Free-body diagrams and support reactions
Introduced| English | Meaning |
|---|---|
| free-body diagram/friː ˈbɒdi ˈdaɪəɡræm/ | the isolated body with every external force drawn on it and nothing else |
| load/ləʊd/ | A force or weight that a structure or part must carry. |
| centre of gravity/ˈsentə ɒv ˈɡrævɪti/ | The point through which an object’s resultant weight acts. |
| reaction/rɪˈækʃn/ | The force or moment a support exerts on a body to hold it in place. |
| roller/ˈrəʊlə/ | A support that stops movement perpendicular to its surface only, giving one reaction force. |
| pin/pɪn/ | A support that stops movement in any direction but allows rotation, giving two force components. |
| fixed support/fɪkst səˈpɔːt/ | A support that stops both movement and rotation, giving two forces and a moment. |
| statically determinate/ˈstætɪkli dɪˈtɜːmɪnət/ | When the equilibrium equations alone are enough to find every unknown force. |
| two-force member/tuː fɔːs ˈmembə/ | A member in equilibrium under exactly two external forces, which must be equal, opposite and collinear. |
What does the table do to the book?
- A $2.0\ \text{kg}$ book rests on a table. Its weight is $2.0 \times 9.81 = 19.6\ \text{N}$ down, and it does not move.
- So the table must push up with $19.6\ \text{N}$. You only see that push when you cut the book free and draw what the table gives it.
A $3.0\ \text{kg}$ toolbox rests on a shelf. How large is the upward force from the shelf, in N? Use $g = 9.81\ \dfrac{\text{m}}{\text{s}^{2}}$.
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The box does not move, so the shelf's push balances the weight: $N = mg = 3.0 \times 9.81 = 29.4\ \text{N}$.
Cut the body free, then draw its world
- A free-body diagram 受力图 shows one body on its own, with every outside force acting on it, and nothing else.
- Each cut through a support, contact or cable is replaced by the force that the cut part was giving.
Three kinds of force go on it
- Applied loads 载荷 (cables, pushes, pressure), with direction and point of application. The weight acts at the centre of gravity 重心.
- Reactions 约束反力: what each support pushes back with. Internal forces stay off: if you did not cut a member, its force is not on this diagram.
You draw the free-body diagram of a beam resting on a pin at A and a roller at B. Which forces belong on it? Select all that apply.
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Only forces acting on the beam from outside appear. The beam's force on the pin belongs on the pin's diagram, and an internal force appears only if you cut there.
Each support speaks a fixed language
- Roller 滚动支座: one force, perpendicular to the surface. Pin 固定铰支座: two force components. Fixed support 固定端: two components and a moment.
- A cable gives one force: a pull along the cable. It can never push.
Match each support to the unknowns it gives.
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Each support supplies a reaction for each motion it prevents. A roller stops one motion, a pin two, a fixed support three.
Count before you calculate
- A particle in 2D gives two equations; a rigid body gives three. When the unknowns equal the equations, the body is statically determinate 静定.
- A beam on a pin and a roller: $2 + 1 = 3$ unknowns, solvable. Fixed at both ends: $3 + 3 = 6$ unknowns, too many. On two rollers: only $2$, and nothing stops it sliding sideways.
A beam has a fixed support at A and a roller at B. How many unknown reactions does its free-body diagram have?
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The fixed support gives $3$ (two components and a moment) and the roller gives $1$. Four unknowns against three equations, so this beam is statically indeterminate.
A beam rests on two rollers on a level floor and carries vertical loads. What is wrong?
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Two rollers give only $2$ vertical unknowns. The vertical loads can be balanced, but the smallest sideways push would make the beam roll away.
Two-force members point the way
- A two-force member 二力杆 is loaded only at its two ends. Its force must act along the line joining those two ends, whatever the member's shape.
- That fact turns two unknown components into one unknown size. A strut meeting a beam at $45^{\circ}$ pushes along the strut, at $45^{\circ}$.
A curved bar pinned at both ends, with no other loads on it, pushes or pulls along the straight line joining its two pins.
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Two forces can only balance if they are equal, opposite and on the same line, and that line runs through both pins. The bar's shape does not matter.
Worked example: the lamp ring
- A $200\ \text{N}$ lamp hangs from a ring held by two cables, at $30^{\circ}$ and $45^{\circ}$ above the horizontal. Draw the ring's free-body diagram and audit it.
- Cut both cables and the lamp's chain. On the ring: $200\ \text{N}$ down, $T_1$ along cable 1, $T_2$ along cable 2. Both tensions pull away from the ring.
- Audit: two unknowns ($T_1$, $T_2$), two particle equations. Determinate. The next lesson solves it.
Put the steps for drawing a free-body diagram in order.
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The count comes last because it needs the finished diagram. A reaction left out at step four is the error the count is there to catch.
Marks that slip away
- Drawing forces the body does not feel, such as the force the beam puts on the wall. On the beam's diagram, only the wall's force on the beam appears.
- Leaving out a reaction because a support "looks rigid". Rigid is exactly why it pushes back.
- Pushing with a cable. A cable only pulls, away from the body.
You've got it
- Isolate one body; replace every cut with the force it gave; draw loads, weight and reactions, never internal forces.
- Roller 1 unknown, pin 2, fixed support 3; a two-force member acts along its own line.
- Count unknowns against equations before any algebra.