--- title: "Force method: frame reactions and BMD" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:force-method-frame-100kn-30knm-reactions" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/force-method-frame-100kn-30knm-reactions site: BCT Engineering --- # Force method: frame reactions and BMD Theory of Structures II · Chapter 2: Force method - Asked in: 2075 Asoj, 10 marks - Video: [Force Method for Frames - Structural Analysis](https://www.youtube.com/watch?v=pVX1AYEEqsE) — Civil Engineering with Tanya J. Laird - 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/force-method-frame-100kn-30knm-reactions - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Determine the horizontal and vertical reaction at the hinged support and also draw BMD using the force method. [Figure: portal frame; beam of 2EI, span 5 m, with 30 kN/m UDL; columns EI of height 4 m; left base fixed, right base hinged; 100 kN horizontal at the top-left joint.] ## Short answer Release hinge at D, treat horizontal and vertical reactions as redundants X1, X2. From compatibility, X1 = –31.44 kN (31.44 kN left) and X2 = 121.79 kN upward. ## Answer Reading of the figure: portal frame ABCD, columns AB and CD are 4 m high ($EI$), the beam BC is 5 m long ($2EI$) with 30 kN/m, a 100 kN horizontal load acts at B (towards C), A is fixed and D is hinged. $r = 3 + 2 = 5$, so $D_s = 2$. Release the hinge at D and take $X_1$ = horizontal reaction ($\rightarrow$) and $X_2$ = vertical reaction ($\uparrow$) at D. ```text 100 kN -> B ============ C 30 kN/m | | | 4 m | 4 m | | ///A o D (hinge) |<---- 5 m --->| ``` #### Primary structure (cantilever fixed at A) - $M_0$ (+ tension on the inner face): at B, $-30\times5\times2.5 = -375$ kNm (the 100 kN load acts at the joint B); at A, $-(375 + 100\times4) = -775$ kNm; the beam falls parabolically from $-375$ at B to 0 at C; column CD has no moment. - $m_1$ (unit horizontal force at D): CD from 0 at D to 4 at C, beam constant 4, column from 4 at B down to 0 at A. - $m_2$ (unit vertical force at D): beam from 5 at B to 0 at C, column AB constant 5. #### Flexibility coefficients (in $1/EI$ units) $$ \begin{aligned} f_{11} &= \frac{82.667}{EI} \\ f_{12} &= \frac{65.000}{EI} \\ f_{22} &= \frac{120.833}{EI} \\ \Delta_{10} &= \frac{-5316.7}{EI} \\ \Delta_{20} &= \frac{-12671.9}{EI} \end{aligned} $$ #### Compatibility ($\Delta_{Dx} = \Delta_{Dy} = 0$) $$ \begin{aligned} 82.667\,X_1 + 65.000\,X_2 &= 5316.7 \\ 65.000\,X_1 + 120.833\,X_2 &= 12671.9 \end{aligned} $$ $$ X_1 = H_D = -31.44\ \text{kN},\qquad X_2 = V_D = 121.79\ \text{kN} $$ #### Final bending moments (kNm) | Section | $M_0$ | $m_1$ | $m_2$ | $M=M_0+\sum X m$ | |---|---|---|---|---| | A | -775.00 | 0.00 | 5.00 | -166.07 | | B (column) | -375.00 | 4.00 | 5.00 | 108.15 | | mid BC | -93.75 | 4.00 | 2.50 | 84.94 | | C (beam) | 0.00 | 4.00 | 0.00 | -125.78 | | D | 0.00 | 0.00 | 0.00 | 0.00 | The maximum sagging moment in the beam is 121.42 kNm at 0.94 m from B. #### Reactions - At D (hinge): $H_D = 31.44$ kN (towards the left), $V_D = 121.79$ kN ($\uparrow$) - At A (fixed): $H_A = 68.56$ kN (towards the left), $V_A = 28.21$ kN ($\uparrow$), $M_A = 166.07$ kNm (anticlockwise) Check: $\Sigma H$: $100 - 68.56 - 31.44 = 0$; $\Sigma V$: $28.21 + 121.79 = 30\times5 = 150$ kN. **Answer: at the hinged support D, the horizontal reaction is $31.44$ kN (towards the left) and the vertical reaction is $121.79$ kN (upward). End moments: $M_A = 166.07$ kNm, $M_B = 108.15$ kNm, $M_C = -125.78$ kNm (signs as in the table).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What redundants are released at the hinge D? A. Moment and shear at A B. Only the horizontal reaction C. The 100 kN load D. Horizontal and vertical reactions at D Answer: D. r = 3 + 2 = 5 so D_s = 2; both reactions at D are released, leaving a cantilever fixed at A. (Key point: Redundants) 2. What is M0 at the fixed base A of the primary cantilever? A. -375 kNm B. -100 kNm C. -775 kNm D. -475 kNm Answer: C. At A: beam UDL moment 30x5x2.5 = 375 plus 100x4 = 400 gives -775 kNm. (Key point: M0 in primary structure) 3. What compatibility conditions are written? A. Horizontal and vertical displacements at D are zero B. Rotations at A and D are zero C. Moments at B and C are equal D. Sum of reactions equals 150 kN Answer: A. D is a hinge, so Δ_Dx = Δ_Dy = 0, giving two simultaneous equations in X1 and X2. (Key point: Compatibility) 4. What are the reactions at the hinge D? A. HD = 68.56 kN left, VD = 28.21 kN up B. HD = 31.44 kN right, VD = 28.21 kN up C. HD = 31.44 kN left, VD = 121.79 kN up D. HD = 100 kN, VD = 150 kN Answer: C. Solving gives X1 = -31.44 (31.44 kN to the left) and X2 = 121.79 kN up; HA = 68.56 and VA = 28.21 satisfy equilibrium. 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