Structures — trusses and frames
Introduit| English | Sens |
|---|---|
| truss/trʌs/ | treillis |
| tension/ˈtenʃn/ | traction |
| compression/kəmˈpreʃn/ | compression |
| method of joints/ˈmeθəd ɒv dʒɔɪnts/ | méthode des nœuds |
| zero-force member/ˈzɪərəʊ fɔːs ˈmembə/ | barre sans effort |
| method of sections/ˈmeθəd ɒv ˈsekʃnz/ | méthode des sections |
| frames and machines/freɪmz ænd məˈʃiːnz/ | charpentes et mécanismes |
Hundreds of slim bars, and none of them bends
- A railway bridge carries a train on a web of thin steel bars. Each bar is only pulled or pushed along its length, never bent.
- The secret is the triangle. A pinned triangle cannot change shape, so every bar is a two-force member, and every bar force can be found.
What makes a truss
- A truss 桁架 is straight members joined by frictionless pins and loaded only at the joints. Every member is a two-force member.
- So each member carries pure tension 拉力 (it is stretched, and pulls on its joints) or compression 压力 (it is squeezed, and pushes on them). Take tension as positive.
A straight member that is pinned at both ends and loaded only at its joints carries only a force along its own axis.
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It is a two-force member, so the two end forces are equal, opposite and along the line through both pins, which is its axis.
Method of joints: two equations at each pin
- Method of joints 节点法: draw one joint as a particle. Two equations, $\Sigma F_x = 0$ and $\Sigma F_y = 0$, find two member forces.
- Draw every unknown member force as tension (pulling away from the joint). A negative answer then means compression.
- A joint at the end of a wall bracket carries $3.6\ \text{kN}$ hanging down. A horizontal member and a member rising on 3-4-5 geometry hold it.
- Vertically: $F \times \tfrac{3}{5} = 3.6$, so the inclined member carries $6\ \text{kN}$ tension: the joint hangs from it.
- Horizontally: the horizontal member takes $6 \times \tfrac{4}{5} = 4.8\ \text{kN}$ compression: it pushes the joint away from the wall.
A $6\ \text{kN}$ load hangs at a truss joint held by a horizontal member and a member rising on 3-4-5 geometry (3 up, 4 across). What is the force in the inclined member, in kN?
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Only the inclined member has a vertical part: $F \times \tfrac{3}{5} = 6$, so $F = 10\ \text{kN}$, in tension.
In the same joint, what does the horizontal member carry?
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The inclined member pulls the joint towards the wall with $10 \times \tfrac{4}{5} = 8\ \text{kN}$. The horizontal member must push back with $8\ \text{kN}$, so it is in compression.
Where to start
- Find the support reactions first, using the whole truss as one rigid body.
- Then start at a joint with no more than two unknown members, and walk joint to joint. Each solved member gives one known force to the next joint.
Zero-force members, found by eye
- At an unloaded joint with only two members that are not in line, both carry zero. At an unloaded joint with three members, two of them in line, the third carries zero.
- These zero-force members 零杆 are not waste. They stop long slim members from buckling and carry load when the loading changes.
In which cases must a member carry zero force? Select all that apply.
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With no load, the two in-line forces cancel each other, so nothing is left to balance the third. A load at the joint, or a support reaction, changes that.
Worked example: a three-panel truss
- A Pratt truss spans $6\ \text{m}$ in three $2\ \text{m}$ panels and is $2\ \text{m}$ tall: bottom joints A, C, D, B; top joints E above C and F above D; diagonals A–E, E–D and F–B. It carries $6\ \text{kN}$ at C and at D.
- By symmetry $R_A = R_B = 6\ \text{kN}$. Joint A: $6 + F_{AE}\sin 45^{\circ} = 0$ gives $F_{AE} = -8.49\ \text{kN}$, so compression.
- Working on: bottom chord $6\ \text{kN}$ tension, top chord E–F $6\ \text{kN}$ compression, verticals $6\ \text{kN}$ tension, inner diagonal E–D zero.
- The truss sags, so its bottom chord stretches and its top chord shortens. A bottom chord in compression under downward loads means a sign slip.
In the three-panel Pratt truss with $6\ \text{kN}$ at C and D, what is the size of the force in end diagonal A–E, in kN?
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At joint A only A–E has a vertical part: $6 + F_{AE}\sin 45^{\circ} = 0$, so $F_{AE} = -8.49\ \text{kN}$, which is $8.49\ \text{kN}$ compression.
Method of sections: cut straight to the member you want
- Method of sections 截面法: cut through at most three members and use one piece as a free body. Three equations find three forces.
- Take moments about the point where two cut members meet. Only the third force is left. For the truss above, cut E–F, E–D and C–D; moments about E on the left piece: $-6 \times 2 + 2F_{CD} = 0$, so $F_{CD} = 6\ \text{kN}$ tension.
Put the method of sections in order.
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The reactions come first because the piece you keep usually includes a support.
Frames come apart into pieces
- Frames and machines 刚架与机构 contain members loaded along their length. Their pin forces do not act along the member, so the joint method fails.
- Take the frame apart: one free-body diagram per member, with equal and opposite forces at each shared pin. Solve any two-force member first, because its direction is known.
Match each structure or task to the best method.
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Choosing the method is half the work. The section method saves walking through every joint to reach one member.
Marks that slip away
- Mixing up the arrow directions. Draw every unknown as tension, pulling away from the joint, and let the sign decide.
- Cutting four members in a section. Three equations can only find three forces.
- Loading a member between its joints. Then it bends, it is not a truss member, and the truss method does not apply.
You've got it
- Truss members carry only tension or compression; take tension as positive.
- Joints: two equations per pin. Sections: cut at most three members and take moments where two of them meet.
- Find zero-force members first; take frames apart member by member.