(Updated 30 May 2009)
Introduction
Now, finally, we look at two more examples of using the method of Superposition for solving statically indeterminate beams. And these are two pretty cool examples. The first is the example of a two-span continuous beam with one of the supports not in alignment with the others. This first example is essentially the same as that of a cambered beam (that wasn't supposed to be cambered) placed over supports that are aligned. The second is one beam placed across another beam.
First Example
Consider a continuous beam placed over two spans of 12 ft each. The beam has a stiffness, EI, of 2200,000 lb-in.2. The center support is 1.00 in. low. The beam is loaded with a uniformly distributed load of 600 plf. We will disregard self-weight (since we don't know what it is), or assume that it is small.
Approach
Assuming that the beam deflects enough to come in contact with the center support (which we will check) the example becomes one of a simple beam spanning between the far ends loaded uniformly, and a (similar) simple beam spanning between the far ends loaded upward with a point load at center such that the net deflection at the center (center support) is 1.00 in.
Solution
Simple Beam Uniform Load
Δ = (5/384) ω L4 / EI … where L = 24 ft (from far end to far end) … call it Δω.
Simple Beam Point Load at Center
Δ = PL3 / 48EI … where, again, L = 24 ft … call it ΔP.
We now add the two as follows:
Δω = ΔP + 1.00 in.
(We can think of it like this: the uniform load causes it to sag the amount Δω and then the center support acts like a jack, `jacking it back up' and amount ΔP to a point 1.00 in. less than the original (unbent) condition.)
Dumping in our values and watching the units,
(5/384) (600 /12) lb/in.(24*12 in.)4 / 2200,000,000 lb-in.2 =
P (24*12 in.)3 / (48*2200,000,000 lb-in.2) + 1.00 in.
2.036 in. = 0.0002262 in. P /lb + 1.00 in.
Gives,
P = 4579 lb.
This value is the center support reaction, acting up.
With P now known we can determine the rest of the reactions, the shear, and the bending moment by statics.
We can also get a curve for the bent shape …
From the `uniform load' part …
v(x) = – ω x (L3 – 2 L x2 + x3) / 24EI
= – 600 x (243 – 2 (24) x2 + x3) / 24EI
= (- 345,600 x + 1200 x3 – 25 x4) / EI
From the `point load' part (acting up) …
v(x) = – Px (3L2 – 4×2)/ 48EI … where P = – 4579 lb.
= 4579 x [3(24)2 - 4 x2] / 48EI
= (164,844 x – 382 x3 ) / EI
Adding,
v(x) = (-180,756 x + 818 x3 – 25 x4 ) /EI …
And this is valid for x from 0 to 12 ft …
Plotting it up, we obtain the following, and, note, it checks, since v (x = 12ft) = -1.00 in.
@ x = 0 ft we get v = 0.00 in.
@ x = 1 ft we get v = – 0.14 in.
@ x = 2 ft we get v = – 0.28 in.
@ x = 3 ft we get v = – 0.41 in.
@ x = 4 ft we get v = – 0.53 in.
@ x = 5 ft we get v = – 0.64 in.
@ x = 6 ft we get v = – 0.72 in.
@ x = 7 ft we get v = – 0.82 in.
@ x = 8 ft we get v = – 0.89 in.
@ x = 9 ft we get v = – 0.94 in.
@ x = 10 ft we get v = – 0.97 in.
@ x = 11 ft we get v = – 0.99 in.
@ x = 12 ft we get v = – 1.00 in. … Checks!
YEAH!
Second Example
Now, finally, for our next and final example.
Instead of a support at mid-span, let's place another beam, say, 12 ft long, perpendicular to the first, supporting the first at its midspan, and with this first beam bearing at the midspan of the additional beam. Assume no initial displacement, that the beams are light compared to the applied load, and when the load is applied to the upper beam, both beams will now deflect. For illustration let's use EI = 2200,000,000 lb-in.2 for the second beam as well.
Approach
The upper beam will deflect due to the uniform load, and be somewhat supported (jacked up) by the lower. So, similar to the first example we will have a total deflection of the upper beam of …
Δ upper = Δupper due to ω – Δ upper due to P
where P is the load of the lower upward at midspan of the upper.
This total deflection will be equal to the deflection of the lower, which is due to the equal and opposite P acting downward,
Δ lower = Δ lower due to P
and these can be set equal.
Therefore,
Δupper due to ω – Δ upper due to P = Δ lower due to P
We'll solve the above equation for P (the reaction between the two beams), and then solve for deflection, or whatever else we want to solve for.
(5/384) ω L4 / EI – PL3 / 48EI = PL3 / 48 EI.
In this example we let the EI of the two beams be the same, so at least for now the EI vanishes from our calcs. But note that the L's for the two beams are different.
So,
(5/384)(600)(24)4 – P (24)3/48 = P (12)3/ 48.
Solving for P gives 8000 lb.
Let's check.
The deflection due to the 600 plf uniform load on the upper beam only is 2.04 in. downward (calcs not shown).
The deflection due to 8000 lb acting upward at the midspan of the upper beam is, 1.81 in. (see Note below).
Hence the net deflection of the midspan of the upper beam is 2.04 – 1.81 = 0.23 in.
We can check this against the deflection of the lower beam due to a point load of 8000 lb at mid-span, not forgetting that we have used 12 ft for the span of this beam.
Δ = PL3 / 48EI = 8000 lb (12*12 in.)3 / (48 * 2200,000,000 lb-in.2) = 0.23 in.
What is interesting is how the total load on the beam `splits out' between the two beams. If we consider a total load of 600 plf * 24 ft we get W = 14,400 lb. The shorter beam carries 8000 of the 14,400 or 56%, while the upper beam carries 44%. The shorter beam is overall `stiffer' due to it being shorter, and `load goes to stiffness'.
Note: the deflection of 1.81 in. for the 8000 lb load is obtained as follows:
In the first example the load of 4579 lb made up the difference between 2.036 and 1.00 in. or 1.036 in. Considering `linear behavior' a load of 8000 lb acting on the same beam in the same place would generate a deflection of 8000/4579'ths as much.
Hence, 1.036 (8000/4579) = 1.81.
Once we understand superposition, we are UNSTOPPABLE!
Closing Remarks
I said we would check the 1.00 in. gap thing. Indeed, under the action of the uniform load, unhindered, the beam deflects more than the 1.00 in., and thus `closes the gap'. Had that not been the case, we would need to look at the actual conditions of the support. If there was indeed just a gap, and under the action of the load the deflection is less than the gap, then the beam never `touches', and it is as though the center support is not even there. On the other hand, if, during installation, the beam is forced to come in contact, such force would (need to) be downward … and actually the beam would be pulling up on the center support. (It would be in tension.) The same would be the case if the supports were originally aligned but the center support settled. If it settles less than the deflection due to load, the support still provides upward reaction; if it settles more, then it pulls the beam down with it.
References
Mechanics of Materials, Fifth Edition, James M. Gere, Brooks/Cole, 2005.
