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Unformatted text preview: 1.3.3 Analysis of frames supporting vertical and lateral loads
• Building frames ⎧Braced frames
two types: ⎨
⎩ Unbraced frames o A braced frame is designed to resist vertical loads only (such as
dead and imposed loads).
o An unbraced frame is designed to resist both vertical and
horizontal (such as wind/earthquake) loads. diaphragm (c) Fig. 1.310 Multistorey building. (a) Plan; (b) rigid transverse frame;
(c) side elevation. • In most of tall buildings, structural walls (shear walls) and cores
provide stability and resistance to lateral loads. In general,
⎧Shear walls and cores are considered to carry lateral loads mainly.
⎨
⎩frames are considered to carry vertical loads only (braced frames). 24 • Subframes for braced frame analysis (BS 8110:Part 1:cl. 3.2.1.2)
For a braced frame, three methods of simplification may be used. Fig. 1.311 Braced frame 1. Subframe type I (Clause 3.2.1.2.1) Fig. 1.312 Subframe type I Analysis of this subframe will give the bending moments and
shear forces in the beams and columns for the floor considered.
The load effects from upper and lower storeys are neglected.
Critical load arrangements (Clause 3.2.1.2.2):
Case I − the maximum design ultimate load (1.4Gk + 1.6Qk) on alternate spans and minimum design ultimate
load (1.0Gk) on all other spans;
Case II − the maximum design ultimate load (1.4Gk + 1.6Qk) on all spans.
25 2. Subframe type II (Clause 3.2.1.2.3) (a) Subframe: for Beam AB (b) Subframe: for Beam BC Fig. 1.313 Subframe type II
Analysis of this subframe will give the bending moments
and shear forces in the beam AB using Fig. 1.313(a) or
beam BC using Fig. 1.313(b).
The moments in an individual column may be found:
(1) For column at A, using Fig. 1.313(a).
(2) For column at B, using Fig. 1.313(a) when AB > BC;
using Fig.1.313(b) when BC > AB.
Critical load arrangements:
Same as those for Subframes type I.
3. Continuousbeam simplification (Clause 3.2.1.2.4) (a) Analogous continuous beam (b) Column moments Fig. 1.314 Continuousbeam simplification 26 This alternative is a more conservative to the subframes types I
and II.
The bending moments and shear forces in the beams are
obtained by considering the beams as a continuous beam
shown in Fig. 1.314(a).
Column moments are calculated form the model in Fig.
1.314(b). The longer of the beams AB and BC would carry the load (1.4Gk + 1.6Qk) and the shorter would carry
the load 1.0Gk . (see text Example 3.4)
Critical load arrangements are the same as those for
subframes type I.
• Simplified analysis for lateral loads
o Portal method − useful for frames of up to 25 storeys with a
heighttowidth ratio not greater than 4.
o Cantilever method − useful for frames of up to 35 storeys with a
heighttowidth ratio of up to 5. (see text Example 3.5)
• BS 8110 specifies that all building structures:
o should be capable of resisting a notional horizontal design
ultimate load equal to 1.5% of the characteristic dead weight of
the structure; and
o this load is to be distributed proportionally to act as point loads
at each floor and roof level.
27 1.3.4 Shear wall structures resisting horizontal loads
(Floor slabs  rigid floor diaphragm) Resultant horizontal
force F Fig. 1.315 Shear wall structure • Symmetrical arrangement of walls
o Relative flexural stiffness of each wall k i = bh 3 (Flexural stiffness: EI = E bh3
)
12 where b is the thickness of the wall and h the length of the wall.
o The force distributed into each wall
Rigid floor diaphragm assumption: Rigid diaphragm action of floor slabs. Pi = F ki
n ∑k
i =1 i where F is the total horizontal force.
o Example of symmetrical arrangement of shear walls
A structure with a symmetrical arrangement of shear walls is
shown in the figure. Calculate the proportion of the 100 kN
horizontal load carried by each of the walls. 28 Solution:
(1) Relative flexural stiffness
Wall A: kA = 0.3 × 203 = 2400
Wall B: kB = 0.2 × 123 = 346
Σk = 2(2400 + 346) = 5492
(2) Force in each wall
PA = F k A / ∑ k = 100 × 2400/5492 = 43.7 kN
PB = F k B / ∑ k = 100 × 346/5492 = 6.3 kN Check: 2 × (43.7+6.3) = 100 kN
• Unsymmetrical arrangement of walls 29 With an unsymmetrical arrangement of shear walls, as shown in
the above figure, there will also be a torsional force on the structure
about the centre of rotation in addition to the direct forces caused
by the translational movement.
o The calculation procedure for this case is:
(1) Determine the location of the centre of rotation by taking
moments of the wall flexural stiffness k about convenient
axes, such that
x= ∑ (k x)
∑k
x and y= x ∑ (k y )
∑k
y y where kx and ky are the stiffnesses of the walls orientated in
the x and y directions, respectively.
(2) Calculate the torsional moment Mt on the group of shear
walls as
Mt = F⋅e
(3) Calculate the force Pi in each wall as the sum of the direct
component Pd and the torsional rotation component Pr Pi = Pd + Pr = F × k i ri
kx
± Mt ×
2
∑ kx
∑ (k i ri ) where ri is the perpendicular distance between the axis of
each wall and the centre of rotation. 30 • Example of unsymmetrical layout of shear walls
(The structure has been shown above.)
Determine the distribution of the 100 kN horizontal force F into the
shear walls A, B, C, D and E. The relative stiffness of each shear wall has been shown in the figure in terms of multiples of k.
Solution:
(1) Centre of rotation ( x , y ) ∑k x = 20 + 5 + 5 = 30 ; Taking moments for kx about yy at wall A,
x= ∑ (k x) = 20 × 0 + 5 × 32 + 5 × 40 = 12.0 m .
30
∑k
x x ∑k y = 6 + 4 = 10 ; Taking moments for ky about xx at wall C
y= ∑ (k y) = 6 × 0 + 4 × 16 = 6.4 m
10
∑k
Y y (2) The torsional moment is
Mt = F × (20 − x ) = 100× (20 − 12) = 800 kNm (3) The remainder of these calculations are conveniently set out in
the following tabular form: 31 As an example for wall A: PA = Pt + Pr = F × k xA
k xA rA
− Mt ×
2
∑ k xA
∑ (ki ri ) 20
20 × 12
− 800 ×
30
9415
= 66.6 − 20.4
= 46.2 kN
= 100 × 32 ...
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 Spring '06
 JSKUANG

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