Combined Loading The steel pipe is fixed at the bottom and h

Combined Loading The steel pipe is fixed at the bottom and has an outside diameter of 100 mm and an inside diameter of 84 mm Sketch the stress element for point H if the tension in the cable is 40 kN. A correct solution will a free-body diagram and equations of equilibrium.

Solution

solution: here pipe of d1=100 mm and d2=84 mm is held in position by cable of tension T=40000N

1) at point D tension is ressolve in two component as Tx and Ty

Tx=Tcos60=20000N

Ty=-Tsin60=-34641.016 N

2) here Tx and Ty will create bending around point H about z axis and about x axis

hence moment around point h about z axis as follows

Mhz=Tx*225+Ty*70=20000*225+34641.016*70=6924871.131 N mm

where intertia of steel pipe as follows

I=pi/64(d1^4-d2^4)

I=2464818.19 mm4

where y=50 mm for point H bending stress

hence bending stress arounh point h about z axis

Sbz=Mhz*y/I=6924871.131*50/2464818.19=140.474 N/mm2

4) where Ty will create moment around H about x axis

Mx=Ty*50=34641.016*50=1732050.8

Sbx=Mx*y/i=1732050.8*50/2464818.19=35.135

5) bending stress around y axis is zero

hence resultant bending stress

Sb=(Sbz2+Sbx2+Sby2)^.5=144.8014 N/mm2

7) where in equillibrium this tension force is balanced by reaction at H

in equillibrium

sum Fy=0

Rv+Ty=0

Rv=-(Ty)=-(-34641.016)=34641.016 N

sum Fx=0

Rh+Tx=0

Rx=-20000 N

7) where this Rv create longitudinal stress in pipe

Sl=Rv/(pi/4)*(d1^2-d2^2)=14.9817 N/mm2

8) where Rh will create tangential or hop stress as follows

St=Rh/(d1-d2)*t=5.555 N/mm2

8) stress at point H are

bending stress=Sb=144.8014 N/mm2

longitudinal stress=Sl=14.98 N/mm2

hoop stress=St=5.55 N/mm2