REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT13/
UNIT 13
DESIGN OF SHORT BRACED COLUMNS
GENERAL OBJECTIVE
To understand how to design short braced reinforced concrete columns
according to BS 8110 requirements.
At the end of this unit you will be able to;
1. calculate the area of longitudinal reinforcement of short braced axially
loaded columns.
2. calculate the area of longitudinal reinforcement of short braced
columns carrying an approximately symmetrical arrangement of
beams.
3. calculate the distance of ties.
4. calculate the diameter of ties.
5. calculate the area of longitudinal reinforcement of short braced
columns carrying axial load and moment.
6. use BS8110 column design charts to get the bar steel area
7. sketch the details of reinforcement.
1
OBJECTIVES
SPECIFIC OBJECTIVES
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT13/
13.1 Short columns
The effect of deflection or bending to a short column is minimal compared to a
slender column. Therefore, the short column normally fails in compression
due to the crushing of the concrete. Short column is usually designed to resist
maximum bending moment about the critical axis only. This is stated in clause
3.8.4.3 of the code. The maximum axial load that can withstand a column is
denoted by Nuz, which is calculated based on the ultimate capacity of concrete
and reinforcements. Nuz is calculated using the equation given below;
where,
Nuz = ultimate axial load
Ac = net area of column
Asc = area of longitudinal reinforcement
fcu = characteristic strength of concrete
fy = characteristic strength of reinforcement
13.2 Short Braced Axially Loaded Columns
A column can be designed as axially loaded when there is no significant
moment in the column. An example of this is in precast column where there is
no continuity among the structural elements. The equation given can be used
2
INPUT 1
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT13/
when the axial load is currently acting on the axis of a column. But this perfect
condition rarely occurs because there will always be some inaccuracy in
alignment of the reinforcement and the formworks. To allow some eccentricity
in the column, the concrete and steel stress is reduced. After the reduction, the
equation becomes;
The area of the main reinforcement, i.e. the longitudinal reinforcement can be
calculated using this equation if fcu, fy and Ac are known. This is shown in the
following example.
13.2.1 Design Example
A short braced column is to carry an axial load of 1700kN. The dimension of
the column is 300 mm square. fcu and fy are 30 and 460 N/mm2 respectively.
Use fyv = 250 N/mm2 for ties. What is the area of main reinforcement?
Solution
Using equation 38 of BS 8110,
=
= 1797 mm 2
Use 4T25 bars (Asc = 1960 mm2)
Ties
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REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT13/
Minimum size =
= 6.25 mm
Maximum size = 12 x 25 = 300 mm
Hence, use R8 at 300 centres.
Details of the reinforcements are shown below in Figure 13.1.
13.3 Braced Short Column carrying an Approximately Symmetrical
Arrangement of Beams
Bending moment in this type of column is very small. This is because the
column supports approximately symmetrical arrangement of beams. The
ultimate axial load is calculated using equation 39 of the code and is
reproduced below;
4
R8-3004T25
Section
R8
T25
T25
T25 T25
Elevation
Figure 13.1: Detail of reinforcement for column.
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT13/
This equation can be used when the following requirements are fulfilled;
a) The beam spans does not differ by 15% of the longest.
b) The beams are designed for uniformly distributed loads.
The application of this equation is shown in the following example.
13.3.1 Example
C1 is a short braced column carrying 4 beams on its sides. The beams are of
equal span of 8.0 m. Refer figure 13.2.
Given the data;
i) N = 1900 kN (from uniformly distributed load)
5
Figure 13.2: Column layout plan
C1
8m
8m
8m 8m
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT13/
ii) b= 300 mm
iii) h= 300 mm
iv) fcu = 30 N/mm2
v) fy = 460 N/mm2
vi) fyv=250N/mm2
Calculate the area of longitudinal reinforcement required to carry the load and
sketch details of the reinforcement.
Solution
The following BS 8110`s requirements are met;
a) column is short and braced
b) loads are uniformly distributed
c) spans are equal
Therefore, equation 39 of BS 8110 can be used.
Asc =
=
= 3099 mm 2
Use 4T20 and 4T25 (Asc = 1257 + 1964 = 3221 mm2 )
Ties
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REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT13/
Minimum size of ties =
= 6.25 mm
Maximum distance of ties = 12 x 20
= 240 mm
Use R8 at 225 centres.
Details of the reinforcement are as shown below;
7
R8 -225
4T20 + 4T25
SECTION
T25 T20 T25
T20 T20
T20 T25T25
R8 - 225
ACTIVITY 13
Figure 13.2: Details Of The Reinforcement
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT13/
Complete these statements.
13.1 The term short column indicates that no correction is needed to take
account for additional bending due to _______________________.
13.2 In practice, a column is never constructed absolutely plumb nor is the
load applied truly _____________________.
13.3 Some ________________________ of the load exists which induces a
degree of bending.
13.4 The degree of ________________________ accepted as permissible
within these limits is 0.05h
13.5 The expression for the capacity of the column to resist load is given by
the equation _____________________________.
13.6 The expression for the capacity of column to resist load and allowing
for the effect of small moments is ____________________________.
13.7 Design a column for N = 2500 kN within a system of beams of
approximately equal spans, if b = 300 mm and h = 500 mm. Use
fcu = 40 N/mm2 and fy = 460 N/mm2.
8
FEEDBACK 13
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT13/
Now, check your answers.
13.1 deflection
13.2 slenderness
13.3 axial
13.4 eccentricity
13.5 eccentricity
13.6
13.7
Asc =
=
= 1298 mm 2
Use 4T25 (Asc = 1964 mm 2 )
“Are your answers the same as the printed answers here? Never mind if your
answers are not correct. Please do them again until you get the right answer”
9
INPUT 2
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT13/
13.4 Short Braced Columns Carrying Axial Load and Moment
In many cases bending moment on a column may be large in relation to the
load or may be applied to a column, which is already carrying a substantial
load. For these cases, design charts are used. The bending moment to be
considered is the largest out-of-balance moment resulting from conditions of
loading. When dealing with moment, h relates to the dimension in the
direction of bending, which is not necessarily the larger dimension.
Column design charts are given in Part 3 of BS 8110. They are for columns
with equal steel in opposite faces. The area of reinforcement obtained from the
chart, Asc is the total area required, half of which is to be placed in each face.
13.5 Design Example (Bending About The Major Axis )
Design a short braced column for N = 3050kN, Mx = 30.6 kNm but the
minimum moment, i.e. 0.05Nh is 61kNm about x-axis. The following data are
known;
b = 300 mm, h = 400 mm, fcu = 35N/mm2, fy = 460N/mm2, fyv = 250 N/mm2,
cover to main reinforcement, c = 30 mm.
Solution
First calculate the ratios and as follows;
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REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT13/
= 25.4 N/mm 2
= 1.27 N/mm 2
To calculate the effective depth, we have to assume a size of bar, say 25 mm
hence,
= 357 mm
= 0.89
For fcu = 35 N/mm2 and fy = 460 N/mm2, we should use Chart No. 34 (Using
)
11
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT13/
From the chart, read the value.
(Minimum value = 0.4, maximum value = 6.0)
Therefore,
= 3840 mm 2
= 1920 mm 2
Provide this area in each opposite face.
(4T25 on each opposite face ) .
Ties
Minimum diameter =
= 6.25 mm
Maximum spacing = 12 X 25 = 300 mm
Provide R8 at 250 centres.
The details are shown below;
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REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT13/
* Note that the longitudinal reinforcement is arranged so that they are
symmetrical about the axis of bending, i.e. the x-axis.
13.6 Design Example (Bending About The Minor Axis)
The term minor axis refers to the y-axis. The bending moment in this direction
is denoted as My. The column in the previous example is to be designed for
My = 5.8kNm and N = 3050 kN.
Now, for bending about the minor axis, (bending in x-x direction);
h = 300 mm, b = 400 mm
d = 300 -30 – 13 (assuming T25 bars are used)
= 257 mm
= 0.86
13
x
x
400
300
2T25
2T25
2T25
2T25
Figure 13.3:Details of longitudinal reinforcement
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT13/
Using Chart No. 33 (for ,
= 25.4 N/mm 2
= 1.27 N/mm 2
= 3960 mm 2
Provide 6T32 (Asc = 4827 mm2)
Ties
Minimum diameter =
= 8 mm
Maximum spacing = 12 x 32 = 384 mm
Hence, use R10 at 350 centres.
Details of the reinforcement are shown in Figure 13.4.
14
y y
300
400
Figure 13.4: Details of reinforcement
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT13/ 15
SUMMARY
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT13/
1. The general method in designing braced and unbraced short columns is
to obtain the axial loads and moments from analysis.
2. The moment from analysis is compared with the minimum moment of
Nemin and the larger value is taken.
3. If certain criteria are met, it is not necessary to calculate the moments.
4. If a column is not subjected to a significant moment (less than Nemin),
equation 38 of BS 8110 can be used.
5. Equation 39 can be used for braced short column when the following
criteria are met;
a) The beams carry a uniformly distributed imposed loads
b) The beam spans do not differ by more than 15% of the longest
beam.
2. Design Charts of Part 3, BS 8110 can be used for a short braced
column carrying axial load and bending moment.
3. When moment is considered, h is taken as the dimension in the
direction of bending.
4. The minimum diameter of ties in a column is one quarter of the
smallest diameter of the main bars.
5. The maximum spacing of the ties in a column is 12 times the smallest
diameter of the main bars.
16
SELF-ASSESMENT
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT13/
Answer all the questions.
1. A 500mm by 300 mm short column is reinforced with 4T32 bars. If
fcu = 35 N/mm2, calculate the design ultimate capacity, design if Nuz of
the section if it is subjected to axial load only.
(3 marks)
2. A short braced axially loaded column 400mm by 300 mm is to be
designed for an axial load of 2500 k. Calculate the area of longitudinal
reinforcement allowing some eccentricity. Use fcu and fy as 40 and 460
N/mm2 respectively.
(4 marks)
3. Design the longitudinal reinforcement for a 450mm by 450 mm short
braced column supporting an approximately symmetrical arrangement
of beams. The axial load, N is 3000 kN. Use fcu and fy as 30 and 460
N/mm2 respectively.
(4 marks)
4. Design the longitudinal reinforcement for a 500mm by 300 mm
column section if N = 2300 kN and Mx = 300 kNm. Mx is the bending
moment about the major axis and bending is in the y – axis. Assume
the column is short and braced. The following information is known ;
fcu = 40 N/mm2
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REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT13/
fy = 460 N/mm2
(8 marks)
5. Design the longitudinal reinforcement for a 500mm by 300 mm short
braced column given that N = 2300 kN , My = 120 kNm, fcu = 40
N/mm2 , fy = 460 N/mm2 and
(8 marks)
6. Design the ties for the column in Question 5.
(3 marks)
Check your score
18
FEEDBACK ON SELF-ASSESSMENT
1
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT13/
1. …………………………………
= N……………..
= 3650 kN. …………………………………………….
(3 marks)
2.
……………………………………….....
= …………………....
= 1681 mm 2 ……………………………………………
Provide 4T25 ( Asc = 1964 mm 2 ) …………………………
(4 marks)
3.
Asc = ………………………………………
= …………………..
= 2835 mm 2 ……………………………………………..
Provide 6T25 ( Asc = 2946 mm 2 ) ………………………….
(4 marks)
4. ………………………………………..
= 15.33 N/mm 2 ………………………………………..
19
1
1
1
1
1
1
11
1
1
11
1
1
1
1
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT13/
= 4.00 N/mm 2 ……………………………………..
(3 marks)
Use chart no. 38 for ………………………….
………………………………………………
……………………………………..
= 3450 mm 2 ……………………………………………
Provide 6T32 ( Asc = 4825 mm 2 )……………………….
(5 marks)
5. In this case, b = 500 mm and h = 300 mm.
……………………………………
= 15.33 N/mm 2 ………………………………..
………………………………….
= 2.67 N/mm 2 ………………………………
From chart no. 38………………………………....
20
1
1
1
1
1
1
1
1
1
1
1
1
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT13/
………………………………………
= 1950 mm 2 …………………………………
Provide 4T25 ( Asc = 1963 mm 2 ) …………………
(8 marks)
6. Minimum diameter =
= 6.25 mm……………………..
Maximum spacing = 12 x 25 = 300 mm centres………
Use R8 at 275mm centres……………………..
(3 marks)
21
1
1
1
1
1
“The power to live with joy and victory,”
says Norman Vincent Peale “is available
to you and me. This power can lead you
to a solution to your problem, help you to
meet your difficulties successfully and fill
your heart with peace and contentment.”
END OF UNIT 13
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