unit 9 ( design of shear reinforcement )
TRANSCRIPT
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
UNIT 9
DESIGN OF SHEAR REINFORCEMENT
GENERAL OBJECTIVE
To know how to design shear reinforcement in accordance with BS8110.
At the end of this unit you will be able to;
1. explain the purpose of providing shear reinforcement.
2. identify the types of shear reinforcement.
3. calculate the ultimate shear stress of a beam section.
4. calculate the area of shear reinforcement.
5. provide shear reinforcement according to Table 3.8 of BS8110.
6. specify spacing of shear reinforcement.
7. check whether the spacing is within the allowable limits.
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OBJECTIVES
SPECIFIC OBJECTIVES
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
9.1 Shear
The unit shear stress on a section is calculated from equation 3 of BS 8110, i.e.
where V is the shear force due to ultimate loads, b is the width of a section and d
is the effective depth.
…………………………Equation 3
Shear failures occur under combined shear forces and bending moments, which
causes diagonal tension in concrete. A typical failure mode is shown
diagrammatically in Figure 9.1 below:
2
Load
Support
INPUT 1
Figure 9.1: Typical failure mode
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
Shear in a reinforced concrete beam without shear reinforcement is resisted by
three main components as follows:
a) concrete in the compression zone
a) dowel actions in the bars where the concrete between the cracks transmits
shear force to the bars.
b) aggregate interlocks along the cracks.
The plane of failure occurs near the support. This is where the maximum shear
occurs. When the diagonal tension exceeds the ultimate strength of concrete,
shear reinforcement must be provided. Shear reinforcement can be provided in the
following forms:
a) links
b) bent-up bars
c) bent-up bars which are used together with links.
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REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
Complete the statements.
9.1 Shear reinforcement should be designed according to clause ______________
of BS 8110.
9.2 Shear failures are caused by shear force and ______________ moments.
9.3 _________________ tension occurs near the beam supports.
9.4 Shear reinforcement must be provided when ___________________ tension
exceeds concrete’s ultimate strength.
9.5 The forms of shear reinforcement that can be provided are links, bent-up bars
or ____________________.
9.6 Shear is resisted by the combined action of concrete in the compression zone,
dowel actions and _________________________.
9.7 Shear stress in a section is calculated using the equation given by
_____________________.
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ACTIVITY
9a
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
Answers.
9.1 Clause 3.4.5
9.2 Bending
9.3 Diagonal
9.4 Diagonal
9.5 combination of links and bent-up bars
9.6 aggregate interlocks
9.7
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FEEDBACK 9a
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
9.8 Shear Reinforcement
Concrete is weak in tension. Shear failure is caused by a failure in diagonal
tension with cracks running at 45 to the beam axis. Shear reinforcement is
provided by bars, which cross the cracks and either vertical links or inclined bars
are used. The following equation is used to calculate the size and spacing of links;
Where, Asv = total cross section of links (this is taken as 2 )
sv = spacing of the links
b = breadth of section
v = design shear stress at a cross-section
vc = design concrete shear stress
fyv = characteristic strength of the links
This equation is given in Table 3.5 in the code while vc can be obtained from
Table 3.9. It depends on the grade of concrete used and percentage of tension
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INPUT 2
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
reinforcement provided. Note that an increase in the amount of tension
reinforcement increases the capacity of the concrete in shear. As is the area of
tension reinforcement, which continues for a distance and that, is equal to the
effective depth beyond the section where the shear is checked.
If v is less than vc or located for the region 0.5vc < v < (vc + 0.4), a minimum shear
reinforcement in the form of links must be provided. This is calculated using the
following equation;
Minimum or nominal links provide a design shear resistance of 0.4 N/mm2.
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REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
Rewrite the two important equations that are used to design shear links as follows;
9.8 Provide design links when (vc+ 0.4) < v < 0.8 , equation to be used
____________________________
9.9 Provide nominal links when v < 0.5vc , and 0.5vc < v < (vc + 0.4)
Equation to be used ____________________________
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ACTIVITY 9b
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
The answers are given below:
9.8
9.9
9
FEEDBACK 9b
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
9.10 Size and spacing of links
When a bar type and diameter of the link is chosen, the spacing sv can be
determined. Values of for various link bar size and spacing can be obtained
from the following table;
The spacing sv is limited to 0.75d in the direction of the span as stated in Clause
3.4.5.5 of the code. This limitation is to ensure that every potential crack is
Sv
8 10 12 16100 1.00 1.57 2.26 4.02125 0.81 1.26 1.81 3.22150 0.67 1.05 1.51 2.68175 0.58 0.90 1.29 2.30200 0.50 0.79 1.13 2.01225 0.45 0.70 1.00 1.79250 0.40 0.63 0.90 1.61275 0.34 0.57 0.82 1.46300 0.33 0.52 0.75 1.34
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INPUT 3
Table 9.1: Link bar size and spacing, Sv
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
crossed by at least one link. Links must be at least one-quarter of the size of the
largest bar at spacing not greater than 12 times the size of the smallest bar. Every
corner and alternate bar is to be supported by a link passing around the corner.
9.10 If , determine the following;
a) size of link
b) link spacing
c) maximum allowable distance of the link if d = 450mm
d) If R8 is used, calculate the area of the links, Asv?
e) Calculate sv for the link in Question 4 if
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ACTIVITY 9c
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
9.10 Let’s check your answers now!
a) R8
b) 200 centres
c) 0.75 x 450 = 338 mm centres (taken as 325 mm centres)
d) = 101 mm2
e) sv = = 139.6 mm (taken as 125 mm centres)
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FEEDBACK 9c
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
9.4 Design example
Figure 9.2 show reinforced concrete beams, which have been designed to carry
bending moments. Design the shear reinforcements for the beams using fyv =250
N/mm2 and fcu = 30 N/mm2.
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INPUT 4
6 m
35 kN/m
105 kN
105 kN
Figure 9.2: Shear Force Diagram
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
Refer to Figure 9.3
Solution:
The ultimate shear stress at support is
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200 mm
d = 300mm
2T16
3T25
(1470 mm2)
Section
Figure 9.3: Cross-section of beam
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
=1.75 N/mm 2
0.8
= 4.38 N/mm 2 .
Since , section the dimensions are satisfactory
= 2.45
= 1.33
=
= 0.97 N/mm 2
= 1.37 N/mm 2
Since design shear links are required.
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REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
= 0.72 mm
Refer to Table 9.1
Provide R10 at 200 centre to centre (c/c).
0.75d = 0.75 x 300
= 225 mm
Therefore 200mm c/c is suitable.
Use R10-200 c/c.
Nominal Link
= 0.37 mm
Refer to Table 9.1 select R8 at 250
0.75d = 225 mm
250 mm distance is not o.k.
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REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
Use R8-200
Shear capacity of R8-200: -
0.50 =
v – vc = 0.54
v = 0.54 + 0.97
= 1.51 N/mm 2
Shear force capacity, V = vbd
= 1.51 x 200 x 300
= 90.8 kN
Arrangement of links;
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105 kN
90.8 kN
x
3 m 90.8 kN 105 kNFigure 9.5: Shear Capacity with Arrangement of Link
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
The distance from support, x where V = 90.8kN is calculated as follows (using the
theorem of equilateral triangle)
x = 0.40 m
The details of the reinforcement are shown below;
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R10-200R8-200 R8-200
0.4 m 0.4 m
Figure 9.6: Details of the reinforcement supported beam
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
`
1. Calculation of the ultimate shear stress,
2. If v > 0.87 or 5 N/mm2, increase the beam dimensions.
3. Calculation of concrete shear resistance, vc using Table 3.9 of the code or
the following equation;
4. If v < vc, shear reinforcement is not required but in all beams of structural
importance, minimum links are to be provided. This is calculated using the
following equation;
5. If 0.5vc < v < (vc + 0.4), minimum links are provided. This is calculated
using the same equation in step 4, i.e. ;
6. If (vc + 0.4) < v < 0.8 , design shear links are provided. This is
calculated using the following equation;
7. Check that sv < 0.75d
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SUMMARY
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
Circle the correct answer. Award 1 mark for every correct answer.
SECTION A (10 marks)
1. The direction of the shear force in a beam is ______________ of
the beam.
A. along the axis
B. perpendicular to the axis
C. on a normal plane to the axis
D. vertically downwards
2. The shear forces causes sheer stress on the____________ of the
beam.
A. transverse plane
B. horizontal plane
C. on the planes transverse and parallel to the axis
D. on all the planes
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SELF-ASSESSMENT
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
3. The shear force is usually associated with _____________
A. bending moment.
B. torsional moment.
C. normal thrust.
D. none of the above.
4. The shear force on a beam is generated by ___________________
A. axial forces only.
B. transverse forces only.
C. bending moment on the beam too.
D. none of the above.
5. Shear stress is caused by __________________
A. pure shear force only.
B. pure bending moment.
C. torsional force only.
D. transverse shear and torsional moment.
2. The shear stress on a beam section is maximum ______________
A. at the centroid of the section.
B. on the extreme free-surfaces fibres.
C. at the neutral axis but not at the centroid.
D. at the free edges.
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REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
3. The shear stress in a beam is zero ___________________
A. at the centroid of the section.
B. on the extreme free surfaces fibres.
C. at the neutral axis but not at the centroid.
D. at the free edges.
4. The shear stress caused by a transverse shear force (V) is given by
the equation ___________________.
A.
B.
C.
D.
5. If V = 316 kN, b = 400 mm and d = 550 mm , the shear stress is __
A. 0.44 N/mm2
B. 1.44 N/mm2
C. 2.44 N/mm2
D. 3.44 N/mm2
6. The area , Asv of R12 links is_________________ .
A. 26 mm2
B. 126mm2
C. 226mm2
D. 326mm2
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REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
SECTION B (27 marks)
A reinforced concrete beam which is 300mm wide and 600mm deep is required to
span 6.0m between the centres of supporting piers 300mm wide. The beam carries
dead and imposed loads of 25 kN/m and 19 kN/m respectively. Design the shear
reinforcement using the following information:
1. Mild exposure
2. Ultimate design load = 428.7 kN
3. Main Steel : 4T25 (As = 1960 mm2 )
4. Effective depth , d = 552 mm
5. Materials characteristic strength :
i) fcu = 30 N/mm2 ,
ii) fy = 460 N/mm2
iii) fyv = 250 N/mm2,
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REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
Section A
1. C
2. C
3. A
4. C
5. D
6. A
7. B
8. A
9. B
10. C
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(1 mark x = 10 marks)
FEEDBACK ON SELF-ASSESSMENT
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
Section B
………………………….
…………………………….
Since the beam is symmetrically loaded,
RA = RB = = 214.4kN………………………………………..
Ultimate shear force, V = 214.4 kN………………………………….
25
RA RB
V
V
1
1
1
1
( 4 marks )
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
Design shear stress,
v = …………………………………………………………….
=
= 1.29 N/mm 2 ………………………………………………………
Permissible shear stress = 0.8 = 4.38 N/mm 2 …………………
Since v < 4.38 N/mm2, beam dimensions are satisfactory. ……………
( 4 marks )
Design concrete shear stress;
………………………………………………..
= 1.18……………………………………………………….
From Table 3.9 of BS 8110; ……………………………………….
…………………………………………………...
= 0.70 N/mm 2 …………………………………………………….
(5 marks)
Diameter and spacing of links;
26
1
1
1
1
1
1
1
1
1
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
Where v < (vc + 0.4) = 0.70 + 0.4 = 1.10 N/mm 2 , nominal links are
required according to; ………………………………………………
=
= 0.552 mm……………………………………………………….
From Table 9.1, …………………………………………………….
Provide R10 at 275 mm centres where v < 1.10 N/mm2…………..
(4 marks)
Distance of links;
…………………………………………………………
x =
= 2.558 mm…………………………………………………….
27
1.10 N/mm2
1.10 N/mm2
x 1.29 N/mm2x
2/29.1 mmN
1
1
1
1
1
1
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
…………………….
R10 – 275 links are provided 2.558 mm on either side of mid-span of the
beam. ……………………………………………………………….
Where
v > (vc + 0.4) = 1.1.0 N/mm 2 …………..………………………….....
Design links are required according to;
=
= 0.81 mm……………………………………………………….
Provide R10 at 175 mm centres where v > 1.10 N/mm2 at 0.442m from
the supports. ………………………………………………………
(8 marks)
Maximum spacing of links;
28
2
1
1
1
1
1
REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT9/
0.75d = 0.75 x 552 = 414 mm……………………………………….
Therefore,
The spacing of R10 – 275 centres and R10 – 175 centres are
O.K. ……………………………………………………….
(2 marks)
END OF UNIT 9
29
1
TOTAL = 27 marks
Yes, I did it. Actually it’s not difficult for you and me to understand. If we are smart and do practise frequently, we will soon get it. For the next unit, I try to do better.