Elastic and plastic behaviour · 弹性与塑性行为
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| elastic/ɪˈlæstɪk/ | 弹性 | tán xìng |
| plastic/ˈplæstɪk/ | 塑性 | sù xìng |
| limit of proportionality/ˈlɪmɪt ɒv prəˌpɔːʃəˈnælɪti/ | 比例极限 | bǐ lì jí xiàn |
| elastic limit/ɪˈlæstɪk ˈlɪmɪt/ | 弹性极限 | tán xìng jí xiàn |
| permanent extension/ˈpɜːmənənt ekˈstenʃn/ | 永久伸长 | yǒng jiǔ shēn cháng |
The paperclip test
- Bend a paperclip a little and it springs back — elastic 弹性.
- Bend it far and it stays bent — plastic 塑性.
- The same material can do both, depending on how hard you push it.
回形针测试
- 把回形针弯一点,它会弹回来——弹性(elastic)。
- 把它弯很多,它就保持弯曲——塑性(plastic)。
- 同一种材料两者都能,取决于你用多大力。
Three stages of stretching
- Elastic and straight — obeys Hooke's law; returns to shape.
- Elastic but curved — past the limit of proportionality 比例极限; still returns to shape.
- Plastic — past the elastic limit 弹性极限; a permanent change is left.
A stress-strain graph, straight up to the limit of proportionality P
拉伸的三个阶段
- 弹性且笔直——服从胡克定律;能恢复形状。
- 弹性但弯曲——超过比例极限;仍能恢复形状。
- 塑性——超过 弹性极限(elastic limit);留下永久改变。
一张应力-应变图,在比例极限 P 以内是直线
Hooke's law · 胡克定律
F = kx
Up to the limit, extension is proportional to force — the gradient is the spring constant k. · 在限度内，伸长量与力成正比——斜率即为弹簧常数 k。
Match each stage of stretching to what happens. · 把拉伸的每个阶段与发生的情况配对。
Elastic = returns to its first length on unloading; plastic = a permanent change once you pass the elastic limit. · 弹性 = 卸载后恢复到原来的长度;塑性 = 一旦超过弹性极限就留下永久改变。
Loading and unloading
- Stretch past the elastic limit, then remove the load.
- The unloading line is parallel to the first line but shifted — leaving a permanent extension 永久伸长.
加载与卸载
- 拉伸超过弹性极限,然后移除载荷。
- 卸载线与第一条线平行但有偏移——留下一个 永久伸长。
Past the elastic limit, an object returns to its original length when the load is removed. · 超过弹性极限后,移除载荷时物体会恢复到原来的长度。
No — in the plastic region a permanent extension stays after unloading. · 不——在塑性区,卸载后会留下永久伸长。
Match each term to the definition the examiner marks. · 把每个术语与评分认可的定义配对。
The elastic limit sits a little beyond the limit of proportionality, so a material can stop obeying Hooke's law while still being elastic. · 弹性极限略在比例极限之后,所以材料可以不再遵守胡克定律却仍是弹性的。
Energy = area under the graph
- The work done stretching a material is the area under its force–extension graph.
能量 = 图线下的面积
- 拉伸材料所做的功是 力–伸长量图线下的面积。
The energy stored in a stretched material equals the ____ under the force–extension graph. · 拉伸的材料中储存的能量等于力–伸长量图线下的 ____。
Work done = area under the force–extension graph (a triangle for a Hooke material). · 所做的功 = 力–伸长量图线下的面积(对胡克材料是一个三角形)。
Stored elastic energy
- For a Hooke's-law material the area is a triangle: $E_{\text{P}} = \tfrac{1}{2}Fx = \tfrac{1}{2}kx^{2}$.
- An equal form is $E_{\text{P}} = \dfrac{F^{2}}{2k}$.
Force-extension past the elastic limit: P and E marked, with a permanent extension B left after unloading
储存的弹性能
- 对于服从胡克定律的材料,这个面积是一个三角形:$E_{\text{P}} = \tfrac{1}{2}Fx = \tfrac{1}{2}kx^{2}$。
- 一个等价的形式是 $E_{\text{P}} = \dfrac{F^{2}}{2k}$。
超过弹性极限的力-伸长量:标出 P 和 E,卸载后留下永久伸长 B
A spring of constant $200\ \dfrac{\text{N}}{\text{m}}$ is stretched by $0.10\ \text{m}$. How much elastic PE is stored? · 一根常数为 $200\ \dfrac{\text{N}}{\text{m}}$ 的弹簧被拉伸 $0.10\ \text{m}$。储存了多少弹性势能?
$E_{\text{P}} = \tfrac{1}{2}kx^{2} = \tfrac{1}{2} \times 200 \times 0.10^{2} = 1.0\ \text{J}$. · $E_{\text{P}} = \tfrac{1}{2}kx^{2} = \tfrac{1}{2} \times 200 \times 0.10^{2} = 1.0\ \text{J}$。
When it's not a straight line
- For rubber, or a spring past its limit, the graph is a curve.
- Find the area by counting squares or using trapezia — the rule is the same.
当它不是直线时
- 对于橡胶,或超过极限的弹簧,图线是一条 曲线。
- 用数方格或梯形法求面积——规则是一样的。
Comparing and releasing
- Same force, softer spring (smaller $k$) → bigger $x$ → more energy stored.
- Released onto a mass, the elastic PE becomes kinetic energy: $\tfrac{1}{2}kx^{2} = \tfrac{1}{2}mv^{2}$.
比较与释放
- 同样的力,更软 的弹簧(更小的 $k$)→ 更大的 $x$ → 储存 更多 能量。
- 释放到一个质量上,弹性势能变成动能:$\tfrac{1}{2}kx^{2} = \tfrac{1}{2}mv^{2}$。
Two springs are pulled with the same force. The softer spring (smaller $k$) stores: · 两根弹簧用 相同的力 拉。更软的弹簧(更小的 $k$)储存:
Same $F$, smaller $k$ → larger extension $x$ → larger area $\tfrac{1}{2}Fx$, so more energy is stored. · 相同的 $F$,更小的 $k$ → 更大的伸长量 $x$ → 更大的面积 $\tfrac{1}{2}Fx$,所以储存更多能量。
When a stretched spring is released onto a mass, its elastic PE becomes ____ energy. · 当被拉伸的弹簧释放到一个质量上时,它的弹性势能变成 ____ 能。
Set · 集合 $\tfrac{1}{2}kx^{2} = \tfrac{1}{2}mv^{2}$ (plus $mgh$ if it rises) to find the speed. · 令 $\tfrac{1}{2}kx^{2} = \tfrac{1}{2}mv^{2}$(如果它上升,再加 $mgh$)来求速率。
Worked example: energy from a graph
- A spring of spring constant $40\ \text{N/m}$ is stretched by $0.10\ \text{m}$. Find the energy stored, and the extra energy needed to stretch it to $0.20\ \text{m}$.
- Energy stored is the area under the force-extension line, a triangle: $E = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(40)(0.10)^2 = 0.20\ \text{J}$.
- At $0.20\ \text{m}$: $E = \tfrac{1}{2}(40)(0.20)^2 = 0.80\ \text{J}$.
- The extra energy is the difference, $0.80 - 0.20 = 0.60\ \text{J}$, not $0.20\ \text{J}$ again.
- Energy goes as $x^2$: doubling the extension gives four times the energy, so the second half of a stretch always costs more than the first.
例题:从图上求能量
- 一根劲度系数 $40\ \text{N/m}$ 的弹簧被拉长 $0.10\ \text{m}$。求储存的能量,以及再拉到 $0.20\ \text{m}$ 还需要多少能量。
- 储存的能量是力-伸长线下面的面积,一个三角形:$E = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(40)(0.10)^2 = 0.20\ \text{J}$。
- 在 $0.20\ \text{m}$ 处:$E = \tfrac{1}{2}(40)(0.20)^2 = 0.80\ \text{J}$。
- 额外的能量是差值,$0.80 - 0.20 = 0.60\ \text{J}$,而不是再一个 $0.20\ \text{J}$。
- 能量按 $x^2$ 变化:伸长加倍能量变为四倍,所以拉伸的后一半总比前一半费力。
Marks that slip away
- On a force-extension graph the area is the energy and the gradient is the spring constant. Reading one for the other loses the whole question.
- Energy stored goes as $x^2$, not as $x$. The work between two extensions is the difference of two $\tfrac{1}{2}kx^2$ values.
- Elastic deformation returns to the original length when the force is removed; plastic deformation leaves a permanent extension.
- On a length-force graph only the part beyond the original length $L_0$ is extension. The area right down to the axis is not work.
- For a curved line the area must be estimated by counting squares, because $\tfrac{1}{2}kx^2$ assumes a straight line.
容易丢掉的分
- 在力-伸长图上,面积是能量,斜率是劲度系数。把两者读反就整道题都丢了。
- 储存的能量按 $x^2$ 变化,不是按 $x$。两个伸长量之间所做的功是两个 $\tfrac{1}{2}kx^2$ 之差。
- 弹性形变在撤去力后回到原长;塑性形变留下永久的伸长。
- 在长度-力图上,只有超出原长 $L_0$ 的那一部分才是伸长。一直算到坐标轴的面积不是功。
- 对弯曲的线,面积必须用数格子来估算,因为 $\tfrac{1}{2}kx^2$ 假定的是直线。
A spring of spring constant 40 N/m is stretched from 0.10 m to 0.20 m. How much extra energy does that take, in J? · 一根劲度系数 40 N/m 的弹簧从伸长 0.10 m 拉到 0.20 m。这额外需要多少能量(J)?
0.80 J minus 0.20 J = 0.60 J. Energy goes as x^2, so the second half of a stretch costs three times the first, not the same again. · 0.80 J 减 0.20 J = 0.60 J。能量按 x^2 变化,所以拉伸的后一半是前一半的三倍,而不是再来一次同样的量。
On a force-extension graph for a spring, which are true? Select all · 所有 that apply. · 关于弹簧的力-伸长图,哪些是对的?选出所有适用的。
Half k x squared assumes a straight line, so it is simply unavailable once the graph curves. Counting squares is then the only route. · 二分之一 k x 平方假定的是直线,所以图一旦弯曲它就用不上了。那时只能数格子。
Doubling the extension of a spring doubles the energy stored in it. · 弹簧的伸长加倍会使储存在其中的能量加倍。
It quadruples it, because the energy goes as x squared. Treating stored energy as proportional to extension is one of this topic's standard errors. · 是变为四倍,因为能量按 x 的平方变化。把储存能量当作与伸长成正比是这个主题的标准错误之一。
You've got it
- elastic returns to shape; plastic leaves a permanent extension (past the elastic limit)
- energy stored = area under the force–extension graph
- for a Hooke material $E_{\text{P}} = \tfrac{1}{2}kx^{2}$
你掌握了
- 弹性 恢复形状;塑性 留下永久伸长(超过弹性极限)
- 储存的能量 = 力–伸长量图线下的 面积
- 对于胡克材料 $E_{\text{P}} = \tfrac{1}{2}kx^{2}$