2: Forces in Statically Determinate Beams and Plane Frames

2.6: Reactions in Beams - Examples

Example 2.6-1:

Figure 2.3-1 shows a structural model of a beam with three supports and an internal hinge.

Fig. 2.3-1: Beam Example 1

In order to analyze it, it must be both stable and statically determinate. We can tell by inspection that it is stable (because it is not possible to find a mechanism, chiefly because member b-c-d-e is constrained to remain straight and without displacement). Counting the unknowns and equations, we have m=3, r=4, j=4, and c=1, so 3m+r=13, and 3j+c=13. These are equal, so the structure is statically determinate as well. We can now proceed to draw the free body diagram of Fig. 2.3-2.

Fig. 2.3-2: Free Body Diagram of Complete Structure

We see there are 4 unknowns, but only 3 equations of equilibrium, so we cannot solve for all unknowns on this FBD. We can solve for one of them, as the resulting equation for $\sum F_x$ has only one unknown:

\[\begin{split} \sum F_x &= 0~~~~{\text{+}}\rightarrow\\ H_c + 0 &= 0\\ H_c &= 0 \end{split}\]

However, that is all we can do. We have three unknowns remaining, but only two equations.

You may be able to "tell by inspection" that $H_c=0$, but that does not help you very much. You use an equation of equilibrium to determine that, and you now have only two remaining.

Obviously, we must draw a different free body diagram. The structure has a release condition, the internal hinge, so we can break the structure there and draw an FBD of the right portion, as shown in by FBD-2 in Fig. 2.3-3.

Fig. 2.3-3: Free Body Diagram of Right Portion

Unfortunately, we can’t solve for much here, either. $H_c$ is known to be 0 from FBD-1, leaving us with 4 unknowns, but only three equations. We could solve for $H_b=0$, but that is all. We have to look at the other portion of the structure, FBD-3 as shown in Fig. 2.3-4.

Fig. 2.3-4: Free Body Diagram of Left Portion

Notice, that we had already chosen directions for $H_b$ and $V_b$ on the previous FBD; we now must now show those same internal forces acting equally and oppositely on the new FBD. Internal forces always occur in equal and opposite pairs.

When you draw two FBDs that share a contact point, all forces at the contact must be shown to be of equal magnitude and opposite direction on each of the FBDs. That is best done by using the same symbols, and showing the arrows in opposite directions.

Using FBD-3, we can solve for all three of the unknown forces acting on it:

\[\begin{split} \sum F_x &= 0~~~~~\text+ \rightarrow\\ &= -H_b + 0\\ H_b &= 0\\ \\ \sum M_b &=0~~~~~\text{+CCW}\\ &= -V_a\times6 + 20\times6\times{6\over2}\\ V_a &= 60~~~~~(\therefore \uparrow)\\ \\ \sum F_y &= 0~~~~~\text+\uparrow\\ &= V_a - 20\times6 - V_b\\ &= 60 - 120 - V_b = 0\\ V_b &= -60~~~~~(\therefore\uparrow) \end{split}\]

We notice that $V_b$ acts upwards on b, not downwards as we had assumed (thats about the only significance of the minus sign in the value computed for $V_b$ – it just tells us we drew the wrong direction on the FBD).

Negative answers from solving equilibrium equations tell you that you had the unknown forces shown in the wrong direction on the FBD.

Knowing $V_a$ (and $H_a$), we can now return to FBD-2, there are now only three unknowns and we can solve them all:

\[\begin{split} \sum M_c &= 0~~~~~\text{+CCW}\\ &= -V_b\times4 + 20\times6\times(4-{6\over2}) + V_d\times6 - 50\times8\\ &=-(-60)\times4 + 120 + 6V_d - 400\\ 6V_d &= 40\\ V_d &= 6.67~~~~~(\therefore\uparrow)\\ \end{split}\]

Note that we very carefully first wrote equation using the direction of the forces as shown on FBD-1, then substituted in the correct numerical values, including sign. This reduces the likelihood of getting the signs wrong.

Minimize the number of steps you 'do in your head'. Write equilibrium equations with signs derived from the direction of forces shown on the FBD, then substitue numerical values for the knowns, with correct signs.

We can use a second moment equation to get the other vertical reaction:

\[\begin{split} \sum M_d &= 0~~~~~\text{+CCW}\\ &= -V_b\times10 + 20\times6\times(10-3) - V_c\times6 - 50\times2\\ &= -(-60)\times10 + 840 - 6 V_c - 100\\ V_c &= 223.33~~~~~(\therefore\uparrow) \end{split}\]

To complete this portion, we redraw the FBD of the whole structure, showing all forces in their proper directions, and rounding numerical values to 3 significant digits (with the largest value setting the number of decimal points). This is seen in Fig. 2.3-5.

Fig. 2.3-5: Result Free Body Diagram

Summary

All of the diagrams above are repeated on a single drawing here for ease of reading:

Fig. 2.3-6: Summary Free Body Diagrams

Example 2.3-2:

Figure 2.3-6 shows Example 2:

Fig. 2.3-6: Beam Example 2

We can quickly tell that it is stable and statically determinate, and that we can solve for the reaction at b if we draw a free body diagram of portion a-b-c-d, as shown in Fig. 2.3-7:

Fig. 2.3-7: Free Body of Left Portion

Note that there is a 40kN load applied at point d. For simplicity we did not show it on FBD-1, above. However, that means if we draw a free body of portion d-e, we must show the 40kN load acting on that one. In cases like this, where a concentrated load is applied to the same point at which we cut, we should think of the cut being just to one side of the load, and it doesn’t matter which side, as long as we are consistent. If we show the load on one FBD of a pair sharing the common point, we must not show it on the other.

Concentrated forces applied to points at which we 'cut' the structure must be shown on only one of the two possible free body diagrams.

\[\begin{split} \sum M_d &= 0~~~~~\text{+CCW}\\ &= 100\times8 + 150 - V_b\times6\\ V_b &= 158.33~~~~~(\therefore\uparrow)\\ \\ \sum F_y &= 0~~~~~\text+\uparrow\\ &= -100 + V_b + V_d\\ &= -100 + 158.33 + V_d\\ V_d &= -58.33~~~~~(\therefore\downarrow) \end{split}\]

If we now draw a free body diagram of just portion d-e, at point d we would now show a force of 58.33 acting upward (the reaction from a-b-c-d) plus a force of 40kN acting upward (because we didn’t show it on the left FBD). However, instead will will show the whole structure, as in Fig. 2.3-8:

Fig. 2.3-8: Free Body of Complete Strcture

\[\begin{split} \sum F_x &= 0~~~~~\text+\rightarrow\\ &= H_e\\ H_e &= 0\\ \\ \sum F_y &= 0~~~~~\text+\uparrow\\ &= -100 + 158.33 + 40 -{ {15+25} \over2}\times8 + V_e\\ V_e &= 61.67~~~~~(\therefore\uparrow)\\ \\ \sum M_e &= 0~~~~~\text{+CCW}\\ &= 100\times16 - 158.33\times14 + 150 - 40\times8 + (15\times8\times{8\over2}) + ((25-15)\times{8\over2}\times{8\over3}) - M_e\\ M_e &= -200 \text{kN-m}~~~~~(\therefore \text{CCW}) \end{split}\]

And finally, we show the summary free body diagram:

Fig. 2.3-9: Results Free Body Diagram

Summary

And all of the above diagrams repeated on one figure here for ease of reading:

Fig. 2.3-9: Summary Free Body Diagrams