Structural Engineering - Deflection

[bbg5000]

I'm not sure the formula to use to calculate the deflection of a beam. Here's the information I have got.

Steel beam is 50 feet long. There is a 6,000 pound load in the middle(all other loads are disregarded. The modulus of elasticity is 29,000,000 psi. Moment of inertia is 850 inches^4. There are simple pin connections at either end.

What would be the formula for calculating this?? :D An answer would be nice to lol

 

[FredGarvin]

The beam deflection for a simply supported beam with a load in the center of the span (neglecting the beam's own mass) is:

\delta = (PL^3)/(48EI)

Where:
\delta = Deflection
P = Load
E = Modulus of Elasticity (Young's Modulus)
I = Moment of Inertia

 

[bbg5000]

The beam deflection for a simply supported beam with a load in the center of the span (neglecting the beam's own mass) is:

\delta = (PL^3)/(48EI)

Where:
\delta = Deflection
P = Load
E = Modulus of Elasticity (Young's Modulus)
I = Moment of Inertia

What's the L mean?! And the 48? Does the length of the beam have no place in the calculation?

So \delta = (PL^3)/(48EI) would be:

6000^3 / 48 x 29,000,000 x 850^4 = 0.000000000297261601 ?

 

[Astronuc]

The equation specified is for the maximum deflection (i.e. at mid point) of this simply supported beam.

L is the beam length between hinges.

48 is a value coming from the solution of the beam equation for this beam, load (pointwise at midpoint) and boundary conditions (hinges).

This problem has been asked in the Homework: College Level forum.

 

[FredGarvin]

My apologies for the omission.

Thanks for the cover Astronuc

 

[95impalaguy]

What's the L mean?! And the 48? Does the length of the beam have no place in the calculation?

So \delta = (PL^3)/(48EI) would be:

6000^3 / 48 x 29,000,000 x 850^4 = 0.000000000297261601 ?

You want your deflection in inches, so your length is 300", not 25'.
The way I read it, it would be (6000 x 300^3) / (48 x 29,000,000 x 850^4)
= (6000 x 27,000,000) / (48 x 29,000,000 x 522,006,250,000)
From there, it's simple math. My kid asked me how to do this. It's a
neopets.com "Lenny Conundrum" puzzle...

 

[noisynik]

deflection

I too am trying to work out the Lenny Conundrum. The length is 50 feet, so 600 inches
And after using all those calculations, I came up with a very incorrect answer! 1.78 inches as the deflection? I don't think I'm doing this correctly!

 

[95impalaguy]

The weight is at the center of the 50 foot beam, not the end...

 

[Astronuc]

There seems to be some confusion regarding the moment of inertia (850^4)as reflected by the use in the forumula above.

FredGarvin posted the equation for the maximum deflection for this problem, which so happens to be at the location of the load, the midpoint of the beam.

P = 6000 lbf
L = 50 ft = 600 in
E = 29,000,000 psi (lbf/in²)
I = 850 in⁴

units must be consistent.

So sustituting the values into the equation above

\delta = (6000\,psi * (600\,in)^3)/(48*29,000,000\,lbf/in^2 * 850\,in^4) = 1.0953 in

To the units should cancel with the result that the deflection \delta is given in units of 'in' or inches.