Question

2. (40 points) Several forces are applied to the pipe assembly shown. Knowing that the pipe has inner and outer diameters equal to 1.61 in. and 1.90 in., respectively, determine the fatigue factor of safety of pipe AB if the forces are synchronously reversed. Transverse shear is not negligible. Neglect any stress concentrations. 200 lb 150 lb B 10 in. 150 lb 6 in 501b o=M;O=OCTOR = (13 – a');} = 21 Center = 0average Radius = 10:50)*+ (try)? Center = average Radius =

Answer 1

Answer

Solution - Draw the figure thal shows the direction of the moments as follows: write the equation fer outer radius Here, the outer radius of the pipe is do Substitute 1.90 in for de 50.95 in woite the equalion for inner radices ci di 1. Here the inner radius of pipe is di di = 1.6lin c; = 16 = 0.805 in

calculate the thickness of pipe t=6-Ci t -0.95 -0.805 0.145 in. Calculate the moment about the x-axis T:200(0) = 2000lb-in calculate the moment about the z-axis at point H M₂ =(150) (10) cisco lb-in Calculate the moment about the y-axis at point k My = -20o) (10) +(15084) + (50 x 10) =-900 lb-in calculate the area A = n(e-c:) 60=6.95io C; = 6.808 in A = 160.95)?- (0.8&st] A :0.7994 in?

Write the equation for moment of inentia 5-17 (6"-C;") 1.5 (0.85". 0.6054) I .0.3099 in calculate the polar moment of inertia J: (6-Cr"). - (0.95". 0.805) -0.6198 iny calculate the extreme fibre distance - lino Y =0.95 in Calculate the normal stresses at it Here, the force in the disection of x à P, area of pipe à A, and moment along Z-axis iü M2 (iso) (500)6.95 CH=57994 (0.3099) -47 85.898 Psi FA 24.78 ksif

Thues , thue almal stsesses at ll in (4.78 Ksi Calculate the sheer stress al i (T.), Here, the moment about & is i, sadius of pipe is C, polar moment of inertia is J. Substitute 2000.lb-in for I and 0.95 in for c, and 0.6198in for I (Tu), -(000)(0.95) (0.6198) = 3065.6os psi 23.06 KSI Thes, the shear stresses at H is 13.06 Ksi Calculale the normal strenses at k. - A l Here, the moment about Yoaxis is My. Briso lh A=0.7994 My = -90016 150 6900) (0.95) 0.7994 . (0.3099) 2-2571.313 8'si - -2.57 ksi Thes, the normal stresses at K is a. 57ksi)

calculate the shear stress at k. (T.), T=2000lb.in oc=0.95in J=0.6198 in 4 (Ex), 2000 (0.95) (0.6.198) 306S. SOS Psi - 3.06 ksi Therefore, the shear sloess at point ki 3.06ksi /