Two Way Design Slab to BS 8110

Basic Structural Element Design to BS 8110-1:1997 

1.2

TWO WAY SPANNING SLAB DESIGN

Reference

Design Calculations

Output

Design of Slab Panel –Between Grids A/B and 1/2) Assume a slab thickness of a 162.5mm. Top and bottom cover to reinforcement is 25mm.

Effective depth Short way

=162.5-25-10/2=132.5mm

Long way

=162.5-25-10-10/2=122.5mm

Loading 2

Self-weight

=0.1625x24=3.9kN/m

Finishes

=1kN/m

Imposed load

=1.5kN/m

So the design load

=1.4x(3.9+1)+1.6x1.5=9.26kN/m

2 2 2

Bending moments ly/lx=7.470/5.570=1.34<2(Hence the slab panel is two-way spanning). Type of slab panel-Two adjacent edges discontinuous Table 3.14

Short way ,continuous edge

= βsxnl2x = 0.071 × 9.26 × 5.5702 = 20.4kNm/m Short way, mid span

= βsxnl2x = 0.053 × 9.26 × 5.5702 = 15.23kNm/m Long way, continuous edge

= βsynl2x = 0.045 × 9.26 × 5.5702 = 12.93kNm/m Long way, span

= βsynl2x = 0.034 × 9.26 × 5.5702 = 9.77kNm/m

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Basic Structural Element Design to BS 8110-1:1997 

Reference

Design Calculations

Output

Design of reinforcement

Short way, continuous edge Cl.3.4.4.4

K=

M 2 f cubd

=

20.43 × 10

2 25 × 1000 × 132.5



z = d0.5 + 0.25 −

0.9

 = 0.95d 

M =

= 0.046

0.046 



Hence A s

6

20.43 × 10

0.95f y Z

=

6

0.95 × 460 × 0.95 × 132.5

=

2 371mm /m

2

UseT10 @175mm(As, provided=449mm ) Cl.3.12.11.2.7

T10 @ 175

Maximum spacing = 3x132.5 =397.5mm Hence maximum spacing is not violated.

Table 3.27

100As

=

bh

100 × 449 1000 × 162.5

= 0.28 > 0.13

Hence minimum steel requirement is not violated. Cl.3.4.4.4

Short way, mid span K=

M 2 f cubd

=

15.23 × 10

2 25 × 1000 × 132.5



z = d0.5 + 0.25 −



Hence A s

= 0.034

  = 0.96dbut should be less than 0.95d 0.9  K

M =

6

0.95f y Z

15.23 × 10 =

6

0.95 × 460 × 0.95 × 132.5

=

2 277mm /m

2

UseT10 @175mm(As, provided=449mm ) Clause

Maximum spacing = 3x132.5 =397.5mm

3.12.11.2.7

Hence maximum spacing is not violated.

Table 3.25

100As bh

=

100 × 449 1000 × 162.5

T10 @ 175

= 0.28 > 0.13

Hence minimum steel requirement is not violated.

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Basic Structural Element Design to BS 8110-1:1997 

Reference Cl.3.4.4.4 BS 8110

Design Calculations

Output

Long way, continuous edge K=

M 2 f cubd

=

12.93 × 10

2 25 × 1000 × 122.5



z = d0.5 + 0.25 −



Hence A s =

6

= 0.034

  = 0.96d > 0.95d 0.9  K

M

=

0.95f y Z

12.93 × 10

6

0.95 × 460 × 0.95 × 122.5

= 254mm2

2

UseT10 @200mm(As, provided=392mm )

T10 @ 200

Clause

Maximum spacing = 3x122.5 =367.5mm

3.12.11.2.7

Hence maximum spacing is not violated.

Table 3.27

100As

=

bh

100 × 392 1000 × 162.5

= 0.24 > 0.13

Hence minimum steel requirement is not violated. C.3.4.4.4 BS 8110

Long way, mid span K=

M 2 f cubd

=

9.77 × 10

2 25 × 1000 × 122.5



z = d0.5 + 0.25 −



Hence A s =

6

= 0.026

  = 0.97d > 0.95d 0.9 

M

K

=

0.95f y Z

9.77 × 10

6

0.95 × 460 × 0.95 × 122.5

= 192mm2

2

UseT10 @200mm(As, provided=392mm ) Table 3.25

100As bh

=

100 × 392 1000 × 162.5

T10@200

= 0.24 > 0.13

Hence minimum steel requirement is not violated. Cl.3.12.11.2.7

Maximum spacing = 3x122.5 =367.5mm Hence maximum spacing is not violated.

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Basic Structural Element Design to BS 8110-1:1997 

Reference

Design Calculations

Output

Checks for deflection

6 M 12.23 × 10 = = 0.85 2 2 bd 1000 × 120 f s =

2fyA s,req 3A s,prov

=

2 × 460 × 277 3 × 449

= 189N/mm2

Modification factor for tension reinforcement (477 - f  ) (477 - 189) s = 0.55 + = 1.91 f  = 0.55 + 120(0.9 + 0.86) 120(0.9 + M ) bd2 Table 3.9

Basic Span/Effective depth=23 Allowable Span/Effective depth=26x1.91=50 Actual Span/Effective depth=5570/132.5=42 Hence the deflection is within allowable limits.

Deflection is O.K

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Basic Structural Element Design to BS 8110-1:1997 

Reference Cl.3.12.11.2.7

Design Calculations

Output

Maximum spacing = 3x110 =330mm Hence maximum spacing is not violated.

Table 3.25

100As bh

=

100 × 392 1000 × 150

= 0.26 > 0.13

Hence minimum steel requirement is not violated. Checks for deflection

6 M 12.23 × 10 = = 0.85 2 2 bd 1000 × 120 f s =

2fyA s,req 3A s,prov

=

2 × 460 × 245 3 × 392

= 192N/mm2

Modification factor for tension reinforcement f  = 0.55 +

Table 3.9

(477 - f  ) (477 - 192) s = 0.55 + = 1.91 120(0.9 + 0.85) 120(0.9 + M ) bd2

Basic Span/Effective depth=26 Allowable Span/Effective depth=26x1.91=50 Actual Span/Effective depth=5570/120=46 Hence the deflection is within allowable limits.

Deflection is O.K

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Basic Structural Element Design to BS 8110-1:1997 

Reference

Design Calculations

Output

Checks for shear Table 3.15

Short way support v=

Table 3.8

26.30 × 1000 1000 × 132.5

= 0.51x9.26x5.570=26.30kN/m

= 0.20N/mm2

Concrete shear stress 1

1

 100As  3  400  4 1 v c = 0.79 ×  bd  ×  d  × δ m 1

1

 100 × 449  3 ×  400  4 × 1 = 0.79 ×   132.5  1.25  1000 × 132.5 

vc

= 0.58N/mm2 > 0.20N/mm2 Hence no shear reinforcement is required.

No shear r/f required

Table 3.15

Long way support v=

Table 3.8

20.63 × 1000 1000 × 122.5

= 0.40x9.26x5.570=20.63kN/m

= 0.17N/mm2

Concrete shear stress 1

1

 100As  3  400  4 1 v c = 0.79 ×  bd  ×  d  × δ m 1

vc

1

 100 × 393  3  400  4 1 = 0.79 ×  × × 1.25  1000 × 122.5   122.5  = 0.58N/mm2 > 0.17N/mm2

Hence no shear reinforcement is required.

No shear r/f required

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