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posted by  MMSP187 on 11/2/2009 6:13:12 AM  |  status: Live  |  Earned Karma: 1127

Problem 7.48

Course Textbook Chapter Problem Needs by
Mechanics of Materials Mechanics of Materials (5th) by Beer, Johnston, DeWolf 7 48P 11/3/2009 at 9:00:00 AM
Question Details:
Solve Prob. 7.26, using Mohr's circle.
Problem 7.26: Several forces are applied to the pipe assembly. shown. Knowing that the inner and outer diameters of the pipe are equal to 1.50 in. and 1.75., respectively, determine (a) the principal planes and the principal stresses at point H located at the top of the outside surface of the pipe, (b) the maximum shearing stress at the same point.
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posted by Gatorbait2008 on 11/20/2009 4:09:21 PM  |  status: Live
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Response Details:
Replace forces on the pipe with equivalent force-couple systems at point C.

T = 8 * 50 = 400 lb-in
M = 16*30 = 480 lb-in
F2 = 50 lb

Solve for J

J = π/2(c24-c14) = 0.42376 in4
I = J/2 = 0.21188 in4
Qy =2/3(c23-c13) = 0.16536 in3
t = c2 - c1 = 0.125 in

At the section containing element H:

T = 400 lb-in.
Mz = 480 lb-in.
Vz = 50 lb

Stresses:
Torsion:  τzx = -Tc/J = -400(0.875)/0.42376 = -825.9 psi
Bending:  σx = -My/I = -480(0.875)/(0.21188) = -1982.3 psi
Transverse Shear:  τzx = VQ/I(2t) = 50(0.16536)/(0.21188*(0.250)) = 156.1 psi
Total:   σz = 0; σx = -1982.3 psi; τzx = -825.9 + 156.1 = -669.8 psi


Now,

Tan (2θp) = 2τzx / σz - σx
θp = -17.00o and 73.00o from z-axis.

σave = (σz + σx)/2 = -991.1 psi
R = = 1196.2 psi

σmax = σave + R = 205 psi at 17.00o away from z-axis
σmin = σave - R = -2187 psi at 73.00o away from z-axis
τmax = R = 1196.2 psi
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