Objectives:

When you finish this module (especially the appropriate textbook sections and problems), you will be able to:
· use geometric properties of the elastic curve to compute the deflection & slope of a beam at a point
· compute slope & deflection using the equation of the elastic curve
· use singularity functions to determine slope & deflection
· compute slope and deflection using the method of superposition
· use the moment-area theorems with cantilever beams & beams with unsymmetric loadings

Reference: §9.1 - 9.14 of Mechanics of Materials, 3rd Edition, Beer, Johnston, & DeWolf, McGraw-Hill

Radius of curvature: A prismatic beam under pure bending has a neutral surface curvature given by (see §4.5, page 190)

Deflection and Slope: Since the bending moment M is a function of x [M(x)], as you have seen in an earlier module, the radius of curvature also varies. From calculus (see section 9.3, page 533) and assuming that dy/dx is very small (it had better be or our beam has serious problems!), the above equation can be shown to be a second order differential equation:

where y is being used a little differently here than earlier. Here, y is a vertical displacement as shown in Figure 9.6b (page 533), whereas in earlier modules y represented the distance from the neutral axis. The two may be the same, but not necessarily. EI is known as the flexural rigidity; it may vary as a function of x if the beam is non-prismatic. However, since this module will deal only with prismatic beams, EI is a constant. Separating variables and integrating once,

This can be integrated once more to obtain y(x) (and a second constant of integration C₂). and y(x) (the slope - or angle - and deflection) are often obtained in this way. The constants of integration are determined from boundary conditions. For example, a cantilever beam (supported at the right end) has a deflection y of zero and a slope (or angle) of zero at the right end. These two values determine the constants C₁ and C₂. A simply supported beam (such as the one shown in Figure 9.8a page 534) has deflections of zero at the two ends (but not slopes of zero there) and a slope of zero at the center. These three conditions are more than sufficient to determine the two constants of integration. You should study Example 9.01 through 9.03 (pages 535-537).

Slope and Deflection from the Load: It is often useful to obtain the slope and deflection directly from the load, rather than in terms of the bending moment. Taking successive derivatives, I get

This can be integrated four times (as shown in equations 9.33 on page 538); four constants of integration must be determined by the boundary conditions. Since two of these are usually the shear and bending moment at the beam's end, it is useful to remember that

In Example 9.04 (page 539) show how easily the constants of integration can be obtained. Rather than use the shear at one end (wL/2) to get C₁, your text uses the simpler conditions that the moment is zero at x = 0 and at x = L. And for the third integration, the fact that the slope is zero at L/2 could have been used. However, it's easier to integrate once more and use the fact that the deflection y is zero at x = 0 and x = L to get both constants C₃ and C₄. Keep this in mind - there are usually several boundary conditions that can be used. Take the easiest path to the objective!

Statically Indeterminate Beams: The free body diagrams of beams are essentially two-dimensional, and so statics provides two force equations and one moment equation:

As you can see from Figure 9.24b (page 540), there are four reactions and the beam is statically indeterminate. We cannot determine all of the reactions using (only) the three equations from statics (above).

But you have seen that we have four more equations (equations 9.33 page 538). While four constants come along with these equations, you have also seen that we usually have a number of boundary conditions (certainly more than four). For example, for the beam shown in Fig. 9.24a, we know (or can easily find) the shear at A, the bending moment at B (zero), the deflections at A and B (zero), and the slope at A (zero). So we have four reactions and four constants of integration (8 unknowns), but the abundance of boundary conditions permits us to determine all of them. I suggest you study the sample problems on pages 542 through 544.

Singularity Functions: The integration method is useful as long as the bending moment can be expressed by a single analytical function. When a concentrated moment exists as in Example 9.06 (page 551), different functions would have to be written for each section of the beam. Singularity functions simplify this by permitting us to write a single function for the entire beam. At the bottom of page 551, the shear V has been written in terms of singularity functions. Before proceeding, make sure you understand how each term has been obtained! As I mentioned in the previous module, since the distributed load does not extend to the right end of the beam, we have to redraw the beam using the superposition of two distributed loads which do extend tot he right end.

The shear can be integrated to obtain the bending moment. Note especially how the concentrated moment of 1.44 kN-m has been included. Note also that the bending moment is satisfied at x = 0, where it is zero (as it should be for a simple support at A). Now check it at x = 3.6, where the bending moment again has to be zero. Thus, you know that the constant of integration has been included. You should follow through the rest of the example (with a pencil - do it, don't simply read it).

Take a look at Example 1.

Method of Superposition: Beams that are subjected to several concentrated and/or distributed loads may be analyzed using superposition. For example (see Figure 9.34, page 559), the 150 kN concentrated load and the 20 kN/m distributed load can be separated (Figures 9.34b and c) as ifthey were acting separately. Since each of these cases were already solved in Examples 9.02 and 9.03 (pages 535 & 536), we use those formulas to compute the slopes and deflections and add them! Appendix D shows many of the commonly encountered cases so that practicing engineers do not need to recompute the slope and deflection for these.

Example 2 demonstrates the use of superposition with statically indeterminate beams.

Moment-Area Theorems: On pages 569 - 571, your textbook derives these relationships:

For an understanding of these, please refer to the Figures on pages 569 - 570. I'll provide a summary here, but the book does it in detail.
First in Fig. 9.41b, M/EI vs x is plotted because M/EI is the rate of change of the deflection angle . Thus you can see that the area between C and D is the change in the deflection angle between those two points on the beam. The subscript can be read as "D relative to C"; that deflection change is positive as shown in Figures 9.41c and 9.41d, since is larger than . By the way, the text's notation means this:

As for the second theorem, you can read through the derivation. Looking at Figure 9.45a, the tangential deviation t_C/D is a representation of how much the tangent line drawn to D deviates from point C; that is, the deviation of point C relative to the tangent drawn to point D (C/D). The derivation shows that the value of this is the area times the distance from C of the centroid of the area. A point with a positive tangential deviation is located above the corresponding tangent, as the example shows. Note that t_C/D is NOT necessarily equal to t_D/C. While the areas are the same, the centroidal distances are not usually the same.

The process is not difficult, as shown by Example 9.09 on page 572 and summarized below:

1. Draw a free body diagram (Fig. 9.49a)
2. Draw a bending moment diagram M vs x (Fig. 9.49b)
3. Draw the graph of M/EI vs x (Fig. 9.50)
4. Use the details from this graph to compute the slope and deflection

The area (of the M/EI vs x plot) from A to B is the sum of negative and positive quantities and is 4.5 x 10^-3 rad. Since the boundary condition at A demands no slope there, this value is also the slope at B, .

The tangential deviation t_B/A equals the first moment of the area about a vertical line through B (notice that the axis always refers to the first letter named in the subscript). The centroidal positions are obtained using the usual one-third rule (for the triangle from A to D, it's 1.2/3 = 0.4 m from A, or 0.8 m from D). Thus, your text obtains t_B/A = -4.50 mm. Now t_B/A = t_B - t_A and, since the tangent at A is horizontal (zero), the deflection y_B = -4.50 mm.

Examples 9.10 and 9.11 (pages 574 - 575) show how to use a superposition method so that the resulting M/EI vs x plots involve geometric figures for which you can use the centroids from tables.  Here's an example using the second theorem and singularity functions, as well as an approximate method for solving a cubic equation.

Unsymmetric Loadings: When the slope of the tangent is zero at some point, if the slope and deflection are calculated relative to that point, the above method is quite easy. However, it is often impossible to determine from inspection at which point the slope of the tangent is zero (for example, see Figure 9.59a, page 582). But it's still pretty easy! From Figure 9.59b, and since the angles are so small that the tangent of the angle is essentially the same as the angle,

Since can be computed using the first equation, and since can be found from the area under the M/EI plot, can be found from the second equation.

As for computing deflections from the tangential deviations, consider Figure 9.63. From similar triangles, your textbook obtains an expression for EF in terms of t_B/A. And in equation 9.63 the authors show how to get the deflection y_D from EF.

Here's an example of a case with unsymmetric loadings.  It's a bit unrealistic, since the beam doesn't weigh anything!  Oh, well!
Here's another example where a force has to be moved, resulting in a moment.