Tutorial 3 Design of two way slabs with solutions - with drop panel PDF

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Tutorial 3 Design of two way slabs with solutions - with drop panel...

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Tutorial 2 – Two-way Slab

Qs 1. Design of moments in x direction for interior design strips:

6.5m

y

x

1.5m

Middle Strip

3.0m

Column Strip

1.5m

Middle Strip 6.1m

Figure 1.1 Design strips (hushed area) Estimation of the depth As a rough estimate start with 𝐷𝑠 = 𝐷𝑠 =

𝐿𝑛

25

=

6100 25

𝐿𝑛

25

= 244mm, use 250mm thick slab

Note that the value of Ds is revised in the following

Load calculations: Dead Load (G) = 24.5 × 0.25 = 6kPa Live Load (Q) = 5kPa Service Load (w) = G + 0.7 Q = 6 + 0.7 × 5 = 9.5kPa Choose flat plate thickness according to Warner et al., pp. 552 (see also lecture notes): 𝐿𝑛 𝐷𝑠

𝐿

≤ 55[( 𝐿′𝑛 ) × 𝑛

1 1 ⁄3 ] 𝑤𝐾1

Where, Ln= longer clear span of the slab, L‘n= shorter clear span of the slab, DS = thickness of the slab

Ln= L‘n= 6100mm K1= 1, for interior panels

6100 𝐷𝑠 𝐿𝑛 𝐷𝑠

≤ 55[( 5600) × 9.5×1 ]1 ⁄3 6100

1

≤ 26.7mm

DS ≥ 200mm

∴ 250mm thickness is OK Ultimate strength design for bending: From AS3600 total static moment for a given span:

Where:

∴

aSup = 260mm

L0 = 6500 –2 (0.7×260) = 6136 mm Lt= 6000mm

Fd = 1.2G + 1.5Q = 1.2 × 6 + 1.5 × 5 = 14.7 kPa ∴ 𝑀𝑜 ≤

14.7×6×6.1362 8

= 415 𝑘𝑁𝑚

Next, calculate the design moments in the strip according to AS3600, 6.10.4.3 (B)

M- = -0.65 × 415 = - 270 kNm M+ = 0.35 × 415 = 145.25 kNm Distribution to column and middle strips according to AS3600, Table 6.10.5.3. M- @ Column strip M- = 0.7 × (-270) = -189 kNm (for the 3m width); or, =- 63 kNm /m M- @ Middle strip M- = 0.3 × (270) = 81 kNm for the 3m width; or, =- 27 kNm /m M+ @ Column strip M+ = 0.5 × 145.25 = 72.63 kNm (for the 3m width); or, = 24.21 kNm /m M+@ Middle strip M+ = 0.5 × 145.25 = 72.63 kNm for the 3m width; or, =24.21 kNm /m Next, calculate the amount of reinforcement for each of the above design moments.

Checking for cover requirements Assume non-residential building, exposure classification A2 (Table 4.3) Minimum cover = 25mm, adopt 30mm (Table 4.10.3.2) For fire resistance period, consider 120min; minimum effective thickness = 120mm (Table 5.5.1) Assume N12 for reinforcement d = 250 – 30 – (12/2) = 214mm

Design for M- @ Column strip M* ≤  Mu, where  = 0.8 Mu = 𝐴𝑠𝑡 𝑓𝑠𝑦 𝑑 (1 −

𝐴𝑠𝑡 𝑓𝑠𝑦

,

1.7𝑏𝑑𝑓𝑐

defined in earlier classes.

), where the terms of the terms of the equation are the same as those

From M* ≤  Mu derive Mu

Mu = M* / 0.8

Mu = 63 / 0.8 = 78.75 kNm

78.75 ×106 = Ast 500 × 214 (1 −

𝐴𝑠𝑡×500

)

1.7×1000×214×32

From the solution of the quadratic equation, the value of Ast is obtained: Ast = 760.84 mm2 /m In N12 reinforcements (AN12 = 113mm2) are used, then, spacing (s) = 1000 / (760.84/113) = 149 mm

Design for M- @ Middle strip (note that the values of M+@ Column and Middle strips are similar, i.e., 25≈28kNm) M* ≤  Mu, where  = 0.8 Mu = 𝐴𝑠𝑡 𝑓𝑠𝑦 𝑑 (1 −

𝐴𝑠𝑡 𝑓𝑠𝑦

,

1.7𝑏𝑑𝑓𝑐

defined in earlier classes.

), where the terms of the terms of the equation are the same as those

From M* ≤  Mu derive Mu Mu = M* / 0.8

Mu = 27 / 0.8 = 33.75 kNm

33.75 ×106 = Ast 500 × 214 (1 −

𝐴𝑠𝑡×500

)

1.7×1000×214×32

From the solution of the quadratic equation, the value of Ast is obtained: Ast = 320 mm2 /m In N12 reinforcements (AN12 = 113mm2) are used, then, spacing (s) = 1000 / (320/113) = 353.13 mm Next, check for minimum reinforcement requirements Cl. 9.1.1a 𝐷 2 𝑓′𝑐𝑡 𝐴𝑠𝑡 ≥ 0.24 ( ) 𝑏𝑑 𝑑 𝑓𝑠𝑦

f’ct = 0.63√𝑓′𝑐 = 2MPa

𝐴𝑠𝑡 250 2 2 ≥ 0.24 ( ) 𝑏𝑑 214 500

𝐴𝑠𝑡 𝑏𝑑

≥ 0.0013, or, 𝐴𝑠𝑡 ≥ 0.0013 × 1000 × 214 ∴ 𝐴𝑠𝑡 ≥ 280𝑚𝑚2

Check for minimum spacing Cl. 9.4.1 Spacing should not exceed the lesser of : 2.0 Ds or 300mm

2.0 × 250 or 300mm, ∴ 𝑆𝑝𝑎𝑐𝑖𝑛𝑔 ≤ 300 𝑚𝑚2

Next, consider the reinforcement distribution according to Cl. 9.1.2 For M- @ Column strip M- = -63 kNm /m 25% M- = 0.25 × 270 = 67.5 kNm This must be resisted by (400 + 2 × 250); where, 400mm is the width of the slab and 250mm is its depth. b = 900mm

M* ≤  Mu, where  = 0.8 Mu = M* / 0.8

Mu = 67.5 / 0.8 = 84.375 kNm

84. 375×106 = Ast 500 × 214 (1 −

𝐴𝑠𝑡×500

)

1.7×900×214×32

From the solution of the quadratic equation, the value of Ast is obtained: Ast = 820.7 mm2 / 900mmm width In N12 reinforcements (AN12 = 113mm2) are used, then, spacing (s) = 900 / (820.7 /113) = 124 mm

Reinforcement arrangement For positive reinforcement (i.e., midspan) @ column strip Required spacing for flexural requirement = 353.13 mm Min spacing requirement = 300mm

∴

Adopt N12@300

∴

Adopt N12@300

For positive reinforcement (i.e., midspan) @ middle strip is similar to column strip

For negative reinforcement @ column strip Required spacing for flexural requirement = 149 mm Min spacing requirement = 300mm Min spacing to meet Cl. 9.1.2 in the middle of column strip (for 900mm width) = 124mm

∴ Adopt N12@120 in the middle 0.9m of the column strip and N12@140 in the remaining of the column strip.

For negative reinforcement @ middle strip Required spacing for flexural requirement = 353.13 mm Min spacing requirement = 300mm

∴

Adopt N12@300

Design of moments in y direction for interior design strip and design of moments in x, y directions respectively for exterior design strip can be calculated in a similar way.

Check for shear Design around an interior column: (1) Determine V* V* = 6.5 x 6 x 14.7 = 573.3 kN (2) Determine Mv* total unbalanced bending moment at interior support 𝑀0 = 415 𝑘𝑁𝑚 as calculated before M = 0.65 × 415 = 269.75 kNm M’ = 0.75 x 415 = 311.25 KNm M*v = M-M’ = 311.25-269.75 = 41.5 KNm AS3600 minimum requirement at an interior support Cl. 6.10.4.5 M*V ≥ 0.06[(1.25g+0.75q)LtLo 2-1.25gLt(L’0)2 ] = 0.06[(1.25x6+0.75x5)x6x6.1362-1.25x6x6x6.1362]= 50.8 kNm Therefore design for 50.8 kNm. (3) Check whether V* and M* can be resisted by the concrete alone. Ds = 250mm, projection of the drop panel below the slab is 60mm. Therefore, the total thickness of slab in the vicinity of columns is 250 +60 = 310 mm. Take dom as the average of dx and dy . dx = 310 - 30 – (12/2) = 274 mm , dy = 274 – 12 = 262 mm

x-bars are closer to the bottom surface of the slab because moments in x direction is larger than moments in y direction. dom = (274 + 262)/2 = 268 mm a=b= 400 + 268 = 668 mm U = 2(a+b) = 2672 mm 𝑓𝑐𝑣 = 0.17 (1 +

2 2 ) √𝑓𝑐′ = 0.17 (1 + ) √𝑓𝑐′ = 0.51√𝑓𝑐′ > 0.34√𝑓𝑐′ 𝛽ℎ 1

∴ 𝑓𝑐𝑣 = 0.34√𝑓𝑐′ = 0.34√32 = 1.92 𝑀𝑃𝑎

𝑉u =

𝑉𝑢𝑜 = 𝑢𝑑𝑜𝑚 𝑓𝑐𝑣 = 2672 × 268 × 1.92 = 1375 𝑘𝑁

𝑉𝑢𝑜 1375 = 1180 𝑘𝑁 = 𝑀𝑣∗ 50.8 × 106 668 𝑎 [1 + ] ] [1 + 8𝑉 ∗𝑑𝑜𝑚 8 × 573.3 × 103 × 268 𝑢 2672 V ∗ = 573.3 𝑘𝑁 < 0.7𝑉𝑢 = 826 𝑘𝑁

Hence 𝑉 ∗ and 𝑀𝑣∗ can be resisted by the concrete alone. Design around an edge column: (1) Determine V* V* = 6.5 x 6 x 14.7 x 0.5 = 286.65 kN

(2) Determine Mv* total unbalanced bending moment at interior support 𝑀0 = 415 𝑘𝑁𝑚 as calculated before 𝑀𝑣∗ = 0.25 × 415 = 103.75 𝑘𝑁𝑚

(3) Check whether V* and M* can be resisted by the concrete alone. dom = 268 mm as calculated before a=b= 400 + 268/2 = 534 mm b= 400+268 = 668 mm u=2a+b=2x534+668=1736 mm 𝑓𝑐𝑣 = 1.92 𝑀𝑃𝑎

𝑉𝑢𝑜 = 𝑢𝑑𝑜𝑚 𝑓𝑐𝑣 = 1736 × 268 × 1.92 = 893.3 𝑘𝑁

𝑉u =

𝑉𝑢𝑜

∗

𝑀𝑎𝑣 8𝑉 ∗𝑑𝑜𝑚

[1 +

103.75 × 106 534 8 × 286893.3 .65 × 103 × 268 = 576.8 𝑘𝑁 = [1 +

V ∗ = 286.65 < 0.7𝑉𝑢 = 403.76 𝑘𝑁] 1736 𝑢 Hence 𝑉 ∗ and 𝑀𝑣∗ can be resisted by the concrete alone. Check for deflection Deemed to comply span-to-depth ratio (Cl.9.3.4):

Check if

(

Lef /𝑑 ≤ 𝑘3 𝐾4 [

∆ )1000𝐸𝑐 𝐿𝑒𝑓

𝐹𝑑.𝑒𝑓

1 3

]

d = 250 – 30 – (12/2) = 214mm

Lef = effective span = lesser of Ln + 𝐷 or L Take longer span L = 6.5m

Ln = clear span = 6500 - 400 =6100 Ln + 𝐷= 6100+250 = 6350 mm

∴ Lef = 6350 mm

K 3 = 1.05 for a two-way slab with drop panels K 4 = 2.1 for interior spans K 4 = 1.75 for end spans 𝑓𝑐′ = 32 𝑀𝑃𝑎

Ec = 24001.5 × 0.043√32 = 28600 , cl.3.1.2 ∆ 1 = 𝐿𝑒𝑓 250

Fd,ef = (1.0 + 𝐾𝑐𝑠 )𝑔 + (𝜓𝑠 + 𝐾𝑐𝑠 𝜓𝑙 )𝑞 𝜓𝑠 = 0.7 short-term

𝜓𝑙 = 0.4 long-term

g = dead load = 0.25 x 24 = 6 KPa q = live load = 5 KPa

]

∴ K cs = 2.0

Fd,ef = (1 + 2) × 6 + (0.7 + 2 × 0.4) × 5 = 18 + 7.5 = 25.5 𝐾𝑃𝑎 Check 6350 214

≤ 1.05 × 2.1 × [

29.67 ≤ 36.4 6350 214

1

1 ( 250)×1000×28600 3

25.5

]

for interior spans

1

1 ( 250)×1000×28600 3

≤ 1.05 × 1.75 × [

29.67 ≤ 30.3

25.5

]

for end spans

∴ deflection is ok Alternatively, use simplified Vanderbilt’s model: ∆ = K1 𝐾2

𝐿𝑛 3 𝜔 + 𝐾𝑐𝑠 𝜔𝑠 𝐿( ) 7𝐸𝑐 𝑑

The short-term live load factor = 0.5 and the long-term live load factor is taken as 0.25. Short-term load 𝜔 = 6 + (0.5 × 5) = 8.5 𝑘𝑃𝑎

Long-term load 𝜔𝑠 = 6 + (0.25 × 5) = 7.25 𝑘𝑃𝑎

Also, 𝐿𝑛 = 6500 − 400 = 6100𝑚𝑚 and 𝐿′𝑛 = 6000 − 2000 = 4000 𝑚𝑚 For the column beam strip:

d of the slab = 250 -30 -12/2 = 214 mm d pf the drop = 214 + 60 = 274 mm 274×2700+214×3800

average: d =

6500

= 238.9 𝑚𝑚

For the middle-beam-strip: d = 214 -12 = 202 mm (reinforcement in y direction is placed above reinforcement in x direction as the moments in the y direction is smaller than those in the x direction) Therefore, average d = (238.9+202)/2 = 220.45 mm K1 = 1.3 for exterior panels with edge beams

K2 =

𝐿′𝑛 4000 = = 0.656 𝐿𝑛 6100

8.5 + 2 × 7.25

6100 ) = 13.5 𝑚𝑚 × 6500 × 10−3 × ( 220.45 7 × 28600 Since Lef = 6350 mm as calculated before, ∆ = 1.3 × 0.656 ×

According to cl. 2.3.2, it is required that ∆ 13.5 = 0.002 < 1/250 = 6350 Lef

3

∆

Lef

≤ 1/250

∴ The deflection is ok.

Q2. The solution shown is for the design of Panel P shown below.

Estimation of the depth Office floor slab thickness = 150mm f’c = 32MPa, fsy = 500 MPa

Load calculations: Dead Load (G) = 24.5 × 0.15 = 3.675kPa Live Load (Q) = 4kPa Service Load (w) = G + ψQ, where, ψ = 0.7 = 3.675 + 0.7 × 5 = 6.475kPa Choose flat plate thickness according to Warner et al., pp. 552 (see also lecture notes): 𝐿𝑥 𝐷𝑠

≤ 105[( 𝐿𝑥 ) × 𝑤𝐾 ]1 ⁄3 𝐿

1

𝑦

1

Where, Ly= longer clear span of the slab, Lx= shorter clear span of the slab, DS = thickness of the slab Lx= 6000mm Ly= 7000mm K1= 1.7, for exterior panels, 1.0 for interior panel

6000 𝐷𝑠

6000 𝐷𝑠

≤ 105[( 7000) × 6.475×1.7 ]1⁄ 3 1

6000

≤ 42.7mm

DS ≥ 141mm The exterior panels is more critical, ie. Ds obtained for interior panel will be larger.

∴ 150mm thickness is OK

Ultimate strength design for bending Ultimate design load: Fd = 1.2 G + 1.5 Q Fd = 1.2 ×3.675+ 1.5× 4 = 10.41 kPa For every panel, Ly/Lx = 1.17; For panel P, two edges discontinuous Type 6 short span βx = 0.045 Long span βy= 0.035

Table 6.10.3.2(A) Table 6.10.3.2

Positive moments for short direction Mx* = βx Fd Lx2

Cl. 6.10.3.2(1)

= 0.045 × 10.41 × 62 Mx* = 17kNm / m

Positive moments for long direction My* = βy Fd Lx2

Cl. 6.10.3.2(2)

= 0.035 × 10.41 × 62 My* = 13.0kNm / m

Next, calculate negative moments Short direction, @ the discontinuous edge AK M- = 0.5 Mx* = 0.5 × (17) = 8.5kNm/m

Cl. 6.10.3.2 (C)

Short direction, @ the continuous edge EH M- = 1.33 Mx* = 1.33 × (17) = 23kNm/m Cl. 6.10.3.2 (B)

Long direction, @ the discontinuous edge AH M- = 0.5 My* = 0.5 × (13) = 6.5kNm/m Cl. 6.10.3.2 (C) Short direction, @ the continuous edge EK M- = 1.33 My* = 1.33 × (13) = 17.4kNm/m

Cl. 6.10.3.2 (B)

Next, calculate the amount of reinforcement for each of the above design moments. First, cover: Assume non-residential building, exposure classification A2 (Table 4.3) Minimum cover = 25mm, adopt 30mm (Table 4.10.3.2) For fire resistance period, consider 120min; minimum effective thickness Ds = 120mm (Table 5.5.1), so the thickness of the slab is adequate. Assume N12 for reinforcement

d = 150 – 30 – (12/2) = 114mm

Reinforcement in the SHORT DIRECTION for flexural @ Continuous edge EH, Midspan and Discontinuous edge A-K Continuous edge EH M* ≤  Mu, where  = 0.8 Mu = 𝐴𝑠𝑡 𝑓𝑠𝑦 𝑑 (1 −

𝐴𝑠𝑡 𝑓𝑠𝑦

1.7𝑏𝑑𝑓𝑐,

defined in earlier classes.

), where the terms of the terms of the equation are the same as those

From M* ≤  Mu derive Mu

Mu = M* / 0.8

29×106 = Ast 500 × 114 (1 −

Mu = 23 / 0.8 = 29kNm /m 𝐴𝑠𝑡 ×500

)

1.7×1000×114×32

From the solution of the quadratic equation, the value of Ast is obtained: Ast = 531mm2 /m In N12 reinforcements (AN12 = 113mm2) are used, then, spacing (s) = 1000 / (531/113) = 217mm Next, check for minimum reinforcement requirements Cl. 9.1.1b 𝐷 2 𝑓′𝑐𝑡.𝑓 𝐴𝑠𝑡 ≥ 0.19 ( ) 𝑓𝑠𝑦 𝑏𝑑 𝑑

f’ct.f = 0.6√𝑓′𝑐 = 3.4MPa

cl. 3.1.1.3

150 2 3.4 𝐴𝑠𝑡 ≥ 0.19 ( ) 𝑏𝑑 114 500

𝐴𝑠𝑡 𝑏𝑑

≥ 0.0022, or, 𝐴𝑠𝑡 ≥ 0.0022 × 1000 × 114 ∴ 𝐴𝑠𝑡 ≥ 255𝑚𝑚2

Check for minimum spacing Cl. 9.4.1 Spacing should not exceed the lesser of : 2.0 Ds or 300mm 2.0 × 150 or 300mm Adopt 200mm spacing

∴ N12 @ 200mm spacing

Midspan

From M* ≤  Mu derive Mu Mu = M* / 0.8

Mu = 17 / 0.8 = 21.25kNm /m

Mu = 𝐴𝑠𝑡 𝑓𝑠𝑦 𝑑 (1 −

𝐴𝑠𝑡 𝑓𝑠𝑦 ), 1.7𝑏𝑑𝑓𝑐,

where the terms of the equation are as defined earlier.

21.25×106 = Ast 500 × 114 (1 −

𝐴𝑠𝑡×500

)

1.7×1000×114×32

From the solution of the quadratic equation, the value of Ast is obtained: Ast = 384mm2 /m In N12 reinforcements (AN12 = 113mm2) are used, then, spacing (s) = 1000 / (384/113) = 295mm Next, check for minimum reinforcement requirements Cl. 9.1.1b

∴ 𝐴𝑠𝑡 ≥ 255𝑚𝑚2

as before

Check for minimum spacing Cl. 9.4.1 Spacing should not exceed the lesser of : 2.0 Ds or 300mm

2.0 × 150 or 300mm, ∴ 𝑆𝑝𝑎𝑐𝑖𝑛𝑔 = 300𝑚𝑚

Adopt 300mm spacing

∴ N12 @ 300mm spacing Discontinuous edge A-K M- = 8.5kNm / m ( as estimated earlier) Since at midspan Mx* = 17kNm / m, minimum spacing governs the design of discontinuous edge A-K as well

∴ 𝑆𝑝𝑎𝑐𝑖𝑛𝑔 =

300𝑚𝑚, Use N12@300mm

Reinforcement in the LONG DIRECTION for flexural @ Continuous edge EK, Midspan and Discontinuous edge A-H Continuous edge EK M- = 1.33 My* = 17.4kNm /m (as calculated earlier)

Check if the reinforcement content provided for continuous edge EH in the short direction can be used:

Reinforcement for long direction, Ø12mm 30mm

Reinforcement for short direction, Ø12mm d = 150 – 30 – 12 – (12/2) = 102mm Mu = 𝐴𝑠𝑡 𝑓𝑠𝑦 𝑑 (1 −

𝐴𝑠𝑡 𝑓𝑠𝑦

1.7𝑏𝑑𝑓𝑐,

), where the terms of the equation are as defined earlier.

Ast = (1000 / 200) × 113 = 565mm2 Mu = 565 × 500 × 102 (1 −

565×500

)

1.7×1000×102×32

Mu = 27kNm

 Mu = 0.8 × 27 = 22kNm

M* ≤  Mu

Adopt N12@200

∴ 𝑆𝑝𝑎𝑐𝑖𝑛𝑔 =

200𝑚𝑚, Use N12@200mm

Midspan

Check if reinforcement content provided for midspan in the short direction can be used Mu = 𝐴𝑠𝑡 𝑓𝑠𝑦 𝑑 (1 −

𝐴𝑠𝑡 𝑓𝑠𝑦

1.7𝑏𝑑𝑓𝑐,

)

d = 102mm, 𝐴𝑠𝑡= (1000/300) × 113 =373mm2 Mu = 373 × 500 × 102 (1 −

373×500

)

1.7×1000×102×32

Mu = 18kNm

 Mu = 0.8 × 18 = 15kNm

M* ≤  Mu

Adopt N12@300

∴ 𝑆𝑝𝑎𝑐𝑖𝑛𝑔 =

300𝑚𝑚, Use N12@300mm

Corner reinforcement For corner A: provide 4 layers of reinforcement following Cl. 9.1.3.3 (e), with Area for each layer ≥ 0.75 Ast Spacing =

𝐴𝑑𝑜𝑝𝑡𝑒𝑑 𝑠𝑝𝑎𝑐𝑖𝑛𝑔 𝑎𝑡 𝑚𝑖𝑑𝑠𝑝𝑎𝑛 0.75

300

= 0.75 = 400mm,

Spacing should be less than 300mm

∴ Use 4 layers N12@300mm For corner K and H:

provide 4 layers of reinforcement following Cl. 9.1.3.3 (e), with Area for each layer ≥ 0.75 Ast Spacing =

𝐴𝑑𝑜𝑝𝑡𝑒𝑑 𝑠𝑝𝑎𝑐𝑖𝑛𝑔 𝑎𝑡 𝑚𝑖𝑑𝑠𝑝𝑎𝑛 0.5

Spacing should be less than 300mm

∴ Use 4 layers N12@300mm Summary Short direction Edge AK – N12@300 Midspan – N12@300 Edge EH – N12@200

Long direction Edge AH – N12@300 Midspan – N12@300 Edge EK – N12@200

Corner A, K and H 4 layers of N12@300

=

300 0.5

= 600mm,

Check for shear Consider it as an equivalent beam spanning in the short direction through the centre of a slab panel, 𝑉 ∗ = 6.475kPa × 1 × 6 = 38.85 KN/m

𝑉𝑢𝑐 = 0.17√𝑓𝑐′ 𝑏𝑑0 = 0.17 × √32 × 1000 × 114 = 109.6 𝐾𝑁/𝑚 𝜙𝑉𝑢𝑐 = 0.7 × 109.6 = 76.72 𝐾𝑁/𝑚 > 𝑉 ∗

∴ Shear failure is not critical in slab. Loads on beams:

Allocation of load to beams EF and EG are shown in the figure below.

E

F

G

Total load on beam EF = 2(0.5x6x3x10.5) = 189 KN 7+1

Total load on beam EG = 2 [

2

× 3 × 10.5] = 252 𝐾𝑁

For the purpose of designing the beams, the calculated load above may be assumed to be uniformly distributed over the full span.

Check for deflection Since live load exceed the dead load in this case, deemed to comply span-to-depth ratio will not be used according to cl.9.3.4.2. Simplified calculation method referring to Cl.9.3.3 is adopted to check the deflection in this case.

Consider a prismatic beam of unit width through the centre of the slab, spanning in the short direction Lx , with the same condition of continuity as the slab in that direction and with the load distributed so that the proportion of load carried by the beam is given by Lx /(𝛼𝐿𝑥4 + 𝐿4𝑦 ). ∴ Lx = 6 𝑚, 𝐿𝑦 = 7 𝑚,

Consider a more critical situation, 𝛼 = 2 for the first span, 𝛼 = 1 for the second span and 𝛼 = 2 for the third span. 6.475 × 7 ×

6.475 × 7 ×

64 = 11.76 𝑘𝑃𝑎 2 × 64 + 74 64 = 16 𝑘𝑃𝑎 64 + 74

Use stiffness method to derive moment curves in the beam. The maximum deflection will then be derived from moment curve by methods like double integration. Demonstration of stiffness methods: 16 kPa

16 kPa

11.76 kPa

6m

6m

6m

= 35.28 kNm

72 kNm

𝑤𝑙2 16 × 62 = = 72 𝑘𝑁𝑚 8 8

72 kNm

𝑤𝑙2 11.76 × 62 = = 35.28 𝑘𝑁𝑚 12 12

36.72 kNm

Bending moment diagram:

35.28 kNm

+ 36.72 kNm

Deflection diagram:

∆𝑚𝑎𝑥 = 18.81 𝑚𝑚

1 ∆ 18.81 = 0.0031 < = 6000 l 250 ∴ 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑖𝑠...