Question

The point loads are placed at the fixed positions shown in the figure and they are live loads. А B D E K 12 44 w * C (centre) A b f' = 32 Mpa far = 3 MPa fs, = 500 MPa E = 200 GP E = 28600 MPa A a Cross section The following values are used for the question. • 3 m load before cracking • h = 2.5 m • P=2 KN • a = 50 mm . G=2 kN/m . b = 500 mm • Q=0 kN/m • C 450 mm • A c = 620 mm? • As = 3200 mm? load after cracking .P=15 kN • G15 kN/m .Q = 15 kN/m Now, the live load increases gradually and the moment at the critical section just exceeds the cracking moment (Mr) but the compressive section of the concrete is still governed by the linear elastic region. It should be noted that in the flexural design, the cracking of the concrete section starts where the tensile stress reaches the tensile strength. It is assumed that the cracks then propagate rapidly up to the entire tension section (up to the neutral axis) and this cracked concrete section cannot resist the tension. In reality, concrete sections between the primary cracks can still resist some tensile stresses as shown in the figure below, which should be considered in the displacement design. However, in the flexure design, we design the critical section where the largest moment occurs and thus it is reasonable to have such primary cracks in this critical section. In the linear elastic region, the transformed area method is still valid. However, we may solve the quadratic equation for calculating the neutral axis distance from the top surface. A A NA b NA A Stress profile concrete Cross section before cracking cracks A Stress profile (concretel Cross section just after cracking Longitudinal Section secondary crack Neutral axis primary cracks M M -concrete in tension 1. The new neutral axis is in the cracked section under the elastic region on the compressive section. dn (distance to the top surface) mm (3 decimals) (2 marks) 2. Cracked second moment of areas ler *10 mm (3 decimals) (2 marks)

Answer 1

Solution based on cracked transformed area of the section is given below.

1) Depth of nential axis (consilered section is Cracked) K 450 Ian (n-1) As' 550 LAS * Modular natio, n = Es Ec 200000 26 600 6.993 * To find neutral axis depth 6 du (en) + (n-1) As' (wa)= nas ( 550 - du) 450 din (456 + (5,993)C620 )(dn-50) 6.993 (3200)(550-dn) 225 dn² + 3715.66 dn = 123076 80 22377.6dn 2 185783 → 2a5dn² + 26093.26 1249 3 463 ro Solving above quadratic equation. da 184, 685 mm

2. 184.685 2 685]? 12 2o Cuacked Second Moment of area Icn= (450 450 x 18416853 :) + [450x184.685) 35] ( 184 + [5.993 x 620][1841685-50]? + (6.993 x 3400) [ 550-184.685] ? Iar (944.9 + 67. HOX + 2986.404) x106 - Ian 3998. 706 x 106 mm