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

REINFORCED CONCRETE STRUCTURAL DESIGN

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UNIT 11 DESIGN OF REINFORCED CONCRETE SLABS (RCS)

OBJECTIVES

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

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.

slab


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INPUT 1

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

Figure 11.1: Simply supported slab

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.


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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.


11.1.2

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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.


11.1.4

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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.


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Ly is the longer side and Lx the shorter side. Ly

Lx

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.


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4.5 m

Figure 11.4 : Simply supported slab

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


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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/m2 Total dead load = 1.0 + 4.8 = 5.8 kN/m2

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


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As 32 .7 ×10 6 =M = 0.87 f y Z 0.87 × 460 ×161

=

508 mm2/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: M 32.7 x 10 6 = bd 2 1000 x 170 2

= 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 1.34 effective depth = 26.8

Span (actual) = 4500 effective depth 170 = 26.5 It shows that d = 170mm and this is satisfactory.


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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 mm2/m

Provide T10- 300c/c

The reinforcement detail is shown as follows:

T10 - 300

T10 - 150 Plan

Figure 11.5: The reinforced details of bar

ACTIVITY 11a Answer the following questions: 11.1

What is the definition of slab?


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___________________________________________________________ ___________________________________________________________ ___________________________________________________________ ___

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

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State the five (5) types of slabs. a)__________________________________________________________ b)__________________________________________________________ c) _________________________________________________________ d) _________________________________________________________ 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? ___________________________________________________________ ___________________________________________________________


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___________________________________________________________ ___

11.8

What is the purpose of distribution steel in a slab? ___________________________________________________________ ___________________________________________________________ ___________________________________________________________ ___

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. ___________________________________________________________ ___________________________________________________________ ___________________________________________________________ ___


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FEEDBACK 11a

Check your answers given below: 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

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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.

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/m3

11.10 Required area of main steel reinforcement, As = M = 0.87fyZ

50 x 106 mm2 0.87 x 250 x 85


=

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2705 mm2

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.

INPUT 2

11.2

Design of one-way continuous slab

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 qk < 1.25 gk 11.13 Qk <5.0 kN/m2

11.2.1 Example:


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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

7.0 m

4.5 m

4.5 m

4.5 m

Figure 11.6: A one-way continuous slab

Solution: Span (ratio) = 26 d d = Span = 4500 = 173 mm 26 26

4.5 m


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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/m2 Total dead load = .0 + 4.08 = 5.08 kN/m2

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) b) c)

Bay area > 30 mm2, Equal span and Qk <1.25 Gk

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 = bd2

20.8 x 106 = 1.06 1000x 1402

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


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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 106 0.87fyZ 0.87 x 460 x 0.95 x 40 = 391 mm2 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: As = 0.13 bh 100 = 0.13 x 1000 x 170 100 = 221 mm2/m Provide T10 bars at 300 mm centres. Refer to Figure 11.7

T10-2-300c/c

7.0 m

T10-1-300 c/c

T10-2-300 c/c

4.5 m

4.5 m

4.5 m

Figure 11.7: Detailed drawing of the Reinforcement bar

4.5 m


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T10-1-300 c/c

ACTIVITY 11b

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. ___________________________________________________________ ___________________________________________________________ ___________________________________________________________ ___

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


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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. ___________________________________________________________ ___________________________________________________________ ___________________________________________________________ ___


FEEDBACK 11b

Compare your answers with the ones given below: -

11.11 Table 3.13, BS 8110: Part 1

11.12 b)

a)

The area of each bay exceeds 30m2

qk does not exceed 1.25 gk c)

qk does not exceed 5 kN/m2

11.13 From Table 3.13, M = 0.063 F 

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= 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 = 610 mm2 per metre width

11.15

As = 0.24 bh 100 = 0.24 x 1000 x 180 100 = 432 mm2 per metre width

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

INPUT 3

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

and its

Ly Lx

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.


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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. Solution: Ly = 6.3 = 1.4 Lx 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: α

sx

= 0.099 and α

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/m2 The ultimate load is, η = 1.4 gk + 1.6 qk = 1.4 x 5.3 + 1.6 x 10.0 = 23.4 kN/m2

For bending in shorter span: For moderate exposure, take d = 185 mm The bending moment in the shorter span, is calculated as follows: Msx = α

sx

η x 2

= 0.099 x 23.4 x 4.52 = 46.9 kNm

The lever arm, Z is, Z = 0.95d = 0.95 x 185 = 176 mm The area of reinforcement required in the shorter direction is, As =

Msx 0.87 fyZ

=

46.9 x 106 0.87 x 460 x 176

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=

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666mm2/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

Effective depth

185

There, d = 185 mm is satisfactory.

For bending in the longer span: Msy = α

sy

η x 2

= 0.051 x 23.4 x 4.52 = 24.2 kNm *Notice that x = 4.5 m and NOT 6.3 m!

= 24.3


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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

=

666m2/m

0.87 x 460 x 176 Provide T12 bar at a distance of 150mm centre to centre. (As = 393 mm2/m).

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)


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The detailed designed reinforcement is shown below T10 - 200

4.5 m

Figure 11.8: The detailed designed reinforcement

ACTIVITY 11c

Answer the following questions: 11.16 What is meant by a two-way spanning slab? ___________________________________________________________ ___________________________________________________________ ___________________________________________________________ ___________________________________________________________ ____

11.17 Given that Lx = 5.5 m and Ly = 6.5 m, can this slab be considered as a twoway spanning slab? State your reasons. ___________________________________________________________ ___________________________________________________________


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___________________________________________________________ ___

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. ___________________________________________________________ ___________________________________________________________ ___________________________________________________________ ___

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 α

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sx

and α

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 y = 4.7 m and x = 3.8 m. This slab is simply supported on four sides. Calculate Msx by using the total ultimate load,η . 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.


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FEEDBACK 11c

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).

11.17 Ly Lx

=

6.5 = 1.18 <2 5.5

Therefore, this is a two-way spanning slab.


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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 α

sx

= 0.0104

α

sy

= 0.046

11.22 η = 1.4 gk + 1.6 qk = 1.4 (7.5) + 1.6 (3.0) = 5.3 kN/m2

Msx

αSXηχ

Y X 3.8 = 0.084 x 15.3 x 3.82

11.23 Msx =

2

;

= 4.7 = 1.2 ;

= 18.56 kNm

11.24 α

sy

= 0.059

Msy = α syη x 2 = 0.059 x 5.3 x 3.82

11.25 Z

= 0.95d = 0.95 x 110 = 104.5 mm

=

13.03 kNm

α

sx

= 0.084


Asx =

Msx

18.56 x 106

=

0.87fyZ

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0.87 x 460 x 104.5

= 444 mm2 For Asy, Z = 104.5 – 8 – 6 = 97.5 mm 2 2

Asy =

Msy

=

0.87fyZ =

13.03 x 106 0.87 x 460 x 97.5

334 mm2

SUMMARY

1.

Reinforced concrete solid slabs are designed according to Clause 3.5, BS 8110: Part 1.

2.

Slabs are easier to design because they behave like beams and their width are limited to 1 metre.


3.

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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 =

6.

M 0.87fyZ

The area of the main reinforcement required in a two-way spanning slab is calculated from the formula given as; Asx =

Msx

and

Asy =

0.87fyZ

Msy 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:for fy = 250 N/mm2,

As = 0.24 bh 100

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


9.

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The maximum spacing of the reinforcement in a slab is three times the effective depth (3d) or 750 mm which ever is the lesser.

SELF-ASSESSMENT

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


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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

5000 mm

3000 mm Figure 11.8: Simply Supported Reinforced Concrete Slab

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 … 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 … A.

173

B.

273

C.

373

D.

473

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

8.

A.

100

B.

200

C.

300

D.

400

R10 is used as the main reinforcement, what is the maximum clear spacing between these bars in (mm) refer to d in question 4.

9.

A.

300

B.

350

C.

400

D.

450

If R10 is used as the main reinforcement, and using all related values calculated earlier, the actual modification factor for tension reinforcement is…

10.

A.

1.00

B.

1.47

C.

1.66

D.

2.00

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

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Part 2: For questions 11-20, refer to figure given below: 6@ 4m

9m

Floor Plan

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/m 2 and the load due to floor finishes and ceiling below it is 0.5kN/m2. This building requires a fire resistance of one hour.


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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…..

13

14

15

12.11.1

118 mm

12.11.2

128 mm

12.11.3

138 mm

12.11.4

148 mm

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

The self-weight of the slab is… 14.11.1

3.96 kN/m2

14.11.2

4.96 kN/m2

14.11.3

5.96 kN/m2

14.11.4

6.96 kN/m2

By using h = 165 mm. The design load η 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 …. area of bay exceed 30 mm2

16.11.1 16.11.2

qk is less than 1.25 gk

17

18

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16.11.3

qk is less than 5.0 kN/m2

16.11.4

all spans are equal

The bending moment near the middle of end span is equal to… 17.11.1

17.39 kNm per metre

17.11.2

18.39 kNm per metre

17.11.3

19.39 kNm per metre

17.11.4

20.39 kNm per metre

The bending moment at the middle of interior spans is equal to… 18.11.1

10.74 kNm per metre

18.11.2

11.74 kNm per metre

18.11.3

12.74 kNm per metre

18.11.4

13.74 kNm per metre


19

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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. 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


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load inclusive of self-weight, cement screed, floor finishes and building services is 6.20kN/m2. 7500 mm

5000 mm

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

23

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

The design load of the slab is… 23.11.1

12.68 kN/m2

23.11.2

13.68 kN/m2

23.11.3

14.68 kN/m2

23.11.4

15.68 kN/m2


24

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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 = 1.5 Lx

24.11.4 25

26

α

sx

All of the above

is equal to… 25.11.1

0.074

25.11.2

0.093

25.11.3

0.104

25.11.4

0.113

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

27.11.2

195 mm

27.11.3

175 mm


27.11.4

28

29

30

C4301/UNIT11/ 46

165 mm

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

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

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

31.11.3

288 N/mm2

31.11.4

285 N/mm2


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FEEDBACK OF SELF-ASSESSMENT

Here are the answers to the self-assessment test. Award 1 mark for every correct Part 2: Part 3: answer. Calculate the total marks that you have scored. Use this formula : 11. B 21. B Total correct answer x 100% = ______ %marks 12. B 30 22. A 13. A

23. D

1. B

14. C

24. C

2. A

15. D

25. B

3. C

16. A

26. A

4. C

17. C

27. B

18. D

28. A

19. C

29. C

20. B

30. D

Part 1:


C4301/UNIT11/ 48

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!

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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