--- title: "Consistent deformation: frame, BMD, SFD, normal thrust" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:consistent-deformation-frame-15kn-10knm-sfd" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/consistent-deformation-frame-15kn-10knm-sfd site: BCT Engineering --- # Consistent deformation: frame, BMD, SFD, normal thrust Theory of Structures II · Chapter 2: Force method - Asked in: 2071 Chaitra, 15 marks - Video: [ANALYSIS OF FRAME SHEAR BENDING & AXIAL THRUST DIAGRAM | EX 03](https://www.youtube.com/watch?v=VhZ1_8AFJrI) — Structural Funda by:- KheNiSiR - 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-15kn-10knm-sfd - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Using the consistent deformation method analyse the frame shown in figure and draw bending moment, shear force and normal thrust diagram. [Figure: frame; left column 2EI with 10 kN/m UDL horizontally, height 3 m + 1 m...; beam EI of 2 m + 2 m with 15 kN downward at mid-span; right column 2EI.] ## Short answer Releasing hinge at D, choosing H_D and V_D as redundants, consistent deformation gives H_D = 3.09 kN (left) and V_D = 10.84 kN (up). Fixed-end reactions: H_A = 26.91 kN (left), V_A = 4.16 kN (up), M_A = 31.63 kNm (anticlockwise). ## Answer Reading of the figure (as assumed): portal frame of height 4 m and span 4 m. The left column AB ($2EI$) is fixed at A and carries a horizontal UDL of 10 kN/m over its lower 3 m (towards the right); the beam BC ($EI$, 2 m + 2 m) carries 15 kN downward at mid-span F; the right column CD ($2EI$) is hinged at D. $r = 3 + 2 = 5$, so $D_s = 2$. By the consistent deformation method, release the hinge at D and take the reactions $X_1$ (horizontal, $\rightarrow$) and $X_2$ (vertical, $\uparrow$) at D as redundants. The displacements of D in the primary structure must be cancelled by the redundants, that is, the total displacement at D along each redundant is zero. ```text B ----F---- C | 15 kN | 10-> | 4 m kN/m | | o D A (fixed) ``` #### Consistent deformation equations $$ \begin{aligned} \Delta_{10} + f_{11}X_1 + f_{12}X_2 &= 0 \\ \Delta_{20} + f_{21}X_1 + f_{22}X_2 &= 0 \end{aligned} $$ where $\Delta_{i0} = \int \dfrac{M_0\,m_i}{EI}dx$ (displacements of D in the primary structure) and $f_{ij} = \int\dfrac{m_im_j}{EI}dx$. $$ \begin{aligned} f_{11} &= \frac{85.333}{EI} \\ f_{12} &= \frac{48.000}{EI} \\ f_{22} &= \frac{53.333}{EI} \\ \Delta_{10} &= \frac{-256.9}{EI} \\ \Delta_{20} &= \frac{-430.0}{EI} \end{aligned} $$ $$ \begin{aligned} 85.333\,X_1 + 48.000\,X_2 &= 256.9 \\ 48.000\,X_1 + 53.333\,X_2 &= 430.0 \end{aligned} $$ $$ X_1 = H_D = -3.09\ \text{kN},\qquad X_2 = V_D = 10.84\ \text{kN} $$ #### Final bending moments $M = M_0 + X_1m_1 + X_2m_2$ (kNm, + tension on the inside) | Section | $M_0$ | $m_1$ | $m_2$ | $M=M_0+\sum X m$ | |---|---|---|---|---| | A | -75.00 | 0.00 | 4.00 | -31.63 | | Q (top of UDL) | -30.00 | 3.00 | 4.00 | 4.10 | | B | -30.00 | 4.00 | 4.00 | 1.01 | | F (load) | 0.00 | 4.00 | 2.00 | 9.33 | | C | 0.00 | 4.00 | 0.00 | -12.35 | | D | 0.00 | 0.00 | 0.00 | 0.00 | #### Shear force (kN) | Member | At lower end | At upper end | |---|---|---| | AQ (UDL zone) | +26.91 | -3.09 | | QB | -3.09 | -3.09 | | BF | +4.16 | +4.16 | | FC | -10.84 | -10.84 | | CD | +3.09 | +3.09 | (Column shears are horizontal forces; positive means towards the right on the left face.) #### Normal thrust (kN, compression negative) | Member | Axial force | |---|---| | AB (left column) | -4.16 | | BC (beam) | -3.09 | | CD (right column) | -10.84 | #### Reactions - D (hinge): $H_D = 3.09$ kN (towards the left), $V_D = 10.84$ kN ($\uparrow$) - A (fixed): $H_A = 26.91$ kN (towards the left), $V_A = 4.16$ kN ($\uparrow$), $M_A = 31.63$ kNm (anticlockwise) Check: $\Sigma H$: $30 - 26.91 - 3.09 = 0$ (UDL resultant $= 10\times3 = 30$ kN); $\Sigma V$: $4.16 + 10.84 = 15$ kN. **Answer: $H_D = 3.09$ kN, $V_D = 10.84$ kN; $H_A = 26.91$ kN, $V_A = 4.16$ kN, $M_A = 31.63$ kNm. The BMD, SFD and thrust values are tabulated above.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the consistent deformation method require at hinge D? A. Displacement of D in the primary structure is cancelled by the redundants, so total displacement is zero B. Zero moment at A C. Equal reactions at A and D D. Zero axial force Answer: A. Real D cannot move along the reactions, so D0 + F X = 0. (Key point: Principle of the method) 2. With r = 3 + 2 = 5, how many redundants? A. 1 B. 3 C. 5 D. 2 Answer: D. D_s = 5 - 3 = 2: H_D and V_D at the released hinge. (Key point: Number of redundants) 3. X1 = H_D = -3.09 kN (X1 assumed to the right). What does this mean? A. H_D = 3.09 kN acting to the left B. H_D = 3.09 kN acting to the right C. H_D is zero D. H_D = 85.33 kN Answer: A. A negative redundant acts opposite to the assumed direction. (Key point: Sign of the redundant) 4. Which is correct for the results? A. V_A + V_D = 30 kN B. Right column CD carries axial thrust of -10.84 kN (compression), and V_A + V_D = 15 kN C. H_A = 3.09 kN D. V_A = V_D Answer: B. The only vertical load is 15 kN: 4.16 + 10.84 = 15, and the right column carries V_D = 10.84 kN in compression. 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