Question

00 Several forces are applied to the pipe assembly shown. Knowing that Q = 230 lb and that the pipe has inner and outer diameters equal to 1.61 in. and 1.90 in., respectively, determine the normal and shearing stresses at point Hand point K. (Round the final answers to decimal places.) 10 points eBook 150 lb Hint 10 in Print 150 lb References 6 in 50 lb The normal stress at point His The shearing stress at point His The normal stress at point Kis The shearing stress at point kis ksi. ksi. ksi. ksi.

Answer 1

of ay 230lb. o di = 1061 in do log in I - Moment of martia cross-section I= Ilde-d") 64 ne ( 1099-1•61") 150 lb co 6u 0.3098 ile. Loin K 1solb x 4 in * 32 5016 bin y 32 I 2300 x 2 23.529 ksi J = polar moment of 2 inertia of cross section (do" diy) (1099-1061) -0.6.19 iu4. T= Torque at kand H due to 23016 = 230 (10) = 2300 l-ele According to torsional equation; 2 J R z 109 IxR 2 3529.8 psi 0.619 The shearing stress at H=3.53ksi The shearing stress at k =3.53 ksi - direct normal stress at both Handki 187.62 psi - (0.18 7 ksi) Tensile 74 (109?_36612) (6) Total bending stresses at H: (150x10x do I = 150XL0 x 10 199 0.3098 = 4600psi 406ksi (Tensile). :. The normal stress at Haga + H=4.787 ksi · a = 150 2

(Opz Total bending stress at k (150 x4)+150X10) do x I 0.3098 -150X4) +(50x10) x 10g 3373.14 psi = 3.37ksi ( Tensile). .: The normal stress at k = carbok = 3.56 ksi.