You are watching: Several forces are applied to the pipe assembly shown
Solution - draw the number thal mirrors the direction of the moments together follows: create the equation fer external radius Here, the outer radius the the pipeline is execute Substitute 1.90 in because that de 50.95 in woite the equalion because that inner radices ci di 1. Here the within radius of pipeline is di di = 1.6lin c; = 16 = 0.805 in
calculate the thickness of pipeline t=6-Ci t -0.95 -0.805 0.145 in. Calculate the moment around the x-axis T:200(0) = 2000lb-in calculate the moment about the z-axis at suggest H M₂ =(150) (10) cisco lb-in calculate the moment about the y-axis at allude 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 minute of inentia 5-17 (6"-C;") 1.5 (0.85". 0.6054) ns .0.3099 in calculation the polar moment of inertia J: (6-Cr"). - (0.95". 0.805) -0.6198 iny calculate the too much fibre distance - lino Y =0.95 in calculation the common stresses at it Here, the pressure in the disection that x à P, area of pipeline à A, and also moment follow me Z-axis iü M2 (iso) (500)6.95 CH=57994 (0.3099) -47 85.898 Psi FA 24.78 ksif
Thues , thue almal stsesses in ~ ll in (4.78 Ksi calculate the sheer tension al i (T.), Here, the moment about & is i, sadius of pipe is C, polar moment of inertia is J. Instead of 2000.lb-in for I and 0.95 in for c, and also 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 regular strenses in ~ k. - A together Here, the moment about Yoaxis is My. Briso lh A=0.7994 mine = -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)
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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 suggest ki 3.06ksi /