Question 6.CS.1: CS Photo 6.1 shows a pin-connected truss that is part of the......
CS Photo 6.1 shows a pin-connected truss that is part of the railroad bridge spanning the Connecticut River at Warehouse Point in East Windsor, Connecticut. Built by the New York, New Haven & Hartford Railroad in the early 1900s, the seven-panel structure has features of both the Baltimore- and Pratt-style trusses. In addition to being connected to the joints of the lowest horizontal members (or chords), the floor system is also supported by hangers suspended from the mid-panel points. The central three panels also employ diagonal counters, giving these panels a characteristic “X” appearance (see Probs. 6.65, 6.66, 6.67, and 6.68).
If we assume the geometry illustrated in CS Fig. 6.1 for one of the two sides of the truss, let’s determine the force developed in chord member CF when a unit 1-kip load is applied at D as shown, and then repeat the analysis with the unit load moved to joint G. For clarity, we will omit the mid-panel hangers, because they are zero-force members for both load cases. (What other zeroforce members are present?)
STRATEGY:
For each load case, determine the reactions by treating the entire truss as a free body. Then, using the method of sections, cut through the second panel to expose the force in member CF and apply equilibrium to determine this force.
MODELING and ANALYSIS:
Case I: Unit 1-kip Load at D. For this load case, CS Fig. 6.2 shows the free-body diagram of the entire truss. From this diagram, find the reaction at A:
+↺ΣM_{U} = 0: + (1 kip)(150 ft) − A(175 ft) = 0 A = 0.8571 kips ↑
Now pass a section nn vertically through the truss (CS Fig. 6.2) and choose the left-hand portion as a free body (CS Fig. 6.3).
Applying equilibrium,
+↺ΣM_{G} = 0: + (0.8571 kips)(50 ft) − (1 kip)(25 ft) + F_{CF} (35 ft) = 0
F_{C F}=-0.5101\mathrm{~kips}\qquad F_{C F}=0.5101\mathrm{~kips~}C ◂
Case II: Unit 1-kip Load at G. Moving the unit load from point D to point G and repeating the process for finding the reaction in Case I:
A = 0.7143 kips ↑
Again, pass a section nn vertically through the truss (CS Fig. 6.2) and choose the left-hand portion as a free body (CS Fig. 6.4). Because the unit load is no longer acting directly on the chosen free body, it is not shown in this figure.
Applying equilibrium,
+↺ΣM_{G} = 0: + (0.7143 kips)(50 ft) + F_{CF} (35 ft) = 0
F_{C F}=-1.0204\,\mathrm{kips}\qquad F_{C F}=1.0204\,\mathrm{kips}\ C ◂
REFLECT and THINK: Repeating the process to consider the force developed in member CF due to the unit 1-kip load applied in order to each of the joints along the lower chord, the following table can be compiled:
These results can also be plotted as shown in CS Fig. 6.5. This plot is known as an influence line, and it displays the force developed in member F_{CF} as a unit load traverses the deck. A valuable tool for bridge design, such influence lines allow engineers to study the structural effects of moving loads. And while they are developed using a single unit force as has been done here, techniques exist that allow these plots to be readily applied in the evaluation of multiple-axle loads, such as the train shown in CS Photo 6.1.
| Location of 1-kip Load | FCF, kips |
| A | 0 |
| D | -0.5101 |
| G | -1.0204 |
| J | -0.8163 |
| M | -0.6123 |
| P | -0.4082 |
| S | -0.2041 |
| U | 0 |