Lecture_2

Lecture_2 - 1.2 Limit State Design (Design to BS 8110)...

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Unformatted text preview: 1.2 Limit State Design (Design to BS 8110) 1.2.1 Introduction • The design of engineering structures must ensure: o The structure is safe under the worst loading. o The deformation of the members does not detract from the appearance, durability or performance of the structure during normal working conditions. • Design purpose – to achieve acceptable probabilities that a structure will not become unfit for its intended use – i.e., that will not reach a limit state. • Two principal types of limit state: o Ultimate limit state − relating to safety: The whole structure or its components should not collapse, overturn or buckle when subjected to design loads. o Serviceability limit state - relating to satisfactory performance: The structure should not become unfit for use due to excessive deflection and cracking. o Other limit states that may be reached include excessive vibration, fire resistance, fatigue, etc. 1.2.2 Characteristic material strengths (fk) • Concrete strength — it is based on the mean cube crushing strength (cube strength) 7 P 150 150×150×150 cube used by BS 8110 and Hong 150 Kong codes 150 • Steel strength – it is based on the mean yield strength or a specified proof stress. σ0.2 σy 2% • Characteristic strength – Limit state design uses the concept of characteristic strength The strengths of materials upon which design is based are those strengths below which results are unlikely to fall. called as ‘characteristic’ strengths. These are This is given by f k = f m − 1.64s where fk - characteristic strength fm - mean strength s - standard deviation 8 Number of test results The formula (fk = fm – 1.64s) ensures that not more than 5% of test specimen results should fall below the characteristic value. concrete - cube strength ⎧ Test results for ⎨ ⎩steel reinforcement - yield strength • Characteristic strength of concrete (fcu) at 28 days o Normal concrete: f cu = 25 ~ 60 N / mm 2 (cube specimen) (grade 25, 30, 35, 40, 45, 50, 60, …) 2 When f cu ≥ 60 N / mm , the concrete may be considered as high-strength concrete. o In ACI code: concrete strength cylinder strength f c′ In practice, taking f c′ = 0.8 f cu 9 o Indirect tensile strength (Splitting tensile strength) For normal concrete: f t = ( 1 1 ~ ) f cu 10 13 • Characteristic strength of steel reinforcement (fy) o mild steel bars (R): f y = 250 N/mm 2 o high yield deformed bars (T or Y): f y = 460 N/mm 2 T – type 2 deformed bar (ribbed bars); most commonly used Y – type 1 deformed bar (square twisted bars) 1.2.3 Characteristic loads (FK) Characteristic load = Mean load ± 1.64 Standard deviations • Not more than 5% of cases will exceed the upper limit. Not more than 5% will fall below the lower limit. • In most cases, o it is the maximum loading on a structural member (+ 1.64s) o minimum value may apply when considering the stability of continuous members (− 1.64s) • The characteristic loads are the actual loads that structure is designed to carry. • In BS 8110: Gk = Characteristic dead load Qk = Characteristic imposed load (live load) Wk = Characteristic wind load 10 o Dead load Gk is the self-weight of the structure and the weight of finishes, ceilings, services and partitions. Density of reinforced concrete: 24 kN/m3 (2400 kg/m3) used in HK, UK, USA, etc. 25 kN/m3 (2500 kg/m3) used in China o Imposed load (live load) Qk is caused by people, furniture, equipment, etc. Imposed loads for various types of buildings are given by design codes, such as Hong Kong code − refer to ‘Building (Construction) Regulations, 1990’ CAP. 123 − Imposed loads ‘Code of practice for dead and imposed loads for buildings’ British code − ‘Loading for Buildings, BS 6399: 1984’ Chinese code − refer to 建築結構荷載規範 GB 50009-2001 • Typical live load values Type of Use Apartment buildings: Private units Public rooms Corridors Office buildings: Offices Lobbies Corridors above first floor Garages (cars only) Stores: First floor Upper floors Warehouse: Light storage Heavy storage Minimum uniformly distributed live loads (kPa = kN/m2) 1.92 4.80 3.84 2.40 4.80 3.84 2.40 4.80 3.60 6.00 12.00 11 Comparison: Minimum live load (kN/m2) Case Floor area usage HK Britain China 1. Domestic building 2. Office for general use 3. Shop floor for the display and sale of merchandise 4. Car park (vehicles not exceeding 2500 kg gross mass) 2.5 3.0 5.0 1.5 2.5 4.0 1.5 2.5 3.5 4.0 2.5 4.0 The imposed loads used in Hong Kong are generally heavier than those used in other national standards. This tends to give more conservative designs. 1.2.4 Design strength Deasign strength = Characteristic strength ( f k ) Partial factor of safety (γ m ) γm (partial factor of safety for materials) takes account of : 1. uncertainties in strength 2. uncertainties in accuracy of the methods of analysis Table 1.1 Partial factors of safety applied to materials (γm) Material Ultimate limit state Concrete Flexure Shear Bond Serviceability 1.5 1.25 1.4 1.0 Steel 1.05 (Hong Kong:1.15) 1.05 (Hong Kong:1.15) 1.0 12 • Design strength for concrete: • Design strength for steel: fy γm f cu ( γm ( f cu = 0.67 f cu ) 1 .5 fy 1.05 = 0.95 f y ) (In Hong Kong, γm for steel = 1.15; then fy / 1.15 = 0.87fy ) 1.2.5 Design loads Design load = Characteristic load (Fk) × Partial factor of safety (γf) For example: design dead load = GK⋅γf design live load = QK⋅γf Table 1.2 Partial factors of safety for loads γf γf (partial factor of safety for loads) takes account of : 1. Possible unusual load increase 2. Design assumptions and inaccuracy of calculation 3. Unforeseen stress redistributions 4. Constructional inaccuracies 13 1.2.6 Global factor of safety γm ×γ f Example: a global factor of safety for structural concrete (γm = 1.5) with a safety factor 1.4 (γf ) is 1.5 × 1.4 = 2.1. Example 1.2.1 Determine the cross-sectional area of a mild steel cable which supports a total dead load of 3.0 kN and a live load of 2.0 kN. Carry out the calculations using rG = 1.4, rQ = 1.6 and γm = 1.05 for the steel strength. Solution (limit state method): (1) Design load = γG × dead load + γQ × live load = 1.4 × 3.0 + 1.6 × 2.0 = 7.4 kN (2) Design stress = fy γm = 250 = 238 N/mm 2 1.05 (3) Required cross-sectional area 7.4 × 103 design load = = = 31 mm 2 238 design stress 14 Example 1.2.2 A beam supported on foundations at A and B. The loads supported by the beam are its dead weight of 20 kN/m and a 170 kN live load at C. Determine the weight of foundation required at A in order to resist uplift. foundation Loading arrangement for uplift at A at the ultimate limit state Solution: The arrangement of loads for the maximum uplift at A is shown in (b). (1) Design dead load over BC = γG × 20 × 2 = 1.4 × 20 × 2 = 56 kN (2) Design dead load over AB = γG × 20 × 6 = 1.0 × 20 × 6 = 120 kN (3) Design live load = γQ × 170 = 1.6 × 170 = 272 kN (4) Taking moments about B for the ultimate loads Uplift RA = ( 272 × 2 + 56 × 1 − 120 × 3) / 6.0 = 40 kN ∴ The weight of foundation required at A is 40 kN. 15 ...
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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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