Crack patterns can offer insight into the most likely cause of a structural failure in reinforced concrete structures. Looking at the picture in the post, what is the most likely cause of the structural failure in the building?
Structures are designed to satisfy ultimate and serviceability limit state requirements. Under ultimate limit state requirements, strength requirements such as bending, beam shear, punching shear, axial compression or tension, fatigue, etc are expected to be satisfied. Punching shear occurs when there is a high concentration of localised load (such as column load) on a flat element such as a flat slab or footing.
In serviceability limit state requirements, phenomena that can affect the appearance or functionality of the structure, or comfort of the occupants/users are addressed. Such phenomenon can include deflection, vibration or cracking.
Curved rafters can be adopted in the construction of industrial buildings, warehouses, churches, etc due to their aesthetic appeal. Quite a significant number of portal frames have been constructed all over the world with curved rafters. The relative simplicity of construction can also be a reason for the adoption of curved rafters given that the need for ridge details in the sheeting will be eliminated. Universal Beams are commonly used for curved rafters.
The curving process for a rafter can be achieved using the roller bending or induction bending process, with the former being more popular than the latter due to the cost implications. A straight portion always remains at the end of each bar after curving a member. If a perfectly curved rafter is required, then the straight lengths at the ends should be cut off.
However, for economical reasons, the straight portions can be included to form the final length of the member. This eliminates off-cut waste and the need for making additional cuts. In practical construction, it is difficult to observe the effect of curvature at the end of rafters, therefore it is acceptable to retain the straight portions at the ends. This can also simplify the fabrication of the haunches.
Design Issues of Curved Rafters
Discussed in the sections below are some of the issues encountered in the design of curved rafters;
Residual Stresses
It is important to know that the process of curving steel members affects the residual stresses in the member. The magnitude of residual stresses remaining after the curving process depends on the section properties in the plane of curvature. Residual stresses do not affect the cross-sectional resistance of a section.
For Universal Beams curved in elevation (such as curved rafters), the ratio Sx/Zx is typically 1.12. Therefore;
pr= (1.12 – 1)py = 0.12py
This level of residual stress is well within the range commonly found in straight beams due to the differential rate of cooling at the mills between the web and the flange tip. Because the curving process induces strains in excess of the yield strain, the curving process will remove the residual stresses from the straight beam. The residual stresses will therefore be limited to those caused by the curving process, and are not in addition to those in a straight beam.
Out-of-plane bending of flanges due to curvature
The flanges of curved rafters subject to in-plane bending or axial loads must also resist the out-of-plane component of loads resulting from the curvature of the member. This applies to both open sections and box sections.
The out-of-plane bending stress of an I-section is given by;
σ2 = 3σ1b2/RT
Frame analysis – elastic or plastic?
According to King and Brown (2001), elastic frame analysis may be used for any frame with members curved in elevation, provided that the geometry of the curved elements is allowed for. The member resistance checks for structures designed elastically. Plastic analysis is common for portal frames with straight members, and may also be used, within limits, to analyse portal frames with curved rafters.
Modelling curved members for analysis
Computer software can normally only model straight lengths of elements. It is possible to use a series of short straight segments to model a curved member but there are several issues that should be considered. The accuracy of the model will improve with the number of analysis elements, provided that these are chosen so that the off-set between the curved member and the segments is minimised.
The choice of the number of elements will depend largely on the curvature of each curved member. For members with a very slight curvature, there is normally no need to divide the model into a large number of analysis elements, as the differences between the analysis model and the actual structure can be accounted for simply.
Procedurefor the design of curved rafters
In common with all portal frames, the following steps must be followed:
· determination of initial sizes · check of in-plane stability · check of member resistances
To show how this can be done, a solved example is presented below. The design procedure and geometry are similar to the same presented by King and Brown (2001) using BS 5950-1:2000 design code.
Initial sizes
Following an initial analysis, the members shown in the general arrangement below were chosen. The purlins are spaced at 1.5m centre to centre.
Section Properties
Rafters – 457 x 191 x 67 UB S275
D = 453.4mm; B =189.9 mm; t = 8.5 mm; T = 12.7mm; r = 10.2 mm; A = 85.5 cm2; J = 31.1 cm4; Zx = 153 cm3; Zy =153 cm3; Sx = 1470 cm3; Iy = 1450 cm4; ry = 4.12 cm; rx = 18.5 cm; H = 0.705 dm6
In-plane stability
The portal geometry must be checked to ensure the (simple) sway-check method may be used to verify that the portal is stable in-plane.
5 × h = 5 × 7.450 = 37.25 m span, L = 36 m < 37.25 m Okay
Rise hr = 4.279 m (based on centre-line dimensions) 0.25 × span = 0.25 × 36 = 9 m > 4.25 m Okay
The sway-check method may therefore be used. The notional horizontal loads are taken as 0.5% of the base reaction and applied horizontally in the same direction at the top of each column. The modelling has been carried out using Staad Pro software with 36 straight members to form the curved rafters as shown below.
With a uniformly distributed ultimate limit state load of 10 kN/m, a vertical reaction of 180 kN was obtained at both supports. The horizontal notional load applied at the top of each column was calculated as (0.005 × 180 = 0.9 kN).
The checks on the columns are no different from those in portal frames with all straight members, and are not included in this example. Between A—B, and C-D, the curved rafter is in hogging. The bending moment produces compression on the concave side of member, and the design resistance is at least that of a straight member.
The regions A-B and C-D should be checked as if straight, with full consideration of restraint positions etc. Between points B and C, the curved rafter is sagging. The bending moment produces compression on the convex side of the member, reducing its capacity. The following checks illustrate how the member in this zone should be checked.
From the analysis, the maximum forces between B and C (assumed to coexist).
Where: MEd = 393.834 kNm NEd = 128 kN VEd = 73.2 kN
Reduced Design Strength
The curvature of the rafter induces bending stresses in the flanges, which combine with the axial stress to reduce the effective yield stress of the section. The reduced design strength, Pyd, is calculated following the procedure given in Section 6.3.2 of King and Brown (2001):
Lateral-torsional buckling between purlin positions must be checked, because the bending moment produces compression on the convex side of the member, and the resistance of the member is reduced.
To check lateral-torsional buckling, Clause 5.5.2 directs the designer to Clause 4.8.3.3. In-plane stability has already been checked by the sway stability provisions (Clause 5.5.4) so, in general, only the second relationship (covering out-of-plane effects) needs to be satisfied.
For out-of-plane buckling:
NEd/PcY + mLTMLT/Mb ≤ 1.0
For Pcx , Le (eaves to apex) = 18.5 m approximately. Assuming a straight member without haunches: λy = L/ry = 1500/41.2 = 36.4; pcy = 250 N/mm2 AgPcy = 8550 × 250 × 10-3 = 2137.5 kN
Over a length between purlins, the moment will be taken as constant, therefore, mLT = 1.0 with λLT = 32.42, and py = 275 N/mm2 pb = 274 N/mm2
The section is at limiting equilibrium between B and C and may be accepted for ULS or a slightly higher section selected. Rafter sections A–B and C–D, and the columns, must be checked as straight members.
References King C. and Brown D. (2001): Design of curved steel. The Steel Construction Institute, UK
The internet space (mostly blogs) is filled with so much information about hidden beams (also called concealed beams) which is defined as a beam whose depth is equal to the thickness of the slab. In other words, the beam is embedded inside/within the reinforced concrete slab. However, in standard civil engineering textbooks, codes, and standards, no mention is usually made of hidden beams. On a very simple explanation, the idea of hidden beams is the provision of more reinforcements with shear stirrups in a banded area of a floor slab.
According to the literature review of a 2020 research carried out in Ankara, Turkey, and published by the Journal of Building Engineering(Elsevier) no previous literature on experimental studies on hidden beams was encountered by the authors. By implication, prior to the year 2020, no experimental studies have been published on the behaviour of hidden beams.
On the issue of hidden beams, the researchers (Özbek et al., 2020) said,
… some of the designers illegally try to remove the beams by arranging the reinforcement with equivalent strength in the slab and call it the hidden beam. In other words, these hidden beams are constructed by placing additional longitudinal reinforcing bars in the slab along the line where the actual beam should have been present.
Figure 1: Typical hidden beam arrangement in a floor slab
In the research by Özbek et al., (2020), reinforcement ratio and slab thickness were used as the test parameters for the 14 specimens used in the pioneering research campaign. The results showed that hidden beams were able to achieve reference strength of equivalent drop beams after excessive deformations (about eight times larger) or in some cases never achieved the reference strength.
A construction concept that comes close to the so-called hidden beams is the idea of a wide-shallow beam or wide-band beams. A wide-shallow beam is a beam whose depth is generally lower than 350 mm with a cross-section having the width over the depth ratio greater than 2. However, the commonly used values in practice are 250–300 mm for depth and 4–5 for the width to depth ratio (Özbek et al., 2020). Wide-shallow beams are usually adopted in the construction of one-way ribbed slabs and maintain the same thickness with the slab for architectural convenience and for supporting partition loads.
Figure 2: Wide shallow beams (Conforti et al, 2013)
In a numerical modelling study published in the Asian Journal of Engineering and Technology by Mohd and Helou (2014) titled ‘Slabs with hidden beams: facts and fallacies‘, the authors concluded that in comparison with the system of drop beams, selected as a comparison basis, hidden beams provide little, if any, added value. Moreover, under monostatic loading, the hidden beams behave more of a slab than a beam.
Furthermore, Helou and Awad (2014) suggested that the presence of hidden can add to the vulnerability of the supporting columns under seismic conditions since they tend to be significantly stiffer than the columns thereby contradicting the preferred strong column-weak beam arrangement.
Numerical Modelling of Hidden Beams
To investigate the design concept of a typical hidden beam, a simple numerical model was carried out using design software, Prota Structures. A simply supported slab of dimensions 7m x 6m and thickness of 200 mm was modelled under the following conditions;
(a) A solid slab without internal beams(CASE 1)
Figure 3: Solid slab with no internal beam
(b) A solid slab with a drop beam (450 x 230 mm) at the midspan(CASE 2)
Figure 4: Solid slab with normal drop beam at the mid-span
(c) A solid slab with a beam of 200 mm depth and width of 800 mm at the midspan (hidden beam) – CASE 3
Figure 5: Solid slab with hidden beam at the mid-span
Case 1 From the analysis and design of the floor system as a solid slab without any internal beam, the maximum elastic deflection under the reference load was observed to be 18 mm. For the design, the use of T12@100 c/c (parallel to the short span) was observed to satisfy strength and deflection requirements with allowable span/effective depth ratio of 37.98 and actual span/effective depth ratio of 35.503.
Figure 6: Deflection profile of the solid slab
Case 2 For the slab with a normal drop beam of 450 x 230 mm at the mid-span, the following analysis and design results were obtained for the normal drop beam. At the bottom of the beam (tension zone), the area of steel required was 1308 mm2, while the area of steel provided was 1384 mm2 (2T25 + 2T16). The allowable span/effective depth ratio was calculated as 18.84 while the actual span/effective depth ratio was calculated as 15.61. By implication, the design requirements were observed to be satisfied as shown in Figure 7. The reinforcement detailing is shown in Figure 8.
Figure 7: Design snippet of the normal drop beam
Figure 8: Reinforcement detailing drawing of the normal beam
The elastic deflection of the slab under the reference load and arrangement (drop beam at the mid-span) as shown in Figure 9 was observed to be 11 mm;
Figure 9: Deflection profile of the solid slab with normal drop beam at the mid-span
Case 3 For a slab with a hidden beam of 200 x 800 mm, the following analysis and design results were obtained for the hidden beam. The area of tension reinforcement required was calculated as 2861 mm2, while the area of compression reinforcement required was calculated as 577 mm2. The area of steel provided in the tension zone was 3140 mm2 (10T20), while the area of steel provided in the compression zone was 1608 mm2 (8T16).
Furthermore, unusually high reinforcement requirement was observed at the supports of the hidden beam. The allowable span/effective depth ratio was calculated as 20.68 while the actual span/effective depth ratio was calculated as 38.22. By implication, despite the heavy reinforcement provided in the band of the hidden beam, it failed in deflection using ‘deemed to satisfy rules’ as shown in Figure 10. The reinforcement detailing is shown in Figure 11.
Figure 10: Design snippet of the hidden beam
Figure 11: Reinforcement detailing of the hidden beam
The deflection of the slab under the reference load and hidden beam arrangement as shown in the Figure 12 was observed to be as 15 mm.
Figure 12: Deflection profile of the solid slab with a hidden beam at the mid-span
Conclusion
From the literature surveyed and from the simple numerical model carried out in this study, the use of hidden beams do not appear to be favourable or significantly beneficial to the behaviour of floor slabs. In other words, the parameter that significantly affects the deflection behaviour of slabs which is the depth will effectively remain constant, while the designer tweaks the area of steel for higher modification factor. In other words, the design of hidden beams can only be based on strength and not deformation.
A lot of blogs suggest that hidden beams are designed as normal beams, with the sole difference of having the same depth with the slab. However, a little study of the hidden beam designed in CASE 3 showed that it failed in deflection and will not likely pass no matter how how the reinforcement is tweaked. By implication, if the hidden beam is assumed as a regular support and the slab designed as two panels as in CASE 2 (lx = 3500 mm), the slab will still likely have serious problems with deflection, unless the satisfactory Y12@100 c/c steel or equivalent area provided in the slab. This therefore makes no economical sense.
Therefore, the use of hidden beams in floor slabs appears to be an imaginary strength enhancing concept without any research based technical backing so far.
References
[1] Conforti A., Minelli F., Plizzari G. A. (2013): Wide-shallow beams with and without steel fibres: A peculiar behaviour in shear and flexure. Composites Part B: Engineering (51):282-290 https://doi.org/10.1016/j.compositesb.2013.03.033 [2] Helou S. H. and Awad R. (2014): Performance-based analysis of hidden beams in reinforced concrete structures. MATEC Web of Conferences (16)100001 (2014) [3] Mohd M. and Helou S. (2014): Slabs with Hidden Beams, Facts and Fallacies. Asian Journal of Engineering and Technology 02(04):316-319 [4] Özbek E., Aykaç B., Bocek M., Cem Yılmaz M. C., Mohammed A. B. K., Er S. B., Aykaç S. (2020): Behavior and strength of hidden Rc beams embedded in slabs. Journal of Building Engineering, 29(2):101130. https://doi.org/10.1016/j.jobe.2019.101130
Earth retaining structures are subjected to a myriad of loads such as earth pressure, water pressure, and earthquake loads (in seismic zones). However, when loads are applied on the earth retained at the back of the retaining wall, surcharge loads are induced on the retaining structure. The magnitude of lateral pressure transferred to the wall depends on the spatial distribution of the load, its magnitude, and location (distance away from the wall). The aim of this article is to evaluate the common sources of surcharge loads on earth retaining structures.
Different forces or loads can be placed on the earth being retained. These loads will produce a surcharge on the retaining structure. Surcharge loads on retaining walls can be permanent or temporary. The most common sources of surcharge are;
Foundations of structures close to the retaining wall
Pavements or compound floorings
Moving vehicles (traffic)
Construction equipment
Compaction process
Variation in ground surfacing/undulations
These loads vary in spatial distribution and will exert different magnitudes and distributions of lateral pressure. For instance, the surcharge pressure exerted by a uniformly distributed concrete flooring will be quite different from the surcharge load due to an isolated footing close to the retaining wall.Let us now review these possible loads/actions in detail;
General Loading Condition
According to BS 8002, the minimum surcharge load that should be applied to retaining walls is 10 kN/m2. For shallower retaining walls, the surcharge may be reduced if the designer is confident that a surcharge of 10 kN/m2 will not occur during the life of the structure.
Additional surcharge loading should be used in the design to take account of incidental loading arising from the construction plant, stacking of materials, and movement of traffic both during construction and subsequently unless the nature or layout of the site precludes the need for such additional surcharge. However, as practicable as possible, stockpiling materials close to retaining walls should be avoided.
Uniform surcharge loads such as that due to HA load and 45 units of HB loads (with values of10 kN/m2 and 20 kN/m2 respectively) will exert a rectangular pressure distribution on the back of the retaining wall. The magnitude of the surcharge pressure (ps) will be given by;
ps= khq (kN/m2)
Where; kh = coefficient of active earth pressure or at rest (as appropriate) ka = coefficient of active pressure = (1 – sin Φ’)/ (1 + sin Φ’) k0 = coefficient of earth pressure at rest = 1 – sin Φ’ q = characteristic value of the surcharge load (kN/m2)
Figure 1: Typical surcharge load and surcharge load lateral pressure distribution
Imposed Surcharge Loads from Traffic: Highways
Where retaining walls give support to highways it was traditional to check for the effects of HA loading at ground level. HA surcharge load is usually taken to be 10 kN/m2. According to BD37/01, the surcharge load for 45 units of HB load should be 20 kN/m2. The equivalent to BS EN 1991-2 and the UK NA is the application of the load model in Figure NA.6 of the National Annex, where the axle loads are 65, 65, 115, and 75 kN (S = 320 kN) at centres 1.2, 3.9 and 1.3 m apart (S = 6.4 m) as shown in Figure 2. Each axle consists of two wheels of equal load at a distance of 2.0 m apart.
Figure 2: Normal traffic surcharge load model (EN 1991-2, UK NA)
Clause 6.6.3 of PD 6694-1 allows an alternative. For global effects on ‘other earth retaining walls’ adjacent to highways, two vertical uniformly distributed transverse line loads of QL, are applied 2.0 m apart on a notional lane of the carriageway, where
QL = 320/(2 × 6.4) = 25 kN/m over a length of 6.4 m
Besides normal γQ factors, axle loads and line loads are subject to an overload factor of 1.5 and, when checking a single vehicle in one notional lane, a dynamic factor of 1.4. The dynamic factor dissipates to 1.0 at 7.0 m depth or when convoys of vehicles are considered in each notional lane (representing a traffic jam situation).
In association with the axle loads or line loads, it is suggested that a surcharge of 5 kN/m2 is applied as an imposed load to pavements adjacent to basements. This figure is in agreement with NA.2.36 of the UK NA to BS EN 1991-2 as a maximum uniformly distributed load for continuous dense crowding (e.g. footbridges serving a stadium) and with BS EN 1991-1-1 for traffic and parking areas with vehicles > 30 kN.
Concentrated Surcharge Loads(Point Loads)
Using Boussinesq equations to determine vertical pressure, lateral pressure at a point O on a wall due to a discrete load Q as shown in Figure 3 may be taken as:
σ’ah = Kh(3QZ3)/(2πR5)
where Kh = Kad or K0d (coefficient of active pressure or at rest as appropriate) Q = load, kN Z = depth, m R = (x2 + y2 + z2)0.5
Figure 3: Schematic representation of point load and line load surcharge
It is recommended that horizontal earth pressures against ‘rigid’ walls determined using Boussinesq’s theory of stresses in an elastic half-space should be doubled for design purposes. Boussinesq’s theory for horizontal pressures assumes horizontal movement. However, with truly rigid walls there is actually no movement. So an identical balancing surcharge on the other side of the wall, i.e. a mirror image surcharge, is required and this in effect doubles the pressure. Doubling pressure is in line with field data by Terzarghi and with French practice.
SurchargeLine Load Parallel to Retaining Wall
Assuming that the length of the load is comparable to that of the wall, lateral pressure at a point O will depend mainly on the depth z. In this case, lateral pressure may be taken as:
σ’ah = Kh(2QZ3)/(πR4)
where Kh = Kad or K0d as appropriate Q = load per metre length of load kN/m Z = depth, m R = (x2 + z2)0.5
As explained earlier, it is recommended that horizontal earth pressures determined using Boussinesq’s theory of stresses in an elastic half-space should be doubled.
Strip Surcharge Load Parallel to Retaining Wall
Assuming that the length of the load is comparable to that of the wall, lateral pressure at a point O will depend only on the depth z as shown in Figure 4. In this case, lateral pressure may be taken as:
σ’ah = Kh(q/π) [α + sinα cos(α + 2β)]
where Kh = Kad or K0d as appropriate q = load per metre width of load α, β = angles (in radians)
Figure 4: Schematic representation of strip load surcharge
Rectangular loads such as pad footings will exert a surcharge load on retaining walls. Standard textbooks provide solutions for vertical stress, σ’v,z at depth z under a corner of a rectangular area carrying a uniform pressure q. It is usually in the form;
σ’v,z = qIr
where; q = uniform pressure Ir = coefficient.
Values of Ir are provided for different aspect ratios of the loaded area to depth and can be read from Fadum’s chart. Lateral pressure at the required depth may be determined as KhqIr. The method of superposition allows the determination of vertical stress under any point within or outside the loaded area. An example of this has been presented on how to apply surcharge load from pad footings to retaining walls.
Compaction Pressure on Retaining Walls
The lateral pressure due to the action of compacting equipment can induce surcharge load on the earth retaining structure. The magnitude of the lateral pressure depends on the characteristic design compaction design force, the unit weight of the soil, and the coefficient of earth pressure adopted. The compaction pressure of machines on earth retaining walls can be read here.
Sometimes, I-beam sections or girders are built up by welding structural steel plates together. This is usually done when the section required is so heavy that it cannot be picked from the standard sections available, or when the section required is not available with local manufacturers or dealers.
The difference between hot-rolled I-section and welded steel sections is shown in Figure 1. The design of a built-up beam (plate girder) involves the selection of adequate individual section sizes, weld sizes, and stiffeners (if required), and verifying their performance as a composite whole in satisfying ultimate and serviceability limit state requirements.
Figure 1: (a)Built-up I-beam (b) Hot rolled I-section
Design Example of Built-Up Beams
Ghosh (2010) presented an example of the design of a welded-plate girder for a crane gantry in an iron melting workshop/industry. The example has been reproduced here to show how to design welded-plate girders according to the requirements of EN 1993-1-1:2005.
Design Forces
Ultimate design vertical moment = Mvu = 29515 kNm Ultimate design vertical shear force = Vvu = 6282 kN Ultimate design horizontal moment = Mhu = 601 kNm Maximum ultimate horizontal longitudinal tractive force = 312 kN Span of girder = 24 m
Design of section The section will be designed as a welded-plate girder. Eurocode 3, Part 1-1 (Eurocode, 2005) will be followed. The tables and figures referred to below can be found in Annex A of the Eurocode (Appendix B of this book), except where otherwise mentioned.
Figure 2: Huge welded I-Section
Design strength By referring to Table 3.1 of Eurocode 3 (“Nominal values of yield strength fyand ultimate tensile strength fu for hot-rolled structural steel”), the design strength (fy) in the ULS method of design for the flanges and web can be obtained; its value varies with the thickness of plate considered.
In our case, we adopt steel grade S275 with a design yield strength fy = 275 N/mm2. So, for a nominal plate thickness t ≤ 40 mm, fy = 275 N/mm2, and for t ≤ 80 mm, fy = 255 N/mm2. If the design strength of the web fyw is greater than the design strength of the flange fyf, then the design strength of the flange should always be used when considering moments or shear.
Initial sizing of section The dimensions of the webs and flanges are assumed to be as given below. Overall depth of girder = h. The depth should be chosen to limit the allowable deflection. In practice, the overall depth should normally be taken to be between 1/10 and 1/12 of the span. In our case, we assume the overall depth h = 1/10 of span = 23.4/10 = 2.34 m = 2340 mm. We assume an overall depth h = 2500 mm (as the girder is subjected to high dynamic wheel loads).
Depth of straight portion of web (d) d = h − (2 × size of weld) − (2 × thickness of flange )= 2500 − 2 × 12 (assumed weld size) − 2 × 55 (assumed) = 2366 mm.
Breadth of flange (b) The breadth of the flange should be at least 1/40 to 1/30 of the span in order to prevent excessive lateral deflection. In our case, we assume a breadth b = 1/30 of span = (1/30) × 23.4 = 0.78 m, say 0.9 m = 900 mm.
Thickness of web (tw) Several tests have shown that the web does not buckle owing to diagonal compression when the ratio d/tw is less than 70, if the web is not stiffened by a vertical transverse stiffener. Referring to Table 5.2 (sheet 1) of Eurocode 3, Part 1-1, the minimum thickness of web required to avoid buckling of the compression flange in the ULS design method with an unstiffened web is as follows.
For class 1 classification, d/tw ≤ 72ε, where ε = stress factor = (235/fy)0.5 = (235/255)0.5 = 0.96; d/tw = 2366/tw = 72 × 0.96 Therefore tw = 2366/(72 × 0.96) = 34 mm
With a stiffened web and a spacing of transverse stiffeners a ≤ d, tw ≥ (d/250)(a/d)0.5. Assuming a spacing of stiffeners a = 2366 mm; tw = 2366/250 × (2366/2366)0.5 = 10 mm. We assume tw = 30 mm.
Thickness of flange (tf) The minimum thickness of the flange required to limit the outstand of the flange is calculated as follows. The approximate flange area required is given by;
Af = Mvu/(hfy) = (29515 × 106)/(2500 × 255) = 46298 mm2 Assuming the width of the flange b = 900 mm tf = 46298/900 = 51.4 mm.
We, therefore, assume tf = 55 mm
Classification of cross-sections Referring to Clause 5.5.2 of Eurocode 3, Part 1-1, the function of cross-section classification is to identify the extent to which the resistance and rotation capacity of the cross-section are limited by its local buckling resistance. In our case, we assume a class 1 cross-section classification without reduction of resistance. Thus, to determine the thickness tf of the flange, we do the following.
For class 1 section classification, c/tf ≤ 9ε,
where; c = outstand of flange plate = [b − (tw + 2 × 12 (weld size))]/2 = [900 − (25 + 24)]/2 = 425.5 mm. Assuming tf = 55 m, c/tf = 425.5/55 = 7.7 and 9ε = 9 × 0.96 = 8.64
Since c/tf (7.7) < 9ε (8.64), the section satisfi es the conditions for class 1 section classification. So we assume tf = 55 mm.
To determine the thickness of the web tw, we do the following; For class 1 section classification, d/tw ≤ 72ε. Assuming tw = 30 mm, d/tw = 2366/30 = 78.9 and 72ε = 72 × 0.96 = 69 < d/tw which does not satisfy the condition. We increase the thickness tw to 35 mm.
d/tw = 2366/35 = 67.6 < 72ε (69) which satisfies the condition. So, we assume tw = 35 mm.
Thus the initial sizing of the section is as follows:
Depth of girder h = 2500 mm.
Breadth of flange b = 900 mm.
Depth of straight portion of web d = 2500 − 2 × 55 − 2 × 12 (weld size) = 2366 mm.
Thickness of web tw = 35 mm.
Thickness of flange tf = 55 mm.
Design strength with flange thickness 55 mm = fy = 255 N/mm2
Design strength of web with thickness 35 mm = fy = 255 N/mm2
Although the design strength of the web is 275 N/mm2 for 35 mm thickness, the lower value of fy (255 N/mm2) of the flange should be considered in calculations for moments and shears.
Moment capacity Total maximum ultimate vertical design moment = Mvu = 29515 kNm Total maximum design shear = Vu = 6281 kN
The moment capacity should be calculated in the following way. When the web depth-to-thickness ratio d/tw ≤ 72ε, it should be assumed that the web is not susceptible to buckling, and the moment capacity should be calculated from the equation;
Mrd = fyWpl
provided the shear force VEd (Vvu) ≤ 0.5Vpl,rd (shear capacity),
where Mrd = moment capacity, Wpl = plastic section modulus Vpl,rd = shear capacity.
In our case, d/tw (67.6) < 72ε (69.1). Thus, the web is not susceptible to buckling.
The ultimate shear force (Vvu) should also be less than half the shear capacity (Vpl,rd) of the section. Referring to equation (6.18) of Eurocode 3, Part 1-1;
Vpl,rd = Av[fy/(3)0.5]/γMo
where Av = shear area = dtw + (tw + 2r)tf = 2366 × 25 + (25 + 2 × 12) × 50 = 61600 mm2
Thus, Vvu (6281 kN) < 0.5Vpl,Rd (6335 kN). So the section satisfies the conditions.
Since the web is not susceptible to buckling, and the lowest shear value in the section is less than half the shear capacity of the section, the moment capacity for this class 1 compact section should be determined by the “flange only” method. In this case, the whole moment will be taken up by the flanges alone and the web takes the shear only.
Therefore moment capacity of section My,Rd = fyAfhs
where; Af = area of compression flange = b × tf = 900 × 55 = 49500 mm2 hs = depth between centroids of flanges = 2500 − 55 = 2445 mm fy = 255 N/mm2
Alternatively, referring to Clause 6.2.5, the moment capacity of the section may be expressed by the following equation:
Mpl,Rd = Wplfy/γMo
The plastic modulus is calculated to be 1.71008 x 108 mm3 Mpl,Rd = Wplfy/γMo = [(1.71008 x 108 × 255)/1.00] × 10-6 = 43607 kNm
where Wpl is the plastic modulus of the section. As the section assumed is built-up welded and of high depth, no rolled I section is available of this depth. The plastic modulus for the assumed depth can be calculated. So the above equation can only be used when the assumed section is manufactured industrially by welding.
In addition, the top flange is also subjected to a stress due to the horizontal transverse moment caused by horizontal crane surges. Therefore, the “flange only” method is suitable in our case. The horizontal transverse ultimate moment Mhu is equal to 601 kN m.
This transverse horizontal moment is resisted by a horizontal girder formed by the connection of the 6 mm plate (acting as a web) of the walking platform (Durbar), between the top flange of the main plate girder and the tie beam at the walking-platform level. Distance between centre line of plate girder and the tie beam = hz = 2.5 m
Horizontal moment of resistance = Mz,Rd = fyAfhz where Af = area of top flange = 900 × 55 = 49500 mm2.
Bearing capacity is the maximum load a soil profile can withstand before undergoing excessive deformation and final shear failure. It is well understood that the depth of the groundwater table can affect the strength of soils, but a high water table does not necessarily indicate that the soil is weak as sometimes misunderstood. However, the presence of groundwater in soils can reduce the strength due to the way water affects the unit weight of soils and the shear strength parameters – cohesion (c) and angle of internal friction (ϕ).
We will discover that the equation contains terms for cohesion (c), angle of internal friction (ϕ), and unit weight of the soil (γ). The bearing capacity factors Nc, Nq, and Nγdepend on the angle of internal friction. When the soil is submerged, the effective unit weight (γ′) is to be used in the computation of bearing capacity.
γ′ = γsat – γw —— (2)
Where; γ′ = Effective unit weight of the soil γsat = saturated unit weight of the soil γw = Where is the unit weight of water
As can be seen from Equation (2), the effective unit weight γ′ is about half the saturated unit weight; consequently, there will be about 50% reduction in the value of the corresponding term in the bearing capacity formula. Similarly, the effective stress parameters, c′ and φ′, obtained from an appropriate test in the laboratory, on a saturated sample of the soil, are to be used. It should be noted that water also affects the shear strength parameters c′ and φ′ but their effects are usually so small that they are ignored.
Foundation construction in an area of high water table
It should be now obvious that the location of the groundwater table and its seasonal fluctuations have an effect on the bearing capacity of a foundation. If the water table is at a great depth from the base of the foundation, there will be no effect or reduction in the bearing capacity.
In the design of footings, the minimum depth below the base of the footing at which the water table is not expected to have an effect on the bearing capacity is set at a value equal to the width of the footing. This is because the maximum depth of the zone of shear failure below the base is not expected to exceed this value ordinarily.
However, if the water table is above this level, there will be a reduction in the bearing capacity. If the water table is at the level of the base of the footing, γ′ is to be used for γ in the third term, which indicates the contribution of the weight of the soil in the elastic wedge beneath the base of the footing, since the entire wedge is submerged.
Therefore, three cases are usually considered in the design of footings for the effect of the water table;
Modification of bearing capacity for water table
Case 1: If the water table is located so that 0 ≤ D1 ≤ Df(water table above the footing level), the factor q in the bearing capacity equations takes the form; q = effective surcharge = D1γ + D2(γsat – γw)
Case 2: If the water table is located so that 0 ≤ d ≤ B (water table at footing level or within the depth d), the factor q in the bearing capacity equations takes the form; q = effective surcharge = γDf
The factor γ in the last term of the bearing capacity equation will have to be replaced by the factor;
ỹ = γ’ + d/B(γ – γ’)
Case 3: When the water table is located such that d ≥ B, the water table will have no effect on the bearing capacity.
Worked Example on Effect of Groundwater Table
A footing 2 m square, subjected to a centric vertical load, is located at a depth of 1.0 m below the ground surface in a deep deposit of compacted sand, φ′ = 30°, and γsat = 19 kN/m3. Determine the allowable bearing capacity using Terzaghi’s theory when the water table is at;
(a) at 5m below the ground surface (b) at the ground surface (c) at the bottom of the base of the footing, and (d) at 1 m below the base
Solution
For a square footing according to Tezarghi’s theory;
qult = γDfNq + 0.4γbNγ
Nq = 22.46 Nγ =19.13
(a) If the water table is at 5m below the surface, d (4 m) > B (2 m), hence the water table will have no effect on the bearing capacity; qult = γDfNq + 0.4γbNγ = (19 × 1 × 22.46) + (0.4 × 19 x 2 x 19.13) = 717.516 kN/m2
(b) If the water table is at the ground surface; γDf = Df(γsat – γw) = 1m (19 – 9.81) = 9.19 kN/m2 γ′ = γsat – γw = (19 – 9.81) = 9.19 kN/m3 qult = γDfNq + 0.4γ’bNγ = (9.19 × 22.46) + (0.4 × 9.19 x 2 x 19.13) = 347.05 kN/m2
(c) If the water table is at the bottom of the footing; γDf = (1m × 19) = 19 kN/m2 γ′ = γsat – γw = (19 – 9.81) = 9.19 kN/m3 qult = γDfNq + 0.4γ’bNγ = (19 × 22.46) + (0.4 × 9.19 x 2 x 19.13) = 567.38 kN/m2
(d) If the water table is at 1m below footing; γDf = (1m × 19) = 19 kN/m2 ỹ = γ’ + d/B(γ – γ’) = 9.19 + [(1/2) × (19 – 9.19)] = 14.095 kN/m3 qult = γDfNq + 0.4ỹbNγ = (19 × 22.46) + (0.4 × 14.095 x 2 x 19.13) = 642.45 kN/m2
From the results above, the following was observed;
(1) When the water table is at the ground surface, the bearing capacity reduces by 51.63%. (2) When the water table is at the bottom of the footing, the bearing capacity reduces by 20.92%. (3) When the water table is at 1m below the footing, the bearing capacity reduces by 10.46%.
During the shear strength test of soils using a triaxial machine, one of the major ways of monitoring stress changes is by showing the Mohr’s stress circles at different stages of loading/unloading. However, this may be difficult to plot as well as confusing when a number of circles are to be shown in the same diagram. In this case, a stress path may be used to show the changes in stress in the soil or material.
Fig 1: Typical triaxial testing machine
A Stress Path is therefore a curve or a straight line which is the locus of a series of stress points showing the changes in stress in a test specimen or in a soil element in-situ, during loading or unloading, engineered as in a triaxial test in the former case or caused by forces of nature as in the latter. This idea that the locus of points depicting the maximum shear stress acting on a soil sample can be drawn as a stress path in place corresponding Mohr’s circle was suggested by Lambe and Whitman (1969). This is shown in Fig 2.
Fig. 2: Stress path for the case of σ1 increasing and σ3 constant
The co-ordinates of the points on the stress path are (σ1 + σ3)/2 and (σ1 – σ3)/2.
When stress paths only are plotted then the axes of the diagram are really particular values of the shear stress τ and normal stress σ. These values are commonly referred to as q and p. Mean stress, p, is the average stress on a body or the average of the orthogonal stresses in three dimensions. Deviatoric stress, q, is the shear or distortional stress or stress difference on a body. p and q are called stress invariants and do not depend on the axis system chosen.
p = mean stress = (σ1 + σ3)/2 q = deviator stress = (σ1 – σ3)/2
This shows that q is the same regardless of whether total stresses or effective stresses are being considered. In other words, the deviatoric (shear) stress is unaffected by pore water pressure. Since q is equal to the radius of the Mohr circle this means that the total and effective Mohr circles must always have the same size
By implication, either the effective stresses or the total stresses may be used for plotting stress paths. The basic types of stress path and the co-ordinates are:
The static pore water pressure is zero in the conventional triaxial test, and (b) and (c) coincide in this case. But if backpressure is used in the test, u0 equals the backpressure. For an in-situ element, the static pore water pressure depends upon the level of the groundwater table. Typical stress paths for triaxial compression and extension tests (loading as well as unloading cases) are shown in Fig. 3.
Fig 3: Typical stress paths for triaxial compression and extension tests (Venkatramaiah, 2006)
A-1 is the effective stress path for conventional triaxial compression test during loading. (Δσv = positive and Δσh = 0, i.e., σh is constant). A typical field case is a footing subjected to vertical loading.
A-2 is the unloading case of the triaxial extension text (Δσh = 0 and Δσv = negative). Foundation excavation is a typical field example.
A-3 is the loading case of the triaxial extension test (Δσv = 0 and Δσh = positive). Passive earth resistance is represented by this stress path.
A-4 is the unloading case of the triaxial compression test (Δσu = 0 and Δσh = negative). Active earth pressure on retaining walls is the typical field example for this stress path.
Figure 4 shows the typical stress paths for a drained test. Point A corresponds to the stress condition with only the confining pressure acting (σ1 = σ3 and τ = 0). Point F represents failure. Stress paths for effective stresses, total stresses, and total stresses less static pore water pressure are shown separately in the same figure.
Fig. 4: Stress paths for drained test (Venkatramaiah, 2006)
Figure 5 shows the stress paths for a consolidated undrained test on a normally consolidated clay.
Fig 5: Stress paths for consolidated undrained test on a normally consolidated clay (Venkatramaiah, 2006)
Figure 6 shows the stress paths for a consolidated undrained test on an overconsolidated clay.
Fig 6: Stress paths for consolidated undrained test on an overconsolidated clay (Venkatramaiah, 2006)
Stress path approach enables the engineer to predict and monitor the shear strength mobilized at any stage of loading/unloading in order to ensure the stability of foundation soil.
Procedure for Plotting Stress Paths
A summary of the procedure for plotting stress paths is as follows:
Determine the loading conditions drained or undrained, or both.
Calculate the initial loading values of p’o, po, and qo.
Set up a graph of p’ (and p, if you are going to also plot the total stress path) as the abscissa and q as the ordinate. Plot the initial values of (p’o, qo) and (po, qo).
Determine the increase in stresses ∆σ1, ∆σ2, and ∆σ3. These stresses can be negative.
Calculate the increase in stress invariants ∆p’, ∆p, and ∆q. These stress invariants can be negative.
Calculate the current stress invariants as p’ = p’o+ ∆p’ , p = po + ∆p, and q = qo + ∆q. The current value of p’ cannot be negative, but q can be negative.
Plot the current stress invariants (p’, q) and (p, q).
Connect the points identifying effective stresses, and do the same for total stresses.
Repeat items 4 to 8 for the next loading condition.
The excess porewater pressure at a desired level of deviatoric stress is the mean stress difference between the total stress path and the effective stress path. Remember that for a drained loading condition, ESP = TSP, and for an undrained condition, the ESP for a linear, elastic soil is vertical.
Sources:
Venkatramaiah, C. (2006): Geotechnical Engineering. New Age Publishers, India Budhu M. (2010): Soil mechanics and foundations (3rd Edition). John Wiley and Sons, USA
It is common knowledge that a well-designed hollow steel section has lesser material weight and higher strength to weight ratio when compared with an equivalent open section profile such as a universal beam (I-section) or universal column (H-section). However, an important aspect of hollow steel section design is in the design of the connections. The design of hollow steel section connection in terms of joint resistance and cost fabrication is affected considerably by member sizing since the sections are usually welded to each other.
The most common types of hollow steel sections employed in design are;
Circular hollow section (CHS)
Square hollow section (SHS)
Rectangular hollow section (RHS)
Joint Geometry
The most common types of hollow section joint configuration are;
X-joints
T- and Y-joints
K- and N-joints with gap
K- and N-joints with overlap
These configurations are shown in Figure 1.
Figure 1: Different configurations for hollow sections
Generally, it is recommended that the angle between the chord and a bracing or between two bracings should be between 30° and 90° inclusive. If the angle is less than 30° then:
a) The designer must ensure that a structurally adequate weld can be made in the acute angle b) The joint resistance calculation should be made using an angle of 30° instead of the actual angle
When K- or N-joints with overlapping bracings are being used, the overlap must be made with: • Partial overlap where the first bracing runs through to the chord, and the second bracing sits on both the chord and the first bracing, or • Sitting fully on the first bracing.
The typical joint geometry symbols used in the design are shown in Figure 2 and 3;
Figure 2: Circular and rectangular chord symbols
Figure 3: I- or H-chord symbols
Multi-planar Joints
Multi-planar joints are typically found in triangular and box girders.To determine whether a joint should be considered as a multi-planar or a single planar joint refer to Figure 4. By applying the multiplanar factor, μ (Figure 5) to the calculated chord face deformation, you can use the same design formulae as planar joints.
Figure 4: Multi-planar joints
The planar factors shown in the figure below have been determined for angles between the planes of 60° to 90°.
Figure 5: Multi-planar factors
Welding of Hollow Steel SectionConnections
When a bracing member is under load, a non-uniform stress distribution is present in the bracing close to the joint, see Figure 6. Therefore, to allow for this non-uniformity of stress, the welds connecting the bracing to the chord must be designed to have sufficient resistance. Normally, the weld should be around the whole perimeter of the bracing using a buttweld, fillet-weld, or a combination of the two.
Figure 6: Typical localised stress distribution at a joint
For bracing members in a lattice construction, the design resistance of a fillet-weld should not normally be less than the design resistance of the member. This is satisfied if the throat size (a) is at least equal to or larger than 1.1 times the width of the section, provided you use electrodes with an equivalent strength grade to the steel (both yield and tensile strength).
The weld resistance can also be verified from the simplified method for design of fillet weld EN 1993:1-8 clause 4.5.3.3;
Fw,Ed < Fw,Rd
Where; Fw,Ed is the design value of the weld force per unit length Fw,Rd is the design weld resistance per unit length
For a more efficient weld, you can use the directional method from EN 1993:1-8 clause 4.5.3.2.
Fabrication
In a lattice-type construction, the largest fabrication cost is the end preparation and welding of the bracings, and the smallest is the chords. As a general rule, the number of bracing members should be as small as possible. The best way to achieve this is using K- type bracings, rather than N-type bracings. Hollow sections are much more efficient in compression than open sections, angles or channels, meaning compression bracings do not need to be as short as possible. This makes the K-type bracing layout much more efficient.
In circular chords, the ends of each bracing in a girder has to be profile-shaped to fit around the curvature of the chord member (see Figure 7), unless the bracing is very much smaller than the chord. Also, for overlap joints with circular bracings and chords, the overlapping bracing has to be profile shaped to fit to both chord and the overlapped bracing.
Figure 7: Connections to a circular chord
Joint Design Parameters and Resistance
The various geometric parameters of the joint have an effect on its resistance. This is dependant on the:
• joint type (single bracing, two bracings with a gap or an overlap) and, • type of forces on the joint (tension, compression, moment).
The following points should be considered in the design of hollow steel section connections.
(1) The joint
(a) The joint resistance will always be higher if the thinner member sits on and is welded to the thicker member, rather than the other way around. (b) Joints with overlapping bracings will generally have a higher resistance than joints with a gap between the bracings. (c) The joint resistance, for all joint and load types (except fully overlapped joints), will be increased if small thick chords rather than larger and thinner chords are used. (d) Joints with a gap between the bracings have a higher resistance if the bracing to chord width ratio is as high as possible. This means large thin bracings and small thick chords. (e) Joints with partially overlapping bracings have a higher resistance if both the chord and the overlapped bracing are as small and thick as possible. (f) Joints with fully overlapping bracings have a higher resistance if the overlapped bracing is as small and thick as possible. In this case, the chord has no effect on the joint resistance. (g) On a size-for-size basis, joints with circular chords will have a higher resistance than joints with rectangular chords.
(2) The overall girder requirements
(a) The overall girder behaviour, e.g. lateral stability, is increased if the chord members are large and thin. This also increases the compression chord strut resistance, due to its larger radius of gyration. (b) Consideration must also be given to the fabrication costs.
Joint Failure Modes
Joints have a number of different failure modes depending on the joint type, the geometric parameters of the joint and the type of loading. These various types of failure are described below;
(a) Chord Face Failure This is the most common failure mode for joints with a single bracing, and for K- and N-joints with a gap between the bracings if the bracing to chord width ratio (β) is less than 0.85.
Figure 8: Chord face failure
(b) Chord sidewall failure This is the yielding, crushing, or instability (crippling or buckling) of the chord sidewall or web under the compression brace member. Also includes sidewall yielding if the bracing is in tension. Usually only occurs when the bracing to chord width ratio (β) ratio is greater than about 0.85, especially for joints with a single bracing.
Figure 9: Chord sidewall failure
(c) Chord shear This is found typically in the gap of a K-joint. The opposite vertical bracing force causes the chord to shear. It does not often become critical, but can if you use rectangular chords with the width (b0) greater than the depth (h0). If the validity limits are met then chord shear does not occur with circular chords.
Figure 10: Chord Shear
(d) Chord punching shear This can be caused by a crack initiation in the chord face leading to rupture failure of the chord. It is not usually critical, but can occur when the chord width to thickness ratio (2γ) is small.
Figure 11: Chord Punching shear
(e) Bracing Effective Width This is non-uniform stress distribution in the brace causing a reduced effective brace width. This reduces the effective area carrying the bracing force. It is mainly associated with rectangular chord gap joints with large β ratios and thin chords. It is also the predominant failure mode for rectangular chord joints with overlapping rectangular bracings.
Figure 12: Bracing Effective Width
(f) Chord or bracing localised buckling Due to the non-uniform stress distribution at the joint, reducing the effective area carrying the bracing forces. This failure mode will not occur if the validity ranges are met.
Figure 13: Chord or bracing localised buckling
(g) Shear of overlapping bracings Due to the bracing’s horizontal force causing shearing at the chord face. This failure mode becomes critical for large overlaps, over 80% or 60%, depending on if the hidden toe of the overlapped bracing is welded to the chord.
Figure 14: Shear of overlapping bracings
Disclaimer: The contents of this article are culled from; TATA Steel: Design of welded joints Celsius®355 and Hybox®355
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Underpinning is a process used to rectify distress caused to a building by excessive movement of its foundation, to extend foundations into the ground to facilitate future construction work, to accommodate additional load applied to an existing building, to allow the adjacent ground to be lowered, or to change the support system (BS 8004-2015).
According to BS 8004-1986, the two major objectives of underpinning are to transfer the load carried on a foundation from its existing bearing level to a new level at a lower depth and to limit the settlement of foundations.
Types of Underpinning
According to BS 8004, the major types of underpinning include:
Extending the depth of strip or pad footings (continuous underpinning constructed in sections usually not exceeding 1.2 m);
Constructing piles that are drilled through the foundations;
Constructing ground beams beneath walls formed by stooling or needling and supported by piers or piles;
Constructing ground beams beside walls supported by piers or piles with needles and/or load transfer by stressing or other details to transfer the load to the new support.
Generically, the types of permanent underpinning can be seen in the chart below;
Before underpinning is considered and resorted to, it is very important for an experienced and competent person to carry out a full investigation to determine whether the purpose of carrying out the underpinning will be achieved. The choice of underpinning system is expected to consider the loads carried, the sensitivity of the ground and the existing structure, nearby structures, and the working conditions.
The ground investigation should determine the conditions responsible for any excessive movement so that appropriate underpinning measures will be taken. Furthermore, the general nature of the ground, load-bearing capacity, and the effects of adjoining foundations should be determined before the full design of the underpinning system can be done.
Furthermore, the structure to be underpinned should be carefully investigated. All indications of distress, differential foundation settlement, and weak planes which may be accentuated during the process of underpinning should be identified and temporarily supported and/or strengthened before the underpinning can commence.
Design for Underpinning
Underpinning systems should be designed in accordance with good soil mechanics practice taking into account all vertical and lateral loads imposed upon them, including particularly transient conditions that may arise during construction. The design of underpinning should be done according to EN 1997-1 and Clause 4 of BS 5400:2015. The structural design of underpinning for spread and pile foundations should conform to relevant standards.
It is important to note that underpinning may alter or deepen only part of a foundation if only part is to be underpinned or is shown to be unstable. The designer has to be satisfied that the whole foundation including that part modified by underpinning will continue to perform satisfactorily.
If the changed support conditions will inevitably give rise to excessive differential movement, jacks may be installed temporarily or permanently to correct this. Such elaborate arrangements are rarely required except when underpinning particularly heavy walls or piers. If the whole of a structure is subsiding unevenly, for example, a building on a bed of made-up ground of variable thickness, partial underpinning will be unsuccessful and the whole structure has to be underpinned.
Construction Processes for Underpinning
Live Load Reduction Before the commencement of excavation, it is important to reduce the live loads on the walls and columns of the structure as practically as possible. This is more important when the live load is large in relation to the dead load.
Excavation The area of open excavations should not exceed 25% of the building’s footprint. This limit should be reduced if the building comprises a number of isolated piers.
Support of Excavations All excavations necessary for underpinning should be properly supported using struts of steel, timber, concrete, or other adequate temporary or permanent works as adequately designed using standard geotechnical processes.
Underpinning legs If the underpinning will not give rise to excessive ground pressure, the underpinning process should be carried out in a series of legs. The length of each leg should depend on the general character and condition of the ground and the structure to be underpinned, the intensity of the loading, and the nature of the ground below. In normal brick or stone walls, each leg should be 1000 mm to 1400 mm in length. This length can be increased in walls that are capable of arching.
Each series of legs should be planned to provide sufficient support between the legs under construction and to ensure that the loads from the unsupported portions of the wall are distributed throughout the length of the wall.
If underpinning is required to take care of excessive settlement, or if the allowable settlement is likely to be exceeded during the underpinning operations, the structure to either side of an underpinning leg should be needled and the load transferred to temporary bearings capable of carrying the additional load. Alternatively, stooled reinforced concrete beam might be introduced before the underpinning can commence.
If the width of the foundation to be underpinned is greater than 1000 mm, it is advisable to leave the outer half more than 150mm below the underside of the foundation to facilitate the pinning up of the furthest part.
