Second-order effects involve the analysis of a structure based on the deformed geometry. In other words, second-order analysis recognizes the deflection in a structure due to an externally applied load, and determines its effect on the internal forces generated thereof. The magnitude at which the internal forces in a structure increase due to second-order effects depend on the geometry, stiffness, and support conditions of the structure. This is usually employed in the verification of the stability of steel structures against phenomena such as buckling.
The sensitivity of a frame to second order effects may be illustrated simply by considering one ‘bay’ of a multi-storey building in simple construction (i.e. with pinned connections between beams and columns); the bay is restrained laterally by a spring representing the bracing system. First and second order displacements are illustrated below.
For first order effects; kδ1 = H1
For second order effects; kδ2 = H1 +V(δ2 / h) = H2
On rearranging the equation for second order effects, the equilibrium condition can be expressed as;
H2 = H1[1 / (1 – V/kh)]
Hence, it can be observed that if the stiffness of the structure k is large, there will be little amplification of the horizontal force, and first order analysis will be adequate for the structure. On the other hand, if the value of vertical force tends toward a critical value Vcr (= kh) then displacements and forces in the restraint tend toward infinity. The ratio Vcr/V, which may be expressed as a parameter αcr, is thus an indication of the second order amplification of displacements and forces in the bracing system due to second order effects.
The amplifier is given by:
[1/(1 – 1/αcr)
This amplification factor applies to horizontal forces such as wind loads and imperfection loads.
Criteria for considering second order effects
According to BS EN 1993-1-1, 5.2.1(2), the effects of the deformed geometry of the structure (second order effects) need to be considered if the deformation significantly increase the forces in the structure or if the deformations significantly modify structural behaviour. For elastic global analysis, clause 5.2.1 says that the second order effects are significant if the parameter αcr < 10, where αcr is determined by first order analysis and for a braced frame is defined by the approximate expression:
αcr = (HEd/VEd) x (h/δh,Ed)
where: HEd is the design value of the horizontal reaction at the bottom of the storey to the horizontal loads and the equivalent horizontal forces1 VEd is the total design vertical force on the structure on the bottom of the storey δh,Ed is the horizontal displacement at the top of the storey, relative to the bottom of the storey, when the frame is loaded with horizontal loads (e.g. wind) and equivalent horizontal forces which are applied at each floor level h is the storey height.
Methods for determining second order effects
Where second order effects need to be evaluated, BS EN 1993-1-1, 5.2.2 says that they may be allowed for by:
An appropriate second-order analysis, taking into account the influence of the deformation of the structure.
Using appropriate (increased) buckling lengths of members
Amplification of an elastic first order analysis using the initial geometry of the structure.
Second Order Analysis
A range of second order analysis software is available. Use of any software will give results that are to some extent approximate, depending on the solution method employed, the types of second-order effects considered and the modelling assumptions. Generally, second-order software will automatically allow for frame imperfections, so the designer will not need to calculate and apply the equivalent horizontal forces. The effects of deformed geometry (second-order effects) will be allowed for in the analysis. The effect of member imperfections and such things as residual stresses are allowed for if verifying members in accordance with the rules in Section 6 of BS EN 1993-1-1.
Use of increased column buckling length
The use of increased column buckling effective lengths is generally not recommended, simply because of the manual effort involved in calculating the effective length factors. However, if this option is chosen, effective length factors can be determined using a source of non-conflicting complementary information (NCCI), such as BS 5950 Annex E or DD ENV 1993-1-1 Annex E.
Amplification of first order effects
Use of amplified first order effects is subject to the limitation that αcr ≥ 3 (if αcr is less than 3, second order analysis must be used). The amplifier is given by:
[1/(1 – 1/αcr)
Only the effects due to the horizontal forces (including the equivalent horizontal forces) need to be amplified. In a braced frame, where the beam to column connections are pinned and thus do not contribute to lateral stiffness, the only effects to be amplified are the axial forces in the bracing members and the forces in columns that are due to their function as part of the bracing system.
Vibration analysis of floors is not entirely new in the field of structural engineering. Numerous studies and researches have been devoted to human-structure interaction, with emphasis on human perception of vibration in civil engineering structures. Recently, the quest for construction of slender structures with large unobstructed areas (with flexible partitioning) has made the idea of vibration serviceability very important in the design of structures. Once a building is constructed, it is usually challenging to ameliorate vibration issues since it involves modifying the mass and stiffness of the structure.
In the UK, the vibration sensitivity of composite slabs has been traditionally checked by ensuring that the vertical natural frequency of secondary and primary beams is greater than 4Hz. However, a new approach has been recommended in the publication SCI P354 (2009) for checking the vibration serviceability of composite slabs. An example using the approach has been presented in this article.
In P354, two modes of vibration have been recommended for checking acceptability. In mode A, alternate secondary spans may be deflecting up and down (assuming simply supported conditions) with the participation of the slabs (as fixed ended) but not the primary beams. The primary beams are assumed to form nodal lines with zero deflection. In mode B, the primary beams may be deflecting in the same manner, but the secondary beams and slabs which are effectively fixed ended contribute to extra deflection. Therefore for mode B, the deflection is a sum of three contributions. The lower frequency of the two modes is taken as the fundamental frequency and should be at least 3Hz to ensure that walking activities will be outside the frequency zone that will cause resonance.
The design procedures for determining the dynamic performance of composite floors is as follows;
Determine the natural frequency
Determine the modal mass of the floor
Evaluate the response of the floor
Verify the response of the floor against the requirements
Solved Example
The figure below shows a part plan of a composite floor. The slab is to be constructed using profiled metal decking and normal weight, grade 30 concrete. The longitudinal beams are of grade S275 steel with a span of 7.5 m and spaced 3 m apart. Check the acceptance of the floor for vibration. 406 x 178 x 67 UKB as the internal beams. T
Member Geometry
Primary beams 610 x 229 x 140 UKB Secondary beams 406 x 178 x 67 UKB
Beam span = 7.5 m Beam spacing = 3.0 m Total depth of slab hs = 130 mm Depth of profile hp = 60 mm Overall height of profile hd = 72 mm Depth of concrete above profile = 58 mm Profile: SMD TR60+ (1.2 mm gauge) Gauge = 1.2 mm Mesh: A142
Floor loading 130 mm deep concrete slab = 2.21 kN/m2 (manufacturer’s data) Self weight of profile decking = 0.131 kN/m2 Ceiling and services = 1 kN/m2 Finishes = 1.2 kN/m2 10% Imposed = (0.1 x 5 kN/m2)= 0.5 kN/m2 Sub total = 5.041 kN/m2
Primary beams = (140 kg/m x 9.81/7.5) x 10-3 = 0.183 kN/m2 Scondary beams = (67 kg/m x 9.81/3) x 10-3 = 0.22 kN/m2 Total = 5.44 kN/m2 = (5.44 x 103)/9.81 = 554.54 kg/m2
Typical SMD TR60+ profile
Calculation of composite slab properties
Profile neutral axis = 33.7 mm Profile area/unit width = 1633 mm2/m Profile moment of inertia = 119.8 cm4/m Height of re-entrant rib = 60 mm
Concrete area/unit width for 130 mm thick slab= 0.096 m2/m Therefore effective slab thickness = 96 mm
For dynamic properties, take the gross uncracked moment of inertia. Dynamic modulus of elasticity of concrete Ecm = 38 kN/mm2.
Modular ratio α = Es/Ecm = 210/38 = 5.526
We can calculate the elastic neutral axis of the composite slab from the table below.
Section
Area A (cm2)
Neutral axis y (cm)
Area x y (cm3)
Concrete
960/α = 173.724
9.6/2 = 4.8
816.5
Profile
16.33
13 – 3.37 = 9.63
157.257
Total
∑A = 190.054
∑Ay = 973.757
Elastic neutral axis of the composite slab NA = ∑Ay/∑A = 973.757/190.054 = 5.123 cm below the top of the slab
The moment of inertia of the composite slab can be calculated from the table below;
Section
Distance from NA (cm)
Area x Distance2 (cm4)
Iy,local (cm4)
Concrete
0.323
18.124
3139
Profile
4.507
331.712
119.8
349.836
3258.8
Second moment of area of composite slab = 349.836 + 3258.8 = 3608.636 cm4 (36.0863 x 10-6 m4)
Moment of inertia of the composite secondary beam
406 x 178 x 67 UKB Span = 7.5m; Weight = 67.1 kg/m Depth = 409.4 mm; Area = 85.5 cm2 Iy = 24300 cm4; Effective breadth beff = 1875 mm
Composite section properties
Section
Width (cm)
Depth (cm)
y (cm)
A (cm2)
Ay (cm3)
Ay2 (cm4)
Ilocal
Slab
187.5
7.0
3.50
1312.5/5.526 = 237.513
831.297
2909.532
969.847
Beam
33.47
85.5
2861.68
95780.596
24300
∑A = 323.013
∑Ay = 3692.977
∑Ay2 = 98690.128
∑Ilocal = 25269.847
Position of elastic neutral axis = ∑Ay/∑A = 3692.977/323.013 = 11.432 cm from the top of the slab
Gross moment of area = 98690.128 + 25269.847 – (11.4322 x 323.013) = 81745.204 cm4 (8.1745 x 10-4 m4)
Moment of inertia of the composite primary beam
610 x 229 x 140 UKB Span = 6.0m; Weight = 139.9 kg/m Depth = 617.2 mm; Area = 178 cm2 Iy = 112000 cm4; Effective breadth beff = 1500 mm
Composite section properties
Section
Width (cm)
Depth (cm)
y (cm)
A (cm2)
Ay (cm3)
Ay2 (cm4)
Ilocal
Slab
150
9.6
4.8
1440/5.526 = 260.586
1250.81
6003.9
2001.3
Beam
43.86
178
7807.08
342418.53
112000
∑A = 438.586
∑Ay = 9057.89
∑Ay2 = 348422.43
∑Ilocal = 114001.3
Position of elastic neutral axis = ∑Ay/∑A = 9057.89/438.586 = 20.652 cm from the top of the slab
Gross moment of area = 114001.3 + 348422.43 – (20.6522 x 438.586) = 275364.5625 cm4 (27.5364 x 10-4 m4)
Fundamental natural frequency
Mode A
(i) Slab – Fixed Ended w = 5.041 kN/m2 x 3 m = 15.123 kN/m δ1 = wL3/384EI = (15.123 x 33)/(384 x 210 x 106 x 36.0863 x 10-6) = 1.403 x 10-4 m = 0.14 mm
(ii) Secondary beam – Simply Supported w = (5.041 kN/m2 + 0.22 kN/m2) x 3 m = 15.783 kN/m δ2 = 5wL4/384EI = (5 x 15.123 x 7.54)/(384 x 210 x 106 x 8.1745 x 10-4) = 3.629 x 10-3 m = 3.629 mm
Total deflection for mode A δ = 0.14 + 3.629 = 3.760 mm
Natural frequency for mode A fA = 18/√δ = 18/√3.760 = 9.282 Hz
Mode B
(i) Slab – As above
(ii) Secondary beam (assume fixed ended) w = (5.041 kN/m2 + 0.22 kN/m2) x 3 m = 15.783 kN/m δ2 = wL4/384EI = (15.123 x 7.54)/(384 x 210 x 106 x 8.1745 x 10-4) = 7.2589 x 10-4 m = 0.725 mm
(iii) Primary beam (assume simply supported) Reactive force from secondary beam = (15.783 x 7.5)/2 = 59.186 kN w = (5.041 kN/m2 + 0.183 kN/m2) x 3.75 m = 19.59 kN/m
δ3 = 5wL4/384EI + PL3/48EI = [(5 x 19.59 x 64)/(384 x 210 x 106 x 27.5364 x 10-4)] + [(59.186 x 63)/(48 x 210 x 106 x 27.5364 x 10-4)] = (5.716 x 10-4) + (4.605 x 10-4) = 1.032 x 10-3 m = 1.032 mm
Therefore total deflection = 0.14 + 0.725 + 1.032 = 1.897 mm
Natural frequency for mode B fB = 18/√δ = 18/√1.879 = 13.13Hz
Therefore the fundamental frequency of the floor is found in Mode A f0 = 9.292 Hz > 3.0 Hz (Okay)
Calculation of Modal Mass
The effective floor length can be calculated from;
Leff = 1.09(1.10)ny – 1(EIb/mbf02)1/4 Leff ≤ nyLy
Where; ny is the number of bays in the secondary beam direction (ny ≤ 4) EIb is the flexural rigidity of the composite secondary beam (expressed in Nm2 when m is expressed in kg/m2) b is the spacing of the secondary beam = 3.0 m f0 is the fundamental natural frequency Ly is the span of the secondary beams
Leff = 1.09(1.10)1 – 1(210 x 109 x 8.1745 x 10-4/554.54 x 3 x 9.2922)1/4 = 7.049 m < (1 x 7.5 = 7.5 m)
The effective width of the slab can be calculated from;
S = η(1.15)nx – 1(EIs/mf02)1/4
η = 0.71 for f0 > 6.0 nx = 2.0
S = 0.75(1.15)2 – 1 x (210 x 109 x 36.0863 x 10-6/554.54 x 9.2922)1/4 = 3.05 m < (2 x 6 = 12)
Modal mass M = mLeffS = 554.54 x 7.04 x 3.05 = 11907 kg
Calculation of Floor Response
As f0< 10 Hz, the response is to be assessed according to the low frequency floor recommendations.
aw,rms = μeμr (0.1Q/2√2Mζ) x Wρ
Q = 746 N based on an average weight of 76 kg Take critical damping ratio ζ as 4.68% Since f0 > 8Hz, take the weighting factor W = 8/f0 = 8/9.292 = 0.860 Conservatively assume that μe = μr = 1.0
Assuming a walking frequency fp of 2Hz in a maximum corridor length Lpof 15 m; v = 1.67fp2 – 4.83fp + 4.5 = 1.52 m/s
ρ = 1 – e(-2πζLpfp/v) = 1.0
aw,rms = 1.0 x 1.0 x [(0.1 x 746)/(2√2 x 11907 x 0.0468)] x 0.86 x 1.0 = 0.0407 m/s2
Response Factor
R = aw,rms/0.005 = 0.0407/0.005 = 8.14
For an office building, the floor is is acceptable for vibration since the response factor is greater than 8.0.
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Structural design of water retaining structures
Serviceability considerations in water retaining structures
Materials and construction methodology of water retaining structures
Interactive question and answer sessions
Detailing of water retaining structures
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Bearings are ancillary bridge components that facilitate the transfer of traffic actions, permanent actions, and other environmental actions from the bridge deck down to the substructure, and ultimately, to the ground. To fulfill this function effectively, bearings must be able to accommodate all anticipated service movements (rotations and translations), while also restraining extraordinary movements induced by extreme load cases. The movements allowed by an adjacent expansion joint must be compatible with the displacement restrictions imposed by a bearing. Therefore, bearings and expansion joints must be designed interdependently and in conjunction with the anticipated behavior of the overall structure.
Types of bridge bearing
The type of bearing to be used in a bridge can be determined by a lot of factors. Some prominent issues that can be considered are the strength and the stiffness of the bearing, cost, ease of installation, maintenance cost, etc. The common types of bearings used in contemporary bridges are;
Reinforced elastomeric bearings
fabric pad sliding bearings
steel pin bearings
rocker bearings,
roller bearings
steel pin bearings
pot bearings
disc bearings
spherical bearings, and
seismic isolation bearings.
Reinforced Elastomeric Bearings
A steel-reinforced elastomeric bearing consists of discrete steel shims vulcanized between adjacent discrete layers of elastomer. This vulcanization process occurs under conditions of high temperature and pressure. The constituent elastomer is either natural rubber or synthetic rubber (neoprene). Reinforced elastomeric bearings are broadly classified into four types:
plain elastomeric pads
fiberglass reinforced elastomeric pads
steel reinforced elastomeric pads, and
cotton duck reinforced elastomeric pads.
Steel reinforced elastomeric pads
Of these four types, steel reinforced elastomeric pads are used most extensively for bridge construction applications. They are commonly used with prestressed concrete girder bridges and maybe used with other bridge types.
Fabric Pad Bearings
Fabric Pad Bearing
Cotton duck or fabric pads are preformed elastomeric pads reinforced with very closely spaced layers of cotton or polyester fabric. The close spacing of the reinforcing fibers, while allowing fabric pads to support large compressive loads, imposes stringent limits upon their shear displacement and rotational capacities. Unlike a steel reinforced elastomeric bearing having substantial shear flexibility, the fabric pad alone cannot accommodate translational movement. Fabric pads can accommodate very small amounts of rotational movement; substantially less than can be accommodated by more flexible steel reinforced elastomeric bearings.
Steel pin bearings are generally used to support high loads with moderate to high levels of rotation about a single predetermined axis. This situation generally occurs with long straight steel plate girder superstructures. Rotational capacity is afforded by rotation of a smoothly machined steel pin against upper and lower smoothly machined steel bearing blocks. Steel keeper rings are typically designed and detailed to provide uplift resistance.
Pin Bearing
Rocker/Roller Bearings
Steel rocker bearings have been used extensively in the past to allow both rotation and longitudinal movement while supporting moderately high loads. Because of their seismic vulnerability and the more extensive use of steel reinforced elastomeric bearings, rocker bearings are now rarely specified for new bridges.
Rocker type of bearing for bridge deck
Steel roller bearings have also been used extensively in the past. Roller bearings permit both rotational and longitudinal movement. Pintles are often used to effect transverse force transfer by connecting the roller bearing to the superstructure above and to the bearing plate below.
Roller bearing
Pot Bearings
A pot bearing is composed of a plain elastomeric disc that is confined in a vertically oriented steel cylinder, or pot.
Elastomeric pot bearing under a steel girder
Vertical loads are transmitted through a steel piston that sits atop the elastomeric disc within the pot. The pot walls confine the elastomeric disc, enabling it to sustain much higher compressive loads than could be sustained by more conventional unconfined elastomeric material. Rotational demands are accommodated by the ability of the elastomeric disc to deform under compressive load and induced rotation. The rotational capacity of pot bearings is generally limited by the clearances between elements of the pot, piston, sliding surface, guides, and restraints.
Disc Bearings
Disc bearing
A disc bearing relies upon the compressive flexibility of an annular shaped polyether urethane disc to provide moderate levels of rotational movement capacity while supporting high loads. A steel shear resisting pin in the center provides resistance against lateral force. A flat PTFE-stainless steel sliding interface can be incorporated into a disc bearing to additionally provide translational movement capability, either guided or nonguided.
Spherical Bearings
Spherical bearing for a bridge
A spherical bearing, sometimes referred to as a curved sliding bearing, relies upon the low-friction characteristics of a curved PTFE-stainless steel sliding interface to provide a high level of rotational flexibility in multiple directions while supporting high loads. Unlike pot bearings and disc bearings, spherical bearing rotational capacities are not limited by strains, dimensions, and clearances of deformable elements. Spherical bearings are capable of sustaining very large rotations provided that adequate clearances are provided to avoid hard contact between steel components.
Spherical bearings are classified into three according to their displacement directions. The three types of spherical bearings are;
Fixed type – Provides only rotation capacity from any direction. Guided type (Uni-directional sliding) – Provides rotation plus movement in one direction Free sliding (multi-directional sliding) – Provides rotation plus movement in all directions
Seismic Isolation Bearings
Seismic isolation bearing
Seismic isolation bearings mitigate the potential for seismic damage by utilizing two related phenomena: dynamic isolation and energy dissipation. Dynamic isolation allows the superstructure to essentially float, to some degree, while substructure elements below move with the ground during an earthquake. The ability of some bearing materials and elements to deform in certain predictable ways allows them to dissipate seismic energy that might otherwise damage critical structural elements.
Source: Chen W. and Duan L. (2014): Bridge Engineering Handbook – Substructure Design (2nd Edition). CRC Press Taylor and Francis Group. International Standard Book Number-13: 978-1-4398-5230-9
Composite slab with profiled metal decking provides economical solutions for floors of steel framed building systems. This is because they are easier to install, lighter in weight, and faster to execute when compared with precast, prestressed, and solid slab system for steel-framed buildings. The composite action of this floor system is achieved by welding steel studs to the top flange of the steel beams and embedding the studs in the concrete during concrete pouring.
Composite construction reduces frame loadings and results in a cheaper foundation system. Cold-formed thin-walled profiled steel decking sheets with embossments on top flanges and webs are widely used as the profiles. The use of this profiled metal decking eliminates the need for mat reinforcement in the slab and acts as the permanent shuttering for the concrete. Props are therefore not usually required during the process of concreting. This support scheme is usually suitable for spans that are less than 4m (spacing of the supporting beams). The supporting beams themselves can, however, span up to 12m.
Structural engineers usually rely on load/span tables produced by metal deck manufacturers in order to determine the thickness of slab and mesh reinforcement required for a given floor arrangement, fire rating, method of construction, etc. The table below shows an example of a typical load/span table available from one supplier of metal decking.
To download the full SMD technical data sheet for different types of profiled metal decking, click below.
Once the composite slab has been designed, the design of the primary and secondary composite beams (i.e. steel beams plus slab) can begin. This is normally carried out in accordance with the recommendations in Part 3: Section 3.1 of BS 5950. In Europe, composite sections are designed according to the requirements of Eurocode 4 (EN 1994 – 1- 1).
The steps in the design of profile metal decking for composite floors are;
Determine the effective breadth of the concrete slab.
Calculate the moment capacity of the section.
Evaluate the shear capacity of the section.
Design the shear connectors.
Assess the longitudinal shear capacity of the section.
Check deflection.
Design Example
The figure below shows a part plan of a composite floor. The slab is to be constructed using profiled metal decking and normal weight, grade 30 concrete. The longitudinal beams are of grade S275 steel with a span of 7.5 m and spaced 3 m apart. Design the composite slab and verify the suitability of 406 x 178 x 67 UKB as the internal beams. The required fire-resistance is 1 hour.
Imposed load = 4 kN/m2 Partition load = 1 kN/m2 Weight of finishes = 1.2 kN/m2 Weight of ceiling and services = 1 kN/m2 Total applied load = 7.2 kN/m2 (to be used for slab design i.e. selection from manufacturer’s span-load table)
SLAB DESIGN From the span-load table above, the configuration below will be satisfactory for the unpropped slab.
Beam span = 7.5 m Beam spacing = 3.0 m Total depth of slab hs = 130 mm Depth of profile hp = 60 mm Overall height of profile hd = 72 mm Depth of concrete above profile = 58 mm Profile: SMD TR60+ (1.2 mm gauge) Gauge = 1.2 mm Mesh: A142
Typical SMD TR60+ profile
From the manufacturer’s technical data sheet; Volume of concrete = 0.096 m3/m2 Weight of concrete (wet) = 2.26 kN/m2 Weight of concrete (dry) = 2.21 kN/m2 Weight of profile = 0.131 kN/m2 Height to neutral axis = 33 mm
Shear connector Connector diameter d = 19 mm Overall welded height of hsc = 95 mm Ultimate tensile strength fu = 450 N/mm2
Permanent Actions Self weight of sheeting = 0.131 kN/m2 x 3 m = 0.393 kN/m Allowance for beam self-weight = 1.0 kN/m Allowance for mesh = 0.05 kN/m2 x 3m = 0.15 kN/m Total gk = 0.393 + 1 + 0.15 = 1.543 kN/m
Variable Actions Self-weight of fresh concrete = 2.26 kN/m2 x 3m = 6.78 kN/m (note that fresh concrete is treated as variable action in the construction stage) Construction load = 0.75 kN/m2 x 3m = 2.25 kN/m Total qk = 6.78 + 2.25 = 9.03 kN/m
At ultimate limit state = 1.35gk + 1.5qk = 1.35(1.543) + 1.5(9.03) = 15.63 kN/m
Design moment MEd = ql2/8 = (15.63 x 7.52)/8 = 109.898 kNm Design shear force VEd = ql/2 = (15.63 x 7.5)/2 = 58.61 kN
Actions at the composite stage
Permanent Actions Self weight of sheeting = 0.131 kN/m2 x 3 m = 0.393 kN/m Allowance for beam self-weight = 1.0 kN/m Allowance for mesh = 0.05 kN/m2 x 3m = 0.15 kN/m Self-weight of dry concrete = 2.21 kN/m2 x 3m = 6.63 kN/m Weight of finishes = 1.2 kN/m2 Weight of ceiling and services = 1 kN/m2 Total gk = 0.393 + 1 + 0.15 + 6.63 + 1.2 + 1 = 10.373 kN/m
Variable Actions Imposed load on floor = 4 kN/m2 x 3m = 12 kN/m Movable partition allowance = 1 kN/m2 x 3m = 3 kN/m Total qk = 12 + 3 = 15 kN/m
At ultimate limit state = 1.35gk + 1.5qk = 1.35(10.373) + 1.5(15) = 36.5 kN/m
Design moment MEd = ql2/8 = (36.5 x 7.52)/8 = 256.6 kNm Design shear force VEd = ql/2 = (36.5 x 7.5)/2 = 136.875 kN
An advanced UK beam S275 is to be used for this design. Fy = 275 N/mm2 γm0 = 1.0 (Clause 6.1(1) NA 2.15 BS EN 1993-1- 1:2005)
From steel tables, the properties of 406 x 178 x 67 UKB are;
Depth h = 409.4 mm Width b = 178.8 mm Web thickness tw = 8.8mm Flange thickness tf = 14.3 mm Root radius r = 10.2 mm Depth between fillets d = 360.4 mm Second moment of area y axis Iy = 24300 cm4 Elastic modulus Wel,y = 1190 cm3 Plastic modulus Wpl,y = 1350 cm3 Area of section A = 85.5 cm2 Height of web hw = h – 2tf = 380.8 mm Es (Modulus of elasticity) = 210000 N/mm2 (Clause 3.2.6(1))
Outstand flange Flange under uniform compression c = (b – tw – 2r)/2 = [178.8 – 8.8 – 2(10.2)]/2 = 74.8 mm
c/tf = 74.8/14.3 = 5.23
The limiting value for class 1 is c/tf ≤ 9ε = 9 × 0.92 5.23 < 8.28 Therefore, outstand flange in compression is class 1
Internal Compression Part (Web under pure bending) c = d = 360.4 mm c/tw = 360.4/8.8 = 40.954
The limiting value for class 1 is c/tw ≤ 72ε = 72 × 0.92 = 66.24 40.954 < 66.24 Therefore, the web is plastic. Therefore, the entire section is class 1 plastic.
Member Resistance Verification – Construction Stage
Moment Resistance For the structure under consideration, the maximum bending moment occurs where the shear force is zero. Therefore, the bending moment does not need to be reduced for the presence of shear force (clause 6.2.8(2))
43.27 < 66 Therefore shear buckling need not be considered.
Design Resistance of Shear Connectors
Shear connector in a solid slab The design resistance of a single headed shear connector in a solid concrete slab automatically welded in accordance with BS EN 14555 should be determined as the smaller of;
PRd = (0.8 x fu x π x 0.25d2)/γv (Clause 6.6.3.1(1) Equ(6.18) or PRd = [0.29 x α x d2 x √(fck x Ecm)]/γv
PRd = (0.8 x 450 x π x 0.25 x 192)/1.25 = 81.7 kN PRd = [0.29 x 1.0 x 192 x √(25 x 31 x 103)]/1.25 = 73.7 kN
Shear connectors in profiled steel sheeting
For profiled sheeting with ribs running transverse to the supporting beams, PRd,solid should be multiplied by the following reduction factor;
kt = (0.7/√nr) x (b0/hp) x (hsc/hp – 1)
b0 = width of a trapezoidal rib at mid height of the profile = (133 + 175)/2 = 154 mm hsc = 95 mm hp = 60 mm nr = 1.0 (for one shear connector per rib)
kt = (0.7/√1.0) x (154/60) x (95/60 – 1) = 1.0
Therefore PRd = ktPRd,solid = 1.0 x 73.7 = 73.7 kN The design resistance per rib = nrPRd = 1 x 73.7 = 73.7 kN
Degree of shear connection
For composite beams in buildings, the headed shear connectors may be considered as ductile when the minimum degree of shear connection given in clause 6.6.1.2 is achieved.
For headed shear connectors with; hsc≥ 4d and 16mm ≤ d ≤ 25 mm
The degree of shear connection may be determined from;
η = Nc/Nc,f
Where; Nc is the reduced value of the compressive force in the concrete flange (i.e. force transferred by the shear connectors) Nc,f is the compressive force in the concrete flange at full shear connection (i.e. the minimum of the axial resistance of the concrete and the axial resistance of the steel)
For steel sections with equal flanges and Le < 25 m;
η ≥ 1 – (355/fy) x (0.75 – 0.03Le) where ≥ 0.4 Le = distance between points of zero moment = 7.5 m η ≥ 1 – (355/275) x (0.75 – 0.03 x 7.5) = 0.322, therefore η = 0.4
Degree of shear connection present
To determine the degree of shear connection present in the beam, the axial resistances of the steel and concrete are required (Npl,a and Nc,f respectively)
Stress block for calculating the resistances of concrete sections
Determine the effective width of the concrete flange
At the mid-span, the effective width of the concrete falnge is;
beff = b0 + ∑bei
Effective flange width of composite section
For nr = 1.0, b0 = 0 mm bei = Le/8 but ot greater than bi
Le = 7.5 m (point of zero moment) bi = distance from the outside shear connector to a point between adjacent webs. Therefore;
b1 = b2 = 1.5 m be1 = be2 = Le/8 = 7.5/8 = 0.9375 m
The effective flange width is therefore beff = b0 + be1 + be2 = 0 + 0.9375 + 0.9375 = 1.875 m = 1875 mm
Compressive resistance of the concrete flange
The design strength of the concrete fcd = 25/1.5 = 16.7 N/mm2
The TR60+ profile has a 12 mm deep re-entrant above the stiffener making the overall profile depth hd = 12 mm + 60 mm = 72 mm
The compressive resistance of the concrete flange is therefore; Nc,f = 0.85fcdbeffhc = 0.85 x 16.7 x 1875 x 58 x 10-3 = 1543.7 kN
Tensile Resistance of the steel member Npl,a = fy.A = 275 x 85.5 x 102 x 10-3 = 2351.25 kN
The compressive force in the concrete at full shear connection is the lesser. Therefore Nc = 1543.7 kN
Resistance of the shear connectors
n is the number of shear connectors present to the point of maximum bending moment. In this example, there are 7.5(2 x 0.333) = 12 ribs available for positioning shear connectors per half span.
Nc = n x PRd = 12 x 73.7 = 884.4 kN
The degree of shear connection present therefore is;
η = Nc/Nc,f = 884.4/1543.7 = 0.572 > 0.40 (Okay)
Design Resistance of the Cross-section at the composite stage
Bending Resistance
According to clause 6.2.1.2, the plastic rigid theory may be used for one connector per trough. With partial shear connection, the axial force in the concrete flange Nc is less than Npl,a (884.4 kN < 2351.25 kN). Therefore, the plastic neutral axis lies within the steel section. Assuming that the plastic neutral axis lies a distance xpl below the top of the flange of the section, where;
xpl = (Npl,a – Nc)/2fyb = (2351.25 – 884.4)/(2 x 275 x 178.8) = 0.0149 m = 14.916 mm > tf (14.3 mm)
Therefore the plastic neutral axis lies below the top flange.
Stress block of composite cross-section
yc = Nc/[0.85fckbeff /γc] ≤ hc yc = (884.4 x 1000)/[0.85 x 25 x 1875/1.5] = 33.28 mm
MRd = Nc(hc + da – yc/2) + 2btffy(da – tf/2) + tw(ya – tf)(fy)(2da – ya – tf) MRd = (884.4 x 103) x (130 + 204.7 – 33.28/2) + 2 x 178.8 x 14.3 x 275 x (204.7 – 14.3/2) + 8.8(14.92 – 14.3) x 275 x (2 x 204.7 – 14.92 – 14.3) = 559669744.2 Nmm = 559.669 kNm
MEd = 256.6 kNm
MEd/MRd = 256.6/559.669 = 0.458 < 1.0 (Okay)
Shear Resistance at the Composite Stage
The shear resistance is therefore; Vc,Rd = Vpl,Rd = [3854 × (275/ √3)/1.0] × 10-3 = 612 kN VEd = 136.875 kN VEd/Vc,Rd = 136.875/612 = 0.223 < 1.0 Ok
Longitudinal shear resistance of the slab
Crack control mesh as transverse reinforcement
Neglecting the contribution of the steel, we ned to verify that;
Asffsd/Sf > vEdhf/cotθ
Where; vEd is the design longitudinal shear stress in the concrete slab fsd is the design yield strength of the reinforcing mesh = 0.87fyk = 0.87 x 500 = 434.8 N/mm2 hf = depth of the concrete above the profiled sheeting = 70 mm θ angle of failure (try 26.5o) Asf/Sf = At (for the plane of failure shown as section a-a) At is the cross-sectional area of transverse reinforcement mm2/m)
The verification equation therefore becomes;
Atfyd > vEdhf/cotθ
The required area of tensile reinforcement At must satisfy the following;
At > vEdhf/fydcotθ
The longitudinal shear stress is given by;
vEd = ∆Fd/hf∆x
Where; ∆x is the critical length under consideration, which is usually taken as the distance between the maximum bending moment and the support = L/2 = 7.5/2 = 3.75m ∆Fd = Nc/2 = 884.4/2 = 442.2 kN
vEd = ∆Fd/hf∆x = (442.2 x 103)/(70 x 3750) = 1.68 N/mm2
vEdhf/fydcotθ = (1.68 x 70)/(434.8 x cot 26.5o) = 0.134 mm2/mm
For the arrangement, the area of tensile reinforcement required is 134 mm2/m Therefore A142 mesh provided is adequate (Asprov = 142 mm2/m)
Crushing of the concrete flange
It is important to verify that vEd < vfcdsinθfcosθf v = 0.6(1 – fck/250) = 0.6 x (1 – 25/250) = 0.54 vfcdsinθfcosθf = 0.54 x 16.67 x sin(26.5) x cos(26.5) = 3.59 N/mm2
(vEd) 1.68 N/mm2 < 3.59 N/mm2 (Okay)
Serviceability limit state
Modular ratios For short term loading, the secant modulus of elasticity should be used. Ecm = 31 kN/mm2. This corresponds to a modular ratio of;
n0 = Es/Ecm = 210/31 = 6.77 (clause 5.4.2.2)
For long term loading; nL = n0(1 + ψLϕt) Where ψL is the creep multiplier taken as 1.1 for permanent loads and ϕt is the creep coefficient taken as 3.0. nL = 6.77 x (1 + 1.1 x 3) = 29.11
When calculating deflection due to variable action, the modular ratio is taken as;
For n0 = 6.77 Ic = 24300 x 104 + [1875(130 – 60)3/(12 x 6.77)] + [85.5 x 102 x 1875(130 – 60) x (409.4 + 130 + 60)2]/4[85.5 x 102 x 6.77 + 1875(130 – 60)] = 78385 x 104 mm4
For nL = 29.11 Ic = 24300 x 104 + [1875(130 – 60)3/(12 x 29.11)] + [85.5 x 102 x 1875(130 – 60) x (409.4 + 130 + 60)2]/4[85.5 x 102 x 29.11 + 1875(130 – 60)] = 50999 x 104 mm4
For n = 14.22 Ic = 24300 x 104 + [1875(130 – 60)3/(12 x 14.22)] + [85.5 x 102 x 1875(130 – 60) x (409.4 + 130 + 60)2]/4[85.5 x 102 x 14.22 + 1875(130 – 60)] = 64543 x 104 mm4
Deflection due to actions on the steel at the construction stage; Actions = self weight of fresh concrete + mesh + sheeting + steel section w1 = 5gl4/384EI = (5 × 8.323 × 75004)/(384 × 210000 × 24300 × 104) = 6.719 mm
Deflection due to permanent action on the steel at the composite stage; Actions = weight of finishes + ceiling and services w2 = 5gl4/384EI = (5 × 6.6 × 75004)/(384 × 210000 × 50999 x 104) = 2.54 mm
Deflection due to variable action on the steel at the composite stage; Actions = Imposed load + partition allowance w3 = 5ql4/384EI = (5 × 15 × 75004)/(384 × 210000 × 64543 x 104) = 4.559 mm
Total deflection = w1 + w2 + w3 = 6.719 + 2.54 + 4.559 = 13.818 mm
Allowable deflection = L/360 = 7500/360 = 20.833 mm. 13.818 < 20.833 Therefore deflection is okay.
Pedestrian wind comfort can be improved when horizontally incoming airflow passes through trees in the urban areas. This is according to recent research carried out in the Department of Environmental Atmospheric Sciences, Pukyong National University, Busan, Republic of Korea, and published in Elsevier – Sustainable Cities and Societies. In the study, the authors applied computational fluid dynamics incorporating tree drag parameters to evaluate how trees improved the wind comfort of pedestrians.
The way wind is perceived at the ground level depends on a lot of factors such as wind direction, wind speed, obstacles, and many other parameters. This experience affects pedestrians’ comfort and safety, and can impact the financial returns or economic viability of an area. According to Lawson-based wind comfort criterion, wind speed exceeding 10 m/s can be uncomfortable for pedestrians at the ground level, while wind speed above 15 m/s is outrightly dangerous. Other wind comfort criteria exist such as that proposed by Davenport.
Trees are known to function as porous obstacles to airflow, and they eventually affect wind speed and direction. The presence of high rise buildings and urban densification has been observed to reduce airflow in city centers and may lead to increased urban heat and poor dispersal of pollutants. However, the effect of wind on tall buildings can amplify wind pressure in the surroundings due to issues like vortex shedding, reverse flow, channeling effects, etc. This can lead to discomforting wind effects on pedestrians. Therefore, by applying computational fluid dynamics model (CFD) with tree drag parameterization scheme, the researchers were able to evaluate the effect of trees on pedestrian wind comfort in Pukyong National University campus. The CFD model used in the study was verified with field observations.
The CFD model used in the study was based on Reynolds-averaged Navier–Stokes (RANS) equations and assumes a three-dimensional, non-rotating, non-hydrostatic, incompressible airflow system. Turbulence was parameterized using the renormalization group (RNG) k-ε turbulence closure scheme. Tree drag terms were introduced into the momentum, turbulent kinetic energy (TKE), and TKE dissipation rates equations to account for the loss of airflow pressure due to winds.
The target area for the research was the Pukyong National University (PKNU) campus (See Fig. 1a) which is located in the downtown area and is surrounded by commercial and residential areas. PKNU boasts relatively high vegetation density for student recreational spaces and ecologically friendly landscaping within the campus. Regions A and B (Fig. 1b) contains a dense forest of trees taller than 10–15 m. Several tree species are planted along the PKNU boundary (region C in Fig. 1b).
Fig. 1. Aerial photograph of (a) the target area, including Pukyong National University campus (from www.daum.net) and (b) high-density vegetation regions (A, B, and C in Fig. 1a) (from www.pknu.ac.kr). The red dot in (a) indicates the location of automated weather station (AWS) 942. (Kang et al, 2020)
The authors adopted the wind comfort criteria proposed by Isyumov and Davenport (1975), which distinguish four sensory levels (good, tolerable, unpleasant, and dangerous) for four categories of human activity or activity location (Table 1). These sensory levels are determined by the Beaufort wind force scale (BWS), represented by wind speeds at 10 m above the ground level. To evaluate wind comfort at the pedestrian level (z =1.75 m), they used the BWS values converted into pedestrian height.
Table 1: Sensory levels in terms of suitability for outdoor activities, represented by the Beaufort wind force scale (BWS) (Isyumov & Davenport, 1975).
From the study, poor wind comfort for outdoor activities (BWSs ≥ 4) was observed in the areas without trees, mainly around the edges of buildings, in the windward regions of buildings, in the spaces between buildings, and in wide, unobstructed areas. This was attributed to venturi effects between the spaces in buildings. In the case of where trees were present, the BWS values declined by one to three levels, improving the overall level of wind comfort within the PKNU campus.
The highest ABWS and TBWS values (≥ 4) were observed near the southeast perimeter of the PKNU campus, where a 10-lane road is located. By contrast, the lowest ABWS and TBWS values (≤ 3) were observed in the southwest and northwest of the campus. Where trees were present, the overall wind speeds inside the campus were reduced due to drag.
The authors concluded that tree arrangement can reduce wind speeds in the lee of the trees by more than half and proposed that trees should be planted at 90° to the dominant wind direction. The presence of trees decrease wind speeds. However, because wind speeds can increase in surrounding areas without trees, the effects of trees on strong winds in such areas should be assessed.
Reference Kang G., Kim J., Choi W. (2020): Computational fluid dynamics simulation of tree effects on pedestrian wind comfort in an urban area. Sustainable Cities and Societies 56 (2020)102086. https://doi.org/10.1016/j.scs.2020.102086
Disclaimer: Contents of this research article have been shown on www.structville.com because it is an open access article under creative commons licence (http://creativecommons.org/licenses/BY/4.0). All other rights belong to the authors and Elsevier.
Pile foundations are slender structures used to transmit superstructure load to firmer sub-soil stratum beneath the natural ground surface. They can also be used for other purposes such as resisting heavy lateral forces, compaction of soils (compaction piles), avoiding excessive settlement, etc. Due to their importance in civil engineering structures, piles are usually subjected to tests such as pile load tests and pile integrity tests before they are loaded. These two tests are completely different and are sometimes confused for one another, even though they do not serve the same purpose. This article aims to highlight the difference between these two tests.
Generically, pile load test can be described as a reliable method of pile foundation design which involves loading constructed piles on-site to determine their load-carrying capacity. A pile load test involves applying increments of static loads to a test pile and measuring the settlement. The load is usually jacked onto the pile using either a large deadweight or a beam connected to two uplift anchor piles to supply reaction for the jack. Generally, an installed pile, weights, deflection gauge, hydraulic jack, and load indicator are required for a pile load test.
Loading of test piles is usually applied in increments of 25% of the total test load which should be 200% of the proposed design load. After the load test, the load-settlement curve is plotted and the failure load determined. Eurocode 7 permits three different methods for the design of pile foundations which are;
It is important that the validity of static load test be checked using calculations.
Pile Integrity Test
Pile Integrity Test setup
Pile integrity test (PIT) is a non-destructive method of testing of piles that is used for qualitative evaluation of the physical dimensions, continuity, and consistency of materials in a bored (cast in-situ) pile. This test is very important for quality control and quality assurance of piles at great depth.
The three most common methods of carrying out pile integrity tests are;
Low-strain pile integrity test
Crosshole sonic logging
Thermal integrity test
In the low-strain impact integrity testing, the head of the pile shaft is subjected to impact using a tool like a simple hammer and the response is determined using a high precision transducer. The transducer can either be an accelerator, or a velocity sensor. Low-strain pile integrity tests can provide information such as embedment length, changes in cross-section (such as bulging), discontinuity (such as voids), and consistency of pile materials (such as soil inclusion and segregation). However, it cannot provide information such as bearing capacity and cannot be applied to pile caps.
Differences and similarities between pile load test and pile integrity test at a glance
Pile Load Test
Pile Integrity Test
Used for determining bearing capacity of piles
Used for determining physical properties of constructed piles
Can evaluate pile settlement under load
Cannot evaluate pile settlement
Expensive to set up
Cost effective
Takes time to complete
Very quick test
Cannot provide the embedment length of the pile
Provides embedment length of the pile
Cannot give information on the quality of the piling job
Provides information on the quality of the piling job
Therefore, pile load test and pile integrity tests should be carried out as soon as piling jobs are concluded on site before the next stage of the construction commences.
Pile caps are rigid plate structures that are used to transfer superstructure load from columns to a group of piles. They are usually subjected to bending and shear forces, and shear considerations usually govern the thickness design of pile caps. The three main approaches that are used in the analysis of pile caps are;
Truss Analogy
Bending analogy, and
Finite element analysis
While truss analogy and bending theory can be easily carried using quick manual calculations, finite element analysis usually require the use of computer models. In this article, we are going to explore the potentials of Staad Foundation Advanced Software in the analysis and design of pile caps.
A quick design of pile caps can be done on Staad Foundation Advanced using the Foundation Toolkit option. This approach does not require importing models and can be used for quick stand-alone design when the column load and geotechnical parameters of the soil are available. To use this option, launch the ‘Staad Foundation Advanced‘, click on ‘New Project‘, and select ‘Foundation Toolkit‘ labelled as shown below.
Step 1: Launch the foundation toolkit
Step 2: Create Pile Cap Job
When the Foundation Toolkit opens, go to ‘Main Navigator‘, and from ‘Project Info‘ drawdown list, select ‘Create Pile Cap Job‘ as shown below.
Step 3: Select design code, units, and pile layout
When the ‘Pile Cap Job’ is launched, select the desired code of practice, unit, and click ‘Next’. The pile layout can be left as predefined.
Step 4: Define the load
On clicking ‘Next’, the dialog box for load comes up. Make sure that the unit is consistent as desired, and for this exercise, I am applying a factored column load of 3500 kN. If there are other forces such as moment and shear coming from the column, you can define them also.
Step 5: Define Load Combination
Since we are dealing with an already factored load, select ‘User Defined‘ from the drawdown list of load combination. If you have defined dead load, live load, wind load etc in Step 4, you can select the desired code of practice for the combination of the loads. Since I defined my factored load as dead load, I assigned a factor of 1.0 to dead load at ULS and SLS (actually I am not interested in SLS in this design). Then click ‘Next‘
Step 6: Define the design parameters
In this case, the column dimensions are taken as 450 mm x 450 mm, and the thickness of the pile cap was taken as 1300 mm. Other design parameters specified are as shown below.
The diameter of the pile was selected as 750 mm, with a spacing of 2250 mm. The safe working load of the pile was taken as 900 kN. You can also input the uplift and lateral load capacity of the pile. The edge distance is taken as (diameter of pile/2 + 150 mm) – where 150 mm is the overhang from the edge of the pile to the edge of the pile cap. Then click on ‘Calculate‘.
This brings the possible pile arrangements based on the safe working load and the superstructure load. For this tutorial, the arrangement below was adopted. The simple idea behind it is simply (Column load/pile safe working load). Note that for practical purposes, serviceability limit state load should be used when selecting the number of piles. Then click ‘Ok‘ and ‘Next‘.
Step 8: Finish the model
Step 9: Carry out the Design
Clicking ‘Finish‘ returns you to the ‘Main Navigator‘ page, where you can click on ‘Design‘ to carry out the design of the pile cap.
Step 10: View the output
The output page is where you can view the geometry drawing, details and schedule drawing, calculation sheet, and graphs.
The design approved the 1300 mm thick pile cap provided, and provided Y16@100 mm c/c reinforcement. You can go ahead and print the calculation sheet which you can download below.
Researchers from the Department of Construction Sciences, Lund University, Sweden have presented isogeometric analysis as an alternative to finite element analysis for modelling of soil plasticity. In a study published in the year 2017 in Geomaterials Journal, the researchers were able to show that isogeometric analysis showed good agreement with finite element method for drained soils in two- and three-dimensions. The research, therefore, suggested that isogeometric analysis is a good alternative to conventional finite element analysis for simulations of soil behavior.
Isogeometric analysis is a numerical method that uses non-uniform rational B-splines (NURBS) as basis functions instead of the Lagrangian polynomials often used in the finite element method. These functions have a higher-order of continuity and therefore makes it possible to represent complex geometries exactly. The basic idea behind isogeometric analysis is to use splines (NURBS) as basis functions for computational analysis by applying them directly. This allows the same basis function to be used for discretzation and for analysis.
Representation of quadratic B-spline basis function with knot vector Ξ = {0, 0, 0, 1, 2, 3, 4, 5, 5, 5} (Spetz et al, 2017)
Representation of quadratic Lagrangian basis function (Spetz et al, 2017)
Since being introduced by Thomas J.R. Hughes at the University of Austin, Texas in the year 2005, isogeometric analysis has found numerous applications in engineering such as analysis of thin plates and shells, soil-structure interaction, fluid-structure interaction, flow through porous media, etc. However, finite element analysis has been used extensively for constitutive modelling of soils for the design of foundations, retaining walls, slope stability problems etc.
For the purpose of the research, Drucker-Prager criterion and the theory of plasticity was used by the researchers to evaluate the influence of the NURBS-basis functions on the plastic strains for granular materials against the conventional finite element analysis approach.
To compare the results of the findings, a 2D model of a strip footing on sandy silt was analyzed. The problem was solved for plane strain conditions using quadratic NURBS based IGA and conventional FEA with 5 different meshes. In order to compare the two methods, the element meshes were constructed using quadratic isoparametric elements for both IGA and conventional FEA.
Load/displacement response at point A and B for the finest mesh (4800 elements) (Spetz et al, 2017)
The first point, A, denotes the center of the footing and the second point, B denotes the edge of the footing. It was observed that the displacements in the model are in good agreement at center of the footing but a minor difference was observed between the displacements from the isogeometric and conventional finite element analysis at the edge of the footing.
References Spetz, A. , Tudisco, E. , Denzer, R. and Dahlblom, O. (2017): Isogeometric Analysis of Soil Plasticity. Geomaterials, 7, 96-116. doi: 10.4236/gm.2017.73008.
Disclaimer: This research article has been featured on www.structville.com because it is an open access article that permits unrestricted use and distribution provided the original source of the article has been cited. See the Creative Commons Attribution International License (CC BY 4.0). http://creativecommons.org/licenses/by/4.0/
Tall masts are used in a variety of applications such as telecommunication, radio/television broadcasting, supporting lighting fixtures on the highway, raising flags, etc. Due to their height above the ground, these structures are usually subjected to wind action, and this makes the knowledge of their behavior under the effect of wind very important.
The dynamic behavior of telecommunication masts under the effect of wind is studied under ‘flow-induced vibration of structures’. This term is used to denote the phenomena associated with the response of a structure immersed in or conveying fluid flow. It covers those cases in which an interaction develops between fluid dynamic forces and the inertia, damping, or elastic forces in the structure. The study of these phenomena draws on three disciplines which are, structural mechanics, mechanical vibration, and fluid dynamics (Ahmad, 2009).
This article investigates the natural frequency of vibration of X-braced and V-braced (Chevron bracing) telecommunication masts. To advance the study, the wind-induced vibration can be investigated using Staad Pro software (finite element analysis) in order to determine the critical wind speed that will potentially lead to resonance in the structure under the effect of wind. Resonance occurs when one of the natural frequencies of the structure coincides with the frequency of the vortices shedding around the mast due to the effect of wind.
Methodology
To carry out this study, two models of telecommunication masts were considered;
MODEL 1
Type: X-Bracing Height: 30 m Width at the base: 3m x 3m Width at the top: 1m x 1m Section of the legs: UA 100 x 100 x 8 Section of the horizontal bracings and diagonals: UA 50 x 50 x 6 Solidity ratio: 20.6%
The first four fundamental natural frequencies of the structure were obtained as follows;
Mode
Frequency (Hz)
Period (Sec)
MPF-X (%)
MPF-Y (%)
MPF-Z (%)
1
2.637
0.379
32.303
0.000
24.153
2
2.637
0.379
24.153
0.000
32.303
3
11.568
0.086
13.073
0.000
9.760
4
11.568
0.086
9.760
0.000
13.073
Table 1: Natural frequency of x-braced telecom masts
Type: V-Bracing (Chevron Bracing) Height: 30 m Width at the base: 3m x 3m Width at the top: 1m x 1m Section of the legs: UA 100 x 100 x 8 Section of the horizontal bracings and diagonals: UA 50 x 50 x 6 Solidity ratio: 19.51 %
The first four fundamental natural frequency of the structure were obtained as follows;
Mode
Frequency (Hz)
Period (Sec)
MPF-X (%)
MPF-Y (%)
MPF-Z (%)
1
2.775
0.360
26.153
0.000
30.132
2
2.775
0.360
30.132
0.000
26.153
3
11.390
0.088
9.021
0.000
14.375
4
11.390
0.088
14.375
0.000
9.021
Table 2: Natural frequency of v-braced telecom masts
N/B: MPF – Mass participation factor
2nd mode of vibration of v-braced (chevron) mast
3rd mode of vibration of v-braced (chevron) mast
Time history analysis can be used to evaluate the response of the mast to periodic vortex shedding excitations. It is a step-by-step analysis that involves solving the dynamic equilibrium equations given by;
Kx(t) + Cẋ(t) + Ṁẍ(t) = r(t) (1)
Where,
K is the stiffness matrix C is the proportional damping matrix Ṁ is the diagonal mass matrix x is the relative displacements with respect to the ground ẋ is the relative velocities with respect to the ground ẍ is the relative accelerations with respect to the ground r is the vector of the applied load. t is time
The load r(t) applied in a given Time-History analysis may be an arbitrary function of space and time. It can be written as a finite sum of spatial load vectors multiplied by time function. The frequency of vortex shedding (fv) is related to the Strouhal number (S) and the relationship is given by;
S = fvW/U —— (2)
Where W is the average outside width of the mast (m). The value of ݂fv is calculated for different values of velocities. The velocity values chosen should result in Reynolds number (Re) values that still fall in ranges of constant Strouhal number S = 0.21 for low speeds and 300 < ܴ݁Re < 2 x 106 or ܵS = 0.27 for high speeds and ܴ݁Re > 3.5 x 106.
From equation (2) we can deduce that;
fv = SU/W ——- (2a)
For resonance to occur, note that fv = fs (where fs is the natural frequency of the structure)
Note that we can calculate the range of velocities based on Reynolds’s number. The Reynolds numbers that can be used for this purpose are 300, 2 x 106 and 3.5 x 106.
Re = UW/vair (3)
Where vair is the kinematic viscosity of flowing air (ν = 1.51 x 10-5 m2/s at standard atmospheric pressure and 20oC).
However, it should be noted that a latticed tower has a complex aerodynamic shape to the wind such that consistent vortex shedding to cause complete oscillation of the structure over a prolonged period is almost impossible. This is because the vortex shed from the different trussed members can rarely have a uniform frequency, thereby reducing the possibility of vortex-induced vibration. However, for towers with circular or tubular cross-section over the whole or part of its body, vortex-induced vibration is a possibility.
