Design Of Two Way Slab..edited

DESIGN OF TWO WAY SLAB (Direct Design Method) Two – Way Slab: Dimension 11m x 10m L S

=

Assume h =

5.5 5

= 1.1; Two-Way Slab

ln[ ( 800+0.73 fy ) ] 36000+9000 β

longclearspan β= shortclearspan

β=

5500−55 0 50000−55 0

= 1.11

Ln = long clear span = 55000 – 550 = 4950 mm h=

( 4950 ) [ ( 800+0.73 ( 414 ) ) ] 36000+9000 (1.11)

h= 118.63 mm say 120mm The value of I is computed from that of an I-beam. This procedure is very tedious. From computer calculation, the author found that for edge beam, with slab on

one side only, I = 1.4

Ld ³ 12

and for interior beam with slab on both sides, I = 1.6

where L is the beam width and d is the total depth of beam. For interior beam: Ld ³ I = 1.6 12

Ld ³ 12

550 (800) ³ ¿ = 3.75 x 10¹⁰ mm⁴ I = 1.6 [ 12 For edge beam: Ld ³ I = 1.4 12 (55 0)(800) ³ ¿ = 3.29 x 10¹⁰ mm⁴ 12

I = 1.4[ Value of α: α=

( Ecb)(Ib) ( Ecs)(Is)

(NSCP 2010, Sec. 413.7.16, pg.

4-88) Ec b=E c s=Ec For edge beam with 11.00 m slab: 10

αa =

3.29 x 10 55 00 ( 120 ) ³ 12

= 41.54

For interior beam with 11.00 m slab:

αb =

3.75 x 1010 550 0 ( 120 ) ³ 12

= 47.35

For interior beam with 10.00 m slab width: (2 beams)

αc =

Average:

3.75 x 1010 5 000 ( 120 ) ³ 12

= 52.08

αm =

αa+ αb+αc 4

αm =

41.54+ 47.35+2(5 2.08) 4

αm = 48.26 βs =

length of continuous edges perimeter

( 5.5 ) +2(5) 2 ( 5.5 ) +2(5)

βs =

= 0.74

Ln = 5500-550 = 4950 mm 36000+5000 β( αm−0.12+ h=

1+1 ) β

ln (800+0.73 fy) ¿ ¿ 495 0( 800+0.73 ( 414 )) h=

(

36000+5000 (1.11)(48.26−0.12 1+

1 ) 1.11

)

h=58.92 but not less than hmin : h > hmin but should not be more than hmax hmax =

ln(800+0.73 fy) 36000

hmax =

(495 0)(800+ 0.73 ( 414 ) ) 36000

= 151.56 mm

h > hmin < hmax Therefore, use h = 200 mm Compute weight of slab = 23.56(0.20) = 4.712 KPa Total DL = weight of slab Total DL = 7.068 KPa LL acting on slab= 12 KPa Wu= 1.2DL + 1.6LL Wu= 1.2(4.712) + 1.6(12) = 24.8544 kPa Using 12mm bars: d = 200 -

1 2 (12) – 20 = 174 mm

Check for depth shear: L2 αc( L1 ¿ ≥ 1.0

( from NSCP

2010 P4-84, 413.2.8) 5 ¿=10.2 ≥ 1.0 (11.25)( .5 11 x=( 2 )-

0.550 2

x = 5.051 m

Taking b = 1 m Vu = qu x Area shaded

- 0.174

(OK!)

Vu = 24.8544 x 5.051 (1) Vu = 125.54 KN Vc =

1 2 (1+ ) λ √ f ' c bd 6 β

Vc =

1 2 (1+ )(1) √27.6 6 1 .11

(from NSCP 2010 P4-67, 411.13.2.1)

(1000) (174)

Vc = 426.864 KN Or Vc =

1 λ √f ' c bd 3

Vc =

1 (1) √ 27.6 (1000)(1 74) 3

Vc = 304.71 KN ФVc = 0.85 (304.71) ФVc = 259.00 KN > Vu

(OK)

Moment along the long span (11-m interior span): Mo =

( Wu L2 ) ( Ln)² 8 Ln = 5.5 – 0.550 = 4.95

Mo =

(125.54 ( 5 ) )(4.9 5)² 8

= 1922.53 KN-m

Negative factored moment = -0.65(848.90) = -551.785 KN.m Positive factored moment = 0.35(848.90) = 297.115 KN.m Distributing these moments to beam and the column strips:

L2 L1

5.5 5

=

= 1.10

α 1 = αb = 1.65 (in the direction of L1) L2 α 1( L1 ¿

= 1.10(10.23) = 11.253

According to NSCP 2010 Sec. 43.7.5.1 pg 4-84, Beams between supports shall be

L2 proportioned to resist 85 percent of column strip moments if (α 1( L1 ¿ ) is equal to or greater than 1.0. Factored Moments in Column Strips

(NSCP 2010, 413.7.4.1, pg 4-84)

(Interior Negative Factored Moments) l 2 /l 1 α 1( α 1(

l 2 /l 1 ¿

=0

l 2 /l 1 ¿ ≥1

By interpolation: Percentage =

0.5

1.0

2.0

75

75

75

90

75

45

75−45 1−2 = 75−x 1−1.1 Percentage = 72%

According to Section 5.13.6.5 (Fundamentals of RCD by DIT Guillesania). Interior Negative Moment: Column Strip: 72%( -1249.64) = -899.74 kN-m Beam: 85% (899.74) = -764.78 kN.m

Slab: 15% (899.74) = -134.96 kN.m Middle Strip: -(1249.64 - 899.74) = -349.9 kN.m Positive Moment: Column Strip: 72% (672.89) = 484.48 kN.m Beam: 85% (484.48) = 411.808 kN.m Slab: 15% (484.48) = 72.672 kN.m Middle Strip: (672.89- 484.48) = 188.41 kN.m Moment along the long span (along edge beam) 5 2

L6 =

.55 2

+

= 2.775 m

Ln = 5.5 – 0.55 = 4.95 m Mo =

( Wu L2 ) ( Ln)² 8 (24.8544 ( 2.775 ) )(4.95)² 8

Mo =

= 211.25 KN-m

Negative factored moment: -0.65(211.25) = -137.3125 KN.m Positive factored moment: 0.35(211.25) = 73.94 KN.m β1 =

EcbC 2 EcsIs

x C = Σ( 1 – 0.63 y

=

(

1−0.63

Note: Ecb = Ecs = Ec 3

x y 3

)

3

55 0 55 0 (800) 800 3

)

+

(

1−0.63

3

2 00 2 00 (55 0) 600 3

)

C = 26309.02 x

Is =

6

10 mm

2500 (2 00) ³ 12 Is = 1666.67 x 10⁶

β1 =

4

mm4

26309.02 x 106 2( 1666.67 x 106 ) β1 = 7.89

L2 L1

=

5 5.5

= 0.91

α 1 = αa = 8.97 ( for edge beam ) L2 α 1( L1 ¿

= 8.97(0.91) = 8.16

Exterior Negative Factored Moments (Fundamentals of RCD by DIT Guillesania): l 2 /l 1 α 1( α 1(

l 2 /l 1 ¿

= 0.80

l 2 /l 1 ¿=0.90

0.9

1.0

2.0

77.40

75

75

77.70

75

45

By interpolation: Percentage =

77.70−75 0.9−1.0 = 77.40−x 0.9−0.91 Percentage = 77.13%

Exterior Negative Moment: Column Strip: 77.13 %( -137.3125) = -105.91 kN-m Beam: 85% (-105.91) = -90.02 kN.m

Slab: 15% (-105.91) = -15.89 kN.m Middle Strip: (-137.125 – 105.91) = -31.4025 kN.m Positive Moment: Column Strip: 77.13% (73.94) = 57.03 kN.m Beam: 85% (57.03) = 48.48 kN.m Slab: 15% (57.03) = 8.55 kN.m Middle Strip: (73.94 – 57.03) = 16.91 kN.m

Moment along the short span (10 m end span)

L2 = 11 m

Ln = 5 – 0.55 = 4.45 m Mo =

( Wu L2 ) ( Ln)² 8

Mo =

(24.8544 ( 5.5 ) )(4.45)² 8

= 338.37 KN-m

According to NSCP 2010 Sec. 413.7.3.3 pg 4-83, In an end span, total factored static moment Mo shall be distributed as follows:

Factored Interior Negative Moment: -0.7(338.37) = -236.859 kN.m Factored Positive Moment: 0.57(338.37) = 192.8709 kN.m Factored Exterior Negative Moment: -0.16(338.37) = -54.1392 kN.m L2 L1

=

5 5.5

= 0.91

α 1 = αa = 8.97 ( for edge beam ) L2 α 1( L1 ¿

= 8.97(1.10) = 9.867

According to NSCP 2010 Sec. 43.7.5.1 pg 4-84, Beams between supports shall be

L2 proportioned to resist 85 percent of column strip moments if (α 1( L1 ¿ ) is equal to or greater than 1.0. Factored Moments in Column Strips

(NSCP 2010, 413.7.4.1, pg 4-84)

(Interior Negative Factored Moments) l 2 /l 1 α 1( α 1(

l 2 /l 1 ¿

=0

l 2 /l 1 ¿ ≥1

By interpolation: Percentage =

0.5

1.0

2.0

75

75

75

90

75

45

75−45 1−2 = 75−x 1−1.1

Percentage = 72% Interior Negative Moment: Column Strip: 72%( -236.859) = -170.54 kN-m Beam: 85% (-170.54) = -144.96 kN.m Slab: 15% (-170.54) = -25.58 kN.m Middle Strip: -(236.859– 170.54) = -66.319 kN.m Exterior Negative Moment: Column Strip: 72%( -54.1392) = -38.98 kN-m Beam: 85% (-38.98) = -33.133 kN.m Slab: 15% (-170.54) = -25.58 kN.m Middle Strip: -(54.1392– 38.98) = -15.1592 kN.m

Positive Moment: Column Strip: 72% (192.8709) = 138.87 kN.m Beam: 85% (138.87) = 118.04 kN.m Slab: 15% (138.87) = 20.83 kN.m Middle Strip: (192.8709 – 138.87) = 54.001 kN.m

Ab =

ρmin =

π 4 (12)² = 113 mm² 1.4 414

= 0.00338

0.85(27.6)(0.85)(600) ρmax =0.75 414(600+ 414)

= 0.02138

Row A & C

Row B

Mo b d Ru ρ Use ρ As N = As/Ab s Use s Position

Mo b d Ru ρ Use ρ As N= As/Ab s Use s Position

Across F -2.92 1100 162 0.112387 401 0.00027 0.00338 602.316 5.330230 088 187.6091 62 180 Top

Across A -75.89 975 174 2.856530 932 0.00738 0.00738 1252.017 11.07979 646 90.25436 556 90 Top

Across E 10.42 1100 174 0.347643 431 0.000846 0.00338 646.932 5.725061 947 174.6705 991 170 Bottom

Row D & F Across B 8.55 975 174 0.321825 53 0.000783 0.00338 573.417 5.074486 726 197.0642 656 190 Bottom

Across D -12.79 1100 162 0.492272 212 0.0012 0.00338 602.316 5.330230 088 187.6091 62 180 Top

Across C -75.89 975 174 2.856530 932 0.00738 0.00738 1252.017 11.07979 646 90.25436 556 90 Top

Across F -15.16 2750 174 0.202313 797 0.000491 0.00338 1617.33 14.31265 487 69.86823 963 60 Top

Across A -206.35 2500 174 3.029168 685 0.00786 0.00786 3419.1 30.25752 212 33.04963 294 40 Top

Across E 54 2750 174 0.720642 813 0.00177 0.00338 1617.33 14.31265 487 69.86823 963 60 Bottom

Row E Across B 111.12 2500 174 1.631215 044 0.00409 0.00409 1779.15 15.74469 027 63.51347 554 60 Bottom

Across D -66.32 2750 174 0.885056 137 0.00218 0.00338 1617.33 14.31265 487 69.86823 963 60 Top

Across C -206.35 2500 174 3.029168 685 0.00786 0.00786 3419.1 30.25752 212 33.04963 294 40 Top