--- title: "Consistent deformation: frame BMD with 20 kN" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:consistent-deformation-frame-bm-20kn" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/consistent-deformation-frame-bm-20kn site: BCT Engineering --- # Consistent deformation: frame BMD with 20 kN Theory of Structures II · Chapter 2: Force method - Asked in: 2079 Baishakh, 10 marks - Video: [Consistent Deformation Method - Analysis of Frames - Problem No 1](https://www.youtube.com/watch?v=yKqGc4EiqtQ) — Stan Academy - 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/consistent-deformation-frame-bm-20kn - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Use the consistent deformation method to draw the BM diagram of the given frame. [Figure: frame with column AB (A fixed at the bottom, B at top, height 4 m) carrying 10 kN/m UDL along its height, beam BCD with BC = CD = 3 m, 20 kN downward at C, EI constant, D supported at the right end.] ## Short answer Release the roller at D, taking its vertical reaction as redundant. Compute primary moment M0 and unit-load moment m, integrate to get ΔD0 = –2530/EI and fDD = 216/EI. Compatibility gives RD = 11.71 kN upward. Final moments: MA = 69.72 kNm (outer face tension), peak sagging 35.14 kNm at C. ## Answer Reading of the figure: A is fixed at the base of the column AB (4 m), the beam BCD has $BC = CD = 3$ m, the column carries a horizontal UDL of 10 kN/m (towards the right), a 20 kN load acts down at C, and D is on a roller (vertical support). EI is constant. The frame is indeterminate to the first degree ($r = 3 + 1 = 4$). Take the vertical reaction $R_D$ at D as the redundant. ```text 20 kN | B ------C------ D (roller) 10 ->| v o kN/m | A (fixed) ``` #### Step 1: Primary structure (D released), loads only It is a cantilever fixed at A. Downward deflection of D under the loads is $\Delta_D$. | Section | $M_0$ (kNm) | $m$ (unit load at D) | |---|---|---| | D | 0 | 0 | | C | 0 | 3.00 | | B (beam) | -60.00 | 6.00 | | B (column) | -60.00 | 6.00 | | mid AB | -80.00 | 6.00 | | A | -140.00 | 6.00 | ($M_0$ is the moment of the loads on the cantilever (hogging, negative) and $m$ is the moment due to a unit upward force at D (sagging, positive), so their product is negative.) For example, at A: $M_0 = -(10 \times 4 \times 2 + 20 \times 3) = -140$ kNm and $m = +6$ kNm per kN. #### Step 2: Deformations by the unit-load method $$ \begin{aligned} \Delta_{D0} &= \int \frac{M_0\, m}{EI}\,dx = \frac{-2530}{EI}\ \ (\text{i.e. }2530/EI\ \text{downward}) \\ f_{DD} &= \int \frac{m^2}{EI}\,dx = \frac{216}{EI} \end{aligned} $$ (Column AB: $6^2 \times 4 = 144$; beam BC: $\frac{3}{3}(36 + 18 + 9) = 63$; beam CD: $\frac{3}{3}(9) = 9$; total 216.) #### Step 3: Consistent deformation (deflection at D is zero) $$ \Delta_{D0} + f_{DD} R_D = 0 \ \Rightarrow\ R_D = \frac{2530}{216} = 11.713\ \text{kN}\ (\uparrow) $$ #### Step 4: Other reactions (statics) - $V_A = 20 - 11.713 = 8.287$ kN ($\uparrow$) - $H_A = 10 \times 4 = 40$ kN ($\leftarrow$) - $M_A$ from $\Sigma M_A = 0$: $M_A = 40 \times 2 + 20 \times 3 - 11.713 \times 6 = 69.72$ kNm (anticlockwise reaction). #### Step 5: Bending moments $M = M_0 + R_D\, m$ | Point | BM (kNm) (+ve = tension on inner/under face) | Tension side | |---|---|---| | A | -69.72 | outer (left) face of column | | Mid AB | -9.72 | outer face | | B | 10.28 | inner (right) face of column / underside of beam | | C | 35.14 | underside | | D | 0 | | The BM in the column is $M(y) = -69.72 + 40y - 5y^2$ (y from A), which changes sign at $y = 2.57$ m from A. In the beam, BM rises linearly from 10.28 kNm at B to 35.14 kNm under the 20 kN load, then falls linearly to zero at D. **Answer: $R_D = 11.71$ kN up; $M_A = 69.72$ kNm; peak sagging BM 35.14 kNm at C.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the consistent-deformation (compatibility) condition used to find R_D? A. The slope at A is zero: theta_A0 + f_AA R_D = 0 B. The net vertical deflection at D is zero: Delta_D0 + f_DD R_D = 0 C. The sum of vertical forces is zero: R_D = 20 kN D. The deflection at D under the loads equals f_DD, so R_D = 1 Answer: B. D is a support that does not move vertically, so the deflection from the loads on the primary structure plus that from R_D must cancel. (Key point: Compatibility condition) 2. What does f_DD = 216/EI represent? A. The deflection at D due to the real loads on the primary structure B. The reaction at D produced by a unit deflection C. The vertical deflection at D of the primary cantilever due to a unit upward load at D, found as the integral of m^2/EI D. The fixed-end moment at A divided by EI Answer: C. f_DD is the flexibility coefficient: the deflection at D per unit load at D, obtained by summing m^2 L/EI over column and beam. (Key point: Flexibility coefficient f_DD) 3. How is the final BM obtained and what is its value at A? A. M = M0 + R_D m; at A it is -140 + 11.713 x 6 = -69.72 kNm B. M = M0 only; at A it is -140 kNm C. M = M0 - R_D; at A it is -151.7 kNm D. M = M0 x R_D; at A it is -1640 kNm Answer: A. By superposition, the primary-structure moment M0 is corrected by R_D times the unit-load moment m, which reduces the hogging at A to -69.72 kNm. 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