Does the equation δ = (PL^3)/(48EI) apply only to maximum beam deflection?

[stoner]

regarding this equation,
\delta = (PL^3)/(48EI)
does it only apply for maximum deflection, or will it give me the corresponding beam deflection with any given load?

if not what other equations are there?

thanks.

 

[FredGarvin]

That is the max deflection of a simply supported beam with a concentrated load at the mid span. It will always be the max deflection which is a function of the load applied. Apply a different load to the same beam, you'll get a different deflection. Makes sense, don't it?

 

[stoner]

ok, if i apply a different load on the beam which doesn't give the maximum deflection, can i still substitute it into the formula and get the corresponding deflection?

 

[PhanthomJay]

ok, if i apply a different load on the beam which doesn't give the maximum deflection, can i still substitute it into the formula and get the corresponding deflection?

As noted, this is the deflection of the beam at midpoint under a concentrated load P applied at midpoint. It is the maximum deflection along the beam length. Beam deflection at the end supports will be 0, and somewhere in between the min and max at other locations between the end and midpoint. Double the load, and deflections at any given point along the beam wil also double.
Now if you apply the same load but not at midpoint, say 1/4 of the way along the beam, the maximum deflection in the beam occurs at the point of the load. It is something different than pL^3/48EI, I don't know what it is without looking it up in a handbook, or doing the calculus involved.
Put a concentrated load at the end of a cantilever beam, and the deflection is max at the end of the beam where the load is applied, and equal to PL^3/3EI. Is your question being answered?

 

[stoner]

sorry for not being clear...here is my more specific question,
if i am only interested in applying a point load at the centre of the beam, can i use the formula to calculate deflection (not maximum) for any given load, for that particular beam?

 

[PhanthomJay]

sorry for not being clear...here is my more specific question,
if i am only interested in applying a point load at the centre of the beam, can i use the formula to calculate deflection (not maximum) for any given load, for that particular beam?

for constant EI, the 2nd derivative of the deflection curve times EI is equal to the bending moment at any point. So for a beam of length L with a concentrated load P at midpoint, you must calculate the moment at any point as a function of x, divide it by EI, then integrate it twice to get the deflection curve equation, which is, simply,

y=P/48EI(4x^3 -3L^2x - 8(x-L/2)^3)
(I had to look it up).
Note that at midpoint, under the load, where x = L/2, the deflection is the familiar PL^3/48EI, which can be shown to be the maximum deflection.
Is this what you are looking for?

 

[stoner]

fthe deflection is the familiar PL^3/48EI, which can be shown to be the maximum deflection.
Is this what you are looking for?

lets say i have a beam which gives me a maximum deflection 5 mm when applied 10N. if i were to insert a value of 2 N into the above equation, will i be able to get the corresponding deflection? or is the formula only true for max and will not work?

(for a constant E)

 

[PhanthomJay]

lets say i have a beam which gives me a maximum deflection 5 mm when applied 10N. if i were to insert a value of 2 N into the above equation, will i be able to get the corresponding deflection? or is the formula only true for max and will not work?

(for a constant E)

Oh heck sure if it deflects 5mm under a 10N force, it will deflect 2/10 of that (1mm) under a 2N force. The formula for deflection is linear with respect to P. The given formula gives the deflection at midpoint for any load P applied at midpoint. Plug in whatever P you are given and chug out the deflection. When we say maximum deflection, we mean the beam deflects the greatest at its center, less toward the ends. the formula calculates that max deflection only, which can be any amount depending on the value of P you put in.

 

[stoner]

ahh ok...thanks