A possible solution to the struct-hub second design assessment. Inspired by the civic centre building 2018 involving wide slab panels of solid slab construction
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Possible solution struct_hub_design assessment
1. StructuresCentre.xyz
1
POSSIBLE SOLUTION TO STRUCT-HUB: DESIGN ASSESSMENT
II
Omotoriogun Victor Femi
Copyright ยฉStructures centre, 2021. All rights reserved
18th February 2021
structurescentre@gmail.com
1.0 DESIGN SOLUTION
Preliminary sizing of the structural element has been carried using the guidance provided
in: Preliminary sizing of structural elements as well as the guidance provided in the
publication by the Concrete Centre- Economic Concrete Frame Elements.
2. StructuresCentre.xyz
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3.1 Actions
3.1 1 Roof
Permanent Actions = 1.5๐๐/๐
Variable Actions = 0.6๐๐/๐
3.1.2 Floors
Permanent Actions:
i. Self weight of slab = 0.20 ร 25 = 5๐๐/๐ (๐ถ๐๐๐ ๐ ๐๐๐๐)
= 0.15 ร 25 = 3.75๐๐/๐ (๐๐๐๐๐ค๐๐ฆ)
ii. Finishes & Services = 1.5๐๐/๐
iii. Partition Allowances = 1.0๐๐/๐
Total Permanent Actions = ๐ = 7.5๐๐/๐ & 6.25๐๐/๐
Variable Actions:
i. Floor Imposed Loading (Classroom & Walkways) = ๐ = 3.0๐๐/๐
Design Value of Actions:
By inspection the permanent actions are less than 4.5 times the variable actions, therefore equation
6.13b of BS EN 1990 will give the most unfavourable results.
Roof
Design Actions= 1.35๐๐ + 1.5๐ =(1.35 ร 0.925 ร 1.5) + (1.5 ร 0.6) = 2.77๐๐๐
Design Permanent Load 1.35๐๐ = 1.35 ร 0.925 ร 1.5 = 1.87๐๐/๐
Classroom Panels
Design Actions= 1.35๐๐ + 1.5๐ =(1.35 ร 0.925 ร 7.5) + (1.5 ร 3) = 13.87๐๐๐
Design Permanent Load 1.35๐๐ = 1.35 ร 0.925 ร 7.5 = 9.37๐๐/๐
Walkway Panels
Design Actions= 1.35๐๐ + 1.5๐ =(1.35 ร 0.925 ร 6.25) + (1.5 ร 3) = 12.30๐๐๐
3. StructuresCentre.xyz
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Design Permanent Load 1.35๐๐ = 1.35 ร 0.925 ร 6.25 = 7.80๐๐/๐
3.2 Slab Panels
3.2.1 Classroom Panels
๐ฟ
๐ฟ
=
9000
7500
= 1.2 (๐๐ค๐ ๐ค๐๐ฆ ๐๐๐๐ โ ๐๐ค๐ ๐ด๐๐๐๐๐๐๐ก ๐๐๐๐๐ ๐ท๐๐ ๐๐๐๐ก๐๐๐๐ข๐ )
Using coefficients from table for two ways slabs with two adjacent sides discontinuous.
Short span coefficients = โ0.064 & 0.047
Long span coefficients = โ0.045 & 0.034
3.2.1.1 Flexural Design
Negative Moment at Support (Short Span)
๐ = โ0.064๐ , ๐ = โ0.064 ร 13.87 ร 7.5 = โ49.93๐๐. ๐/๐
Assuming cover to reinforcement of 20mm, 12mm bars
d = h โ c + links +
โ
= 200 โ 20 + = 174mm; b = 1000mm
k =
M
bd f
=
49.93 ร 10
1000 ร 174 ร 20
= 0.0825
z = d 0.5 + โ0.25 โ 0.882k โค 0.95d
=d 0.5 + 0.25 โ 0.882(0.0825) โค 0.95d
=0.92d = 0.92 ร 174 = 160.08mm
A =
M
0.87f z
=
49.93 ร 10
0.87 ร 410 ร 160.08
= 874.42mm /m
Try Y12mm bars @ 125mm Centres (As, prov = 904mm2
)
Positive Moment at Midspan (Short Span)
๐ = 0.047๐ , ๐ = 0.047 ร 13.87 ร 7.5 = 36.67๐๐. ๐/๐
Assuming cover to reinforcement of 20mm, 12mm bars
4. StructuresCentre.xyz
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d = h โ c + links +
โ
= 200 โ 20 + = 174mm; b = 1000mm
k =
M
bd f
=
36.67 ร 10
1000 ร 174 ร 20
= 0.0606
z = d 0.5 + โ0.25 โ 0.882k โค 0.95d
=d 0.5 + 0.25 โ 0.882(0.0606) โค 0.95d
=0.94d = 0.94 ร 174 = 163.56mm
A =
M
0.87f z
=
36.67 ร 10
0.87 ร 410 ร 163.56
= 628.54mm /m
Try Y12mm bars @ 125mm Centres (As, prov = 904mm2
)
Negative Moment at Support (Long Span)
๐ = โ0.045๐ , ๐ = โ0.045 ร 13.87 ร 7.5 = โ35.11๐๐. ๐/๐
Assuming cover to reinforcement of 20mm, 12mm bars
d = h โ c + links +
โ
+ โ = 200 โ 20 + + 12 = 162mm; b = 1000mm
k =
M
bd f
=
35.11 ร 10
1000 ร 162 ร 20
= 0.067
z = d 0.5 + โ0.25 โ 0.882k โค 0.95d
=d 0.5 + 0.25 โ 0.882(0.067) โค 0.95d
=0.94d = 0.94 ร 162 = 152.28mm
A =
M
0.87f z
=
35.11 ร 10
0.87 ร 410 ร 152.28
= 646.38mm /m
Try Y12mm bars @ 150mm Centres (As, prov = 753mm2
)
Positive Moment at Midspan (Long Span)
๐ = 0.034๐ , ๐ = 0.034 ร 13.87 ร 7.5 = 26.53๐๐. ๐/๐
Assuming cover to reinforcement of 20mm, 12mm bars
d = h โ c + links +
โ
+ โ = 200 โ 20 + + 12 = 162mm; b = 1000mm
5. StructuresCentre.xyz
5
k =
M
bd f
=
26.53 ร 10
1000 ร 162 ร 20
= 0.051
z = d 0.5 + โ0.25 โ 0.882k โค 0.95d
=d 0.5 + 0.25 โ 0.882(0.051) โค 0.95d
=0.95d = 0.95 ร 162 = 153.9mm
A =
M
0.87f z
=
26.53 ร 10
0.87 ร 410 ร 153.9
= 483.28mm /m
Try Y12mm bars @ 150mm Centres (As, prov = 753mm2
/m)
3.2.1.2 Deflection Verification
Deflection verification can be carried using either of the two alternative method provided in
section 7.4 of Eurocode 2 (Part 1). The conservative span-effective method and the rigorous
calculation approach which involve using theoretical expression to estimate the actual
deflection of the slab. The span-effective method is used here, the rigorous method is not
suitable for hand calculations, spreadsheets or finite element softwareโs are required for fast
calculations.
Basic Requirement: โฅ
= ๐ ร ๐พ ร ๐น1 ร ๐น2 ร ๐น3
๐ =
๐ด ,
๐ด
=
๐ด ,
๐๐
=
628.54
(1000 ร 174)
= 0.36%
ฯ = 10 f = 10 ร โ20 = 0.45% since ฯ < ฯ
N = 11 +
1.5 f ฯ
ฯ
+ 3.2 f
ฯ
ฯ
โ 1 = 11 +
1.5โ20 ร 0.45
0.36
+ 3.2โ20
0.45
0.36
โ 1
= 21.20
F1 = 1.0
K = 1.3 (end spans)
6. StructuresCentre.xyz
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F2 =
7.0
l
=
7.0
7.5
= 0.93
F3 =
310
ฯ
โค 1.5
ฯ =
f
ฮณ
g + ฯq
n
A ,
A ,
โ
1
ฮด
=
410
1.15
ร
7.5 + 0.6(3)
13.87
ร
628.54
904
= 166.21Mpa
F3 =
310
166.21
= 1.87 > 1.5
L
d
= 21.20 ร 1.3 ร 1.0 ร 0.93 ร 1.50 = 38.45
L
d
=
span
effective depth
=
7500
174
= 43.10
Since actual span-effective depth ratio is greater than the limiting span-effective depth ratio. It
shows that we might have a deflection problem. Hence, the options available includes, increasing
the slab thickness, the concrete class or even the steel bars grade. However, readers are reminded
that, the span/effective depth approach is only a fast and conservative method of verifying
deflection. In this case the rigorous calculation method was used and the slab found to perform
satisfactory with respect to deflection.
3.2.1.3 Detailing Checks.
The minimum area of steel required in panel:
A , = 0.26
f
f
๐ d โฅ 0.0013bd
f = 0.30f = 0.3 ร 20 = 2.21Mpa
A , = 0.26 ร
2.21
410
ร 1000 ร 174 โฅ 0.0013 ร 1000 ร 174
= 243.85mm . By observation it is not critical anywhere in slab. Hence adopt all steel
bars.
3.2.2 Walkway Panels
The walkways panel are one-way slabs which can be idealized as propped cantilevers.
7. StructuresCentre.xyz
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Moment in Spans
๐ =
๐ , ๐
10
=
12.30 ร 2.25
10
= 6.23๐๐. ๐/๐
Since moment is relatively low. Provide Y12 @ 200mm Centres (As, prov = 565mm2
/m)
3.3 Concrete Beams
3.3.1 Beam C-C (300x750)
3.3.1.1 Actions on Beam
Permanent Actions:
Span 1
a. Equivalent uniformly distributed load transferred from slab to beam
= 2 ร
๐ ๐
6
3 โ
๐
๐
= 2 ร
7.5 ร 7.5
6
3 โ
7.5
9.0
= 43.31kN/m
b. self-weight of beam = (0.75 โ 0.2) ร 0.3 ร 25 = 4.125kN/m
c. Walls = (3.75 โ 0.75) ร 3.5 = 10.5kN/m
Permanent Actions G = 43.31 + 4.125 + 10.5 = 57.94kN/m
Span 2
a. Equivalent uniformly distributed load transferred from slab to beam = 0kN/m
b. self-weight of beam = (0.45 โ 0.15) ร 0.3 ร 25 = 2.25kN/m
Permanent Actions G = 2.25kN/m
Variable Actions:
Span 1
a. Equivalent uniformly distributed load transferred from slab to beam
= 2 ร
๐ ๐
6
3 โ
๐
๐
= 2 ร
3.0 ร 7.5
6
3 โ
7.5
9.0
= 17.32kN/m
Variable Actions Q = 17.32kN/m
Design Value of Actions on Beam
8. StructuresCentre.xyz
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By inspection the permanent actions are less than 4.5 times the variable actions, therefore
equation 6.13b of BS EN 1990 will give the most unfavourable results.
Span 1:
Design Load = 1.35๐๐บ + 1.5๐ = (1.35 ร 0.925 ร 57.94) + (1.5 ร 17.32) = ๐๐. ๐๐๐ต/๐
Design Permanent Load 1.35๐๐บ = 1.35 ร 0.925 ร 57.94 = ๐๐. ๐๐๐ต/๐
Span 2:
Design Load = 1.35๐๐บ + 1.5๐ = (1.35 ร 0.925 ร 2.25) + (1.5 ร 0) = ๐. ๐๐๐๐ต/๐
Design Permanent Load 1.35๐๐บ = 1.35 ร 0.925 ร 2.25 = ๐. ๐๐๐๐ต/๐
3.3.1.2 Analysis of Beam
Approximate methods of analysis could be used, this requires, the use of simple coefficients
reflecting the approximate value of the internal forces within the structural element.
However, an analysis of the subframe will be carried out here, with the basic aim of obtaining
the loads & moment transferred to the column as well as the internal forces on the beam.
Only the shear and bending moment diagrams are presented here. The reader is expected to
already have a basic knowledge on analysis of subframes. However, guidance can be obtained
from: How to Analyse Element in Frames.
The load cases considered are:
๏ท All spans loaded with the maximum design loads
๏ท Alternate spans loaded with the maximum design load while the other spans are loaded
with the design permanent actions
10. StructuresCentre.xyz
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3.3.1.3 Flexural Design
End Support (3-2)
M = 200.6kN. m
Assuming cover to reinforcement of 25mm, 25mm tensile bars, 16mm compression bars & 8mm
links
d = c + links +
โ
2
= 25 + 10 +
16
2
= 43๐๐
d = h โ c + links +
โ
= 450 โ 25 + 10 + = 402.5mm; b = 300mm
k =
M
bd f
=
200.6 ร 10
300 ร 402.5 ร 20
= 0.021 > 0.168 (Section is doubly reinforced)
๐ด =
(๐ โ ๐ )๐ ๐๐
0.87๐ (๐ โ ๐ )
=
(0.21 โ 0.168) ร 20 ร 300 ร 404.5
0.87 ร 410 ร (402.5 โ 43)
= 321.54๐๐
Try 3Y16mm bars Bottom (As, prov = 602mm2
).
z = d 0.5 + โ0.25 โ 0.882k โค 0.95d
=d 0.5 + 0.25 โ 0.882(0.21) โค 0.95d
=0.75d = 0.75 ร 402.5 = 301.88mm
A =
๐๐ ๐๐
0.87f z
+ ๐ด =
0.168 ร 20 ร 300 ร 402.5
0.87 ร 410 ร 301.88
+ 321.54 = 1838.08mm
Try 4Y25mm bars Top (As, prov = 1962mm2
).
Interior Support (2-1)
M = 531.4kN. m
Assuming cover to reinforcement of 25mm, 25mm tensile bars, 16mm compression bars & 8mm
links
d = c + links +
โ
2
= 25 + 10 +
16
2
= 43๐๐
d = h โ c + links +
โ
= 750 โ 25 + 10 + = 702.5mm; b = 300mm
k =
M
bd f
=
531.4 ร 10
300 ร 702.5 ร 20
= 0.018 > 0.168 (Section is doubly reinforced)
11. StructuresCentre.xyz
11
๐ด =
(๐ โ ๐ )๐ ๐๐
0.87๐ (๐ โ ๐ )
=
(0.18 โ 0.168) ร 20 ร 300 ร 702.5
0.87 ร 410 ร (702.5 โ 43)
= 151.05๐๐
Try 3Y16mm bars Bottom (As, prov = 602mm2
) .
z = d 0.5 + โ0.25 โ 0.882k โค 0.95d
=d 0.5 + 0.25 โ 0.882(0.18) โค 0.95d
=0.80d = 0.80 ร 702.5 = 562mm
A =
๐๐ ๐๐
0.87f z
+ ๐ด =
0.168 ร 20 ร 300 ร 702.5
0.87 ร 410 ร 562
+ 151.05 = 2632.55mm
Try 4Y25mm + 4Y20mm bars Top (As, prov = 3218mm2
). To be spread across the effective width.
b = b + b , + b , โค b
b , = 0.1l = 0.1 ร 0.15(l + l ) = 0.1 ร 0.15(2250 + 7500) = 146.25mm
b = 300 + 146.25 + 146.25 = 592.5mm
Therefore, this reinforcement will be spread across a width of 592.5mm.
Span (2-1)
M = 524.2kN. m
Assuming cover to reinforcement of 25mm, two layers of 25mm tensile bars, 16mm compression
bars & 8mm links
d = c + links +
โ
2
= 25 + 10 +
16
2
= 43๐๐
d = h โ c + links +
โ
= 750 โ 25 + 10 + = 677.5mm;
b = b = b + b , + b , โค b
b , = b , = 0.2b + 0.1๐ โค 0.2๐
๐ =
7500 โ 150 โ 150
2
= 3600๐๐
๐ = 0.85๐ = 0.85 ร 9000 = 7650๐๐
b , = b , = (0.2 ร 3600) + (0.1 ร 7650) โค (0.2 ร 7650) = 1485๐๐
b = 300 + 1485 + 1485 = 3270mm โค 3600mm
k =
M
bd f
=
524.2 ร 10
3270 ร 677.5 ร 20
= 0.017
12. StructuresCentre.xyz
12
z = d 0.5 + โ0.25 โ 0.882k โค 0.95d
=d 0.5 + 0.25 โ 0.882(0.017) โค 0.95d
=0.95d = 0.95 ร 677.5 = 643.63mm
We have to verify the position of the neutral axis:
x = 2.5(d โ z) = 2.5(677.5 โ 643.63) = 84.68mm
Therefore x < h = 84.68 < 200 (neutral axis is within the flange)
Hence, we can design as a rectangular section.
A =
M
0.87f z
=
524.2 ร 10
0.87 ร 410 ร 643.63
= 2283.27mm
Try 5Y25mm bars Bottom in two layers (As, prov = 2452.5mm2
).
End Support (1-2)
M = 412.6kN. m
Assuming cover to reinforcement of 25mm, 25mm tensile bars, 16mm compression bars & 8mm
links
d = c + links +
โ
2
= 25 + 10 +
16
2
= 43๐๐
d = h โ c + links +
โ
= 750 โ 25 + 10 + = 702.5mm; b = 300mm
k =
M
bd f
=
412.6 ร 10
300 ร 702.5 ร 20
= 0.14 < 0.168 (Section is singly reinforced)
z = d 0.5 + โ0.25 โ 0.882k โค 0.95d
=d 0.5 + 0.25 โ 0.882(0.14) โค 0.95d
=0.86d = 0.86 ร 702.5 = 604.15mm
A =
M
0.87f z
=
412.6 ร 10
0.87 ร 410 ร 604.15
= 1914.6mm
Try 3Y25mm +3Y20mm bars Top (As, prov = 2413mm2
). To be spread across the effective width.
3.3.1.4 Shear Design
By inspection, the critical section for shear occurs at the interior support 2, hence can be used to
conservatively size the shear reinforcement for the whole beam.
14. StructuresCentre.xyz
14
= ๐ ร ๐พ ร ๐น1 ร ๐น2 ร ๐น3
๐ =
๐ด ,
๐ด
=
๐ด ,
๐ ๐ + (๐ โ ๐ )โ
=
2283.27
(300 ร 677.5) + (3270 โ 300)200
= 0.29%
ฯ = 10 f = 10 ร โ20 = 0.45% since ฯ < ฯ
N = 11 +
1.5 f ฯ
ฯ
+ 3.2 f
ฯ
ฯ
โ 1 = 11 +
1.5โ20 ร 0.45
0.29
+ 3.2โ30
0.45
0.29
โ 1
= 29.81
F1 = 0.82
K = 1.3 (end spans)
F2 =
7.5
๐
=
7.5
9.0
= 0.83
F3 =
310
ฯ
โค 1.5
ฯ =
f
ฮณ
g + ฯq
n
A ,
A ,
โ
1
ฮด
=
410
1.15
ร
57.94 + 0.6(17.32)
98.3
ร
2283.27
2453
= 230.7Mpa
F3 =
310
230.7
= 1.34
L
d
= 29.81 ร 1.3 ร 0.82 ร 0.83 ร 1.34 = 35.34
L
d
=
span
effective depth
=
9000
677.5
= 13.28
Since the limiting span-effective depth ratio is greater than the actual span-effective depth ratio. It
therefore follows that deflection has been satisfied in the end spans.
3.3.1.6 Detailing Checks
Minimum Area of Steel
A , = 0.26
f
f
๐ d โฅ 0.0013bd
f = 0.30f = 0.3 ร 20 = 2.21Mpa
15. StructuresCentre.xyz
15
Hogging at Supports
A , = 0.26 ร
2.21
410
ร 300 ร 702.5 โฅ 0.0013 ร 300 ร 702.5
= 295.4mm . By observation it is not critical anywhere at the supports.
Sagging in Spans
A , = 0.26 ร
2.21
410
ร 300 ร 677.5 โฅ 0.0013 ร 300 ร 677.5
= 284.99mm . By observation it is not critical anywhere in the spans. Hence adopt all
attempted bars.
3.3.2 Beam 2-2 (225x750)
3.3.2.1 Actions on Beam
Permanent Actions:
Span 1 & 2
a. Equivalent uniformly distributed load transferred from slab to beam
=
๐ , ๐ ,
3
+
๐ , ๐ ,
2
=
(7.5 ร 7.5)
3
+
(6.25 ร 2.25)
2
= 25.8kN/m
b. self-weight of beam = (0.75 โ 0.2) ร 0.225 ร 25 = 3.1kN/m
c. Walls = (3.75 โ 0.75) ร 3.5 = 10.5kN/m
Permanent Actions G = 25.8 + 3.1 + 10.5 = 39.4kN/m
Variable Actions:
Span 1 & 2
a. Equivalent uniformly distributed load transferred from slab to beam
=
๐ , ๐ ,
3
+
๐ , ๐ ,
2
=
(3.0 ร 7.5)
3
+
(3.0 ร 2.25)
2
= 10.88kN/m
Variable Actions Q = 10.88kN/m
Design Value of Actions on Beam
By inspection the permanent actions are less than 4.5 times the variable actions, therefore
equation 6.13b of BS EN 1990 will give the most unfavourable results.
16. StructuresCentre.xyz
16
Span 1 & 2:
Design Load = 1.35๐๐บ + 1.5๐ = (1.35 ร 0.925 ร 39.4) + (1.5 ร 10.88) = ๐๐. ๐๐๐๐ต/๐
Design Permanent Load 1.35๐๐บ = 1.35 ร 0.925 ร 39.4 = ๐๐. ๐๐๐๐ต/๐
3.3.2.2 Analysis of Beam
Coefficients for beam analysis can be used to determine the internal forces in this beam, since
the spans are equal and uniformly loaded. However, in-order to determine the column
moments, analysis of the entire subframe 2-2 will be carried out
The load cases considered are:
๏ท All spans loaded with the maximum design loads
๏ท Alternate spans loaded with the maximum design load while the other spans are loaded
with the design permanent actions
Figure 4: Subframe 2-2
17. StructuresCentre.xyz
17
Figure 5: Bending Moment Envelope
Figure 6: Shear Force Envelope
3.3.2.3 Flexural Design
End Support (A & D)
M = 172.5kN. m
Assuming cover to reinforcement of 25mm, 20mm tensile bars, 16mm compression bars & 8mm
links
d = c + links +
โ
2
= 25 + 10 +
16
2
= 43๐๐
d = h โ c + links +
โ
= 750 โ 25 + 10 + = 705mm; b = 225mm
18. StructuresCentre.xyz
18
k =
M
bd f
=
172.5 ร 10
225 ร 705 ร 20
= 0.078 < 0.168 (Section is singly reinforced)
z = d 0.5 + โ0.25 โ 0.882k โค 0.95d
=d 0.5 + 0.25 โ 0.882(0.078) โค 0.95d
=0.93d = 0.93 ร 705 = 655.7mm
A =
M
0.87f z
=
172.5 ร 10
0.87 ร 410 ร 655.7
= 737.53mm
Try 4Y16mm bars Top (As, prov = 1608mm2
).
Span (2-1)
M = 206.0kN. m
Assuming cover to reinforcement of 25mm, 20mm tensile bars, 16mm compression bars & 8mm
links
d = c + links +
โ
2
= 25 + 10 +
16
2
= 43๐๐
d = h โ c + links +
โ
= 750 โ 25 + 10 + = 705mm;
b = b = b + b , + b , โค b
b , = 0.2๐ + 0.1๐ , โค 0.2๐ ,
๐ =
9000 โ 112.5 โ 112.5
2
= 4387.5๐๐
๐ , = 0.85๐ = 0.85 ร 7500 = 6375๐๐
b , = (0.2 ร 4387.5) + (0.1 ร 6375) โค (0.2 ร 6375) = 1275๐๐
b , = 0.2๐ + 0.1๐ , โค 0.2๐ ,
๐ =
2250 โ 112.5 โ 112.5
2
= 1012.5๐๐
๐ , = 0.85๐ = 0.85 ร 7500 = 6375๐๐
b , = (0.2 ร 1012.5) + (0.1 ร 6375) โค (0.2 ร 6375) = 840๐๐
b = 225 + 1275 + 840 = 2340mm โค 4387.5mm
19. StructuresCentre.xyz
19
k =
M
bd f
=
206 ร 10
2340 ร 705 ร 20
= 0.009
z = d 0.5 + โ0.25 โ 0.882k โค 0.95d
=d 0.5 + 0.25 โ 0.882(0.009) โค 0.95d
=0.95d = 0.95 ร 705 = 669.75mm
We have to verify the position of the neutral axis:
x = 2.5(d โ z) = 2.5(705 โ 669.75) = 88.13mm
Therefore x < h = 88.13 < 200 (neutral axis is within the flange)
Hence, we can design as a rectangular section.
A =
M
0.87f z
=
206 ร 10
0.87 ร 410 ร 669.75
= 862.3mm
Try 3Y20mm bars Bottom in two layers (As, prov = 942mm2
).
Interior Support (C)
M = 373.1kN. m
Assuming cover to reinforcement of 25mm, two layers of 20mm tensile bars, 16mm compression
bars & 8mm links
d = c + links +
โ
2
= 25 + 10 +
16
2
= 43๐๐
d = h โ c + links +
โ
= 750 โ 25 + 10 + = 705mm; b = 225mm
k =
M
bd f
=
373.1 ร 10
225 ร 685 ร 20
= 0.167 < 0.168 (Section is singly reinforced)
z = d 0.5 + โ0.25 โ 0.882k โค 0.95d
=d 0.5 + 0.25 โ 0.882(0.167) โค 0.95d
=0.82d = 0.82 ร 705 = 578.1mm
A =
M
0.87f z
=
373.1 ร 10
0.87 ร 410 ร 578.1
= 1809.3mm
Try 6Y20mm bars Top (As, prov = 1884mm2
) Spread across effective width of the beam.