Question

The point loads are placed at the fixed positions shown in the figure and they are live loads. A B D E k w C (centre) a b f' = 32 Mpa fas= 3 MPa fsy = 500 MPa E = 200 GPa E = 28600 MPa A a Cross section The following values are used for the question. • 1 = 3 m load before cracking • 12 = 2.5 m • P = 2 kN • a = 50 mm G = 2 kN/m . b = 500 mm • Q=0 kN/m • c = 450 mm • Asc = 620 mm • Ast = 3200 mm? load after cracking • P=15 kN . G = 15 kN/m • Q=15 kN/m Now, the live load increases gradually and the moment at the critical section just exceeds the cracking moment (Mc) 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 do Por is still valid. However, we may solve the quadratic equation for calculating the neutral axis distance from the top surface. a A a A NA b NA. b A a Stress profile (concrete) Cross section before cracking cracks A, a Stress profile (concrete) 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 *106 mm (3 decimals) (2 marks)

Answer 1

(2n-1) AC Som 1 620 mm 500 K 450mm - 3200 mm - * 450 mm MAS Es 2008103 28600 location of from top NA 450 ** +((2x2) –] (620) - 50 ) = (7x3200) (50-01) +8060 X - 40 3000 - 12.32x10 + 22400 N ñ 225x2 + 3046082 12.723X106 co on solving 179.55 mm

of inentia Ir cocking moment 1 +1 Iar (450) (179.55) [(3x7) -1] (620) (129.55-50)? + 7 x 3200 (590-179.55 )? 868.268100 + 135.27X100 + 10 3074,023 X 106 Irr 4077.553 X 106 mmt

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