Question

An I-section has the dimensions shown in Fig. Q4 and is subjected to an axial torque, T = 5000 Nm A) Determine the maximum value of the shear stress (Tmax) in the material. (8 marks) B) Find the twist angle per meter length, (0/L). Assume the modulus of rigidity G for the material is 82 GN/m2. d1= 266 mm — b1 = 30 b2= 16.5 mm d2= 779 mm b3 = 30 Fig. Q4: An I-Section Structure.

Answer 1

Yi ခု mm gomme ta ņ 16.5 mm х o 79mm 2 30 min! T IT A,8, tA2 Zz +A3% ý = A14 A2t Az lo 30x266x15719x66.5* (30+719) 42664308764 30 x 266 + 30X266 + 719x 1615 889.5 mm ll # For polar moment of inertia, J= Iz- Ix+Iy [perpendicular axi's theoren] Ix = Ist Iz fI3 . To - MOL about 'G! parallel Axis theaven I, - It Ab 2 А Area ha distence between

center of gravity and Ý. 30x466 Ix = + 30x266 (389.5-15) + 12 719 & 16.5 1 2 + 16-58719 x(389.5-38915) 303x266 + 80X266 ( 38915-15)2 12 2750671598.6 mos 4 I ya I lx +Izrtlay 2663 30 719x{16:5)3 2266330 + t 12 12 12 - 94374633.15 mm 4 polar moment of inertia Iz- Ix +Iy mint Iza 2750671598-6 +9437 4633.15 2.84 504 xros minst

ca) Imat = ? GO = T J 2 Y + Torgianal equation T= I x x 389.5 5000x103 Tomat 22.84506x109 0.6845 Mra Kla 11 684.5 b) Frum Toasimul equation GE = T J 3 TL 1) is 5000x10° x1 x 106 0.84 5047109X 82x109 JG on rad. 2.145 = 20 (4322x100 rad Angle of twist per mit length

Angle of twist per meter o Flenght 21433 x 108x103 sad 2:1433 Xco хсоб rad =