Beam stresses and deflection
Введено| English | Значение |
|---|---|
| flexure formula | flexure formula |
| transverse shear | transverse shear |
| elastic curve | elastic curve |
Two stresses, one section
- The diagrams price the beam: the flexure formula 弯曲公式 $\sigma = Mc/I$ on the outer fibres, and transverse shear 横向剪切 $\tau = VQ/(It)$ peaking on the neutral axis.
Rectangle 40 x 80 mm, V = 3 kN: maximum transverse shear (two decimals, MPa)?
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
1.5 V over A.
The flexure route
- For the $40 \times 80\ \mathrm{mm}$ rectangle at $M_{max} = 4.5\ \mathrm{kN\,m}$: $I = bh^3/12 = \mathbf{1.707 \times 10^{-6}}\ \mathrm{m^4}$, so $\sigma = Mc/I = \mathbf{105.5}$ MPa - and the catalogue route $S = I/c = \mathbf{42.7 \times 10^{-6}}\ \mathrm{m^3}$ gives $\sigma = M/S$, same number, one lookup.
- The rectangle's shear peak: $\tau_{max} = 1.5V/A = 1.5 \times 3000/0.0032 = \mathbf{1.41}$ MPa at $V = 3$ kN - small beside 105.5, exactly what long slender beams show.
Which distribution, which case?
Route each beam question to its stress law or deflection case.
sigma_x = 100, sigma_y = 40, tau_xy = 25 MPa: Mohr radius (one decimal, MPa)?
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
sqrt(30^2 + 25^2).
Principal stress sigma_1 (one decimal, MPa)?
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
70 + 39.1.
Principal stress sigma_2 (one decimal, MPa)?
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
70 - 39.1.
Principal angle from the x-face (one decimal, degrees)?
Сложность 2/5 · 140 монет + небольшой бонус за первое завершение сегодня
atan(50/60)/2.
With both principal stresses positive, the absolute-maximum shear is:
Сложность 1/5 · 100 монет + небольшой бонус за первое завершение сегодня
The out-of-plane plane carries sigma_1/2 when both share a sign.
On Mohr's circle, a physical 20° rotation of the element appears as 40° on the circle.
Сложность 1/5 · 100 монет + небольшой бонус за первое завершение сегодня
The circle angle is twice the physical angle.
The sag
- The elastic curve 挠曲线 $EIv'' = M(x)$ integrates to the standard cases: central load $\delta = PL^3/(48EI) = \mathbf{9.89}$ mm mid-span; the same section as a $1.2\ \mathrm{m}$ cantilever with 800 N sags $PL^3/(3EI) = \mathbf{1.35}$ mm.
- Superposition sums the standard cases - point plus uniform is the sum of the two sags; and depth enters as $h^3$: double it, cut the sag eightfold.
Three stress slips
- Full depth for $c$ on a symmetric section. It is the half-depth.
- Judging stiffness by stress. A calm beam can still sag past tolerance - a serviceability limit.
- Mixing mm and m inside $I$. Three decades vanish silently.
Recap
- $\sigma = Mc/I = M/S = 105.5$ MPa; $\tau_{max} = 1.5V/A = 1.41$ MPa.
- $\delta = PL^3/48EI = 9.89$ mm; $PL^3/3EI = 1.35$ mm; superpose when loads combine.
- Depth is the cheap stiffness ($h^3$ in $I$).