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Two beams, one open section (like an I-beam with flanges and web) and the other closed section (like a hollow square section) are cantilevered and have the same overall cross sectional moment of inertia, material and length. Which beam will deflect more under concentrated load and twist more under concentrated torque, at the free end?

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  • $\begingroup$ Sounds like if you punch the numbers you will then know. $\endgroup$
    – Solar Mike
    Commented Mar 13 at 17:41
  • $\begingroup$ Why wouldn't you actually answer the question instead of downvoting? $\endgroup$ Commented Mar 13 at 22:49
  • $\begingroup$ Sorry, I did not downvote, but I can easily correct that if you wish to accuse me. $\endgroup$
    – Solar Mike
    Commented Mar 14 at 6:22
  • $\begingroup$ This question sounds like homework. I'm willing to help you with your homework, but not just feed you an answer. The purpose of homework is to think through it all yourself so that, in practice, when presented with a similar problem, you can think through it all yourself. $\endgroup$
    – TimWescott
    Commented Mar 16 at 16:55
  • $\begingroup$ @SolarMike, my bad. Pardon me for that. $\endgroup$ Commented Mar 26 at 0:22

1 Answer 1

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For a transverse force on the tip of the beam the deflection will be the same because M/I=E/R.

For a torque on the end of the beam about the long axis (call it the X axis) of the beam the I beam will have a lower torsional constant than the box section, so it will twist more.

For a torque at the end of the beam about the Y or Z axis the deflection will be the same because M/I=E/R.

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  • $\begingroup$ Excuse me, @GregLocock in the last sentence, "the deflection will... "what? I'll agree with the first sentence since in the problem statement, they have the same moment. $\endgroup$ Commented Mar 13 at 23:14
  • $\begingroup$ I appreciate your answer. Could you also specify any relation which proves that an I beam will have a lower one as compare to a box section? Plus, how is the torsional constant and twist related? $\endgroup$ Commented Mar 26 at 0:41
  • $\begingroup$ 1) only by googling, and you can google as well as I can. The approximate argument would go that for a given cross sectional area and maximum dimension a thin walled circular tube is optimum, and each deviation from that makes it worse. 2) en.wikipedia.org/wiki/Torsion_constant $\endgroup$ Commented Mar 26 at 3:32

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