Question

Learning Goal: To determine the angle of twist on a composite rod given the geometry and externally applied torques, to properly apply a sign convention to determine the angle of twist, to use a torque diagram to aid in determining the angle of twist, and to determine the maximum applicable torque given a maximum allowable angle of twist. The rod shown below is made of two different materials. Segment AB is made of aluminum (G=27 GPa). Segment BC is bonded to segment AB and is made of brass (G=39 GPa). Segment BC is hollow and has an inner diameter of d; = 25.0 mm . External torques are applied to the rod at point B and point C. Let LA = 1.250 m, LB = 1.025 m, da = 100.0 mm, dp = 62.5 mm, TB = 1.1 kNm, and Tc = 1.3 kN.m. B

Answer 1

Given data GAA = 27 apa GBC = 39 apa di= 25 mm LA = 1,25 m lo = 1.025m da = 100 mm da = 62.5 mm Tc= TB= l l kom 1.3 kNm draw the torque diagram as shown in the figure A B Torgu al BC Shaft is Tc Tac = tea 1.1 kNm 138106 mm TATA - torque of AA shaft TAB - Tot tn 1.21 pils 2.9 kem. = 2.98 106 pm onge of twist at poin A IA find the a do = of AIA TAA LA CAR TAA

do = TABLA GAB X ( I dar) 2.9x10 (1.254/000) 278 101x da - 0.011217 gad find the ange of twist of с de deint & AIA de a = TAC L TAALA + GBC JAC GAA JAA $c- TACLB + TAALA GBC X Tu (do ²-dit) 12 GARY TO da 12

de linxlorX.1025 ། 62-54-25+) 998167x Tu 32 + 2.9 X los X 1250 278 102 x x loot 12 de = 0.0219 + 0.011917 do - 0.09.17 rad find the angle of twist etc if Segment BC solid. de = dunt dAIA de = Tacla TAA LA + CAC JAC GAA TAA

de a paylor A 1025 POLI sest 798101x/T X 62.5 12 7 2.98 10 6 Wide x 1250 27 Xolx x loat de a 0228 + 0.011317 de o.039 12 reed