Beam Deflections by Method of Integration: Statically Indeterminate Beams

Now we will look at using the method of integration to `solve' for a statically indeterminate beam. By `solve' I mean `solve for the deflection equation' and maximum deflection, AND ALSO the shear, bending moment, and reactions, for, by definition, the shear, moment, and reactions cannot be determined by `statics' alone. Recall that for the statically determinate beam, we could solve for the reactions, shear, and bending moment prior to the solution of deflections. Now we must bring in what we know of the bent (deflected or curved) shape to solve for them.

Background.

We will, as before, assume that the deflections are small, and thus the slopes of the curved shape are very shallow, and thus the Euler-Bernoulli Beam Equation applies, namely,

- w / EI = d⁴ v / dx^4,,

where,

w is the distributed loading,

E and I are (in this case) constants (the Modulus of Elasticity and the Moment of Inertia of the beam),

v is the amount of deviation (positive upward) of the axis of the beam from the straight (un-bent) shape, and

x is the distance along the beam (starting from the left end).

In this case w is taken to be positive downward (and thus the minus sign in the above equation).

It is assumed that the beam is `flexing' within the `linear elastic range' of the material (no plastic hinges, etc.).

And, for us to `integrate', the load is some algebraic expression in terms of `x'.

Procedure.

We will, as before, integrate the load equation to get shear, shear to get moment, moment to get slope, and slope to get curved shape. Unlike before, however, the boundary conditions on shear and moment are unknown, and will need to be carried along in our integrations as (unknown) constants of integration. These constants of integration will (hopefully) be solved `later in the problem' by using the boundary conditions on slope and displacement.

Example.

As an example let us use the beam with a triangular load starting with 0 plf at one end and ending at 500 plf at the other, with span of 10 ft, except now fixed with respect to displacement and rotation at the ends. We will use a beam with stiffness (EI) = 810,000,000 lb-in². The self weight of the beam will be neglected.

(The solution to this beam simply supported is ... here. In that solution, for example, we determined the maximum deflection to be 0.70 in. at a location 5.2 ft from the left support.)

Solution.

Step 1 - Algebraic Expression for the Load

The load may be expressed as

w(x) = 50 x,

where the 50 has units of plf / ft. For convenience the units will not be carried through all the calculation steps.

Step 2 - Integrate Load to get Shear

V(x) = ∫ (-) w(x) dx + C₁

= - ∫ 50 x dx + C₁

= - 25 x ² + C₁

We do not know the boundary conditions on shear (statically indeterminate beam), so we keep integrating.

Step 3 - Integrate Shear to Get Bending Moment

M(x) = ∫ V(x) dx + C₂

=∫ ( - 25 x ² + C₁) dx + C₂

= - 8.33 x³ + C₁x + C₂

And again we're `stuck' in that we don't know the boundary conditions on Moment, so we keep going.

Step 4 - Integrate Moment to get Slope (actually, EI times Slope)

EI θ (x) = ∫ M(x) dx + C₃

= ∫ (- 8.33 x³ + C₁x + C₂) dx + C₃

= - 2.0833 x⁴ + 0.50 C₁ x² + C₂ x + C₃

We get a bit of relief here in that we can solve for C₃, being the initial slope (in this case zero).

EI θ = 0 (since θ = 0) at x = 0; solving the above equation for C₃ gives C₃ = 0.

Step 5 - Integrate Slope to get Curved Shape

EI v (x) = ∫ EI θ (x) dx + C₄

= ∫ ( - 2.0833 x⁴ + 0.50 C₁ x² + C₂ x ) dx + C₄

= - 0.4167 x⁵ + 0.1667 C₁ x³ + 0.5 C₂ x² + C₄

And we can also solve quickly and easily for C₄ by knowing that the displacement at the left end is zero. Hence, since v = 0 at x = 0, then C₄ = 0.

Our shape of the bent beam is, therefore,

EI v(x) = - 0.4167 x⁵ + 0.1667 C₁ x³ + 0.5 C₂ x²

We still have two constants of integration, presently unknown. However, we can employ our other two boundary conditions, namely, at the other end of the beam the slope and displacement are zero.

Step 6 - Impose additional boundary conditions to solve for remaining unknown constants of integration

For displacement, EI v (x = 10) = 0 gives,

0 = - 0.4167(10)⁵ + 0.1667 C₁ (10) ³ + 0.5 C₂ (10)²

41,670 = 166.7 C₁ + 50 C₂

For slope, EI θ ( x = 10) = 0 gives,

0 = - 2.0833 (10)⁴ + 0.50 C₁ (10)² + C₂ (10)

20,833 = 50 C₁ + 10 C₂

Solving these two equations simultaneously for C₁ and C₂ gives,

C₁ = 750 and C₂ = - 1667

Therefore,

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

All of these terms have units of lb-ft³. (We started with lb/ft and integrated four times with respect to length.)

We could also write,

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

And we could also substitute our value(s) for EI into the equation, and actually calculate displacements. However, first, let's check equilibrium.

Step 7 - Check equilibrium

Substituting the constants of integration back into the equations for Shear and Moment we get,

For Shear,

V(x) = -25 x² + 750.

At x = 0 we get V = 750.

From our study of Shear and Moment diagrams this is also the left hand reaction.

At x = 10 we get,

V(x = 10) = -2500 + 750 = - 1750.

And this is the negative of the reaction at the left end.

Checking force equilibrium,

Σ F_y = 0?

750 - ½ (500) (10) + 1750 = 0?

0 = 0 ... Good!

We might have noticed by now that C₁ is thus the left hand reaction and the starting shear (V₀).

For Moment,

M(x) = - 8.33 x³+ 750 x - 1667

At x = 0 we get,

M(0) = - 1667

Thus, again, we might notice that C₂ is our starting moment, and the left hand moment reaction.

At x = 10 we get,

M(10) = - 2500

And if we check moment equilibrium (not shown), it checks out, also; good!

From the above equations we may now get our critical values for shear and moment, if we want them.

They are:

V_max = 1750 lb at the right support,

M _{positive max} = approx 1070 lb-ft at about x = 5.5 ft, and

M _{neg max} = 2500 lb-ft at the right support.

These critical values can be found by calculus, or by simply dumping in various values of x and examining the values of V and M. (I don't show them all here.)

Step 8 - Solve for the Deflection(s)

Now, with confidence (since everything checks), we can look at the deflection equation.

We get a maximum deflection of about - 6540 lb-ft³ / EI at about x = 5.25 ft.

(Note: so far the units have been in lb and ft.)

Substituting for EI, and watching our units (since EI is in lb-in.²) ...

- v _max = Δ = 6540 lb-ft³ (12 in. / ft)³ / (810,000,000 lb-in.²)

Δ = 0.0140 in.

Note that this is considerably less than the (similar) example with ends that were free to rotate (Δ = 0.070 in.), like I said it would be.

Concluding Remarks

In real life we would need to check to make sure we have not exceeded the shear and moment `strengths' of the beam, and along with this make sure we're still in the elastic range (of behavior).

In this example the ends were fixed ... and were fixed with zero displacement and zero rotation (slope).

In the next example (here) we will deal with one end having non-zero displacement and rotation.

References

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

Beam Deflections by Integration: Statically Indeterminate Beam with One End Displaced and Rotated, Jeff Filler, Associated Content.