Generic Solution for a Fixed Fixed Beam with Triangular Load

In earlier lessons we found the deflection (equation and maximum value) for a beam fixed at both ends and subject to a triangular load from one end to the other. We used the method of Integration (here), and Superposition (here). We solved for a particular case (beam length 10 ft, distributed load varying from zero to a maximum of 500 plf, and EI = 81,000,000 psi). Suppose now we want a general (generic) solution for the same essential problem (any beam fixed at both ends with triangular load). There are two ways we can go about getting such a solution. One would be to re-do any of our solutions from before, using `variables' instead of specific values. (Actually, solving in terms of `variables' is how we did the Superposition example.) The other, which I like, since it derives from what we have already done, is to take our specific solution and make it generic. I have already illustrated this for the case of a `simple' beam with triangular load (here). But I wanna do it again, here, for a couple reasons. First, because it's fun, and, second, I want to also illustrate getting the Reaction, Shear, and Moment values. Recall that in our simple beams the Reactions, Shears, and Moments can be determined by Statics (and so in the present sense are less interesting). Let's do it.

Deflection Equation and Maximum

From before (here or here) our deflection equation is ...

v(x) = ( - 0.4167 x⁵ + 125 x³ - 833 x² ) / EI ... units of lb-ft³.

I guess in some sense this is partly generic in that we left it in terms of EI.

We then went on to find the maximum deflection of ... Δ = 6542 lb-ft³ / EI ... at x = 5.25 ft.

And, dumping EI = 81,000,000 lb-in.² in ... we get (got) ... 0.14 in. (0.1396 in.).

Generic Expression for Maximum Deflection

Now let's make this generic (following our other example, here) ...

Δ = C w_o L⁴ / EI ...

C = Δ EI / w_o L⁴ = 0.1396 in. (81,000,000 lb-in.²) / [(500/12 lb/in.)(120 in.)⁴] =

C = 0.00131 ...

So,

Δ = 0.00131 w_o L⁴ / EI ... cool!

Generic Expressions for Reactions, Shears, and Moments

Back to our solution by Integration we had for our Shear ...

V(x) = - 25 x² + C₁ ... and then we found C₁ = 750.

So,

V(x) = - 25 x² + 750.

We dump in x = 0 and we get V = 750 lb at the left end.

We dump in x = 10 and we get V = -1750 lb at the right end.

Making them `generic' ...

V(x=0)/(w_o L) = 750 lb / [(500 lb/ft)(10 ft)] = 0.15 ...

So, V _{left end} = 0.15 w_o L ...

Similarly,

V_{right end} = - 0.35 w_o L.

From what we know about Shear and Moment Diagrams ... our Reactions are ...

R _{left end} = 0.15 w_o L ... and ...

R _{right end} = 0.35 w_o L.

Here it would be fun to do a force equilibrium check.

The whole downward load is ½ w_o L ...

Σ F_y = 0 ???

Let's see ...

0.15 w_o L - 0.5 w_o L + 0.35 w_o L = 0? ... YES!

Moments ...

Our Moment equation is (was) ...

M(x) = -8.33 x³ + C₁ x + C₂.

C₁ we have already said we found to be 750 ... and also in that solution we found C₂ ... C₂= -1667.

M(x) = -8.33 x³ + 750 x - 1667.

M _{left end} (we found to be) = - 1667 lb-ft.

M _{right end} ... - 2500 lb-ft.

And we also found a maximum positive moment of 1070 lb-ft at about x = 5.5 ft.

And now let's make these generic ... (Dimensional Analysis ... ha, ha! ... you'll get the hang of it!)

M _{left end} / (w_o L²) = - 1667 lb-ft / [(500 lb/ft)(10 ft)²] = - 0.0333 ...

So,

M _{left end} = - 0.0333 w_o L².

M _{right end} = (2500/1667) (-0.0333) (w_o L²) = - 0.050 w_o L².

And, similarly ...

M _{max pos} = + 0.0214 w_o L².

Let's finally get our deflection equation (curved shape) completely generic ...

v(x) = ( - 0.4167 x⁵ + 125 x³ - 833 x² ) / EI ... = `C' w_o L⁴ / EI ...

`C' = [ ( - 0.4167 x⁵ + 125 x³ - 833 x² ) / EI ] / ( w_o L⁴ / EI ) ... where I put the C in single quotes because it is not necessarily (or necessarily not) equal to any previous C's ...

`C' = [ - 0.4167 (x/L)⁵ L⁵ + 125 (x/L)³ L³ - 833 (x/L)² L² ] / ( w_o L⁴ ) ... where see how I am multiplying and dividing by L to get the distance term dimensionless ...

`C' = [- 0.4167 (x/L)⁵ L⁵ + 125 (x/L)³ L³ - 833 (x/L)² L² ] / ( w_o L⁴ ) = ...

... = - 0.4167 (100,000) (x/L)⁵ + 125 (1000) (x/L)³ - 833(100)(x/L)² ] / [(500)(10,000)] ...

Fierce!!!

... = - 0.00833 (x/L)⁵ + 0.025 (x/L)³ - 0.0167 (x/L)²

So,

... v(x) = [ - 0.00833 (x/L)⁵ + 0.025 (x/L)³ - 0.0167 (x/L)² ] w_o L⁴ / EI.

Absolutely CRAZY!!!

Obviously, let's check.

Let's dump in x = 5 ft ... so x/L = 0.50 ...

... v(x=5) = - 0.139 in. ...

That's what we got last semester ...

YE-HAH!!!

Summary

Total load ...

R = W = ½ w_o L

Reactions ...

R _left = 0.15 w_o L

R _right = 0.35 w_o L

Shears

V _{left end} = R _left = 0.15 w_o L

V _{right end} = - R _right = - 0.35 w_o L ... and is the Maximum absolute value Shear, (-) V _max

Moment

M _{left end} = - 0.0333 w_o L² ... negative ... or `tension on top'

M _{right end} = - 0.050 w_o L² ... negative ... and also the maximum absolute value Moment, (-) M _max.

And, ...

M _{max pos} = + 0.0214 w_o L².

And, ...

... v(x) = [ - 0.00833 (x/L)⁵ + 0.025 (x/L)³ - 0.0167 (x/L)² ] w_o L⁴ / EI ...

And,

Δ = 0.00131 w_o L⁴ / EI ... or ... Δ / L = 0.00131 w_o L³ / EI

Concluding Remark(s)

So, we could use these for any beam subject to the same end and loading conditions (with constant EI). Maybe these should be published in a book. Maybe they already ARE!

References

Deflections of Beams by Integration: Statically Indeterminate Beams, Jeff Filler, Associated Content.

Solution of a Statically Indeterminate Beam by Superposition, Jeff Filler, Associated Content.

Generic Results of Solving for the Deflection of a Statically Determinate Beam, Jeff Filler, Associated Content.