A reinforced cement concrete beam section has a size of 300 mm width and 400 mm effective depth, the grade of concrete used is M20 and grade of steel is Fe 415. If the ultimate bending moment of 150 kN-m acts at the section, the beam has to be designed as

Approach: Check the ULTIMATE and limiting bending moment.Concept:To know which type of beam need to be used is determined by checking Ultimate bending moment and Limiting Bending moment.Need for Doubly reinforced reinforcementi) When depth is restrictedii) Reversal of stresses.iii) Subjected to Impact loadsiv) MU > MUlim\({{\RM{M}}_{{\rm{Ulim}}}}{\rm{\;}} = {\rm{compressive\;force}} \times {\rm{lever\;arm\;\;}}\)\({{\rm{M}}_{{\rm{Ulim}}}}{\rm{\;}} = {\rm{C}} \times \left( {{\rm{d}} - 0.42{{\rm{x}}_{{\rm{umax}}}}} \right){\rm{\;\;}}\)\({{\rm{M}}_{{\rm{Ulim}}}}{\rm{\;}} = 0.36{\rm{\;}}{{\rm{F}}_{{\rm{ck}}}} \times {\rm{B}} \times {{\rm{x}}_{{\rm{umax}}}} \times \left( {{\rm{d}} - {{\rm{x}}_{{\rm{umax}}}}} \right){\rm{\;}}\)Where,Limiting bending moment (MUlim)Ultimate bending moment (MU )d is the effective distanceSteel XUmaxFe2500.53×dFe4150.48×dFe5000.46×d Calculation: Given,Grade of concrete M20Width of beam = 300mmDepth of the beam = 400mmGrade of steel is Fe415Ultimate bending moment (MU) = 150kN-mLimiting bending moment (MUlim )\({{\rm{M}}_{{\rm{Ulim}}}}{\rm{\;}} = 0.36{\rm{\;}}{{\rm{f}}_{{\rm{ck}}}} \times 0.48 \times {\rm{d}} \times \left( {{\rm{d}} - 0.48 \times {\rm{d}}} \right){\rm{\;}}\)\({M_{Ulim}} = 0.138{f_{ck}} \times b \times {d^2}\) kN-m \({M_{Ulim}} = 0.138 \times 20 \times 300 \times {400^2}\) \(= 132.48\) kN-m∴ MU > MUlim ∴ Doubly reinforced beam is used.