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posted by  NoleBrain on 9/7/2009 12:46:50 AM  |  status: Closed  |  Earned Karma: 11196

strength of materials **lifesaver guaranteed**

Course Textbook Chapter Problem Needs by
Mechanics of Materials Mechanics of Materials, 5e, Beer Chapter 2 2.13 9/7/2009 at 11:00:00 PM
Question Details:

Rod BD is made of a steel (E=29x106psi) and is used to brace the axially compressed member ABC.  The maximum force that can be developed in member BD is 0.02P.  If the stress must not exceed 18ksi and the maximum change in BD must not exceed 0.001 times the length of ABC, determine the smallest - diameter rod that can be used for member BD.

The answer is 0.429 in

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Oracle
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posted by PhysicsPhanatic on 9/7/2009 7:42:20 PM  |  status: Live
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Response Details:
FBD = 0.02P = 0.02(130) = 2.6 kips = 2.6 x 103 lb
Now consider the stress:
σ = 18 ksi = 18 x 103 psi
σ = FBD / A ---> A = FBD / σ
                                = 2.6 / 18
                                = 0.14444 in2 --- (Area 1)
Next consider the deformation:
δ = 0.001(144) = 0.144 in
δ = FBDLBD / AE ---> A = FBDLBD / Eδ
                                        = (2.6 x 103)(54) / (29 x 106)(0.144)
                                        = 0.03362 in2 --- (Area 2)
We know that the larger area governs so we use Area 1 to get
A = 0.14444 in2
A = (π/4)d2  ---> d = √[4A / π]
                               = √[4(0.14444) / π]
                               = 0.429 in
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