Design of a singly reinforced concrete beam as per IS 456:2000

Design of a singly reinforced concrete beam

Example: Design of a singly reinforced concrete beam for the following data.

•          Clear Span of the beam – 3 m
•          Width of supports – 200 mm
•          Working live load – 6 KN/m
•          Grade of concrete – M20
•          Grade of steel – Fe-415 HYSD bars

Solution:

Given Data –   f_ck  = 20 N/mm².

                        f_y    = 415 N/mm².                       

                        Load factor = 1.5 for Dead load & Live load.

• Determination of the dimensions of the beam:

              Let us adopt the span/depth ratio of 20 for the given span & loading

              Effective depth = d

                                          = (Span/20)

                                          = 3000 / 20 = 150 mm.

              Adopt d = 160 mm.

              Overall depth (D) = 200 mm.

              Width of beam (b) = 200 mm.

              Therefore, Effective span = (Clear span + Effective depth)  = 3 + 0.160 = 3.16 m.

              Center to center of supports = 3 + (2 x (.2/2))  = 3.2 m.

              Hence Effective span of the beam (L) = 3.16 m.

• Load calculation:

            Dead load : Self weight of the beam = (0.2 x 0.2 x 25) = 1 KN/m.

            Live Load : as given = 6 KN/m.

            Total designed load = 1 + 6 = 7 KN/m.

            Design Ultimate Load (Factored Load) = (1.5 x 7) = 10.5 KN/m.

• Calculation for Ultimate Moment & Shear force:

            Moment of UDL on a simply supported beam is given by – wl² / 8.

                                M_u = (10.5 x (3.16)²) /8 = 13.1 KN-m.

            Shear force on a simply supported beam is given by – wl / 2

                                V_u = (10.5 x 3.16) /2 = 16.59 KN.

• Calculation for tension reinforcement:

            Calculate the limiting moment of the RCC section, which is given by 0.138 f_ck x b x d².

                                M_u,lim = (0.138 x 20 x 200 x 160²) x 10^-6 = 14.13 KN-m

            As, M_u,lim is greater than M_u the section is under reinforced.

            Therefore, M_u = 0.87 x f_y x A_st x d [1- (A_st x f_y)/ (b x d x f_ck)].

                From solving equation for A_st we get: A_st = 275 mm².

                Check: Minimum A_st,min required = (0.85 x b x d)/ f_y = 65.5 mm².

                As, A_st > A_st,min Hence, Ok.

            Provide 3 bars of 12 mm diameter which gives 3 x 113 = 339 mm².

• Check for Shear stress:

i_v = (V_u / bd) = {(16.59 x 10³)/ (200 x 160)}

                         = 0.51 N/mm².

                                P_t = (100 A_st / bd) = {(100 x 339) / (200 x 160)}

                                                                   = 1.05.

Design Shear Strength of concrete as per IS 456:2000

From above table we get

                            i_c = 0.62 N/mm^2, which is greater than i_v so nominal shear reinforcement is provided. Using 6 mm diameter two legged stirrups.

                             S_v = (A_sv x 0.87 x f_y) / (0.4 x b)

                                    = (2 x 28 x 0.87 x 415)/ (0.4 x 200)

                                    = 253 mm.

But, spacing for stirrup shall not be greater than 0.75 x d = 0.75 x 160 = 120 mm & 300 mm.

                                Therefore, adopting spacing of 120 mm.