Compatibility condition; portal frame temperature variation

Physical meaning of the compatibility condition

A statically indeterminate structure has more supports or members than are needed for equilibrium. When the redundants are removed, the primary structure deforms freely, and its displacements at the released points generally contradict the real conditions of the structure (for example, a fixed or hinged support cannot move). The compatibility condition states that the redundants must be of such a size that the displacements of the real structure at the releases agree with the real support conditions (usually zero, or equal to the settlement). It ensures that the deformed shape is continuous and fits the supports.

Derivation (one redundant $X_1$)

Let the primary structure carry the loads and the unknown $X_1$. By superposition, the displacement at the release point in the direction of $X_1$ is

$$\Delta_1 = \Delta_{10} + f_{11}X_1$$

where $\Delta_{10}$ = displacement of the primary structure due to the loads (and temperature, settlement of other supports, lack of fit) and $f_{11}$ = displacement due to a unit $X_1$. Compatibility requires $\Delta_1$ to equal the actual displacement $\delta_1$ of the support (zero for a rigid support):

$$\Delta_{10} + f_{11}X_1 = \delta_1 \quad\Rightarrow\quad X_1 = \frac{\delta_1 - \Delta_{10}}{f_{11}}$$

For $n$ redundants: $\{\Delta_0\} + [F]\{X\} = \{\delta\}$, with $f_{ij} = \int \dfrac{m_im_j}{EI}dx$ and $\Delta_{i0} = \int \dfrac{M_0m_i}{EI}dx$. For a temperature change, the load term is

$$\Delta_{it} = \int m_i\,\frac{\alpha\,(t_1 - t_2)}{d}\,dx + \int n_i\,\alpha\,t_m\,dx,\qquad t_m = \frac{t_1+t_2}{2}$$

where $d$ is the depth, $m_i$ the unit BM and $n_i$ the unit axial force.

Portal frame with temperature variation

Reading of the figure: hinged portal ABCD, height $h = 6$ m, span $L = 3$ m, EI constant. The beam is warmer on the outer (top) face by $t_1 = 20^\circ$C against $t_2 = 10^\circ$C on the inner face. Section $0.3\times0.6$ m ($d = 0.6$ m). Temperatures are changes from the original state.

$D_s = 4 - 3 = 1$. Release D horizontally and take the thrust $H$ at D (positive outward) as the redundant.

   B ____________ C   t1 = 20 (outer)
   |              |   t2 = 10 (inner)
   |              |
   | 6 m          | 6 m
   |              |
  A o            o D <- H
   |<--- 3 m --->|

Elastic constants

$$E = 5000\sqrt{f_{ck}} = 5000\sqrt{20} = 22360.7\ \text{MPa} = 22360680\ \text{kN/m}^2$$ $$I = \frac{0.3\times0.6^3}{12} = 0.0054\ \text{m}^4,\qquad EI = 120748\ \text{kNm}^2$$

Flexibility coefficient

A unit outward force at D gives $m = y$ in each column (from the hinge) and $m = 6$ in the beam:

$$f_{11} = \frac{1}{EI}\left[2\cdot\frac{6^3}{3} + 6^2(3)\right] = \frac{144 + 108}{EI} = \frac{252}{EI} = 2.087\times10^{-3}\ \text{m/kN}$$

Displacement of D due to temperature (primary structure)

Mean temperature of the beam $t_m = (20+10)/2 = 15^\circ$C; temperature difference $t_1 - t_2 = 10^\circ$C.

• Axial expansion of the beam (moves D outward): $\alpha\,t_m\,L = 11\times10^{-6}\times15\times3 = 4.950\times10^{-4}$ m
• Curvature $\kappa = \dfrac{\alpha(t_1-t_2)}{d} = \dfrac{11\times10^{-6}\times10}{0.6} = 1.833\times10^{-4}$ per m. The beam bends concave downward (hot face on top), which moves D inward: $-\,\kappa\,(h)(L) = -1.833\times10^{-4}\times6\times3 = -3.300\times10^{-3}$ m

$$\Delta_{1t} = 4.950\times10^{-4} + (-3.300\times10^{-3}) = -2.805\times10^{-3}\ \text{m}\ (\text{inward})$$

Compatibility ($\delta_1 = 0$)

$$\Delta_{1t} + f_{11}H = 0 \ \Rightarrow\ H = \frac{2.805\times10^{-3}}{2.087\times10^{-3}} = 1.344\ \text{kN}$$

The support reaction at D is 1.344 kN acting outward (away from the frame) and an equal and opposite reaction acts at A. The vertical reactions are zero.

Bending moments

$M_B = M_C = H\,h = 1.344\times6 = 8.06$ kNm, with tension on the inside (sagging in the beam). The columns have a linear BM from 0 at the hinge to 8.06 kNm at the joint.

Answer: flexibility coefficient $f_{11} = 2.087\times10^{-3}$ m/kN, thermal displacement $\Delta_{1t} = -2.805\times10^{-3}$ m, redundant $H = 1.344$ kN (outward at each hinge), corner moment $8.06$ kNm.