Question

1. Predictions of elastic modulus, yield and ultimate loads (attach additional sheets not enough space): Given: - steel tube outer diameter: 165.1mm; wall thickness: 3.3mm; no gap between concrete and steel. - Young's modulus of steel (200 GPa). Young's modulus of concrete (30 GPa). - strengths of concrete (45 MPa) and steel (yield stress q=350 MPa, ultimate stress 0,450 MPa). Calculate/Predict (see questions 9 & 10, in "stress-strain_deformation_16Questions_solutions.pdf): -Young's Modulus E of the composite material (concrete-filled steel tube). - Yield and ultimate compressive loads of the concrete-filled steel tube. 450mm 13.3mm K 165.1mm 2. Sketch of Graph - 1 Load (kN) Load (kN) P =1717.19 kN -- Pyield=1362.89 KN Strain gauge 249.891 450m 450m 49.32 Deformation over the Strain & measured whole length of the by strain gauge composite column (mm) 1315.95 1544.94 (micro-strain) 3. On the compression graph (above, load - deformation, for large deformation) write down the load at yield and the ultimate loads. Explain the difference, if any, with your predictions. 4. From the compression graph (above, load - strain gauge measurement, for small deformation) calculate the value of Young's Modulus E of the composite material (concrete-filled steel tube). (E = slope of elastic stress-strain curve). Discuss the difference, if any, with your prediction. Please note that the strain unit recorded using strain gauge is in micro-strain (micro-strain = strain * 10%).

plz anser q1 3 4

Answer 1

K = 3.3mm - wosh So, •3.3ml mm Answer:- O Given a steel tube and concrete arrangement outer diameter s steel tube do = 165.1 mm Thickness t internal diameer at tube di edo-at - 165.1 -2(33) = 165.1 - 6.6 2 di = 158.5mm and diameter & concrete Cored=d; = 158.5mm In this type of problem, composite elatic modulus can be found using A, E + A₂E2 Ecomposite A+Az need to know where it from Concrete let Pc be the load taken steel tube Po be the load taken by So total load P= Pct Ps 0 from compatibility condition A concrete A: Astel concrete is equal deformation in steel in PL PEL Azzel Es ACEC TA CEC from this we Pc P Ps = PE -ASES comes But first we PS2 AE composite can write A E composite f CS A Ecomposite

P = P(A Ecomposite re) + P put this in equation Ac Ec ASES A Ecomposite Hore A. total A- Actas AcEc +AGES E composite A Ec Зобро Given Asteel 1 Esteel - 200 Gpa Now calculating Asted = 1 ( do do? ) (165.4- 158-5) Askel = 1677.42 mm A concrete 2 Id ū A concrete T (158.9² 19730.96mm? As Es + AcEc As + Ac 50 Ecomposite (1671.42) (200) + (19730-96) (30) 1677.42 + 1930.96 Ecomposite = 43.32 Gpa yoking's modules & composite method is 43 32 Gpa Now, part (2) yeild and ultimate composite loads for composite tube concrete strength fa - 45Mpa yeild strength of steel Ty = 350 mpa

B So We Now material ultimate strength of steel Tu = 450 MPa Let the stress in steel be steel in concrete be & concrete total loads- & Steel Asteel) + (Concrete A concret) have to determine the stresses in each from computability Askel A concrete steel L Concrete L Estel Econcrete Concrete (E) & Steel 30 steel Concrete a & concrete = (0-15 ) Isteel Zoo (in steel) at geild point ry 350 MPa so I concrete 0-15x350 = 52.5MPa > 45MPa in concrete it will be limited to 45 MPa 50 Aconcrete pa (steel Asteel) + concrete 415 mpc)/1730-96) (350 N/mm²) (167742 mm) + P p = 1474999.IN P 1475KN at yeild of steel point in steel & Steel - Ty - 45ompa & concrete: 45 MPa at ultimate

At this load both steel and concrete will not P = (450 X1677-42) +45X 19730-96) p=1642732.2N P= 1642-73 kN at ultimate point If we want to prevent concrete from failut 45 MPa take Concrete 45 300 mpa so steel 015 P: (300 x 1677.42) + C45x1950.96) p=1391119. 2 N p. 1391119 KN tail