--- title: "Why force method; continuous beam with settlements" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:flexibility-force-method-settlement-continuous" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/flexibility-force-method-settlement-continuous site: BCT Engineering --- # Why force method; continuous beam with settlements Theory of Structures II · Chapter 2: Force method - Asked in: 2071 Shrawan, 3+7 marks - Video: [force method -support settlement](https://www.youtube.com/watch?v=B8bH_CXgyCo) — battle structures - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/flexibility-force-method-settlement-continuous - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Explain why the flexibility method is called a Force Method. Using the force method determine the reactions in the continuous beam shown in figure below, if support B settles 18 mm and support C settles 12 mm. Given EI is constant, $E = 232\ \text{kN/mm}^2$ and $I = 112.5\times10^{6}\ \text{mm}^4$. [Figure: beam ABC, all supports; AB = BC = 4.8 m.] ## Short answer The method's primary unknowns are redundant forces, hence called force method. For the beam with B settling 18 mm and C 12 mm, select R_B as redundant; compatibility with δ_B = -0.018 m gives R_B = -16.99 kN (downward), R_A = R_C = 8.50 kN upward. ## Answer #### 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).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the flexibility method called a force method? A. Its primary unknowns are joint displacements B. Its primary unknowns are redundant forces, applied as loads on a determinate primary structure C. It only works for structures carrying point forces D. It uses equilibrium equations only and ignores displacements Answer: B. The redundants (forces) are the unknowns; compatibility equations with flexibility coefficients find them, and displacements are found afterwards. (Key point: Why it is a force method) 2. For the two-span beam ABC with support B removed as the redundant, what is the primary structure? A. A cantilever fixed at A B. Two separate simply supported beams AB and BC C. A simply supported beam AC of span 9.6 m D. A fixed-fixed beam of span 9.6 m Answer: C. Removing the roller at B leaves A (hinge) and C (roller) supporting a 9.6 m simply supported beam. (Key point: Primary structure) 3. When C settles 12 mm, how far does B move in the primary structure (straight beam moving rigidly)? A. 12 mm downward B. 18 mm downward C. 6 mm downward D. Zero Answer: C. B is at mid-span of the rigid-body-moving beam, so it moves half of C's settlement, 6 mm. (Key point: Delta10 from settlement of C) 4. Compatibility gives R_B = (-0.018 + 0.006)/f11 = -16.99 kN. What does the negative sign mean? A. The reaction at B acts downward B. The calculation is wrong C. B has lifted off the support D. R_B acts upward with magnitude 16.99 kN Answer: A. R_B was assumed upward, so a negative value means the support actually pulls the beam down by 16.99 kN. 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