6.4-Restrained (Two-Way).pdf

Restrained (Two-Way Spanning Slab)

The figure below shows part of the first floor plan of a reinforced concrete office building. During construction, slabs and beams are cast together. The finishes, ceiling and services form a characteristic permanent action of 1.5kN/m 2  (excluding self-weight). The characteristic variable action is 4.0kN/m 2. The floor is inside enclosed building and subjected to 1 hour fire resistance requirements. The construction materials consist of grade C25 concrete and 500 steel. Based on the information provided, design slab panel B-C/1-2.

A

B 7000

C 7000

D 7000

1         0         0         0         9

2         0         0         0         9

3

Bending moment

When the slab are provided with different edge conditions like fixed or continuous edges, the maximum moments per unit width are given by:

      

Ly

where;

 = total ultimate load per unit area  = length of shorter side  = length of longer side  and  are the moment coefficient from Table 3.14 BS8110: Part 1: 1997

 

 

Lx

There are nine different types of support conditions to be considered which relate to the particular support/restraint provided on edge of individual slabs. Slab

Case 4

Case 2

Case 3

Case 1

Case 5

Case 6

Case 7

Void

Case 8

Case 9

Case 1: Four edges continuous Case 2: One short edge discontinuous Case 2: One long edge discontinuous Case 4: Two adjacent edge discontinuous Case 5: Two short edge discontinuous

Case 6: Two long edges discontinuous Case 7: Three edges discontinuous (one long edge continuous) Case 8: Three edges discontinuous (one short edge continuous) Case 9: Four edges discontinuous

Shear force and actions on supporting beams

The design shear forces of slab or loads on beams which supported the slab can be evaluated using the equations below:

      where;

 = total ultimate load per unit area  = length of shorter side  = length of longer side  and  are the shear coefficient from Table 3.115 BS8110: Part 1: 1997

 

 

1.0 SPECIFICATION

Long span, L y Short span, L x Ly/Lx

 Two-way slab Characteristic actions: ----Permanent, gk ----Variable, qk Design life Fire resistance Exposure classes Materials: ----Characteristic strength of concrete,  f ck  ----Characteristic strength of steel,  f yk  ----Unit weight of reinforced concrete Assumed: Øbar

= 7000mm = 4000mm = 7000/4000 = 1.75 < 2.0

= 1.5kN/m2 (excluding selfweight) = 4.0kN/m2 = 50 Years = R60 = XC1 = 25N/mm2 = 500N/mm2 = 25kN/m3 = 10mm

2.0 SLAB THICKNESS

Min. thickness for fire resistance Estimated thickness for deflection control, h

= 80mm = Lx/35 (Based on Table 5.4N, L/30) = 4000/35 = 114mm

Use, h = 125mm

3.0 DURABILITY, FIRE AND BOND REQUIREMENTS

Min. concrete cover regard to bond, C min,b Min. concrete cover regard to durability, C min, dur Min. required axis distance for R60, a Min. concrete cover regard to fire Cmin,fire = a-Øbar/2

= 10mm = 15 = 15mm

= 15-(10/2) = 10mm



Allowance in design for deviation, Cdev

= 10mm

Nominal cover, Cnom = Cmin+ Cdev



 C

nom =

= 15+10 = 25mm 25mm

4.0 LOADING AND ANALYSIS

Slab self-weight Permanent load (excluding self-weight) Characteristic permanent action, G k Characteristic variable action, Q k Design action, n d

= 1.35Gk+1.5Q k = 1.35(4.63) + 1.5(4.00) = 12.24kN/m2

Moment: Short span, M sx1 =

2 sx1ndLx

Short span, Msx2

2 sx2ndLx

  = 

  = 

Long span, M sy1 =

sy1ndLx

2

Long span, Msy2

sy2ndLx

2

Shear: Short span, V sx1 =

vx1ndLx

Short span, Vsx2

vx2ndLx

  = 

  = 

= 0.125x25 = 3.13kN/m2 = 1.50kN/m2 = 4.63kN/m2 = 4.00kN/m2

Long span, V sy1 =

vy1ndLx

Long span, Vsy2

vy2ndLx

= 0.065x12.24x4.02 = 12.7kNm/m = 0.087x12.24x4.02 = 17.0kNm/m = 0.034x12.24x4.02 = 6.66kNm/m = 0.045x12.24x4.02 = 8.82kNm/m

= 0.57x12.24x4.0 = 27.9kNm/m = 0.38x12.24x4.0 = 18.6kNm/m = 0.40x12.24x4.0 = 19.6kNm/m = 0.26x12.24x4.0 = 12.7kNm/m

5.0 MAIN REINFORCEMENT

Effective depth: dx = h-Cnom-0.5Øbar dy  = h-Cnom-1.5Øbar

Short span (Midspan): Msx K = M/bd 2f ck

= 125-20-(0.5x10) = 95mm = 125-20-(1.5x10) = 85mm

= 12.7kNm/m = 12.7x106/(1000x952x25)

= 0.056 < Kbal=0.167

 Compression reinforcement is not required z = d[0.5+  ]

= 0.95d

 0.95d

Use 0.95d

= 12.7x106/(0.87x500x0.95x95) = 325mm2/m

As = M/0.87f ykz H10-200 (bot) (393mm2/m)

Short span (Support): Msx K = M/bd 2f ck

= 17.0kNm/m = 17.0x106/(1000x952x25) = 0.076 < Kbal=0.167

 Compression reinforcement is not required z = d[0.5+  ]

= 0.93d

 0.95d

Use 0.93d

= 17.0x106/(0.87x500x0.93x95) = 444mm2/m

As = M/0.87f ykz H10-175 (top) 2 (449mm /m)

Long span (Midspan): Msy K = M/bd 2f ck

= 6.7kNm/m = 6.7x106/(1000x852x25) = 0.037 < Kbal=0.167

 Compression reinforcement is not required z = d[0.5+  ]

= 0.97d

 0.95d

Use 0.95d

= 6.7x106/(0.87x500x0.95x85) = 190mm2/m

As = M/0.87f ykz H10-350 (bot) (224mm2/m)

Long span (Support): Msy K = M/bd 2f ck

= 8.8kNm/m = 8.8x106/(1000x852x25) = 0.039 < Kbal=0.167

 Compression reinforcement is not required z = d[0.5+  ]

Use 0.95d

= 0.96d

 0.95d

= 8.8x106/(0.87x500x0.95x85) = 225mm2/m

As = M/0.87f ykz H10-325 (top) 2 (242mm /m)

Min. and max. reinforcement area, As,min = 0.26(f ctm/f yk)bd

= 0.26x(2.56/500)bd Use 0.0013bd = 0.0013bd > 0.0013bd = 0.0013x1000x95 = 127mm2/m = 0.04x1000x125 = 5000mm2/m

As,max = 0.04Ac = 0.04bh H10-425 (185mm2/m)

6.0 SHEAR

Max. design shear force, V Ed

= 27.9kN

Design shear resistance, k = 1+(200/d)1/2  2.0



1

ρ = As1/bd

= 1+(200/95)1/2 = 2.45  2.0



 0.02

= 449/(1000x95) = 0.0047  0.02



VRd,c = [0.12k(100ρ 1f ck)1/3]bd

= [0.12x2.0(100x0.0047x25)1/3]x1000x95 = 51925N = 51.9kN

Vmin = [0.35k3/2f ck1/2]bd

= [0.0035x2.03/2x251/2]x1000x95 = 47023N = 47.0kN VRd,c ; Vmin > VEd (OK)

7.0 DEFLECTION

Percentage of required tension reinforcement, =As,req/bd



Reference reinforcement ratio, =(f ck)1/2x10-3



= 325/(1000x95) = 0.0034

= (25)1/2x10-3 = 0.0050

Percentage of required compression reinforcement  =As’,req/bd



= 0/(1000x139) = 0.000

Factor for structural system, K = 1.3  use equation (2)

 

               

= 1.3(11+11+5) = 35

Modification factor for span less than 7m

= 1.0

Modification factor for steel area provided

= As,prov/As,req = 393/325 = 1.21  1.5



Therefore allowable span-effective depth ratio

= (l/d)allowable = 35x1.0x1.21 = 42.3

Actual span-effective depth

= (l/d)actual = 4000/95 = 42.1  (l/d)allowable (OK)



8.0 CRACKING

h = 125mm < 200mm Main bar, Smax,slabs = 3h or 400mm Max. bar spacing = 350mm

= 375mm < 400mm

 S

max,slab =

375mm

(OK)

Secondary bar, Smax,slabs = 3.5h or 450mm Max. bar spacing = 425mm

= 438mm < 450mm

 S

max,slab =

438mm

(OK)