Example 2.2 [Ribbed slab design]

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Randolf Haynes
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1 Example 2.2 [Ribbed slab design] A typical floor system of a lecture hall is to be designed as a ribbed slab. The joists which are spaced at 400mm are supported by girders. The overall depth of the slab without finishing materials is 300mm. Imposed load of 1.5KN/m 2 for partition and fixture is considered in the design. In addition, the floor has a floor finish material of 3cm marble over a 2cm cement screed and it ha 2cm plastering as ceiling. Take the unit weight of ribbed block to be 2KN/m2. Use: C 20/25 S 300 Class 1 works a) Analyze the ribbed slab system, considering the effects of loading pattern b) Design the ribbed slab system 1

2 Solution: Step 1:Material property Concrete: fctk, 0.05 = 1.5Mpa fctm = 2.2Mpa ɤc = 1.5 Rebar fck = 20Mpa, fcu = 25Mpa fcd = = 11.33Mpa fyk = 300Mpa fyd = fyk 1.15 = Mpa ɛyd = fyd Es = = 1.74 Step 2: Verify if the general requirements for Rib slab are met using Euro Code 2 1. The centers of the ribs should not exceed 1.5 m: - This is satisfied, as the center-to-center spacing between the ribs is 400mm. 2. The depth of ribs excluding topping should not exceed four times their average width. - Also satisfied as 80 x 4 > 240 mm. 3. The minimum rib width should be determined by consideration of cover, bar spacing and fire resistance - BS 8110 code - recommends 125 mm, - Assume for this example the conditions are satisfied hence assume requirement satisfied. 4. The thickness of structural topping or flange should not be less than 50mm or one tenth of the clear distance between ribs mm satisfies this requirement. Step 3: Loading Dead load: Joist 0.2 * 0.08 * 25 = 0.4 Topping 0.4 * 0.06 * 25 = 0.6 Floor finish 0.4 * 0.03 * 27 = 0.32 Cement Screed 0.4 * 0.02 * 23 =

3 Plastering 0.4 * 0.02 * 23 = Partition and fittings 0.4*1.5 = 0.6 Ribbed block 0.4 * 2 = 0.8 Live load: G k = KN/m Q k = 4KN/m 2 * 0.4 = 1.6 KN/m Design load: Gd = 1.35 Gk = = 4.174KN/m Qd = 1.5 Qk = = 2.4KN/m Step 4: Analysis (for Ribs) i) Full design load ii) Maximum support moment [at B and C] 3

4 iii) For maximum span moment [ at span AB and CD] iv) Maximum span moment [ at BC] 4

5 v) Only dead load acting - Moment envelope diagram for the rib - Maximum reaction envelope - Minimum reaction envelope 5

Step 5.Loading on Girders - Assume Width of girders A, D.W=300mm B, C.W=600mm For all girders...d=300mm - Note: the section should be checked for serviceability Self-weight: A & D = 0.3 x 0.

6 Step 5.Loading on Girders - Assume Width of girders A, D.W=300mm B, C.W=600mm For all girders...d=300mm - Note: the section should be checked for serviceability Self-weight: A & D = 0.3 x 0.3 x 25 = 2.25 KN. m B&C = 0.6 x 0.3 x 25 = 4.5 KN. m Design loads: A & D Gd = 1.35 x2.25 = 3.04 KN. m B & C. Gd= 1.35 x4.5 = 6.08 KN. M Step 6. Analysis of Girders i. For Girder on axis A and D - To get to maximum support moment [at 2] From the maximum reaction of the Ribs divided by the rib spacing at A and D =10.67 KN.m Total load W= = KN.m 6

- To get maximum span moment on girder A & D at 12 & 23 From the maximum reaction of the Ribs divided by the rib spacing A & D 10.67 0.

7 - To get maximum span moment on girder A & D at 12 & 23 From the maximum reaction of the Ribs divided by the rib spacing A & D =10.67 KN.m From the minimum reaction of the Ribs divided by the rib spacing 6.24 =15.52 KN.m Moment envelop for girder A and D 7

8 = 74.5 KN. mand18.467 = 46.15 KN. m 0.4 0.

8 ii. For Girder on axis B and C - Loading: Self-weight = 6.08 KN.m Reactions from the ribs (divided by the rib spacing) 29.8 = 74.5 KN. mand = KN. m To get maximum support moment [at 2 ] - To get maximum span moment [at 12 or 23 ] 8

9 - Moment envelop diagram for girders on axis B and C Step 7.Loading on the Beam. Axis 1, 2 and 3 - Self-weight width= 200 mm Depth= 300 mm N.B: cross section should be checked for serviceability. Since there are columns at the intersection of the beams and girders, the beams will only support their own loads. DL = 0.2 x 0.3 x 25 = 1.5 KN/m Gd= 1.35 x 1.5 = KN/m - Beam analysis 9

mm h = 260 mm Take cover 15 mm d = 260 15 6 12 2 =

10 Step 8. Design 1. Rib design Cross section at span h f = 60 mm b w = 80 mm h = 260 mm Take cover 15 mm d = = 233 mm - Effective width computation b eff,i = 0.2b i + 0.1l o 0.2l o I. For end span(sagging moment) l o = 0.85l 1 l o = = 3400mm b 1 = b 2 = 160 mm b eff 1 = b eff 2 = 372 < 680 < b 1 10

b eff = b eff,i + b w b b eff = 824 400 NOT OK b eff = 400 mm II. For interior sagging moment (+ve) l o = 0.7l 2 l o = 0.

11 b eff = b eff,i + b w b b eff = NOT OK b eff = 400 mm II. For interior sagging moment (+ve) l o = 0.7l 2 l o = = 2800mm b 1 = b 2 = 160 mm b eff 1 = b eff 2 = 312 < 560 < b 1 b eff = b eff,i + b w b b eff = NOT OK III. For support hogging moment (-ve) l o = 0.15(l 1 + l 2 ) l o = 1200mm b 1 = b 2 = 160 mm b eff 1 = b eff 1 = 152 < 240 < b 1 b eff = b eff,i + b w b b eff = OK b eff = 400 mm b eff = 384 mm Note: However since it is a negative moment the width of the compression zone will be, b= 80mm - Design of the T-section A. Positive span moment AB and CD 11

12 M sd = KNm b eff = 400 mm d = 233mm f cd = mpa f yd = mpa µ sd = M sd f cd bd 2 = Nmm = μ sd < μ sd,lim = Singly reinforced K x = X = K x d = mm < h f design as a rectangular section K z = Z = K z d = mm A s = M sd f yd Z = Nmm = mm A smin = 0.26f ctm f yk b t d where b t = b w d = 233mm f ctm = 2.2 mpa f yk = 300 mpa A smin = mm 2 < A s OK! using 12 a s = 113.1mm 2 n = A s a s = use 2 12bottom bars B. Negative moment on the rib support B and C M sd = KNm b w = 80 mm d = 233mm f cd = mpa f yd = mpa µ sd = M sd f cd bd 2 = Nmm = μ sd < μ sd,lim = Singly reinforced K z = 0.88 Z = K z d = mm A s = M sd f yd Z = Nmm = mm2 A smin = 0.26f ctm f yk b t d where b t = b w d = 233mm f ctm = 2.2 mpa f yk = 300 mpa A smin = mm 2 < A s OK! 12

13 using 12 a s = 113.1mm 2 C. Span moment between B and C n = A s a s = use 2 12 bars at the top M sd = 4.63 KNm b eff = 400 mm d = 233mm f cd = mpa f yd = mpa µ sd = M sd f cd bd 2 = Nmm = μ sd < μ sd,lim = Singly reinforced K x = 0.07 X = K x d = mm < h f design as a rectangular section. K z = Z = K z d = mm A s = M sd f yd Z = Nmm = 77.33mm2 A smin = 0.26f ctm f yk b t d where b t = b w d = 233mm f ctm = 2.2 mpa f yk = 300 mpa A smin = mm 2 < A s OK! using 12 a s = 113.1mm 2 n = A s a s = use 2 12 bottom bars Note: For the shear design of the ribs, refer to Example 2.3 for ribbed slabs. 2. Girder design a. Girder at A and D - Positive span moment M sd = KNm b w = 300 mm D = 300 mm f cd = mpa f yd = mpa d = = 259 mm µ sd = M sd f cd bd 2 = Nmm = μ sd < μ sd,lim = Singly reinforced K z = Z = K z d = mm A s = M sd f yd Z = Nmm = mm2 13

14 A smin = 0.26f ctm f yk b t d where b t = b w d = 259 mm f ctm = 2.2 mpa f yk = 300 mpa A smin = mm 2 < A s OK! using 16a s = mm 2 n = A s a s = use 4 16 bottom bars - Negative support moment M sd = KNm b w = 300 mm D = 300 mm f cd = mpa f yd = mpa d = = 259 mm µ sd = M sd f cd bd 2 = Nmm = μ sd < μ sd,lim = Singly reinforced K z = Z = K z d = mm A s = M sd f yd Z = Nmm = mm2 A smin = 0.26f ctm f yk b t d where b t = b w d = 259 mm f ctm = 2.2 mpa f yk = 300 mpa A smin = mm 2 < A s OK! using 16 a s = mm 2 n = A s a s = use 6 16 bottom bars b. Girder at B and C Positive span moment M sd = KNm b w = 600 mm D = 300 mm f cd = mpa f yd = mpa d = = 257 mm µ sd = M sd f cd bd 2 = Nmm = μ sd < μ sd,lim = Singly reinforced K z = Z = K z d = mm A s = M sd f yd Z = Nmm = mm2 14

15 A smin = 0.26f ctm f yk b t d where b t = b w d = 259 mm f ctm = 2.2 mpa f yk = 300 mpa A smin = mm 2 < A s OK! using 20a s = 314mm 2 n = A s a s = use 6 20 bottom bars Negative support moment M sd = KNm b w = 600 mm D = 300 mm f cd = mpa f yd = mpa d = = 257 mm µ sd = M sd f cd bd 2 = Nmm = μ sd > μ sd,lim = Doubly reinforced section K z,lim = M sd,lim = µ sd,lim f cd bd 2 = = KNm Z = K z,lim d = = mm A s1 = M sd,lim Zf yd use Compression reinforcement design + M sd,s M sd,lim ( ) 106 = + f yd (d d 2) (257 35) = mm 2 - Check if the reinforcement has yielded d 2 d = = 0.14ε s2 = 2.6 (read from chart) ε s2 = 2.6 > ε yd use f yd = Calculate the stress in the concrete at the level of compression reinforcement to avoid double counting of area. ε cs2 = 2.6 2, Therefore, we take ε c = 3.5 and σ cd,s2 = mpa 15

16 A 1 s2= (σ s2 σcd,s2 ) (M sds M sd,lim d d2 = 1 ) ( ) (( ) 106 (257 35) = mm 2 use Beams Design i) Negative support moment M sd = 3.491KNm b w = 200 mm D = 300 mm f cd = mpa f yd = mpa d = = 261 mm µ sd = M sd f cd bd 2 = Nmm = μ sd < μ sd,lim = Singly reinforced K z = Z = K z d = mm A s = M sd f yd Z = Nmm = 52.04mm2 A smin = 0.26f ctm f yk b t d where b t = b w = 200 d = 261 mm f ctm = 2.2 mpa f yk = 300 mpa A smin = 99.52mm 2 > A s Not OK! using 12a s = mm 2 n = A s,min a s = 0.88 use 2 12 bottom bars Use 2 12 bottom and top bar for the total lenth of the beam. Step 8. Transverse reinforcement Secondary reinforcement is required for temperature and shrinkage. A s2 = 20% A s,min A s2 = 0.12% A topping Spacing S = ba s A s use 8 c c 200 mm 16

17 Step 9. Detailing 17

18 18

CHAPTER 4. Design of R C Beams

CHAPTER 4. Design of R C Beams CHAPTER 4 Design of R C Beams Learning Objectives Identify the data, formulae and procedures for design of R C beams Design simply-supported and continuous R C beams by integrating the following processes