Determine the state of stress at point f on the cross sectio

Determine the state of stress at point f on the cross section of the pipe at section a-a.

Solution

GIVEN-->

CROSS SECTION a-a IS SUBJECTED TO TORSION STRESS CAUSED BY TORQUE OF SIN(60) COMPONENT OF 50 lb FORCE AND BENDING STRESS CAUSED BY COS(60) COMPONENT OF 50 lb FORCE.

OUTER RADIUS OF PIPE Ro = 0.5 in

INNER RADIUS OF PIPE Ri = 0.375 in

LENGTH OF PIPE L = 10 in

WRENCH LENGTH Lw = 12 in

F = 50 lb [(Fv = F SIN(60) = 50x0.87 = 43.3 lb) AND (Fh = F COS(60) = 50x0.5 = 25 lb)]

TO FIND-->

BENDING STRESS Sb = ? AND TORSION SHEAR STRESS St = ?

SOLUTION-->

Sb = (Fh x L) / Z WHERE Z = 0.067 FOR THIS PIPE [Z = 0.78 (Ro^4 - Ri^4)/Ro]

Sb = (25 x 10) / 0.067 = 3731.34 PSI

BENDING STRESS Sb = 3731.34 PSI (ANSWER)

TORQUE ACTING ON PIPE AT SECTION a-a = T = Fv x Lw = 43.3 x 12 = 519.6 lbin

MAXIMUM TORSION SHEAR STRESS St = (16xT) / [3.142 x (2Ro)^3] = (16x519.6) / [3.142x(2x0.5)^3]

St = 8313.6 / 3.142 = 2645.96 PSI (ANSWER)