = 0.75 × 33 = 24.75°
(tan 24.75°) = 0.461
Effective stress at the layer = 186.5 kN/m2
fu3.As3 = 1.5 × 0.461 × (0.5 × 186.5) × (π × 0.6 × 11) = 1337.184 kN

Therefore, total load from shaft resistance Rs;k = 144.945 + 1337.184 = 1482.129 kN

Design Approach 1

Combinations of sets of partial factors
DA1.C1  —–     A1 + M1 + R1
DA1.C2  ——-   A2 + M1 or M2 + R4

Partial factors for actions;
A1   γG = 1.35   γQ = 1.5
A2   γG = 1.0     γQ = 1.30

Partial factors for materials
M1 and M2 not relevant (γϕ’ = 1.0, not used)

Partial Resistance factors
R1    γb = 1.0    γt = 1.0
R4    γb = 1.3      γt = 1.3

DA1.C1   Fc,d = 1.35Gk + 1.5Qk = (1.35 × 665) + (1.5 × 275) = 1310.25 kN
DA1.C2   Fc,d = 1.0Gk + 1.3Qk = (1.0 × 665) + (1.3 × 275) = 1022.5 kN

By UK National Annex (with no SLS verified);

DA1.C1   
Rc,d = Rb;kb + Rs;ks = [896.55/2 + 1482.129/1.6] = 1374.6 kN

Also check;
Rc,d = Rc,kt = [896.55 + 1482.129] / 2.0 = 1189.3395 kN

Rc,d (1189.3395 kN) < Fc,d (1310.25 kN)

DA1.C2   
Rc,d = Rb;kb + Rs;ks = [896.55/(1.3 × 2) + 1463.681/(1.3 × 1.6)]  = 1048.518 kN
Rc,d (1048.518 kN) > Fc,d (1022.5 kN)

Therefore, Design Approach 1 Combination 1 (DA1.C1) is more critical, and pile length will need to extend beyond 20m length. Therefore from Design Approach 1, Combination 1, the length of 20m is rejected.

Design Approach 2

Combinations of sets of partial factors
DA2  —–     A1 + M1 + R2

Partial factors for actions;
A1   γG = 1.35   γQ = 1.5

Partial factors for materials
M1 not relevant (γϕ’ = 1.0, not used)

Partial Resistance factors
R2    γt = 1.1 (Total/combined compression)

Fc,d = 1.35Gk + 1.5Qk = (1.35 × 665) + (1.5 × 275) = 1310.25 kN

DA2
Rc,d = Rb;kb + Rs;ks = [896.55/(1.1 × 2) + 1482.129/(1.1 × 1.6)]  = 1249.641 KN
Rc,d (1249.641 kN) > Fc,d (1310.25 kN)

Hence Length of 20m is rejected using Design Approach 2

Conclusion

We have seen from the design that the pile length of 20m at a diameter of 600mm is inadequate to support the design load specified. An alternative is to increase the diameter of the pile, but the reader should pay close attention to the assumptions made in arriving at the ultimate values.

The probable shaft resistance of the rejected two layers was shown, so that it could serve as a guide for judgement and further decision making. However, the closeness of the answers offer very intellectually stimulating ideas about the assumptions made However, I will also be happy if I can get an alternative design based on the soil properties, which might include the SPT and CPT values reported for the various layers.

As a matter of fact, it took a lot of time to arrive at some decisions especially regarding the implementation of the Eurocode 7. I more or less agree with the conclusions of Raisons (2017) regarding Eurocode 7 as follows;

  1. EC7 does not tell the Designer how to design piles but does give rules and procedures to be followed
  2. EC7 has complicated pile design with the introduction of numerous partial factors; load factors, combination factors, material factors, resistance factors, model factors and correlation factors
  3. More design effort is required in EC7, and ultimately,
  4. Engineering judgement cannot be suspended

Thanks for reading this post. You can contact the author via mail;

ubani@structville.com

To download this post as a PDF file, click HERE

References
[1] Wrana B. (2015): Pile Load Capacity – Calculation Methods. Studia Geotechnica et Mechanica, Vol. 37, No. 4, 2015 DOI: 10.1515/sgem-2015-0048
[2] Faseki O.E., Olatinpo O.A., Oladimeji, A.R. (2016): Assessment of Sub-Soil Geotechnical Properties for Foundation Design in Part of Reclaimed Lekki Pennisula, Lagos, Nigeria. International Journal of Advanced Structures and Geotechnical Engineering ISSN 2319-5347, Vol. 05, No. 04, October 2016
[3] Raison Chris (2017): Pile Design to BS EN 1997-1:2004 (EC7) and the National Annex. Raison Foster Associates, University of Birmingham, UK
[4] NAVFAC DM 7.2 (1984): Foundation and Earth Structures, U.S. Department of the Navy.
[5] EN 1997-1:2004:Geotechnical design – Part 1: General rules, European Committee for Standardization 
[6] Meyerhof,  G.G.  (1951):  The  ultimate  bearing  capacity  of  foundations,  Géotechnique,  2, 301-332 
[7] Warmate T. and Nwankwoala H. O. (2019): Geotechnical Indications and Shallow Bearing Capacity Analysis within Lekki Peninsula, Lagos using Direct Shear Analysis. Cur Trends Civil & Struct Eng. 1(4): 2019. http://dx.doi.org/10.33552/CTCSE.2019.01.000516

Linear Interpolation for Structural Engineers (Gregory-Newton Forward Difference Formular)

1.0 Introduction
In the design of structures, engineers are always faced with the task of carrying out interpolations as supported by various codes of practice. These interpolations are often encountered when we are carrying out wind load analysis, designing our columns, designing our two-way slabs, etc.

Some people may not be very familiar with the process of linear interpolation, and this is the problem this post attempts to address, showing how to use  Gregory-Newton forward difference formular for carrying out interpolations. I am not going to go through the process of deriving the formular, but I will present it exactly the way it is applied, and use practical examples to consolidate it.


2.0 Gregory-Newton Forward Difference Formular
Let x0, x1, x2,…, xn be equally spaced values, so that xi = x0 + ih, for i = 1, 2,…, n.

If the values f0, f1, f2,…, fn are known, where fi = f(xi), for some function f.

The Gregory–Newton forward difference formula is a formula involving finite differences that gives an approximation for f(x), where x = x0 + ph, and f(x) ≈ f0 + pΔf0 gives the result of linear interpolation. For each pair of consecutive function values f(x0) and f(x1), the forward difference is obtained by subtracting f(x0) from f(x1).   However, when the series is terminated after one more term provides an example of quadratic interpolation.

In this post, we are considering linear interpolation only, and cases where it can be applied.

Quickly, let us use the table below to show how this process is carried out. The relationship between the values is fairly linear.

linear%2Binterpolation%2Btable

For the table given below, we are required to calculate the value of f(3.5).
Note that the value we seek (3.5) is between 3 and 4
So;
 x0 = 3; x1 = 4; xp = 3.5
h = x1 –  x0 = 4 – 3 = 1
p = (xp – x0)/h = (3.5 – 3)/1.0 = 0.5
Δf0 =   f(x1– f(x0) = 59 – 47 = 12

Therefore; f(3.5) = 47 + 0.5(12) = 53

This one is just as simple as that.

Now let us go into practical scenarios encountered in structural design.


3.0 Applications in Structural Design

Example 1
Calculation of bending moment coefficient for two-way slabs

Table 3.14 of BS 8110-1:1997 gives coefficients that are used to obtain the bending moment coefficients of rectangular slabs subjected to uniform pressure under various support conditions. This same coefficient is also used in the Eurocodes, and an excerpt is given below.

derrr

Let us assume that we have a slab with one short edge discontinuous with an aspect ratio of 1.46.  We wish to obtain the positive bending moment coefficient for the mid-span by using linear interpolation.

x0 = 1.4; x1 = 1.5; xp = 1.46
h = x1 –  x0 = 1.5 – 1.4 = 0.1
p = (xp – x0)/h = (1.46 – 1.4)/0.1 = 0.6
Δf0 =   f(x1– f(x0) = 0.043 – 0.041 = 0.002

Therefore; f(1.46) = 0.041 + 0.6(0.002) = 0.0422

I guess that was very simple and straightforward.

Example 2
External Wind Pressure coefficient on Buildings

Table 7.1 of EN 1991-1-4:2004 gives the recommended values of external pressure coefficients for buildings that are rectangular in plan. Design engineers usually encounter this table when carrying out wind load analysis. The code permits us to determine intermediate values by interpolation.

Let us assume that we are to determine Cpe,10 for a building in zone A, having a h/d ratio of 2.3

x0 = 1; x1 = 5; xp = 2.3
h = x1 –  x0 = 5 – 1 = 4
p = (xp – x0)/h = (2.3 – 1)/4 = 0.325
Δf0 =   f(x1– f(x0) = -1.2 – (-1.2) = 0

Therefore; f(2.3) = -1.2 + 0.325(0) = -1.2

Example 3
Compressive Strength of Steel to BS 5950-1:2000

Table 24 of BS 5950-1:2000 gives values for compressive strength of steel members. We are also permitted to obtain intermediate values through interpolation.

strut%2Bcurve

The slenderness ratio of a steel stanchion is 17.5, the grade of  the steel is S275, and the thickness of the section is less than 40mm. So we are to interpolate from the table above to obtain the compressive strength.

x0 = 15; x1 = 20; xp = 17.5
h = x1 –  x0 = 20 – 15 = 5
p = (xp – x0)/h = (17.5 – 15)/5 = 0.5
Δf0 =   f(x1– f(x0) = 271 –  275 = -4

Therefore; f(2.3) = 275 + 0.5(-4) = 273 N/mm2

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