--- title: "Consistent deformation: frame with 12 kN, SFD and BMD" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:consistent-deformation-frame-12kn-sfd-bmd" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/consistent-deformation-frame-12kn-sfd-bmd site: BCT Engineering --- # Consistent deformation: frame with 12 kN, SFD and BMD Theory of Structures II · Chapter 2: Force method - Asked in: 2070 Chaitra (old course), 8 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-12kn-sfd-bmd - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Compute the reactions and draw shear force and bending moment diagram for the frame shown in figure below. Use the consistent deformation method. [Figure: frame ABC; horizontal member AB (2I) of 6 m (3 m + 3 m) with 12 kN at mid-span, A hinged; vertical member BC (I) of 4 m with C hinged.] ## Short answer Take H_A as redundant; compatibility Δ10+f11H_A=0 gives H_A=1.446 kN. Then H_C=1.446 kN, V_A=5.036 kN, V_C=6.964 kN; bending moments are M_D=15.11 kNm sagging and M_B=-5.79 kNm hogging. ## Answer Reading of the figure: the horizontal member AB is 6 m long ($2I$) with a 12 kN load at mid-span D (3 m from A), A hinged; the vertical member BC is 4 m long ($I$) with C hinged at its foot. $E$ is constant. $r = 4$, so $D_s = 1$. Take the horizontal reaction $H_A$ at A as the redundant. Release A horizontally (roller): the primary structure is a simply supported frame. ```text 12 kN v A o-----D-----B H_A-> | | 4 m o C |<-- 3 -->|<-- 3 -->| ``` #### Primary structure (A on a roller) $V_A = V_C = 6$ kN. A unit horizontal force at A is balanced by a horizontal reaction at C and a vertical couple: $V = 4/6 = 0.667$ kN. | Section | $M_0$ (kNm) | $m$ (unit $H_A$) | |---|---|---| | A | 0 | 0 | | D | 18 | -2.0 | | B | 0 | -4.0 | | C | 0 | 0 | (With $m$ negative at D and B because a unit force at A produces hogging in AB: $m_D = -0.667\times3 = -2$, $m_B = -4$.) #### Deformations $$ \begin{aligned} f_{11} &= \frac{1}{2EI}\left[\frac{3}{3}(4) + \frac{3}{3}(4 + 8 + 16)\right] + \frac{1}{EI}\left[\frac{4}{3}(16)\right] = \frac{16 + 21.333}{EI} = \frac{37.33}{EI} \\ \Delta_{10} &= \frac{1}{2EI}\left[\frac{3}{3}(18)(-2) \cdot 1 + \frac{3}{6}\left\{2(18)(-2) + 18(-4)\right\}\right] = \frac{-54.00}{EI} \end{aligned} $$ #### Consistent deformation (A does not move horizontally) $$ \Delta_{10} + f_{11}H_A = 0 \ \Rightarrow\ H_A = \frac{54.00}{37.33} = 1.446\ \text{kN}\ (\rightarrow) $$ #### Reactions - A: $H_A = 1.446$ kN ($\rightarrow$), $V_A = 5.036$ kN ($\uparrow$) - C: $H_C = 1.446$ kN ($\leftarrow$), $V_C = 6.964$ kN ($\uparrow$) #### Bending moment $M = M_0 + H_A\,m$ (kNm) | Point | BM | Tension side | |---|---|---| | A | 0 | | | D | 15.11 | bottom | | B | -5.79 | top of beam / outside of corner | | C | 0 | | #### Shear force and axial force (kN) | Member | SF | Axial force | |---|---|---| | AD | +5.04 | -1.446 (compression) | | DB | -6.96 | -1.446 (compression) | | BC | +1.45 | -6.964 (compression) | **Answer: $H_A = H_C = 1.446$ kN; $V_A = 5.036$ kN and $V_C = 6.964$ kN; $M_D = 15.11$ kNm (sagging), $M_B = -5.79$ kNm (hogging).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The frame has r = 4 reactions, so D_s = 1. What is the redundant and the primary structure used? A. Vertical reaction at C; cantilever B. Horizontal reaction H_A; A released to a roller C. Moment at B; hinge at B D. Horizontal reaction at C; C released to a hinge Answer: B. Releasing A horizontally (roller) leaves a stable determinate frame; H_A is the redundant. (Key point: Choice of redundant) 2. What is the consistent-deformation equation used? A. Delta10 + f11 H_A = 0, as A cannot move horizontally B. Delta10 = f11 H_A + 12 C. f11 H_A = 12 kN D. Delta10 - f11 H_A = V_A Answer: A. The real support A is a hinge, so its horizontal displacement is zero. (Key point: Consistent deformation condition) 3. With Delta10 = -54/EI and f11 = 37.33/EI, what is H_A? A. 0.69 kN B. 3.5 kN C. 5.04 kN D. 1.446 kN Answer: D. H_A = 54.00/37.33 = 1.446 kN. (Key point: Value of H_A) 4. With M = M0 + H_A m, M_D = 18 + 1.446(-2) is about what? A. 20.9 kNm B. 15.11 kNm sagging C. 18 kNm D. -5.79 kNm Answer: B. 18 - 2.892 = 15.11 kNm; at B the moment is 1.446(-4) = -5.79 kNm. 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