扭转：剪应力线性增长到表面
引入| English | 释义 | 拼音 |
|---|---|---|
| torsion/ˈtɔːʃn/ | 扭转 | niǔ zhuǎn |
| angle of twist | 扭转角 | niǔ zhuǎn jiǎo |
Torque on a shaft
- The drive shaft is the assembly's torsion 扭转 member: $d = 30\ \mathrm{mm}$, $T = 300\ \mathrm{N\,m}$, $L = 1.2\ \mathrm{m}$. Shear runs $\tau = T\rho/J$ - zero on the axis, worst at the surface - with $J = \pi d^4/32 = 7.952 \times 10^{-8}\ \mathrm{m^4}$ for a solid shaft.
Stress or stiffness?
Sort each shaft question by what it asks.
Solid shaft d = 30 mm, T = 300 N·m: surface shear (one decimal, MPa)?
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
16T over pi d cubed.
Same shaft, L = 1.2 m, G = 80 GPa: twist angle (two decimals, degrees)?
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
TL over GJ, then x 180/pi.
Shear stress in a solid circular shaft under torsion is:
难度 1/5 · 今日首次完成可获 100 金币及少量额外奖励
tau = T rho / J grows linearly with radius.
Stress and twist
- Surface stress: $\tau_{max} = 16T/\pi d^3 = \mathbf{56.6}$ MPa. The angle of twist 扭转角 accumulates along the length: $\phi = TL/(GJ) = \mathbf{0.0566}$ rad $= \mathbf{3.24°}$ with $G = 80\ \mathrm{GPa}$ - stiffness, not strength, sizes most shafts.
T = 400 N·m at tau_allow = 60 MPa: minimum diameter (one decimal, mm)?
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
cube root of 16T over pi tau.
T = 300 N·m at 180 rpm: transmitted power (two decimals, kW)?
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
T omega with omega = 18.85 rad/s.
Sizing and selling
- A $T = 400\ \mathrm{N\,m}$ shaft at $\tau_{allow} = 60$ MPa needs $d^3 = 16 \times 400 \times 10^3/(\pi \times 60) = 33{,}960\ \mathrm{mm^3}$, so $d = \mathbf{32.4}$ mm - round up to stock size.
- The same 300 N·m at $180\ \mathrm{rpm}$ ($\omega = 18.85$ rad/s) transmits $P = T\omega = \mathbf{5.65}$ kW: stress sizes the shaft, power sells it.
A longer shaft of the same diameter twists more.
难度 1/5 · 今日首次完成可获 100 金币及少量额外奖励
phi = TL/GJ is linear in L.
Three torsion slips
- Using $I$ for $J$. Torsion wants $\pi d^4/32$, bending $\pi d^4/64$.
- Mixing N·mm with m⁴ in $\phi = TL/GJ$. The radians vanish by six decades.
- Solid-shaft formulas on tubes. Thin walls wrinkle and crush.
Recap
- $\tau_{max} = 16T/\pi d^3 = 56.6$ MPa at $d = 30$ mm, $T = 300$ N·m.
- $\phi = TL/GJ = 3.24°$ over 1.2 m; $P = T\omega = 5.65$ kW at 180 rpm.
- Design: $d = 32.4$ mm for 400 N·m at 60 MPa.