Statics · Trusses
Condition: The theoretical fundamentals develop idealisations and determinacy, the formation rules, the three zero-force-member rules, and the method of joints and Ritter's method on the roof truss.
Open theoretical foundations →First draw every truss on paper: strike out zero-force members, isolate joints, set up the balances. The part steps here are your check.
Exercise 1 · Zero-force members and determinacy
face — the count works out exactly, so no reaction is redundant. The truss is therefore statically determinate by count; whether it is also immovable must be checked separately.
Assign each situation to the zero-force-member rule that describes it.
Rule 1 · both members are zero-force members (unloaded two-member joint)
Rule 2 · the other member is a zero-force member
Rule 3 · the third member is a zero-force member
No rule applies — this one has to be calculated
For each zero-force member rule, check the number of members, the loading and the alignment of the member axes. Collinear means lying on the same straight line, including in opposite directions. At an unloaded joint with two collinear members, non-zero member forces can balance each other; rule 1 does not apply there. The situation described here, with three members in different directions at a loaded joint, does not directly match any of the three rules.
Under the tension assumption a member yields . What does that mean?
By convention tensile forces are positive: positive member forces act away from the joints, they pull on them. A negative value indicates compression — the force acts towards the joint. The sign is therefore not an error message but the result.
Conclusion: strike out first, then calculate
Striking out the zero-force members before calculating saves half the work on larger trusses. And keeping the tension assumption consistently means the type of loading is simply read off from the sign at the end.
Exercise 2 · Method of joints
A bracket consists of two members: the horizontal member runs from the wall at to the jib end , the inclined one from — that is 3.0 m above — likewise to . The load hangs at the bracket end: . Dimensions in metres.
— a 3-4-5 triangle. The direction cosines are therefore 0.8 and 0.6, and the calculation stays feasible without a calculator.
At which joint do you start the method of joints?
On only the load and the two member forces act — two unknowns, two equations. At and the support forces would come on top.
Force in the inclined member, symbolically from the vertical balance at joint .
Allowed symbols: F, l, hb (height of B above A)
Notation help: fraction with /, brackets allowed.
The vertical component of the inclined member is . From it follows that — positive, so the inclined member is in tension.
Under the tension assumption both member forces point of away: the one to towards , the one to towards .
— negative: the horizontal member is in Pressure. That is exactly how it must be, because the inclined member pulls the bracket end towards the wall.
Assume both member forces as tension — then their arrows point away from the joint, and you read the type of loading off the sign later.
For joint write down the two unit vectors first. That turns both balances into simple sums.
Conclusion: the right joint first
Two equations per joint means: at most two unknowns. Starting there, one works through without a system of equations — starting at the wrong joint, one immediately faces three unknowns.
Exercise 3 · Ritter section
A parallel-chord girder of span and height consists of four panels of . The lower-chord joints lie at 0, 3, 6, 9 and 12 m, the upper-chord joints at 3, 6 and 9 m. At the lower-chord joint at acts acts vertically downwards — the load therefore stands off-centre. Sought are only the upper- and lower-chord forces in the second panel.
The load stands off-centre — symmetry does not help here. Moment balance about :
As a check: — the support closer to the load carries more.
Assign to each sought member force the reference point that leaves it alone in the moment balance.
The reference point belongs at the intersection of the two other cut members. For that is the lower-chord joint at 6.0 m — diagonal and lower chord meet there. For it is the upper-chord joint at 3.0 m, where upper chord and diagonal meet.
The load does not appear in this balance — it lies to the right of the section. Negative means: the upper chord is in compression, as on every simply supported girder.
Positive — the lower chord is in tension. Both values follow from one moment balance each; not a single joint had to be calculated for them.
The section must sever the truss completely and may cut at most three members.
Calculate on the left part — only the support force stands there, the load lies to the right of the section.
Place the reference point where the two members intersect that you are not looking for.
Conclusion: one balance instead of a chain
With the method of joints one would have had to work across three joints to reach the second panel. The Ritter section delivers the same values from two moment balances — and therefore also serves as an independent check on a member force found by the method of joints.
Report
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| Exercise | Status |
|---|---|
| Exercise 1: Zero-force members and determinacy | Open |
| Exercise 2: Method of joints | Open |
| Exercise 3: Ritter section | Open |