Rigid-body equilibrium and supports
Introducido| English | Significado |
|---|---|
| pin/pɪn/ | articulación fija |
| roller/ˈrəʊlə/ | apoyo móvil |
| fixed support/fɪkst səˈpɔːt/ | empotramiento |
| cantilever | cantilever |
| statically indeterminate | statically indeterminate |
Why is one end of a bridge left free to slide?
- A steel bridge grows several centimetres longer on a hot day. If both ends were bolted down, the bridge would push hard on its supports and buckle.
- So engineers pin one end and put the other on rollers. The supports give exactly the reactions the equations can find, and no more.
Three equations for a rigid body
- A rigid body in a plane gives three equations: $\Sigma F_x = 0$, $\Sigma F_y = 0$ and $\Sigma M = 0$. They can find three unknown reactions.
- Each support gives one unknown for each motion it stops: roller 滚动支座 one, pin 固定铰支座 two, fixed support 固定端 three (two forces and a moment).
Moments first, forces second
- Take moments about one support. The other support's unknown is then alone in the equation. The force equations finish the job.
- Keep one equation you did not use. It is your check.
- A $6\ \text{m}$ beam rests on a pin at A and a roller at B. It carries $12\ \text{kN}$ at $2\ \text{m}$ from A.
- Moments about A: $R_B \times 6 = 12 \times 2$, so $R_B = 4\ \text{kN}$. Vertically: $R_A = 12 - 4 = 8\ \text{kN}$.
- Check with moments about B: $8 \times 6 = 48 = 12 \times 4$. The support nearer the load carries more.
Balance the moments
The beam is level only when the moments on each side of the pivot match, force times distance. A small load far out can balance a large one close in.
An $8\ \text{m}$ beam rests on a pin at A and a roller at B. It carries $20\ \text{kN}$ at $3\ \text{m}$ from A. What is the roller reaction $R_B$, in kN?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
Moments about A: $R_B \times 8 = 20 \times 3$, so $R_B = 7.5\ \text{kN}$.
For the same beam, what is the pin's vertical reaction $R_A$, in kN?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
Vertically $R_A = 20 - 7.5 = 12.5\ \text{kN}$. Check with moments about B: $12.5 \times 8 = 100 = 20 \times 5$.
A slanted load needs the pin's sideways force
- Resolve a slanted load first. Its vertical part enters the moment equation; its horizontal part can only be held by the pin, because a roller cannot push sideways.
- A $10\ \text{kN}$ load at the middle of a $6\ \text{m}$ beam, at $60^{\circ}$ to the beam: the vertical part is $8.66\ \text{kN}$, so $R_B = R_A = 4.33\ \text{kN}$, and the pin holds $A_x = 5.00\ \text{kN}$.
An $8\ \text{kN}$ load acts on a beam at $30^{\circ}$ to the beam. The beam sits on a pin and a roller (the roller on a level surface). What horizontal force must the pin supply, in kN?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
The load's horizontal part is $8\cos 30^{\circ} = 6.93\ \text{kN}$. The roller can only push vertically, so the pin takes all of it.
Overhangs flip a reaction
- The roller moves to $4\ \text{m}$ from A, and a $10\ \text{kN}$ load hangs at the $6\ \text{m}$ tip.
- Moments about A: $R_B \times 4 = 10 \times 6$, so $R_B = 15\ \text{kN}$, more than the load.
- Vertically: $R_A = 10 - 15 = -5\ \text{kN}$. The minus sign means the pin pulls down: it holds the left end on.
- Check with moments about B: $R_A \times 4 + 10 \times 2 = -20 + 20 = 0$.
You draw the pin reaction at A pointing up and solve to get $R_A = -5\ \text{kN}$. What does this mean?
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
A negative answer means the real force points opposite to your arrow. A pin can pull; a plain roller resting on a surface could not, so this end must be bolted.
A fixed end holds alone
- A cantilever 悬臂梁 is held only by a fixed support. Its three reactions come one from each equation.
- A $2\ \text{m}$ cantilever with $3\ \text{kN}$ at its tip: $A_y = 3\ \text{kN}$ up, and the wall supplies a moment $M_A = 3 \times 2 = 6\ \text{kN}\cdot\text{m}$ counterclockwise.
A $1.5\ \text{m}$ cantilever carries $4\ \text{kN}$ at its free end. What moment must the fixed support supply, in $\text{kN}\cdot\text{m}$?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
$M_A = 4 \times 1.5 = 6\ \text{kN}\cdot\text{m}$, turning against the load. The vertical reaction is $4\ \text{kN}$ up.
More supports than equations
- With more than three unknown reactions, equilibrium alone cannot find them: the body is statically indeterminate 超静定. A fixed end plus a roller gives four unknowns.
- That is not a harder problem to solve with more effort. It needs another idea, such as how much the beam bends, which comes later in the degree.
Which beams can be solved by the three equilibrium equations alone? Select all that apply.
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
Pin + roller and a lone fixed end each give exactly $3$ unknowns. Fixed + roller gives $4$ and two pins give $4$, so both are statically indeterminate.
Marks that slip away
- Assuming every reaction points up. Overhangs and offset loads flip them. Draw it up, and let a minus sign tell you.
- Using the wrong arm. Mark each distance from the moment point on the diagram before you sum.
- Checking with the same equation that gave the answer. Only an unused equation is a real check.
Put the method for finding a beam's reactions in order.
Dificultad 3/5 · 200 monedas + un pequeño bono en el primer intento de hoy
Moments about a support remove two of the three unknowns at once, which is why it comes before the force equations.
You've got it
- Three equations: $\Sigma F_x = 0$, $\Sigma F_y = 0$, $\Sigma M = 0$. Roller 1 unknown, pin 2, fixed support 3.
- Take moments about a support to find the other one alone; check with an unused equation.
- A negative reaction means the support pulls; more than three unknowns means statically indeterminate.