Ch2_2W&rib_Slab_HKC

CSE3032 STRUCTURAL & TALL BUILDING

Chapter 2 1

HD in Civil Engineering

│CHAPTER 2│

Design of R.C. Slabs

Learning Objectives

Acquire the knowledge of the requirements of the new HK Code in designing RC slabs.

Appreciate the structural actions of different types of slab elements.

Apply HK Code to design 2-way slab and rib slab (fcu ≤ 40 N/mm2, fy = 460 N/mm2 and fyv = 250 N/mm2)


Chapter 2 2


2. R.C. Slabs 2.1 Types of Slab Slabs are plate elements forming floors and roofs in buildings which normally carry uniformly distributed loads. Slabs may be simply supported or continuous over supports. Slabs can be classified as:

• One-way spanning slabs supported by beams or walls. The slabs can be simply supported or continuous as shown in Figure 2.1(a) & (b). In drawing, one-way spanning slab is shown as . Figure 2.2 shows a typical one-way spanning slab.

• Two-way spanning slab supported by beams or walls. Two-way

spanning slabs can be simply supported or continuous as shown in Figure

2.3. In drawing, two-way spanning slab is shown as .

• Ribbed is shown in Figure 2.4.

• Waffle slabs are shown in Figure 2.5, Photo 2.1, 2.2 & 2.3.

• Flat slabs carried on columns and edge beams or walls with no interior beams (see Figure 2.6, Figure 2.7, Photo 2.4 and 2.5).

hd

Span L

Distribution SteelMoment Steel

Moment Steel Distribution Steel

Span LBeam

(a)

(b) Figure 2.1 – (a) Simply Supported Slab (b) Continuous One-Way Slab


Chapter 2 3


Figure 2.2 – One-way Spanning Slab6

Figure 2.3 – Two-Way Spanning Slab6

Figure 2.4 – Ribbed Slab6


Chapter 2 4


Figure 2.5 – Waffle Slab Designed as a Flat Slab6

Figure 2.6 – Flat Slab without Drop

Figure 2.7 – Flat Slab with Drop and Column Head6


Chapter 2 5


Photo 2.1 – Waffle Slab



Chapter 2 6



Photo 2.4 – Flat Slab


Chapter 2 7


Photo 2.5 – Flat Slab


Chapter 2 8


2.2 Design of Two-way Spanning Solid Slab When floor slabs are supported on four sides two-way spanning action occurs as shown in Figure 2.8. In square slabs, the action is equal in each direction. In narrow slabs where the length is greater than twice the breath the action is effectively one way. However the end beams always carry some slab load.

Edge Beam

Strips spanningbetween supports

(a) Two-way action

Strip spans acrossshort span

l

(l > 2 l )

(b) One-way action

x

y

l

y x

Figure 2.8 – Two-way and One-way Action

Two-way slabs may be classified according to the edge conditions. They can be defined as follows:-

• Simply supported one panel slabs where the corners can lift away from the supports.

• A one panel slab held down on four sides by integral edge beams (the stiffness of the edge beam affects the slab design)

• Slabs with all edges continuous over supports • A slab with one, two or three edges continuous over supports - the

discontinuous edge(s) may be simply supported or held down by integral edge beams (see Figure 2.9)


Chapter 2 9


C

C

C

C

C

C

C

CC

C

C

C C

D D

D

D

D

D

D

DDD

D

D

D

DC - Continuous EdgeD - Discontinuous Edge

Figure 2.9 – Edge Conditions

2.2.1 Two-way Spanning Simply Supported Slabs The design of simply supported slabs may be made in accordance with HK Code, Cl.6.1.3.3. This cl. gives the following equations for max. moment msx and msy at mid-span on strips of unit width for spans lx and ly respectively, where lx is the length of shorter span, ly is the length of the longer span, n = 1.4gk + 1.6qk per unit area and αsx, αsy are the moment coeff. from Table 6.55 of the code. msx = αsxn lx

2, msy = αsyn lx2

Table 6.5 Bending moment coefficients for slabs spanning in two directions at right angles, simply supported on four sides ly / lx

1.0 1.1 1.2 1.3 1.4 1.5 1.75 2.0

αsx

0.062 0.074 0.084 0.093 0.099 0.104 0.113 0.118

αsy

0.062 0.061 0.059 0.055 0.051 0.046 0.037 0.029


Chapter 2 10


The centre strips and location of max. moments are shown in Figure 2.10

Load on beam A

Load on beam BB

A

1m

1m

Load causing shear

ly1m

1m

lx

msy

msx

(a)

(b)

Figure 2.10 – (a) Centre Strips (b) Loads on Beams and Slab Shears

2.2.1.1 Design Requirements for Two-way Slabs

Bending Reinforcement (For fcu ≤ 45 N/mm2, Normally K ≤ 0.156)

⎥⎦

⎤⎢⎣

⎡−+=

9.025.05.0 ) Kdzii (z ≤ 0.95d and z ≥ 0.775d)

zfMAiii

ys 87.0

) =

The min. area of main reinf. is:- For rectangular sections and solid slabs:-

Mild steel, fy = 250 N/mm2, 24.0100=

c

s

AA

High yield steel, fy = 460 N/mm2, 13.0100=

c

s

AA

156.0 ) 2 ≤=cufbd

MKi


Chapter 2 11


Distribution Steel The distribution steel runs at right angles to the main moment steel and serves the purpose of tying the slab together. The area of distribution steel is the same as the min. area for main reinforcement (i.e. 0.13%) and the distribution reinforcement of not less than 20% of the principal reinforcement should be provided.

Shear Reinforcement (For fcu = 40 N/mm2) Under normal loads shear stresses are not critical and shear reinforcement is normally not required. Shear reinforcement is provided in heavily loaded thick slabs but should not be used in slabs less than 200 mm thick. In design, the average shear stress v is given by v = V/bd The design procedures for slab are essentially the same as that for beams. However, in the design of slab in shear, the following points should be noted. i) v should not exceed cuf8.0 or 7 N/mm2 whichever is less. ii) If v < vc, then no shear reinforcement is required. iii) If vc < v ≤ (vc + 0.4), then provide min. links as:-

yvv

sv

fb

sA

87.04.0

≥

iv) if v > (vc + 0.4), provide shear links as

( )yv

c

v

sv

fbvv

sA

87.0−

≥


Chapter 2 12


Deflection In slab design, deflections are usually controlled by limiting the span/effective depth ratio. i) Basic span/effective depth ratio (Table 7.3):- Cantilever slab 7 Simply supported slabs 20 Continuous slabs 26 End Span 23 * For deflection control, it is necessary to check the span/effective depth

ratio of the shorter span. i) The basic span/effective depth ratio is multiplied by a modification factor

for tension reinforcement (Table 7.4). Only the tension steel at the centre of span is taken into account.

ii) Determination of the allowable span/effective depth ratio.

Allowable l/d ratio = Basic l/d ratio * M.F for tension reinforcement. iii) Determination of the actual span/effective depth ratio.

Actual l/d ratio = (Actual span) / (Effective depth)

iv) If allowable l/d ratio ≥ actual l/d ratio, ⇒ Deflection O.K. v) If allowable l/d ratio < actual l/d ratio, ⇒ Deflection NOT O.K.

(Redesign the slab)

Crack Control To control cracking in slabs, max. values for clear spacing between bars are set out in HK Code, Cl.9.3.1.1 (b). The maximum spacing of bars should comply with the following requirements:

For the principal reinforcement, 3h ≤ 400 mm; and For the secondary reinforcement, 3.5h ≤ 450 mm.

In areas with concentrated loads or areas of maximum moment those provisions become respectively:

For the principal reinforcement, 2h ≤ 250 mm; and For the secondary reinforcement, 3h ≤ 400 mm.


Chapter 2 13


In addition, unless crack widths are checked by direct calculation, the following rules will ensure adequate control of cracking for slabs subjected to normal internal and external environments: No further check is required on bar spacing if either:

h ≤ 250 mm (grade 250 steel) h ≤ 200 mm (grade 460 steel) or the percentage of required tension reinforcement (100As / bd) is less

than 0.3%. Where none of these three conditions apply, the bar spacings should be limited to the values calculated in cl. 9.2.1.4 which will NOT be covered here.


Chapter 2 14


Example 2.1 – Two-Way Spanning Simply Supported Slab Figure 2.11 shows a simply supported slab 220 mm thick and spanning in two directions. The effective span in each direction of a two-way spanning slab is 4.5 m and 6.3 m. The slab supports a live load of 10 kN/m2. Design the bending and shear reinforcement for the slab. Check the adequacy of the slab section against deflection by span-effective depth ratio approach. The characteristic material strengths are fcu = 35 N/mm2 and fy = 460 N/mm2. Nominal cover = 25 mm.

6.3m4.

5m

(220)

Figure 2.11 Solution ly / lx = 6.3/4.5 = 1.4 From design table 6.5, αsx = 0.099 and αsy = 0.051 Self-weight of slab = 0.22*24 = 5.28 kN/m2 Imposed load = 10 kN/m2 Design load n = 1.4*5.28 + 1.6*10 = 23.4 kN/m2


Chapter 2 15


Bending (Short Span) msx = αsx*n*lx

2 = 0.099*23.4*4.52 = 46.9 kNm d = 220 -25-12/2 = 189 mm

0.156 < 038.035*189*10

10*9.4623

6

2 ==cu

sx

fbdm

From design table 1, z = 0.95d = 0.95*189 = 179 mm.

26

655179*460*87.0

10*9.4687.0

mmzf

MAy

s ===

Provide T12 - 150 c/c (As = 754 mm2)

O.K. 4.0<

0.13 > 34.0220*10754*100100

3

∴

==bh

As

Bending (Long Span) msy = αsy*n*lx

2 = 0.051*23.4*4.52 = 24.2 kNm d = 220 -25-12 - 10/2 = 178 mm

0.156 < 022.035*178*10

10*2.2423

6

2 ==cu

sy

fbdm


26

358169*460*87.0

10*2.2487.0

mmzf

MAy

s ===

Provide T10 - 200 c/c (As = 393 mm2)

O.K. 4.0<

0.13 > 18.0220*10393*100100

3

∴

==bh

As


Chapter 2 16


Deflection (Check Short Span Only) Basic l/d ratio = 20 (table 7.3)

31.1189*1010*9.46

23

6

2 ==bdM sx

2

prov

req N/mm 266754655*460*

32

32

===s

sys A

Aff

From design table 7.4, modification factor = 1.35 Allowable l/d ratio = 20*1.35 = 27.0 Actual l/d ratio = 4500 / 189 = 23.8 < allowable ratio = 27.0 ∴ O.K. Shear (Most Critical in Long Span Direction) Max. shear force per 1 m width slab V= 23.4*4.5/2 = 52.7 kN Shear stress v = 52.7*103 / (103*178) = 0.30 N/mm2

Span) (Long 22.0178*1000393*100100

==bd

As

From “ design table 6.3”, vc = 0.47 31

2535

⎟⎠⎞

⎜⎝⎛x N/mm2 > 0.30 N/mm2

⇒ No shear reinforcement required.


Chapter 2 17


2.2.2 Restrained Two-way Spanning Solid Slabs2 2.2.2.1 Design and Arrangement of Reinforcement The design method for restrained slabs is given in HK Code Cl.6.1.3.3(c). In these slabs the corners are prevented from lifting and provision is made for torsion. The maximum moments at mid-span on strips of unit width for spans lx and ly are given by: 2

xsxsx lnm ⋅⋅= β 2

xsysy lnm ⋅⋅= β The moment coefficients βsx and βsy in the equations above are given in HK Code Table 6.65. The clause states that these equations may be used for continuous slabs when the following provisions are satisfied.

• The characteristic dead and imposed loads are approximately the same on adjacent panels as on the panel being considered.

• The spans of adjacent panels in the direction perpendicular to the line of

the common support are approximately the same as that of the panel considered in that direction.

middle strip

Edge strip

Edge strip

Edge

strip

Edge

strip

lyly8

ly8

xl

xl 8xl 8

Figure 2.12 – Division of Middle and Edge Strips


Chapter 2 18


The design rules for restrained TWO-WAY slabs are as follows: • The slabs are divided in each direction into middle and edge strips as

shown in Figure 2.12. • The maximum moments defined above applied to the middle strip. The

bars are spaced uniformly across the middle strip.

• At the discontinuous edge, it is a usual practice to have the top steel equals to one-half the area of the bottom steel at mid-span to be provided to control cracking.

• The minimum tension reinforcement is to be provided in the edge strip

together with the torsion reinforcement specified below.

• Torsion reinforcement is to be provided at corners where the slab is simply supported on both edges meeting at the corners. Corner X and Y shown in Figure 2.13 require torsion reinforcement. This consists of a top and bottom mesh with bars parallel to the sides of the slab and extending from the edges a distance of ONE-FIFTH of the shorter span. The area of bars in each of the four layers should be, at X, THREE-QUARTERS of the area of bars required for the maximum mid-span moment and, at Y, ONE-HALF of the area of the bars required at corner X. Note that no torsion reinforcement is required at the internal corners Z.

X

Y

Y

Z

Figure 2.13 – Torsional Reinforcement for Two-Way Slabs


Chapter 2 19


Table 6.6 – βsx and βsy from HK Code5


Chapter 2 20


2.2.2.2 Adjacent Panels with Markedly Different Support Moments. The moment coefficients in Table 6.65 of the code apply to slabs with similar spans and loads giving similar support moments. If the support moments for adjacent panels differ significantly, adjustment by means of moment distribution is required. The detailed procedures are given in cl. 6.1.3.3 (e) of the code. An example can be found in ref. 3. Detailed description will not be given here. 2.2.2.3 Loads on Supporting Beams of Two-way Slabs Shear force coefficients βvx and βvy for various support cases for continuous slabs are given in Table 6.75 of the code. The design loads on supporting beams per unit width are given by: xvxsx lnV ⋅⋅= β xvysy lnV ⋅⋅= β The load distribution on a beam supporting a two-way slab is shown in Figure 2.14.

V kN/ms

0.75l

l

Figure 2.14 – Distribution of Load on a Beam Supporting a Two-Way Spanning Slab

2.2.2.4 Shear Resistance, Deflection & Cracking Similar to those of simply supported two-way spanning slabs.


Chapter 2 21


Table 6.7 –βvx and βvy from HK Code5


Chapter 2 22


Example 2.2 – Continuous Two-Way Spanning Slabs The part floor plan for an office building is shown in Figure 2.15. It is a monolithic construction and the slabs are restrained by the integral beams. The slab is 175 mm thick and is subjected to the following loading. Characteristic D.L. Characteristic L.L.

Slab self-weight L.L. on slab = 3 kN/m2 Finishes = 25 mm thick screed (Unit weight of screed = 20 kN/m3)

Partition load = 1 kN/m2 Services = 0.5 kN/m2 Ceiling load = 0.5 kN/m2 Design the corner slab using grade 35 concrete and grade 460 reinforcement. Nominal cover = 25 mm.

6m 6m

5m5m

Figure 2.15 - Part Plan of an Office Building


Chapter 2 23


Solution D.L. Slab self-weight = 0.175*24 = 4.2 kN/m2 Finishes = 0.025*20 = 0.5 kN/m2 Partition = 1 kN/m2 Services = 0.5 kN/m2 Ceiling load = 0.5 kN/m2 Total =6.7 kN/m2 L.L. Slab = 3 kN/m2 Design Load n = 1.4*6.7 + 1.6*3 = 14.2 kN/m2

2.156==

x

y

ll

From table 6.6, the moment coefficients are: Negative moment at continuous edge βsx = 0.063, βsy = 0.045 Positive moment at mid-span βsx = 0.047, βsy = 0.034 The corner slab is divided into middle strips and edge strips as shown below.

XY

1

2

3

4

5

6 (-0.045)

(-0.

063)

(0.034)

(0.0

47)

6m0.75m 4.5m 0.75m

0.62

5m0.

625m

3.75

m

Y


Chapter 2 24


Effective depth of short and long spans: Short span direction: d1 = 175 – 25 –10/2 = 145 mm Long span direction: d2 = 175 – 25 – 10 – 10/2 = 135 mm Bending in Short Span Direction (lx): Positive moment (position 3):

msx = βsx*n*lx2 = 0.047*14.2*52 = 16.7 kNm

156.0023.035*145*10

10*7.1623

6

2 <==cu

sx

fbdm

From design table 1 z = 0.95d = 0.95*145 = 137 mm

mzf

MAy

s /mm 305137*460*87.0

10*7.1687.0

26

===

Provide T10-225 c/c Bottom (As = 349 mm2)

O.K. 4.0

0.13 20.0175*1000349*100100

∴<

>==bh

As

Negative moment (position 1):

msx = βsx*n*lx2 = -0.063*14.2*52 = -22.4 kNm

156.0030.035*145*10

10*4.2223

6

2 <==cu

sx

fbdm

From design table 1 z = 0.95d = 0.95*145 = 137 mm

mzf

MAy

s /mm 409137*460*87.0

10*4.2287.0

26

===

Provide T10-175 c/c Top (As = 449 mm2)

O.K. 4.0

0.13 26.0175*1000449*100100

∴<

>==bh

As


Chapter 2 25


Bending in Long Span Direction (ly): Positive moment (position 2):

msy = βsy*n*lx2 = 0.034*14.2*52 = 12.1 kNm

156.0019.035*135*10

10*1.1223

6

2 <==cu

sy

fbdm

From design table 1, z = 0.95d = 0.95*135 = 128 mm

mzf

MAy

s /mm 236128*460*87.0

10*1.1287.0

26

===

Provide T10-325 c/c Bottom (As = 242 mm2)

O.K. 4.0

0.13 14.0175*1000242*100100

∴<

>==bh

As

Negative moment (position 4):

msy = βsy*n*lx2 = -0.045*14.2*52 = -16.0 kNm

156.0025.035*135*10

10*1623

6

2 <==cu

sy

fbdm

From design table 1, z = 0.95d = 0.95*135 = 128 mm

mzf

MAy

s /mm 312128*460*87.0

10*1687.0

26

===

Provide T10-250 c/c Top (As = 314 mm2)

O.K. 4.0

0.13 18.0175*1000314*100100

∴<

>==bh

As

Bending reinforcement at positions 5 & 6 A minimum of 50% bending reinforcement at the mid-span of position 3 and position 2 is required for the top reinforcement at position 5 and position 6 respectively. However this amount of reinforcement is less than 0.13%, T10 – 325 c/c top is provided at position 5 & 6.


Chapter 2 26


Shear forces and shear resistance Shear forces coefficients are obtained from Table 6.7 Position 1, d = 145 mm xvxsx lnV ⋅⋅= β = 0.47*14.2*5 = 33.4 kN/m

23

3

N/mm 23.0145*1010*4.33

==v

31.0145*1000449*100100

==bd

As

From table 6.3, 2231

N/mm 0.23 v N/mm 62.02535*55.0 =>=⎟

⎠⎞

⎜⎝⎛=cv

No shear reinforcement required. Position 4, d = 135 mm xvysy lnV ⋅⋅= β = 0.40*14.2*5 = 28.4 kN/m

23

3

N/mm 21.0135*1010*4.28

==v

23.0135*1000314*100100

==bd

As


N/mm 0.21 v N/mm 57.02535*51.0 =>=⎟

⎠⎞

⎜⎝⎛=cv

No shear reinforcement required. Position 5 (d = 145 mm) , Position 6 (d = 135 mm) xvxsx lnV ⋅⋅= β = 0.31*14.2*5 = 22.0 kN/m (Position 5) xvysy lnV ⋅⋅= β = 0.26*14.2*5 = 18.5 kN/m (Position 6)

As positions 5 & 6 have the same amount of bending reinforcement i.e. T10-325 c/c (based on the top reinforcement), shear check for position 5 only is required.


Chapter 2 27


23

3

N/mm 15.0145*1010*22

==v

17.0145*1000242*100100

==bd

As


N/mm 0.15 v N/mm 50.02535*45.0 =>=⎟

⎠⎞

⎜⎝⎛=cv

No shear reinforcement required. Torsion Steel Torsion steel of length 5/5 = 1.0 m is to be provided in the top and bottom of the slab at the three external corners marked X and Y. (i) Corner X The area of torsion steel is 0.75*305 = 229 mm2 Provide the min. steel T10-325 c/c (As = 242 mm2) (ii) Corner Y The area of torsion steel is 0.5*229 = 114.5 mm2 Provide the min. steel T10-325 c/c (As = 242 mm2) Edge Strips Provide min. steel T10-325 c/c. Deflection Use steel at mid-span and d = 145 mm Basic l/d ratio (Table 7.3) = 23 M.F. for tension reinforcement (table 7.4)

794.0145*100010*7.16

2

6

2 ==bdmsx

2

,

, N/mm 268349305460

321

32

=⋅⋅=⋅⋅⋅=bprovs

reqsys A

Aff

β

∴ mt = 1.58 Allowable l/d ratio = 23*1.58 = 36.3 Actual l/d ratio = 5000/145 = 34.5 < 36.3 ∴ O.K.


Chapter 2 28


2.3 Ribbed Slabs2 Ribbed slabs are more economical than solid slabs for long spans with relatively light loads. Two principal methods of construction are:

• Ribbed slabs without permanent blocks • Ribbed slabs with permanent hollow or solid blocks

The topping or concrete floor panels between ribs may or may not be considered to contribute to the strength of the slab. The hollow or solid blocks may also be counted in assessing the strength using the rules given in the code. The design of slabs with topping taken into account but without permanent blocks is discussed here. 2.3.1 Ribbed Slab Proportions Proportions for ribbed slabs without permanent blocks are given in Clause 6.1.4 of the code.

• In situ ribs should be spaced at centers not exceeding 1.5m. • The depth of rib, excluding any topping should not exceed four times their

width. • The minimum width of rib will be determined by considerations of cover,

bar spacing and fire resistance requirements. From practical point of view, the minimum rib width is 125 mm.

• The minimum thickness of topping shall not be less than 50 mm or one-tenth of clear distance between ribs whichever is greater.

> 1.5 m

b < 125 mm

< 50

> 4b

v

v

Figure 2.16 – Proportions of Ribbed Slab


Chapter 2 29


2.3.2 Design Procedure and Reinforcement 2.3.2.1 Analysis of Ribbed Slabs The moments and forces due to design ultimate loads on continuous slabs may be obtained by any of the methods given in cl. 6.1.3.2 of the code for solid slabs. The coefficients of Table 6.4 may be used for the design of one-way spanning continuous rib slab.

Table 6.4 – Ultimate Bending Moment and Shear Forces in One-Way Spanning Slab5

2.3.2.2 Bending The mid-span of a ribbed slab shall be designed as a T-beam, to resist sagging moment, with flange width determined in accordance with cl. 5.2.1.2 (a). At the support, the ribbed slab shall be designed as a rectangular section to resist the hogging moment. Moment reinforcement consisting of one or more bars is provided in the top and bottom of the ribs. 2.3.2.3 Shear The design of shear reinforcement is similar to that of solid slabs. 2.3.2.4 Reinforcement in the Topping A single layer of welded steel fabric, having a cross-sectional area of not less than 0.12% of the topping, shall be provided in each direction. The spacing between wires should not be greater than half the center-to-center distance between ribs.


Chapter 2 30


Figure 5.2 – Effective Flange Width Parameters5


Chapter 2 31


Example 2.3 – Design of Ribbed Slab The ribbed floor is constructed without permanent blocks. It is continuous over several equal spans of 5.5 m. The effective section is shown in Figure 2.17. The characteristic dead load including self-weight and finishes is 4.5 kN/m2 and the characteristic imposed load is 3 kN/m2. Grade 35 concrete and grade 460 steel are used for the design. Design the end span of the ribbed slab.

70

230

130 130400

195

Figure 2.17 Solution Design load = (0.13 + 0.4*2/2)(1.4*4.5 + 1.6*3) = 5.88 kN/m Ultimate load F on ONE span = 5.88*5.5 = 32.3 kN Bending: Bending moment coefficients are obtained from Table 6.4 At end span (Design as T-beam) M = 0.086FL = 0.086*32.3*5.5 = 15.3 kNm

∑ += wieffeff bbb ,

beff, 1 = 0.2*200 +0.1*(0.85*5500) = 507.5 mm < 0.2*(0.85*5500) = 935 mm and beff, 1 ≤ 200 mm, ∴ beff, 1 = beff, 2 = 200 mm Hence beff = 200 + 200 + 130 = 530 mm

)8.0( 0.132 022.035*195*530

10*3.15b2

6

2 =<== βcufbd

M


Chapter 2 32



d– z = 195 – 185 = 10 mm < mmhf 35

270

2==

Thus stress block lies within the flange. 2

6

mm 207185*460*87.0

10*3.1587.0

===zf

MAy

s

Provide 2T12 bars in the ribs (As = 226 mm2) (Design table 2) At 1st interior support (Design as rectangular beam with 130 mm width) M = -0.086FL = -0.086*32.3*5.5 = -15.3 kNm

)8.0( 0.132 088.035*195*130

10*3.15b2

6

2 =<== βcufbd

M


26

mm 221173*460*87.0

10*3.1587.0

===zf

MAy

s

Provide 2T12 bars in the ribs (As = 226 mm2) (Design table 2) Deflection: At end span: Basic l/d ratio (Table 7.3) = 18.5 (bw / b = 130/530 = 0.245 < 0.3) M.F. for tension reinforcement (table 7.4)

759.0195*53010*3.15

2

6

2 ==bdM

N/mm 281226207*460*

32 2==sf

∴ mt = 1.53 Allowable l/d ratio = 18.5*1.53 = 28.3 Actual l/d ratio = 5500/195 = 28.2 < 28.3 ∴ O.K.


Chapter 2 33


Shear Shear force coefficients are obtained from table 6.4. Max. shear force V= 0.6*32.3 = 19.4 kN Shear stress v = 19.4*103 / (130*195) = 0.77 N/mm2

891.0195*130226*100100

==bd

As

From “ design table 6.3”, vc = 0.73 31

2535

⎟⎠⎞

⎜⎝⎛x N/mm2 = 0.81 N/mm2

> 0.77 N/mm2 ⇒ No shear reinforcement required.


Chapter 2 34


Revision

Read reference 1 on P.178 - 209.

Main Reference

1. Reinforced Concrete Design, 5th Edition, W.H. Mosley, J.H. Bungey & R.

Hulse, Palgrave. 2. Reinforced Concrete Design Theory and Examples, 2nd Edition, T.J.

MacGinley & B.S. Choo, E & FN Spon. 3. Reinforced Concrete Design to BS 8110, Simply Explained, A.H. Allen, E &

FN Spon. 4. Design of Reinforced Concrete Flat Slabs to BS 8110, CIRIA Report 110,

CIRIA. 5. Code of Practice for Structural Use of Concrete 2004, Buildings

Department, HKSAR. 6. Tall Building Structures Analysis and Design, Bryan Stafford Smith, Alex

Coull, John Wiley & Sons, Inc.


Chapter 2 35


│TUTORIAL 2A│

Q1. A simply supported R.C. slab 175 mm thick is spanning in two directions. The effective span in

each direction of the two-way spanning slab is 4 m and 5 m. The slab supports the following loading:-

Characteristic Dead Load Self-weight 20 mm thick finishes (unit weight = 20 kN/m3) Ceiling load = 0.5 kN/m2 Partition = 1.0 kN/m2 Characteristic Imposed Load = 5 kN/m2

Design the bending and shear reinforcement for the slab. Check the adequacy of the slab section against deflection by span-effective depth ratio approach. The characteristic material strengths are fcu = 35 N/mm2 and fy = 460 N/mm2. Nominal cover = 30 mm. Main bars assumed to be 10 mm.

Q2. The continuous two-way spanning slabs in Figure Q2 are 150 mm thick and are restrained by integral beams. Design the slab groups 1, 2, 3 & 4. The slabs are subjected to loading as in Q1. The characteristic material strengths are fcu = 35 N/mm2 and fy = 460 N/mm2. Nominal cover = 25 mm. Main bars assumed to be 10 mm.

Q3. A four-span R.C. ribbed slab, with spans of 6.5 m, has an effective section as shown in Figure Q3.

The ribbed slab is constructed without permanent blocks. The slab is subjected to the following loadings:-

Characteristic Dead Load Self-weight 25 mm thick finishes (unit weight = 20 kN/m3) Services = 0.75 kN/m2 Partition load = 1.0 kN/m2 Characteristic Imposed Load = 3 kN/m2

Grade 35 concrete and grade 460 steel are used for the design. Nominal cover = 30 mm. Design the four-span ribbed slab.


Chapter 2 36


│TUTORIAL 2A│

Q5. A slab 250 mm thick is constructed with grade 35 concrete and is reinforced with T12-200 c/c

in one direction and T12-300 c/c in the other direction. Determine the maximum load that can be carried on an area 300 * 300 mm, without the exceeding the ultimate shear stress. Nominal cover = 25 mm

Q6. A 250 mm thick slab of grade 35 concrete is reinforced with T12-200 c/c in one direction and

T12-300 c/c in the other direction. The slab is subjected to an ultimate concentrated load of 600 kN over a square area of 300*300 mm. Determine the shear reinforcement required for fyv = 250 N/mm2 and the nominal cover to reinforcement = 25 mm.


Chapter 2 37


│TUTORIAL 2A│

4m 4m 4m

5m5m

5m

(150)

1 2 1

343

1 2 1

Figure Q2


Chapter 2 38


│TUTORIAL 2A│

175 500 175 175500

275

80

Figure Q3

InternalPanel

6.4m 6.4m 6.4m

5.76

m5.

76m

5.76

m

140 140500 500

250

75

Figure Q4

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