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Basic Structural Element Design to BS 8110-1:1997
2.0
Beam Design
Bending Moment Diagramm 1000 800 ) m N 600 k ( 400 t n 200 e m 0 o M -200 g n -400 i d n -600 e B -800 -1000 0
2
4
6
8
10
12
Length (m)
Bending Moment Diagramm
Figure 2.1: Bending Moment Diagram (Span 4-6)
Shear Force Diagramm 500 ) 400 m 300 N k ( 200 t n 100 e m 0 o M -100 g n i -200 d n e -300 B -400 -500
443kN
438kN
0
2
4
6
8
10
12
Length (m)
Shear Force Diagramm
Figure 2.2: Shear Force Diagram (Span 4-6)
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Basic Structural Element Design to BS 8110-1:1997
Reference
Calculations
Out Put
Design of Hogging Moment at LHS Support
M=804Nm Beam Size 350mmx750mm and Span of the Beam in 12.9m Cover to reinforcement is 25mm. Assume T25 bars and T10 Links are to be used. Hence Effective Depth (d) d = 750 − 30 − 10 − Cl.3.4.4.4 BS 8110:1-1997
Provide 4T25+4T25 at support. (See Figure 7.3 ) 2 A s,provided = 3928mm
Table 3.25 BS 8110:1-1997
4T25+4T25 At support 4
Checks for Minimum area of reinforcement
100A
s = 100 × 3928 = 1.49 > 0.13 bh 350 × 750
Hence minimum steel requirement is satisfied. Clause 3.12.6
Checks for Maximum area of reinforcement
BS 8110:1-1997
Neither the area of tension reinforcement nor compression
Minimum r/f Ok.
Maximum r/f Ok
reinforcement should exceed 4% of cross sectional area of the concrete.
Figure 7.3: Beam Section at support 4/6
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Figure 6.4: Beam Section at mid span of 4/6
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Basic Structural Element Design to BS 8110-1:1997
Reference
Calculations
Out Put
Design of Hogging Moment at RHS Support
M=784.8kNm Beam Size 350mmx750mm Cover to reinforcement is 25mm. Assume T25 bars and T10 Links are to be used. Hence Effective Depth (d) d = 750 − 30 − 10 − Cl.3.4.4.4 BS 8110
k=
M f bd2 cu
=
25 2
= 697.5mm
784.8 × 106 30 × 350 × 697.52
= 0.154 < 0.156
Hence the beam can be designed as a singly reinforced beam.
0.9
0.154
z = d0.5 + 0.25 −
z = d0.5 + 0.25 −
As =
M 0.95f y z
=
k
0.9
= 0.79d < 0.95d
784.8 × 10
6
0.95 × 460 × 0.79 × 697.5
= 3259mm2
Provide 4T25+4T25 at support. (See Figure 7.3 ) 2 A s,provided = 3928mm
Table 3.25 BS 8110:1-1997
4T25+4T25 At support 6
Checks for Minimum area of reinforcement
100A
s = 100 × 3928 = 1.49 > 0.13 bh 350 × 750
Hence minimum steel requirement is satisfied. Cl. 3.12.6
Checks for Maximum area of reinforcement
BS 8110:1-1997
Neither the area of tension reinforcement nor compression
Minimum r/f Ok.
Maximum r/f Ok
reinforcement should exceed 4% of cross sectional area of the concrete.
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Reference
Calculations
Out Put
Design for Sagging Bending Moment between supports
M=859.5kNm(as per analyzed results with SAP 2000) At Mid Span Top of the beam in compression and can be designed as a flange beam. (“T” Beam) Hence effective width of the flange beam (“T” beam) Cl.3.4.1.5 BS 8110:1-1997
lz = 0.7 × 12900 = 9030mm b w = 350 +
9030 5
= 2156mm
Moment of resistance of the Flange “T” beam when the Neutral Axis is at the slab bottom level,
h = 0.45f cubwhf × d − f
M
2
= 0.45 × 30 × 2156 × 125 × 697.5 −
125 2
= 2310.29kNm > 859.5kNm Hence Neutral axis is within the bottom flange. Cl.3.4.4.4 BS 8110
k=
M f bd2 cu
=
859.5 × 106 30 × 2156 × 697.52
= 0.026 < 0.156
Hence the beam can be designed as a singly reinforced beam.
0.9
0.026
z = d0.5 + 0.25 −
z = d0.5 + 0.25 −
k
0.9
= 0.97d > 0.95d
Hence z =0.95d As =
M 0.95f y z
=
859.5 × 10
6
0.95 × 460 × 0.95 × 697.5
= 2947mm2
Provide 4T25+4T25 at mid span. (See Figure 7.4 )
4T25+4T25
2 A s,provided = 3928mm
At mid span
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Basic Structural Element Design to BS 8110-1:1997
Reference
Table 3.25 BS 8110:1-1997
Calculations
Out Put
Checks for Minimum area of reinforcement
100A
s = 100 × 3928 = 1.49 > 0.13 bh 350 × 750
Hence minimum steel requirement is satisfied.
Clause 3.12.6
Checks for Maximum area of reinforcement
BS 8110:1-1997
Neither the area of tension reinforcement nor compression
Minimum r/f Ok.
reinforcement should exceed 4% of cross sectional area of the concrete.
Maximum r/f Ok.
Checks for shear
Maximum shear at LHS Support= 443kN Cl.3.4.5.2 BS 8110:1-1997
Cl.3.4.5.2 BS 8110:1-1997
vmax =
443 × 10
3
(350 × 697.5)
= 1.81N/mm2
Maximum permissible shear stress is lesser of 0.8 f cu or 5N/mm2
0.8 30 = 4.38N/mm > 1.80N/mm 2
2
Hence maximum shear is O.K
Maximum Shear Ok
Shear force at distance “d” from the face of the support V = 430kN d
Hence , shear stress at distance “d” from the face of the support V = 430kN d
v=
430 × 10
3
(350 × 697.5)
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= 1.75N/mm2
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Basic Structural Element Design to BS 8110-1:1997
Reference
Cl. 3.4.5.4 BS 8110:1-1997
Calculations
Out Put
Concrete shear Stress
The area of tension reinforcement at a distance “d” from the
Hence slenderness limit for lateral stability is satisfied.
Cl.3.4.6
Checks for Deflection
BS 8110:1-1997
Mid span of the section is designed as a flanged section.
b b Table 3.9 BS 8110:1-1997
350
=
2156
w
= 0.16 < 0.3
For simply supported beam,
= 16 Effective Depth Span
Basic
For a Continuous beam,
= 20.8 Effective Depth Span
Basic
Cl.3.4.6 .4
Span of the beam is 12.9m.
BS 8110:1-1997
10 Hence modification factor for span is 12.9 In this beam end condition at support 6 can be regarded as simply supported and end condition at support 4 can be regarded as continuous.
Hence deflection check is satisfied The deflection of the beam is checked in SAP 2000 under Serviceability limit state and found to be 13.6mm which is less than 20mm , the allowable deformation in general to avoid cracking of brittle partition walls Deflection Ok
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Basic Structural Element Design to BS 8110-1:1997
Reference
Calculations
Out Put
Detailing of the reinforcement
Curtailment of Bars Curtailment of bars is done according to the guidelines specified in BS 8110-1:1997 and standard method of detailing manual. Procedure of curtailment of bars at right hand side of support 4 is described here. At mid span provided reinforcement for maximum bending is 4T25+4T25. It is desired to curtail 4T25 bars. As per Figure 3.24 of the BS 8100-1:1997 at least 30% of reinforcement for maximum sagging moment should be continued throughout the beam. In this case curtailment percentage is 50%. Hence area of continuing reinforcement is ,
3928 × 0.5 = 1964mm2 For the sectional equilibrium of simplified stress block of a rectangular section at ultimate state (See Figure 2.3) Compression of concrete =Tension of bottom steel
0.67f cu δm
× b × 0.9x = 0.95 × f y × A s
0.67 × 30 1.5
× 350 × 0.9x = 0.95 × 460 × 1964
Hence X =203.3mm x
=
d
203.3 697.5
Z = d−
= 0.291 < 0.5
0.9x 2
= 697.5 −
0.9 × 203.3 2
= 606.02kNm
Hence moment of resistance (MOR) is,
M = 0.95f y AsZ = 0.95 × 460 × 1964 × 606.02 = 524.42kNm
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Basic Structural Element Design to BS 8110-1:1997
Figure 2.3: Simplified stress block of a rectangular section at ultimate state
Curtailment of Reinforcement 1000 800 600 ) m 400 N k ( 200 t n e m 0 o M g -200 n i d n e -400 B
262.2kNm
-600
524.4kNm
-800 -1000 0
2
4
6
8
10
12
Length (m)
Bending Moment Diagramm
MOR of 4T25
1/2 MOR of 4T25
Figure 7.4: Bending moment diagram and MOR
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Basic Structural Element Design to BS 8110-1:1997
Reference
Calculations
Out Put
Bending moment diagram of the beam segment and Moment or Resistance corresponds to 4T25 is drawn in Figure 6.6 The point where MOR line cuts the bending moment diagram is referred to as the Theoretical Cut Off Point. (TCP). Cl.3.12.9.1
In a flexural member , every bar should be extended beyond
BS 8110:1-1997
the TCP for a distance equal to greater of 1.0
Effective depth of the member =702.5mm
2.0
Twelve times the bar size
=12x25 = 300mm
The point of physically cutting off the bars is referred as (PCP) In addition the least distance of, 3.0 Full anchorage bond length =40x25 = 1000mm 4.0 At PCP actual shear < Half shear capacity 5.0 At PCP moment< Half moment at TCP Condition 4 is not critical and hence not checked. Distance to TCP from centerline of LHS is 3.6m. Distance to PCP as per condition “5” from TCP is 1m. Hence Distance to PCP from centre line of Support is 3.61=2.6m Hence 4T 25 is curtailed 2m from the centre line of the support.
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Reference
Calculations
Out Put
Anchorage
Anchorage length required beyond the face of the column is given by Table 3.27
L = 40 × φ = 40 × 25 = 1000mm
BS 8110:1-1997
Stress in reinforcement is,
= 0.95 × 460 ×
3315 3928
= 368.80N/mm2
Assume bend starts at the centre line of column. Stress at the start of the bend is,
= 368.80 ×
(1000 − 350) 1000
= 239.72N/mm2
3.12.8.25.2
Bending stress inside the bend
BS 8110:1-1997
2f Fbt cu ≤ rφ 1 + 2( φ
a b
)
a = 30 + 25 = 55mm b 239.72 × 491 r × 25
≤
2 × 30 1 + 2(25 ) 55
r = 149.80mm Hence use a radius of 150mm. The bend is started as close as possible to the edge of the column. Hence length from the face of the column to start of the bend is
Minimum distance between bars should not be less than
= 5+h
agg
= 5 + 20 = 25mm
C.l.3.12.11.2.3
Maximum Clear Horizontal distance between bars in tension
BS 8110:1-1997
with 0% of moment redistribution is 155mm. Hence maximum spacing of bars in tension is satisfied.
C.l.3.12.11.2.6
Bars near side faces of beams exceeding 750mm overall depth
BS 8110:1-1997
Spacing in not exceeding 250mm near the face of the beam. The distribution should be a distance of two third of the beam’s overall depth measured from its tension face.
Cl.3.12.5.4
Minimum size of bars in side faces of beams to control cracking