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

Đồ án này thực hiện theo cuốn Sàn sườn bê tông toàn khối (GS. Nguyễn Đình Cống). Ngôn ngữ sử dụng là tiếng anh chuyên ngành xây dựng. Cuốn đồ án này giúp cho các bạn sinh viên có thêm một nguồn tham khảo cách trình bày đồ án, và cũng là một nguồn thú vị cung cấp các thuật ngữ tiếng anh chuyên ngành xây dựng

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

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: b c = h c = 30 cm

1.2 Characteristics of the construction.

– Building construction slab which has 4 levels as shown in the figure

– Nominal live load: p tc = 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 R b = 7,5 MPa, R bt = 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:

– One-way slab, establish the simplified model for a 1 m wide strip of the plate which isperpendicular to the secondary beams, regard strip of slab like a continuous beam.

– The design span length of the slab:

+ Outer span length:

2.3 The design load.

– Dead load is calculated as following:

coefficient Designvalue

Following the plastic hinge scheme model:

– Bending moment at the outer span and the secondary support:

Concrete is satisfied shear resistance.

Sheme of calculation and internal force diagram in slab 2.5 Bending moment reinforcement.

– Data: Concrete grade B12,5, R b = 7,5 MPa

Reinforced steel C-I, R s = 225 MPa

– Calculate the internal force based on plastic hinge scheme, with α pl = 0,255

Assume a = 15 mm for all section: h0 = h b – a = 80 – 15 = 65 mm.

– At the outer bearing and outer span, with M = 3,863 kN.m:

Thus, Select diameter of reinforced steel 6 mm, spacing of adjacent steels is 100 mm (ϕ6a100).ϕ6a100).

– At the middle bearings and middle spans, with M = 2,784 kN.m:

– Check the working height h0 with thickness of reinforcement protection cover c = 10 mm:

h0 = h b – c –

1

2 ϕ = 80 – 10 – ×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:

2 ×180 ≈ 665 mm (ϕ6a100).seen from axis of the secondary beam).

+ Extension of the shorter reinforcement is:

2 1

3 1

Select ϕ6, s = 250 mm, area per 1 m plate is 131 mm2, higher than 20% area of reinforcement atthe outer bearing of plate which is 0,2×282,6 = 56,52 mm2, and higher than 20% area ofreinforcement at the middle bearing of plate which is 0,2×199,5 = 39,9 mm2

Region which is allowed to reduce 20% reinforcement

Arrangement of reinforcement in slab

Arrangement of reinforcement in slab

Arrangement of reinforcement in slab

Some cross section in secondary beam

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 S d = 220 mm

– The design span length of the secondary beam:

+ Outer span length:

where γ = 25 kN/m2, is unit weight of concrete

n = 1,1, is reliability coefficient of self-weight of beam.

Where, β1, β2 is looked up in Appendix 11, with p p /g p = 2,7084 and the factor k = 0,2763.

The results are shown in the bellow table:

Span, section Value β1, β2 Ordinate of moment M (ϕ6a100).kN.m)

Q2p = 0,5q

p l p = 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, R b = 7,5 MPa

Reinforced steel C-II, R s = 280 MPa, R sc = 280 MPa

– Calculate the internal force based on plastic hinge scheme model, with α pl = 0,255

a Longitudinal reinforcement subjected to negative moment

– Design with rectangular cross section: b = 180 mm, h = 450 mm.

0

58,852.10280.10 0,851 0,415

s

s

M A

51, 444.10280.10 0,873 0, 415

s

s

M A

b Longitudinal reinforcement subjected to positive moment

– Design with T-section, the flange is in compressive zone, flange thickness h' f = 80 mm.

N.mm

6 max 76,786.10

M

Mf> Mmax+

, the neutral axis is in the flange

– Design as in rectangular section with b=b' f=1900 mm, h = 450 mm, h

0

76,786.10280.10 0,984 0, 415

s

s

M A

A

b h



span Bearing 2 Middle span Bearing 3

Arrangement of the longitudinal reinforcement in secondary beam

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

h0 = h – c – 2



= 450 – 20 –

18 2 = 421 mm > 415

h0 = h – c – 2



= 450 – 20 –

16 2 = 422 mm > 415

h0 = h – c – 2



= 450 – 20 –

14 2 = 423 mm > 415

h0 = h – c – 2



= 450 – 20 –

16 2 = 422 mm > 415

The values of h0 is greater than the calculating height h0 – safety

3.6 Design of lateral reinforcement

a Condition of calculating shear reinforcement

– From the shear force diagram, have: Q1 = 64,107 kN, Q2t = 96,160 kN, Q2p = 79,144 kN.

Choose Q = 96,160 kN for calculation of stirrups.

Diameter of rein 2ϕ18 + 1ϕ16 2ϕ16 + 1ϕ16 2ϕ14 + 1ϕ14 2ϕ14 + 1ϕ16

Materials: R b = 7,5 MPa, R bt = 0,66 MPa, R sw = 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 < 0,3R b bh0 = 0,3×7,5.103×0,18×0,421 = 170,505 kN – satisfaction

– Check shear strength of concrete:

Q bmin = 0,5R bt bh0 = 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 Rbtbh0 2( 0,75qsw+ qp−0,5 pp)

where q p = 30,44 kN/m, is total load acting on the secondary beam.

p p = 22,23 kN/m, is live load acting on the secondary beam.

q sw= Q2

4,5 R bt bh 0 2−q p−0,5 pp

0 ,75 =

96 ,1624,5×0 , 66 103×0 ,18×0 , 4212−30 , 44−0,5×22 , 23

0 , 75

= 71,82 N/mm

q swmin = 0,25R bt b = 0,25×0,66×180 = 29,7 N/mm < q sw = 71,82 N/mm

Recalculate c0 with q sw = 71,82 N/mm:

0 0

– Select stirrups: diameter ϕ6, number of legs n = 2.

Cross-sectional area of stirrups perpendicular to member axis:

A sw=n πφ w 2

2×3 , 14×62

4 =56 , 52 mm2.Calculation spacing of stirrups:

Take s1 = 130 mm for segment near the bearings.

Spacing of stirrups at remaining segment:

s 2 ≤ min(ϕ6a100).0,75h; 500) = min(ϕ6a100).0,75×450; 500) = 337,5 mm.

Take s2 = 250 mm for remaining segment

– Length of segment near the bearings

1

1

175 56,52

76,08130

sw sw sw

R A q

sw sw sw

R A q

3.7 Checking for anchorage of reinforcement.

After cutting, amount of remaining positive reinforcement when anchoring to the bearing must begreater 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

breadth b = b'f

= 1900 mm, reinforced steels 2ϕ18 + 1ϕ16, A s = 710 mm2

Thickness of the protection cover, c = 20 mm, h0 = 450 – 20 – 0,5×18 = 421 mm

' 0

Thickness of the protection cover, c = 20 mm, h0 = 450 – 20 – 0,5×16 = 422 mm.

' 0

The results of calculations of bearing capacity are shown in the bellow table:

mm

kN.mCenter of outer span 2ϕ18 + 1ϕ16, A s = 710 421 0,033 0,983 82,308Outer span adjacency Cut 1ϕ16, remain 2ϕ18, A s =

Center of bearing 2 2ϕ16 + 1ϕ16, A s = 603,2 422 0,296 0,852 60,709Bearing 2 adjacency Cut 1ϕ16, remain 2ϕ16, A s =

Center of middle span 2ϕ14 + 1ϕ14, A s = 461,8 423 0,021 0,989 54,109

Middle span adjacency Cut 1ϕ14, remain 2ϕ14, A s =

Center of bearing 3 2ϕ14 + 1ϕ16, A s = 509 422 0,250 0,875 52,620Bearing 3 adjacency Cut 1ϕ16, remain 2ϕ14, A s =

33,715

b Determine theoretical cutting position and extension length W.

– Reinforcement no.2 (ϕ6a100).the left part)

After cutting reinforced steel no.2 (ϕ6a100).1ϕ16), section near the bearing 1, the outer span remainsreinforced steel no.1 (ϕ6a100).2ϕ18) Bearing capacity is 59,277 kN.m

The moment capacity envelope diagram meet the moment envelope diagram at a point which istheoretical 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 is25,32 kN

Determine the extension W:

175 56,52

76,085 130

sw sw sw

25,32 0

5 5 0,016 0.246

s inc t

sw

Q Q W

Reinforced steel Theoretical cutting position Extension length

RS no.2 – left part Distances edge of the wall 1164 mm

2

t

W = 320 mm

RS no.2 – right part Distances left edge of the 2-bearing 2015 mm W2p = 520 mm

RS no.3 – left part Distances left edge of the 2-bearing 395 mm

RS no.6 – right part Distances left edge of the 3-bearing 1716 mm W6p = 411 mm

RS no.7 – left part Distances left edge of the 3-bearing 644 mm

Moment capacity envelope and shear force diagram

Arrangement of reinforcement in secondary beam

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 (ϕ6a100).not include plate 80 mm):

G0 = b dc (ϕ6a100).h dc – h b )γnl1

where: γ = 25 kN/m2, is unit weight of concrete

n = 1,1, is reliability coefficient of self-weight of beam.

– Determine the critical case of the internal force diagram of the beam

+ Determine moment diagram caused by dead load G.

M G = αGl = α×52,179×5,7 = α×297,420 kN.m.

+ Determine moment diagram caused by live load Pi

M P = αPl = α×121,154×5,7 = α×690,578 kN.m.

α is looked up in Appendix 12.

Diagram of M P3 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

Auxiliary diagram to calculate moment at some sections

From that, we can calculate moment M P3

1

1230,193 221,675 156,301

3

kN.m

Similarly for diagram of M P4 , M P5 and M P6, calculate moment at sections 1, 2, 3, 4

The results are shown in the below table:

Moment scheme of calculation in primary beam

– Moment envelope diagram

Ordinate of the moment envelop diagram:

M i = M G + M Pi

Mmax and Mmin at each section is written in the above table

The results are shown in the below table:

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 ofthe bearing has higher magnitude

3 1

(ϕ6a100) )2

4 1

(ϕ6a100) ) 2

c g C

Shear force at center of span is calculated by section method

The results are shown in the bellow table:

Shear force Right of

bearing A Center ofouter

span

Left of bearing B Right of bearing B Center of span 2 Left of bearing C

Shear force envelope diagram in primary beam

4.4 Calculation of longitudinal reinforcement.

– Data: Concrete grade B12,5, R b = 7,5 MPa

Reinforced steel C-II, R s = 280 MPa, R sc = 280 MPa

With work condition factor γ b2 = 1,0, look up in Appendix 9, α R = 0,442

, the neutral axis is in the flange.

– Design as in rectangular section with b = b' f = 2150 mm, h = 650 mm, h

4.5 Selection and arrangement of longitudinal reinforcement.

Outer span Bearing B Middle span Bearing C

Arrangement of reinforcement in primary beam

Constructing the primary beam reinforcement into 2 layers which are laid under the secondary

beam reinforcement, the upper layer 4ϕ22 (ϕ6a100).A s1 = 1520,5; a1 = 38 + 22/2 = 49 mm), the lower layer

2ϕ25 (ϕ6a100).A s2 = 981,8; a2 = 38 + 22 + 30 + 25/2 = 102,5 mm)

a= 49×1520,5+102,5×981,8

2502,3 =69,99 mm.

h0 = h – a = 650 – 70 = 580 mm ≥ 580 mm – safety.

+ At middle span, select thickness of protection cover c = 25 mm.

Diameter of rein 2ϕ25 + 2ϕ22 4ϕ22 + 2ϕ25 2ϕ22 + 1ϕ22 4ϕ20+ 2ϕ20

h0 = 650 – 25 – 22/2 = 614 mm > 610 mm – safety.

+ At bearing C, thickness of protection cover calculated to edge of reinforcement of secondarybeam is 20 + 18 = 38 mm

Constructing the primary beam reinforcement into 2 layers which are laid under the secondary

beam reinforcement, the upper layer 4ϕ20 (ϕ6a100).A s1 = 1256,6; a1 = 38 + 20/2 = 48 mm), the lower layer

2ϕ20 (ϕ6a100).A s2 = 628,3; a2 = 38 + 20 + 30 + 20/2 = 98 mm)

a= 48×1256 ,6+98×628,3

h0 = 650 – 64,7 = 585,3 ≈ 585 mm > 580 mm – safety

– Check clear distance between adjacent bars (ϕ6a100).4ϕ25 in a layer)

Thickness of protection cover c = 25 mm, clear distance:

250−2×25−4×25

2 =50 mm > 30 mm – satisfaction.

4.6 Design of lateral reinforcement.

– From shear force diagram of primary beam, have:

+ Right side of bearing A: Q A p = 141,085 kN, shear force is constant in segment l

1.+ Left side of bearing B: Q B t = 227,146 kN, shear force is constant in segment l

1.+ Right side of bearing B: Q B p = 205,790 kN, shear force is constant in segment l

1.+ Left side of bearing C: Q C t = 196,091 kN, shear force is constant in segment l

1

– Materials: R b = 7,5 MPa, R bt = 0,66 MPa, R sw = 175 MPa

a Calculate with the shear force in the right of bearing A: Q A p = 141,085 kN.

Dimensions of the beam: b = 250 mm, h = 650 mm, h0 = 612,5 mm

– Condition of calculating shear reinforcement

+ Check condition to resist principal compressive stress of beam’s web:

Q = 141,085 kN < 0,3R b bh0 = 0,3×7,5.103×0,25×0,6125 = 344,531 kN – satisfaction

+ Check shear strength of concrete:

Q bmin = 0,5R bt bh0 = 0,5×0,66.103×0,25×0,6125 = 50,531 kN < Q = 141,085 kN.

Concrete is not capable to resist all of shear force, need to design stirrups

– Calculation of stirrups without inclined reinforcement

– Select stirrups: diameter ϕ8, number of legs n = 2.

Cross-sectional area of stirrups perpendicular to member axis:

A sw=n πφ w 2

2×3 , 14×82

4 =100 , 53 mm2.Calculation spacing of stirrups:

Take s1 = 210 mm for segment near the bearing

Spacing of stirrups at remaining segment:

s 2 ≤ min(ϕ6a100).0,75h; 500) = min(ϕ6a100).0,75×650; 500) = 487,5 mm.

Take s2 = 250 mm for remaining segment

b Calculate with the shear force in the left of bearing B: QB t = 227,146 kN.

Dimensions of the beam: b = 250 mm, h = 650 mm, h0 = 580 mm

– Condition of calculating shear reinforcement

+ Check condition to resist principal compressive stress of beam’s web:

Q = 227,146 kN < 0,3R b bh0 = 0,3×7,5.103×0,25×0,580 = 326,25 kN – satisfaction

+ Check shear strength of concrete:

Q bmin = 0,5R bt bh0 = 0,5×0,66.103×0,25×0,580 = 47,85 kN < Q = 227,146 kN.