It is desired to find the eigenfrequencies and eigenvectors (modal parameters) of concrete water tank support when filled with water. This is a case of undamped free vibration, treating the stored water as a lumped mass on the tank stand. The tank will be treated as a system with 2 degrees of freedom, considering lateral displacement only. Data and Load Analysis;
Density of concrete = 24 KN/m3
Thickness of slab = 150mm
Dimension of columns = 230 x 230mm
Supporting beams = 300 x 230mm
Weight of water = 10 KN/m3
Size of tank = 2000 litres
Modulus of elasticity of concrete = 21.7 × 106 KN/m2
Load Analysis
Self weight of slab = 24 KN/m3 × 0.15m × 2.5m × 2.5m = 22.5 KN = 2293.75 kg
Weight of supporting beams on the four sides = 4 × 24 KN/m3 × 2.27m × 0.23m × 0.3m = 15.036 KN = 1532.77kg
Weight of water = 10 KN/m3 × 2m3 = 20 KN = 2038.73 kg
Weight of tank (assume) = 75kg
At the 1st level = 2293.75 + 1532.77 + 2038.73 + 75 = 5940.25 kg At the 2nd level = 2293.75 + 1532.77 + 2(2038.73) + 2(75) = 8053.98 kg
We are assuming that the floor slab is very stiff compared to the columns. So the flexural rigidity is taken as infinity.
Geometrical properties of the sections
Columns = 23cm x 23cm = 0.23m x 0.23m
Moment of inertia of column IC = (0.23 × 0.233)/12 = 2.332 × 10-4 m4
EIC = (2.332 × 10-4 m4) × (2.1 × 107 KN/m2) = 4897.2 KN.m2
Several codes of practice in the world allow us to idealise structures into 2-dimensional frames for simplified analysis. For sub-frames, it is obvious that the force method becomes less handy due to the high number of redundants, and the next best alternative is the displacement method (stiffness method), where we solve for the unknown displacements.
The displacement method, also known as the stiffness method, has become a cornerstone of structural analysis. It offers a powerful and versatile framework for solving equilibrium problems in various structures, from trusses and beams to complex frames and continua.
At its core, the stiffness method focuses on determining the unknown displacements of points within a structure under the action of applied loads. This is achieved by establishing a relationship between the displacements and the internal forces generated within the structure. This relationship is expressed through the stiffness matrix, which encodes the inherent rigidity of the structure and its resistance to deformations.
Solved Example Using the Stiffness Method
In this article, we have a typical example where a problem that would have generated a (21 x 21) matrix using the force method, has been solved using a (4 x 4) matrix by the stiffness (displacement) method. Another approach to the solution of this problem is the moment distribution method. But in this case, the stiffness method remains the fastest.
For the frame loaded as shown above, we have to start by drawing the kinematic basic system of the structure. This has been achieved by fixing the nodes 1, 2, 3 and 4 against rotation as shown below.
We will now have to evaluate the basic system for different cases of a unit rotation Zi = 1.0, applied at the fixed nodes. Analysis of Case 1 Z1 = 1.0; Z2 = Z3 = Z4 = 0
This article is aimed at providing the procedure for the design of column base plates subjected to axial and shear forces according to Eurocode 3. Find a design example below.
Problem Statement
It is required to specify the appropriate thickness of a base plate to support a UC 203 x 203 x 60 subjected to the following loads. The connection is assumed to be pinned with four bolts outside the column profile. The supporting concrete is to have a grade fck = 25 N/mm2
(Image Copyright belongs to Albion Sections Limited, UK)
Roof purlins are members used to directly support roof sheeting materials, and could be made of timber or steel. In timber construction, purlins are nailed to the rafter or supporting trusses, while in steel roof construction, they are welded or bolted to the rafters or trusses by the means of cleats. As structural members, they resist loads, and provide lateral restraints for truss members, therefore it is important to design them properly against forces such as bending, shear, torsion, buckling etc.
In their design life, purlins are subjected to dead load (e.g self weight of sheeting materials and accessories), live load (e.g. during maintenance services and repairs), and environmental loads (e.g. wind and snow load). Therefore, a purlin should be adequately strong to withstand the loads it will encounter during its design life, and should not sag in an obvious manner thereby giving the roof sheeting an undulating and/or unpleasant appearance. This article will be focusing on design of steel purlin using cold formed sections.
Arrangement of Roof Purlins
By default, purlin sections assume the slope of the roof they are supporting. The spacing of roof purlins on rafters usually calls for careful arrangement, in the sense that it should follow the nodal pattern of the supporting trusses. By implication, purlins should ideally be placed at the nodes of trusses and not on the members themselves so as not to induce secondary bending and shear forces in the members of the truss. Furthermore, if manual analysis is employed to analyse a truss loaded in such a manner, such secondary stresses may not be captured especially if nominally pinned connections are assumed.
Cold formed Z (Zed) and C (channel) sections are normally specified for roof purlins in steel structures (see their form in image below).
When compared with thicker hot-rolled sections, cold-formed sections normally offer the advantages of lightness, high strength and stiffness, easy fabrication and installations, easy packaging and transportation etc. The connection of purlins can be sleeved or butted depending on the construction method adopted.
In terms of arrangement of roof purlins, we can have single spans with staggered sleeved/butt arrangement, single/double span with staggered sleeve arrangement, double span butt joint system, and single span butt joint system. The choice of the arrangement to be adopted can depend on the supply length of the sections as readily available in the market, the need to avoid wasteful offcuts, the loading and span of the roof, the arrangement of the rafters etc.
Therefore the roof designer must plan from start to finish. However, single and double span butt joint system are the most popular in Nigeria, due to their simplicity, and the culture of adopting shorter roof spans in the country. However, they are less structurally efficient than sleeved connections.
Design Exampleof Roof Purlins
We are to provide a suitable cold formed channel section for the purlin of the roof arrangement shown below.
Partial Factor for loads (BS EN 1990 NA 2.2.3.2 Table NA.A1.2(B)) Permanent action γG = 1.35 (unfavourable) Variable action γG = 1.5 Combination factor for Roofs, ψ0 = 0.7 Wind loads, ψ0 = 0.5 Reduction factor ξ = 0.925
Initial Sizing of Sections The maximum spacing of the trusses is at 3000 mm c/c Limiting the deflection to L/200; For continuous purlins, minimum depth of section (preliminary guide) = L/45 3000/45 = 66.667 Minimum width = 0.5L/60 = (3000/2)/60 = 25 Try C120-15 section (C – purlin) (Section picked from Albion Technical Manual, 2010).
Section Properties In the design of purlins using EN 1993-1-3:2006, we normally utilise the effective section properties. This calculation of effective section properties can be very tedious and prone to error, hence it is very advisable to obtain information from the manufacturer’s details or you can write a program using Microsoft Excel or MATLAB for such calculations. However, we are going to make a sample calculation for the section that we are considering.
Thickness (tnom) = 1.5 mm Depth = 120mm; Flange width = 50 mm; Lips/Edge Fold = 15 mm; Steel core thickness (t) = 1.5 – 0.05 = 1.45 mm (Note that EN 1993-1-3:2006 recommends a thickness of 0.04 mm for zinc coated sections, but we are using 0.05 mm here) Unit weight = 2.8 kg/m2; Web height hp = h – tnom = 120 – 1.5 = 118.5 mm Width of flange in tension = Width of flange in compression bp1 = bp2 = b – tnom = 50 – 1.5 = 48.5 mm Width of edge fold cp = C – tnom/2 = 15 – 1.5/2 = 14.25 mm
Effective section properties of the flange and lip in compression (clause 3.7.2) Effective width of the compressed flange; The stress ratio ψ = 1.0 (uniform compression) kσ = 4 for internal compression element (clause 3.7.2 Table 3.5)
ε = √(235/fyb) = √(235/350) = 0.819
The effective width is; beff = ρbb1 = 0.966 × 48.5 = 46.851 mm be1 = be2 = 0.5beff = 0.5 × 46.851 = 23.4255 mm
Effective Width of the edge fold (lip) Clause 3.7.3.2.2 Equation 3.47 The buckling factor is;
cp ⁄ bp1 = 14.25/48.5 = 0.2938 < 0.35 So kσ = 0.5
Therefore since 1.149 < 1.0, take reduction factor as 1.0
The effective width is therefore ceff = ρcp = 1.0 × 14.25 = 14.25mm
The effective area of the edge stiffener; As = t(be2 + ceff) = 1.45 (23.4255 + 14.25) = 54.629 mm2
We now have to use the initial effective cross-section of the stiffener to determine the reduction factor, allowing for the effects of the continuous spring restraint. The elastic critical buckling stress for the edge stiffener is;
Where K is the spring stiffness per unit length;
b1 is the distance of the web to the centre of the effective area of the stiffener in compression flange (upper flange);
kf = 0 for bending about the y-y axis
Therefore;
= 175882.211 × 0.00000491305 = 0.8641 N/mm
So the elastic critical buckling stress for the edge stiffener is;
The relative slenderness factor for the edge stiffener;
In our own case; χd = 1.47 – 0.723(0.840)= 0.862
As the reduction factor for buckling is less than 1.0, we can optionally iterate to refine the value of the reduction factor for buckling of the stiffeners according to clause 5.5.3.2(3). But we are not iterating in this post.
Position of the neutral axis with regard to the flange in tension; Zt = hp – Zc
Second moment of area:
Effective section modulus; With regard to the flange in compression; Weff,y,c = Ieff,y/zc = (777557.517) / 60.903 = 12767 mm3
With regard to the flange in tension; Weff,y,t = Ieff,y/zt = (777557.517) / 57.697 = 13476 mm3
LOAD ANALYSIS Permanent loads Employing long span aluminium roofing sheet (gauge thickness = 0.55mm) Load due to sheeting = 0.019 kN/m2 Other permanent accessories and fittings = 0.15 kN/m2 Total = 0.169 kN/m2
At a spacing of 1.2m, = 0.169 kN/m2 × 1.2m = 0.2028 kN/m Self weight of purlin = 2.8 kg/m = 0.0275 kN/m
Total Gk = 0.2028 KN/m + 0.0275 KN/m = 0.230 kN/m
Live load For a roof with 20° slope and no access except for normal repairs and maintenance, let us adopt a live load of 0.75 kN/m2 At a spacing of 1.2m, Qk = 0.75 kN/m2 × 1.2m = 0.9 kN/m
Wind Load Taking a dynamic wind pressure of 1.5 kN/m2 When the wind is blowing from right to left, the resultant pressure coefficient on a windward slope with positive internal pressure is; cpe = −0.90 upwards
Therefore the external wind pressure normal to the roof is; pe = qpcpe = 1.5 × − 0.90 = −1.35 kN/m2 The vertical component of the wind pressure is; pev = pecosθ = −1.35 × cos 20° = −1.268 kN/m2 acting upwards. At a spacing of 1.2m; Wk = −1.268 kN/m2 × 1.2m = 1.522 kN/m
STATIC SYSTEMS We are adopting two possible systems that will offer us continuous and single span systems. The 6m span is based on supply length.
Static Model 1
Static Model 2
Load Case 1 When Dead load and live load are acting alone; q = 1.35Gk + 1.5Qk = 1.35(0.230) + 1.5(0.9) = 1.6605 kN/m
Model 1
Load Case 2 When Dead load, live load and wind load are acting together; q = 1.35Gk + 1.5Qk + 0.9Wk Where the live load is the leading variable action q = 1.35(0.230) + 1.5(0.9) – 0.9 (1.522) = 0.2907 kN/m
Model 1
Model 2
Load Case 3 When Dead load and wind load are acting alone; q = 1.0Gk – 1.5Wk Where dead load is favourable q = 1.0(0.230) – 1.5(1.522) = -2.053 kN/m
Model 1
Model 2
Maximum span design moment MEd = 2.31 kNm Maximum shear force VEd = 3.85 kN
Verification of Bending Design moment resistance MC,Rd = Weff,y.Fy / γm0 [Clause 6.1.4.1(1) of EN 1993-1-3:2006]
Check for shear (using maximum shear force); VEd / Vb,Rd = (3.85)/(28.867) = 0.1333 < 1.0 Shear is ok
Deflection Check Maximum deflection under SLS (1.0gk + 1.0qk) = 0.07 mm Limiting deflection = L/200 = 3000/200 = 15 mm Since 0.07 < 15mm, deflection is Ok!
Therefore, the channel Z120-15 section is adequate for the applied load.
An indeterminate truss is supported and loaded as shown above, using the direct stiffness method, obtain the displacements, support reactions, and internal forces that are induced in the members due to the externally applied loads, (EA = Constant, dimensions in mm).
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Summary of Procedure
(1) Establish the x and y global coordinate system. The origin is usually located at the joint for which the coordinates for all the other joints are positive.
(2) Identify each joint and member numerically, and arbitrarily specify the near and far ends of each member symbolically by directing an arrow along the member with the head directed towards the far end.
(3) Specify the two code numbers at each joint, using the lowest numbers to identify unconstrained degrees of freedom, followed by the highest numbers to identify the constrained degrees of freedom.
For the truss that is loaded as shown above, there are four unconstrained degrees of freedom which occur at nodes ② and ③. The displacements are the horizontal and vertical translations which have been labelled from 1 – 4 and represented by the arrow heads at those nodes. The members have also been labelled accordingly with their numbers enclosed in a box (see figure above). Also notice that the unconstrained degrees of freedom have labelled first. The aim of this is to simplify the arrangement of the structure’s stiffness matrix.
Computation of member global stiffness matrix
Without much attention to the derivation, the stiffness matrix is given by;
[k] =[TT][k’][T] ————— (1)
Where;
K’ = member stiffness matrix which is of the same form as each member of the truss. It is made of the member stiffness influence coefficients, k’ij
T = Displacement transformation matrix
TT = This transforms local forces acting at the ends into global force components and it is referred to as force transformation matrix which is the transform of the displacement transformation matrix [T].
When the equation is solved, the global member stiffness matrix is obtained which is given by;
Where; l = cos θ; m = sin θ;
L = Length of member
The global stiffness of each member is given below;
The general stiffness matrix of the structure [KT] is given by;
[KT]= [K1] + [K2] + [K3] + [K4] + [K5] + [K6]
This now yields an 8 x 8 matrix which represents all the degrees of freedom in the truss both unconstrained (1-4) and constrained (5-8).
Since the unconstrained degrees of freedom are at points 1-4, we can therefore compute the deformation at such nodes using the relation below;
[P] = [K][u]
Where [P] is the vector of joint loads acting on the truss, [u] is the vector of joint displacement and [k] is the global stiffness matrix. On partitioning the above stiffness matrix, the relationship for this problem is as given below;
This is then given below;
On solving;
In the same vein, this same relationship can help us compute the support reactions;
On solving;
You can also realise that the obtained support reactions satisfy equilibrium requirements in the structure.
When foundations are constructed, they increase the net pressure in the soil. The knowledge of this stress increment is very important for estimating the settlement of foundations, and for the evaluation of lateral pressure on adjacent structures.
Boussinesq in 1883 solved the problem of stresses produced at any point in a homogeneous, elastic, and isotropic medium as the result of a point load applied on the surface of an infinitely large half-space. This has been extended into solving most of the problems of stresses that foundations impose on soils and for different types of loading, despite the fact that soils are practically neither homogeneous nor elastic.
In this example, it is required to obtain the stress that a soil mass experiences at a depth of 3m under the pad foundation shown below.
The increment in the stress at any point below a pad foundation can be obtained by following the relationship below.
For example, let us consider the plan of a pad footing shown below. The loaded area can be divided into four rectangles A, B, C, and D, as shown. It is desired to determine the stress at a point below point 1 at depth z. Note that this point is very common to all four rectangles. The increment in stress at a depth z below point 1 due to each rectangular area can be calculated using the equation above.
The total stress increase caused by the entire loaded area is given by equation (2) below;
∆σz = q [I3(A) + I3(B) + I3(C) + I3(D)] ———- (2)
At the centre of a pad footing, this is given by;
∆σz = q I4 ————– (3)
Where;
For the increment in stress at the centre of a rectangular footing, we can obtain the value of I4 from the Table below (derived from the relationship above);
Therefore, for the example under consideration; The service pressure (q) = P/A = (1350 kN) / (1.5m × 1.5m) = 600 kN/m2
Therefore, apart from the overburden pressure (geostatic stress), the stress at a depth of 3m under the pad foundation increased by 64.8 kN/m2 due to the foundation load.
The values given for the following triangles were obtained using a consistent relationship. I really do not expect a lot of people to be able figure it out….
Cohesive soil profiles due to their nature are partially self supporting up to an extent called the critical depth (the critical depth is given by 2Zo, kindly see figure above). As a result, such soils would not exert much pressure on a retaining wall as a granular material would (especially considering active earth pressure). Tropical laterites (especially those found in Nigeria) usually possess angle of internal friction and some cohesion. Their usage for construction purposes is also widespread.
The walls retaining such soils are subjected to active and passive pressure. In this example, we are going to consider active pressure only (all passive pressure neglected), but note that passive pressure could be more critical and hence govern the design (I recommend you read standard geotechnical engineering textbooks for more knowledge on this subject).
WorkedExample
The cantilever retaining wall shown below is backfilled with tropical lateritic earthfill, having a unit weight, ρ, of 18 kN/m3, a cohesion C of 8 kN/m3 and an internal angle of friction, φ, of 26°. The allowable bearing pressure of the soil is 150 kNm/2, the coefficient of friction is 0.5, and the unit weight of reinforced concrete is 24 kN/m3. Water level is at a great depth and adequate drainage is provided at the back of the wall.
We are to; 1. Determine the factors of safety against sliding and overturning for the active pressure. 2. Calculate ground bearing pressures. 3. Design the wall and base reinforcement assuming fcu = 30 N/mm2, fy = 460 N/mm2 and the cover to reinforcement in the wall and base are, respectively, 40 mm and 50 mm.
Geotechnical Design
Wall Pressure Calculations
Coefficient of active pressure KA Using Rankine’s theory KA = (1 – sinφ) / (1 + sinφ); KA = (1 – sin 26°) / (1 + sin 26°) = 0.39046;
Pressure at the top of wall = 2C√KA = 2 × 8 × √0.39046 = 9.998 kN/m2 Z0 = 2 C/γ√KA = (2 × 8)/(18 × √0.39046) = 1.423m
Pressure at the bottom of retaining wall = H.γ.KA – 2C√KA = (5.5 × 18 × 0.39046) – (2 × 8 × √0.39046) = 32.408 kN/m2
Resistance to Sliding
Consider the forces acting on a 1 m length of wall. Horizontal force on wall due to backfill, PA, is; PA = 0.5 × 32.408 × 4.077 = 66.063 kN/m
Weight of wall (Ww) = 0.35 × 5.1 × 24 = 42.84 kN/m Weight of base (Wb) = 0.4 × 4 × 24 = 38.4 kN/m Weight of soil (Ws) = 2.85 × 5.1 × 18 = 261.63 kN/m Total vertical force (Wt) = 324.87 kN/m
Neglecting all passive pressure force (FP) = 0. Hence factor of safety against sliding is; F.O.S = FF/PA = 171.435/66.063 = 2.595 The factor of safety 2.595 > 1.5. Therefore the wall is very safe from sliding.
Resistance to Overturning
Taking moments about the toe, sum of overturning moments (MO) is; MO = (66.063 kN/m) × (4.077/3) = 89.779 kN.m
F.O.S = MR / MO = 792.266/89.779 = 8.824 The factor of safety 8.824 > 2.0. Therefore the wall is very safe from overturning.
Bearing Capacity Check
Bending moment about the centre line of the base; M = [66.064 × (4.077/3)] + (Ww × 1.025m) – (Ws × 1m) M = [66.064 × (4.077/3)] + (42.84 × 1.025m) – (261.63 × 0.575m) = -16.745 kNm
Total vertical load = 324.87 kN
The eccentricity (e) = M/N = (16.745/324.87) = 0.0515m
Let us check D/6 = 4/6 = 0.666 Since e < D/6, there is no tension in the base
The maximum pressure in the base; qmax = P/B (1 + 6e/B) = 324.87/4 [1 + (6 × 0.0515)/(4)] = 87.491 KN/m2 87.491 KN/m2 < 175 KN/m2 Therefore, the bearing capacity check is satisfied.
Structural Design
Design of the Wall
Conservatively taking the moment at the base of the wall due to the active force; M = [66.064 × (4.077/3)] = 89.781 kN.m At ultimate limit state M = 1.4 × 89.781 = 125.693 kN.m
Thickness of wall = 350mm
Effective depth = 350 – 40 – (8) = 302mm (assuming Y16mm bars will be employed for the construction)
DETAILING I will need my readers to do the detailing sketches. I will accept both manual (hand) and CAD details. But your manual sketches must be very neat and well scaled. You should show the plans and at least one section. If your detailing is good enough, I will publish it on this post with your Facebook, LinkedIn, or G+ URL attached. So, I will keep updating the post with as many good pictures as I receive. Good knowledge for all to flourish @ Structville Send your sketches and URL profile to info@structville.com
An arch is supported on frame with different levels of supports as shown above. The frame is supported on a hinge at support A and a roller at support C. A tie beam is used to connect node B and support C. Plot the bending moment, shear force, and axial force diagram die to the externally applied load.
A section of a beam profile is shown above. BASED ON THE GIVEN DATA ONLY AND WITHOUT MAKING ANY ASSUMPTIONS, identify the technical error(s) in the section detailing using BS 8110-1:1997 recommendations? Read the hints below carefully!
Data
Figured dimensions are in mm
Concrete cover at the sides and the soffit = 40 mm
Concrete cover at the top = 30 mm
Strength of reinforcement = 460 N/mm2
Strength of concrete = 35 N/mm2
Section is at mid-span (sagging moment)
Area of tension reinforcement required = 2960 mm2
Hints: (1) This post is for educational purposes, hence, your reply is better dropped here on Structville as a comment so that every person accessing the post from different platforms can see the reply. (2) Please, do not guess. If you identify the error(s), kindly drop your comment, otherwise, keep checking back for other people’s replies. Do not post your guesses. (3) If you are not familiar with BS 8110-1:1997, but you identify any error based on the code that you are familiar with, kindly state the code, and the error you discovered. Try to be technical in your explanations, or quote the requirement as given in the code. (4) The diagram is not to scale, so use the figured dimensions only. Also, do not call on any data or information that is not provided. (5) Let your explanations be offered with love. We believe that knowledge is power.
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