--- title: "Flexibility matrix: reactions at support D" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:flexibility-matrix-frame-2knm-5kn-coords" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/flexibility-matrix-frame-2knm-5kn-coords site: BCT Engineering --- # Flexibility matrix: reactions at support D Theory of Structures II · Chapter 2: Force method - Asked in: 2069 Asar, 10 marks - Video: [Flexibility Matrix Method - Analysis of a Frame](https://www.youtube.com/watch?v=BrlxaC7AWX0) — 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/flexibility-matrix-frame-2knm-5kn-coords - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Generate the flexibility matrix for the coordinates shown in figure below and use this to determine the reactions at support D. Take EI is constant for all members. [Figure: portal frame ABCD, height 5 m, span 4 m; A fixed, D hinged; 2 kN/m UDL on the left column AB (printed near B) and a 5 kN horizontal load at C; coordinates 1 (vertical) and 2 (horizontal) at D.] ## Short answer Remove the hinge at D; treat the frame as a cantilever fixed at A. Compute moment diagrams for unit loads at D and for applied loads, form the flexibility matrix, calculate load displacements, and solve [F]{R} = -{Δ}. Vertical reaction at D = 5.95 kN upward, horizontal = 2.07 kN to the left. ## Answer **Assumptions:** A is fixed, D is hinged. AB = CD = 5 m high, BC = 4 m, EI constant. The 2 kN/m UDL acts horizontally (to the right) on AB and the 5 kN load acts to the right at C. Axial deformations are ignored. The hinge D gives two unknown reactions, so the frame is indeterminate to the 2nd degree. The redundants are the coordinates at D: **1 = vertical reaction (up +)** and **2 = horizontal reaction (right +)**. #### Step 1: Released structure Remove the hinge at D. The frame A-B-C-D becomes a cantilever fixed at A with a free end D. #### Step 2: Moment diagrams (positive = tension on the inside face) | Case | on AB (A to B) | on BC (B to C) | on CD (C to D) | |---|---|---|---| | Unit vertical force at D ($m_1$) | 4 (constant) | 4 at B to 0 at C | 0 | | Unit horizontal force at D ($m_2$) | 0 at A to 5 at B | 5 (constant) | 5 at C to 0 at D | | Given loads ($M_L$) | -50 at A to 0 at B (parabolic) | 0 | 0 | For the loaded case, $M_L=-\left[(5-y)^2+5(5-y)\right]$ on AB (y up from A): the UDL gives $2(5-y)^2/2$ and the 5 kN load at C gives $5(5-y)$, both causing tension on the outside. #### Step 3: Flexibility matrix ($f_{ij}=\sum\int m_im_j\,dx/EI$) $$ f_{11}=\frac{1}{EI}\left[4^2(5)+\frac{4^2(4)}{3}\right]=\frac{101.33}{EI},\quad f_{22}=\frac{1}{EI}\left[\frac{5^3}{3}+5^2(4)+\frac{5^3}{3}\right]=\frac{183.33}{EI} $$ $$ f_{12}=f_{21}=\frac{1}{EI}\left[4\cdot\frac{5^2}{2}+5\cdot\frac{4\times4}{2}\right]=\frac{90.00}{EI} $$ $$ [F]=\frac{1}{EI}\begin{bmatrix}101.33&90.00\\90.00&183.33\end{bmatrix} $$ #### Step 4: Displacements of D in the released structure due to the loads $$ \Delta_{1L}=\frac{1}{EI}\int m_1M_L\,dx=-416.67/EI,\qquad \Delta_{2L}=\frac{1}{EI}\int m_2M_L\,dx=-156.25/EI $$ (negative: D moves down and to the left in the released frame.) #### Step 5: Compatibility (D cannot move) $$ [F]\begin{Bmatrix}R_1\\R_2\end{Bmatrix}=-\begin{Bmatrix}\Delta_{1L}\\\Delta_{2L}\end{Bmatrix} \;\Rightarrow\; \begin{Bmatrix}R_1\\R_2\end{Bmatrix}=\begin{Bmatrix}5.95\\-2.07\end{Bmatrix}\ \text{kN} $$ #### Reactions at A (equilibrium) $H_A=-(2\times5+5)-(-2.07)=-12.93$ kN (to the left), $V_A=-5.95$ kN, $M_A=26.21$ kNm. Check $\sum F_x$: loads $15$ kN to the right, reactions -12.93 and -2.07 kN, total $=0$ ✓. Final BM (inside tension +): $M_A=-26.21$, $M_B=13.45$, $M_C=-10.34$, $M_D=0$ kNm. **Answer: reactions at D: vertical $R_1=5.95$ kN upward and horizontal $R_2=2.07$ kN directed to the left.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the portal frame released in this method? A. Fixed support A is removed B. The hinge at D is removed, leaving a cantilever fixed at A, with vertical and horizontal reactions at D as redundants C. A hinge is inserted at B D. The 5 kN load is removed Answer: B. The hinge D gives two unknown reactions, so the frame is 2nd degree indeterminate and these two reactions are taken as redundants. (Key point: Redundants and released structure) 2. How is f_ij computed here (axial deformation ignored)? A. Sum of the integral of m_i m_j dx / EI over all members B. Sum of the product of the loads and the unit forces C. The inverse of the stiffness of column AB D. Sum of m_i dx / EI without multiplying by m_j Answer: A. Flexibility coefficients come from the unit-load method: f_ij = sum of integral(m_i m_j dx)/EI. (Key point: Computing flexibility coefficients) 3. Which equation gives the reactions at D? A. [F]{R} = {Delta_L} B. [F]{R} = -{Delta_L}, because D cannot move C. [F]{R} = 0 D. {R} = [F]{Delta_L} Answer: B. The redundants must bring the displacement of D back to zero, so [F]{R} + {Delta_L} = 0. (Key point: Compatibility equation) 4. What are the reactions at D? A. Vertical 5.95 kN up, horizontal 2.07 kN to the left B. Vertical 2.07 kN up, horizontal 5.95 kN to the right C. Vertical 5.95 kN down, horizontal 2.07 kN to the right D. Vertical 12.93 kN up, horizontal 5.95 kN to the left Answer: A. Solving [F]{R} = -{Delta_L} gives R1 = 5.95 kN (up) and R2 = -2.07 kN (to the left). 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