Solution of a Statically Indeterminate Beam by Superposition

We return to the example of a beam loaded with a triangular load, zero at the left end and maximum of ω₀ at the other, with end conditions fixed against rotation and translation, the `generic' solution for which is not readily available (`in the back of the book'). (I'm not saying it is not in the back of any book - it's just not in the back of ours.) We solved this problem using the Method of Integration (here). We obtained ...

v (x) = ( -0.4167 x⁵ + 125 x³ - 833 x² ) / EI

and

v _max = - 0.014 in.

for the case of w_o = 500 plf, L = 10 ft, and EI = 81,000,000 lb-in.².

Approach

We will develop a solution for this example by `superimposing' solutions that are available, first in a generic sense, and then with the particulars of the specific example at hand. In this example both ends are (in the `end') fixed with respect to both translation and rotation. The principles used, however, can be used for the additional example of one end rotated and displaced, but we will not go through the labor for that additional example here.

The `cases' that are available to us in `the back of the (Gere) book' are:

Case 9: Cantilever beam with triangular loading,

Case 4: Cantilever beam with point load at the far (free) end, and

Case 6: Cantilever beam with applied end moment at far (free) end.

Note that in all three cases we have non-zero (far) end deflections and (far) end angles.

We will superimpose the solution (information) of these three cases with the following conditions:

1. The total displacement (deflection) at the far end is (adds to) zero.

2. The total angle at the far end is (adds to) zero.

In so doing we will `solve' for the vertical reaction at the far end (the `P' in Case 4) and the moment reaction at the far end (the M₀ of Case 6).

Solution

Case 9:

Δ _{(far) end} = 11ω₀ L⁴ / 120 EI

θ _{(far) end} = ω₀ L³ / 8 EI

Case 4:

Δ _end = P L³ / 3 EI

θ _end = P L² / 2 EI

Case 6:

Δ _end = M₀ L² / 2 EI

θ _end = M₀ L / EI

Condition 1 ... all the Δ _end 's add to zero.

(Note: they could add to some initial displacement if we wanted them to.)

11 ω₀ L⁴/120 EI + PL³/3EI + M₀ L²/ 2 EI = 0.

And ...

Condition 2 ... all the θ _end 's add to zero.

ω₀ L³/ 8EI + PL²/ 2 EI + M₀L/ EI = 0.

(And, yes, here also we we could put in a `number' other than zero.)

Solving these two equations simultaneously we get,

P = - 21ω₀L/60, and

M₀ = ω₀L²/20.

The minus sign on the solution for P makes sense, as P is actually the far end vertical reaction, acting up.

The positive M₀ also makes sense, as it indicates tension on top from where we started with Case 6.

Now we know enough to solve for the Reaction, Shear, and Moment Diagrams, by statics.

We can also manufacture the equation for the curved shape by `adding' the curves of the individual cases with the values for P and M₀ from above.

In so doing we get,

From Case 9:

v(x) = - ω₀ x² (20L³ - 10L² + x³ ) /120 L EI

From Case 4:

v(x) = - Px² (3L - x) /6EI ... where P = - 21ω₀L/60

From Case 6:

v(x) = - M₀ x² / 2EI ... where M₀ = ω₀L²/20.

Which gives,

v(x) = ω₀ (-x²L²/60 + x³L/40 - x⁵/120L) /EI.

Sweet!

This is the `generic' solution for the curved shape.

If we dump in our values for ω₀, L, E, and I, we get,

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

Answer

The equation for the curved shape of the beam is,

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

... Yeah, Baby! ... Exactly as before, except that the terms are in reverse order (no big deal).

We could take derivatives of this equation to obtain slope, EI M(x), EI v(x), and even EI ω(x), if we wanted to.

References

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

Mechanics of Materials, 5th Edition, James M. Gere, Brooks/Cole, 2005.