Unformatted text preview: 6.3 Twoway Solid Slabs (ly / lx ≤ 2)
(Solid slabs spanning in two directions) 6.3.1 Failure behaviour
• Crack pattern at failure Lifting
(simplysupported slab) bottom face Fig. 6.31 Crack pattern at failure • Loading on supporting beams yield line Beam B Fig. 6.32 Load patterns on supporting beams 216 6.3.2 Simply supported slabs
• Maximum moments per unit width (1 m) at midspan: lx 1m 2
msx = α sx nl x (6.1) 2
msy = α sy nl x (6.2) where n = 1.4Gk + 1.6Qk; α sx = (l y / l x ) 4
8 [1 + (l y / l x ) 4 ] and α sy = (l y / l x ) 2
8 [1 + (l y / l x ) 4 ] Table 6.31 • Asx = msy
m sx
and Asy =
(per meter width), or
0.95 f y z
0.95 f y z use the design charts to determine Asx and Asy .
Text Example 8.5: Design of a simply supported twoway slab,
pp.194195.
Reinforcement details 217 6.3.3 Restrained slabs
• The slabs are divided in each direction into middle and edge strips as shown in the following figure. The maximum moments msx and
msy will be applied to the middle strips only. • The maximum design moments in directions of spans lx and ly are
2
msx = β sx nl x (6.3) 2
msy = β sy nl x (6.4) The moment coefficients βsx and βsy are given in Table 3.14 of the
code (see the table in page 219 of the Notes).
• The steel bars are uniformly spaced across the middle strips.
• At the discontinuous edge, top steel of onehalf the bottom steel at midspan should be provided for cracking control.
• In the edge strips, only nominal reinforcement (ρmin = 0.13% for highyield steel; ρmin = 0.24% for mild steel) is necessary, which is
to be provided together with torsion reinforcement specified below. 218 BS 8110: Part 1: 1997, Table 3.14 219 • Torsion reinforcement is required to be provided at discontinuous corners: such as corners X and Y shown in Fig. 6.33(a).
o This is to consist of top and bottom meshes, each having bars in both directions of spans and extending from the edges a
minimum distance of 1/5 of the shorter span (lx /5).
o At the corner X (discontinuous in both directions), the area of steel in each of the four layers = 3/4 of that required for
maximum midspan moment (0.75Asx).
o At the corner Y (discontinuous in one direction only), the area of steel = 1/2 of that required at the corner X (0.375Asx). (a) (b) (c)
Fig. 6.33 (a) Floor plan; (b) slab with two adjacent edges discontinuous; (c)
moment coefficients from the code
220 Text Example 8.6: Moments in a continuous twoway slab, p.197.
The panel considered is an edge panel, as shown in the figure and the
uniformly distributed load, n = (1.4Gk + 1.6Qk) = 10 kN/m2.
lx = 5.0 m Continuous panel
spanning in two
directions Support b d c Support ly = 6.0 m a Discontinuous
supported edge Support The moment coefficients are taken from case 3 of Table 3.14 of BS 8110
(p.219 of the Notes).
ly / lx = 6/5 = 1.2.
Positive moments at midspan, m sx = β sx nl x = 0.042 × 10 × 5 2 = 10.5 kNm in direction of l x
2 msy = β sy nl x = 0.028 × 10 × 5 2 = 7.0 kNm in direction of l y
2 Negative moments at supports,
2
Support ad, m x = 0.056 × 10 × 5 = 14 kNm 2
Supports ab and dc, m y = 0.037 × 10 × 5 = 9.3 kNm The moments calculated are for a meter width of the slab.
The design of reinforcement to resist these moments would follow the
usual procedure. Torsion reinforcement is required at corners b and c. A
check would also be required on the spaneffective depth ratio of the slab.
221 Example: Twoway restrained solid slab
1. Specification
ο The part floor plan for an office building is shown in the figure. It consists of restrained slabs poured monolithically with the edge
beams. The slab is 180 mm thick; total dead load = 6.2 kN/m2 and
imposed load = 2.5 kN/m2.
ο Design the corner slab using grade 35 concrete and grade 460 reinforcement. Show the reinforcement on sketches.
6m 6m
4.5m 6m 6m 0.75 m Y (a) (b) 2. Slab division, moments and reinforcement
ο The corner slab is divided into middle and edge strips shown in (a). The moment coefficients are taken from Table 6.33(c) (p.220) for
a square slab for the case with two adjacent edges discontinuous.
The values of the coefficients and locations of moments are shown
in (b). Design load = (1.4 × 6.2) + (1.6 × 2.5) = 12.68 kN/m2. 222 ο Assuming 10 mm diameter bars and 20 mm cover from Table 3.4 of the code, the effective depth of the outer layer is
180 – 20 – 5 = 155 mm
The effective depth of the inner layer is
180 – 20 – 5 – 10 = 145 mm
The moments and steel areas for the middle strips are calculated.
Because the slab is square, only one direction needs to be considered.
(i) Positions 1 and 4, d = 155 mm
m sx = −0.047 × 12 .68 × 6 2 = −21 .45 kNm
21.45 × 10 6
= 0.026 < 0.156
K=
35 × 1000 × 1552 ( ) z = 155 0.5 + 0.25 − 0.026 / 0.9 = 150.4 > 0.95d = 147.3 mm
As = 21.45 × 106
= 364 mm 2 / m
0.87 × 460 × 147.3 Provide [email protected] (As = 392 mm2/m).
(ii) Position 2, d = 145 mm. (Note that the smaller value of d is used.) msx = 0.036 × 12.68 × 62 = 16.43 kNm
Repeat the above calculations (or using the design charts) to give
As = 293 mm2/m
Provide [email protected] (As = 314 mm2/m).
Check the minimum area of steel in tension: ρmin = 0.13 × 1000 × 180 /100 = 234 mm 2 / m 223 (iii) Positions 3 and 5
As = 0.5 × 293 = 149 mm 2 /m < ρmin = 234 mm2/m Provide [email protected] (As = 251 mm2/m).
In detailing, the moment steel will not be curtailed because both
negative and positive steel would fail below the minimum area if
50% of the bars were cut off.
3. Shear forces and shear resistance
(i) Positions 1 and 4, d = 155 mm
Vsx = 0.4 × 12.68 × 6 = 30.43 kN / m
30.43 ×103
v=
= 0.196 N/mm 2
3
10 × 155
100 As 100 × 393
=
= 0.253
bd
1000 ×155
vc = 0.79(0.253)1/ 3 (400 /155)1/ 4 (32 / 25)1/ 3 /1.25 = 0.567 N/mm 2 No shear reinforcement is required.
(ii) Positions 3 and 5, d = 145 mm
The bottom tension bars are to be stopped at the centre of the
support. The shear resistance is based on the top steel with As =
251 mm2/m.
Vsx = 0.26 × 12.68 × 6 = 19.78 kN/m
v = 0.136 N/mm 2
vc = 0.49 N/mm 2
The shear stress is satisfactory. 224 4. Torsion steel
Torsion steel of length 6m/5 = 1.2 m is to be provided in the top and
bottom of the slab shown in figure (b) of the Notes p. 222 at the three
external corners marked X and Y.
(i) Corner X: The area of torsion steel is 0.75 × 293 = 220 mm2/m.
This is provided by the minimum steel of [email protected]
(ii) Corner Y: The area of torsion steel is 0.5 × 220 = 110 mm2/m.
This is provided by the minimum steel of [email protected]
5. Edge stripsProvide the minimum reinforcement, [email protected], in the edge strips.
6. Deflection
Check using steel at midspan with d = 145 mm.
Basic span/d ratio = 26
msx 16.43 × 106
5 × 460 × 293
=3
= 0.78 and f s =
= 268.3 N/mm 2
2
2
10 × 145
8 × 314
bd The modification factor is
0.55 + 477 − 268.3
= 1.59
120(0.9 + 0.78) Allowable span/d ratio = 1.59 × 26 = 41.34
Actual span/d ratio = 600/145 = 41.37
The slab can be considered to be just satisfactory. The deflection
could be based on the average value of d and the slab would be
satisfactory.
225 7. Cracking
The bar spacing does not exceed 3d = 3 × 145 = 435 mm; in addition, for
grade 460 steel the depth of the slab is less than 200 mm. No further
checks are required as stated in Clause 3.12.11.2 of the code.
8. Sketch of the slab
The arrangement of reinforcement is shown in following figure. The
top and bottom bars are shown separately for clarity.
ο The moment steel in the bottom of the slab is stopped at the support at the outside edges and lapped with steel in the next bays
at the continuous edges.
ο Secondary steel is provided in the top of the slab at the continuous edges to tie in the moment steel. Summary of Steel Arrangement
(i) Moment reinforcement
Positions 1, 4: [email protected] 2: [email protected] 3, 5: [email protected] (ii) Edge strip reinforcement:
[email protected] Y (iii) Torsion reinforcement
Corners X, Y: [email protected]
226 200 Bottom steel × Top steel 227 ...
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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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