Lecture_3 - 1.3 Analysis of the Structure 1.3.1 Load...

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Unformatted text preview: 1.3 Analysis of the Structure 1.3.1 Load combinations (1) Load combinations for the ultimate limit state The arrangement of loads should be such as to cause the most severe effects, i.e. the most severe stresses. The design ultimate load should be obtained as follows: 1. Dead load + imposed load 1.4 Gk + 1.6 Qk Load combination 1 should also be associated with a minimum design dead load of 1.0 Gk applied to such parts of the structure as will give the most unfavourable condition. 2. Dead load + wind load 1.0 Gk + 1.4 Wk or 1.4 Gk + 1.4 Wk 3. Dead load + imposed load + wind load 1.2 Gk + 1.2 Qk + 1.2 Wk • For Load Combination 1, a three-span continuous beam would have the loading arrangement shown in Fig. 1.3-1, in order to cause the maximum sagging moment in the outer spans and the maximum possible hogging moment in the centre span. A study of the deflected shape of the beam would confirm this to be the case. 16 Fig. 1.3-1 Three-span beam • Fig. 1.3-2 shows the arrangements of vertical loading on a multi-span continuous beam to cause: (i) maximum sagging moments in alternate spans and maximum possible hogging moments in adjacent spans, and (ii) maximum hogging moments at support A. Fig. 1.3-2 Load arrangements of a multi-span beam 17 • Load arrangement to be considered for stability of a structure Fig. 1.3-3 Load combination: dead plus wind loads The combination shown in the figure gives a critical condition. (2) Load combination for the serviceability limit state In general, a partial factor of safety γf = 1.0 is usually applied to all load combinations at the serviceability limit state. 1.3.2 Analysis of beams (1) Non-continuous beams: For the ultimate limit state, it only needs to consider the maximum load of 1.4Gk +1.6Qk on the span. Example: The one-span simply supported beam shown in Fig. 1.3-4 carries a distributed dead load including self-weight of 25 kN/m, a permanent concentrated partition load of 30 kN at mid-span, and a distributed imposed load of 10 kN/m. Fig. 1.3-4 shows the values of the ultimate load required in calculations of the bending moments and shear forces: 18 Maximum bending moment = Maximum shear force = 42 × 4 204 × 4 + = 144 kNm 4 8 42 204 = 123 kN + 2 2 The analysis is completed by drawing the shear-force and bending-moment diagrams, which would later be used in the design and detailing of the bending and shear reinforcement. Fig. 1.3-4 Analysis of one-span simply supported beam (2) Continuous beams A continuous beam should be analysed for critical loading arrangements, as shown in Fig. 1.3-2, which give the maximum stresses at each cross-section. 19 • Analysis of continuous beam The continuous beam (Fig. 1.3-5) has a constant cross-section and supports a uniformly distributed dead load including its self-weight of Gk = 25 kN/m and an imposed load Qk = 10 kN/m. Fig. 1.3-5 Continuous beam loading arrangement The critical load arrangements for the ultimate limit state are shown in Fig. 1.3-5 where the ‘stars’ indicate the region of maximum moments, sagging or possible hogging. Bending moments: Fig. 1.3-6 Bending moment diagrams for different load arrangements (kNm) 20 Shear force diagrams: Fig. 1.3-7 Shear force diagrams for different load arrangements (kN) Bending moment and shear force envelopes: Fig. 1.3-8 Bending moment and shear force envelopes 21 • Bending moments and shear forces of uniformly-loaded continuous beams with approximately equal spans: ο The ultimate bending moments and shear forces in continuous beams with three or more approximately equal spans can be obtained from BS 8110, Part 1: Clause 3.4.3, provided that (i) The spans differ by no more that 15% of the longest one; (ii) Loads should be substantially uniformly distributed over three or more spans; and (iii) Characteristic imposed load Qk may not exceed characteristic dead load Gk. ο Design ultimate bending moments and shear forces (BS 8110, Part 1: Clause 3.4.3: Table 3.5) 22 ο The values from BS 8110 are shown in diagrammatic form in Fig. 1.3-9 for beams (equivalent simplified values for slabs are given in Chapter 6). Fig. 1.3-9 Bending-moment and shear-force coefficients for beams • The possibility of hogging moments in any of the spans should not be ignored, even if it is not indicated by these coefficients. For example, a beam of three equal spans will have a hogging moment in the centre span if Qk exceeds Gk /16. 23 ...
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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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