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

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450 K. X 50 don 550 m As WWW Es Ec 200000 28600 +(5.993x620)(184.685-50)+(6.993x3200) 550-184-685) New meutral axis of cracked section and comment of imestia of cracked transformed section is calculated as below. 17 Depth of neutral axis (comsidered section is cracked). (2) AS * Modular ratio, m = 6.993 * To find neutral axis depth bdo (am) + Con-DA s' (an-a) - MA (550-don) (4 50 db ) + (5443) (620) (dw-50)= 6-498 (2209) (550- d) ->22.5 do + 3715.66 dm-185783=12307680-22377-6do - 2250,3 +26093.26 - 12493463 = 0 Solving above quadratic equations Jan= 184.685mm 2) Cracked second moment of Icr=1450X484-685) Icp = 1944.9+67-402+2986.404) x106 Ico 3998.706 X106 corron' apea 29+ sox195ces) (89-685)