A corbel is a very short structural cantilever member projecting from a wall or a column for the purpose of carrying loads. In reinforced concrete structures, corbels are cast monolithically with the walls or columns supporting them. They are found mainly in bridges, industrial buildings, and commercial buildings with precast construction.
Corbels in precast construction
If a structure is subjected to in-plane forces, then the stress distribution will consist of normal stresses in the two planes (σx and σy) and an accompanying shear stress τxy. These stresses will lead to two principal stresses σ1 and σ2. If for instance σ1 is tensile and σ2 is compressive, then the external load on the structure can be idealized as being resisted by a combination of tensile and compressive stresses.
In reinforced concrete structures, the compressive stresses can be resisted by the concrete, while the tensile stresses can be resisted by the steel reinforcement. Generally, these are idealized by a series of concrete struts and steel ties, and forms the underlying principle behind the strut-and-tie method of design. Structures that can be designed using this method are pile caps, deep beams, corbels, half-joints, etc.
3D model of a corbel
Under certain conditions, Eurocode 2 permits the use of strut and tie method for the analysis and design of corbels. According to clause 6.5.1 of EN 1992-1-1:2004, strut-and-tie method can be used where non-linear stress distribution exists (for example at supports, near concentrated loads or plain stress).
Strut-and-tie models utilize the lower-bound theorem of plasticity which can be summarized as follows: for a structure under a given system of external loads, if a stress distribution throughout the structure can be found such that;
(1) all conditions of equilibrium are satisfied and (2) the yield condition is not violated anywhere,
Then the structure is safe under the given system of external loads.
When using the strut−and-tie method of design, because the given structure is replaced by a pin-jointed truss, different types of nodes occur where members meet. These nodal zones need to be carefully designed and detailed.
(a) CCC Node: If three compressive forces meet at a node, it is called a CCC node. According to Eurocode 2 equation (6.60), the compression stress in each strut is restricted to a maximum value of 1.0(1 – fck/250)fcd.
(b) CCT node: If two compressive forces and a tie force anchored in the node through bond meet, it is called a CCT node. According to Eurocode 2 equation (6.61), the compression stress in each strut is restricted to a maximum value of 0.85(1 – fck/250) fcd.
(c) CTT node: If two tensile forces at a compressive force meet at a node it is called a CTT node. According to Eurocode 2 equation (6.62), the compression stress in each strut is restricted to a maximum value of 0.75(1 – fck/250)fcd.
Strut−tie model for a corbel with different types of nodes.
The following conditions are to be applied in the design of corbels;
(1) Corbels with 0.4hc < ac < hc may be designed using a simple strut and tie model. (2) For deeper corbels (ac < 0.4hc), other adequate strut and tie models may be considered. (3) Corbels for which ac > hc may be designed as cantilever beams (4) Unless special provision is made to limit horizontal forces on the support, or other justification is given, the corbel should be designed for the vertical force Fv, and a horizontal force Hc > 0.2Fv acting at the bearing area. (5) The overall depth (hc) of the corbel should be determined from considerations of shear. (6) The local effects due to the assumed strut and tie system should be considered in the overall design of the supporting member. (7) The detailing requirements shall be met.
Design Example of a thick short corbel (ac < hc/2)
Design the corbel shown below. The bearing area of the support is 200 x 300 mm. The width of the corbel is 150 mm.
Design strength of concrete fcd = αccfck/γc = (0.85 x 30)/1.5 = 17 N/mm2 Design strength of steel fyd = fyk/γs = 500/1.15 = 435 N/mm2
Compressive strength of nodes;
CCC nodes: σ1Rd,max = 1.0 x (1 – fck/250)fcd = 1.0 x (1 – 30/250) x 17 = 14.96 N/mm2
CCT nodes: σ2Rd,max = 0.85 x (1 – fck/250)fcd = 0.85 x (1 – 30/250) x 17 = 12.716 N/mm2
CTT nodes: σ3Rd,max = 0.75 x (1 – fck/250)fcd = 0.75 x (1 – 30/250) x 17 = 11.22 N/mm2
Actions FEd = 600 kN
Load eccentricity with respect to the column side: e = 125 mm The beam vertical strut width is evaluated by setting the compressive stress equal to σ1Rd,max:
x1 = FEd/σ1Rd,maxb = (600 x 103)/(14.96 x 350) = 114.6 mm Node 1 is therefore located at x1/2 = 114.6/2 = 57.3 mm
Let the cover from the top of the cantilever to the reinforcements be 30 mm. Assuming 16 mm bars;
The effective depth d = 450 – 35 – (16/2) = 407 mm
The distance y1 of the node 1 from the lower border is evaluated setting the internal drive arm z equal to 0.8⋅d (z = 0.8 x 407 = 325.6 mm):
y1 = 0.2d = 0.2 x 407 = 81.5 mm
Rotational equilibrium: FEda = Fcz a = ac + x1/2 = 125 + 57.3 = 182.3 mm 600 x 103 x 182.3 = Fc x 325.6 Fc = Ft = 335933.66 N = 335.93 kN
Node 1 verification; σ = Fc / (b∙2y1) = (335.93 x 103)/(350 x 2 x 81.5) = 5.89 N/mm2 < σ1Rd,max (14.96 N/mm2)
Main Top Reinforcement Design As1 = Ft/fyd = (335.93 x 103)/435 = 772 mm2 Provide 5H16 (As,prov = 1005 mm2)
Shear reinforcement design The beam proposed in EC2 is indeterminate, then it is not possible to evaluate the stresses for each single bar by equilibrium equations only, but we need to know the stiffness of the two elementary beams shown below in order to make the partition of the diagonal stress [Fdiag = Fc/cosθ = FEd/sinθ] between them.
Based on the trend of main compressive stresses resulting from linear elastic analysis at finite elements, some researchers from Stuttgart have determined the two rates in which Fdiag is divided, and they have provided the following expression of stress in the secondary reinforcement.
Strut-and-tie model resolution in two elementary beams and partition of the diagonal stress Fdiag.
Fwd = [(2z/a – 1)/(3 + FEd/Fc)] x Fc = [(2(325.6/182.3) – 1)/(3 + 600/335.93)] x 335.93 = 180.525 kN
Asw = Fwd/fyd = (180.525 x 103)/435 = 415 mm2 > k1As,prov = 0.25 x 1005 = 251.25 mm2
Trying 2 legs of H10 mm, Provide 4 number of H10 mm stirrups (Asw,prov = 628 mm2)
Node 2 verification, below the load plate The node 2 is a tied-compressed node, where the main reinforcement is anchored; the compressive stress below the load plate is; σ = FEd /(200 x 300) = (600 x 103)/(200 x 300) = 10 N/mm2 < σ2Rd,max (12.716 N/mm2) Okay
Detailing Requirements
(1) The reinforcement, corresponding to the ties considered in the design, should be fully anchored beyond the node under the bearing plate by using U-hoops or anchorage devices, unless a length lb,net is available between the node and the front of the corbel, lb,net should be measured from the point where the compression stresses change their direction.
(2) In corbels with hc > 300 mm, when the area of the primary horizontal tie As is such that (where Ac denotes the sectional area of the concrete in the corbel at the column), then closed stirrups, having a total area not less than |0.4|As, should be distributed over the effective depth d in order to cater for splitting stresses in the concrete strut. They can be placed either horizontally or inclined.
Typical detailing of a corbel when the diameter of the main bars is 20mm or greater
Typical detailing of a corbel when the diameter of the main bars is 16mm or smaller
P-delta is a geometric non-linear effect that occurs in structures that are subjected to compressive loads and lateral displacement. Under the action of compressive loads and lateral displacement, tall slender structures will experience additional stresses and deformations due to the change of position of the structure. First-order structural analysis will normally consider small displacements and will compute the equilibrium of the structure and internal stresses based on the undeformed geometry. However, second-order analysis considers the deformed geometry of the structure and may require an iterative approach for the computation of equilibrium.
During the simultaneous action of vertical and horizontal loads, the structure deflects due to the action of the horizontal load. As the structure deflects, the position of the vertical load P shifts by a distance ∆ such that the vertical load instead of acting axially along the column now induces a moment reaction at the base P∙∆. The interaction of the compressive force (P) and lateral displacement (∆) to produce additional secondary effects in a structure is handled under P-delta analysis.
Solved Example
Let us quickly investigate the effect of P-delta on the cantilever column loaded as shown below.
Let; P = 150 kN H = 20 kN
1st iteration First-order moment at the base of the cantilever M = H x L = 20 kN x 7.5m = 150 kNm So, the lateral displacement (Δ1) at the column tip = ML2/3EI = (150 x 7.52)/(3 x 210 x 106 x 4.09 x 10-4) = 0.0327 m
Also, the vertical load P acting on displaced Δ1 column tip generates additional moment M1 at the base. M1 = P x Δ1 = 150 x 0.0327 = 4.905 KNm
The total moment at the base MT1 = M + M1 = 150 + 4.905 = 154.905 kNm
The modified horizontal displacement “Δ2” undergone for the first modified moment MT1 Δ2 = (MT1L2)/ (3EI) =(154.905 x 7.52)/(3 x 210 x 106 x 4.09 x 10-4) = 0.0338 m
2nd iteration The vertical load P acting on the newly displaced column tip (Δ2 – Δ1) results in additional moment M2 at the base;
M2 = P x (Δ2 – Δ1) = 150 x (0.0338 – 0.0327) = 0.165 KNm Now, the total moment at the base, MT2 = (M + M1 + M2) = (MT1 + M2) = 154.905 + 0.165 = 155.07 KNm
The second modified horizontal displacement Δ3 against the second modified Moment Mt2 Δ3 = (MT2L2)/3EI = (155.07 x 7.52)/(3 x 210 x 106 x 4.09 x 10-4) = 0.0338 m
For all practical purposes, the analysis above will be taken to have converged.
There are two types of P-Delta effects which are;
P-“large” delta (P-∆) – a structure effect
P-“small” delta (P-δ) – a member effect
While large delta is associated with the global displacements and effects on the structure, small delta takes into account the local effects and displacements of the members. The sensitivity of a structure to P-Delta effect is related to the magnitude of the axial load P, stiffness/slenderness of the structure as a whole, and slenderness of individual elements. Second-order effects are prominent in structures under high compressive load with low stiffness or high slenderness. The closer the compressive load is to the critical buckling load, the more sensitive the structure is to P-delta effects.
P-delta analysis combines two approaches to reach a solution which are;
Large-displacement theory, and
Stress stiffening
In large-displacement theory, the resulting forces and moments take full account of the effects due to the deformed shape of both the structure and its members. The large-displacement theory identifies the fact that there is a clear distinction between the deformed and undeformed shapes of the structure. In small displacement theory, the strains and the displacements are small, while large-displacement theory assumes small strains in the structure but large displacements.
Stress stiffening is concerned with the effect of stress state on a structure’s stiffness. For instance, tensile loads straighten the geometry of an element thereby stiffening it. On the other hand, compressive loads accentuate deformation thereby reducing the stiffness of the element. The effect of stress stiffening is accounted for by generating and then using an additional stiffness matrix (stress stiffness matrix) for the structure. The stress stiffness matrix is added to the regular stiffness matrix in order to give the total stiffness.
Second-order effects have been incorporated in many codes of practice for the design of concrete, timber, and steel structures. This has usually been included in the design equations and verifications to ensure that the behaviour of the structure stays within the prescribed limits. However, with the availability of advanced computational power of software, engineers are now encouraged to carry out full second-order analysis for design purposes.
P-Delta Analysis on Staad Pro
The procedure adopted by Staad Pro for P-delta analysis are as follows;
First, the primary deflections are calculated based on the provided external loading.
Primary deflections are then combined with the originally applied loading to create the secondary loadings. The load vector is then revised to include the secondary effects.
A new stiffness analysis is carried out based on the revised load vector to generate new deflections.
Element/Member forces and support reactions are calculated based on the new deflections.
For proper analysis of P-delta effect in Staad, it is important that the vertical and horizontal load cases be combined under one single load case. This can be achieved using the ‘REPEAT LOAD CASE’, which allows the user to combine previously defined primary load cases to create a new primary load case. A REPEAT LOAD is treated as a new primary load which enables the P-Delta analysis to reflect the correct secondary effects. It is important to note that ‘LOAD COMBINATIONS’ command will algebraically combine the results such as displacements, member forces, reactions, and stresses of previously defined primary loadings evaluated independently. Therefore, it is not suitable for P-Delta analysis.
Furthermore, one can perform the P-Delta Analysis in Staad directly by considering the Geometric Stiffness Matrix [KG]. In this approach, the stress stiffening effect due to the axial stress is used directly to modify the actual Stiffness Matrix [K]. In the view of this approach, the compressive force (depending on its magnitude) reduces the lateral load-carrying capacity of the structure. This ultimately modifies the geometric stiffness of the member and is referred to as ‘stress softening effect’. In Staad, the change in the GEOMETRIC STIFFNESS from [K] to [K + KG] can be achieved by using the KG option as shown below.
So, by this approach the P-Delta stiffness equation is directly linearized by the [K + KG] matrix and the solution can be obtained directly and exactly, without iteration.
According to Staad Pro technical manual, there are two options in carrying out P-Delta analysis on Staad;
(1) When the CONVERGE command is not specified: The member end forces are evaluated by iterating “n” times. The default value of “n” is 1 (one).
(2) When the CONVERGE command is included: The member end forces are evaluated by performing a convergence check on the joint displacements. In each step, the displacements are compared with those of the previous iteration in order to check whether convergence is attained. In case “m” is specified, the analysis will stop after that iteration even if convergence has not been achieved. If convergence is achieved in less than “m” iterations, the analysis is terminated. (DO NOT ENTER “n” when CONVERGE is provided).
(3) To set convergence displacement tolerance, enter SET DISPLACEMENT f command. Default is maximum span of the structure divided by 120.
Solved Example
Let us consider the portal frame loaded as shown shown below.
Action Effect
1st order linear analysis
P-delta (1 iteration)
P-delta (2 iterations)
P-delta (5 iterations)
P-delta (10 iterations)
P-delta (15 iterations)
Displacement (mm)
126.204
143.634
146.602
147.235
147.239
147.239
Maximum bending Moment (kNm)
406.704
425.321
427.837
428.322
428.325
428.325
VB (kN)
128.343
128.343
128.343
128.343
128.343
128.343
HB (kN)
58.101
58.101
58.101
58.101
58.101
58.101
As can be seen from the table above, 5 to 10 iterations is sufficient for second-order analysis of portal frames for practical purposes. While internal forces and displacements varied with second-order analysis, the equilibrium of the structure was maintained.
Many problems of practical importance can be related to the solution of plates resting on an elastic foundation. Reinforced concrete pavements of highways and airport runways, raft foundation slabs of buildings, bases of water tanks and culverts etc., are well-known direct applications. Just like beams on elastic foundation, it is also based on the assumption that the foundation’s reaction q(x, y) can be described by the following relationship;
q(x, y) = kw
Where k is a constant termed themodulus of subgrade reaction, which has the unit (kN/m2/m), and q(x, y) is the resisting pressure of the foundation, and w is the deflection of the plate.
Rigid pavement can be idealised as plate on elastic foundation
When the plate is supported by a continuous elastic foundation, the external load acting in the lateral direction consists of the surface load p(x, y) and of the reaction of the elastic foundation q(x, y). Thus, the differential equation of the plate becomes the following:
In this differential equation, the reactive force, q(x, y) exerted by the elastic foundation is also unknown, because it depends on the deflection, w(x, y) of the plate.
D∇2 ∇2w + kw = p
This equation can be solved using the classical methods developed by Navier and Levy. Note that D is the flexural rigidity of the plate, and it is given by;
D = Eh3/[12(1 – μ2)]
The different methods that can be employed in the analysis of plates on elastic foundation are;
Classical methods
Finite difference methods
Finite element method
Cylindrical bending of plates on elastic foundation
Let us consider the cylindrical bending of a thin plate on an elastic foundation, and rigidly supported at the edges as given in Timoshenko.
Cylindrical plate on elastic foundation
Cutting out an elemental strip, we may consider it as a beam on an elastic foundation. Assuming that the reaction of the foundation at any point in time is proportional to the deflection w at that point, we can obtain the equation given below;
D(∂4w)/∂x4 = q – kw
Introducing the notation; β =L/2 ∜(k/4D)
The general solution to the equation above can be written as;
Where the four constants of integration must be determined from the end conditions of the strip. With the case under consideration, we can assume that the deflection is symmetrical with respect to the middle strip. Taking the coordinate axis as shown in the figure above, we can conclude that C2 = C3 = 0. The constants C1 and C4 are found from the conditions that the deflection and bending moment of the strip are zero at the end (x = L/2). Hence;
w(x=l/2) = 0
(d2w)/dx2 )(x = l/2) = 0
Substituting expressions (a) for w and observing that C2 = C3 = 0, we obtain;
The bending moment at any cross-section of the strip is obtained from the equation;
M = -D(d2w/dx2)
Substituting expression (D) for w, we find for the middle strip;
M(x=0) = (qL2/8)∙ φ2(β) Where φ2(β) = 2/β2 [1 – (sinhβ – sinβ)/(cosh2β + cos 2β)]
To simplify the calculation of deflection and stresses, numerical values of functions of φ, φ1, and φ2 are presented in the Table below;
Plates on Elastic Foundation Using Finite Element Analysis
Software like Staad Pro can be used in the analysis of plates on elastic foundation. The general approach to solving such problems is to sub-divide the slab into several plate elements. Each node of the meshed slab will then have an influence area or a contributory area, which is to say that soil within the area surrounding that node acts as a spring. The influence area is then multiplied by the subgrade modulus to arrive at the spring constant. Subgrade modulus has units of force per length3. So, the spring will have units of force/length.
Typical model of a plate on grade supported on soil springs
The influence area is calculated automatically in Staad Pro using the ‘Foundation’ type of support. The two options available for doing this are of doing this are;
ELASTIC MAT OPTION, and
PLATE MAT OPTION
The elastic mat method calculates the influence area of the various nodes using the Delaunay triangle method. The distinguishing aspect of this method is that it uses the joint-list that accompanies the ELASTIC MAT command to form a closed surface. The area within this closed surface is then determined and the share of this area for each node in the list is then calculated. Without a properly closed surface, the area calculated for the region may be indeterminate and the spring constant values may be erroneous.
If the foundation slab is modeled using plate elements, the influence area can be calculated using the principles used in determining the tributary area of the nodes from the finite element modeling standpoint. In other words, the rules used by the program in converting a uniform pressure load on an element into fixed end actions at the nodes are used in calculating the influence area of the node, which is then multiplied by the subgrade modulus to obtain the spring constant.
Solved Example
A 150 mm thick rectangular slab of dimensions 6m x 5m is resting on a soil of modulus of subgrade reaction ks = 30000 kN/m2/m. Evaluate the response of the slab when subjected to a full pressure load of 35 kN/m2 all over the surface and a concentrated load of 200 kN at the centre.
Solution
The meshing of the plate has been carried out as follows; In the longer direction: 24 divisions In the shorter span: 20 divisions Size of each finite element = 250 mm x 250 mm
The modelling of the foundation is shown below;
The loading of the plate is shown below.
Analysis Result
(a) Settlement
Deformed profile under concentrated load
Maximum settlement under concentrated load = 1.912 mm
Deformed profile under uniformly distributed pressure load
Maximum settlement under uniform pressure load = 1.167 mm
(b) Base Pressure
Base pressure under concentrated load
Base pressure under UDL
(c) Bending Moment
Transverse bending moment under concentrated load
Transverse bending moment under UDL
Longitudinal bending moment under concentrated load
The modulus of subgrade reaction ks (also called the coefficient of subgrade reaction of soil) is the ratio of the pressure against a flat surface on soil and the settlement at that point. Mathematically, this is expressed as;
ks= q/δ ——- (1)
Where; ks = Coefficient of subgrade reaction expressed in force/length2/length q = pressure on the surface at the given point δ = settlement at the same point
Against many popular opinions, the modulus of subgrade reaction is not an exclusive property of soil but depends mainly on the loaded area (size of the footing or mat). Other factors that affect the modulus of subgrade reaction are the shape of footing, depth of foundation, type of soil, and type of foundation. Different approaches have been suggested for the evaluation of the modulus of subgrade reaction, but the most practical approach is to carry out an in-situ plate load test on the soil.
Plate bearing test is an in-situ load-bearing test that is used to evaluate the ultimate bearing capacity and likely settlement of soil under a given load. The test can be carried out in accordance with BS 1377 Part 9: 1990. It basically consists of loading a steel plate of known diameter and recording the settlements corresponding to each load increment.
The test load is gradually increased till the plate starts to settle at a rapid rate. The total value of the load on the plate divided by the area of the steel plate gives the value of the ultimate bearing capacity of the soil. A factor of safety is applied to give the safe bearing capacity of the soil.
Fig 1: Typical plate load test set-up
By implication, the coefficient of subgrade reaction is the unit pressure required to produce a unit settlement in soil. In saturated clay soils, the settlement under load will take time due to consolidation, so the coefficient of subgrade reaction should be determined on the basis of the final settlement.
On granular soil deposits, the settlement should take place immediately after the application of load. Therefore, the modulus of subgrade reaction is premised on two simplified assumptions which are;
The value of ks is independent on the magnitude of the pressure
The value of ks has the same value for every point on the surface of the footing
These two assumptions are however not strictly correct.
Terzaghi in 1955 presented empirical relationships for determining the coefficient of subgrade reactions (ksf) for full-scale foundations, based on results from plate load tests. This is based on a rigid 1 ft x 1 ft (0.305 m x 0.305m) slab placed on a soil medium.
(1) For a square footing on cohesionless soil (B x B);
ksf = ks[(B + 0.305)/2B]2 ——- (2)
(2) For a rectangular footing on cohesionless soil (B x L)
ksf,r = [ksf(1 + 0.5B/L)]/1.5 ——- (3)
(3) For a long strip footing of width B, the coefficient of subgrade reaction is approximately 0.67ksf
(4) For clay soils, it has been observed that the value of ks varies with the length of the footing. Therefore, for clays;
ksc = ks[(L + 0.152)/1.5L)] ——- (4)
Where;
ks = plate-load test value of modulus of subgrade reaction kN/m2/m, using square plate (1 × 1) ft or circular plate with diameter = 0.305 m ksc = corrected plate-load test value of modulus of subgrade reaction kN/m2/m, using square plate (1 × 1) ft or circular plate with diameter = 0.305 m for clays ksf = desired value of modulus of subgrade reaction for full-sized square footings B × B, kN/m2/m ksf,r = desired value of modulus of subgrade reaction for rectangular full-sized footings B × L, kN/m2/m B = footing width, or least dimension of rectangular or strip. L = Length of footing
Some formulas suggested by different authors for evaluation of the modulus of subgrade reaction are given in the Table below;
Author
Year
Suggested formula
Terzaghi
1955
ksf = ks[(B + B’)/2B]2
Vesic
1961
ks= [0.65Es/B(1 – μ2)] x (EsB4/EI)1/12
Meyerhof and Baike
1965
ks= Es/[B(1 – μ2)]
Selvadurai
1985
ks= 0.65/B x [Es/(1 – μ2)]
Bowles
1998
ks = Es/[B'(1 – μ2) x mIsIf]
ks = the coefficient of subgrade reaction. B’ = side dimension of square base used in the plate load test. B = width of footing. ks = the value of subgrade reaction for 0.3 × 0.3 (1 ft wide) bearing plate. ksf= value of modulus of subgrade reaction for the full-size foundation. Es = modulus of elasticity. μ = Poisson’s ratio. EI = flexural rigidity of footing, m = takes 1, 2 and 4 for edges, sides and center of footing, respectively. Is and If= influence factors depend on the shape and depth of footing.
One of the most popular relationships between allowable bearing capacity and modulus of subgrade reaction is given in equation (5) according to Bowles (1996);
ks = 40.(FS).(qa) ——- (5)
Where qa is the allowable bearing capacity of the soil, and FS is the factor of safety that was used in converting the ultimate pressure (qult) to allowable pressure (qa). It is important to note that in equation (5), the author assumed a 25 mm settlement value for the soil.
Ping-Sien Lin, Li-Wen Yang, and C. Hsein Juang (1998) made a series of plate-load tests to investigate the load settlement characteristics of a gravelly cobble deposit and estimated the value of modulus of subgrade reaction ks as follows:
Floors in buildings have a primary function of carrying loads and supporting the activities of the occupants. In addition to carrying loads, floors in buildings also provide the needed rigid diaphragm action for transmitting horizontal loads to the stabilising vertical components. Furthermore, floors also support additional superimposed loads such as ceilings, building services, and finishes such as screeds and tiles.
Composite slab supporting building services
The types of floor systems that used in steel-framed buildings are;
Short-span composite beams and composite slabs with metal decking.
Slimdek®.
Cellular composite beams with composite slabs and steel decking.
Slimflor® beams with precast concrete units.
Long-span composite beams and composite slabs with metal decking.
Composite beams with precast concrete units.
Non-composite beams with precast concrete units.
Short span composite beam and composite slab with metal decking
In this floor system, shear connectors are welded through the metal decking to the top flange of downstand beams to enable it act compositely with an in-situ composite slab. For short span floor systems, the secondary beams are typically spaced between 3 – 4m and are supported by the primary beams. The primary and secondary beams act compositely with the composite slab, but the edge beams are usually non-composite. At 3-4m spacing, secondary beams will span about 6 – 7.5m when positioned orthogonally to the slab, while the primary beams will span about 6-9m (positioned parallel to the slab).
Short span composite decking
The floor slab consists of composite profiled metal decking with a typical depth of about 130 mm thick with in-situ concrete topping. The profiles may be re-entrant decking or trapezoidal. Re-entrant decking uses more concrete than trapezoidal decking, but has increased fire resistance for a given slab depth. Trapezoidal decking generally spans further than re-entrant decking, but the shear stud resistance is less with trapezoidal decking than with re-entrant decking. The profiles are usually between 0.5 to 1.2mm thick.
Re-entrant profile composite decking
Trapezoidal profile composite decking
Mesh reinforcement is provided at the top of the slab to help reduce cracking, spread localised loads, enhance fire resistance, and act as shear reinforcement around the shear connectors. The decking is normally designed to support the wet weight of the concrete and construction loading as a continuous member over at least two spans, but the composite slab is normally designed as simply supported between beams (but some continuity reinforcement is required). The design of the decking is usually picked from the manufacturer’s technical data sheet.
Advantages (1) Shallower beams than non-composite floors (2) More economical (3) Light weight
Disadvantages (1) More columns needed than with long-span systems. (2) Deeper overall floor zone than shallow floor systems. (3) Generally, beams require fire protection.
Slimdek
Slimdek is a shallow floor system comprising asymmetric floor beams (ASBs) supporting heavily ribbed composite slabs with 225 mm deep decking. ASBs are proprietary beams with a wider bottom flange than top. The section has embossments rolled into the top flange and acts compositely with the floor slab without the need for additional shear connectors.
Typical slimdek floor
The decking spans between the bottom flanges of the beams and acts as permanent formwork to support the slab and other loads during construction. The in-situ concrete acts compositely with the decking and encases the beams so that they lie within the slab depth – apart from the exposed bottom flange. The floor normally spans between 6 – 9m grid with floor depth of about 280 – 350 mm. Reinforcements of 16mm to 25mm bars are placed in the ribs of the slab to improve strength in the fire condition, while mesh reinforcement is paced above ASB.
Advantages (1) Shallow floor zone – reduction in overall building height and cladding. (2) Virtually flat soffit allows easy service installation and offers flexibility of internal wall positions.
Disadvantages (1) Steel weight is often greater than other floor systems. (2) Connections require careful detailing due to the width of the bottom flange.
Cellular composite beams with composite slabs and steel decking
Cellular beams are beams with openings at short regular intervals along their length. The beams are either fabricated from 3 plates or made from rolled sections. Openings, or ‘cells’, are normally circular, which are ideally suited to circular ducts, but can be elongated, rectangular or hexagonal. Cells may have to be filled in to create a solid web at positions of high shear, such as at supports or either side of point loads along the beam.
Cellular beam composite slab construction
Cellular beams can be arranged as long-span secondary beams, supporting the floor slab directly, or as long-span primary beams which are aligned parallel to the span of the slab supporting other cellular beams or conventional rolled section secondary beams. The secondary beams are typically placed at 3 – 4 m spacing, supported by primary beams on a 6 m, 7.5 m or 9 m column grid. The decking and slab can be designed using decking manufacturer’s design tables or software.
Cellular beam composite slab construction
Advantages Long, column-free floor spans. Relatively lightweight beams compared with other long-span systems. Economic long-span solution. Precamber can be accommodated during the fabrication of the members. Regular openings in the web allow ducts and other services to pass through the beams.
Disadvantages Increased fabrication costs compared with plain sections.
Slimflor® beams with precast concrete units
In slim floor system where the beams are contained within the structural floor depth. A steel plate (typically 15 mm thick) is welded to the underside of a UC section to make the Slimflor beam. This plate extends beyond the bottom flange by 100 mm either side, and supports the precast floor units. A structural concrete topping with reinforcement is recommended to tie the units together and the topping thickness should cover the units by at least 30 mm.
Typical slimflor arrangement
A composite Slimflor beam can be achieved by welding shear connectors (normally 19 mm diameter by 70 mm long) to the top flange of the UC. Reinforcement is then placed across the flange into slots prepared in the precast units, or on top of shallow precast units. If the steel beams are to be designed compositely, the topping should cover the shear connectors by at least 15 mm, and the precast units by 50 mm.
Typical section through a slimflor slab
Only 152 UCs and 203 UCs are normally suitable as composite beams because the overall depth of the floor slab becomes impractical for larger serial sizes. Precast units are usually cambered to cancel out dead load deflections between beams, and the floor spans are typically between 4.5m to 7.5m even though spans of 10m can be achieved.
Advantages (1) Beams normally require no fire protection for up to 60 minutes fire protection. (2) Shallow floor zone – reduction in overall building height and cladding. Virtually flat soffit allows easy service installation and offers flexibility of internal wall positions. (3) Shear connectors can be welded off-site, enabling larger stud diameters to be used and reducing site operations.
Disadvantages (1) The steelwork is relatively heavy. (2) Extra fabrication is involved in welding the plate to the UC. Connections require more detailing as the plate is wider than the column. (3) Precast units involve more individual lifting operations than decking, which is delivered and erected in bundles. The erection sequence requires access for installation of the concrete units.
Long-span composite beams and composite slabs with metal decking
This system consists of composite beams using rolled steel sections supporting a composite slab in a long-span arrangement of, typically, 10 to 15 m. Grids are either arranged with long-span secondary beams at 3 m to 4 m spacing supporting the slab, supported by short-span primary beams, or with short- span secondary beams (6-9 m span) supported by long-span primary beams.
Long span composite cellular beam floor
The depth of the long-span beams means that service openings, if required, are provided within the web of the beam. Openings can be circular, elongated or rectangular in shape, and can be up to 70% of the beam depth. They can have a length/depth ratio of up to 2.5. Web stiffeners may be required around holes. Shear studs are normally positioned in pairs, with reinforcing bars placed transversely across the beams to act as longitudinal shear reinforcement.
Advantages (1) Large column-free areas. (2) Service ducts pass through openings in the web of the beams
Disadvantages (1) Deeper floor zones. (2) Heavier steelwork than some short-span solutions. (3) Fire protection required for 60 minutes fire resistance and above.
Composite beams with precast units
This system consists of rolled steel beams with shear studs welded to the top flange. The beams support precast concrete units with a structural concrete infill over the beam between the ends of the units, and often with an additional topping covering the units. The precast units are either hollow core, normally 150 – 260 mm deep, or they are solid planks of 75 mm to 100 mm depth.
Different configurations if composite floors with precast units
The shear studs and transverse reinforcement allow the transfer of the longitudinal shear force from the steel section into the precast units and the concrete topping, so that they can act together compositely. Composite design is not permitted unless the shear connectors are situated in an end gap (between the concrete units) of at least 50 mm. Minimum flange widths are crucial for providing a safe bearing for the precast units and room for the shear studs.
Advantages (1) Fewer secondary beams, due to long-span precast units. (2) Shear connectors for most beams can be welded off site, enabling larger stud diameters to be used and fewer site operations. It is usually convenient to weld studs to edge beams on site.
Disadvantages (1) The beams are subject to torsion and may need stabilising during the construction stage. (2) The precast units need careful detailing for adequate concrete encasement of shear connectors and installation of transverse reinforcement. (3) More individual lifting operations compared to the erection of decking, and the erection sequence requires access for installation of the concrete units.
Non-composite beams with precast units
This system consists of rolled steel beams supporting precast concrete units. The precast units may be supported on the top flange of the steel beams, or, to reduce construction depth, supported on ‘shelf’ angles. Shelf angles are bolted or welded to the beam web, with an outstand leg long enough to provide adequate bearing of the precast units and to aid positioning of the units during erection. Precast concrete units are generally grouted in position. The units may have a screed (which may be structural), or may have a raised floor. The precast units are either hollow core, normally 150-260 mm deep, or they are solid planks of 75 mm to 100 mm depth.
Floor construction with precast concrete units in non-composite construction
In order to meet robustness requirements, mesh and a structural topping may be required, or reinforcement concreted into hollow cores and passed through holes in the steel beam web. Tying may also be required between the concrete units and the edge beams.
Advantages (1) Fewer secondary beams, due to long-span precast units. (2) A simple solution involving basic member design.
Disadvantages The beams are subject to torsion and may need stabilising during the construction stage. More individual lifting operations compared with the erection of decking, and the erection sequence requires access for installation of the concrete units.
Source: Brown D.G., Iles D.C., Yandzio E. (2009): Steel building design: Medium rise braced frames (In accordance with Eurocodes and the UK National Annexes). The Steel Construction Institute, UK
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
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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.
