Design bond stress 1.2 x 1.60 x 1.25 2.4N/mm2 Therefore, development length of bar (20x 435) / (4 x 2. so 10mm weight = 72.10*0.617 = 44.48 KG Design bond stress of concrete as per IS 456, 1.2N/mm2 The design bond stress should be increased 60 for deformed bars and 25 for compression bar.3 number of bottom bar of 20mm with 3.56m with straight length = 3*3.56 = 10.68m.Verify for minimum shear encouragement, optimum spacing of shear reinforcement is completed as per the Code. Check for Minimum and Maximum Reinforcement criteria is regarded for Flexure as per the Program code. so 16mm weight = 66.96*1.58 = 105.80 KG The style of support for Control State of Collapse in Flexure and Shear for Capital t Beam Section is transported out as per IS-456.So length of bar with hooks = 5.50+0.256 = 5.756 m Lets bar bending schedule of beam with development length without hooks Stirrups/Ring of 10mm size = 0.54×0.39 m with bend hook Length of Ring = 2x(0.54+0.39+0.10) = 2.06 Hook length for 20mm bar = 20×16 = 320 mm.Here hook length for 16mm bar = 16×16 = 256 mm.We can also provide the hook of U shape, which is x16 times of diameter of bar.

It should also be kept in mind while designing the beam that its’ width should not be greater than the width of the columns on which it is supported. Using M20 concrete and Fe250 steel design the beam for shear as per IS 456-2000. However, the width of the beam should not be lesser than 200mm. The beam is subjected to shear force as shown in Fig.2 below.

Length of Bar = total length – cover + 2x development length 2. As per IS 13920:1993, the depth of the beam should not be more than 1/4 th of the clear span and the width to depth ratio should be more than 0.3.

Rccbeam sizingas per is 456