3: Forces in Statically Determinate Trusses

3.3: Member Forces by Method of Joints

3.3.1: Zero-force members

Figure 3.3-1: Zero-force members

It is convenient before starting any analysis to identify all of the zero-force members.
These are members known by “inspection” to carry no force. Fig. 3.3-1 shows several common arrangements.

In the first, three members attach to joint a, two of them co-linear. If there are no external forces applied to the joint, then the third member carries no force. This can be seen by summing forces in a direction perpendicular to the co-linear members; the component of the third member force in that direction must be zero and therefore the force must be zero. As it is a zero-force member, it can be “removed” from the truss (for analysis purposes).

In the second example, two members connect at joint b. An applied force is co-linear with one of the members, leaving the other a zero-force member.

In the third example, two members connect at joint c. If there are no external forces applied to the joint, both members will be zero-force.

In all cases, the member arrangement at the other end of the zero-force members is irrelevant.

Identifying a zero-force member requires use of an equation of equilibrium, even though it is trivially easy to setup and solve. It doesn't allow you to solve for more unknowns than you normally could.

If the conditions at one end of a member suggest that it is zero-force, it doesn't matter what the conditions are at the other end.

3.3.2: Example J-1

Figure 3.3-2: Example J-1

Before beginning the numerical calculations for the truss in Fig. 3.3-2, we should identify the zero-force members. We quickly see that members bh and df are zero-force, because of the conditions at joints h and f respectively. As they can be ‘removed’ from the structure with no ill effects, member gd is also zero-force because of the conditions at joint d (with member df removed). However, bg is not zero-force, because of the external load at joint b.

The three zero-force members are marked on the figure to help us keep track of the solved members.

Calculating the reactions using external equations of equilibrium:

\[\begin{split} &\sum M_a = 0~~~~~(\text{+CCW})\\ &-120\times4.5 - 80\times9.0 + V_e\times18=0\\ &\underline{V_e = 70~ \text{kN}}\\ &\sum F_y = 0~~~~~(+\uparrow)\\ &V_a - 120 - 80 + V_e = 0\\ &\underline{V_a = 130~ \text{kN}} \end{split}\]

To start the analysis by the method of joints, choose one of the joints on the truss into which two or fewer unknown members frame and also on which some known forces act. Joints a and e are the only possibilities, so we choose joint a. The free body diagram of that joint is shown in Fig. 3.3-3.

Figure 3.3-3: Joint a

We choose to show all unknown member forces as if they were tension forces in the members. That means, when they are shown on the same side as they member, they pull away from the joint toward the other end of the member, as we see in the figure. When we solve the equilibrium equations, the sign will tell us if we assumed correctly or not.

To avoid solving simultaneous equations, we try to write equilibrium equations involving only one unknown, where ever possible. On joint a, that involves summing forces in the vertical direction. Because that direction is perpendicular to $F_{ah}$, that won’t appear in the equation.

\[\begin{split} &\sum F_y = 0~~~~~(+\uparrow)\\ &130 + \frac{3}{5.408} \times F_{ab}= 0\\ &\underline{F_{ab} = -234.4}~~~~~(\therefore C) \end{split}\]

The negative sign tells us that we showed the force in the wrong direction on the free body diagram. Therefore, the member force must be compression and we carefully label the result with that observation.

Engineering results of this kind are normally shown to no more than three significant figures. We sometimes carry four significant figures in intermediate results, as done here, to reduce the round-off errors.

On the same joint a, we can now solve for the remaining unknown by summing forces in the horizontal direction:

\[\begin{split} &\sum F_x = 0~~~~~(+\rightarrow)\\ &F_{ab}\times\frac{4.5}{5.408} + F_{ah} = 0\\ &-234.4\times\frac{4.5}{5.408} + F_{ah} = 0\\ &\underline{F_{ah} = 195.0}~~~~~(\therefore T) \end{split}\]

In the above, we were careful to first write the equilibrium equations in terms of the forces shown on the free body diagram - that figure shows $F_{ab}$ having a positive component in the x-direction. Then we substituted the numerical value for that force, using the proper sign as determined in the previous solution.

Equilibrium equations should first be written using the directions shown on the free body diagram. Attempting to "correct" the signs mentally before writing the equation is a recipe for disaster.

Figure 3.3-4: Joint e

Now we can do a similar solution for joint e (see Fig. 3.3-4):

\[\begin{split} &\sum F_y = 0~~~~~(+\uparrow)\\ &70 + \frac{3}{5.408} \times F_{ed}= 0\\ &\underline{F_{ed} = -126.2}~~~~~(\therefore C) \end{split}\] \[\begin{split} &\sum F_x = 0~~~~~(+\rightarrow)\\ &-F_{ed}\times\frac{4.5}{5.408} - F_{ef} = 0\\ &126.2\times\frac{4.5}{5.408} - F_{ef} = 0\\ &\underline{F_{ef} = 105.0}~~~~~(\therefore T) \end{split}\]

Figure 3.3-5: Joints f, h

By inspection, from Fig. 3.3-5, we can determine:

\[\begin{split} &\underline{F_{hg} = 195.0}~~~~~(\therefore T)\\ &\underline{F_{fg} = 105.0}~~~~~(\therefore T) \end{split}\]

Figure 3.3-6: Joint d

Also by inspection (i.e. from $\sum F_{x’} = 0$) from Fig. 3.3-6, we can determine:

\[\begin{split} &\underline{F_{cd} = -126.2}~~~~~(\therefore C) \end{split}\]

Figure 3.3-7: Joint c

Fig. 3.3-7 now shows joint c at which two unsolved members terminate. By inspection (or by $\sum F_x=0$) we can see:

\[\begin{split} &\underline{F_{cb} = -126.2}~~~~~(\therefore C) \end{split}\]

Also:

\[\begin{split} &\sum F_y = 0~~~~~(+\uparrow)\\ &-F_{cb}\times\frac{3}{5.408} -80 + 126.2\times\frac{3}{5.408} - F_{cg} = 0\\ &126.2\times\frac{3}{5.408} -80 + 126.2\times\frac{3}{5.408} - F_{cg} = 0\\ &\underline{F_{cg} = 60.02}~~~~~(\therefore T) \end{split}\]

Figure 3.3-8: Joint b

We have one remaining unknown, $F_{bg}$ and we can use joint b to solve for that. See Fig. 3.3-8.

\[\begin{split} &\sum F_x = 0~~~~~(+\rightarrow)\\ &234.4\times\frac{4.5}{5.408} - 126.2\times\frac{4.5}{5.408} + F_{bg}\times\frac{4.5}{5.408} = 0\\ &\underline{F_{bg} = -108.2}~~~~~(\therefore C) \end{split}\]

We haven’t used all equilibrium equations for this joint, yet, so we can use $\sum F_y$ as a statics check:

\[\begin{split} \sum F_y &= 234.4\times\frac{3}{5.408} - 120 - 126.2\times\frac{3}{5.408} - F_{bg}\times\frac{3}{5.408}\\ &= 234.4\times\frac{3}{5.408} - 120 - 126.2\times\frac{3}{5.408} + 108.2\times\frac{3}{5.408}\\ &= 0.04 \approx 0~~~~~\text{O.K.} \end{split}\]

Statics checks of joints should be performed whenever possible.

Figure 3.3-9: Joint g

Joint g hasn’t been used at all, yet. We will use it now to perform two more statics checks:

\[\begin{split} \sum F_x &= -195.0 + 108.2\times\frac{4.5}{5.408} + 105.0\\ &= 0.03 \approx 0~~~~~\text{O.K.}\\ \sum F_y &= -108.2\times\frac{3}{5.408} + 60.02\\ &= -0.002 \approx 0~~~~~\text{O.K.} \end{split}\]

Figure 3.3-10: Summary of Forces

Finally, a summary of all forces is shown in Fig. 3.3-10. Note that all values are rounded to three significant figures, and that we clearly show the magnitude of each force and the sense of each as tension or compression.

There is no clear convention as to whether to label a compression force as negative or positive. Do not rely on the +/- signs to convey that crucial information.