Sign convention used here is downward deflection = negative, upward deflection = positive. Note that deflection fbb calculated due to upwards unit load instead of support is positive and delta bo is negative (downward).
Hello! Thank you so much for this video it helped me out so much. I was just wondering why 144.33 came out as positive. When I tried it on my own I got fbb = (1/6EI)*((7^3)-(3*7*7^2)) = -144.33 Thank you again and please keep up the good work!!
Thanks for the feedback, glad you're enjoying the video! The negative and positive sign are simply sign convention, we assigned downward deflection as negative and positive as upwards. You could switch them if you like as long as you're consistent. Hope that helps.
@@loolick8307 sorry I'm not really sure why you're confused, just take sum of forces in y and solve for Ay. Whatever Ay ends up being (negative or positive) that's what it is. Remember there's also a moment reaction at A.
deflection tables are located in a number of places, if you've purchased or have a pdf of the structural analysis textbook for your course they should be in the appendix in the back. Also, the steel or concrete building code for your country will have all the relevant deflection tables, if you're in civil engineering in university you're probably obligated to buy these. If you don't have any of these, google beam deflection tables and you should find some.
Also, remember there are many different formula derived for deflection..you don't have to use this one. In this particular formula a refers to the length of the distributed load starting from the fixed support.
Sign convention used here is downward deflection = negative, upward deflection = positive. Note that deflection fbb calculated due to upwards unit load instead of support is positive and delta bo is negative (downward).
Unknowns are referring to the reaction forces (what were trying to solve for). We have a fixed support (3 unknown reaction force) and a roller (1 unknown reaction force).
+THE STRUCTURAL ANALYSIS MATRIX FORMULATION If you return to strengths of materials 1 you'll see that the reaction vector is broken up into two forces, reaction in x and reaction in y and considered two reactions when analyzing supports such as a fixed support or pinned support.
What happens if you try to add in consistent deformation compatibility equations into an already statically determinate scenario? Would you wind up with more equations than unknowns and an unstable structure?
Click here to try using the method of consistent deformations to solve a 2nd degree indeterminate beam! ruclips.net/video/avU_YnKP3M8/видео.html
Awesome video...clear and straight to the point!
+A DC thanks so much!
Great video! Didnt see this subject around RUclips, plz do more hahaha
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This will help me so much with my computer aided analysis course!
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shouldn't Ybb be negative since the force is upward and the formula is for downward force.Hence sign switch?
Sign convention used here is downward deflection = negative, upward deflection = positive. Note that deflection fbb calculated due to upwards unit load instead of support is positive and delta bo is negative (downward).
which formula did the used formula came from?
Hello! Thank you so much for this video it helped me out so much.
I was just wondering why 144.33 came out as positive.
When I tried it on my own I got
fbb = (1/6EI)*((7^3)-(3*7*7^2)) = -144.33
Thank you again and please keep up the good work!!
Thanks for the feedback, glad you're enjoying the video! The negative and positive sign are simply sign convention, we assigned downward deflection as negative and positive as upwards. You could switch them if you like as long as you're consistent. Hope that helps.
I got -22,509.375/EI for Bo. w=15 a=14 x=7. did i get the values wrong?
degree 2.. i want to learn plz make video
ruclips.net/video/avU_YnKP3M8/видео.html
Just wonder if you get your By at such big value with small point load / uniform load, wouldn't Ay be negative value? I'm facing the same problem
Why are you finding Ay here?
@@AFMathandEngineering After the force method, using the equilibrium equations
@@loolick8307 sorry I'm not really sure why you're confused, just take sum of forces in y and solve for Ay. Whatever Ay ends up being (negative or positive) that's what it is. Remember there's also a moment reaction at A.
My country didn’t have a deflection table they use some kind of intergrade like equation BO = 0 = |m1m2ei
Structural analysis basics
ruclips.net/channel/UCfZBo0T5G0W_rGYHAXfYIRwvideos
where do you find the table of deflections
deflection tables are located in a number of places, if you've purchased or have a pdf of the structural analysis textbook for your course they should be in the appendix in the back. Also, the steel or concrete building code for your country will have all the relevant deflection tables, if you're in civil engineering in university you're probably obligated to buy these. If you don't have any of these, google beam deflection tables and you should find some.
How you get fbb= 114.38/EI ?when i do with same equation i got a negative value.
Structural analysis basics
ruclips.net/channel/UCfZBo0T5G0W_rGYHAXfYIRwvideos
How do you get those deflection equations by just formulae...... P/6EI(X3-3ax2)
You can derive them, or just use the already derived table from your textbook, which will most likely be available to you during the exam.
anybody tell me...does a refer to entire span or the span other than x
Knowledge Easy if you refer to your deflection tables in your code book or textbook it should be very clear which one is which.
Also, remember there are many different formula derived for deflection..you don't have to use this one. In this particular formula a refers to the length of the distributed load starting from the fixed support.
Hey I think your fbb should be negative
Sign convention used here is downward deflection = negative, upward deflection = positive. Note that deflection fbb calculated due to upwards unit load instead of support is positive and delta bo is negative (downward).
4 unknowns?? Which unknowns?
Unknowns are referring to the reaction forces (what were trying to solve for). We have a fixed support (3 unknown reaction force) and a roller (1 unknown reaction force).
3 reactions, not 4 ?
+THE STRUCTURAL ANALYSIS MATRIX FORMULATION there are 4 reactions on the original beam, 3 reactions after the redundant support is removed.
by removing the support @B leaves two only Ra and Ma ........
Ma, Ax, Ay
let me tell you something Ax and Ay are component of a F vector.
+THE STRUCTURAL ANALYSIS MATRIX FORMULATION If you return to strengths of materials 1 you'll see that the reaction vector is broken up into two forces, reaction in x and reaction in y and considered two reactions when analyzing supports such as a fixed support or pinned support.
What happens if you try to add in consistent deformation compatibility equations into an already statically determinate scenario? Would you wind up with more equations than unknowns and an unstable structure?
I think statically determinate structures already satisfies equilibrium equations.
Structural analysis basics
ruclips.net/channel/UCfZBo0T5G0W_rGYHAXfYIRwvideos
sir your calculations are wrong
Knowledge Easy thanks for the post...which calculation is incorrect?
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