Question

9 Find the Moment of inertia of the given section about X-X axis passing through its center of gravity. Take A= 60 mm, B=10 mm, C= 40 mm and D= 70 mm 10 Find The Tension in cable RP & RQ, where PR =6m, AR = 5m, RQ = 14 m & RB = 5 m 60KN 100KN

Answer 1

CRIVEN DATA Abomm Blomm C. 40mm D-70mm ATY SOLUTION - XX SECTTION PASSES DISTANCE bomm lomm y, lom 2 Somm Y2 usb 3 10mm k SECTIONO A la Cox lo-beomm? Yi 10+ 30+ 10/2 y = 45mm 0-70mm SECTION SECTION Ag- 30 XIDE 300mm A3 = 70X10 = 700mm? Y 2 = 10 + (30/2) 10-5mm. 9 2 = 25 mm Y = 10 y Ai Yi+ Azyz+A343 (600x 45) +( 900825) + ( 700x5) Ait A2 + A3 600 + 300 + 700 9. = 23.75 mm 11x - ( 50x103) + 60X10 X 210253) Y=23.75 Ilxs 275937.5 mm 4 12x = ( 10x303) + (300 x 1-253) 12* = 22968-75 mm 4 * 70x103)+( 400818-753) I 3X = 251927.083 TSX a = 45 – 23:15 = 9-4 9.21.25mm 92=ye-g = 25-23,75 92: 1.25mm a3 y - y3 = 23.79-5 a3 = 18.76 Ixx = 11x + 12x + 13x CS Scannee 275937-4 +22968.75 + 251927.0 Txxscan 550 833.33 mm 4 mmal

(010CTIVEN DATA PR= om RQ=14m AR= 5m RB = 5m. R HookN A B GORN SOLUTION coso = 5 5 AT JOINT R 6m 0 - 33.55 A 5m R TRO TPRI . 14m Cosa- Ja A d: 69.07° R 69.07 9 33-5° om BOKN FOR STATIC EQILIBRIUM : Cos 33.55 + Trocos 69.01 co. EFxzo -TPR ToR = 0.4286 TRO & fq=0 TPR Sin 33.5 + Tro. Sin 69.07 360. Co.4286 TRA) Sin 33.5+ Tro sin 69.09 = 60 TRO = 51024 KN TRR: 21.96 KN RESULT 51.24kn RO: IW CABLE TENSION 21. 96kN IN CABLE PR TENSION CS Scanned with CamScanner