Combined stress and Mohr's circle
Введено| English | Значение |
|---|---|
| Mohr's circle | Mohr's circle |
| principal stress | principal stress |
Stresses at an angle
- A bracket point carries $\sigma_x = 100$, $\sigma_y = 40$, $\tau_{xy} = 25$ MPa - but a weld or gauge plane sits at an angle. Mohr's circle 莫尔圆 draws the whole state: centre $\sigma_{avg} = (\sigma_x + \sigma_y)/2$, radius $R = \sqrt{((\sigma_x - \sigma_y)/2)^2 + \tau_{xy}^2}$.
Deflection checks are:
Сложность 1/5 · 100 монет + небольшой бонус за первое завершение сегодня
A calm beam can still sag past tolerance.
The principals
- For the point above: centre $\mathbf{70}$, $R = \sqrt{30^2 + 25^2} = \sqrt{1525} = \mathbf{39.1}$ MPa, so the principal stresses 主应力 $\sigma_1 = 70 + 39.1 = \mathbf{109.1}$ and $\sigma_2 = 70 - 39.1 = \mathbf{30.9}$ MPa sit at $\theta_p = \tan^{-1}(50/60)/2 = \mathbf{19.9°}$.
- The circle turns $2\theta$ as the element turns $\theta$ - halve carefully.
Which case, which cube?
Match each deflection request to its standard case.
Simply supported 3 m, central P = 6 kN, EI = 341.3 N·m^2 x 10^6: mid-span deflection (two decimals, mm)?
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
PL^3/48EI.
Cantilever 1.2 m, tip P = 800 N, same EI: tip deflection (two decimals, mm)?
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
PL^3/3EI.
Halving a cantilever span cuts its tip deflection to one eighth.
Сложность 1/5 · 100 монет + небольшой бонус за первое завершение сегодня
delta goes with L^3.
Which shear is maximum
- Both principals positive: the absolute-maximum shear is $\sigma_1/2 \approx \mathbf{54.5}$ MPa.
- Swap the second normal stress to $-40$ MPa: centre 40, $R = \sqrt{80^2 + 30^2} = \mathbf{85.4}$, so $\sigma_1 = \mathbf{125.4}$ and $\sigma_2 = \mathbf{-45.4}$ MPa - now the principals straddle zero and the absolute-maximum shear IS the radius, $\mathbf{85.4}$ MPa ✓.
Doubling the section depth multiplies the deflection by (a factor, three decimals)?
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
I carries h^3.
Three rotation slips
- Halving the wrong angle. The circle angle is twice the physical one.
- In-plane $R$ as the absolute maximum for same-sign principals. Use $\sigma_1/2$.
- $Q$ as a force. It is a first moment of area.
Recap
- Centre $\sigma_{avg}$, radius $R$; principals on the $\tau = 0$ axis.
- (100, 40, 25): $R = 39.1$, $\sigma_1 = 109.1$, $\sigma_2 = 30.9$ MPa at 19.9°.
- Same sign: max shear $\approx \sigma_1/2 = 54.5$ MPa; straddling: it is $R$.