Question

305 mm The assembly shown is restrained at both ends, knowing that the assembly is initially unstressed, find: a) The normal stresses induced in portion AB and BC by a temperature rise of 18°C b) The corresponding deflection at point B A C T Se mers) st of = X AT K = (1.78100) (18)/886) = 31.75-mm diameter steed B of = QATL = (20,9x109/18) = Sp. = 5.668x10* 5.15 mm diameter → 87 = dp = y(24. = (a + c) at & brass >$=(10(swit* (15) (951) -P[rias * 27.54] =>dp = P6.00477) Steel Ex=200 GPa . = 11.7x10-6 °C 8p = dp → 5.848716* = Pl.com Brass ../1793 - (48976? P= 11793 Ost 79,73 ² E.=105 GPa ab = 20.9 x 10-6 °C 351 mm

Answer 1

SO3 OT = 10°C Stue 305 mm. 231-75 31.75mm 66 = 62 = Stress in sted Stess in Brass 30 mm Brass - S7-15mm a) Total Elong cetions =0 La LG ATE AT 6B LB 50 69 Ls at / LS&SAT- Bross Steel 301420.9410 X16 65 7305 200*107 305x117x10 -O 68 x 381 105X103 = 207.565 Os x 305 x 63 X30 105. 200 152566 + 3.6368 = 207.585 til

WAR also - 7 Esteel = f Brass Gs As = 6B AB 65 x 11 x 39.752 = 60% ] *57.152 u 65 = 3-24 Gen - ich 9 fli from lis -24.22Mpac 66 = 78.46 MPa (c) Elongation of for speed B = Ls as AT - osts E 305 x117x10-6x10 - 78.46x3057 o 200x103 § As = -0.055mm (contraction) L means point ® will go up by 0.055mm