Manual Design Of Doubly Reinforced Continuous Beam
In this blog, you will learn step by step about the design of doubly reinforced beam of first floor of G+5 building procedure using I.S. 456:2000. For daily blogs, subscribe to our blog page and learn complete information about the structural engineering industry Given:- 1) Grade of Concrete = M202) Grade of Steel = Fe5003) Clear Cover to Reinforcement (c) = 25 mm4) Length of Beam :- B24 & B25 = 4060 mm5) Unit Weight of Concrete = 25 kN/m26) Slab Dimensions :- S3 = (1975 mm X 4060 mm) & S4 = (3050 mm X 4060 mm)7) Floor to Floor Height (H) = 3000 mm Step 1 :- Trial Dimension of The Beam :- Assume the Width of The Beam (b) = 230 mmEffective Depth of Beam (d) = (L/10) to (L/15) = (4290/10) to (4290/15) = 429 mm to 286 mm Take, d = 325 mmAssume, 20 mm Diameter Bars are to be Provided at a Clear Cover of 25 mm.Therefore, Overall Depth (D) = 325 + (20/2) + 25 = 360 mm ≈ 375 mm…(Rounded off on Higher Side)Therefore, Effective Depth of Beam Provided (d) = 375 – (20/2) – 25 = 340 mm Step 2 :- Effective Span (Le) :- Minimum of The Following , i) Le = L + d = 4060 + 340 = 4400 mmii) Le = L + Bearing = 4060 + 230 = 4290 mmTherefore, Effective Span (Le) = 4290 mm Step 3 :- Load Calculations :- Load on Beam B24 & B25 :-i) Super Imposed Dead Load (SIDL) :-Wall Load = Wall Thickness X Floor to Floor Height X Unit Weight of Bricks = (0.150 X 3 X 20) = 9 kN/m …(No Deduction of Depth of Beam is Made From Floor to Floor Height)ii) Slab Load Transferring on Beam :-Slab S3 :-a) Self Weight of Slab (DL) = D X Unit Weight of Concrete = 0.125 X 25 = 3.125 kN/m2b) Live Load (LL) = 2 kN/m2 …[Refer Table No. 1 , Page No. 7 , I.S. 875 (Part 2) : 1987]c) Super Imposed Dead Load (SIDL) =Floor Finish = Wt. of Screeding (50 mm Thk.) + Flooring (10 mm Thk.) = (0.05 X 24) + (0.01 X 22) …[Refer I.S. 875 (Part 1) : 1987] = 1.42 kN/m2 ≈ 1.5 kN/m2Total Load of Slab S3 (w) = 3.125 + 2 + 1.5 = 6.625 kN/m2 Trapezoidal Load of Slab S3 is Transferring on Beam B24 & B25 Because S3 Slab is Two Way Slab Which is Given by,WS3 = [{(w. Lx)/2} X {1- (1/ 3. β2)}] Where, β = l_y/l_x = {(6.625 X 1.975 ) / 2} X [1- {1/ 3 X (4.06/1.975)2}] = 6.02 kN/mSlab S4 :-a) Self Weight of Slab (DL) = D X Unit Weight of Concrete = 0.125 X 25 = 3.125 kN/m2b) Live Load (LL) = 2 kN/m2 …[Refer Table No. 1, Page No. 7, I.S. 875 (Part 2) : 1987]c) Super Imposed Dead Load (SIDL) =Floor Finish = Wt. of Screeding (50 mm Thk.) + Flooring (10 mm Thk.) = (0.05 X 24) + (0.01 X 22) … (Refer I.S. 875 (Part 1) : 1987) = 1.42kN/m2 ≈ 1.5 kN/m2Total Load of Slab S4 (w) = 3.125 + 2 + 1.5 = 6.625 kN/m2 Trapezoidal Load of Slab S4 is Transferring on Beam B24 & B25 Which is Given by,WS4 = [{(w. Lx)/2} X {1- (1/ 3. β2)}] ={(6.625 X 3.05) / 2} X [1- {1/ 3 X (4.06/3.05)2}] = 8.2 kN/mTherefore, Total Load of Slabs Transfers on Beam B24 & B25 = WS1 + WS4 = 6.02 + 8.2 = 14.22 kN/miii) Self Weight of Beam (DL) = b X D X Unit Weight of Concrete = 0.230 X 0.375 X 25 = 2.156 kN/m Total Load on Each Beam B24 & B25 (W) = 9 + 14.22 + 2.156 = 25.376 kN/m ≈ 25.4 kN/mUltimate Load (Wu) = 25.4 X 1.5 = 38.1 kN/m Step 4 :- Bending Moment (Mu) :- Max. Moment on ContinuousBeam (B24 & B25) is (M) = W . Le2 / 8 = (25.4 X 4.292) / 8 = 58.43 kN.mUltimate Moment on ContinuousBeam (B20 & B21) is (Mu) =58.43X 1.5 = 87.65 kN.m CONTINUOUS BEAM HAVING TWO EQUAL SPAN CARRYING UNIFORMALLY DISTRIBUTED LOAD (UDL) OVER ENTIRE SPAN :- Step 5 :- Limiting Moment of Resistance :- Mulim= 0.133 fck.b.d2Mulim= 0.133 X 20 X 230 X 3402Mulim= 0.133 X 20 X 230 X 3402Mulim= 70724080 NMulim= 70724080 / 106Mulim= 70.72 kN.mAs, Mulim< Mu Note :- Either We Have to Increase Depth and Design as a Singly Reinforced Beam or if We Are Not Allowed to Increase Depth of Beam Then We Will Design the Beam as a Doubly Reinforced Beam. So We Design Assumed Beam of 230 X 375 mm As Doubly Reinforced Beam. Step 6 :- Balanced Moment (Mu2) :- Equate, Mu2 = Mu – Mulim Mu2 = 87.65 – 70.72 Mu2 = 16.93 kN.m Step 7 :- Area of Steel Calculation (Ast) :- [Refer Cl. No. 26.5.1.1 (a) & (b), Page No. 46 & 47,I.S. 456 : 2000] Astmin = As/bd = 0.85/ FyAstmin= As = [(0.85 X 230 X 340) / 500]Astmin= As = 132.94 mm2Astmax = 0.04 b.D = 0.04 X 230 X375 = 3450 mm2 i) Steel in Tensile Zone :-Astreq= Ast1+Ast2 [For kumax & fsc Refer above Tables From, Illustrated Design of Reinforced Concrete Buildings by S. R. Karve & V. L. Shah] ii) Steel in Compression Zone :-dc / d= (D – d) / d = (375 – 340) / 340 = 0.1By Interpolation,fsc = 412 N/mm2 …Refer Table 4.2.2 Step 8 :- Check For Shear :- Codal Provisions for Shear :- [Refer Cl. No. 26.5.1.6, Page No. 48 & Cl. No. 40.1, 40.3, Page No. 72 of I.S. 456 : 2000] i) Maximum Shear Force (Vu) :-V = (5.W . Le/ 8) = [(5 X 25.4 X 4.29) / 8] = 68.1 kNVu= 68.1 X 1.5 = 102.15 kN ii) Nominal Shear Stress (τv) :-τv = Vu / b.d
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