Unit 11 ( DESIGN OF REINFORCED CONCRETE SLABS (RCS) )

REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/

UNIT 11

DESIGN OF REINFORCED CONCRETE SLABS (RCS)

GENERAL OBJECTIVE:

To understand how to design reinforced concrete solid slabs according to BS 8110

specifications and guidelines.

SPECIFIC OBJECTIVES

By the end of this unit, you should be able to: -

1. state the different types of slabs.

2. define one-way and two-way spanning slabs.

3. differentiate between simply supported and continuous slab

reinforcements.

4. use BS 8110 given in bending moments.

5. calculate the required area of main reinforcements.

6. calculate the required area for the distribution of steel reinforcement.

7. check whether the effective depth of slabs are adequate.

8. sketch the details of the reinforcement based on the calculations.

1

OBJECTIVES


11.1 Introduction

A slabs is a structural element in which its overall depth (h) is relatively less than

its width (b). An almost simple type of slab is shown in Figure 11.1

A concrete slabs behaves like a flexural member like a beam. The design of a

Reinforced Concrete Slab (RCS) is similar to the design of reinforced concrete

beams.

2

INPUT 1

Figure 11.1: Simply supported slab


The design of RCS is generally much easier because of the following reasons:

a) The of the slab width is fixed at 1m in the design calculation.

b) The shear stress in the slab is normally negligible and is not that critical

except when there is a large point load.

c) Compression reinforcement is rarely needed.

d) Normally a slab supports a uniformly distributed load.

11.1.1 Types of slabs

There are various types of slab in the construction industry. They are as follows:

a) Solid slabs

b) Flat slabs

c) Ribbed slabs

d) Waffle slabs

e) Pre-stressed and post-stressed slabs

Solid slabs are widely used because they are easier to construct, easier to form

into various shapes and cheaper compared to the other types of slabs. In the

following lesson, we will focus our attention on the design of solid slabs only.

3


11.1.2 Analysis method.

In determining the shear force and bending moment for slabs, we are going to use

BS 8110’s bending moment and shear force coefficients. The use of this method

is limited to rectangular slabs, which are supported on four of its sides. For other

types of slabs, the method taught in the Theory of Structural Design could be

used.

11.1.3 Design

Reinforced Concrete Solid Slabs (RCSS) can be designed into a one-way

spanning slab or two-way spanning slab, depending on the number and

arrangement of the supports, which could be either beams or walls.

4


11.1.4 One-Way spanning slab

In a one-way spanning slab, the main reinforcement is designed to span in one

direction only. This can only happen when the slab is supported only on its two

sides as shown in Figure 11.2 below;

Figure 11.2:Elevation of a one-way spanning slab (supported on 2 slabs)

For slabs supported on four sides as shown in Figure 11.3 below, it is considered

as a one-way spanning slab if the ratio Ly / Lx is greater than 2.

5


Ly is the longer side and Lx the shorter side.

Figure 11.3: One-way spanning slab (supported on 4 sides)

For these types of slabs, the main reinforcement is in the direction of span

because the slab is spanning in one direction. Reinforcement, which is

perpendicular to the direction of span, is also known as distribution bars. They act

as ties to the main reinforcement and help to distribute any stress caused by any

change in temperature and shrinkage of concrete.

The analysis and design of simply supported one-way spanning slab is similar to

the analysis and design of simply supported beams. You have successfully passed

the limit on the design of this type of beam.

The design of this type of slab is given in the following example:

Example.

6

Lx

Ly


The slab shown in Figure11.4 is to be designed to support the following loads;

Live load = 3.0 kN/m2

Floor finishes and ceiling load = 1.0 kN/m2

The characteristic strength of materials used is:

fcu = 30 N/mm2 and fy = 460N/mm2.

The span-effective depth ratio is 20 and a nominal cover of 25mm is to be

provided.

Solution.

The minimum effective depth is calculated as follows:

d = span

20 x modification factor

The modification factor can be taken as 1.3 for reinforced slab that is not

reinforced with too much steel.

Therefore, d = 4500mm 20 x 1.3

= 173mm

7

Figure 11.4 : Simply supported slab

4.5 m


Let’s say d = 70 mm (round off value). Assume that we use a decimeter of the

slab; h is calculated as follows;

h = 170 + 25 +5

= 200 mm

The self-weight of slab = 200 x 24 x 10-3

= 4.8 kN/m 2

Total dead load = 1.0 + 4.8 = 5.8 kN/m 2

Consider a width of 1 m: -

Ultimate load = (1.4 gk + 1.6 qk ) x 4.5

= (1.4 x 5.8 + 1.6 x 3.0) x 4.5

= 58.1 kN

Therefore,

M = 58.1 x 4.5 8

= 32.7 kN/m

Main Reinforcement:

Lever arm, Z = 0.95d

= 0.95 x 70

= 161 mm

8


= 508 mm 2 /m

Provide T10 at 150 mm centre to centre

i.e. T10 – 150 c/c (As = 523 mm2/m)

Check whether d = 170 mm provided is enough by calculating the following:

= 1.13

From BS8110 Table 3.11,

fs = 288 N/mm2 and the modification factor for tension reinforcement, is 1.34.

Therefore,

Span (limit) = 20 x 1.34effective depth

= 26.8

Span (actual) = 4500effective depth 170

= 26.5

It shows that d = 170mm and this is satisfactory.

9


The distribution steel is calculated using the minimum percentage of

reinforcement from Table3.27, BS 8110 as shown below:

As = 0.13bh 100

= 0.13 x 1000 x 200 100

= 260 mm 2 /m

Provide T10- 300c/c

The reinforcement detail is shown as follows:

Plan

Answer the following questions: -

11.1 What is the definition of slab?

10

T10 - 150

T10 - 300

ACTIVITY 11a

Figure 11.5: The reinforced details of bar


____________________________________________________________

____________________________________________________________

____________________________________________________________

11.2 Why are the designed of slabs similar to the design of beams?

____________________________________________________________

____________________________________________________________

____________________________________________________________

11.3 State the four (4) reasons why slabs are easier to design.

a) _________________________________________________________

_________________________________________________________

b) _________________________________________________________

__________________________________________________________

c) _________________________________________________________

__________________________________________________________

d) _________________________________________________________

__________________________________________________________

11.4 State the five (5) types of slabs.

a)__________________________________________________________

b)__________________________________________________________

c) _________________________________________________________

d) _________________________________________________________

11


e) _________________________________________________________

11.5 Given Ly = 6.0 m and Lx = 5.0 m. Determine whether this is a one-way or

two-way spanning slab and whether it is supported at all four of its sides.

____________________________________________________________

____________________________________________________________

____________________________________________________________

11.6 What is meant by ‘main reinforcement’?

____________________________________________________________

____________________________________________________________

____________________________________________________________

11.7 Where is steel distributed and placed in a slab?

____________________________________________________________

____________________________________________________________

____________________________________________________________

11.8 What is the purpose of distribution steel in a slab?

____________________________________________________________

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____________________________________________________________

12


11.9 If the thickness of a slab is 100 mm. What is its self-weight in kN/m2?

____________________________________________________________

____________________________________________________________

____________________________________________________________

11.10 Given that M = 50 kN/m, fy = 250 N/mm2 and Z = 85mm, calculate the

required area of the main reinforcement.

____________________________________________________________

____________________________________________________________

____________________________________________________________

Check your answers given below:

13

FEEDBACK 11a


11.1 A slab is a structural element where the dimension of its thickness is far

less than its width.

11.2 A slab is a flexural structural element similar to that of a beam.

11.3 a) A slab’s width is fixed to 1m

b) A shear stress is very small and not critical.

c) Compression reinforcement not needed.

d) A slab supports only uniformly distributed loads.

11.4 a) Solid slab

b) Flat solid

c) Ribbed slab

d) Waffle slab

e) Pre-stressed and post-stressed slab

11.5 Ly = 6.0 = 1.0 Lx 5.0

This is a two-way spanning slab because Ly/Lx is less than two.

11.6 Main reinforcement is the steel reinforcement provided in a slab, which is

placed in the direction of the span. This is referred to as a one-way

spanning slab.

11.7 Distribution bars are placed perpendicular to the direction of the main

reinforcement.

14


11.8 They are meant to distribute the loads from the main reinforcement to the

supporting beam or walls and to counter any effect of temperature change

and shrinkage of concrete.

11.9 Self-weight = thickness x concrete density

= 0.10m x 24 kN/m2

= 2.4 kN/m 3

11.10 Required area of main steel reinforcement,

As = M = 50 x 10 6 mm2

0.87fyZ 0.87 x 250 x 85

= 2705 mm 2

You should get all the answers correct. If not, please go through this unit again

until you are satisfied with your answers. Good students always produce

maximum effort.

11.2 Design of one-way continuous slab

15

INPUT 2


For one-way continuous slab, reinforcements are required at the bottom section at

mid-span while support reinforcements are required at the top section. The

effective span is the distance between the centerline of the supports. The ratio of

span-effective depth is equal to 26. For this type of slab, the bending moments

and shear forces are obtained from Table 3.13, BS 8110 provided that the

following guidelines are fulfilled:

11.11 The area of each bay is greater than 30 m2

11.12 q k < 1.25 gk

11.13 Qk <5.0 kN/m2

11.2.1 Example:

A one-way continuous slab carries a live load of 3.0 kN/m2 and loads due to floor

finishes and ceiling load of 1.0 kN/m2. The characteristic strength of concrete and

reinforcement are 30 N/mm2 and 460 N/mm2 respectively. Determine the

reinforcement required.

Refer to the slab shown below in Figure 11.6

16

7.0 m


Solution:

Span (ratio) = 26 d

d = Span = 4500 = 173 mm26 26

Try d = 40 mm. For mild exposure, the cover is 25 mm, therefore the slab

thickness; h is equal to 170mm.

Self-weight = 170 x 24 x 10-3 = 4.08 kN/m 2

Total dead load = .0 + 4.08 = 5.08 kN/m 2

Ultimate load for each span,

F = (1.4gk + 1.6qk) x 4.5

= (1.4 x 5.08 + 1.6 x 4.5)

= 53.6 kN per metre width

Because a) Bay area > 30 mm2,

17

4.5 m 4.5 m 4.5 m 4.5 m

Figure 11.6: A one-way continuous slab


b) Equal span and Qk <1.25 Gk

c) qk < 5.0 kN/m2

Therefore we can use the bending moments given in Table 3.13, BS8110. This is

shown below: -

For the first outer panel;

M = 0.086 F

= 0.086 x 53.6 x 4.5

= 20.8 kNm

M = 20.8 x 10 6 = 1.06 bd2 1000x 1402

From Table 3.11, BS 8110; Modification factor for tension reinforcement is 1.36.

Therefore,

Span (limit) = 26 x 1.36 = 35.3 d

Span (actual) = 4500 = 32.1 d 140

This shows that d = 140 mm, thus it is satisfactory.

The area of reinforcement required is calculated as follows: -

As = M = 20.8 x 10 6 0.87fyZ 0.87 x 460 x 0.95 x 40

= 391 mm 2 per metre width

Provide T10 bars with a distance of 200 mm centre to centre (As = 393 mm2/m)

The area of distribution steel reinforcement is calculated as shown below: -

18


As = 0.13 bh 100

= 0.13 x 1000 x 170 100

= 221 mm 2 /m

Provide T10 bars at 300 mm centres. Refer to Figure 11.7

Now answer the following questions: -

11.11 State the table in BS 8110 that can be used to calculate the bending

moments and shear forces for one-way spanning continuous slab.

19

ACTIVITY 11b

T10-1-300 c/c

T10

-2-3

00c/

c

T10-2-300 c/c

T10-1-300 c/c


____________________________________________________________

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____________________________________________________________

11.12 What are the three requirements you need before you use Table 3.13?

___________________________________________________________

___________________________________________________________

___________________________________________________________

11.13 Given that each span of a continuous slab is 5.0 m and the ultimate load is

60 kN per metre width. What is the bending moment at the middle of the

interior spans assuming all the requirements in question 2 are fulfilled?

____________________________________________________________

____________________________________________________________

____________________________________________________________

11.14 If d = 150 mm, fy = 250 N/mm2, use bending moment in question 3 to

calculate the area of the main reinforcement required.

____________________________________________________________

____________________________________________________________

____________________________________________________________

11.15 Calculate the area of the distribution steel required for the slab in question

4 if the thickness of the slab is 180 mm.

20


____________________________________________________________

____________________________________________________________

____________________________________________________________

Compare your answers with the ones given below: -

11.11 Table 3.13, BS 8110: Part 1

21

FEEDBACK 11b


11.12 a) The area of each bay exceeds 30m2

b) q k does not exceed 1.25 gk

c) qk does not exceed 5 kN/m2

11.13 From Table 3.13,

M = 0.063 F

= 0.063 x 60 x 5.0

= 18.9 kNm per metre width

11.14 As = M

0.87 fyZ

= 18.9 x 106

0.87 x 250 x 0.95 x 50


11.15 As = 0.24 bh

100

= 0.24 x 1000 x 180

100


22

Now proceed, if all your answers are correct or otherwise go back to the input


11.3 Two-way spanning slab .

The main reinforcement of a two-way spanning slab is designed to carry bending

moments in both directions. This occurs when a slab is supported at its four sides

23

INPUT 3


and its ratio is less or equal than 2. Two-way spanning slabs can be designed

either as a simply supported slab or as a restrained slab. Restrained slab are slab

where the corners are prevented from lifting and provision for torsion are made.

This will depends on the condition of connections between the slab and the

supporting elements.

11.3.1 Simply supported two-way spanning slab

In this type of slab, the slab has only one panel and the side of slab is not

restrained from the uplifting effect of the loads. This occurs when steel beams

support the slab or if there is a non-monolithic connection between the slab and

beam. The bending moments for this type of slab can be calculated by the method

as described in clause 3.5.3.3, BS 8110 and is illustrated in the example on the

next page.

11.3.2 Example

Given that Ly = 6.3

L x = 4.5

h = 220 mm

fcu = 30 N/mm2

fy = 460 N/mm2

qk = 10 kN/m3

Design this slab by considering it as a simply supported slab.

24


Solution:

Ly = 6.3 = 1.4Lx 4.5

Because this ratio is less than 2, therefore this slab is designed as two-way

spanning slab. The bending moment coefficient is obtained from Table 3.15,

BS8110 and the values are as follows:

sx = 0.099 and sy = 0.051

Now calculate the self-weight of the slab as follows:

Self-weight of slab = thickness x concrete density

= 220 mm x 24 x 10-3 kN/m3

= 5.3 kN/m 2

The ultimate load is,

= 1.4 gk + 1.6 qk

= 1.4 x 5.3 + 1.6 x 10.0

= 23.4 kN/m 2

For bending in shorter span:

For moderate exposure, take d = 185 mm

The bending moment in the shorter span, is calculated as follows:

Msx = sx

= 0.099 x 23.4 x 4.52

= 46.9 kNm

The lever arm, Z is,

25


Z = 0.95d

= 0.95 x 185

= 176 mm

The area of reinforcement required in the shorter direction is,

As = Msx = 46.9 x 106

0.87 fyZ 0.87 x 460 x 176

= 666mm 2 /m

Provide T12 bar at a distance of 150mm centre to centre, i.e. T12 – 150 c/c

From Table 3.11, BS 8110 Part 1,

fs = 288 N/mm2, the modification factor for tension reinforcement is equal

to 1.25.

Therefore,

Span (limiting) = 20 x 1.25 = 25.0

Effective depth

Span (actual) = 4500 = 24.3

Effective depth 185

There, d = 185 mm is satisfactory.

26


For bending in the longer span :

Msy = sy

= 0.051 x 23.4 x 4.52

= 24.2 kNm

*Notice that = 4.5 m and NOT 6.3 m!

The effective depth in this longer span is smaller than the shorter span,

Therefore;

Z = 176 – 12 mm = 164 mm

The area of the reinforcement required in this span is calculated as follows:

As = Msx

0.87fyZ

= 46.9 x 106 = 666m 2 /m

0.87 x 460 x 176

Provide T12 bar at a distance of 150mm centre to centre. (As = 393 mm2/m).

27


Check that the area provided is greater than the specified minimum. This is shown

below;

100 As = 100 x 393 bh 1000 x 220

= 0.18 %

This value is greater than the minimum i.e. 0.13 % (From Table 3.27, BS 8110)

The detailed designed reinforcement is shown below

Figure 11.8: The detailed designed reinforcement

Answer the following questions: -

11.16 What is meant by a two-way spanning slab?

____________________________________________________________

____________________________________________________________

28

4.5 m

T10 - 200

ACTIVITY

11c


____________________________________________________________

____________________________________________________________

11.17 Given that Lx = 5.5 m and Ly = 6.5 m, can this slab be considered as a two-

way spanning slab? State your reasons.

____________________________________________________________

____________________________________________________________

____________________________________________________________

11.18 State the two types of two-way spanning slabs.

___________________________________________________________

____________________________________________________________

____________________________________________________________

____________________________________________________________

11.19 Give an example where a slab can be designed as a two-way simply

supported slab.

____________________________________________________________

____________________________________________________________

____________________________________________________________

29


11.20 State the table in BS 8110, which gives the values of bending moment

coefficients.

____________________________________________________________

____________________________________________________________

11.21 If Ly/Lx = 1.5, what is the value for sx and sy if this is a two-way simply

supported slab?

____________________________________________________________

____________________________________________________________

11.22 If gk = 7.5 kN/m2 and qk = 3.0 kN/m2, calculate the total ultimate load per

metre square.

11.23 If = 4.7 m and = 3.8 m. This slab is simply supported on four sides.

Calculate Msx by using the total ultimate load,. Refer to question7.

11.24 For the same slab in question 8, calculate Msy.

11.25 Calculate Asx and Asy for the slab in question 9 if d = 0 m. Assume that the

size of bar calculating, Asx as 8 mm and Asy as 6mm.

30


11.16 A slab is said to be two-way spanning slab when it is supported on all four

(4) of its sides and also if Ly / Lx is less or equal to two (2).

31

FEEDBACK 11c


11.17 Ly = 6.5 = 1.18 <2

Lx 5.5

Therefore, this is a two-way spanning slab.

11.18 a) Simply supported slab

b) Restrained slab

11.19 A slab which is supported on all four of its sides by steel beams

11.20 Table 3.14

11.21 sx = 0.0104

sy = 0.046

11.22 = 1.4 gk + 1.6 qk

= 1.4 (7.5) + 1.6 (3.0)

= 5.3 kN/m 2

11.23 Msx = ; = 4.7 = 1.2 ; sx = 0.084

Msx = 0.084 x 15.3 x 3.82

= 18.56 kNm

11.24 sy = 0.059

Msy = sy

32


= 0.059 x 5.3 x 3.82 = 13.03 kNm

11.25 Z = 0.95d

= 0.95 x 110

= 104.5 mm

Asx = Msx = 18.56 x 106

0.87fyZ 0.87 x 460 x 104.5

= 444 mm 2

For Asy,

Z = 104.5 – 8 – 6 = 97.5 mm

2 2

Asy = Msy = 13.03 x 106

0.87fyZ 0.87 x 460 x 97.5

= 334 mm 2

1. Reinforced concrete solid slabs are designed according to Clause 3.5, BS

8110: Part 1.

33

SUMMARY


2. Slabs are easier to design because they behave like beams and their width

are limited to 1 metre.

3. Solid slabs can be categorised as one-way or two-way spanning slab

depending on the :

Ratio Ly. Lx

4. Two-way spanning slabs can be divided into simply supported slab and

restrained slab.

5. The area of main reinforcement required in a one-way spanning slabs is

As = M 0.87fyZ

6. The area of the main reinforcement required in a two-way spanning slab is

calculated from the formula given as;

Asx = Msx and Asy = Msy

0.87fyZ 0.87fyZ

7. Deflection of slab is a critical factor, thus it has to be checked.

8. The area of the distribution steel is calculated as follows:-

34


for fy = 250 N/mm2, As = 0.24 bh 100

for fy = 460 N/mm2, As = 0.13 bh 100

9. The maximum spacing of the reinforcement in a slab is three times the

effective depth (3d) or 750 mm which ever is the lesser.

Part 1:

Refer to the question given below to answer questions 1-10.

35

SELF-ASSESSMENT


A simply supported reinforced concrete slab spans between two brick walls on

both sides. The span is 3.0 m from centre to centre. The slab is designed to carry a

dead load of 0.35 kN/m2 (including its self-weight) and a live load of 2.5 kN/m2.

The construction materials to be used are concrete of grade 25 and steel

reinforcement of grade 250. The cover to main reinforcement to be provided is 20

mm. Assume that the size of bar is 10mm. Assume that the modification factor for

tension reinforcement as 1.5. The slab is shown below in Figure 11.8

1. What is the total ultimate load on the slab per unit width, if h = 125 mm?

A. 0.87 kN/m2

B. 8.7 kN/m2

C. 86.7 kN/m2

D. 869.0 kN/m2

2. The maximum moment in kN/m if h = 125 mm is …

36

5000 mm

3000 mm

Figure 11.8: Simply Supported Reinforced Concrete Slab


A. 9.78

B. 7.8

C. 978

D. 9,780

3. The self-weight of this slab (in kN/m2) if h = 125 mm is …

A. 1.0

B. 2.0

C. 3.0

D. 4.0

4. The effective depth, d for this slab, if h = 125 mm is …

A. 80 mm

B. 90 mm

C. 100 mm

D. 110 mm

5. The lever arm, Z (in mm) with reference to d in question 4 is…

A. 85

B. 95

C. 105

D. 115

6. The area of main reinforcement required in mm2/m is …

37


A. 173

B. 273

C. 373

D. 473

7. The area of the distribution steel in mm2/m required is …

A. 100

B. 200

C. 300

D. 400

8. R10 is used as the main reinforcement, what is the maximum clear spacing

between these bars in (mm) refer to d in question 4.

A. 300

B. 350

C. 400

D. 450

9. If R10 is used as the main reinforcement, and using all related values

calculated earlier, the actual modification factor for tension reinforcement

is…

A. 1.00

B. 1.47

C. 1.66

D. 2.00

38


10. Using the effective depth, d in question 4, and modification factor in

question 9, the span (limit) is equal to …

D

11.25.1 20

11.25.2 26

11.25.3 40

11.25.4 36

Part 2:

For questions 11-20, refer to figure given below: -

39

6@ 4m

9 m


The figure above is the first floor plan of an office building. It is estimated that

the live load on the floor slab is 4.0kN/m2 and the load due to floor finishes and

ceiling below it is 0.5kN/m2. This building requires a fire resistance of one hour.

By using concrete of grade 30 and high yield steel reinforcement, answer

questions 11 to 20.

12 If the modification factor for tension reinforcement is 1.2, the effective

depth, d is…..

12.11.1 118 mm

12.11.2 128 mm

12.11.3 138 mm

12.11.4 148 mm

13 The nominal cover for the slab is…

13.11.1 20 mm

13.11.2 25 mm

13.11.3 30 mm

13.11.4 35 mm

14 The self-weight of the slab is…

14.11.1 3.96 kN/m2

14.11.2 4.96 kN/m2

40

Floor Plan


14.11.3 5.96 kN/m2

14.11.4 6.96 kN/m2

15 By using h = 165 mm. The design load is…

15.11.1 10.64 kN/m2

15.11.2 11.64 kN/m2

15.11.3 12.64 kN/m2

15.11.4 13.64 kN/m2

16 We can use Table 3.13 BS 8110 to calculate the bending moments for this

slab EXCEPT when ….

16.11.1 area of bay exceed 30 mm2

16.11.2 q k is less than 1.25 gk

16.11.3 qk is less than 5.0 kN/m2

16.11.4 all spans are equal

17 The bending moment near the middle of end span is equal to…

17.11.1 17.39 kNm per metre




18 The bending moment at the middle of interior spans is equal to…



41




19 The area of main reinforcement is required at the middle of end spans and

the first interior support is equal to… (use d = 135 mm)

19.11.1 39 mm2/m

19.11.2 139 mm2/m

19.11.3 239 mm2/m

19.11.4 339 mm2/m

20 The area of main reinforcement is required at the middle of interior span

and interior support is equal to…

20.11.1 228 mm2/m

20.11.2 238 mm2/m

20.11.3 248 mm2/m

20.11.4 258 mm2/m

21 Using bending moment in question 16 and d = 135 mm, the modification

factor for tension reinforcement is equal to…

21.11.1 0.16

21.11.2 1.58

21.11.3 1.68

21.11.4 1.78

Part 3:

For questions 21 – 30, refer to this question.

42


Steel beams on its four sides as shown below support a slab’s panel of an office

building, which measures 5m x 7m. The thickness is 225 mm. The estimated dead

load inclusive of self-weight, cement screed, floor finishes and building services

is 6.20kN/m2.

The live load is estimated to be 2.5kN/m2. Concrete of grade 30 and high yield

steel reinforcement are used. The exposure condition is mild and the slab needs a

fire resistance of 1.5 hours.

Question:

22 The nominal cover of reinforcement of this slab is…

22.11.1 20 mm

22.11.2 25 mm

22.11.3 30 mm

22.11.4 35 mm

23 The design load of the slab is…

23.11.1 12.68 kN/m2

23.11.2 13.68 kN/m2

43

7500 mm

5000 mm


23.11.3 14.68 kN/m2

23.11.4 15.68 kN/m2

24 This slab is designed as a two-way spanning slab because…

24.11.1 Ly < 2.0 Lx

24.11.2 Simply supported on four sides

24.11.3 Ly = 1.5Lx

24.11.4 All of the above

25 sx is equal to…

25.11.1 0.074

25.11.2 0.093

25.11.3 0.104

25.11.4 0.113

26 Msy is equal to…

26.11.1 13.57 kN/m

26.11.2 14.58 kN/m

26.11.3 15.58 kN/m

26.11.4 16.58 kN/m

27 If d in the x-direction is 195 mm and a bar of diameter 10 mm is used.

Calculate d in the y-direction if a bar used in this direction is 10 mm. Use

the cover in question 2.

27.11.1 185 mm

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27.11.2 195 mm

27.11.3 175 mm

27.11.4 165 mm

28 The x-direction using d = 195 mm is equal to...

28.11.1 345 mm2/m

28.11.2 445 mm2/m

28.11.3 545 mm2/m

28.11.4 645 mm2/m

29 Using d in question 26, the y-direction is equal to..

29.11.1 207 mm2/m

29.11.2 307 mm2/m

29.11.3 407 mm2/m

29.11.4 507 mm2/m

30 The area of the nominal reinforcement required in this slab is equal to ..

30.11.1 93 mm2

30.11.2 193 mm2

30.11.3 293 mm2

30.11.4 393 mm2

31 Is Asreq = 445 mm2 and Asprov = 449 mm2, the service stress, fs is equal

to….

31.11.1 200 N/mm2

31.11.2 250 N/mm2

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31.11.3 288 N/mm2

31.11.4 285 N/mm2

Here are the answers to the self-assessment test. Award 1 mark for every correct

answer. Calculate the total marks that you have scored. Use this formula :

Total correct answer x 100% =

30

Part 1:

1. B

2. A

46

Part 3:

21. B

22. A

23. D

24. C

25. B

26. A

27. B

28. A

29. C

30. D

Part 2:

11. B

12. B

13. A

14. C

15. D

16. A

17. C

18. D

19. C

20. B

______ %marks

FEEDBACK OF SELF-ASSESSMENT


3. C

4. C

5. B

6. D

7. C

8. A

9. D

10. C

You should have scored 80% or better to pass this unit. If you have scored less

than 80%, you should work through this unit or parts of the unit again. Good

Luck!

47

Now, I understand, it’s easy like ABC. I like this subject very much.

“Not all those who wonder are lost”

END OF UNIT 11

Unit 11 ( DESIGN OF REINFORCED CONCRETE SLABS (RCS) )

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