Generic Results of Solving for the Deflection of a Statically Determinate Beam

Okay, in the above example (here) we solved for the deflection of a specific beam (specific material, support conditions, length, and load distribution). It was more interesting (in my opinion), but limiting. Of value (one might argue) would be a more generic equation giving us the deflection (especially the maximum deflection) for, say, any beam subject to the same kind of loading (triangular) and support conditions (free to rotate at both ends). Yeah, that would be cool, and (more) useful. In fact, there are pages in books with such `generic' solutions. And it is probably published somewhere, but let's pretend it (still) is not, and come up with the solution ourselves.

Procedure

There are really two main ways we can approach this. The first would be to go back to our starting place and make our beam `generic'. In other words, leave the length `L', the maximum load at the right end w_o, E as E, and I as I. And then solve, leaving all this stuff as variables. We would come up with `expressions' for the Shear, Moment, Slope, Curved Shape ... in terms of w<sub>o</sub>, L, E, I. It would be a good exercise. But there is another way ... kind of going `backward' from the result of a particular solution, like the one we have just finished. It is this latter approach I will employ here.

The Backward Generic Approach

From experience (or Dimensional Analysis) I propose that the answer to the maximum deflection equation will be of the form ...

... Δ = C w_o L⁴ / EI ... or ... C W L³ / E I

where

C is a dimensionless `constant' that involves (is determined by) the support conditions and how the load (either w_o or W) is distributed ... and will be different depending on whether we go the w_o or the W route;

... wo is the maximum value of distributed load in this case at the far (right) end;

(or ...) W is the `whole' load (in this case ½ w_o L);

L of course is the Length of the beam (strictly speaking the distance between the supports);

E is the Modulus of Elasticity, and

I is the Moment of Inertia of the beam.

If we did a Dimensional Analysis on the problem, we would probably come up with something like this ...

... (Δ/L) = C (w_o / LE) (L⁴ / I) ... (for the first option).

Solving for the Coefficient C

Now let's do it.

From before we obtained Δ = 0.70 in. (actually 0.696 in.) with w_o = 500 lb/ft, L = 10 ft, and EI = 81,000,000 lb-in.².

... Δ = C w_o L⁴ / EI ...

C = ... Δ E I / w_o L⁴ ...

... 0.696 in. (81,000,000 lb-in.²) / [(500/12 lb/in.)(120 in.)⁴] =

... 0.006525.

Let's check!

... Δ = C w_o L⁴ / EI = 0.00653 (500/12 lb/in.) (120 in.)⁴ / 81,000,000 lb-in.² = ...

... 0.696 ... SWEET!

If we had used the option of casting our equation in terms of W ...

W = ½ w_o L = ½ 500 lb/ft (10 ft) = 2500 lb.

C = Δ E I / W L³ ...

C = 0.696 in. (81,000,000 lb-in.²) / [(2500 lb)(120 in.)³] =

... = 0.01305.

Check ...

Δ = C W L³ / EI = 0.01305 (2500 lb)(120 in.)³ / 81,000,000 lb-in.² =

... 0.696 in. ... good!

SO,

For these particular loading and support conditions ...

... Δ = 0.0065 w_o L⁴ / EI = 0.0130 W L³ / EI,

where

W = ½ w_o L.

(And these are probably published somewhere. I am not pretending to be the inventor of the solution to this particular problem - the solution is illustrative. We can do this with essentially any loading-support condition.)

Concluding Remarks

In the above example we have dealt with the load as expressed in terms of (in this case maximum) distributed load (w_o), and the `whole' load (W). Sometimes the distributed load will be `q' ... especially in more `academic' stuff. It seems that real practitioners in this stuff, and especially over in the Architect camp, use the `W' form, while in academic circles and some engineering circles (or books), the `w' is used. Probably you should be comfortable with both, and certainly not get the two confused! The one piece of advice to give to people starting out in this stuff is to ... ALWAYS follow along with your units. If you mistake w for W your units won't come out right, and if your units are wrong, it is almost certain that your NUMBER is wrong also!

Update

Ah, yes ... In the Timber Construction Manual, 5^th ed., page 433, Case 2 ... there it is ... the 0.0130 with the `W' form expression.

And in the back of our Gere Text ... Page 896, Case 11 ... the 0.00652 ... with the `q' form of the equation.

References

Deflection of a Statically Determinate Beam by Integration, Jeff Filler, Associated Content.

Timber Construction Manual, 5^th Edition, American Institute of Timber Construction, Wiley, 2005.

Mechanics of Materials, 5^th Edition, James M. Gere, Brooks/Cole, 2001.