Question

For the section and material properties shown in Figure 1, calculate the moment and corresponding curvature given a compressive strain at the top fibre of 0.0036 and a neutral axis depth of 48 mm. [6 marks] Perform only one iteration regardless of convergence. 0.0036 E 25 mm 48 mm 220 mm 325 mm 350 mm 675 mm You 250 mm E = 200 GPS Es - 7500 MPa f = 500 MPa f = 40 MPa E = 0.0018 & = 0.0054 KE € & € E concrete f steel reinforcement Figure 1

Answer 1

o 8617e 0.0036 mot 25mm Esc 48 mm 48-25 220 mm 325 mm FORN 675 350mm mm 325-48 2010 Est 230 mm Eutest Currature Prosop tan Est (325-48) Ec 325 Esc Cuerature 0.0036 48 -S 7.5X10 a Est 0.0036 x (325_48) (from similar) 48 Triangles 10342 0.020775 Et зоо anase Esc 0.0034 (48-25) Ex= 1.725x 10

Stresses in Sted & - soo MPa Esh- 7500m fy- q soo +(0.02078- st = 200 17500 2x105 200GPa = Es 1.765 10 500 0-090755 200 Gpa 11 500 + (0.020725 st exios) 7roo 637.0625 N/mm mmi 1.725 x 10 x 2 X 10- se Sc 345 N/mmy & 28oo.o gravel 8P faces in Sted :- ZEFOGO force in steel Cs Compressive Sex 220 mm Asc - 220mm 75.9 KM

To = Tensiune force in Steel 3 no area of Since sted in Tension ut & 675 mm2 side is 675 mm) Is 637.0625 x 675 Is 430.017 KN + 3100.0roux Compressive force in Concretein 0.0036 20 MPa 0.0036 2) Yo ma o.ool 8 (18-25) Steain Lagram Steen at the stean of 0.0036 с yo *10.00ly -0.0036) (0.0054 -0.001s) 0.0018 0.0036 obory a=20 MPa

Compeenive force in concrete wao tronom of Jer se = heea of steen X - block Width beam C +44*X40X * 40X (48-25)) + (40420) x>5 )x200 rorna todo trenos zvilst = 302.5 KN Centroid of Compeinsive force of concrete from Top 20 20 Octopce (340.1002) Az az & Aznz + Apai Acta,+AX do CG = AL A 23 (as X20 )* 25 + *20*25*(3425)+ +*23*40x25:23) = 302, 5x103 25 C G from top 21.038 mm

Moment Calculation :o) meet2 I 21.028] 25 Cs Moment force x Lever dem (325-48) Taking Moment about Tension Sted Moment GX (325-21-028) + G (325-29) M = 302.5 (325-21.028) + 75.9 (325-25) M - 14.722 KNM od morze 1x.co