--- title: "Flexibility matrix: reactions at E and A, BMD" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:flexibility-matrix-frame-75knm-20kn-e" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/flexibility-matrix-frame-75knm-20kn-e site: BCT Engineering --- # Flexibility matrix: reactions at E and A, BMD Theory of Structures II · Chapter 2: Force method - Asked in: 2072 Chaitra, 13 marks - Video: [Force method example #2: two degrees indeterminate (part 1/3)](https://www.youtube.com/watch?v=V4C7XXYYwWY) — Engineer4Free - 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-75knm-20kn-e - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Determine the reactions at support E and A and draw bending moment diagram of the frame shown in figure below by using the flexibility matrix method (force method). [Figure: frame; column AB (2EI) of 10 m with A fixed, 20 kN horizontal at B, with a 7 m dimension marked; beam BC-D (2EI) of 10 m carrying 75 kN/m UDL; column DE (EI) of 5 m, E hinged.] ## Short answer Release hinge at E, take horizontal and vertical reactions as redundants; flexibility matrix compatibility yields H_E = 74.68 kN (left), V_E = 369.24 kN (up). At fixed base A: H_A = 54.68 kN (right), V_A = 380.76 kN (up), M_A = 175.80 kNm (clockwise). ## Answer Reading of the figure: A (0, 0) is fixed; column AB is 10 m high ($2EI$) with a 20 kN horizontal load (towards the right) at 7 m above A; beam BD is 10 m long ($2EI$) with 75 kN/m over its whole length; column DE is 5 m long ($EI$), so E is 5 m above the level of A, and E is hinged. $D_s = 3 + 2 - 3 = 2$. Release the hinge at E: $X_1$ = horizontal reaction ($\rightarrow$), $X_2$ = vertical reaction ($\uparrow$). The primary structure is a cantilever fixed at A. ```text B ================= D 75 kN/m | | | 20 kN -> | 5 m | E (hinge) | 10 m A (fixed) ``` #### Primary-structure moments (+ tension on the inner face) - $M_0$: at B, $-75\times10\times5 = -3750$ kNm; at the 20 kN level, $-3750$; at A, $-(3750 + 20\times7) = -3890$ kNm; the beam falls parabolically from $-3750$ at B to 0 at D. - $m_1$ (unit horizontal force at E): DE from 0 at E to 5 at D; beam constant 5; column from 5 at B to $-5$ at A. - $m_2$ (unit vertical force at E): beam from 10 at B to 0 at D; column constant 10. #### Flexibility matrix $[F]$ and load vector $\{\Delta_0\}$ (in $1/EI$ units) $$ \begin{aligned} f_{11} &= \frac{208.333}{EI} \\ f_{12} &= \frac{125.000}{EI} \\ f_{22} &= \frac{666.667}{EI} \\ \Delta_{10} &= \frac{-30596.7}{EI} \\ \Delta_{20} &= \frac{-236825.0}{EI} \end{aligned} $$ #### Compatibility: $[F]\{X\} = -\{\Delta_0\}$ $$ \begin{aligned} 208.333\,X_1 + 125.000\,X_2 &= 30596.7 \\ 125.000\,X_1 + 666.667\,X_2 &= 236825.0 \end{aligned} $$ $$ X_1 = H_E = -74.68\ \text{kN},\qquad X_2 = V_E = 369.24\ \text{kN} $$ #### Final bending moments $M = M_0 + X_1m_1 + X_2m_2$ (kNm) | Section | $M_0$ | $m_1$ | $m_2$ | $M=M_0+\sum X m$ | |---|---|---|---|---| | A | -3890.00 | -5.00 | 10.00 | 175.80 | | Q (20 kN level) | -3750.00 | 2.00 | 10.00 | -206.96 | | B (column) | -3750.00 | 5.00 | 10.00 | -431.00 | | mid BD | -937.50 | 5.00 | 5.00 | 535.30 | | D (beam) | 0.00 | 5.00 | 0.00 | -373.40 | | E | 0.00 | 0.00 | 0.00 | 0.00 | Maximum sagging moment in the beam: 535.52 kNm at 5.08 m from B. #### Reactions - E (hinge): $H_E = 74.68$ kN (towards the left), $V_E = 369.24$ kN ($\uparrow$) - A (fixed): $H_A = 54.68$ kN (towards the right), $V_A = 380.76$ kN ($\uparrow$), $M_A = 175.80$ kNm (clockwise) Check: $\Sigma H$: $20 + 54.68 - 74.68 = 0$; $\Sigma V$: $380.76 + 369.24 = 75\times10 = 750$ kN. **Answer: at E, $H_E = 74.68$ kN (towards the left) and $V_E = 369.24$ kN (upward); at A, $H_A = 54.68$ kN (towards the right), $V_A = 380.76$ kN (upward) and $M_A = 175.80$ kNm (clockwise).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are X1 and X2 and the primary structure? A. Moments at B and D; a simply supported beam B. H_A and V_A; a cantilever fixed at E C. Axial forces in the columns; a pin-jointed frame D. Horizontal and vertical reactions at hinge E; a cantilever fixed at A Answer: D. D_s = 2; releasing the hinge at E leaves the structure fixed only at A, which is determinate. (Key point: Primary structure and redundants) 2. What is M0 at B due to the 75 kN/m UDL on the 10 m beam? A. -3750 kNm B. -7500 kNm C. -750 kNm D. -375 kNm Answer: A. UDL resultant 750 kN acts 5 m from B, so M0 = -750 x 5 = -3750 kNm. (Key point: Primary moments) 3. The columns AB and beam BD are 2EI while DE is EI. How is this handled? A. EI is taken equal everywhere B. Integrals of m_i m_j and M0 m_i are divided by the stiffness of each member (2EI or EI) C. Only DE is considered D. The stiffness changes the loads Answer: B. Flexibility integrals are computed member by member with that member's flexural rigidity. (Key point: Varying stiffness) 4. With V_E = 369.24 kN and V_A = 380.76 kN, what vertical check holds? A. V_A = V_E B. V_A + V_E = 20 kN C. V_A + V_E = 750 kN, the total UDL load D. V_A - V_E = 750 kN Answer: C. The only vertical load is 75 x 10 = 750 kN, and 380.76 + 369.24 = 750. 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