Design a rectangular beam over 4 column YASH PAL, May 24, 2021July 12, 2025 Given Design a Beam – Design a rectangular beam continuous over 4 column supports of effective shap 6m. The beam is subjected to an imposed dead load of 10 KN/m and a live load of 15 KN/m. use M20 and Fe415 steel.GivenEffective shap (L eff. ) = 6m(f ck ) = 20 N/mm 2(f y ) = 415 N/mm 2Imposed dead load = 10 KN/mImposed live load = 15 KN/ma. Effective depth:-Assuming effective depth = = = 400 mmTrying a total depth of D = 500 mmAnd width (b) = to i.e,Nearly 250mmTrying a width (b) of 300mmRectangular Beam DesignAssuming an effective cover of 30mm⸫ effective depth (d) = 500 – 30 = 470mm⸫ the dimensions are:-Width(b) = 300mmEffective depth (d) = 470mmTotal depth (D) = 500mmb. Load calculation:-Self weight of beam = 0.3×0.5×25 = 3.75 KN/mImposed dead load = 10 KN/m⸫ total dead load = 10 + 3.75 = 13.75 KN/m⸫ factored dead load (W d )= 13.75 x 1.5 = 20.625 KN/m⸫ factored live load (W L ) = 15 x 1.5 = 22.5 KN/mc. B.M. & S.F. using coefficient:-(ref. table- 12,13, IS 456:2000)Now,M = WL 2the maximum negative moment occurs at support next to end support,M = – (20.625 x 6 2 ) + (22.5 x 6 2 )= -164.25 KNmThe maximum positive moment occurs near the middle of the end shap,M = (20.625 x 6 2 ) + (22.5 x 6 2 )= 142.875 KNmAnd,V = WLMaximum shear force occurs at the outer side of the support next to the end support,V u = 0.6 ( 20.625×6 ) + 0.6 ( 22.5×6 )= 155.25 KNd. Depth check:-( Ref. CL. 31.1.1, IS 456 : 2000 )M u = 0.36 bd 2 f ck 1- 0.42= 0.36 x 0.48 x 1- ( 0.42 x 0.48 )= 0.138 f ck bd 2d req. == 446 mm < 470 mme. Area of steel:-(Ref. CL. 31.1.1 IS 456: 2000 )Calculating the limiting moment of resistance,We know,M u = 0.138 f ck bd 2= 0.138 x 20 x 300 x 470 2= 182.9 KNm > M u (164.28KNm)Hence, the singly reinforced section can be designed,Ast = 0.5 1 –Now,Designing the section at intermediate support,M u = 0.87 f y Ast d 1-164.25 x 10 6 = 0.87 x 415 x Ast x 470 1 –Solving we get,Ast = 1014 mm 2No o bars = 4Using 4 – 20 mm ϕ bars (Ast Pro. = 1256 mm 2 ) at intermediate support as negativeReinforcement.At mid-span –M u = 142.875 KNmM u = 0.87 f y Ast d 1-142.875 x 10 6 = 0.87 x 415 x Ast x 470 1-Solving we get,Ast = 876 mm 2No. of bars = 3Using 3 – 20 mm ϕ bars (Ast = 942 mm 2 ) at mid-span as positive Reinforcement.Ast = (Ref. CL 26.5.1.1, IS 456:2000)Ast = = 289mm 2 > AstHence, Ast provided is OK.f. Shear design:-At support next to end support,V u = 155.25KNτ v = (Ref. CL 40.1 IS 456:2000)τ v = = 1.1 N/mm 2from table 20, IS 456:2000, for M20,τ c = 2.8 N/mm 2 > τ vAt support,P t = 100 x = = 0.89%From table 19, IS 456:2000, For M20, concrete & P t = 0.89%P t = 0.89 N/mm 2 < τ vHence, shear reinforcement is necessaryV us = V o – τ c bd (Ref. of CL 40.4)V us = 155.25 x 10 3 – 0.02 x 300 x 470 = 67830 NSpacing of 2 legged 8mm ϕ bars stirrupsS v = (Ref. CL 40.4(a) IS 456:2000)S v = = 251 mmMaximum spacing of 2-legged 8 mm dia stirrups should not existI. 0.75d = 0.75 x 470 = 352 mmII. 300 mmIII. S v max == = 302 mmHence provide 2-legged 8mm ϕ @ 250 mm c/c at end support,V u = ( 0.4 x 20.625 x 6) +( 0.45 x 22.5 x 6 ) = 110.25 KNτ v = = = 0.78 N/mm 2From table 20, IS 456: 2000;τ c max = 2.8 N/mm 2 > τ vP t at end support = 100 x = = 0.66 %From table 19, IS 456 2000, For M20 concrete,P t = 0.66%τ c = 0.48 + (0.66 – 0.5 ) = 0.53 N/mm 2 < τ vHence, shear reinforcement is necessaryV us = V u – τ c bd (Ref. CL 40.4 IS 456:2000 )[br]V us = (110.25 x 10 3 ) – (0.53 x 300 x 470 )V us = 35520 NSpacing of 2-legged 8 mm ϕ stirrupsS v = (Ref. CL 40.4(a) IS 456:2000)S v = = 480 mm > 300mmProvide 2-legged 8 mm ϕ stirrups @ 300 mm c/c. hence, provide 2-legged 8 mm ϕ stirrups @250 mm c/c at intermediate support and gradually increasing the spacing to 300 mm c/c near themid-span & at the end support.Design a Rectangular BeamAlso, readDesign step for cantilever retaining wallDesign of spherical domeCompiler Design Interview Questions and Answers civil engineering civil engineeringengineering subjects