剪力与弯矩：载荷的传记
引入| English | 释义 | 拼音 |
|---|---|---|
| shear force | 剪力 | jiǎn lì |
| bending moment | 弯矩 | wān jǔ |
The load's biography
- The bracket arm is a simply supported beam, $L = 3\ \mathrm{m}$, central $P = 6\ \mathrm{kN}$. Cut it and balance each side: reactions $3\ \mathrm{kN}$, the shear force 剪力 $V$ steps $+3 \to -3$ kN at the load, and the bending moment 弯矩 $M$ ramps to $M_{max} = PL/4 = \mathbf{4.5}\ \mathrm{kN\,m}$ mid-span.
Diagram, convention, or relation?
Route each request to the diagram rule that answers it.
Simply supported 3 m span, central P = 6 kN: maximum moment (one decimal, kN·m)?
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PL/4.
Under a central point load on a simply supported span, the moment diagram is:
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Linear ramps to PL/4 at the load point.
Sections, with signs
- Sagging is positive - the sign convention is part of the answer, not decoration. The $3\ \mathrm{kN}$ left reaction lifts the left cut, so $V = +3$ kN on the left half and $M = 3x$ grows to $\mathbf{4.5}$ kN·m at the load.
Rectangle 40 x 80 mm (b x h): I (four significant figures, 10^-6 m^4)?
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b h cubed over 12.
M_max = 4.5 kN·m on that section: maximum bending stress (one decimal, MPa)?
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M c over I.
For a symmetric section, c is the half-depth.
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The farthest fibre sits half the depth from the neutral axis.
The differential chain
- Load, shear and moment tie together: $dV/dx = -w(x)$ and $dM/dx = V$. A point load steps $V$ and kinks $M$; a uniform load tilts $V$ and bends $M$ parabolically - $w = 2\ \mathrm{kN/m}$ over the same span gives $M_{max} = wL^2/8 = \mathbf{2.25}\ \mathrm{kN\,m}$.
- Read the shape back: hinges carry no moment, free ends no shear, and $M$ peaks exactly where $V$ crosses zero.
Section modulus I/c (four significant figures, 10^-6 m^3)?
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1.707e-6 over 0.04.
Three diagram slips
- Dropping the sign convention. Sagging vs hogging changes every downstream sign.
- Peaking $M$ at a support. It peaks at the $V = 0$ point.
- Point-load shortcuts under distributed loads. Only reactions and the peak survive $wL^2/8$.
Recap
- Cut and balance: $V$ steps $\pm 3$ kN, $M_{max} = PL/4 = 4.5$ kN·m.
- $dV/dx = -w$, $dM/dx = V$; uniform load: $M_{max} = wL^2/8 = 2.25$ kN·m.
- $M$ peaks where $V$ crosses zero.