Length of the secondary beam loaded on primary beam = half the effective distance in left and half in right So, weight of slab = unit weight * thikness of slab Since the slab are one-way, the secondary and primary beam along y-axis takes all the loads. So, weight of partition wall = 8 * 3.67 * 0.15 = 4.4 KN/m Unit weight of partition material = 8 kN/m 3 Height of wall = floor to floor height – depth of beam So effective depth = overall depth – clear cover – half of diameter of bar Since it has to resist secondary beam also and for the construction simplicity,Īdopt same size as others beam designed before. Where l is effective span which can be calculated from clause 22.2 of IS 456:2000.Īssume diameter of bar = 20 mm and clear cover = 25 mm The preliminary design is based on the serviceability condition of deflection criteria. If you want to go there, then click here. The detail about the preliminary design is discussed before. The reason is that it have to resist half load from slab of both side (10.02 m and 8.73 m) For this, we know, the beams of row 8 shown in architectural plan are critical. But the critical one is that which have to support greater load and for this the beam which have wider slab should be consider. Since all the beam have equal span, we can’t decide from this which one to choose. We can see that the primary beams along x-axis are supporting the secondary beams.įor the design of beam along x-axis, we have to choose the critical beams. The SAP modeling plan is also attached below in which the light green thick line is secondary beam and the pink one is primary beam. Live load = 3 KN/m 3 and floor finish = 1.5 KN/m 3.
Slab thickness = 130 mm and Partition material= AAC block of thickness 150 mm having unit weight 8 KN/m 3. Grade of concrete = M30 and grade of steel = 500D. Let’s understand all about the primary beam supporting secondary beam from an example.ĭesign a primary beam supporting secondary beam of the given institutional building plan which is aligned in x-axis supporting the secondary beam at the mid-span. Are you thinking what that special reinforcement arrangement is? Just wait, you will know it after completing the whole content. But the arrangement of reinforcement is different. The only difference is concentrated load at the intersection point for the calculation of moment. The design of primary beam with secondary beam is almost same as normal beam supporting on column in both ends. Design of primary beam supporting secondary beam: Thus, primary beam experiences both uniformly distributed load and concentrated load if it support secondary beam. The secondary beam only transfer loads to primary beam so during calculation, we should take point load at the junction of these two types of beam. In spite of assuming as simply supported, it gives satisfactory result for the design and analysis. But in reality, there is development of torsion in primary beam due to secondary and this can be simply solved by the special arrangement of reinforcement. Since simply supported, there is no moment developed at the joint and no torsion developed. As it is already mentioned in secondary beam that it is treated as simply-supported despite it is casted monolithically. If not, click here for primary beam and for secondary beam click here. Do you know what is primary beam and secondary? Hope you have view my previous post.
Primary beam supporting secondary beam means that the secondary beam is resting on the primary beam.