Why force method; continuous beam with settlements

Why the flexibility method is a force method

In the flexibility method the unknowns are redundant forces (reactions or internal forces). The redundants are applied as loads on a statically determinate primary structure, and their values are found from the compatibility equations $\sum f_{ij}X_j + \Delta_{i0} = \delta_i$, in which the flexibility coefficients $f_{ij}$ convert forces into displacements. Since the primary unknowns are forces, the method is called the force method (the displacements are found afterwards).

Continuous beam with settlements

Reading of the figure: beam ABC on three supports, A a hinge and B, C rollers, $AB = BC = 4.8$ m. $E = 232$ kN/mm$^2 = 232\times10^6$ kN/m$^2$ and $I = 112.5\times10^6$ mm$^4 = 1.125\times10^{-4}$ m$^4$, so

$$EI = 232\times10^6\times1.125\times10^{-4} = 26100\ \text{kNm}^2$$

There is no load, so the reactions arise only from the settlements: B settles 18 mm and C settles 12 mm.

$D_s = 1$. Remove the support B and take $R_B$ (upward) as the redundant. The primary structure is a simply supported beam AC of span $9.6$ m.

Displacement of B in the primary structure

C settles by 12 mm, so the straight primary beam moves as a rigid body, and the point B moves down by half of this:

$$\Delta_{10} = 0.5\times12 = 6\ \text{mm downward} = -0.006\ \text{m (in the direction of } R_B)$$

Flexibility coefficient

Deflection at mid-span of a simply supported beam due to a unit load there:

$$f_{11} = \frac{L^3}{48EI} = \frac{9.6^3}{48\times26100} = 7.062\times10^{-4}\ \text{m/kN}$$

Compatibility

In the actual beam B is displaced downward by 18 mm, that is, $-0.018$ m in the direction of $R_B$:

$$\Delta_{10} + f_{11}R_B = -0.018 \ \Rightarrow\ R_B = \frac{-0.018 + 0.006}{7.062\times10^{-4}} = -16.99\ \text{kN}$$

The negative sign means that the reaction at B acts downward (the support pulls the beam down).

Other reactions

$$R_A = R_C = -\frac{R_B}{2} = 8.50\ \text{kN}\ (\uparrow)$$

Check: $R_A + R_B + R_C = 0$.

Bending moment

$M_B = R_A\times4.8 = 40.78$ kNm (sagging, tension at the bottom); $M_A = M_C = 0$. The SF is $+8.50$ kN in AB and $-8.50$ kN in BC.

Answer: $R_A = R_C = 8.50$ kN (upward) and $R_B = -16.99$ kN (i.e. 16.99 kN downward).