Question

15 Section 5, Problem 15 (25 pts.). A member is loaded by forces a, cand d as shown. The member is a hollow member with an outer radius of ro and a wall thickness of t. The member is fixed at ground level where points Fand G occur. Determine: a) the normal and shear stress at point F. b) the normal and shear stress point 02:46:29 Forces: a = 30 (k),c=16 (k), and -17 (K) Dimensions: e = 32 (in.), f = 16 (in.), r0 = 4 (in.) and t = 0.125 (in.) (ksi) (ksi) Point F: The normal stress at point Fisſ The shear stress at point Fis Point G: The normal stress at point G is The shear stress at point G is (ksi) (ksi) Front View Right Side View d с a a C yy-axis 22-ais G F: G F Top View of Cross-Section

Answer 1

Date Page © С Given a = 30k 6 = 16 K d = 17K e=32 in f = 16 in %= 4in a d = 8 in. t = 0.125 in 2t= 0.25.in. ЕЕ le Point W/ G F Z Bending 나 stress (16632) shear stress due 17 k force. Compressive stress due to 30k.

Normal compressive Stress === o=-P T(22-8-0-25) -30 2 -9.7 Ksi. Shear stress =<= v Ay ТЫ [17x102x 3.09x1942 11 x (849-254)xo 25 To 16.gg Ksi 64

Date Page Due to moment, nomal stress- M f y I 16x32 f u osofi 85.4K si. Ţ (84_7.754) 64 Nommal stress at F = 85.4 - 9.7= 75.7ksi Shear stress at F=116.99 ksi Point G M I Compressive normal stress = -30 =-9.7ksi. II (82-7.752) Que to moment, (dxe) y (17x32) fi • f = 90.75ksi I (84-7754) 4 64 i) Que to moment = (axf) = fa = (Box 16) if = 80.07 ksi 7 (84_7754) u Normal stress at G = -9.7-90.75-80.07 = - 180.52 Ksi 1 Shear stress = 166 3.09% 1.94 II (84-7.25")x0.25 15.gg ksi