2: Forces in Statically Determinate Beams and Plane Frames

2.7: External Reactions in Frames (Part 2)

2.7.4: Reactions in Statically Determinate Frames

We will now look at examples of statically determinate flexural type structures and the calculation of reactions. This is preparation for the next chapter in which we will calculate and plot internal forces throughout these structures. Generally we will have to solve for reactions first before applying the method of sections and solving for internal forces so that this is a skill we must acquire before going on to Chapt. 3. Later we will consider statically indeterminate frames. One basic skill required in the analysis of statically indeterminate structures is the ability to analyze statically determinate ones - the process in fact breaks down into a series of such analyses.

In these, we may often find in the beginning that the calculation of reactions is a challenge. We may sometimes have to try to resist a tendency to see the answer (the wrong one) too quickly. There is also a tendency for it too look easy when the answer is in front of us. So try all of the problems at the end of the this chapter, or if time does not permit, at least sketch the FBD’s and satisy yourself that you do know how to do the numbers when required.

The general procedure is already well established from earlier courses. We either sketch or try to imagine free body diagrams of the structure or parts of it. The strategy will be to establish equations to solve for reactions, preferably uncoupled equations having only one unknown.

Example 2.7.4-1 Solve for Reactions in a Beam

Figure 2.7.4-16: Example 2.7.4-1

An FBD of the whole beam in Fig.2.7.4-16 or an FBD of the structure to the left of the hinge, will make it obvious that we want to take moments (about the hinge) of forces to the left.

Figure 2.7.4-17: Example 2.7.4-1

The forces $V_p$ and $H_p$ in Fig. 2.7.4-17 represent the reaction of the structure to the right of the hinge on the FBD shown. There are no moments at the hinge. We can sum moments and set these equal to zero about either the roller or the hinge.

\[\begin{split} &\sum M_p = 0\\ &-12 \times 6 \times (4 + 3) + V_r \times 4 = 0.\\ &V_r = 126.~~~~~\text{(directed as shown)} \end{split}\]

The sign convention for summing moment vectors in the moment equilibrium equation is clockwise positive. In our later discussions of internal forces we might stop here and begin plotting the bending moment and shear forces directly since these are available throughout by taking FBD’s of the structure with cuts isolating left hand portions. From the solution of $\sum F_x = 0.$, and $\sum F_y = 0.$

\[\begin{split} &V_p = -54.~~~~~\text{(acting opposite direction shown above)}\\ &H_p = 0. \end{split}\]

To complete this problem in reaction solving, and as an exercise, FBD’s of both sections are shown. The reader should now spend some time looking at the complete solution to determine whether the numbers look correct. Check the total applied load; check to see if forces balance in both directions; and check to see if the applied reaction moment is counterbalanced by applied loads. Never assume that, even for simple problems, the job is done when the numbers have been computed.

Figure 2.7.4-18: Example 2.7.4-1

The next example is one variation of a type of statically determinate frame that we will see more of in this text. This is called a simply supported frame.

Example 2.7.4-2: Solve for Reactions in Simply Supported Frame

We will solve for the 3 base reaction forces in the given frame in Fig. 2.7.4-19.

Figure 2.7.4-19: Example 2.7.4-2

The procedure is fairly obvious here since this structure is supported by pin and roller combination. There is only one FBD to be considered and it is shown in Fig. 2.7.4-20 and indicates the general approach.

Figure 2.7.4-20: Example 2.7.4-2

There are 2 unknowns at the pin, and one at the roller. This suggests taking moments about the pin. So we will establish the routine that generally when we have a pin and roller combination, we take moments about the pin. The reader will note that the horizontal reaction is available by inspection and has the value given on the free body diagram, therefore moments could be taken about the right hand reaction. We will save that equation for the calculation of $V_l$ and use the simple $\sum F_y$ equation as a check.

\[\begin{split} &\sum M_l = 0\\ & 40 \times 2.5 + 25 \times 5^2/2 - V_r \times 5 = 0.\\ & Vr = 82.5~\mathrm{kN}\\ &\sum M_r = 0\\ & 40 \times 2.5 + V_l \times 5 - 25 * 5^2/2 = 0.\\ & V_l = 42.5~\text{kN}~~~~~\text{ (as shown)}\\ \\ &\mathrm{check} \sum F_y: 25 * 5 = 125 = 82.5 + 42.5~~~~~\text{O.K.}\\ \end{split}\]

In the second equation for $\sum M_r$ we have combined the two 40 kN forces as a single couple, taking into account the couple’s direction (CW).

2.7.4-2: Three-Hinged Frames

The next example is another generic type of statically determinate frame. This is a three hinged frame, and once again we can adopt routine procedures for the calculation of reactions.

With experience at solving these problems, we start making educated guesses about the directions of unknown forces on free body diagrams. In other words putting these forces on the FBD exactly the way we know they go. Mathematics will always take care of the signs if directions are taken incorrectly. When the signs change from those originally assumed in the FBD, it slows us down by requiring additional FBD’s to show the correct orientation of forces. It also creates confusion, since we will have in our notes the same forces going in different directions. An experienced analyst will not feel comfortable with forces on FBD’s whose direction clearly indicates that equilibrium is not satisfied.

Example 2.7.4-3 Solve for Reactions in Three-Hinged Frame

Figure 2.7.4-21: Example 2.7.4-3

There are four base reactions in Fig. 2.7.4-21, indicating that we must take advantage of the zero moment condition at the internal hinge. Before we do this, however, we note from the FBD in Fig. 2.7.4-22 that the pinned supports at the base align horizontally. This may not always be the case, but we will come across this arrangement often enough for this type of structure that we should watch for it. This provides the condition that the horizontal reactions are in-line and do not provide a contribution to moments about either reaction.

Figure 2.7.4-22: Example 2.7.4-3

The first step in calculating the reactions, therefore, will be to take moments about either the left or right support and calculate the vertical reaction at the other support. Following this we could calculate the other vertical reaction.

Taking moments about the left support:

\[\begin{split} &\sum M_l = 0\\ & 10 \times 8.5 - V_r \times 10 = 0.\\ & V_r = + 8.5~~~~~\text{(plus indicates directed as shown)} \end{split}\]

Now we take moments about the interior hinge. It may be easier to sketch an FBD of the partial structure to the left (or the right) of the interior hinge. After some experience this step can be avoided. Inspection of the previous FBD indicates that Hr is not in the direction shown. We can see this in two ways from the FBD in Fig. 2.7.4-23.

Figure 2.7.4-23: Example 2.7.4-3

First the member is a two-force member, having a pin at each end and no loads between the joints. The directions of components $H_r$ and $V_r$ at the base, if correct, creates a transverse shear, which is not possible. Therefore $H_r$ must be opposite the direction shown. The second way to see this is to imagine moments about the upper hinge. Both forces give rise to the same direction of moment about the top pin which again is not possible for equilibrium.

The complete solution of this three-hinged arch is given below, showing the true directions of all external and internal reactions.

Figure 2.7.4-24: Example 2.7.4-3

Once the forces in the right hand strut have been calculated we transfer these with equal and opposite directions to the point on the other section in accordance with Newton’s Third Law of action/reaction. From this second FBD we can obtain the two remaining lower left reactions using equilibrium equations. An appropriate way to do these calculations is to apply $\sum F_x$ and $\sum F_y$ and then apply a moment equation to confirm the results. As an exercise it is recommended that once again you pause and try some moment calculations about any of the 3 exterior member edges in the second FBD, or take some time to reflect on whether the forces seem to be creating moment equilibrium in the Fig. 2.7.4-24.

2.7.4-3: Propped Cantilever Frames

In this last example, the ability to look at the computed results as presented in the final FBD and determine by inspection whether they make sense is indispensible. The presentation of results for reactions on FBD’s is desirable for 4 reasons. If these are quiz problems being marked, the marker knows exactly what was computed, including directions. Secondly, when we look at diagrams like this, equilibrium errors tend to reveal themselves. Thirdly, Newton’s Third Law of action/reaction must be adhered to and the internal forces at hinges for example clearly shows this. Finally, if the objective of the exercise is moment and shear diagrams, a diagram such as this is an excellent place to start and makes the plotting of these diagrams easier, and having the load/reaction FBD on the same page as the bending moment and shear diagrams shows clearly the mathematical relationships between them.

A final problem in this chapter will introduce a third generic statically determinate frame, the propped cantilever with internal articulation.

Example 2.7.4-4: Solve for Reactions in a Propped Cantilever Frame

Figure 2.7.4-25: Example 2.7.4-4

The condition of zero moment at the internal hinge described earlier in Example 2.7.4-3 is again the key to starting our calculations in this problem. We can do this immediately because the roller in Fig. 2.7.4-25 has a single vertical reaction. In a manner similar to Ex. 2.7.4-1 we can start the analysis with the simple supported portion of the structure.

Figure 2.7.4-26: Example 2.7.4-4

We first calculate the vertical reactions in the simply supported portion and then, sketching both FBD’s as shown in Fig. 2.7.4-26, we take care, as we did in the previous problem, to adhere to Newton’s Third Law of action/reaction. The computations are not complete in this case until, as example, the student has confirmed the direction of the base moment in the left hand FBD and the reversal of direction in the shear at the pin between the right and left segments of the structure.

2.7.4-5: Summary and Advice

Once again the student is strongly encouraged to prepare neat sketches such as the one shown above. Then look at it carefully in order to get an intuitive feel for the subject which goes beyond the crunching of numbers.

Some of the exercise problems at the end of this chapter are variations of the example problems. Watch for these variations. The mastery of these calculations will be important throughout the course, and especially important in the next chapter, and a later chapter dealing with indeterminate structures. Before proceeding the student should feel confident that the material presented to this point is understood.