Relation between w, V and M

For a beam carrying a distributed load of intensity $w$ (taken positive downward), the shear force $V$ and the bending moment $M$ at a section $x$ are related by

$$\frac{dV}{dx} = -w,\qquad \frac{dM}{dx} = V,\qquad \frac{d^2M}{dx^2} = -w$$

Derivation

Consider a small element of length $dx$ with load $w$ (acting down). Shear $V$ and moment $M$ act on the left face; $V + dV$ and $M + dM$ act on the right face.

        w (per unit length)
   ↓ ↓ ↓ ↓ ↓ ↓ ↓
  M ┌───────────┐ M + dM
 V ↑│           │↓ V + dV
    └───────────┘
        dx

Vertical equilibrium: $V - w\,dx - (V + dV) = 0 \Rightarrow \dfrac{dV}{dx} = -w$.

Moments about the right face (ignoring the second-order term $w\,dx^2/2$): $M + V\,dx - M - dM = 0 \Rightarrow \dfrac{dM}{dx} = V$.

Integrated forms between sections 1 and 2

$$V_2 - V_1 = -\int_{x_1}^{x_2} w\,dx\;(\text{= minus the area of the load diagram}),\qquad M_2 - M_1 = \int_{x_1}^{x_2} V\,dx\;(\text{= area of the shear force diagram})$$

Consequences

• The slope of the SFD at any point equals minus the load intensity there; the slope of the BMD equals the shear force.
• The bending moment is a maximum or minimum where $V = 0$ (or where $V$ changes sign).
• No load: the SFD is horizontal and the BMD is linear. UDL: the SFD is linear and the BMD is parabolic. Triangular load: the SFD is parabolic and the BMD is cubic.
• A point load causes a sudden jump in $V$ equal to the load; a couple causes a jump in $M$ equal to the couple, while the SFD is unaffected.