DESIGN OF BEAMS:Design of beam section for member numbers- 514,527 from STAAD Size of beam = 230 x 300 mm.Effective cover = 25 mm.Effective depth (d) = 300-25 = 275 mm.Using M20 grade concrete and Fe 415 grade steel.Design Constants:

For Fe 415 steel = 0.48Max BM get from Staad Pro Analysis = 52.232 x 106 N-mmh

Mulim = 0.36 fck = 0.36 x 20x 0.48 [1-0.42 x 0.48] 230 x 2752 = 47.9 x 106 n-mMu lim < Mu maxHence the section is designed as Doubly Reinforced Section.As per IS 456-2000 code for doubly reinforced sectionMu max Mu lim = fsc Asc (d-d)Where d is effective top cover i.e d = 35mmfsc is approximate we taken as 0.87 fy = 0.87 x 415 = 316.05 N/mm2 52.232 x 106 47.9 x 106 = 316.05 Asc(275-35) Asc = 55.92 mm2Adopt 4-8 mm bars Provide Asc = 201.06 mm2

Design of steel reinforcemnt: Ast = Ast1 + Ast2

Ast1 = = = 605.43 mm2

Ast2 = = = 176.0 mm2 Ast = Ast1 + Ast2 Ast = 605.43+176 = 781.43 mm2Provide 4 nos of 16 + 4 nos of 10 = 1005.3 mm2Check for min area of steel to be provided.

=

As = = 129.54 mm2Hence our provided Ast is in safer sideProvided Ast = 1005.30 mm2Design of shear reinforcement:Max. shear force = Vu = 29.063 kN

Nominal shear force = v = = = 0.46 N/mm2

Percentage of steel Pt = = = 1.033%From IS 456-2000 Table No-19Shear stress in concrete = c = 0.62 N/mm2 v < c 0.46 < 0.62 Hence safeProvide Nominal Shear ReinforcementMinimum Shear Reinforcement :

Adopt 4-legged 8 mm stirrups the area Asv = 201.06 mm2

Sv = = = 394mm c/cProvide 8mm bars at 300 mm c/c.

DESIGN OF SLAB

Design of One-way slab :- S4 & S19 Panel

Size = 3.02 1.26 mLy / Lx = 3.020/ 1.260 = 2.39 < 2.0

It is a one-way slab.

Cross Sectional Dimensions:1) Effective Span (L eff ) : Which should be least of the following as per IS456-2000, Pg 34, Clause 22.20, a&b.

i) Leff = Clear span + Effec. Depth = 1.145 + 0.10 = 1.245 mii) Leff = Clear span +

= 1.245+ = 1.36 mAdopting the least value as (Leff) = 1.36 m

2) Effective Depth (d eff ) : Let us assume the thickness of the slab as per IS456-2000, Pg : 39 , Clause 24.10-2. For Continuous slabs, Leff / D = 32

D = = = 42.5 Overall Depth (D) : 125mm

Effective Cover (dc ) : 20 + = 25 mmEffective Depth (deff ) : 125 -25 = 100 mm

3) Design loads per m2 of slab

Dead load of slab = = 1.875 KN/m2Live load (As/IS456-2000 part-III)= 2.00 KN/m2 Floor finish= 1.00 KN/m2= 4.875 KN/m2 Factored load ( Wud) = 1.5 x 4.875 = 7.3125 KN/m2

4) Bending MomentB.M = WL/ 8 = ( 7.3125 * 1.36) / 8=1.69 KN/mDepth from moment considerationMinimum Depth required for resisting Bending Moment Mu = 0.138fk.b.d 1.69 x 106 d = = 24.745 mm < available depth 75mm 0.138 x 20 x 1000

5) Depth from stiffness considerationD = 25 mm ~ 100 mmd = 100 25 = 75 mm

6) Calculation of Area of steel

0.50 x 20 4.6 x 1.69 x 106 Ast = 1- 1- 1000 x 100 415 20 x 1000 x 752 = 85 mm2

Hence Provide 100 mm2

x 82 Provide 8 mm dia. Tor steel ast = = 50.26 mm2 4 ast Spacing of Steel = X 1000 Ast(50.26 / 85) X 1000 = 592 ~ 300 mm c/c

Provide 8 mm @ 300 mm c/c

7) Transverse reinforcementAsty = 0.12% of bD

=

Spacing of 8 mm bars

= 418.83 ~ 250 mm c/cSpacing should be lesser of(1) 5d = 5 100 = 500 mm(2) 450 mm(3) 260 mm

Provide 8 mm @ 250 mm c/c

8) Check for shear

Shear force = 7.315 x 1.26 / 2 = 4.60 KN.

Nominal shear stress

c = Vu / bd = 4.93 x10 / 1000X100 =0.05 N/mm

Percentage of steel =

From code IS 456 tables 19

Slab is safe in shear.

DESIGN OF TWO-WAY SLAB

S1 & S16 Panel1) Grades of Materials Grade of Concrete : M20 (1: 1.5 : 3) Grade of Steel : Fe415 for both main and secondary steel

2) Classification of Slab(Two adjacent edges discontinuous)

Ly = 3.715 m 3.13m Lx = 3.130 m Ly / Lx = 3.715 / 3.13 = 1.18 < 2.0 3.715m Hence the Slab has to be designed as Two way Slab

3) Calculation of depth of Slab

The Slab : Two way continuous Shortspan : Lx = 3.130 Steel of Grade : Fe415 Span / Overall depth = 32 Lx / D = 32 D = 3010 / 32 = 97.81 mm Say 125 mm Assume effective Cover (d) = 15 + 8/2 = 19 mm Effective depth = 125 19 = 106 mm

4) Estimation of Loads

Assume 1.00 m width of the slabTotal depth of slab = 0.10 mSelf weight of the slab = 1.00 x 0.125 x 25 = 3.125 KN / mWeight of the floor finish = 1.00 x 1.75 = 1.75 KN / mLive Load is taken as = 2.00 KN / m

Total Working Load (W) = 6.875 KN / mTotal Ultimate Load (Wu) = 1.5 x 6.875 = 10.3125 KN / m

5) Calculation of effective span

Effective Span of the slab is the least of the following() Centre to centre distance between the supports Lx = 3.130 m , Ly = 3.715 m () Clear span + Effective depth of slab Lx = 2.9 + 0.08 = 2.98 m , Ly = 3.485 + 0.08 = 3.565m Effective Span Lx = 2.98 m , Ly = 3.565 m

6) Calculation of Design momentsEdge conditions :- Two adjacent edges discontinuousThe design coefficients are taken from table 26 of IS 456 2000 a)Negative moment at continuous edge x = 0.0586 y = 0.047 b) Positive moment at midspan x = 0.044 y = 0.035

Moments (According to IS 456 2000 , D 1.1)

Mx = x wlx2 My = y wlx2

Negative moment at continuous edge MX = 0.0586 x 10.3125x 2.982 = 5.36 KN-m MY = 0.047 x 10.3125 x 2.982 = 4.3 KN-m

Positive moment at midspan MX = 0.0446 x 10.3125 x 2.982 = 4.029 KN-m MY = 0.035 x 10.3125 x 2.982 = 3.205 KN-m

7) Effective depth of slab (For max.BM) Mu lim = 0.138fckbd2

Mu lim d = 0.138fckb

5.36 x 106 d = = 44.068 mm < available depth 106mm0.138 20 x 1000Hence OK8) Calculation of Area of steel

Here the slab is two way slabEdge conditions : Two adjacent edges discontinuosThe slab must consists of the required area of steel at the following zones.i) At middle strip along short span : (Astxm)ii) At middle strip along longer span : (Astym)iii) At support along short span : (Astxs )iv) At supports along longer span : (Astys)v) Torsion steel at corners

i) At middle strip along short span (Astxm) MD = 4.029 KN-m

0.50 fCK 4.6 x MD Ast = 1- 1- bd fy fCK bd2

0.50 x 20 4.6 x 4.029 x 106 Ast = 1- 1- 1000 x 106 415 20 x 1000 x 1062

= 113.7 mm2 < 120 mm2 (Min.Ast 0.12% of bD as per Clause 26.5.2.1 of IS 456-2000)

Hence Provide 120 mm2

x 82 Provide 8 mm dia. Tor steel ast = = 50.26 mm2 4 ast Spacing of Steel = X 1000 Ast

50.26 Spacing of Steel = x 1000 = 418 mm 120

But maximum spacing of Tension Reinforcement as per Clause 26.3.3 b 1 from

IS 456 2000 is 3d or 300 mm whichever is less.

3 x 106 = 318 mm or 300 mm

Hence provide 8 mm dia. @ 300 mm C/C

Revised Astxm = (1000 x 50.2) / 300 = 167.53 mm2

ii) At middle strip along Longer span (Astym) MD = 3.205 KN-m

0.50 fCK 4.6 x MD Ast = 1- 1- bd fy fCK bd2

0.50 x 20 4.6 x 3.205 x 106 Ast = 1- 1- 1000 x 106 415 20 x 1000 x 1062

= 91.038 mm2 < 120 mm2 (Min.Ast 0.12% of bD)

Hence Provide 120 mm2 x 82 Provide 8 mm dia. Tor steel ast = = 50.26 mm2 4

ast Spacing of Steel = X 1000 Ast

50.26 Spacing of Steel = X 1000 = 418 mm 120

But maximum spacing of Tension Reinforcement as per Clause 26.3.3 b 1 from

IS 456 2000 is 3d or 300 mm whichever is less.

3 x 106 = 318 mm or 300 mm

Hence provide 8 mm dia. @ 300 mm C/C

Revised Astxm = (1000 x 50.2) / 300 = 167.53 mm2

iii) At Support along short span (Astxs)MD = 5.36 KN-m

0.50 fCK 4.6 x MD Ast = 1- 1- bd fy fCK bd2

0.50 x 20 4.6 x 5.36 x 106 Ast = 1- 1- 1000 x 106 415 20 x 1000 x 1062

= 150 mm2 > 120 mm2 (Min.Ast 0.12% of bD)

x 82 Provide 8 mm dia. Tor steel ast = = 50.26 mm2 4

ast Spacing of Steel = X 1000 Ast

50.26 Spacing of Steel = X 1000 = 335 mm 150

But maximum spacing of Tension Reinforcement as per Clause 26.3.3 b 1 from

IS 456 2000 is 3d or 300 mm whichever is less.

3 x 106 = 318 mm or 300 mm

Hence provide 8 mm dia. @ 300 mm C/C

Revised Astxm = (1000 x 50.2) / 300 = 167.33 mm2

iv) At Support along Longer span (Astys) MD = 4.3 KN-m

0.50 fCK 4.6 x MD Ast = 1- 1- bd fy fCK bd2

0.50 x 20 4.6 x 4.3 x 106 Ast = 1- 1- 1000 x 106 415 20 x 1000 x 1062

= 121.09 mm2 < 120 mm2 (Min.Ast 0.12% of bD)

x 82 Provide 8 mm dia. Tor steel ast = = 50.26 mm2 4 ast Spacing of Steel = X 1000 Ast

50.26 Spacing of Steel = X 1000 = 418.83 mm 120

But maximum spacing of Tension Reinforcement as per Clause 26.3.3 b 1 from IS 456 2000 is 3d or 300 mm whichever is less. 3 x 106 = 318 mm or 300 mmHence provide 8 mm dia. @ 300 mm C/CRevised Astxm = (1000 x 50.2) / 300 = 167.3 mm2

v) Torsion steel

i) At corners that have both edges discontinuous = x Astxm = x 160 = 120 mm2In each layer provide two layer of size Lx/5 , LY/5 = 0.62m, 0.74 m two at top and two at bottom.ii) At corners that have one edge continuous and other edge discontinuous = 3/8 x 160 = 60 mm2 in each layer.iii) No Torsion steel is required for continuous edges

SpanPositionMu KN-mD mmReq. AstDia. mmSpacing mm

ShortSupport5.361251608300

Mid span4.0291251608300

LongSupport3.604.31251608300

Mid span3.2051251608300

9) Check for deflection

Area of Steel (Ast) = 210 mm2Percentage of steel Pt = ( Ast /bd) x 100 = (160 / 1000 x 106) x 100 = 0.15% From Fig-4 of IS 456 2000Modification factor = 1.50 Basic value for Span to effective depth ratio = 40 x 1.06 = 38 Permissible ratio = 1.50 x 38 = 57 Actual ratio = 2980/106 = 28.11 < 57 Hence OK

10) Check for shear

For continuous short edge Vu = WL/2Shear Vu = 10.3125 x 2.98 / 2 = 15.365 KNArea of steel available at this edge Astxs = 160 mm2Pt% = (100 x 160 / 1000 x 106) = 0.15% From Table 19 of IS 456 2000 c = 0.28 N/ mm2 K = 1.30 Shear resisted by concrete Vc = c bdK = 0.28 x 1000 x 106 x 1.30 = 38.58 KN > 15.365 Hence OK