Lecture_19

Lecture_19 - Chapter Four Serviceability 4.1 Serviceability Limit States of Deflection and Cracking 4.1.1 Serviceability limit states • Review

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Unformatted text preview: Chapter Four _____________________________________________________________________ Serviceability 4.1 Serviceability Limit States of Deflection and Cracking 4.1.1 Serviceability limit states • Review: The limit states (states at which a structure becomes unfit for its intended function) are generally divided into two groups: 1. those leading to collapse, referred to as ultimate limit states (ULS); and 2. those which disrupt the use of structures, but do not cause collapse, referred to as serviceability limit states (SLS). • Two major serviceability limit states for RC structures: o excessive deflection. The appearance or efficiency of a structure or any part of a structure must not be adversely affected by deflections. o excessive crack width. The appearance or durability of a structure or any part of a structure must not be adversely affected by any cracking of the concrete. 155 o The serviceability limit states of excessive flexural deflection and excessive flexural cracking are the two that normally must be considered in design. • Other serviceability limit states that may be reached include o Durability – this must be considered in terms of the proposed life of a structure and its conditions of exposure. o Fire resistance – this must be considered to collapse, flame penetration and heat transfer. o Excessive undesirable vibration – causing discomfort or alarm as well as damage. o Fatigue – this may be considered if cyclic loading is likely. o Special circumstances – any special requirements, such as earthquake, explosion (nuclear ~, thermal ~), impact, etc. 4.1.2 Approaches for design requirement • Two approaches are commonly adopted for design requirements of the serviceability limit states: 1. by deemed-to-satisfy provisions, such as detailing rules and limiting span-depth ratios; 2. by analysis whereby the calculated values of effects of loads, e.g. deflections and crackwidths, are compared with acceptable values. 156 • In most cases of day-to-day practical design, the serviceability limit state requirements are normally met by deemed-to-satisfy provisions, such as (1) detailing rules and (2) limiting span-depth ratios. 4.2 Service Loads • The terms service loads and working loads refer to loads encountered in the everyday use of structures. • Service loads are generally taken to be the specified loads without load factors. • Load combinations Traditionally, the load combinations for serviceability limit states use a load factor of 1.0 on all service loads. For example, the SLS load combination for dead, imposed and wind loads is 1.0Gk + 1.0Qk + 1.0Wk (the USL load combination = 1.2Gk + 1.2Qk + 1.2Wk) • In most cases, o design of reinforced concrete will be based on ultimate limit state requirements, and o serviceability behaviour will be considered as secondary check to ensure satisfactory performance under working conditions. 157 4.3 Detailing Rules BS 8110 recommends simple rules to ensure that a structure has a satisfactory serviceability and durability performance under normal circumstances. These rules concern exposure conditions, fire resistance, and quantity and spacing of steel reinforcement. 4.3.1 Minimum concrete mix and cover for different exposure condition • Exposure conditions in environmental classification Mild – protected against weather Moderate – sheltered from severe rain or under water Severe – exposed to severe rain or alternate wetting and drying Very severe – exposed to sea water spray, de-icing salts or corrosive fumes Extreme – exposed to abrasive action • Mix and cover against corrosion Nominal cover: The code states that the actual cover should never be less than the nominal cover minus 5 mm, i.e. Actual cover ≥ Nominal cover – 5 mm Fig. 4.3-1 Concrete cover in a beam section 158 Table 4.3-1 Nominal cover and mix requirements for normal weight 20 mm maximum size aggregate concrete Example: widely used in buildings Concrete mix • Cover for durability requirements (IStructE design manual) Table 4.3-2 Cover for durability requirements for beams 159 • Cover for fire protection Nominal cover to all reinforcement to meet a given fire resistance period for various structural elements in a building: Table 4.3-3 Nominal cover for fire resistance 4.3.2 Minimum dimensions of members for fire protection Table 4.3-4 Minimum dimensions of RC members for fire resistance 160 4.3.3 Spacing of reinforcement • Maximum spacing for cracking control o The maximum clear spacings given in Table 4.3-5 apply to bars in tension in beams when a maximum likely crack width of 0.3 mm is acceptable. Table 4.3-5 Maximum clear spacing (mm) for tension bars in beams Moment redistribution (%) fy -30 250 460 -20 -10 0 +10 +20 +30 210 115 240 130 270 145 300 160 300 180 300 195 300 210 NOTE 1. Any bar of diameter less than 0.45 times that of the largest bar in a section must be ignored when apply these spacing. NOTE 2. Bars adjacent to corners of a beam must not be more one-half of the clear distance given in the table from the corner. o The maximum clear spacing between bars in slabs should not exceed 750 mm or 3d under the specified conditions as follows: 1. If h ≤ 200 mm with high yield steel; or 2. If h ≤ 250 mm with mail yield steel; or 3. if As/bd ≤ 0.3%. o If none of these specified conditions applies, the maximum spacing in slabs should be taken as that given in Table 4.3-5. o The amount of moment redistribution is unknown when using Table 4.3-5, zero should be assumed for span moments and 15% for support moments. 161 • Minimum spacing for construction quality o To permit concrete flow around reinforcement during construction, the minimum clear gap between bars, or group of bars, should exceed 1. (hagg + 5 mm) horizontally, and 2. (2hagg /3) vertically where hagg is the maximum size of the coarse aggregate. o The gap must also exceed the bar diameter. 4.3.4 Areas of reinforcement • Minimum area of reinforcement for cracking control For most purposes, thermal and shrinkage cracking may be controlled by the use of minimum reinforcement quantities. Table 4.3-6 Minimum reinforcement areas NOTE. At least four bars with a diameter not less than 12 mm are required in a rectangular section and six in a circular section. 162 • Maximum area of reinforcement for construction quality The maximum areas of reinforcement are determined largely from the practical need to achieve adequate compaction of the concrete around reinforcement. o For a slab or beam, longitudinal steel 1. 100 As 100 Asc or bh bh ≤ 4% each . 2. Main bars in beams are normally not less than size 16. 3. Where bars are lapped, sum of the bar sizes in a layer must not be greater than 40% of the section breadth. 100 Asc o For a column, bh ⎧6% if cast vertically ⎪ ≤ ⎨8% if cast horizontally ⎪10% at laps if either case ⎩ 4.3.5 Side face reinforcement in beams for cracking control • When beams exceed 750 mm in depth, longitudinal bars should be provided near side faces, as shown in Fig. 4.3-2. Fig. 4.3-2 Reinforcement at side face of beams 163 • These side face bars, which may be used in calculating the moment of resistance, must have a diameter > √(sbbw / fy), where sb is the bar spacing and bw the breadth of the section (or 500 mm if greater). • I.Struct.E. suggests that the size of side face bars should not be less than 0.75√bw for high yield bars and 100√bw for mild steel bars, where bw needs not to be assumed to be > 500 mm. 4.4 Deflection 4.4.1 Methods for deflection control • Excessive deflections may lead to o sagging floors, and to roofs that do not drain properly o damaged partitions and finishes, and to other associated troubles • BS 8110 states that the final deflection (including the effects of creep and shrinkage) should not exceed either of the following limits (for horizontal members): o Span/250 (in general for beams) o Span/500 or 20 mm whichever is the lesser for brittle materials o Span/350 or 20 mm whichever is the lesser for non-brittle partitions or finishes • Excessive response to wind load o Excessive accelerations that may cause discomfort or alarm to occupants should be avoided. 164 o Storey drift (relative lateral defection in any one storey), which may damage to non-structural elements, should not exceed h/500, where h is storey height. o For tall building design in Hong Kong, the top drift ≤ H/ 500 , where H is the total height of the structure. • Two methods are given in BS 8110 for checking that deflection is not excessive: 1. Limiting the span/effective depth ratio – This method is used in all normal cases (BS 8110: Part 1). 2. Calculation of deflection from the curvature of the crosssection subjected to appropriate moments (BS 8110: Part 2). In day-to-day practical design, deflections are controlled by simply limiting the span/depth ratios. 4.4.2 Deflection control of beams by limiting the span /depth ratio (BS 8110: Part 1, clause 3.4.6) • Basic span/effective depth ratios for beams o The code gives a set of basic span/effective depth ratios for rectangular and flanged beams to limit the total deflection to span/250 and the part of deflection occurring after construction of finishes and partitions will be limited span/500 or 20 mm, whichever is the lesser, for spans up to 10 m. 165 Table 4.4-1 o To obtain the span/effective depth ratios, the basic span/effective depth ratios should be modified according to the actual span and amounts of reinforcement provided. • Span/effective depth ratio = Basic span/effective depth ratio × span modification factor × tension reinforcement modification factor × compression reinforcement modification factor o Span modification factor = 10/actual span (4.1) (4.2) o Tension reinforcement modification factor = 0.55 + 477 − f s ≤ 2.0 M⎞ ⎛ 120⎜ 0.9 + 2 ⎟ bd ⎠ ⎝ (4.3) where the service stress is estimated by fs = 5 f y As ,req 8 As , prov ⋅ 1 βb (4.4) in which As,req is the area of tension steel required at mid-span to support ultimate loads, and As,prov the area of tension steel provided at mid-span (at support for a cantilever). 166 Table 4.4-2 Tension reinforcement modification factors (Clause 3.4.6.5) o Compression reinforcement modification factor 100 As , prov ⎤ ⎡100 As , prov = 1+ ⎢ /(3 + )⎥ ≤ 1.5 bd bd ⎣ ⎦ (4.5) Table 4.4-3 Tension reinforcement modification factors (Clause 3.4.6.6) 167 4.4.3* Deflection calculations (BS 8110: Part 2, clauses 3.7) (1) Background o The difficulties concerning the calculation of deflections of concrete members arise from the uncertainties regarding: Stiffness EI (cracked and uncracked sections) Effects of creep and shrinkage o In the elastic theory, the general deflection can be expressed as Δ = ∑∫ NN kv v MM ds + ∑ ∫ ds + ∑ ∫ ds EA GA EI o For a flexural member, such as a beam d2y M =− Deflection: 2 EI dx (Moment–area theorems express deflections in terms of the M/EI diagram.) Curvature: 1 M =− r EI ∴ d2y 1 = dx 2 r Curvature–area theorems express deflections in terms of the 1/r diagram. The curvature–area theorems can be used for deflection calculations even where the deflections are caused by other effects than bending, e.g. creep and shrinkage. o The method of calculation for deflections in BS 8110: Part 2 is based on the calculation of curvature of sections, with allowance for creep and shrinkage effects. * Optional course materials for CIVL 232 168 (2) Calculation of deflection from curvatures (Part 2, clause 3.7) o Deflection: a = Kl 2 1 rb (4.6) where K is constant which depends on the shape of the bending moment diagram in Table 4.4-4, l the effective span, and 1/ rb the curvature at midspan of a beam or at support of a cantilever. Table 4.4-4 Values of K for various bending moment diagrams 169 (3) Calculation of short-term curvatures o Cracked section MR 1 = rb E c I x (4.7) where MR – reduced moment (to be defined in Eq. 4.12) Ec – modulus of elasticity for concrete Ix – moment of inertia for transformed section about neutral axis For a cracked section, the stiffening effect of concrete in tension zone is taken into account by assuming that the concrete develops some stress in tension (tension stiffening) represented by a triangular stress distribution (Fig. 4.4-1) Fig. 4.4-1 Tension stiffening of cracked section In BS 8110, fct = 1 N/mm2 for short–term loading fct = 0.55 N/mm2 for long–term loading The force of the concrete in tension Fct = f ct (h − x )2 b 2(d − x ) (4.8) 170 The neutral axis depth can approximately be determined for the cracked section only using the transformed area method (see Fig. 4.4-2). In Fig. 4.4-2, the modular ratio α e = E s / Ec . Fig. 4.4-2 Cracked section − Taking moments about the neutral axis leads to 12 ′ bx + α e As (x − d ′) = α e As (d − x ) 2 (4.9) This is solved to give x (Fig. 4.2-3). − The moment of inertia of the transformed section about N.A. (axis x-x) is given (Fig. 4.2-4) by 1 2 2 ′ I x = bx 3 + α e As (x − d ′) + α e As (d − x ) 3 (4.10) − From Fig. 4.4-2, the moment of resistance of the concrete in tension M ct = Fct ⋅ 2(h − x ) / 3 , or M ct = f ct ⋅ (h − x )3 ⋅ b 3(d − x ) (4.11) − The reduced moment M R = M − M ct (4.12) 171 where M is the applied moment. determined by Eq. (4.7), i.e. The curvature is then MR 1 = . rb Ec I x Fig. 4.4-3 Neutral axis depth of cracked section (d’/d = 0.1) Fig. 4.4-4 Second moment of area of cracked section (d’/d = 0.1) 172 o Uncracked section o Fig. 4.4-5 Uncracked section Referring to the transformed section in Fig, 4.4-5(a), the equivalent area is ′ Ae = bh + α e ( As + As ) (4.13) The location of the centroid is found by taking moments of all areas about the top face and dividing by Ae , ′ bh 2 / 2 + α e ( As d ′ + As d ) x= Ae (4.14) The moment of inertia Ix about the neutral axis (x-x) is 2 bh 3 ⎛h ⎞ 2 2 ′ Ix = + bh⎜ − x ⎟ + α e As (d − x ) + α e As ( x − d ′) 12 ⎝2 ⎠ (4.15) The curvature is 1 M = rb E c I x (4.16) where M is the applied moment. 173 o The short-term curvature at a section can be calculated using assumptions set out an uncracked or cracked section. The larger value is used for the deflection calculations, i.e. ⎧⎛ 1 ⎞ 1 ⎪ , = max ⎨⎜ ⎟ rb ⎪⎜ rb ⎟ uncracked ⎩⎝ ⎠ ⎫ ⎛1⎞ ⎪ ⎜⎟ ⎬ ⎜r ⎟ ⎝ b ⎠ cracked ⎪ ⎭ (4) Calculation of long-term curvatures In calculation long–term curvatures it is necessary to take into account the effects of creep and shrinkage in addition to the reduced tensile resistance of the cracked concrete. o Creep (BS 8110: Part 2: Clause 7.3) Load on concrete causes an immediate elastic strain and a long-term time-dependent strain knows as creep. This is allowed for generally by reducing the effective modulus of elasticity of the concrete: E eff = Ec 1+ φ (4.17) − The creep coefficient φ is used to evaluate the effect of creep. − The values of φ depend on 1. age of loading 2. effective section thickness 3. ambient relative humidity 174 Fig. 4.4-6 Creep coefficients o Shrinkage (BS 8110: Part 2, Clause 7.4) Concrete shrines as it dries and hardens. The curvature can be caused by shrinkage. In general, more shrinkage will occur at the top because the steel area is less at top than at bottom (Fig. 4.4-7). Fig. 4.4-7 Concrete shrinkage Curvature due to shrinkage must be estimated and added to that due to applied moments, such that 175 1 ε csα e S s = rcs I (4.18) where εcs – free shrinkage strain αe – modular ratio Es / Eeff Ss – first moment of area of reinforcement about the centroid of the cracked or gross cross–section. Fig. 4.4-8 Drying shrinkage εcs o Total long-term curvature (BS 8110: Part 2, Clause 3.6) = long-term curvature under the permanent load + (short-term curvature under the total load − short-term curvature under the permanent load) + shrinkage curvature (4.19) where the permanent load = Gk + 0.25Qk, which is recommended in BS 8110: Part 2, clause 3.3. 176 (5) Procedure for deflection calculations 1. Moments γm = 1, and γf = 1 moments due to: the total load (normally Gk + Qk) the permanent load (Gk + 0.25Qk) 2. Short-term curvatures due to the total load and the permanent load, respectively for cracked section use the greater value for uncracked section 3. Long-term curvature due to the permanent load (considering creep effect) 4. Shrinkage curvature 5. Final curvature (total long-term curvature) using Eq. (4.19) 6. Total long-term deflection using Eq. (4.6): a = K l2 1 rb Examples in References Calculation of a deflection (Text Example 6.2, pp. 119-123) Deflection calculation for a T-beam (Reference by MacGinley and Choo, Example 6.2, pp. 137-144) 177 ...
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This note was uploaded on 12/24/2011 for the course CIVL 232 taught by Professor Jskuang during the Spring '06 term at HKUST.

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