Moment Coefficients for Continuous Beams and Slabs

DESIGN OF SLAB ONE - WAY SLAB

 A

B

C

5.0

5.0

1  A

 A

 S       4     

 S        3     

 S       2     

 S       1     

       0        0  .        0        1

2 WHOLE CROSS SECTION OF SLAB  A

C

B 2.5 2.5

2.5

wl²/14

2.5

wl²/16

2

wl²/16

 

-wl²/12

wl²/14 8

6

4 3

1

5

-wl²/12

 

9

7

-wl²/12

 

-wl²/12

 

ONE END CONTINUOUS

B

C 2.5

2.5

2.5

wl²/16

wl²/16

4

-wl²/12

wl²/14 8

6

3

5

 

-wl²/12

9

7

 

-wl²/12

 

-wl²/12

-wl²/12

Moment Coefficients for Continuous Beams and Slabs

Limitations: Adjacent spans should not differ by more than 20% of the shorter span. Ratio of live load to deal load should not exceed 3. The loads should be uniformly distributed Positive moment: End spans a) If discontinuous end is unrestrained

wL²/11

b) If discontinuous end is integral with support

wL²/14

Interior spans

wL²/16

Negative moment at exterior face of first interior support Two spans

wL²/9

More than two spans

wL²/10

Negative moment at other faces of interior supports

wL²/11

Negative moment at face of all supports a) For slabs with spans less than 3m

wL²/12

b) Beams and girders where

wL²/12

    

Negative moment at interior faces of exterior supports for Members built integrally with their supports a) When the supports is a spandrel beam or girder

wL²/24

b) When the support is a column

wL²/16

Shear in end members at first interior support

1.15wL/2

Shear at all other supports

wL/2



Live load

Dead load

Commercial = 4.8 kPa

Concrete = 23.56 kN/m³ Floor finish = 1.53 kPa Ceiling = 0.24 kPa

Fc` = 27 mPa Fy = 414 mPa

A) Verify if one  – way slab La = 2.5 - .3 = 2.2m Lb = 10  – 0.3 = 9.7m M = la / lb = 2.2 / 9.7 = 0.29 < 0.5 therefore one way slab

B) Compute the minimum slab thickness H = L/24 = 2200 / 24 = 91.67mm say 150 mm

C) Compute the factors load (consider 1 meter strip in short direction) Dead load = 23.56 (0.15) (1) + 0.24(1) + 1.53 (1) = 5.3 kN/m Live load = 4.8 (1) = 4.8 kN/m W = 1.4 (5.3) + 1.7 (4.8) = 15.58 kN/m

D) Check ACI limitations a) LL/DL = 4.8/5.3 = 0.91 < 3 ok b) L-s /s = 0/2.2 = 0< 0.2 ok c) Load is uniformly distributed ok

E) Compute the moments using ACI coefficients -

 = wL² / 12 = 15.58 (2.2)² /12 = 6.29 kN . m  = wL² / 14 = 15.58 (2.2)² / 14 = 5.39 kN. m  = wL² / 16 = 15.58 (2.2)² / 16 = 4.71 kN. m

Compute the steel reinforcement Main bar: Pmin = 1.4 / fy = 1.4 / 414 = 0.00338 Effective depth = 150  – 20 – 12/2 = 124 mm For Mu = 6.29 kN. m

Ru = Mu / Øbd² = 6.29x P=

 / 0.9 (1000)(124)² = 0.46 mPa

 [  √    ] =  [  √    ] = 0.00112    

P < Pmin

therefore

use Pmin

As = Pbd = 0.00338 (1000)(124) = 419.12 mm² Spacing , use 12mm Ø

Ab =

/4 (12)² = 113mm²

S = 1000(113)/419 = 269.67 mm

say 250 mm

Use 12mm Ø @ 0.25m O.C. top bars For Mu = 5.39 kN.m

Ru = Mu / Øbd² = 5.39x P=

 / 0.9 (1000)(124)² = 0.39 mPa

 [  √    ] =  [  √    ] = 0.00095    

P < Pmin

therefore

use Pmin

As = Pbd = 0.00338 (1000)(124) = 419.12 mm² Spacing , use 12mm Ø

Ab =

/4 (12)² = 113mm²

S = 1000(113)/419 = 269.67 mm

say 250 mm

Use 12mm Ø @ 0.25m O.C. bottom bars For Mu = 4.7 kN.m

Ru = Mu / Øbd² = 4.71x P=

 / 0.9 (1000)(124)² = 0.34 mPa

 [  √    ] =  [  √    ] = 0.00083    

P < Pmin

therefore

use Pmin

As = Pbd = 0.00338 (1000)(124) = 419.12 mm² Spacing , use 12mm Ø

Ab =

/4 (12)² = 113mm²

S = 1000(113)/419 = 269.67 mm

say 250 mm

Use 12mm Ø @ 0.25m O.C. bottom bars Temperature Bars : Ast = 0.0018 bt = 0.0018 (1000)(150) = 270 mm² Using 10mm Ø bar Ast =

/4 (10)² = 78.5 mm²

S = 1000(78.5)/270 = 291 mm say 280mm Use 10mm Ø @ 0.28m O.C. temperature bars.

DESIGN OF SLAB ONE - WAY SLAB

 A

B

C

5.0

5.0

1  A

 A

 S        4      

 S         3      

 S        2      

 S        1      

        0         0   .         0         1

2

WHOLE CROSS SECTION OF SLAB  A

C

B 2.5 2.5

2.5

wl²/14

2.5

wl²/16

2

wl²/16

wl²/14 8

6

4 3

5

-wl²/12

 

-wl²/12

 

9

7

1

-wl²/12

 

-wl²/12

 

ONE END CONTINUOUS  A

2.5

wl²/14 2 3 1

-wl²/12

 

-wl²/12

-wl²/12

A. Verify if one  – way slab La = 2.5 - .3 = 2.2m Lb = 10  – 0.3 = 9.7m M = la / lb = 2.2 / 9.7 = 0.29 < 0.5 therefore one way slab

B. Compute the minimum slab thickness H = L/24 = 2200 / 24 = 91.67mm say 150 mm

C. Compute the factors load (consider 1 meter strip in short direction) Dead load = 23.56 (0.15) (1) + 0.24(1) + 1.53 (1) + 1.0(1) = 6.3 kN/m Live load = 4.8 (1) = 4.8 kN/m W = 1.4 (6.3) + 1.7 (4.8) = 16.98 kN/m

D. Check ACI limitations d) LL/DL = 4.8/6.3 = 0.762 < 3 ok e) L-s /s = 0/2.2 = 0< 0.2 ok f) Load is uniformly distributed ok

E. Compute the moments using ACI coefficients -

 = wL² / 12 = 16.98 (2.2)² /12 = 6.85 kN . m  = wL² / 14 = 16.98 (2.2)² / 14 = 5.87 kN. m

Compute the steel reinforcement Main bar: Pmin = 1.4 / fy = 1.4 / 414 = 0.00338 Effective depth = 150  – 20 – 12/2 = 124 mm For Mu = 6.85 kN. m

Ru = Mu / Øbd² = 6.85x P=

 / 0.9 (1000)(124)² = 0.495 mPa

 [  √    ] =  [  √   ] = 0.0012    

P < Pmin

therefore

use Pmin

As = Pbd = 0.00338 (1000)(124) = 419.12 mm² Spacing , use 12mm Ø

Ab =

/4 (12)² = 113mm²

S = 1000(113)/419 = 269.67 mm

say 250 mm

Use 12mm Ø @ 0.25m O.C. top bars For Mu = 5.87 kN.m

Ru = Mu / Øbd² = 5.87x P=

 / 0.9 (1000)(124)² = 0.424 mPa

 [  √    ] =  [  √   ] = 0.00103    

P < Pmin

therefore

use Pmin

As = Pbd = 0.00338 (1000)(124) = 419.12 mm² Spacing , use 12mm Ø

Ab =

/4 (12)² = 113mm²

S = 1000(113)/419 = 269.67 mm

say 250 mm

Use 12mm Ø @ 0.25m O.C. bottom bars Temperature Bars : Ast = 0.0018 bt = 0.0018 (1000)(150) = 270 mm² Using 10mm Ø bar Ast =

/4 (10)² = 78.5 mm²

S = 1000(78.5)/270 = 291 mm say 280mm Use 10mm Ø @ 0.28m O.C. temperature bars.

TABULATE FOR STEEL REINFORCEMENT ( SLABS @ SLAB 1-4) As (mm²)

SECTION

Mu (kN.m)

Ru (MPa)

1

6.85

0.495

0.0012

0.00338

419.12

269 say 250

2

5.87

0.424

0.00103

0.00338

419.12

269 say 250

3

6.85

0.495

0.0012

0.00338

419.12

269 say 250

4

5.39

0.39

0.00095

0.00338

419.12

269 say 250

5

6.29

0.46

0.00112

0.00338

419.12

269 say 250

6

4.71

0.34

0.00083

0.00338

419.12

269 say 250

7

6.29

0.46

0.00112

0.00338

419.12

269 say 250

8

5.39

0.39

0.00095

0.00338

419.12

269 say 250

9

6.29

0.46

0.00112

0.00338

419.12

269 say 250

REQ'D P

USE P

SLAB DETAILS:

 A

B

C

2.5

2.5 2.5

"a"

"a"

2.5

"a"

"a"

"a"

150mm

"b"

"b"

"b"

SCHEDULE OF REINFORCEMENT: bar "a" = 12mm @ 0.25m o.c. top bars bar "b" = 12mm @ 0.25m o.c. bottom bars use 10mm Ø temperature bars @ 0.28m o.c. note: extend 1 of 2 midspan bottom rebars full length

"b"

SPACING   (mm)