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Sunday, December 8, 2019

December 08, 2019

Analysis of Beams on Elastic Foundation


Introduction
When a beam lies on 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.




Fig 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)                (1)

Where kis 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 opposing 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 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)                    (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                 (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 to analyse a beam on elastic foundation and compare the results with 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 carrying a concentrated load of W =  300 kN 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 modulus of elasticity of concrete Ec = 21.7 x 106 kN/m2.



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 – MAβF4WFa1

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.


(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 elastic mat foundation option will not work since the support will not form a closed loop (analysing this way will give ‘colinear support error’). Therefore, the soil spring was modelled using ‘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 subgrade reaction was multiplied with 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;



(4) Analysis and Results
When analysed using static check option, the following results were obtained;

(a) Soil deformation


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

(b) Bending Moment
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 


Monday, August 26, 2019

August 26, 2019

Analysis and Design of Cantilever Retaining Walls on Staad Pro


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.

Thursday, August 22, 2019

August 22, 2019

Manual Design of Beam and Raft Foundation to Eurocode 2 (free PDF download)


In our last post, we considered the analysis of beam and raft foundation of a G+1 building using Staad Pro software. In this particular post we are going to consider the analysis and design of the same foundation using manual calculations according to the principles of the Eurocodes. If you missed the last post, you can read it by clicking on the link below;

Thursday, July 18, 2019

July 18, 2019

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

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.

Wednesday, June 26, 2019

Monday, June 24, 2019

June 24, 2019

Loading and Design of Box Culverts to Eurocodes


Introduction
A culvert is a drainage structure designed to convey storm water 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 such span over 6 m is not a culvert but can be treated as a bridge. Box culverts consist of two horizontal slabs, and two or more vertical side walls which are built monolithically.

Thursday, June 6, 2019

June 06, 2019

Preliminary Plastic Analysis of Portal Frames


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

Thursday, May 16, 2019

May 16, 2019

Structural Design of Flat Slabs to Eurocode 2


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

Saturday, May 4, 2019

May 04, 2019

Simplified Design of Welded Connection to Eurocode 3


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

Tuesday, April 30, 2019

April 30, 2019

Example on Structural Design of Waffle Slab


Waffle slabs can be described as the equivalent of 2-way solid slabs especially when the spans are large, and ribbed system is to be adopted. The design of waffle slabs is the same as that of ribbed slabs, with the difference being that waffle slabs have ribs spanning in both directions, and the coefficients used for analysing the slab is similar to those used for two-way restrained slab. Waffle slabs are supported on beams or columns, where the support zones are made to be uniformly thick.

Friday, April 26, 2019

April 26, 2019

8 Top Civil Engineering Questions (Part 2)

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


(A)
(B)
(C)




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

(A)
(B)
(C)

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

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




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

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

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

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

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

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

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

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

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

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



Thursday, April 25, 2019

April 25, 2019

Use of Polystyrene in Ribbed Slabs: Structural Design Example


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

Sunday, April 21, 2019

April 21, 2019

Advise on the Stability of this Structure


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

Height - 11.28 m
Length - 20.00 m






Note: This is an existing observatory structure in Tielt-Winge Belgium. Report has it that it was recently vandalised and is looking to be rebuilt...
April 21, 2019

Comparative Analysis of Cylindrical Water Tanks


Water tanks are usually rectangular or cylindrical in shape. Cylindrical tanks are usually employed for surface or elevated water tanks. Some of the challenges involved with cylindrical tanks are the difficulty associated with construction such as setting out, formwork preparation/installation, reinforcement installation, increased labour cost, etc.




Cylindrical tanks are subjected to radial pressure from the stored water, and/or from the retained earth when they are buried under the ground. Just like rectangular water tanks, the analysis of cylindrical tanks is easily done with the use of coefficients picked from already made tables (see Fig 2.0). The values of the coefficients are usually based on the support condition of the wall relative to the base.

The internal forces normally analysed for are;

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

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

Table 1.0: Coefficients for Analysis of Cylindrical Tanks Fixed at the Base (Reynolds and Steedman, 2008)
Analysis Example
Determine the maximum service values for circumferential (hoop) tension, vertical moment and radial shear in the wall of a cylindrical tank that is free at the top edge and fixed at the bottom. The wall is 400 mm thick, the tank is 4 m deep, the diameter is 8 m, and the water level is taken to the top of the wall.



Hydrostatic pressure at the base of the tank (n) = 10 kN/m3 × 4 m = 40 kN/m2
From Table 1.0;
lz2/Dh = 42/(8 × 0.4) = 5
Maximum hoop tension (t) = αtnr = 0.477 × 40 × 4 = 76.32 kN/m



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

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


Fig 2.0: Finite Element Meshing of Cylindrical Tank




Fig 3.0: Horizontal Moment on the Tank Walls

Fig 4.0: Vertical Bending Moment on the Tank Walls
Table 2.0: Analysis Result from Staad Pro

Maximum vertical bending moment = 10.714 kNm/m
Maximum vertical shear stress = 0.073 N/mm2 = 29.2 kN/m
Maximum hoop stress (membrane) = 0.195 N/mm2 = 78 kN/m
Let us compare the results from Staad Pro with results from the use of coefficients (see Table 3.0).

Table 3.0: Comprison of Analysis Result from Staad Pro and use of coefficients

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


Tuesday, February 26, 2019

February 26, 2019

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


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

Saturday, February 23, 2019

Thursday, January 10, 2019

January 10, 2019

Transfer Structures: Design Example on 5-Storey Building


Transfer structures can be described as structures in which the loads from above (usually from columns or walls) are transferred to other structures (such as beams or plates) for distribution to another supporting structure which can resist the load. The prominent issue in transfer structures is that the load path is unconventional, and this is usually found in high rise buildings where floor arrangements differ.

Saturday, December 29, 2018

December 29, 2018

Uplift Verification for Underground Structures: Solved Example

It is widely recognised that an object will float in water, if the weight is less than the upthrust. Upthrust is an upward force exerted by a fluid that opposes the weight of an immersed object. This also applies to structures that are buried under the ground, and subjected to ground water action. Structures such as basements, foundations, underground tanks, and swimming pools are at risk if the dead weight is less than the upthrust, especially when the structure is empty.

Monday, December 24, 2018

December 24, 2018

Merry Christmas from all of us at Structville


Christmas is a special season of the year that offers us an atmosphere filled with love and happiness. From all of us at Structville, we wish you and your family a Merry Christmas filled with happiness, love, and fulfilled living.

Merry Christmas and Happy new year in advance. 🎊🎊🎊🎉🎉🎉🎄


Tuesday, December 18, 2018

Saturday, December 15, 2018

December 15, 2018

Plastic Analysis of Framed Structures


The fully plastic moment of a section (Mp) is the maximum moment of resistance of a fully yielded cross-section. The yielded zone due to bending where infinite rotation can take place at constant plastic moment (Mp) is called a plastic hinge. In order to find the fully plastic moment of yielded section, we normally employ the force equilibrium equation by saying that the total force in tension and compression at that section are equal.
December 15, 2018

How to Analyse Retaining Walls for Trapezoidal Load

In the analysis of retaining walls subjected to earth pressure, it is very common to observe trapezoidal load distribution on the walls. Normally, earth pressure on a retaining wall is assumed to adopt a triangular load distribution, but due to surcharge which is usually assumed to act on the ground surface, the top of the wall experiences some degree of lateral pressure.

Tuesday, November 27, 2018

November 27, 2018

Structural Design of Swimming Pools and Underground Water Tanks



I spent a large part of the last few months developing the contents of this booklet on 'Structural Design of Underground Water Tanks and Swimming Pools' (According to the Eurocodes). Water is necessary for survival of mankind, but in one way or another, water is relatively scarce. We all know that rain does not fall continuously, and for water to be available for usage in homes, it will have to be fetched/pumped from the stream, or harvested during rainfall, or dug up from the ground. As a result, survival instincts made man to create different means of storing water in order to face the periods of scarcity.