Question

stion 1 Determine the total vertical displacement of joint A (due to the load and tempreature change). Each bar is made of steel and has a cross-sectional area of 600 mm². Take E - 2001 GPa. Use the method of virtual work. Members AB and BC are subjected to temperature raise AT = +60 °C use a= 1.2 (10-5)/°C —15 m- 1.5m 10 KN

Answer 1

Forces in member due to given loading (F) 2m 15m 15my "Smi OKN Consider joint A FAB A PODPAD ( tamo Į. Simo = 415 P. COSO = 315 10 KN {Fx=0 FADE FAB cost=0 > FAD + FAB Bo -tio {fy=0 FAB SMO =10 = FAB X 4 =10 FAB. = 125 km ] putting inos we get FAD = - ** 12.5 TFAD = -7.5 kN]

Consider joint B - -> FBC ->FBD {Fy zo FAB cos 90-0) +FBD Simo=0 » FAB SPRO TEBO Simo =0 → FBD = - FAB -> FBD =-12-5 kW Ź Fx=0 FBC + FBD COSO - FAB Sim (90-6)=0 >> FBC – 12.5*3-1255*3=0 FBC = 15 KN | NOWO Apply unit load at joint a VAKN 2m

Forces in members due to unit load at A (P) It should be noted that instead of 10 kN given toad, 1 KN load is applied. so all the member forces done to len will be (11th due to the given loading * PAB = FAB = 1.25 KN ? PAD = FRP = -0-25kW PBD - Fpp = -1.25 KN PBC = FBC = 1.5 KN New, Reflection by unit load method is given by A= SEPL + FKL where AT = 60°C i, AE = 2x100x60=12002 A = 600mm2 , E = 200 gra = 200x10”mpa Member! FekN) P() 241 LAAT PPL em) 18X10-3 3-258410-4 AB ДЕ. m 12.5 11025 12.5 X60 = 7.2410-4 1-40604 - -1.753 AD BD -7.5 12.15 -1.25 BC 15 15 2.5 - 3.255X104 15 7-28109 108x10-3 281257104 5 2.88X10m 31.073X10-3 A = 3 + 3 LAAT - 2-88X10-3 + 1.073X103 = 3.953X10-3 m 33.953 mm