--- title: "Force method: portal frame, 50 kN/m" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:portal-frame-force-method-6m-5m" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/portal-frame-force-method-6m-5m site: BCT Engineering --- # Force method: portal frame, 50 kN/m Theory of Structures II · Chapter 2: Force method - Asked in: 2075 Asoj, 6 marks - Video: [Problem 7 Analyse Frame By Force Method / Flexibility Method [ FRAME ]](https://www.youtube.com/watch?v=Mlmp9s0tlgc) — Ganesh Shegar - 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/portal-frame-force-method-6m-5m - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question A portal frame of span 6 m and height 5 m is hinged supported at both ends. The beam of the frame carries a uniformly distributed gravity load of intensity 50 kN/m. Use the force method to solve the frame considering the flexural stiffness EI to be constant. Determine the reactions at both supports. ## Short answer Take inward horizontal reaction H as redundant. Compatibility gives H = 19.286 kN at each hinge. Vertical reactions are 150 kN upward at both supports. ## Answer Portal frame: span $L = 6$ m, height $h = 5$ m, both bases hinged, EI constant, beam load $w = 50$ kN/m. A hinged portal has $r = 4$, so $D_s = 1$. Take the horizontal reaction $H$ at D as the redundant (D is released into a roller). ```text 50 kN/m B vvvvvvvvvvvv C | | | 5 m | 5 m | | A o o D <- H |<---- 6 m --->| ``` #### Primary structure (D on a roller) Reactions: $V_A = V_D = 50\times6/2 = 150$ kN. - $M_0$: columns 0; beam $M_0 = 150x - 25x^2$ (maximum $225$ kNm at mid-span). - $m$ (unit horizontal force at D): column CD: $m = y$ (from D); beam: $m = 5$; column AB: $m = y$ (from A). #### Flexibility coefficient and load term $$ \begin{aligned} f_{11} &= \frac{1}{EI}\left[2\int_0^5 y^2dy + 5^2(6)\right] = \frac{1}{EI}\left[83.33 + 150\right] = \frac{233.33}{EI} \\ \Delta_{10} &= \frac{1}{EI}\int_0^6 M_0\,(5)\,dx = \frac{5}{EI}\cdot\frac{wL^3}{12} = \frac{5\times900}{EI} = \frac{4500}{EI} \end{aligned} $$ (The loads push D outwards by $\Delta_{10}$ in the primary structure, and the inward thrust $H$ brings it back by $f_{11}H$.) #### Compatibility (D does not move horizontally) $$ \Delta_{10} - f_{11}H = 0 \ \Rightarrow\ H = \frac{4500}{233.33} = 19.286\ \text{kN} \left(= \frac{135}{7}\right) $$ The horizontal reaction acts inwards (towards the frame) at each support. #### Final results | Support | Horizontal (inwards) | Vertical (up) | |---|---|---| | A | 19.286 kN | 150 kN | | D | 19.286 kN | 150 kN | Bending moments: $M_A = M_D = 0$; $M_B = M_C = -H h = -19.286\times5 = -96.43$ kNm (tension on the outside); mid-span $M = 225 - 96.43 = 128.57$ kNm (sagging). **Answer: at each hinge, $V = 150$ kN (upward) and $H = 19.286$ kN (towards the other support).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. A portal frame with both bases hinged has what degree of indeterminacy? A. 0 B. 2 C. 3 D. 1 Answer: D. r = 4 reactions, so D_s = 4 - 3 = 1; the horizontal reaction is taken as redundant. (Key point: Degree of indeterminacy) 2. What is f11 for H at D? A. 233.33/EI B. 83.33/EI C. 150/EI D. 4500/EI Answer: A. f11 = (1/EI)[2∫y^2dy + 5^2 x 6] = (83.33 + 150)/EI = 233.33/EI. (Key point: Flexibility and load terms) 3. What is the compatibility condition and the thrust H? A. Rotation at A is zero, H = 150 kN B. D does not move horizontally, H = 19.286 kN C. Beam deflection is zero, H = 96.43 kN D. D settles, H = 4500 kN Answer: B. Δ10 - f11H = 0 gives H = 4500/233.33 = 19.286 kN, acting inwards. (Key point: Compatibility) 4. What is the moment at B and C? A. +225 kNm B. -19.286 kNm C. -96.43 kNm D. 0 Answer: C. M_B = M_C = -H h = -19.286 x 5 = -96.43 kNm, with tension on the outside. 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