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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 threespan continuous beam would
have the loading arrangement shown in Fig. 1.31, 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. 1.31 Threespan beam
• Fig. 1.32 shows the arrangements of vertical loading on a
multispan 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. 1.32 Load arrangements of a multispan beam
17 • Load arrangement to be considered for stability of a structure Fig. 1.33 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) Noncontinuous beams: For the ultimate limit state, it only needs
to consider the maximum load of 1.4Gk +1.6Qk on the span.
Example:
The onespan simply supported beam shown in Fig. 1.34 carries a
distributed dead load including selfweight of 25 kN/m, a
permanent concentrated partition load of 30 kN at midspan, and a
distributed imposed load of 10 kN/m. Fig. 1.34 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 shearforce and
bendingmoment diagrams, which would later be used in the
design and detailing of the bending and shear reinforcement. Fig. 1.34 Analysis of onespan simply supported beam (2) Continuous beams
A continuous beam should be analysed for critical loading
arrangements, as shown in Fig. 1.32, which give the maximum
stresses at each crosssection. 19 • Analysis of continuous beam The continuous beam (Fig. 1.35) has a constant crosssection and
supports a uniformly distributed dead load including its selfweight
of Gk = 25 kN/m and an imposed load Qk = 10 kN/m. Fig. 1.35 Continuous beam loading arrangement The critical load arrangements for the ultimate limit state are
shown in Fig. 1.35 where the ‘stars’ indicate the region of
maximum moments, sagging or possible hogging.
Bending moments: Fig. 1.36 Bending moment diagrams for different load arrangements (kNm)
20 Shear force diagrams: Fig. 1.37 Shear force diagrams for different load arrangements (kN) Bending moment and shear force envelopes: Fig. 1.38 Bending moment and shear force envelopes 21 • Bending moments and shear forces of uniformlyloaded
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. 1.39 for beams (equivalent simplified values for
slabs are given in Chapter 6). Fig. 1.39 Bendingmoment and shearforce 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.
 Spring '06
 JSKUANG

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