unit 11 ( design of reinforced concrete slabs (rcs) )
Post on 10-Mar-2015
5.621 Views
Preview:
TRANSCRIPT
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
= 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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
____________________________________________________________
____________________________________________________________
____________________________________________________________
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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?
____________________________________________________________
____________________________________________________________
____________________________________________________________
12
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
____________________________________________________________
____________________________________________________________
____________________________________________________________
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
____________________________________________________________
____________________________________________________________
____________________________________________________________
Compare your answers with the ones given below: -
11.11 Table 3.13, BS 8110: Part 1
21
FEEDBACK 11b
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
= 610 mm 2 per metre width
11.15 As = 0.24 bh
100
= 0.24 x 1000 x 180
100
= 432 mm 2 per metre width
22
Now proceed, if all your answers are correct or otherwise go back to the input
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
____________________________________________________________
____________________________________________________________
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
= 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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
17.11.2 18.39 kNm per metre
17.11.3 19.39 kNm per metre
17.11.4 20.39 kNm per metre
18 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
41
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
18.11.3 12.74 kNm per metre
18.11.4 13.74 kNm per metre
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
44
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
45
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT11/
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
top related