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
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.
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.
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
Checking of geometrical proportions
b/t ≤ 60 b1/t = 48.5/1.45 = 33.448 < 60 – OK
c/t ≤ 50 c/t = 15/1.45 = 10.344 < 50 – OK
h/t ≤ 500 c/t = 120/1.45 = 82.758 < 500 – OK
0.2 ≤ c/b ≤ 0.6 c/b1 = 15/48.5 = 0.309 – OK
The effect of rounding corners due to root radius has been neglected in this design.
Gross section properties
Abr = t(2cp + bp1 + bp2 + hp) = 1.45[(2 × 14.25) + 48.5 + 48.5 + 118.5)] = 353.8 mm2
Position of the neutral axis with respect to the flange in compression;
Zb1 = 1.45[14.25(118.5 – 7.125) + (48.5 × 118.5) + 118.52/2 + 14.252/2] / 353.8 = 59.253mm
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;
σcr,s = (2√(0.8641 × 210000 × 1007.801))/(54.629 ) = 495.901 N/mm2
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.
Therefore;
χd = 0.862
be2 = 23.4255 mm
ceff = 14.25mm
The buckling factor kσ = 7.81 – 6.29ψ + 9.78ψ2
kσ = 7.81 – 6.29(-0.9346) + 9.78(-0.9346)2 = 22.231
Therefore, take ρ as 1.0 since 1.137 > 1.0.
Therefore, the effective width of the zone in compression of the web is;
heff = ρhc = 1.0 × 61.252 = 61.252mm
Near the flange in compression;
he1 = 0.4heff = 0.4(61.252) = 24.5mm
he2 = 0.6heff = 0.6(61.252) = 36.7512mm
The effective width of the web is;
Near the flange in compression;
h1 = he2 = 24.5mm
Near the flange in tension;
h2 = hp – (hc – he2) = 118.5 – (61.252 – 36.7512) = 93.992mm
Effective section properties;
Aeff = 1.45 × [14.25 + 48.5 + 24.5 + 93.992 + 23.4255 + (23.4255 + 14.25) 0.862]
Aeff = 343.858 mm2
Position of the neutral axis with regard to the flange in compression;
zc = 1.45[(14.25 × 111.375) + (48.5 × 118.5) + (93.992 × 71.504) + 300.125 + 87.5199] / 343.858 = 60.903 mm
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]
From our calculations;
Weff,y = min(Weff,y,c , Weff,y,t) = 12767 mm3
Therefore;
MC,Rd = Weff,y.Fy/γm0 = [(12767 mm3) × (350) × 10-6] / 1.0 = 4.46845 KNm
Check of span and single support;
MEd/MC,Rd = 2.31/4.46845 = 0.5169 < 1.0 OK!
Check of shear resistance at ULS
The design shear resistance is given by;
Vb,Rd = (hw/sinφ.t.fbv) / γM0 (Clause 6.1.5)
The shear buckling strength (fbv) which is based on the relative web slenderness can be obtained from the table below (Table 6.1 of EN 1993-1-3).
Where λw ̅ is the relative slenderness for webs without longitudinal stiffeners.
Since 1.154 < 0.83 but less than 1.40;
fbv = 0.48fyb = 0.48 × 350 = 168 N/mm2
Vb,Rd = [(118.5 / sin90°) × 1.45 × 168] / 1.0 = 28866.6 N = 28.867 KN
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;
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.
∑Fy = 0 ( i.e 12 + 13 = 25 KN)
∑Fx = 0 ( i.e 7.5662 + 2.4338 = 10 KN)
Internal Axial Forces
For computing the internal forces in the members, the following relationship is used;
[N] = [T][K’][u]
Which when expanded yields;
The axial forces in the members are now as given below;
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
b = B/2 = 1.5/2 = 0.75m
m1 = L/B = 1.5m/1.5m = 1.0
n1 = z/b = 3m/0.75m = 4.0
Reading from the table above;
I4 = 0.108
∆σz = q I4
∆σz = 600 × 0.108 = 64.8 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.
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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).
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.
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
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
Friction force, FF, is
FF = μWt = 0.5 × 324.87 = 171.435 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.
Taking moments about the toe, sum of overturning moments (MO) is;
MO = (66.063 kN/m) × (4.077/3) = 89.779 kN.m
Sum of Restoring Moment MR;
MR = (Ww × 0.975m) + (Wb × 2m) + (Ws × 2.575m)
MR = (42.84 × 0.975m) + (38.4 × 2m) + (261.63 × 2.575m) = 792.266 kNm
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.
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.
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)
k = M/(Fcubd2) = (125.693 × 106) / (30 × 1000 × 3022) = 0.0459
k < 0.156, no compression steel needed
la = 0.5 + (0.25 – k/0.9)0.5 = 0.5 + (0.25 – 0.0459/0.9)0.5 = 0.946
ASreq = M/(0.95Fy.la.d ) = (125.693 × 106) / (0.95 × 460 × 0.946 × 302) = 1006 mm2/m
ASmin = (0.13bh)/100 = (0.13 × 1000 × 350) / 100 = 455 mm2
Provide Y16 @ 150mm c/c (Asprov = 1340 mm2/m)
Check for shear
Design shear force on wall at ultimate limit state V = (1.4 × 66.064) = 92.4896 kN/m
Shear stress v = V/bd = (92.4896 × 1000) / (1000 × 302) = 0.3062 N/mm2
Vc = 0.632 (100As/bd)1/3 (400/d)1/4
Vc = 0.632 × [(100 × 1340)/(1000 × 302)]1/3 × (400/302)1/4
Vc = 0.632 × 0.7627 × 1.0727 = 0.517 N/mm2
For Fcu = 30N/mm2, Vc = 0.517 × (30/25)1/3 = 0.5494 N/mm2
Since v < Vc, no shear reinforcement required.
The pressure distribution diagram on the base at serviceability limit state is as shown below;
qmax = P/B (1 + 6e/B) = 324.87/4 [1 + (6 × 0.0515)/(4)] = 87.491 kN/m2
qmin = P/B (1 – 6e/B) = 324.87/4 [1 – (6 × 0.0515)/(4)] = 74.927 kN/m2
At ultimate limit state;
qmax = 1.4 × 87.491 = 122.487 KN/m2
qmin = 1.4 × 74.927 = 104.898 KN/m2
qC = 104.898 + [2.85 (122.487 – 104.898)]/4 = 117.430 kN/m2
qB = 104.898 + [3.2 (122.487 – 104.898)]/4 = 118.968 kN/m2
On investigating the maximum design moment about point C;
Back fill = 1.4 (18 × 5.1 × 2.85 × 2.85/2) = -521.952 kNm
Heel Slab = 1.4 (24 × 0.4 × 2.85 × 2.85/2) = -54.583 kNm
Earth Pressure = (104.898 × 2.85 × 2.85/2) + [(117.430 – 104.898) × 0.5 × 2.85 × 2.85/3] = 426.017 + 16.965 = 442.982 KNm
Net moment = -521.952 – 54.583 + 442.982 = -133.553 kNm
On investigating the maximum design moment about point B;
Back fill = (conservatively neglected)
Heel Slab = 1.4 (24 × 0.4 × 0.8 × 0.8/2) = – 4.3008 kNm
Earth Pressure = (118.968 × 0.8 × 0.8/2) + [(122.487 – 118.968) × 0.5 × 0.8 × (2 × 0.8/3)] = 37.9776 + 0.751 = 38.727 kNm
Net moment = -4.3008 + 38.727 = 34.426 kNm
Design of the toe
MB = 133.553 kN.m
Thickness of base = 400 mm
Effective depth = 400 – 50 – (8) = 342 mm (assuming Y16mm bars will be employed for the construction)
k = M/(Fcubd2) = (133.553 × 106) / (30 × 1000 × 3422) = 0.0385
k < 0.156, no compression steel needed
la = 0.5 + (0.25 – k/0.9)0.5 = 0.5 + (0.25 – 0.0385/0.9)0.5 = 0.95
ASreq = M/(0.95Fy.la.d ) = (133.553 × 106) / (0.95 × 460 × 0.95 × 342) = 941 mm2/m
ASmin = (0.13bh)/100 = (0.13 × 1000 × 400) / 100 = 520 mm2
Provide Y16 @ 175 mm c/c Top (Asprov = 1148 mm2/m)
Design of the heel
MC = 34.426 KNm
Thickness of base = 400 mm
Effective depth = 400 – 50 – (6) = 344 mm (assuming Y12mm bars will be employed for the construction)
k = M/(Fcubd2) = (34.426 × 106) / (30 × 1000 × 3442) = 0.00981
k < 0.156, no compression steel needed
la = 0.5 + (0.25 – k/0.9)0.5 = 0.5 + (0.25 – 0.00981/0.9)0.5 = 0.95
ASreq = M/(0.95Fy.la.d ) = (34.426 × 106) / (0.95 × 460 × 0.95 × 344) = 242 mm2/m
ASmin = (0.13bh)/100 = (0.13 × 1000 × 400) / 100 = 520 mm2
Provide Y12 @ 175mm c/c (Asprov = 646 mm2/m)
Check for shear at the base
The maximum shear force at ‘d’ from the face of the wall is investigated at the either side;
Maximum shear force at point B
Earth fill = 1.4 × 18 × 5.1 × 2.85 = -366.282 kN/m
Wall base = 1.4 × 24 × 0.4 × 2.85 = -38.304 kN/m
Base Earth Pressure = 0.5 (104.898 + 117.430) × 2.85 = 316.817 kN/m
Net shear force = -366.282 – 38.304 + 316.817 = -87.769 kN/m
A little investigation will show that this is the maximum shear force at any section on the base.
Taking the maximum shear stress at the base v = (87.769 × 1000) / (1000 × 342) = 0.2566 N/mm2
Vc = 0.632 (100As/bd)1/3 (400/d)1/4
Vc = 0.632 × [(100 × 1148) / (1000 × 342)]1/3 × (400/342)1/4
Vc = 0.632 × 0.694 × 1.0399 = 0.456 N/mm2
For Fcu = 30N/mm2, Vc = 0.456 × (30/25)1/3 = 0.484 N/mm2
Since v < Vc, no shear reinforcement required.
DETAILING
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Support Reactions
∑MC = 0
30Ay – 10Ax – (10 × 302)/2 = 0
30Ay – 10Ax = 4500 ————- (1)
∑FX = 0
Ax + 25 = 0
Ax = -25 kN (pointing in the reverse of assumed direction)
Plugging the value of Ax into equation (1);
30Ay – 10(-25) = 4500
30Ay = 4250
Ay = 141.667 kN
∑MA = 0
30Cy – (25 × 10) – (10 × 302)/2 = 0
30Cy = 4750
Cy = 158.333 kN
∑MDR = 0
The ordinate of the arch at any given horizontal length section is given by;
y = [4yc (Lx – x2)] / L2
Where yc is the height of the crown of the arch
y = [(4 × 5) × (30x – x2)]/302 = (2/3)x – (x2/45)
dy/dx = y’ = 2/3 – 2x/45
At x = 0; y = 0
y’ = 2/3 – 0/8 = 0.667
sinθ = y’/√(1 + y’2) = 1/√(1 + 0.6672) = 0.5546
cosθ = 1/√(1 + y’2) = 1/√(1 + 0.6672) = 0.8319
At x = 7.5m;
y = (2/3)x – (x2/45) = (0.667 × 7.5) – (7.52/45) = 3.75m
y’ = 2/3 – 2(7.5)/45 = 0.333
sinθ = y’/√(1 + y’2) = 0.333/√(1 + 0.3332) = 0.43159
cosθ = 1/√(1 + y’2) = 1/√(1 + 0.3332) = 0.9487
At x = 15 m;
y = (2/3)x – (x2/45) = (0.667 × 15) – (152/45) = 5.00m
y’ = 2/3 – 2(15)/45 = 0
sinθ = y’/√(1 + y’2) = 0/√(1 + 02) = 0
cosθ = 1/√(1 + y’2) = 1/√(1 + 02) = 1.0
At x = 22.5m;
y = (2/3)x – (x2/45) = (0.667 × 22.5) – (22.52/45) = 3.75m
y’ = 2/3 – 2(22.5)/45 = -0.333
sinθ = y’/√(1 + y’2) = -0.333/√(1 + 0.3332) = -0.43159
cosθ = 1/√(1 + y’2) = 1/√(1 + 0.3332) = 0.9487
At x = 30m; y = 0
y’ = 2/3 – 2(30)/45 = -0.667
sinθ = y’/√(1 + y’2) = -0.667/√(1 + 0.6672) = -0.5546
cosθ = 1/√(1 + y’2) = 1/√(1 + 0.6672) = 0.8319
Internal Stresses in the Arch Structure
Bending Moment
MA = 0
MBB = (25 × 10) = 250 kN.m
MBUP = (25 × 10) = 250 kN.m
M1 = (141.667 × 7.5) + (25 × 13.75) – (25 × 3.75) – (10 × 7.52)/2 – (250 × 3.75) = 93.7525 kN.m
MD = (141.667 × 15) + (25 × 15) – (25 × 5) – (10 × 152)/2 – (250 × 5) = 0
Coming from the right
M2 = (158.333 × 3.75) – (250 × 5) – (10 × 122)/2 = -31.252 kN.m
MC = 0
Shear force
Section A – B (Column)
QA – Ax = 0
QA – 25 = 0
QA – QBB = 25 kN
Qi = ∑V cosθ – ∑H sinθ (for arch section)
Note that the horizontal forces of 25 kN will eliminate each other;
QBUP = (141.667 × 0.8319) – (250 × 0.5546) = -20.797 kN
Q1R = Q1L = [141.667 – (10 × 7.5)] × 0.9487 – (250 × 0.43159) = -44.651 kN
QD = [141.667 – (10 × 15)] × 1.0 – (250 × 0) = -8.333 kN
Q2R = Q2L = [141.667 – (10 × 22.5)] × 0.9487 – (250 × –0.43159) = 28.839 kN
QC = [141.667 – (10 × 30)] × 0.8319 – (250 × -0.5546) = 6.932 kN
Axial force
Member AB = 141.667 kN (compression)
Member BC = 250 kN (Tension)
For the arch section;
Ni = -∑V sinθ – ∑H cosθ
NBUP = -(141.667 × 0.5546) – (250 × 0.8319) = -286.543 KN (Compression)
N1L = -[141.667 – (10 × 7.5)] × 0.43159 – (250 × 0.9487) = -265.848 KN
ND = -[141.667 – (10 × 15)] × 0 – (250 × 1) = -250 KN (Compression)
N2L = N3R = -[141.667 – (10 × 22.5)] × -0.43159 – (250 × 0.9487) = -273.141 KN
NC = -[141.667 – (10 × 30)] × -0.5546 – (250 × 0.8319) = -295.786 KN
Internal Stresses Diagram
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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:
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Beams that are curved in plan are often found in buildings, circular reservoirs, bridges, and other structures with curves. Curved beams always develop torsion (twisting) in addition to bending moment and shear forces because the center of gravity of loads acting normal to the plane of the structure lies outside the lines joining its supports. Therefore to maintain equilibrium in the structure, the supports of a curved beam must be fixed or continuous.
In this post, we are going to show in the most simplified manner, how to analyse continuous circular (ring) beams.
For ring beams;
Maximum Negative Moment at any support = K1wr2
Maximum Positive Moment at any span = K2wr2
Maximum Torsional Moment = K3wr2
Total load on each column (support reaction) R = wr(2θ)
Shear force at any support = R/2 = wrθ
The coefficients are given in the table below;
Solved Example
A cylindrical reservoir with a diameter of 6m is supported by a ring beam, which is supported on 8 equidistant columns. It is desired to analyse and design the ring beam to support the load from the superstructure.
The plan view of the structural disposition of the reservoir is shown below;
Load Analysis
(a) Geometry of sections
Dimension of beams = 450mm x 300mm
Dimension of column = ϕ300mm circular columns
Thickness of reservoir walls = 250mm
Thickness of reservoir slab = 250mm
(b) Density of materials
Density of stored material = 10 kN/m3
Density of concrete = 25 kN/m3
(c) Dead Loads
Weight of the walls = (25 kN/m3 × 0.25m × 4.75m × 18.849m) = 559.579 kN
Weight of bottom slab = (25 kN/m3 × 0.25m × 28.274m2) = 176.7125 kN
Weight of water stored = (10 kN/m3 × 4.5m × 23.758m2) = 1069.11 KN
Total = 1805.4015 kN
Let us transfer this load to the the ring beam based on the perimeter.
Perimeter of ring beam = πd = π × 6 = 18.849m
w = 1805.4015 kN / Perimeter of ring beam = 1805.4015 kN / 18.849m = 95.782 kN/m
Self weight of the beam = 25 kN/m3 × 0.3m × 0.45m = 3.375 kN/m
Total dead load on beam = 95.782 kN/m + 3.375 kN/m = 99.157 kN/m
Factoring the load on the beams at ultimate limit state = 1.35 × 99.157 kN/m = 133.862 kN/m
From the table above;
Number of supports (n) = 8
θ = π/n = 45°
K1 = 0.052
K2 = 0.026
K3 = 0.0040
Radius (r) = 3m
Maximum Negative Moment at the supports = K1wr2 = -0.052 × 133.862 × 32 = -62.647 KN.m
Maximum Positive Moment at the spans = K2wr2 = 0.026 × 133.862 × 32 = 31.323 KN.m
Maximum Torsional Moment = K3wr2 = 0.0040 × 133.862 × 32 = 4.819 KN.m
Shear force at the supports = R/2 = wrθ = 133.862 × 3 × (π/8) = 157.7 KN
Structural Design
Design strength of concrete fck = 35 N/mm2
Yield strength of reinforcement fyk = 500 N/mm2
Nominal cover to reinforcement = 30 mm
Span
MEd = 31.323 kN.m
Effective depth (d) = h – Cnom – ϕ/2 – ϕlinks
Assuming ϕ12 mm bars will be employed for the main bars, and ϕ8mm bars for the stirrups (links)
d = 450 – 30 – 6 – 8 = 406 mm
k = MEd/(fckbd2) = (31.323 × 106)/(35 × 300 × 4062) = 0.0181
Since k < 0.167, no compression reinforcement required
z = d[0.5+ √(0.25 – 0.882K) ] = z = d[0.5+ √(0.25 – 0.882(0.0181))] = 0.95d
As1 = MEd/(0.87yk z) = (31.323 × 106)/(0.87 × 500 × 0.95 × 406) = 186.69 mm2
Provide 3H12 mm BOT (ASprov = 339 mm2)
Supports
MEd = 62.647 KN.m
Effective depth (d) = h – Cnom – ϕ/2 – ϕlinks
Assuming ϕ16 mm bars will be employed for the main bars, and ϕ8mm bars for the stirrups (links)
d = 450 – 30 – 8 – 8 = 404 mm
k = MEd/(fckbd2) = (62.647 × 106)/(35 × 300 × 4042) = 0.0365
Since k < 0.167, no compression reinforcement required
z = d[0.5+ √(0.25 – 0.882K) ] = z = d[0.5+ √(0.25 – 0.882(0.0365))] = 0.95d
As1 = MEd/(0.87fyk z) = (62.647 × 106)/(0.87 × 500 × 0.95 × 404) = 375.23 mm2
Provide 2H16 mm TOP (ASprov = 402 mm2)
Interaction of shear and torsion
According to clause 6.3.2(2) of Eurocode 2, the effects of torsion and shear for both hollow and solid members may be superimposed, assuming the same value for the strut inclination θ. The limits for θ given in 6.2.3 (2) of EC2 are also fully applicable for the case of combined shear and torsion.
According to clause 6.3.2(4) of EC2, the maximum resistance of a member subjected to torsion and shear is limited by the capacity of the concrete struts. In order not to exceed this resistance the following condition should be satisfied:
TEd /TRd,max + VEd /VRd,max ≤ 1 ———- Equation (6.29 of EC2)
VEd = 157.7 KN
TEd = 4.819 kN.m
TRd,max = design torsional resistance moment
VRd,max = maximum shear resistance of the cross-section
Geometrical Properties for analysis of torsion
Area (A) = 300mm × 450mm = 135000 mm2
Perimeter (U) = 2(300) + 2(450) = 1500mm
Equivalent thickness = tef,i = A/U = 135000/1500 = 90mm
The equivalent thin wall section for the rectangular section is given below;
Ak = the area enclosed by the centre-lines of the connecting walls, including inner hollow areas = (450 – 90) × (300 – 90) = 75600 mm2
Uk = is the perimeter of the area Ak = 2(450 – 90) + 2(300 – 90) = 1140 mm
TRd,max = 2 v αcwfcdAktef,i sinθ cosθ
Assuming θ = 21.8° (cot θ = 2.5)
v = 0.6(1 – fck/250) = 0.6(1 – 35/250) = 0.516
fcd = (αcc × fck)/γc = (1 × 35)/1.5 = 23.33 N/mm2
TRd,max = 2 × 0.516 × 1.0 × 23.33 × 75600 × 90 × cos 21.8° × sin 21.8° × 10-6 = 56.485 KNm
VRd,c = [CRd,c.k.(100ρ1 fck)(1/3) + k1.σcp]bw.d
Where;
CRd,c = 0.18/γc = 0.18/1.5 = 0.12
k = 1 + √(200/d) = 1 + √(200/404) = 1.704 > 2.0, therefore, k = 1.702
ρ1 = As/bd = 402/(300 × 404) = 0.003317 < 0.02; K1 = 0.15
VRd,c = [0.12 × 1.704(100 × 0.003317 × 35 )(1/3)] 300 × 404 = 65469.358 N = 65.469 KN
Since VRd,c (65.469 KN) < VEd (157.7 KN), shear reinforcement is required.
The compression capacity of the compression strut (VRd,max) assuming θ = 21.8° (cot θ = 2.5)
VRd,max = (bw.z.v1.fcd)/(cotθ + tanθ)
V1 = 0.6(1 – fck/250) = 0.6(1 – 35/250) = 0.516
fcd = (αcc ) fck)/γc = (1 × 35)/1.5 = 23.33 N/mm2
Let z = 0.9d
VRd,max = [(300 × 0.9 × 404 × 0.516 × 23.333)/(2.5 + 0.4)] × 10-3 = 452.863 KN
Since VRd,c < VEd < VRd,max
Hence, Asw/S = VEd/(0.87Fykzcot θ) = 157700/(0.87 × 500 × 0.9 × 404 × 2.5 ) = 0.3988
Minimum shear reinforcement;
Asw/S = ρw,min × bw × sinα (α = 90° for vertical links)
ρw,min = (0.08 × √(fck))/fyk = (0.08 × √35)/500 = 0.0009465
Asw/S (min) = 0.0009465 × 300 × 1 = 0.2839
Since 0.2839 < 0.3988, adopt 0.3988
Maximum spacing of shear links = 0.75d = 0.75 × 404 = 303mm
Provide H8mm @ 250mm c/c (Asw/S = 0.402) Ok
Note that this link must be properly closed with adequate anchorage length, because it will assist in resisting torsion.
Design Considerations for Torsion
TEd /TRd,max + VEd /VRd,max ≤ 1
(4.891/56.485) + (157.7/453.863) = 0.434 < 1.0 Hence it is ok
However, note that the actual shear force at the point where torsion is maximum is actually less than the shear force at the support. The relationship above is an error but on the safe side.
The maximum torsion occurs at an angle of 9.5° from the support (see Table above). Therefore, the actual shear force at that section ( VEd) = Shear force at support – wrα
VEd = 157.7 – (133.862 × 3 × (9.5/180) × π) = 91.114 KN
Therefore for educational purposes, this is the shear force that should be used to check the shear-torsion interaction. A little consideration will show that VRd,max is constant all through the section, but VRd,c might vary depending on the longitudinal reinforcement provided at the section.
Rechecking the interaction above;
(4.891/56.485) + (91.114/453.863) = 0.287< 1.0 Hence it is ok
Area of transverse reinforcement to resist torsion
Asw/s = TEd/2Ak fyw,d cotθ
Asw/s = (4.819 × 106) / (2 × 135000 × 0.87 × 500 × 2.5) = 0.0164 < Asw/S (min)
Therefore, links provided for shear will be adequate for resisting torsion.
Area of longitudinal reinforcement to resist torsion
As1 = TEdUk cot θ / 2Ak fyd
As1 = (4.819 × 106 × 1140 × 2.5) / (2 × 75600 × 0.87 × 500) = 208 mm2
According to clause 6.3.2(4), in compressive chords, the longitudinal reinforcement may be reduced in proportion to the available compressive force. In tensile chords the longitudinal reinforcement for torsion should be added to the other reinforcement. The longitudinal reinforcement should generally be distributed over the length of side, but for smaller sections it may be concentrated at the ends of this length.
However, for the avoidance of doubt since no definition was given for what could be regarded as a ‘smaller section’, Provide 1H12mm bar at the middle of the section at both faces. The tensile and longitudinal reinforcement provided at the top and bottom of the section should be able to take care of the rest.
Further information
Nominal shear reinforcement is required in rectangular sections when;
TEd /TRd,c + VEd /VRd,c ≤ 1 ————– Equation (6.31 of EC2)
Where;
TRd,c is the value of the torsion cracking moment:
VRd,c is as defined above.
TRd,c = fctd⋅t⋅2Ak
τ = fctd = fctk /γc = 2.2/1.5 = 1.466 MPa (fctk deducted from Table [3.1 – EC2]).
It results therefore:
TRd,c = fctd⋅t⋅2Ak = 1.466 × 90 × 2 × 75600 × 10-6 = 19.949 kNm
From above calculations;
VRd,c = [0.12 × 1.704(100 × 0.003317 × 35 )(1/3)] 300 × 404 = 65469.358 N = 65.469 KN
(4.819/19.949) + 91.114/65.469) = 1.633 > 1.0 (Shear force at point of maximum torsion was used)
Therefore obviously this showed that shear reinforcement calculations were required.
References
Hassoun M.N., Al-Manaseer A. (2008): Structural Concrete Theory and Design (4th Edition). John Wile and Son Inc., New Jersey
Bar bending schedule is an important structural working document that shows the disposition, bending shape, total length, and quantity of all the reinforcements that have been provided in a structural drawing. It is often provided in a separate sheet (usually A4 paper) from the structural drawing. The bar marks from structural detailing drawing are directly transferred to the bar bending schedule. We normally quantify reinforcements based on their total mass in tonnes or kilograms. For smaller projects, you can quantify based on the number of lengths needed.
Bar bending schedule is prepared for floor slabs to show the quantity, size, and shape of rebars needed during the construction. This document is very important for pre-contract and post-contract operations. The information needed for the preparation of bar bending schedule for floor slab is picked from the reinforcement detailing drawings. One important parameter in the preparation of bar bending schedule is the quantity of steel required (in kilograms or tonnes). This is based on the unit mass and size of the rebars.
Unit mass of rebars
The unit mass of the reinforcements are derived from the density of steel. The density of steel normally used for this purpose is 7850 kg/m3.
For example, let us consider 12mm bar;
The area is given by (πd2)/4 = (π × 122)/4 = 113.097mm2 = 0.0001131m2
Considering a unit length of the bar, we can verify that the volume of a metre length of the bar is 0.0001131m3;
Density = Mass/Volume = 7850 kg/m3 = Mass/0.0001131
Therefore, the unit mass of 12mm bar = 7850 × 0.0001131 = 0.888 kg/m
Therefore for any diameter of bar;
Basic weight = 0.00785 kg/mm2 per metre
Weight per metre = 0.006165 ϕ2 kg
Weight per mm2 at spacing s(mm) = 6.165ϕ2/s kg
Where;
ϕ = diameter of bar in millimetres
The unit weight of different types of reinforcement sizes is given in the Table below;
| Diameter of bar (mm) | Weight per metre (kg) | Length per tonne (m) |
| 6 | 0.222 | 4505 |
| 8 | 0.395 | 2532 |
| 10 | 0.616 | 1623 |
| 12 | 0.888 | 1126 |
| 16 | 1.579 | 633 |
| 20 | 2.466 | 406 |
| 25 | 3.854 | 259 |
| 32 | 6.313 | 158 |
| 40 | 9.864 | 101 |
There are some basic shape codes in the code of practice (BS 8666:2005). But these days, it is common to sketch the bending shape on the BBS document to avoid the confusion and extra effort that comes with extracting the shape from a standard document.
To obtain the length of reinforcement bars in a structural drawing, use the following relation;
Length of bar = Effective Length + Width of Support – Concrete cover (s) – Tolerances
The typical values of tolerances (deductions) are given in the table below;
To illustrate how this is done, consider the general arrangement of the first floor of a building as shown below;
Bar Bending Schedule Calculations
Cutting Length of reinforcement = A + B + C – r – 2d (Table 2.19, Reynolds, Steedman, and Threlfall, 2008)
Where;
r = radius of bend (r = 24 mm for high yield 12 mm bars; and 20 mm for Y10mm bars)
d = diameter of bar
Bar Mark 01:
A = 4000 + 230 – 35 = 4195 mm
B = 150 – 2(25) – 10 = 90 mm (including 10 mm tolerance)
C = 230 + 800 – 35 = 995 mm (from detailing considerations 0.2L)
r = 24 (for 12 mm bars)
L = A + B + C – r – 2d = 4195 + 90 + 995 – 24 – 2(12) = 5235 mm
Bar Mark 02:
L = 2230 mm
Bar Mark 03:
A = 3600 + 230 – 35 = 3795 mm
B = 150 – 2(25) – 12 – 10 = 78 mm (including 10 mm tolerance)
C = 230 + 720 – 35 = 915 mm (from detailing considerations 0.2L)
r = 24 (for 12 mm bars)
L = A + B + C – r – 2d = 3795 + 78 + 915 – 24 – 2(12) = 4740 mm
Bar Mark 04:
A = 1080 + 1200 + 230 – 25 = 2485 mm
B = 150 – 2(25) – 10 = 90 mm (including 10 mm tolerance)
C = 1200 + 230 – 25 = 1405 mm (from detailing considerations 0.2L)
r = 24 (for 12 mm bars)
L = A + B + C – r – 2d = 2485 + 1405 + 90 – 24 – 2(12) = 3932 mm
Bar Mark 05:
A = 6000 + 230 – 35 = 6195 mm
B = 150 – 2(25) – 12 – 10 = 78 mm (including 10 mm tolerance)
C = 1200 + 230 – 35 = 1395 mm (from detailing considerations 0.2L)
r = 20 (for 10 mm bars)
L = A + B + C – r – 2d = 6195 + 1395 + 78 – 20 – 2(10) = 7628 mm
Bar Mark 06:
L = 4630 mm
Bar Mark 07:
L = 3830 mm
Bar Mark 08:
A = 1200 + 230 – 35 – 25 – (15) = 1355 mm (including 15 mm tolerance)
B = 150 – 2(25) – 10 = 90 mm (including 10 mm tolerance)
r = 24 (for 12 mm bars)
L = 2(A) + 2(B) + C + D – 3r – 6d = 2(1355) + 2(90) + 2(125) – 3(24) – 6(12) = 2996 mm
Bar Mark 09:
L = 2030 mm
Bar Mark 10:
L = 1830 mm
The final table for the bar bending schedule can be prepared as shown below. However, it is important to include all details in the schedule to avoid confusion.
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