Design of Ribbed slab system with one-way slab (Đồ án bê tông 1 - Thiết kế sàn sườn bê tông toàn khối)

RIBBED SLAB SYSTEM WITH ONE WAY SLAB
1. GIVEN DATA
Plan of slab structure as shown in the bellow figure.

Plan of slab

Primary beams are orientated axis 1, 2, 3, 4, 5.
Secondary beams are orientated axis A, B, C, D, E.

Slab structure

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1.1. Selection of dimensions of components of slab.
– Dimensions of aperture of plate.
+ Distance between adjacent secondary beams: l1 = 1,9 m.
+ Distance between adjacent primary beams: l2 = 5,45 m.
– Load bearing brick wall thickness: t = 34 cm. Beams are directly supported on the wall.
– Dimensions of columns: bc = hc = 30 cm.
1.2. Characteristics of the construction.
– Building construction slab which has 4 levels as shown in the figure.
– Nominal live load: ptc = 9,75 kN/m2, reliability coefficient of live load: n = 1,2.
1.3. Selection of materials.
– Concrete: Heavy concrete, Grade of concrete with compressive strength: grade B12,5.
With Rb = 7,5 MPa, Rbt = 0,66 MPa.
– Reinforcement:
+ For plate: Constructive reinforcement C-I. Main reinforcement C-I.
+ For beam: Constructive reinforcement C-I. Main reinforcement C-II.
2. DESIGN OF PLATE

2-way slab. Have: l1 = 1,9 m, l2 = 5,45 m.
l2 > 2l1, thus design approximately as in 1-way slab.
2.1. Selection of dimensions of components of slab.
– Thickness of the slab:

hb =
hb =

Dl1
m

, where: D = 1.4 with big load, m = 35 with 1-way slab.

Dl1 1,4 ×1900
=
m
35

= 76 mm.

Thus, select hb = 80 mm.
– Dimensions of the secondary beams:

hdp =

1
1
l2 = × 5450 = 454,2
mdp
12

mm.
Thus, select hdp = 450 mm, bdp = 180 mm.
– Dimensions of the primary beams: with span is 3l1.
1
1
hdc =
3l1 = × 3 × 1900 = 633,3
mdc
9
mm.
Thus, select hdc = 650 mm, bdc = 250 mm.
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2.2. The design scheme.
– One-way slab, establish the simplified model for a 1 m wide strip of the plate which is
perpendicular to the secondary beams, regard strip of slab like a continuous beam.
– The design span length of the slab:
+ Outer span length:
bdp t
0,18 0,34 0,08
l0b = l1b −
− + Cb = 1,9 −
−
+
= 1,68
2 2
2

2
2
+ Middle span length:
l0 = l1 − bdp = 1,9 − 0,18 = 1,72

m.

m.

Span difference:
l0 − l0b 1,72 − 1,68
=
× 100% = 2,33 % < 10%.
l0b
1,72
2.3. The design load.
– Dead load is calculated as following:
Layer

Nominal value

Tile 10 mm, γ = 20 kN/m3
0,01×20 = 0,200
Lining mortar 30 mm, γ = 18 kN/m3
0,03×18 = 0,540
Reinforced concrete 80 mm, γ = 25 kN/m3 0,08×25 = 2,000
Plaster mortar 10 mm, γ = 18 kN/m3
0,01×18 = 0,180
Total
2,920

Dead load:
gb = 3,356 kN/m2.
– Live load:
pb = ptc×n = 9,75×1,2 = 11,7 kN/m2.
– Total load:
qb = gb + pb = 3,356 + 11,7 = 15,056 kN/m2.
Calculate in strip of slab b1 = 1 m:
qb = 15,056×1 = 15,056 kN/m.

Reliability
coefficient
1,1
1,3
1,1
1,3

Design
value
0,220
0,702
2,200
0,234
3,356

2.4. Internal force.
Following the plastic hinge scheme model:
– Bending moment at the outer span and the secondary support:

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M nh = M g 2

2
 15,056 × 1,682 
qblob
 = ±3,863
=±
= ±
11
11



kN.m.
– Bending moment at the middle spans and the middle supports:
M nhg1 = M g1 = ±

 15,056 × 1,722 
qblb2
 = ±2,784
= ±
16
16



kN.m.

– The magnitude of the maximum shear force:

QBt = 0,6qblob = 0,6 × 15,056 × 1,68 = 15,176

kN.

Check the shear strength:

QBt = 15,176 < 0,5Rbtb1h0 = 0,5 × 0,66 × 1× 65 = 21,45

kN.

Concrete is satisfied shear resistance.

Sheme of calculation and internal force diagram in slab

2.5. Bending moment reinforcement.
– Data: Concrete grade B12,5, Rb = 7,5 MPa.
Reinforced steel C-I, Rs = 225 MPa.
– Calculate the internal force based on plastic hinge scheme, with αpl = 0,255.
Assume a = 15 mm for all section: h0 = hb – a = 80 – 15 = 65 mm.
– At the outer bearing and outer span, with M = 3,863 kN.m:

M
3,863.103
αm =
=
= 0,122 < α pl = 0,255
Rbbh02 7,5 × 106 × 1× 0,0652
.

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ζ =

1 + 1 − 2α m
2

As =

µ=

=

1 + 1 − 2 × 0,122
= 0,935.
2

M
3,863.103
=
Rsζh0 225.106 × 0,935 × 0,065

= 2,826.10-4 m2 = 282,6 mm2.

As
282,6
=

× 100% = 0,435%.
bh0 1000 × 65

Thus, Select diameter of reinforced steel 6 mm, spacing of adjacent steels is 100 mm (ϕ6a100).
– At the middle bearings and middle spans, with M = 2,784 kN.m:

αm =

ζ =

M
2,784.103
=
= 0,0879 < α pl = 0,255
Rbbh02 7,5 × 106 × 1× 0,0652

1 + 1 − 2α m 1 + 1 − 2 × 0,0879
=
= 0,954.
2
2

M
2,784.103
As =
=
Rsζh0 225.106 × 0,954 × 0,065

µ=

.

= 1,995.10-4 m2 = 199,5 mm2.

As
199,5
=
× 100% = 0,307%.
bh0 1000 × 65

Thus, select diameter of reinforced steel 6 mm, spacing of adjacent steels is 140 mm (ϕ6,a140).
– Check the working height h0 with thickness of reinforcement protection cover c = 10 mm:

h0 = hb – c –

1
2

ϕ = 80 – 10 –

1
2

×6 = 67 mm > 65 mm – safety.

– Positive moment longitudinal reinforcement: is placed alternatively.
Distance from the end of the shorter reinforcement to edge of the secondary beam:

1
8

1
8

l0 = ×1720 = 215 mm.
– With apertures of slab which are pinned to beams by 4 edges, the main reinforcement is allowed
to reduce 20% area.
As 80% = 80% As = 80% × 199,5 = 159,6
mm.
Check reinforcement amount again:

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µ80% =

As80%
159,6
=
× 100% = 0,246% > µ min = 0,05%
bh0
1000 × 65

.
Thus, select diameter of reinforced steel 6 mm, spacing of adjacent steels is 170 mm (ϕ6,a170).
– Negative moment longitudinal reinforcement: is placed alternatively.
pb
11,7
1

=
= 3, 486 > 3 ⇒ ν =
gb 3,356
3
.
+ Extension of reinforcement is:
1
3
νl0 = ×1720 = 573,3 mm ≈ 575 mm (seen from edge of the secondary beam).
1
1
2
2
νl0 + bdp = ×1720 + ×180 ≈ 665 mm (seen from axis of the secondary beam).

1
3

+ Extension of the shorter reinforcement is:
1
1
6
6
l0 = ×1720 = 286,7 mm ≈ 290 mm (seen from edge of the secondary beam).
1
1
1
1
6
2

6
2
l0 + bdp = ×1720 + ×180 = 380 mm (seen from axis of the secondary beam).
2.6. Constructive reinforcement.
– Negative moment reinforcement is placed in direction which is perpendicular to the primary
beams.
+ Select ϕ6, s = 200 mm, area per 1 m plate is 141 mm2, higher 50% than area of reinforcement at
the middle bearing of plate which is 0,5×199,5 = 99,75 mm2.
+ Extension is:
1
1
4
4
l0 = ×1720 = 430 mm (seen from edge of the primary beam).
1
1
1
1
4
4
4
2
l0 + bdc = ×1720 + ×250 = 555 mm (seen from axis of the primary beam).
+ To slab can subjected to negative moment at the wall when slab is supported to the wall, we
place reinforcement with an extension seen from edge of the wall:
1
1
6
6
l0 = ×1720 = 286,7 mm ≈ 290 mm.

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– Distribution reinforcement is placed in direction which is perpendicular to moment
reinforcement:
Select ϕ6, s = 250 mm, area per 1 m plate is 131 mm 2, higher than 20% area of reinforcement at
the outer bearing of plate which is 0,2×282,6 = 56,52 mm2, and higher than 20% area of
reinforcement at the middle bearing of plate which is 0,2×199,5 = 39,9 mm 2 .

Region which is allowed to reduce 20% reinforcement

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Arrangement of reinforcement in slab

Arrangement of reinforcement in slab
Arrangement of reinforcement in slab

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Some cross section in secondary beam

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3. DESIGN OF SECONDARY BEAM
3.1. The design scheme.
– The secondary beams is symmetrically continuous beams with 4 spans.
– The secondary beams is placed on wall Sd = 220 mm.

S d l2
;
2 40

Cd = min(
) = min(110; 136,25) = 110 mm.
– The design span length of the secondary beam:
+ Outer span length:
b
t
0,25 0,34
l pb = l2 − dc − + Cd = 5,45 −
−
+ 0,11 = 5,265
2 2
2
2
+ Middle span length:
l p = l2 − bdc = 5,45 − 0,25 = 5,2

m.

m.

Span difference:
l pb − l p 5,265 − 5,2
=
× 100% = 1,23% < 10%.
l pb
5,265
3.2. The design load.
– Dead load.
+ Self-weight of the secondary beam (not include plate 80 mm):
g0p = bdp(hdp – hb)γn
where γ = 25 kN/m2, is unit weight of concrete.
n = 1,1, is reliability coefficient of self-weight of beam.
g0p = 0,18×(0,45 – 0,08)×25×1,1 = 1,8315 kN/m.
+ Dead load transferred from the plate:
gpl1 = 3,356×1,9 = 6,3764 kN/m.
– Live load.
pp = pbl1 = 11,7×1,9 = 22,23 kN/m.
– Total load:
qp = 1,8315 + 6,3764 + 22,23 = 30.4379 ≈ 30,44 kN/m.

pp
gp

=

22,23
= 2,7084
1,8315 + 6,3764
.

3.3. Internal force.
Calculate the internal force based on plastic hinge scheme model
a. Bending moment.
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– Ordinate of the moment envelop diagram (positive part):
+ At outer span:
2
M + = β1q p l pb
= β1 × 30 ,44 × 5,2652 = β1 × 843,804

kN.m.

+ At middle span:
2
M + = β1q p l pb
= β1 × 30 ,44 × 5,2652 = β1 × 843,804

kN.m.
– Ordinate of the moment envelop diagram (negative part):
M − = β 2 q p l p2 = β1 × 30,44 × 5,2 2 = β 2 × 823,098

kN.m.
Where, β1, β2 is looked up in Appendix 11, with pp/gp = 2,7084 and the factor k = 0,2763.
The results are shown in the bellow table:
Span, section

Value β1, β2
β1

Ordinate of moment M (kN.m)
β2

Outer span
Bearing 1

0

0

1

0,065

54,847

2

0,09

75,942

0,425l

0,091

76,786

3

0,075

63.285

4

0,02

16.876

Bearing 2 – section 5

–0,0715

–58.852

Secondary span
6

0,018

–0,0338

14.816

–27.821

7

0,058

–0,0137

47.740

–11.276

0,5l

0,0625

8

0,058

–0,0111

47.740

–9.136

9

0,018

–0,0278

14.816

–22.882

51.444

Bearing 3 – section 10
–0,0625
–51.444
– The section which has zero negative moment distances left edge of the second support:
x = klpb = 0,2763×5265 = 1454,5 mm.
– The section which has zero positive moment distances edge of the support:
+ At outer span:
0,15lpb = 0,15×5265 = 789,75 mm.
+ At middle span:
0,15lp = 0,15×5200 = 780 mm.

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b. Shear force.
Q1 = 0,4qplpb = 0,4×30,44×5,265 = 64,107 kN.

Q2t
Q2p

= 0,6qplp = 0,6×30,44×5,265 = 96,160 kN.
= 0,5qplp = 0,5×30,44×5,2 = 79,144 kN.
Scheme of calculation and internal force diagram in secondary beam

3.4. Calculation of longitudinal reinforcement.
– Data: Concrete grade B12,5, Rb = 7,5 MPa.
Reinforced steel C-II, Rs = 280 MPa, Rsc = 280 MPa.
– Calculate the internal force based on plastic hinge scheme model, with αpl = 0,255.
a. Longitudinal reinforcement subjected to negative moment.
– Design with rectangular cross section: b = 180 mm, h = 450 mm.
Assume a = 35 mm, h0 = 450 – 35 = 415 mm.
– At bearing 2, M = 58,852 kN.m.

M
58,852.103
αm =
=
= 0,253 < α pl = 0,255
Rbbh02 7,5 × 106 × 0,18 × 0,4152

ζ =

1 + 1 − 2α m
2

=

.

1 + 1 − 2 × 0, 253
= 0,851.
2

M
58,852.103
As =
=
Rsζ h0 280.106 × 0,851× 0, 415

= 5,949.10-4 m2 = 594,9 mm2.

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µ=

As
594,9
=
× 100% = 0,796%.
bh0 180 × 415

– At bearing 3, M = 51,444 kN.m.

M
51, 444.103
αm =
=
= 0, 221 < α pl = 0, 255
Rb bh02 7,5 × 106 × 0,18 × 0, 4152
ζ =

1 + 1 − 2α m
2

As =

µ=

=

.

1 + 1 − 2 × 0, 221
= 0,873.
2

M
51, 444.103
=
Rsζ h0 280.106 × 0,873 × 0, 415

= 5,069.10-4 m2 = 506,9 mm2.

As
506,9
=
× 100% = 0,678%.
bh0 180 × 415

b. Longitudinal reinforcement subjected to positive moment.
– Design with T-section, the flange is in compressive zone, flange thickness

Assume a = 35 mm, h0 = 450 – 35 = 415 mm.

h 'f

= 80 mm.

h 'f

– Because =80 mm > 0,1×450 = 45 mm = 0,1hdp, the extension of flange Sf is not greater than
bellow values:

1
6

1
1
l p = × 5200
6
6

+ span of secondary beam =
= 866,7 mm.
+ A haft of clear distance of two secondary beams 0,5×1720 = 860 mm.
Thus, Sf = min(866,7; 860) = 860 mm.
Flange breadth:

b 'f = b + 2S f = 180 + 2 × 860 = 1900

mm.
– Determine the location of the neutral axis.

M f = Rb b'f h 'f (h0 − 0,5h 'f ) = 7,5 × 1900 × 80 × (415 − 0,5 × 80) = 427,5.10 6
+
M max
= 76,786.106
+
M f > M max

N.mm.

N.mm.

, the neutral axis is in the flange.

– Design as in rectangular section with

b = b 'f = 1900

mm, h = 450 mm, h0 = 415 mm.

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– At outer span, M = 76,786 kN.m:

αm =

ζ =

M

76, 786.103
=
= 0, 031 < α pl = 0, 255
Rb bh02 7,5 × 10 6 × 1,9 × 0, 4152

1 + 1 − 2α m
2

=

1 + 1 − 2 × 0,031
= 0,984.
2

M
76, 786.103
As =
=
Rsζ h0 280.106 × 0,984 × 0, 415
µ=

.

= 6,715.10-4 m2 = 671,5 mm2.

As
671,5
=
× 100% = 0,899%.
bdp h0 180 × 415

– At middle spans, M = 51,444 kN.m:
M
51,444.103
αm =
=
= 0,021 < α pl = 0,255
Rbbh02 7,5 × 106 × 1,9 × 0,4152

ζ =

1 + 1 − 2α m 1 + 1 − 2 × 0,021
=
= 0,989.
2
2

As =

µ=

.

M
51,444.103
=
Rsζh0 280.106 × 0,989 × 0,415

= 4,476-4 m2 = 447,6 mm2.

As
447,6
=
× 100% = 0,599%.
bdp h0 180 × 415

3.5. Selection and arrangement of longitudinal reinforcement.
Section
e
As (mm2)
Diameter of rein
Area of rein
ring 2

Outer span
671,5
2ϕ18 + 1ϕ16
710,0
Middle span

Bearing 2
594.9
2ϕ16 + 1ϕ16
603,2

Middle span
447,6
2ϕ14 + 1ϕ14
461,8

Bearing 3
506.9
2ϕ14 + 1ϕ16
509,0

Out
r
span
Bea

Bearing 3

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Arrangement of the longitudinal reinforcement in secondary beam

Select thickness of protection cover c = 20 mm, check h0 again:

h0 = h – c –

φ
2

h0 = h – c –

18
2

φ
2

h0 = h – c –

16
2

φ
2

h0 = h – c –

14
2

φ
2
16
2

= 450 – 20 –
= 450 – 20 –
= 450 – 20 –
= 421 mm > 415.
= 422 mm > 415.
= 423 mm > 415.
The values of h0 is greater than the calculating height h0 – safety.

= 450 – 20 –

= 422 mm > 415.

3.6. Design of lateral reinforcement.
a. Condition of calculating shear reinforcement.
Q1

Q2t

Q2p

– From the shear force diagram, have: = 64,107 kN,
= 96,160 kN,
= 79,144 kN.
Choose Q = 96,160 kN for calculation of stirrups.
Materials: Rb = 7,5 MPa, Rbt = 0,66 MPa, Rsw = 175 MPa.
Dimensions of the beam: b = 180 mm, h = 450 mm, h0 = 421 mm.
– Check condition to resist principal compressive stress of beam’s web:
Q = 96,160 kN < 0,3Rbbh0 = 0,3×7,5.103×0,18×0,421 = 170,505 kN – satisfaction.
– Check shear strength of concrete:
Qbmin = 0,5Rbtbh0 = 0,5×0,66.103×0,18×0,421 = 25 kN < Q = 96,16 kN.
Concrete is not capable to resist all of shear force, need to design stirrups.
b. Calculation of stirrups without inclined reinforcement.
– Assume c ≤ 2h0.
With assumption that concrete and stirrups resist all of shear force:
Q = 96,160 kN =

QDB = 6 Rbtbh02 ( 0,75qsw + q p − 0,5 p p )

where qp = 30,44 kN/m, is total load acting on the secondary beam.
pp = 22,23 kN/m, is live load acting on the secondary beam.

q p − 0,5 p p
Q2
96,162
30,44 − 0,5 × 22,23
qsw =
−
=
−
0,75
0,75
4,5 Rbtbh02
4,5 × 0,66.103 × 0,18 × 0,4212
= 71,82 N/mm.
qswmin = 0,25Rbtb = 0,25×0,66×180 = 29,7 N/mm < qsw = 71,82 N/mm.
Recalculate c0 with qsw = 71,82 N/mm:
c0 =

1,5Rbt bh02
=
0,75qsw + q p − 0,5 p p

1,5 × 0,66.103 × 0,18 × 0, 4212
0,75 × 71,82 + 30, 44 − 0,5 × 22, 23

= 0,6569 m = 656,9 mm.
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c0 = 656,9 mm < 2h0 = 2×421 = 842 mm, go to select stirrups.
– Select stirrups: diameter ϕ6, number of legs n = 2.
Cross-sectional area of stirrups perpendicular to member axis:

nπφ w2 2 × 3,14 × 6 2
Asw =
=
= 56,52
4
4
Calculation spacing of stirrups:
R A
175 × 56,52
stt = sw sw =
= 137,72
qsw
71,82

mm2.

mm.

Structural spacing of stirrups:
sct < min(0,5h; 150) = min(0,5×450; 150) = 150 mm.
Maximum spacing of stirrups:

sma x =

Rbtbh02 0,66 × 180 × 4212
=

= 218,97
Q
96160

mm.
The final spacing of stirrups is chosen as minimum of {stt; sct; smax}:
s1 = min(137,72; 150; 218,97) = 137,72 mm.
Take s1 = 130 mm for segment near the bearings.
Spacing of stirrups at remaining segment:
s2 ≤ min(0,75h; 500) = min(0,75×450; 500) = 337,5 mm.
Take s2 = 250 mm for remaining segment.
– Length of segment near the bearings.
R A
175 × 56,52
qsw1 = sw sw =
= 76,08
s1
130
N/mm.
R A
175 × 56,52
qsw2 = sw sw =
= 39,56
s2
250
N/mm.
qsw1 − qsw 2 = 76, 08 − 39,56 = 36,52 > q1 = 30, 44 − 0,5 × 22, 23 = 19,325


2 Rbt bh02

Q −  Qb min + qsw 2

qsw1

l1 =
q1

N/mm.


÷
2
÷
 − 2 Rbt bh0
qsw1


2 × 0,66 × 180 × 4212
96160 −  25000 + 39,56 ×

76,08

=
19,325


÷
÷
2
 − 2 × 0,66 × 180 × 421 = 1415

76,08

mm.

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Take l1 = 1420 mm.
3.7. Checking for anchorage of reinforcement.
After cutting, amount of remaining positive reinforcement when anchoring to the bearing must be
greater 1/3 of area of positive at the center of span.
Outer span: 2ϕ18 + 1ϕ16, cut 1ϕ16 and remain 2ϕ18, reach 71,67%.
Middle span: 2ϕ14 + 1ϕ14, cut 1ϕ14 and remain 2ϕ14, reach 66,7%.
3.8. Calculation and drawing moment capacity envelop diagram.
a. Calculation of bearing capacity.
– At outer span, positive moment, T-section with the flange is in compressive zone, the flange
b 'f
breadth b =
= 1900 mm, reinforced steels 2ϕ18 + 1ϕ16, As = 710 mm2.
Thickness of the protection cover, c = 20 mm, h0 = 450 – 20 – 0,5×18 = 421 mm.

ξ=

Rs As
280 × 710
=
= 0,033.
Rb b'f h0 7,5 × 1900 × 421
h 'f

x = ξh0 = 0,033×421 = 13,893 mm < = 80 mm – the neutral axis is in flange.
ζ = 1 – 0,5ξ = 1 – 0,5×0,033 = 0,983.
Mtd = RsAsζh0 = 280×710×0,983×421 = 82,308.106 N.mm = 82,308 kN.m.
– At bearing 2, negative moment, rectangular section with b×h = 180×450 mm, reinforced steels
2ϕ16 + 1ϕ16, As = 603,2 mm2.
Thickness of the protection cover, c = 20 mm, h0 = 450 – 20 – 0,5×16 = 422 mm.
R A
280 × 603, 2
ξ = s' s =
= 0, 296.
Rb b f h0 7,5 × 180 × 422
ζ = 1 – 0,5ξ = 1 – 0,5×0,296 = 0,852.
Mtd = RsAsζh0 = 280×603,2×0,852×422 = 60,709.106 N.mm = 60,709 kN.m.
The results of calculations of bearing capacity are shown in the bellow table:
Section

Quantity of reinforcement

Center of outer span
Outer span adjacency

2ϕ18 + 1ϕ16, As = 710
Cut 1ϕ16, remain 2ϕ18, As =
508,9
Center of bearing 2
2ϕ16 + 1ϕ16, As = 603,2
Bearing 2 adjacency
Cut 1ϕ16, remain 2ϕ16, As =
402,1

Center of middle span 2ϕ14 + 1ϕ14, As = 461,8
Middle span adjacency Cut 1ϕ14, remain 2ϕ14, As =
307,9

h0
mm
421

ξ

ζ

0,033

0,983

Mtd
kN.m
82,308

421
422

0,024
0,296

0,988
0,852

59,277

60,709

422
423

0,198
0,021

0,901
0,989

42,817
54,109

423

0,014

0,993

36,207

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ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING ()


Center of bearing 3
Bearing 3 adjacency

2ϕ14 + 1ϕ16, As = 509

422 0,250 0,875 52,620
Cut 1ϕ16, remain 2ϕ14, As =
33,715
307,9
423 0,151 0,925
b. Determine theoretical cutting position and extension length W.
– Reinforcement no.2 (the left part).
After cutting reinforced steel no.2 (1ϕ16), section near the bearing 1, the outer span remains
reinforced steel no.1 (2ϕ18). Bearing capacity is 59,277 kN.m.
The moment capacity envelope diagram meet the moment envelope diagram at a point which is
theoretical cutting position of the reinforced steel no.2.
By geometrical relation between similar triangles, distance from this point to the outer bearing is
1274 mm. At this area, stirrups are arranged ϕ6a130. The magnitude of shear force at this point is
25,32 kN.
Determine the extension W:

qsw =
W2t =

Rsw Asw 175 × 56,52
=
= 76,085
s
130
Q − Qs ,inc
2qsw

+ 5φ =

N/mm.

25,32 − 0
+ 5 × 0, 016 = 0.246
2 × 76, 085

m < 20ϕ = 20×0,016 = 0,32 m.

W2t

Take
= 320 mm.
– The results of determining theoretical cutting position and extension length are shown in the
bellow table:
Reinforced steel
RS no.2 – left part
RS no.2 – right part
RS no.3 – left part
RS no.3 – right part
RS no.4 – left part
RS no.4 – right part
RS no.6 – left part

Theoretical cutting position
Distances edge of the wall 1164 mm
Distances left edge of the 2-bearing 2015 mm
Distances left edge of the 2-bearing 395 mm
Distances right edge of the 2-bearing 537 mm

Extension length

W2t
W2p
W3t
W3p

Distances right edge of the 2-bearing 1455
mm

W4t

Distances left edge of the 2-bearing 842 mm

W4p

Distances right edge of the 2-bearing 1716
mm

W6t

= 320 mm
= 520 mm
= 630 mm
= 498 mm
= 736 mm
= 433 mm
= 411 mm

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ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING ()

RS no.6 – right part
RS no.7 – left part

Distances left edge of the 3-bearing 1716 mm
Distances left edge of the 3-bearing 644 mm

W6p
W7t

= 411 mm
= 471 mm

3.9. Constructive reinforcement.
Reinforced steel no.9 (2ϕ12): it is used as constructive reinforcement at the outer span where has
not negative moment.
Area of reinforcement is 226,2 mm2, is not lower than 0,1%bh0 = 0,1%×180×421 = 75,78 mm2.

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Moment capacity envelope and shear force diagram

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ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING ()


Arrangement of reinforcement in secondary beam

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4. DESIGN OF PRIMARY BEAM
4.1. Scheme of calculation.
– The primary beams is symmetrically continuous beams with 4 spans.
– The design span lengths of all spans are 3l1 = 3×1,9 = 5700 mm.

Scheme of calculation of primary beam

4.2. The design scheme.
– Dead load.
+ Self-weight of a segment of the primary beam (not include plate 80 mm):
G0 = bdc(hdc – hb)γnl1
where: γ = 25 kN/m2, is unit weight of concrete.
n = 1,1, is reliability coefficient of self-weight of beam.
G0 = 0,25×(0,65 – 0,08)×25×1,1×1,9 = 7,446 kN.
+ Dead load transferred from the secondary beam:
G1 = gdpl2 = (1,8315 + 6,3764)×5,45 = 44,733 kN.
Dead load: G = G0 + G1 = 7,446 + 44,733 = 52,179 kN.
– Live load transferred from the secondary beam.
P = pdpl2 = 22,23×5,45 = 121,154 kN.
4.3. Internal force.
Calculate the internal force based on elastic scheme model.
a. Bending moment.
– Determine the critical case of the internal force diagram of the beam.
+ Determine moment diagram caused by dead load G.
MG = αGl = α×52,179×5,7 = α×297,420 kN.m.
+ Determine moment diagram caused by live load Pi.

MP = αPl = α×121,154×5,7 = α×690,578 kN.m.
α is looked up in Appendix 12.
Diagram of MP3 doesn’t have α to calculate moment at sections 1, 2, 3, 4. To calculate, we
separate spans AB, BC. Span AB and BC, calculate M0 of simple beam borne by two supports
M0 = Pl1 = 121,154×1,9 = 230,193 kN.m.
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Auxiliary diagram to calculate moment at some sections

From that, we can calculate moment MP3

1
M 1 = 230,193 − × 221,675 = 156,301
3
2
M 2 = 230,193 − × 221,675 = 82, 409
3

kN.m.

kN.m

2
M 3 = 230,193 − × (221,675 − 33,148) − 33,148 = 71,360
3

kN.m.

1
M 4 = 230,193 − × (221,675 − 33,148) − 33,148 = 134, 203
3

kN.m.
Similarly for diagram of MP4, MP5 and MP6, calculate moment at sections 1, 2, 3, 4.
The results are shown in the below table:
Moment (kN.m)
MG
α
M
MP1
α
M
MP2
α
M
MP3
α
M
MP4
α
M
MP5
α
M

1
0,238
70,786

0,286
197,505
-0,048
-33,148

2
0,143
42,531
0,238
164,358
-0,095
-65,605

156,301
-0,031
-21,408

82,409
-0,063
-43,506

186,456

142,720

B
-0,286
-85,062
-0,143
-98,753

-0,143
-98,753
-0,321
-221,675
-0,095
-65,605
-0,190
-131,210

3
0,079
23,496
-0,127
-87,703
0,206
142,259

4
0,111
33,014
-0,111
-76,654
0,222
153,308

71,360

134,203

120,621

76,655

-65,605

0

C
-0,190
-56,510
-0,095
-65,605
-0,095
-65,605
-0,048
-33,148
-0,286
-197,505
0,095
65,605

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MP6
Mmax
Mmin

α

M

8,287
268,291
37,638

16,574
206,889
-23,074

0,036
24,861
-60,201
-306,737

-16,374
165,755
-64,207

-57,563
186,322
-43.640

-0,143
-98,753
9,095
-254,015

Moment scheme of calculation in primary beam

– Moment envelope diagram.
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Ordinate of the moment envelop diagram:
Mi = MG + MPi.
Mmax and Mmin at each section is written in the above table.
The results are shown in the below table:
Moment
1
2
B
3
4
C
MG + MP1
268,291
206,889
-183,815
-64,207
-43.640
-121,115
MG + MP2
37,638
-23,074
-183,815
165,755
186,322
-121.115

MG + MP3
227,087
124,940
-306,737
94,856
167,217
-89,658
MG + MP4
49,378
-0,975
-150,667
144,117
109,669
-254,015
MG + MP5
257,242
185,251
-216,272
-42,109
33,014
9,095
MG + MP6
79,073
59,105
-60,201
7,122
-24,549
-155,263
Because the beam is symmetrical, so we just need to perform moment diagram of the left part of
the beam.

Draw moment envelope diagram.

Moment envelope diagram in primary beam

– Determine moment at the edge of the bearing.
At bearing B: From the moment envelope diagram at bearing B, we see that the slope of the right
part of Mmin is smaller than that of the left part. Thus, calculation of moment at the right edge of
the bearing has higher magnitude.
hc ( M g + M 3 )
B
M mg
= Mg −
2l1
.
M3 = 94,856 kN.m.
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ĐOÀN CÔNG ĐAI – 57XE2 – NATIONAL UNIVERSITY OF CIVIL ENGINEERING ()