Neutral point in unloaded span of continuous beam

Neutral point

When one span of a continuous beam is loaded, the bending moment diagram of every unloaded span is a straight line with moments of opposite sign at its two ends (the beam is hogging at one support and sagging at the other end of the span). The point on the unloaded span where this line crosses the axis, that is, where the bending moment is zero, is the neutral point (also called the focal point). It is a point of contraflexure in that span.

Properties:

• Its position depends only on the spans, the stiffnesses ($EI$) and the end condition, not on the load: it is the same for any loading on the other spans lying on the same side of it.
• Once it is known, the whole BMD of the unloaded span follows from one support moment, since the line passes through the neutral point.
• There is one neutral point (left) for loading on the right, and another (right) for loading on the left.

Recurrent formula

Consider a continuous beam with spans numbered from the left; span $n$ lies between supports $n-1$ and $n$, with length $L_n$ and moment of inertia $I_n$. Let the load be on a span to the right of span $n+1$ (all spans from the left end up to $n+1$ unloaded), and let moments be taken with one sign convention (hogging or sagging) at all supports.

Three-moment equation at support $n$ for the unloaded spans $n$ and $n+1$ (the load terms are zero):

$$M_{n-1}\frac{L_n}{I_n} + 2M_n\left(\frac{L_n}{I_n}+\frac{L_{n+1}}{I_{n+1}}\right) + M_{n+1}\frac{L_{n+1}}{I_{n+1}} = 0$$

Because the moment in an unloaded span is linear, the moments at its two ends have opposite signs and a fixed ratio. Define the left focal ratio of span $n$:

$$k_n = -\frac{M_{n-1}}{M_n}$$

and similarly for span $n+1$: $M_n = -k_{n+1}M_{n+1}$.

Substitute $M_{n-1} = -k_nM_n$ in the equation:

$$M_n\left[(2-k_n)\frac{L_n}{I_n} + 2\frac{L_{n+1}}{I_{n+1}}\right] = -M_{n+1}\frac{L_{n+1}}{I_{n+1}}$$

so that $M_n/M_{n+1} = -k_{n+1}$ gives the recurrence formula

$$\boxed{\,k_{n+1} = \frac{L_{n+1}/I_{n+1}}{(2-k_n)\,L_n/I_n + 2\,L_{n+1}/I_{n+1}}\,}$$

For equal spans with constant $EI$ this becomes $k_{n+1} = \dfrac{1}{4-k_n}$.

Starting value: if the left end of the beam is hinged, $M_0 = 0$ so $k_1 = 0$; if it is fixed, the slope there is zero, $2M_0 + M_1 = 0$, so $k_1 = \tfrac12$.

Focal point: the straight BMD of unloaded span $n$ is $M(x) = M_n\left[1 - (1+k_n)\dfrac{x}{L_n}\right]$, with $x$ measured from support $n$, so it crosses the beam axis at $x = \dfrac{L_n}{1+k_n}$ from support $n$. This point is the left focal point, and it is the same for any loading on the spans to the right of span $n$.

Example (equal spans, constant $EI$, hinged end)

$k_1 = 0$, $k_2 = \dfrac{1}{4} = 0.25$, $k_3 = \dfrac{1}{4-0.25} = 0.2667$. In the second span the neutral point lies $L/(1+k_2) = 0.8L$ from its right support (0.2L from the left support).