ANALYSIS OF A STATICALLY DETERMINATE TRUSS (CIVIL/ENVIRONMENTAL ENGINEERING) An important problem in structural engineering is that of finding the forces and reactions associated with a statically determinate truss. Figure 12.4 shows an example of such a truss. The forces (F) represent either

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ANALYSIS OF A STATICALLY DETERMINATE TRUSS (CIVIL/ENVIRONMENTAL ENGINEERING)

An important problem in structural engineering is that of finding the forces and reactions associated with a statically determinate truss. Figure 12.4 shows an example of such a truss.

The forces (F) represent either tension or compression on the members of the truss. External reactions ([latex]H_2, V_2[/latex], and [latex]V_3[/latex]) are forces that characterize how the truss interacts with the supporting surface. The hinge at node 2 can transmit both horizontal and vertical forces to the surface, whereas the roller at node 3 transmits only vertical forces. It is observed that the effect of the external loading of 1000 lb is distributed among the various members of the truss.

Accepted Answer

This type of structure can be described as a system of coupled linear algebraic equations. Free-body force diagrams are shown for each node in Fig. 12.5. The sum of the forces in both horizontal and vertical directions must be zero at each node, because the system is at rest. Therefore, for node 1,

[latex]\Sigma F_{H}=0=-F_{1} \cos 30^{\circ}+F_{3} \cos 60^{\circ}+F_{1, h}[/latex] (12.3)

[latex]\sum F_{V}=0=-F_{1} \sin 30^{\circ}-F_{3} \sin 60^{\circ}+F_{1, v}[/latex] (12.4)

for node 2,

[latex]\sum F_{H}=0=F_{2}+F_{1} \cos 30^{\circ}+F_{2, h}+H_{2}[/latex] (12.5)

[latex]\sum F_{V}=0=F_{1} \sin 30^{\circ}+F_{2, v}+V_{2}[/latex] (12.6)

for node 3,

[latex]\sum F_{H}=0=-F_{2}-F_{3} \cos 60^{\circ}+F_{3, h}[/latex] (12.7)

[latex]\sum F_{V}=0=F_{3} \sin 60^{\circ}+F_{3, v}+V_{3}[/latex] (12.8)

where [latex]F_{i, h }[/latex]is the external horizontal force applied to node i (where a positive force is from left to right) and [latex]F_{1, ν}[/latex] is the external vertical force applied to node i (where a positive force is upward). Thus, in this problem, the 1000-lb downward force on node 1 corresponds to F1, y 5 21000. For this case all other [latex]F_{i, ν} [/latex]’s and [latex]F_{i, h} [/latex]’s are zero. Note that the directions of the internal forces and reactions are unknown. Proper application of Newton’s laws requires only consistent assumptions regarding direction. Solutions are negative if the directions are assumed incorrectly. Also note that in this problem, the forces in all members are assumed to be in tension and act to pull adjoining nodes together. A negative solution therefore corresponds to compression. This problem can be written as the following system of six equations and six unknowns:

[latex]\left[\begin{array}{cccrrr}0.866 & 0 & -0.5 & 0 & 0 & 0 \0.5 & 0 & 0.866 & 0 & 0 & 0 \-0.866 & -1 & 0 & -1 & 0 & 0 \-0.5 & 0 & 0 & 0 & -1 & 0 \0 & 1 & 0.5 & 0 & 0 & 0 \0 & 0 & -0.866 & 0 & 0 & -1\end{array} ight]\left\{\begin{array}{c}F_{1} \F_{2} \F_{3} \H_{2} \V_{2} \V_{3}\end{array} ight\}=\left\{\begin{array}{c}0 \-1000 \0 \0 \0 \0\end{array} ight\}[/latex]

Notice that, as formulated in Eq. (12.9), partial pivoting is required to avoid division by zero diagonal elements. Employing a pivot strategy, the system can be solved using any of the elimination techniques discussed in Chap. 9 or 10. However, because this problem is an ideal case study for demonstrating the utility of the matrix inverse, the LU decomposition can be used to compute

[latex]F_{1}=-500 \quad F_{2}=433 \quad F_{3}=-866[/latex]

[latex]H_{2}=0 \quad V_{2}=250 \quad V_{3}=750[/latex]

and the matrix inverse is

[latex][A]^{-1}=\left[\begin{array}{ccrrrr}0.866 & 0.5 & 0 & 0 & 0 & 0 \0.25 & -0.433 & 0 & 0 & 1 & 0 \-0.5 & 0.866 & 0 & 0 & 0 & 0 \-1 & 0 & -1 & 0 & -1 & 0 \-0.433 & -0.25 & 0 & -1 & 0 & 0 \0.433 & -0.75& 0 & 0 & 0 & -1\end{array} ight][/latex]

Now, realize that the right-hand-side vector represents the externally applied horizontal and vertical forces on each node, as in

[latex]\{F\}^{T}=\left\lfloor\begin{array}{llllll}F_{1, h} & F_{1, v}& F_{2, h} & F_{2, v} & F_{3, h} & F_{3, v}\end{array} ight][/latex] (12.10)

Because the external forces have no effect on the LU decomposition, the method need not be implemented over and over again to study the effect of different external forces on the truss. Rather, all that we have to do is perform the forward- and backward-substitution steps for each right-hand-side vector to effi ciently obtain alternative solutions. For example, we might want to study the effect of horizontal forces induced by a wind blowing from left to right. If the wind force can be idealized as two point forces of 1000 lb on nodes 1 and 2 (Fig. 12.6a), the right-hand-side vector is

[latex]\{F\}^{T}=\left\lfloor\begin{array}{llllll}-1000 & 0 & 1000 & 0 & 0 & 0\end{array} ight][/latex]

which can be used to compute

[latex]F_{1}=866 \quad F_{2}=250 \quad F_{3}=-500[/latex]

[latex]H_{2}=-2000 \quad V_{2}=-433 \quad V_{3}=433[/latex]

For a wind from the right (Fig. 12.6b), [latex]F_{1, h}=-1000, F_{3, h}=-1000 [/latex] and all other external forces are zero, with the result that

[latex]F_{1}=-866 \quad F_{2}=-1250 \quad F_{3}=500[/latex]

[latex]H_{2}=2000 \quad V_{2}=433 \quad V_{3}=-433 [/latex]

The results indicate that the winds have markedly different effects on the structure. Both cases are depicted in Fig. 12.6.

The individual elements of the inverted matrix also have direct utility in elucidating stimulus-response interactions for the structure. Each element represents the change of one of the unknown variables to a unit change of one of the external stimuli. For example, element [latex]a^{-1}_{32}[/latex] indicates that the third unknown ( [latex]F_3 [/latex]) will change 0.866 due to a unit
change of the second external stimulus [latex](F_{1, ν})[/latex]. Thus, if the vertical load at the first node were increased by 1, F3 would increase by 0.866. The fact that elements are 0 indicates
that certain unknowns are unaffected by some of the external stimuli. For instance [latex]a^{-1}_{32} = 0[/latex] means that [latex]F_1[/latex] is unaffected by changes in [latex]F_{2, h} [/latex]. This ability to isolate interactions has a number of engineering applications, including the identification of those components that are most sensitive to external stimuli and, as a consequence, most prone to failure. In addition, it can be used to determine components that may be unnecessary (see Prob. 12.18).

The foregoing approach becomes particularly useful when applied to large complex structures. In engineering practice, it may be necessary to solve trusses with hundreds or even thousands of structural members. Linear equations provide one powerful approach for gaining insight into the behavior of these structures.

Question 12.2: ANALYSIS OF A STATICALLY DETERMINATE TRUSS (CIVIL/ENVIRONMEN...

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