Recent content by LFRIII

I am working with superposition equations to solve a beam bending application. I am not a structural engineer and did not work with this for many years. I have an application which is a circular valve blade, supported by a shaft, with a bearing at each end. This system is subject to a UDL on...

Greg: Good idea. I have read about McCaulay but have not worked with it at all so I would have to do some work with it. I had thought of using multiple point loads but thought if I could integrate the semicircle it would be easier to model. I will probable work on the point load idea. This...

Rb1957: I am not sure what you are referring to above. I am trying to solve this by hand, Any direction you could give me would be appreciated.

Gentlemen: First of all I want to thank you all for your help with solving my problem above. I was not understanding how to apply the beam deflection formulas and you have taught me how for which I am grateful. I now have the calculations worked out and we are applying them to our product...

BAretired: One word: Elegant. I had all the pieces the first time I looked at this but could not see the forest for the trees. Thank you.

rb1957: Unit load is not something I have come across in my Elements of Strength of Materials Textbook I am working through. And I have no experience with structural engineering. I posted to this site with two goals. First to find a solution to the statically indeterminant triangular load...

Doug: For the full sinusoidal load applied from 6" to 68" I get Reactions of 1,634.37 lb. and deflection at 6" of .2529" from my elastic line equations. So we agree. As far as your second part, I am not familiar with the term unit load. What do you mean by that? And how is it applied in...

To all: The total length of the shaft is 74". The sine load is only applied between supports B & C which are 62" apart. The total force applied to the 62" diameter round blade is F = 30 "H20 x 0.03609 lb/In^2/"H20 x PI (62"/2)^2 = 3,268.75 Lb. W = 3,268.75 lb * PI / 62" / 2 = 82.815 lb/in...

BAretired: I noticed that and thought something was amiss but checking page 5 a 2" round cross section is shown. Thanks,

4 support beam deflection reporthttps://files.engineering.com/getfile.aspx?folder=40159615-6d44-4106-b843-c2f870928dab&file=Beam_Analysis_4_Supports.pdf

2 support beam deflection reporthttps://files.engineering.com/getfile.aspx?folder=408fb971-d9dc-4df0-8b66-f200b8c1790a&file=Beam_Analysis_2_Supports.pdf

BAretired: I worked through the information contained in your most recent post and have attached your diagram with my notes. I used your equation for theta 0 due to Ra and Rb and added q0 from the applied load. Solving for Ra resulted in 1,711 lb. which does not equal the result of 3,214 lb...

BAretired: Attached is the derivation of the beam formulas for the simply supported beam with the sine load. From this theta is equal to W*L^3/π^3.https://files.engineering.com/getfile.aspx?folder=c9f90554-7f2f-47c8-8ef1-1f1363a5aa18&file=Beam_Freely_Supported_at_Ends_-_Sine_Wave_Load_20240923.pdf

BAretired: Thank you for your response above. To answer your questions: 1) Too short to be considered a beam. A and B together make one fixed support. Same with C and D - I'm not sure what you mean here. The real world example is a shaft for a valve. The support locations, from left to right...

Gentlemen: I thought I had this. I understand how to solve my original problem with the VDL using superposition setting angles equal to each other. After working it out it seemed simple. But I am now trying to apply this to a real world problem where I know what the answers are but am unable...