Just want to understand this review example Engr225Lecture Examples, Sections 4.6Lecture Example 2: The solenoid current heats rod CD from T= 30°C to T=180 °C. The steel end rodsAB and EF are heated from T1= 30 °C to Ty=50 °C. At T, the gap between Cand the rigid bar is 0.7 mm.Determine the force in rods AB and EF caused by the heating.0.7 mmAB & EF: steelCD: aluminumTask:BLOsteelEst=200 GPaEaF70 GPaCARFind Fur Ceax=12(10)/°Caa-23(10/°C on steel odson Steel rodsaad F on AR steelund FAL240 mm 300 mmA=125mmAF375mm?Giver:ATSE = 50- 80 = 20°cAlestATal = 180 - 30 = + 150° 4FBD rigid bulSulve, Est =F= 1- 85 KAFac = 3it KM+| EFy=0= Fl - 2 F OFAeCompatibility ConditionSst = Sae - 0.7 × 10-3- Fal LalK st A 'st Lst +FLit = tdal ATal Lal-0.7x10 e%3DEul AalEst Ast= 0.072 A 0 m12 x (0°662 (4 20°a)0gm= 72x 106mAgrafes ancat

given; lets take force in steel rod=Fslets take force in aluminium rod=Fal

ons 4.6 Lecture Example 2: The solenoid current heats rod CD from T= 30°C to Ty-180 °C. The steel end rods AB and EF are heated from T3= 30 °C to Ty-50 °C. At T1, the gap between Cand the rigid bar is 0.7 mm. Determine the force in rods AB and EF caused by the heating. 0.7 mm AB & EF: steel CD: aluminum Task: BLO Est=200 GPa EaF70 GPa Find fur ce steel AR as=23(10)/"C on steel rods Bad F or Ae EE und FAL a=12(10/°C 240 mm 300 mm Steel As=125mm? AF375mm? Giver: ATSE = 50- 80 = 20°G ATal z 180 - 3o = + 150' 4 FBD rigid buf Sulwe; Fst =F= 1-85 KM Fac = 3it KN +| EFy=0= Fil -2F/O FAR Compatibiuity Cendition Sst = Sae - 0.7 × 10-3 - Fal Lal Eul Aal K 5t A Tot hst + FLit = aalATal Lal t -0.7x10 e Est Ast - 3 = 0. 072 A lO 12 x(5 6/04 (4 20°C)a.gm= 7ex 106.

ons 4.6 Lecture Example 2: The solenoid current heats rod CD from T= 30°C to Ty-180 °C. The steel end rods AB and EF are heated from T3= 30 °C to Ty-50 °C. At T1, the gap between Cand the rigid bar is 0.7 mm. Determine the force in rods AB and EF caused by the heating. 0.7 mm AB & EF: steel CD: aluminum Task: BLO Est=200 GPa EaF70 GPa Find fur ce steel AR as=23(10)/"C on steel rods Bad F or Ae EE und FAL a=12(10/°C 240 mm 300 mm Steel As=125mm? AF375mm? Giver: ATSE = 50- 80 = 20°G ATal z 180 - 3o = + 150' 4 FBD rigid buf Sulwe; Fst =F= 1-85 KM Fac = 3it KN +| EFy=0= Fil -2F/O FAR Compatibiuity Cendition Sst = Sae - 0.7 × 10-3 - Fal Lal Eul Aal K 5t A Tot hst + FLit = aalATal Lal t -0.7x10 e Est Ast - 3 = 0. 072 A lO 12 x(5 6/04 (4 20°C)a.gm= 7ex 106.

Mechanics of Materials (MindTap Course List)

9th Edition

ISBN:9781337093347

Author:Barry J. Goodno, James M. Gere

Publisher:Barry J. Goodno, James M. Gere

Chapter1: Tension, Compression, And Shear

Section: Chapter Questions

Problem 1.6.7P: A wine of length L = 4 ft and diameter d = 0.125 in. is stretched by tensile forces P = 600 lb. The...

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