Example Solutions of Indeterminate Beams Using Superposition

A Lesson in Mechanics of Materials

(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 Δ_ω.

Simple Beam Point Load at Center

Δ = PL³ / 48EI ... where, again, L = 24 ft ... call it Δ_P.

We now add the two as follows:

Δ_ω = Δ_P + 1.00 in.

(We can think of it like this: the uniform load causes it to sag the amount Δ_ω and then the center support acts like a jack, `jacking it back up' and amount Δ_P 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 ...