BMD of portal frame with support settlement

Settlement of a support in an indeterminate frame produces bending moments even without any external load, because the frame is forced to fit the new support position. The BMD is drawn from the redundant reactions caused by the settlement.

Procedure

1. Find $D_s$ and choose the redundants (for example the reactions of the settling support). Remove them to get the primary structure.
2. The primary structure moves as a rigid body when the support settles, so $M_0 = 0$ and $\Delta_{i0} = 0$.
3. Find the flexibility coefficients $f_{ij}$ from unit-load moment diagrams $m_i$.
4. Write compatibility: the displacement at each redundant must equal the prescribed settlement $\delta_i$ (measured positive in the direction of $X_i$): $\sum_j f_{ij}X_j = \delta_i$.
5. Solve for $X_j$ and compute $M = \sum X_j m_j$. Since there are no member loads, the BM varies linearly between the joints, so the BMD is a set of straight lines joining the joint moments. Draw each ordinate on the tension side.

Note that the moments are proportional to $EI\,\delta/L^2$, so a stiffer frame develops larger moments for the same settlement.

Example

Portal frame ABCD, $h = L = 4$ m, $EI = 20000$ kNm$^2$ constant, A fixed, D hinged. D settles vertically by $\delta = 10$ mm.

Redundants: $X_1$ = $H_D$ ($\rightarrow$), $X_2$ = $V_D$ ($\uparrow$). The unit moment diagrams are: $m_1$ = 0 at A rising to 4 at B, 4 along BC, falling to 0 at D; $m_2$ = 4 along AB, falling from 4 at B to 0 at C.

$$f_{11} = \frac{106.67}{EI},\quad f_{12} = \frac{64}{EI},\quad f_{22} = \frac{85.33}{EI}$$ $$\begin{aligned} 106.67X_1 + 64X_2 &= 0 \\ 64X_1 + 85.33X_2 &= -0.01\times20000 = -200 \end{aligned}$$ $$X_1 = H_D = 2.56\ \text{kN}\ (\rightarrow),\qquad X_2 = V_D = 4.26\ \text{kN}\ (\text{downward, since D is pulled down})$$

Bending moments (kNm; + tension on the inner face)

BM is linear in every member; it is hogging (outer tension) in the left column and at B, and sagging at C (inner tension).