Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Engineering and Comp Sci Homework Help
Castigliano's Second Theorem on Curved Structures
Reply to thread
Message
[QUOTE="Triskelion, post: 4989698, member: 540338"] [h2]Homework Statement [/h2] A semi-circular ring of stiffness EI and radius R is supported on an anchored hinge and on a roller hinge. A vertical load F is applied at the center. Determine the horizontal displacement of the roller support.[ATTACH=full]175713[/ATTACH] [h2]Homework Equations[/h2] [ATTACH=full]175714[/ATTACH] [h2]The Attempt at a Solution[/h2] So I apply a horizontal force H (to the right) at the roller. Sectioning the first quadrant gives [tex]M_1=HR sinθ_1[/tex] (moment is taken as positive clockwise). So [tex]\frac{∂M_1}{∂H}=R sinθ_1[/tex]. Similarly, [tex]M_2=HR cosθ_2-FRsinθ_2[/tex] and [tex]\frac{∂M_2}{∂H}=R cosθ_2[/tex]. Because H is an imaginary load and setting it to 0, [tex]M_1=0[/tex], [tex]\frac{∂M_1}{∂H}=R sinθ_1[/tex], [tex]M_2=-FRsinθ_2[/tex] and [tex]\frac{∂M_2}{∂H}=R cosθ_2[/tex]. So the integral is [tex]\int_0^ \frac{π}{2} \frac{M_1}{EI} \frac{∂M_1}{∂H}R dθ_1 + \int_0^ \frac{π}{2} \frac{M_2}{EI} \frac{∂M_2}{∂H}R dθ_2[/tex]. The first integral is zero. The second gives [tex]\int_0^ \frac{π}{2} \frac{-FR^3}{2EI} sin2θ_2 dθ_2=\frac{-FR^3}{2EI}[/tex]. I checked my workings, and I have no idea why there is a negative sign there. If my understanding is correct, the negative sign implies roller movement to the left, and this is really counter intuitive. Can anyone shed some light on this? [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Engineering and Comp Sci Homework Help
Castigliano's Second Theorem on Curved Structures
Back
Top