# chapter 3 slab - skyscrapers slab 16 3.3 design of one-way slab step 1: estimation of slab thickness

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• CHAPTER 3

SLAB

3.1 INTRODUCTION

Reinforced concrete slabs are one of the most widely used structural elements. In many structures, in addition to providing a versatile and economical method of supporting gravity loads, the slab also forms an integral portion of the structural frame to resist lateral forces. Usually a slab is a broad, flat plate, with top and bottom surfaces parallel or nearly so. It may be supported by reinforced concrete beams, by masonry or reinforced concrete walls, by structural steel members, directly by columns, or continuously by the ground.

3.2 TYPES OF SLAB

• One-way slab : Independent of support condition. (Figure 3.1a 3.1b)

2

1

l l

> 2;

• Two-way slab : Depends on support condition. (Figure 3.1c)

2

1

l l

≤ 2

(a) One- way slab

Figure 3.1: Types of slab

2

1 l

l >2

1l 2l

2

1

l l > 2

• SLAB

15

Two-way slabs are classified as:

• Two-way edge supported slab or slab with beams. ( Figure 3.1d ) • Two-way column supported slab or slab without beams. ( Figure 3.1e,3.1f, 3.1g )

(f) Flat slab (Column supported slab)

(d) Slab with beams (Edge supported slab)

(b) One- way slab

(g) Grid slab (Column supported slab)

(e) Flat plate (Column supported slab)

(c) Two- way slab

1l 2l

1l2l

Figure 3.1: Types of slab (continued)

• SLAB

16

3.3 DESIGN OF ONE-WAY SLAB

Step 1: Estimation of Slab Thickness (h)

Slab thickness is determined according to ACI Code 9.5.2 as given in Table 3.1

Table 3.1: Minimum thickness of non-prestressed one- way slab

Members wc = 145 pcf

fy = 60.000 psi

wc =90∼120 pcf fy 60.000 psi

Rounding up the thickness

Simply supported l/20

One end continuous l/24

Both end continuous l/28

Cantilever l/10

Multiply by (1.65-0.005wc) but > 1.09

Multiply by

0.4 + 000,100 yf

(1) h≤6 in next higher ¼ in

(2) h > 6 in next higher ½ in

# Span length l is in inches, as defined by ACI Code 8.7 given in Fig. 3.2(a), (b), & (c)

Step 2 : Calculation of Factored Load (wu)

wu = 1.4 D+ 1.7 L psf

wc = Unit weight of concrete (145 ~ 150 pcf for normal weight concrete )

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17

Step 3: Determination of Design Moment

Design moment is determined by using ACI Moment Coefficient (ACI Code 8.3.3) as given in Table 3.4.

Step 4 : Checking the Design Thickness

d = )59.01(

c

y y

u

f f

bf

M

′ − ρφρ

putting ρ = ρmax = 0.75ρb.

Where, ρ b = 0.85*β 1 y

c

f f /

* yf+87000

87000

Values of β 1 is given in Table 3.2

Table3.2 : Values of β 1 (ACI Code 10.2.7.3)

f c / ≤ 4000 psi β 1 = 0.85

f c / > 4000 psi β 1 shall be reduced at a rate of 0.05 for each 1000 psi of strength

in excess of 4000 psi.

0.65 ≤ β 1≤ 0.85

Table 3.3: Clear cover for slab ( ACI Code 7.7.1)

No 14 & No 18 bars .................................... .... 1 1/2 in No 11 bar & smaller .................................... ¾ in

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18

t

al

h

(a) Slabs not built integrally with the support (ACI Code 8.7.1)

l= la + h ≤ la + t t

al

h

l= la + t

(b) Slabs are continuous (ACI Code 8.7.2)

t al

Figure 3.2 : Span length

(c) Slabs built integrally with support

If (d + clear cover) < h; Design is ok.

Otherwise redesign the thickness.

• SLAB

19

3. Discontinuous ends are

Table 3.4 : ACI moment coefficient

Support Condition

1 9

1 1 1

0 0

Moment Coefficent

For two span:

1. Discontinuous ends are

1 2 4

1 9

1 1 4

1 1 4

0 11 0

1 1 1

1 1 1

1 2 4

1 1 0

1 1 4

1 2 4

1 9

1 1 6

1 1 4

1 1 6

1 9

1 9

1 1 4

unrestrained

built integrally with support 2. Discontinuous ends are

(spandrel beam or girder)

1 1 6

1 1 1

1 1 1

1 1 6

1 1 1

1 1 1

1 1 1

1 1 4

1 1 6

1 1 6

1 1 0

1 1 1

1 1 1

1 1 1

built integrally with support (when support is a column only)

For continuous Span: 1. Discontinuous ends are unrestrained.

(spandrel beam or girder) built integrally with support 2. Discontinuous ends are

3. Discontinuous ends are built integrally with support

(when support is a column only)

w = Total factored load per unit length of beam or per unit area of slab

l = Clear span for positive moment and the average of two adjacent clear spans for negative moment.

1. Shear in end members at first interior support

2. Shear at all other supports 1.15 w2

w 2

1 9

1 1 1

• SLAB

20

Step 5 : Determination of Steal Area (As)

Reinforcement for 1 ft. strip towards shorter distance is calculated by Iteration. Details shown in Figure 3.3 (here b = 12 in)

Figure 3.3: Iteration process to determine the steel area

Calculate a for next trial with (As) corrected

(a) corrected = bf fA

c

ys

′′′85.0

(As) trial = ( )2adf M

y

u

−φ

(As) corrected

A corrs , ≈

As, trial

Yes

OK

Assume a (hints: a = 0.3d)

No

• SLAB

21

Generally # 3 or # 4 bars are used for slab main reinforcement.

Spacing: ACI Code 7.6.5 specifies that

Spacing ≤ 3h or 18 in, whichever is smaller

but > 1.5 h

Finding out bar spacing: Let us chose # 3 bar (0.11 2in )

Spacing = sA

12*11.0 in c/c

Step 6 : Temperature and Shrinkage Reinforcement

Reinforcement is provided normal to main reinforcements. ACI Code 7.12.2.1 provides required area of temperature and shrinkage reinforcement as given in Table 3.5.

Table 3.5 : Minimum ratio of temperature and shrinkage reinforcement in slabs.

Slabs where grade 40 or 50 deformed bars are used

0.0020

Slabs where grade 60 deformed bars or welded wire fabric are used

0.0018

Slabs where reinforcement with yield strength exceeding 60,000 psi measured at a yield strain of 0.35 percent is used

yf 000,600018.0 ×

But should be:

ρ > 0.0014

Required steel area, As = ρbh 2in per 1 ft. strip

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22

Spacing: ACI Code 7.12.2.2 specifies that

Spacing ≤ 5h

or ≤ 18 in , whichever is smaller

Using # 3 or # 4 bar required spacing can be obtained.

Step 7 : Shear Check

According to ACI shear coefficient given in Table 3.2

Shear at end members at first interior support is 2

15.1 nulw

Critical shear at a distance d from support, Vu = (1.15 122 dWlW unu − )

Design strength for shear, φ Vc = φ ′cf2 bd ; φ = 0.85

If φVc > Vu, slab design for shear is OK

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