Load Shear Moment Diagram with Mathcad.avi

[LIEUKENHAWK]

there is a better and more easier solution for that ....



[Nauttos]

Can I use this for my Current Compression Test Experiment?

[diflection1]

I am the expert for this particular problem,ok?

[diflection1]

look for final results,then how it's done(procedure or method being used)

[diflection1]

and dW/dX=W())/L=W'(x)=1000=EI d^5y/dx^5

[diflection1]

Load formula:W(x)=W(X-L)/L

[diflection1]

shear formula:V(x)=W(2L^2-6LX+3X^2)/­6L

[diflection1]

moment formula:  M(x)=WX(2L^2-3LX+X^2)/6L

[diflection1]

have learned something yet? you still haven't said anything about my formula I invented.

[diflection1]

then W(x)=-3000+1000X or L(X)=-3000+1000X

[diflection1]

W(x)=C(1)+1000x  intial condition Q(3,0)

[diflection1]

integrating,W(x)=EI d^4y/dx^4=C(1)+1000x

[diflection1]

@diflection1 

[diflection1]

so EI d^5y/dx^5=1000

[diflection1]

the two variable are solution poits of the graph of the load equation

[diflection1]

condition:working from left to right i,and below the x-axis.ok?

[diflection1]

now  the load equation is W=-3000+1000X by defenition ,This is something about realizing the equation and load function has the same graph by defenition

[diflection1]

as for me, it means the load function,right?

[purdueMET]

@diflection1 Yes, L(x) is the load as a function of distance along the beam.

[diflection1]

ok, what does L(X)=-30000+1000X means? 

[diflection1]

this a new way of thinking,or what?

[purdueMET]

@diflection1 I'm afraid not. You are correct that the 5th derivative will give the slope of the distributed load, but you will still have to integrate repeatedly to get the deflection. Also, a truly new method would have to be generally applicable and it looks like this one would apply only to very specific loading conditions. That said, it's great that you are really studying this problem. I'm sure you now understand it thoroughly.

[diflection1]

That's why EI d^5y/dX^5=+/- W(0)/L in your case

[diflection1]

if you differentiate the deflection,you obtain the rotation or angle which arch is such a number,stand for letter theta or something, they call it slope or first derivative,so the second derivative of deflection gives the moment,the third derivative of the deflection gives the shear,the fourth derivative of the deflection gives the load,the fifth derivative of the deflection gives the slope to a graph or linear function ,in particular ,load function in your case

[diflection1]

W is not deflection is distributed load.........=3000#/Lf

[diflection1]

this is something on my own choosing ,or invention.

[diflection1]

so it'll work for any given data,give it a go.

[diflection1]

with this formula you can change the inputs.

[diflection1]

.that' s what I am after.

[diflection1]

thank you for getting back to me,but I wonder if my formula for the moment get your attention.

[diflection1]

R(1)=Fa=wL/3 and  R(2)= Fb=-wL/6

[purdueMET]

@diflection1 I'm not sure why you start with the derivative of the deflection and then integrate it again. You appear to just be getting the expression you started with. Perhaps you could do your own video and post it so the larger community could see your method.

[diflection1]

Fa=wL/3 and Fb=-wL/6

[diflection1]

I hope you read me?

[diflection1]

END



[diflection1]

or M(x)=3000x-1500X^2+500X^3/3

[diflection1]

Ilustration:M(x)=500X/3(18-9x+­X^2)

[diflection1]

or M(x)=WX/6L(2L^2-3LX+X^2)  where given w=3000 and L=3

[diflection1]

we can formulate are moment: M(X)=W0)L/3-W(0)X^2/2+W(0)X^3/­6L.

[diflection1]

then C(2)=W(0)L/3=3000 it means V(0)=3000 on the shear diagram

[diflection1]

use the condition M(L,0) to finf C(2)

[diflection1]

moment: M(X)=C(3)+C(2)X-W(0)X^2/2+W(0)­X^3/6L .

[diflection1]

shear: V(X)=C(2)-W(0)X+W(0)X^2/2L

[diflection1]

in formula:W(x)=-W0)+W(0)X/L

[diflection1]

L(X)=W(X)

[diflection1]

understand why?L(x)=-3000+1000X 

[diflection1]

or just do a linear regreassion line to find the best fit with these variable ,again you 'll get the load function with slope 1000n/m^2....

[diflection1]

the coefficient W(0)/L is the slope to a graph or linear function in this particular case

is 10000n/m^2(slope of the slope of load function that relate two variable:P(0,-w(0))

and Q(L,0).

[diflection1]

by defenition the load function and equation has the same graph.

[diflection1]

or load function if W(x)=-W(0)+W(0)X/L

[diflection1]

so to get a particcular solution for the expression we will plug C(1) =-W(0) into

 W=-W(0)+W(0)X/L

this is the load equation

[diflection1]

but C(1)=-W(0)  because w=0 when x=L

[diflection1]

integrating we'll get ; W=C(1)+W(0)X/L

[diflection1]

dw/dx=w'(x)=w(0)/L

[diflection1]

dw/dx=w'(x)

[abcarin0O0]

fantastic

[diflection1]

@abcarin0O0