2.0 ANALYSIS AND DESIGN

2.2 STRUCTURAL ELEMENT Reinforced Concrete Slabs

Rearrangement by : NOR AZAH BINTI AIZIZ KOLEJ MATRIKULASI TEKNIKAL KEDAH

INTRODUCTION

• Concrete slabs are similar to beams in the way they span horizontally between supports and may be simply supported, continuously supported or cantilevered.

• Unlike beams, slabs are relatively thin structural members which are normally used as floors and occasionally as roof systems in multi-storey buildings.

INTRODUCTION

• Slabs are constructed of reinforced concrete poured into formwork. • The formwork defines the shape of the final slab when the concrete is cured (set). • Concrete slabs are usually 150 to 300 mm deep.

INTRODUCTION

• Slabs transmit the applied floor or roof loads to their supports. • Slabs may be classified into two main groups depending on whether they are supported on the ground or suspended in a building.

GROUND SLABS

• Ground slabs are those slabs that are poured directly into excavated trenches in the ground. • They rely entirely on the existing ground for support. • The ground must be strong enough to support the concrete slab. • Normally, a minimum bearing capacity for slab sites is 50 kPa.

GROUND SLABS

• • Diagram of slabs with arrow representing applied floor and roof load pointed down to the slab.

The slab is supported by foundation and the slabs transmit its load to foundation.

SUSPENDED SLABS

• Suspended slabs are slabs that are not in direct contact with the ground. • They form roofs or floors above ground level.

SUSPENDED SLABS

Suspended slabs are grouped into two types: i) i) One way slabs - which are supported on two sides two way slabs - which are supported on all four sides. The way a slab spans its supports has a direct impact on the way in which the slab will bend.

ONE-WAY SLAB

• • One way slabs are usually rectangular where the length is two or more times the width. These slabs are considered to be supported along the two long sides only even if there is a small amount of support on the narrow ends.

ONE-WAY SLAB

A diagram of a concrete slab with two supporting sides is shown. The width of the slab is also the short span.

Rule of Thumb

: For l

y

/l

x

> 2, design as one-way slab l

y

= the length of the longer side l

x

=the length of the shorter side

ONE-WAY SLAB

ONE BEND SLAB

• It is assumed that; one way slabs bend only in the direction of the short span, so; the main steel reinforcement runs in this direction across the slab.

ONE BEND SLAB

• • • A diagram of a concrete slab with two supporting sides is shown. Compression on the slab pushes towards the middle of the slab which causes the slab to bend inwards. Tension is distributed across the supporting sides.

TWO WAY SLAB

• Two way slabs are approximately square where the length is less than double the width and the slab is supported equally on all four sides.

Rule of Thumb

: For l

y

/l

x

≤ 2, design as two-way slab l

y

= the length of the longer side l

x

=the length of the shorter side

TWO WAY SLAB

Spans equally both direction • • A diagram of a concrete slab with four supporting sides is shown. The pressure spans equally across the width and length of the concrete slab.

TWO WAY SLAB

TWO BEND SLAB

• These slabs are assumed to bend in both directions, so main steel reinforcement of equal size and spacing is run in both directions.

TWO BEND SLAB

• A diagram of the compression that occurs in a two bend slab is shown. • The pressure runs to the middle of the slab which causes all four sides to bend equally.

Compression

Example:

Figure shows three floor layouts of a monolithic beam and slab construction.

a) State whether the floor panels are one-way or two-way spanning.

b) Sketch the tributary areas for all the beams

3050mm 3050mm 7650mm 2 7050mm 1 A A1 B C

Answer

: 3050mm 3050mm 2 7650mm 1 A A1 Panel A-A1/1-2 l y  /l x =7050/3050 = 2.3 > 2 one-way slab B Panel B-C/1-2 l y  /l x =7650/7050 = 1.1 < 2 two-way slab C 7050mm

EXAMPLE : The beams supporting the floor panel A-A1/1-2 are 350 mm deep and 150 mm thick, and the floor slab is 150 mm thick, given the density of concrete as 24 kN/m3.

a) Calculate the self-weight of the beam A/1-2, considering only the rib of the beam in kN/m b) Calculate the self-weight of the slab in kN/m 2 c) Calculate ultimate load on beam A/1-2 in kN/m d) Calculate reaction force at column A/1

3050mm 3050mm 7650mm 2 7050mm 1 A A1 B C

ANSWER :

150 rib 350 A/1

a) Self-weight of the rib in kN/m = 0.150 x (0.350-0.150) x 24 = 0.72kN/m

7050

b) Self-weight of the slab in kN/m 2 = 0.15 x 24 = 3.6kN/m 2 c) rib self-weight = 0.72kN/m slab self-weight = 0.5 x

w

x l

x

= 0.5 x 3.6 x 3.05

= 5.49kN/m

A/2

Ultimate load on beam A/1-2 in kN/m = rib self-weight + slab self- weight = 1.4 x 0.72 + 1.4 x 5.49

= 1.008 + 7.686

= 8.694 kN/m d. Reaction force at column A/1 =8.694 x 7.05/2 =30.64kN

One-way slab Design

• Design a one way slab supported on two brick wall spanning 3 m c-c. The characteristic dead load ( excluded self weight slab) and characteristic live load supports by the slab are 0.35 kN/m 2 and 2.5 kN/m 2 . ( f cu =25 N/mm2 , f y =250 N/mm 2 , concrete cover=25 mm and assume diameter of main bars at 10 mm)

One-way slab Design

• • • • Is designed as a shallow rectangular beam Consider a strip 1 m wide for design An upper limit to the value of the lever arm,

z

= 0.95 d The reinforcement area evaluated from; M ult = 0.87 A s f y

z

1a 2 3000 mm 3000 mm 3000 mm 3000 mm 7500 mm 1 A A 1 B A Figure 2: Building layout plan

F st A s

One-way slab Design

d F cc Concrete compression x F st 0.9x

F cc (d-0.9x/2) a



Steel tension b Equation F cc = 0.45f

cu A F st = 0.95A

s ∑M a F cc = 0 (d-0.9x/2) – M = 0 F cc = F st Where: f cu - Characteristic of concrete strength (30N/mm2) f y - Characteristic of reinforcement strength (460N/mm2) A – area of beam cross section A S – area of reinforcement cross section M – Moment

Section A-A

h=125 7500 Characteristic Dead load,g

k

= slab self weight + weigh of services, finishing & ceiling = 24kN/m 3 x h + 0.35 kN/m 2 = 24 x0.125 + 0.35

= 3.35 kN/m 2 Live load, q

k

= 2.5kN/m 2

Gk= 3.35 x 7500 = 25.125 kN/m Qk= 2.5 x 7500 = 18.75 kN/m 7500

Factored load on the slab = 1.4 x 25.125 + 1.6 x 18.75

= 65.175 kN/m

TABLE: Ultimate bending moment and shear force coefficients in one-way spanning

Refer Table: Ultimate bending moment and shear force coefficients in one-way spanning As a continues beam, it is not easy to find shear force and bending moment, so we use diagram given.

Use middle interior span & interior support F= 65.175

kN/m x 3.00 m = 195.53 kN Use M = 0.063 F

L

=0.063x 195.53 x 3.00

=36.96 kNm

F cc H=125 d= 125 - 25-10/2 = 95mm F st 7500 ∑Ma = 0 F cc =F st Fcc (d-0.9x/2) – M = 0 0.45

x

fcu

x

A

x

(d-0.9x/2) – M

= 0

0.45

x

25

x

0.9

x

x

7500

x

(95 - 0.9

x

/2)

–

75937.5

x

(95-0.45

x

)

-

36.96 x 10 6

= 0

7214062.5

x

-

34171.875

x 2

-

36.96 x 10 6

x = 205.86 @ 5.25

36.96 x 10

= 0

6

= 0

F F cc cc =F st = 0.45 x 25 x 0.9(5.25 ) x 7500 = 398671.875 N 398671.875 = 0.95 x f y xAs Where f y =250 (mild steel) A s = 398671.875 / 237.5

= 1680 mm 2 Lets say for 35 rods; 2223/35 = 48 mm 2 (1 rod) So size rebar A = Πj 2 = ΠD 2 /4 = 48 mm 2 D = √ 48 x 4 / Π D = 8 mm for 1 bar Spacing = 7500 – 25 (2) / 34 = 219 mm So use 35 R10 - 219 , (35 mild steel bar 10mm dia. with 219 spacing)