Cantilever Slab Design

UNIT 4 SLABS

Slabs

Structure 4.1

Introduction Objectives

4.2

General Principles of Design and Detailing of Slabs

4.3

Design and Detailing of a Cantilever Slabs

4.4

Design and Detailing of One-way Simply Supported Slabs

4.5

Design and Detailing of Two-way Simply Supported Slabs

4.6

Design and Detailing of Two-way Restrained Slabs

4.7

Summary

4.8

Answers to SAQs

4.1 INTRODUCTION A slab is like a flat plate loaded transversely and supported on its edges. Under the loads, it bends and the directions of its bending depend on its shape and support conditions. A beam bends only in one direction, i.e. in its own plane; whereas a slab may have multidirectional bending. Therefore, slabs may have different names depending upon its bending, support conditions and shapes. For example, a slab may be called (a)

One-way simply supported rectangular slab,

(b)

Two-way simply supported or restrained rectangular slab,

(c)

Cantilever rectangular slab,

(d)

Fixed or simply supported circular slab, etc.

One-way slab means it bends only in one direction and, therefore, reinforcement for bending (i.e. main reinforcement) is provided only in that direction. A slab supported on all sides bends in all the directions so the main reinforcements provided shall be such that they may be effective in all directions. For ease of analysis and convenience of reinforcement detailing, the bending moments in a slab are calculated in two principal directions only and, therefore, such a slab is called a two-way slab. A slab is designed as a beam of unit width in the direction of bending. In this unit, only the most commonly used rectangular slabs, with uniformly distributed load is described.

Objectives After studying this unit, you should be able to •

describe the design and detailing of cantilever slabs,

•

design and explain detailing of one-way and two-way simply supported slabs, and

•

explain the design and detailing of two-way restrained slabs. 75

Theory of Structures-II

4.2 GENERAL PRINCIPLES OF DESIGN AND DETAILING OF SLABS Following are the general principles for design and detailing applicable to all types of slabs. (a)

The maximum diameter of reinforcing bars shall not exceed

1 th of 8

total thickness (D) of the slab. (b)

Normally, shear reinforcement is not provided in slabs. The shear resistance requirements may, then, be complied either by increasing the percentage of tensile reinforcement or by increasing the depth of slab, but the latter is preferred as it is economical. For solid slabs, the design shear strength for concrete slab shall be τc K, where K has the values given below : Overall Depth of Slab (mm)

300 or more

275

250

225

200

175

150 or less

K

1.00

1.05

1.10

1.15

1.20

1.25

1.30

(c)

To take care of temperature and shrinkage stresses, minimum reinforcement in either direction shall not be less than 0.15 percent and 0.12 percent of total cross section area of concrete section for mild steel and high strength deformed bars, respectively.

(d)

To meet the requirement for limit state of cracking the following two rules are observed: (i)

The horizontal distance between parallel main reinforcement shall not be more than 3 times the effective depth of slab or 300 mm whichever is smaller.

(ii)

The horizontal distance between parallel bars provided against temperature and shrinkage shall not be more than 5 d or 450 mm, whichever is smaller.

4.3 DESIGN AND DETAILING OF CANTILEVER SLABS Design and detailing of a cantilever slab is the same as that of a cantilever beam (Section 3.3) of unit width. Temperature and shrinkage reinforcement is provided along the direction perpendicular to the span. This is illustrated through the following example. Example 4.1

Design the cantilever slab of a bus stand shown in Figure 4.1. Load data and design parameters are given below : Load Data

Lime terrace topping of 100 mm thickness is provided over the slab. 76

Imposed load = 0.75 kN/m2.

Slabs

Design Parameters fck = 25 N/mm2; fy = 415 N/mm2 and Nominal cover = 30 mm.

Figure 4.1 : Cantilever Roof Shade for Bus Stop

Solution Effective Span (lef) lef = 3 + 0.3* = 3.3 m Depth of Slab (D) From Deflection Control l ef d

≤ k B k1 k 2 k 3 k 4 , where kB = 7;

For k2,

fs = 0.58 fy

k1 = 1 as lef < 10 m

* Estimate of effective depths (d) : l ef 300 d≈ ≈ = 428.57 mm 7 7 d ∴ ≈ 300 mm = 0.3 m added 2 for evaluating lef (Cl. 22.2c).

Area of cross section of steel required Area of cross section of steel provided

Assuming area of cross section of steel required = Area of cross section of steel provided = Area of balanced tensile steel for M 25 concrete and Fe 415 steel (pt% = 1.19%). According to above assumptions fs = 0.58 × fy × 1 = 0.58 × 415 ≈ 240 N/mm2 and k2 = 0.96 (Figure 1.2) k3 = k4 = 1 as the slab is singly reinforced and it is not a flanged section. Substituting all these values lef d

≤ 7 × 1 × .96 × 1 × 1 = 6.72

3.3 × 10 3 = 491 mm 6.72

or

d≥

∴

D=d+

Taking

D = 550 mm

∴

d =D−

10 φ − Nominal cover = 550 − − 30 = 515 mm 2 2

∴

lef = 3 +

0.515 = 3.26 m 2

φ 10 + nominal cover = 491 + + 30 = 526 mm 2 2 (assuming φ = 10)

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Theory of Structures-II

From Moment of Resistance Consideration Loads

Self

= 1 × 1 × 0.550 × 25

= 13.75 kN/m2

Lime Terrace

= 1 × 1 × 0.1 × 18.8

= 1.88 kN/m2

Ceiling Plaster

= 1 × 1 × 0.01 × 20.4

= 0.204 kN/m2

Total (Dead Load)

= 15.834 kN/m2

Imposed Load (IL)

= 0.750 kN/m2

Total Load, w

= 16.584 kN/m2 wu lef 2

Maximum BM, Mu = M u , lim = 0.36

2

xu ,

max

d

=

1.5 × 16.584 × 3.26 2 = 132.2 kNm 2

x ⎛ ⎜1 − 0.42 u ,max ⎜ d ⎝

⎞ ⎟ bd 2 f ck ⎟ ⎠

132.2 × 10 6 = 0.36 × 0.48 × (1 − 0.42 × 0.48) × 1000 × d 2 × 25

or

d = 195.8 mm

Hence, provided D = 550 mm and d = 515 mm as above. Tensile Reinforcement (Ast) Ast f y ⎛ M u = 0.87 f y Ast d ⎜⎜1 − bd f ck ⎝

⎞ ⎟ ⎟ ⎠ ⎛

132.2 × 106 = 0.87 × 415 × Ast × 515 × ⎜1 − ⎝

or, or

Ast × 415 ⎞ ⎟ 1000 × 515 × 25 ⎠

132.2 × 10 6 = 185940.75 Ast − 5.993 Ast2

Ast2 − 31026.322 Ast + 22059068.9 = 0

Solving the above equation, we get Ast = 728.06 mm2/m width

i.e.

φ 10 bars @ 105 mm c/c. Ast , min =

0.12 × 1000 × 515 = 618 mm 2 /m < 728.06 mm 2 /m width 100

Spacing of φ 8 bars =

1000 × 50 = 80 mm c/c perpendicular to the main 618

reinforcement. Maximum Spacing 3 d = 3 × 515 = 1545 > 105 mm c/c 300 > 105 mm c/c

78

Hence, provided φ 10 @ 105 mm c/c as main reinforcement and φ 8 @ 80 mm c/c as temperature and shrinkage reinforcement.

Slabs

Check for Shear

SF at critical section, i.e. at d from the face of support = Vu = 1.5 × 16.584 × (3 – 0.515) = 61.817 kN τυ =

Vu 61.817 × 10 3 = bd 1000 × 515

= 0.12 N/mm 2

τc, min for M 25 concrete = 0.29 N/mm2 > 0.12 N/mm2 Hence, O.K. Detailing of Reinforcement

As Ast, min (= 618 mm2) is more than 50% of Ast (742.857 mm2/m) provided, hence, all the tensile reinforcement shall be extended up to free end of the slab. Development Length Ld =

φσ s 10 × 0.87 × 415 = = 403 mm 4τbd 4 × 1.6 × 1.4

Value of a 90o bend is 8 φ = 8 × 10 = 80 mm.

∴

Length of bar required = 403 – 80 = 323 mm say 325 mm.

The detailing of the reinforcement has been shown in Figure 4.2.

Figure 4.2 : Detailing of Reinforcement of Cantilever Roof

4.4 DESIGN AND DETAILING OF ONE-WAY SIMPLY SUPPORTED SLABS Design and Detailing of One-way Slab Simply Supported on all Edges

If the ratio Long span l y Short span l x

> 2,

the design is same as that for simply supported beam of unit width, as the slab bends mainly along the short span (Figure 4.3). Only temperature and shrinkage reinforcement is provided along the long span.

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Theory of Structures-II

Figure 4.3 : Bending Profile of an One-way Slab Supported on all Edges

Example 4.2

Design a roof slab simply supported on all its four edges of effective spans 3 m × 7 m. The top of slab is covered with 100 mm lime terrace. Imposed load may be taken as 1.5 kN/m2. Design parameters are : fck = 20 N/mm2; fy = 415 N/mm2 and Nominal Cover = 20. Solution ly lx

=

7 = 2.33 > 2 3

Hence the slab will be designed as one-way simply supported slab. Depth of Slab (D)

From Deflection control l ef d

where

< k B k1 k 2 k 3 k 4

kB = 20; and k1 = 1 as lef < 10 m

Assuming Ast required = Ast provided f s = 0.58 f y

A st required = 0.58 × 415 = 240 N/mm2 A st provided

Assuming balanced section, for M 25 and Fe 415 pt% = 0.96%

80

Therefore, k2 = 1

k3 = k4 = 1 as the slab is singly reinforced and it is not a flanged section. Substituting all the values in the above equation.

Slabs

3 × 1000 ≤ 20 × 1 × 1 × 1 × 1 d 3 × 1000 = 150 mm 20 × 150

or

d≥

∴

D = d + Nominal Cover +

Taking

D = 185 mm D = 185 −

φ 8 = 150 + 20 + = 174 mm 2 2

8 − 20 = 161 mm 2

From Moment of Resistance Consideration Loads

∴

Self

= 1 × 1 × 0.185 × 25

= 4.625 kN/m

Lime concrete

= 1 × 1 × 0.1 × 18.8

= 1.880 kN/m

Ceiling plaster

= 1 × 1 × 0.01 × 20.4

= 0.204 kN/m

Total DL

= 6.709 kN/m

IL

= 1.500 kN/m

Total (DL+IL)

= 8.209 kN/m

wu = 1.5 × 8.209 = 12.31 kN/m M u , max =

wu l 2 ef 12.31 × 3 2 = = 13.849 kN/m 8 8

M u , lim = 0.36

X u , max ⎛ X ⎜1 − 0.42 u , max ⎜ d d ⎝

⎞ ⎟ bd 2 f ck ⎟ ⎠

13.849 × 10 6 = 0.36 × 0.48 × (1 − 0.42 × 0.48) × 1000d 2 × 20

d = 70.85 < 161 mm

Hence, provided D = 185 mm and d = 161 mm. Tensile Reinforcement (Ast) Ast f y ⎛ M u = 0.87 f y Ast d ⎜⎜1 − ⎝ bd f ck

or

⎞ ⎟ ⎟ ⎠

Ast × 415 ⎞ ⎛ 13.849 × 106 = 0.87 × 415 Ast ×161× ⎜1 − ⎟ ⎝ 1000 ×161× 20 ⎠

or

7.492 Ast2 − 58129.05 Ast + 13.849 × 106 = 0

or

Ast2 − 7758.816 Ast + 1848505.07 = 0

Solving the above equation, we get Ast = 246.05 mm 2 /m

i.e. φ 8 @ 200 mm c/c (Ast = 250 mm2/m)

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Theory of Structures-II

Ast , min =

0.12 × 1000 × 161 = 193.2 mm 2 /m < 250mm 2 /m width 100

Hence, provided φ 8 @ 200 mm c/c. Maximum Spacing (a)

3d = 3 × 170 = 510 > 200 mm c/c

(b)

300 > 200 mm

Hence, O.K. Check for Shear

As the width of support is not given, clear span (lc) may be taken as (lef − d) = 3 – 0.161 = 2.839 m Critical section for shear force is at d from the face of the support (Figure 4.4).

Figure 4.4 : Explaining Critical Section for SF

∴

⎛l ⎞ ⎛ 2.839 ⎞ Vu = wu ⎜ c − d ⎟ = 12.31 × ⎜ − 0.161⎟ = 15.492 kN 2 2 ⎝ ⎠ ⎝ ⎠ τυ =

Vu 15.492 × 10 3 = = 0.096 N/mm 2 bd 1000 × 161

For Ast,min and M 20 concrete k τ c min = k × 0.28 = 0.28k > 0.096 N/mm 2 (k ≥ 1)

Hence, O.K. Detailing of Reinforcements Ld =

(a)

The positive main reinforcement shall extend into the support a distance of

(b) where

φσ s 8 × 0.87 × 415 = = 376 4τbd 4 × 1.6 × 1.2

Ld ≤

Ld 376 = = 125 3 3

1.3M 1 + Lo V ⎛

M1 = 0.87 fy Ast d ⎜⎜1 − ⎝

Ast f y ⎞ ⎟ bd f ck ⎟⎠

Taking Ast as 50% of total reinforcement at mid span extending into the support. i.e. 82

Ast =

250 = 125 mm2/m < Ast, min (= 193.2 mm2/m) 2

Hence, all the reinforcement shall extend into the support

Slabs

250 × 415 ⎞ ⎛ M1 = 0.87 × 415 × 250 × 161 × ⎜1 − ⎟ ⎝ 1000 × 161 × 20 ⎠ or

M1 = 14.06 kN-m/m

Vu at simple support =

wu lef 2

=

12.31 × 3 = 18.465 kN 2

⎞ ⎛ 1.3M 1 Ld ≤ ⎜ + Lo ⎟ ⎠ ⎝ V or

1.3M 1 ⎞ ⎛ Lo ≥ ⎜ Ld − ⎟ V ⎠ ⎝ ⎛ ⎝

Now, Lo ≥ ⎜ Ld − i.e.

1.3M 1 ⎞ 1.3 × 14.06 × 106 = − 614 mm ⎟ = 376 − V ⎠ 18.465 × 103

Lo = 0.

Distribution Steel Astd =

0.12 0.12 × 1000 × 161 bd = 100 100

= 193.2 mm 2 /m

Hence, provided φ 8 @ 255 mm c/c. Maximum spacing (a)

5d = 5 × 161 = 805 > 255

(b)

450 > 255

Hence, O.K. Reinforcement detailing has been shown in Figure 4.5.

Figure 4.5 : Reinforcement Detailing of the Designed Slab

83

Theory of Structures-II

* Two-way Slab means the slab bends about both the axes x and y like a saucer (Figure 4.6).

4.5 DESIGN AND DETAILING OF TWO-WAY* SIMPLY SUPPORTED SLABS For a rectangular slab, simply supported on all its edges, having no provisions to resist torsion at corners and to prevent corners from lifting, the maximum bending moment per unit width are calculated by the following equations : M x = α x w l x2 M y = α y w l x2

and

where w = Design loads per unit area and lx and ly = Short and Long spans, respectively, Mx and My = moments of strips of unit width spanning lx and ly, respectively and α x and α y are the coefficients given in Table 4.1. Table 4.1 : Bending Moment Coefficients for Slab 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

2.5

3.0

αx

0.062

0.074

0.084

0.093

0.099

0.104

0.113

0.118

0.122

0.124

αy

0.062

0.061

0.059

0.055

0.051

0.046

0.037

0.029

0.020

0.014

It is evident from the above table that if

ly lx

> 2 , the slab will be treated as

one-way and designed as in Section 4.4. At least 50% of the reinforcement provided at mid span shall extend into the supports and the remaining 50% may extend to within 0.1 lx or 0.1 ly of support as appropriate. The other specifications for design and detailing for the slab are the same as those for simply supported beams (Section 3.2).

Figure 4.6 : Bending of Two-way Slab like a Saucer

SAQ 1

84

(a)

Define one-way and two-way slabs.

(b)

How shear resistance of a slab can be increased economically?

(c)

How temperature and shrinkage stresses is taken care of?

(d)

Why the maximum horizontal distance between parallel bars are limited? Describe the maximum horizontal distance between parallel main bars and that between parallel temperature and shrinkage bars.

Slabs

4.6 DESIGN AND DETAILING OF TWO-WAY RESTRAINED SLABS A rectangular slab supported on beams on all sides and monolithically cast with them and having ly lx

≤2

behaves as a two-way slab. Being monolithic with the beams, the corners are prevented from lifting and, therefore, torsional reinforcements are provided to resist the resultant torsional moments. Main reinforcements are provided along both the principal axes to resist corresponding moments Mx and My, respectively. Detailing of reinforcement is done in the following manner : (a)

A slab is divided in edge strips and a middle strip in both directions as shown in Figure 4.7.

Figure 4.7 : Division of Slab into Middle and Edge Strips

(b)

The maximum moments Mx and My both for spans and supports are determined as M x = α x w l x2

and

M y = α y w l x2

The values of αx and αy for different edge conditions are given in Table 4.2. Table 4.2 : Bending Moment Coefficients for Rectangular Panels Supported on Four Sides with Provision for Torsion at Corners

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Theory of Structures-II

Case No.

(1) 1.

2.

3.

(2)

1.2

1.3

1.4

1.5

1.75

2.0

ly / lx

(4)

(5)

(6)

(7)

(8)

(9)

(10)

(11)

0.037

0.043

0.047

0.051

0.053

0.060

0.065

0.032

Positive moment at mid-span

0.024

0.028

0.032

0.036

0.039

0.041

0.045

0.049

0.024

Negative moment at continuous edge

0.037

0.043

0.048

0.051

0.055

0.057

0.064

0.068

0.037

Positive moment at mid-span

0.028

0.032

0.036

0.039

0.041

0.044

0.048

0.052

0.028

0.037

0.044

0.052

0.057

0.063

0.067

0.077

0.085

0.037

0.028

0.033

0.039

0.044

0.047

0.051

0.059

0.065

0.028

0.047

0.053

0.060

0.065

0.071

0.075

0.084

0.091

0.047

0.035

0.040

0.045

0.049

0.053

0.056

0.063

0.069

0.035

One Short Edge Discontinuous :

One Long Edge Discontinuous

Two Adjacent Edges

Positive moment at mid-span

7.

1.1

(3) 0.032

Discontinuous : Negative moment at continuous edge

6.

1.0

Negative moment at continuous edge

Positive moment at mid-span

5.

Long Span Coefficients, αy for All Values of

Interior Panels :

Negative moment at continuous edge

4.

Short Span Coefficients, αx (Values of ly/lx)

Type of Panel and Moments Considered

Two Short Edges Discontinuous : Negative moment at continuous edge

0.045

0.049

0.052

0.056

0.059

0.060

0.065

0.069

----

Positive moment at mid-span

0.035

0.037

0.040

0.043

0.044

0.045

0.049

0.052

0.035

Negative moment at continuous edge

----

----

----

----

----

----

----

----

0.045

Positive moment at mid-span

0.035

0.043

0.051

0.057

0.063

0.068

0.080

0.088

0.035

0.057

0.064

0.071

0.076

0.080

0.084

0.091

0.097

----

0.048

0.053

0.057

0.060

0.064

0.069

0.073

0.043

Two Long Edges Discontinuous

Three Edges Discontinuous (One Long Edge Continuous) : Negative moment at continuous edge Positive moment at mid-span

8.

0.043

Three Edges Discontinuous (One Short Edge Continuous) :

9.

Negative moment at continuous edge

----

----

----

----

----

----

----

----

0.057

Positive moment at mid-span

0.043

0.051

0.059

0.065

0.071

0.076

0.087

0.096

0.043

0.056

0.064

0.072

0.079

0.085

0.089

0.100

0.107

0.056

Four Edges Discontinuous : Positive moment at mid-span

A continuous edge is that on which slab extends on its both sides whereas on discontinuous edge the slab extends only on one side. 86

(c)

Tension reinforcement provided at mid-span in the middle strip shall extend in the lower part of the slab to within 0.25 l of a continuous edge (Figure 4.8), or 0.15 l of discontinuous edge.

Slabs

Figure 4.8 : Curtailment Rules for Two-way Slab

(d)

(e) (f) (g)

Over the continuous edge of the middle strip, tension reinforcement shall extend in the upper part of the slab a distance equal to 0.15 l from the support and at least 50% shall extend a distance of 0.3 l (Figure 4.8). At discontinuous edge, tension reinforcement equal to 50% of that provided at mid-span shall extend 0.1 l from the support (Figure 4.8). Minimum reinforcement in edge strip shall be sufficient (Figure 4.7). Torsion reinforcement at corners are provided as follows (Figure 4.9). (i) Where corners are bound by discontinuous edges, torsion reinforcement both at top and bottom and in both directions equal to 3/4th the area of short span reinforcement and of length lx/5 shall be provided. (ii) If the corner is bound by continuous edge on one side and discontinuous on the other, half of the area of reinforcement that prescribed in (a) shall be provided. (iii) Torsion reinforcement need not be provided at corners for which both the edges are continuous.

(a) Corner with Two Discontinuous Ends

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Theory of Structures-II

(b) Corner with One Discontinuous End Figure 4.9

Example 4.3

Design slab S1 and S2 of a roof (Figure 4.10) cast monolithically with beams. The slab has a topping of 120 mm thick lime terrace and 10 mm thick ceiling plaster. Design parameters are as follows : Imposed load on roof = 1.5 kN/m2; fck = 20 N/mm2; fy = 415 N/mm2 and Nominal cover = 20 mm.

Figure 4.10 : Plan of Slab System

Solution Type of Slab

Outer edges are taken as discontinuous edges whereas inner edges are continuous. Slab S1 Short span, lx = 3.5 m Long span, ly = 5.5 m ly

Ratio,

lx

= 1.57 < 2,

Hence, the slabs are two-way slabs. Depth of Slab (D)

From Deflection Control l ef d

88

≤ k B k1 k 2 k 3 k 4

kB = (20 + 26)/2 = 23 (for one edge continuous and other discontinuous).

Slabs

For M 20 and Fe 415, pt% = 0.96 % for balanced steel k2 = k3 = k4 = 1 d ≥

l ef k B k1 k 2 k 3 k 4

=

3.5 × 10 3 = 152.17 mm 23 × 1 × 1 × 1 × 1

D = d + Nominal cover +

Taking

D = 180 mm

∴

d = 180 − 20 −

φ 8 = 152.17 + 20 + = 176.17 mm 2 2

8 = 156 mm 2

From Moments of Resistance Consideration

Loads

Self

= 0.180 × 1 × 1 × 25

= 4.5 kN/m2

Lime concrete

= 0.12 × 1 × 1 × 18.8

= 2.26 kN/m2

Ceiling plaster

= 0.01 × 1 × 1 × 20.4

= 0.20 kN/m2

Total DL

= 6.96 kN/m2

IL

= 1.50 kN/m2 = 8.46 kN/m2

Values of coefficients α x and α y and Bending Moments Mx and My are given in Figure 4.11.

Figure 4.11 : Values of Moment Coefficient and Moment along X and Y-axes

Illustration for calculation of α x , α y and M for three edges discontinuous and one short edge continuous (S1). For (+)ve moment at mid-span α x = 0.076 +

(0.087 − 0.076) × (1.57 − 1.5) = 0.079 (1.75 − 1.5)

M x = α x wl x2 = 0.079 × 1.5 × 8.46 × 3.5 2 = 12.28 kNm

For (−)ve moment at continuous edge, α y = 0.057 M y = − α y wl x2 = − 0.057 × 1.5 × 8.46 × 3.52 = − 8.86 kN-m M u = 0.36

xu , max ⎛ x ⎜1 − 0.42 u , max ⎜ d ⎝ d

⎞ ⎟ f ck bd 2 ⎟ ⎠

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Theory of Structures-II

12.28 × 10 6 = 0.36 × 0.48 × (1 − 0.42 × 0.48) × 20 × 1000 × d 2

or

d = 66.7 mm < 156 mm

Hence, provided D = 180 mm and d = 180 – 20 –

8 = 156 mm. 2

Tensile Reinforcement (Ast) ⎛

Mu = 0.87 fy Ast d ⎜⎜1 − ⎝

Ast f y ⎞ ⎟ bd f ck ⎟⎠

Ast × 415 ⎞ ⎛ 12.28 × 10 6 = 0.87 × 415 × Ast × 156 × ⎜1 − ⎟ ⎝ 1000 × 156 × 20 ⎠

or

12.28 × 106 = 56323.8 Ast − 7.492 Ast2

or

Ast2 − 7517.86 Ast + 1639081.69 = 0

Solving the above equation, we get Ast = 24.74 mm 2

Hence, provided φ 8 @ 220 mm c/c. Check

(a)

d = 3 × 156 = 468 c/c > 220 mm c/c

(b)

300 > 220 mm c/c.

Negative reinforcement in long span. Average of BM on both sides of the edge =

(8.86 + 7.00) = 7.93 kN-m/m width 2

Ast f y ⎛ M u = 0.87 f y Ast d ⎜⎜1 − ⎝ bd f ck

⎞ ⎟ ⎟ ⎠

Ast × 415 ⎛ ⎞ 7.93 × 10 6 = 0.87 × 415 Ast × 156 × ⎜1 − ⎟ 1000 156 200 × × ⎝ ⎠

7.93 × 106 = 56323.8 Ast − 7.492 Ast2 Ast2 − 7517.86 Ast + 1058462.36 = 0

or

Solving the above equation, we get Ast = 143.53 mm 2 /m width

Provided φ 8 @ 345 mm c/c. Check

(a)

3d = 3 × 156 = 468 c c > 345 mm c/c

(b)

300 < 345 mm c/c

Hence, provided φ 8 @ 300 mm c/c. 90

Slabs

SAQ 2 (a)

Sketch the detailing of tensile reinforcement at support of a cantilever slab.

(a)

Sketch the plan and section showing detailing of reinforcement of a two-way simply supported slab.

(b)

Sketch the section of a two-way restrained slab of two spans, l1 and l2 continuous at interior support and discontinuous at ends.

(c)

Design an RC slab 4.5 m × 6.5 m in plan supported on 25 mm wide beams. Two of its adjacent edges are discontinuous. Imposed load is 2 kN/m2. Mix M 20 concrete and Fe 415 bars are used.

4.7 SUMMARY Four types of slab have been described in this unit : (a)

Cantilever slab

(b)

One-way simply supported slab

(c)

Two-way simply supported slab

(d)

Two-way restrained slab

Cantilever Slab

This type of rectangular slab has one edge fixed and the other three edges free. Therefore, under gravity loads, it bends about its axis of support causing hogging bending moment and requiring main reinforcement only on top face perpendicular to its support axis. One-way Simply Supported Slab

This type of slab may be either simply supported on two opposite faces or, if supported on all four edges the ratio

ly lx

> 2 . Therefore, under gravity

loads, it bends only in one direction. Two-way Simply Supported Slab

A slab simply supported on all its four edges having

ly lx

≤ 2 , bends about

both of its principal axes under gravity loads. Hence, main reinforcements are provided in both directions. Two-way Restrained Slab ⎛ ly

When a slab ⎜⎜

⎝ lx

⎞ ≤ 2 ⎟⎟ is monolithic with its supporting beams and the ⎠

corners are prevented from lifting under bending, additional reinforcements are provided at the corners to resist torsion. Hogging bending moments at the supports are taken care of by providing reinforcement at the top face.

91

Theory of Structures-II

4.8 ANSWERS TO SAQs SAQ 1

(a)

Refer Section 4.1.

(b)

Refer Section 4.2.

(c)

Refer Section 4.2.

(d)

Refer Section 4.2.

SAQ 2

92

(a)

Refer Section 4.3.

(b)

Refer Section 4.5.

(c)

Refer Section 4.6.