--- title: "Force method: fixed portal frame, 30 kN/m" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:portal-fixed-force-method-4x4-30knm" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/portal-fixed-force-method-4x4-30knm site: BCT Engineering --- # Force method: fixed portal frame, 30 kN/m Theory of Structures II · Chapter 2: Force method - Asked in: 2074 Asoj, 10 marks - Video: [FE Exam Review - FE Civil - Structural Engineering - Indeterminate Structures - Force Method](https://www.youtube.com/watch?v=xxA4M3Z6yZc) — DIRECTHUB FE EXAM PREP - 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-fixed-force-method-4x4-30knm - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question A portal frame of span 4 m and height 4 m is fixed at both supports. The beam of the frame carries a uniformly distributed gravity load of intensity 30 kN/m. Use the force method to solve the frame considering the cross-sectional stiffness (EI) to be constant. Draw bending moment, shear force and normal thrust diagrams for the frame. ## Short answer Using force method with released fixed support D, redundants HD, VD, MD. Solution gives support reactions: H=10 kN inward, V=60 kN upward, M=13.33 kNm at both bases. BMD: +13.33 at base, -26.67 at corners, +33.33 kNm midspan. Shear and thrust diagrams provided. ## Answer Portal frame: span 4 m, height 4 m, both bases fixed, EI constant, UDL $w = 30$ kN/m on the beam BC. $r = 6$, so $D_s = 3$. Release the fixed support at D and take $X_1 = H_D$ ($\rightarrow$), $X_2 = V_D$ ($\uparrow$) and $X_3 = M_D$ (anticlockwise) as redundants. The primary structure is a cantilever fixed at A. ```text 30 kN/m B vvvvvvvvvv C | | | 4 m | 4 m | | A/// ///D (both fixed) |<-- 4 m -->| ``` #### Unit and load moment diagrams (+ tension on the inside) | Section | $M_0$ | $m_1$ | $m_2$ | $m_3$ | |---|---|---|---|---| | A | -240 | 0 | 4 | 1 | | B | -240 | 4 | 4 | 1 | | mid BC | -60 | 4 | 2 | 1 | | C | 0 | 4 | 0 | 1 | | D | 0 | 0 | 0 | 1 | ($M_0 = -30\times4\times2 = -240$ kNm at A and B, falling parabolically to 0 at C.) #### Flexibility coefficients (in $1/EI$ units) $$ [f] = \begin{bmatrix} 106.67 & 64 & 32 \\ 64 & 85.33 & 24 \\ 32 & 24 & 12 \end{bmatrix},\qquad \{\Delta_0\} = \begin{Bmatrix} -3200 \\ -4800 \\ -1280 \end{Bmatrix} $$ #### Compatibility ($[f]\{X\} = -\{\Delta_0\}$) $$ \begin{aligned} 106.67X_1 + 64X_2 + 32X_3 &= 3200 \\ 64X_1 + 85.33X_2 + 24X_3 &= 4800 \\ 32X_1 + 24X_2 + 12X_3 &= 1280 \end{aligned} $$ $$ X_1 = H_D = -10\ \text{kN},\qquad X_2 = V_D = 60\ \text{kN},\qquad X_3 = M_D = 13.33\ \text{kNm} $$ (The symmetric structure and load give $V_A = V_D = 60$ kN and $|H| = 10$ kN inwards at both supports.) #### Reactions | Support | $H$ (inwards) | $V$ (up) | Moment | |---|---|---|---| | A | 10 kN | 60 kN | 13.33 kNm | | D | 10 kN | 60 kN | 13.33 kNm | #### Bending moment diagram (kNm, + tension inside) | Point | BM | |---|---| | A, D | +13.33 (inner face tension) | | B, C | -26.67 (outer face tension) | | Mid-span of BC | +33.33 (bottom tension) | The BM in each column is linear, $M = 13.33 - 10y$, zero at $y = 1.33$ m above the base. In the beam $M = -26.67 + 60x - 15x^2$, zero at $0.51$ m and $3.49$ m from B, with the maximum $+33.33$ kNm at mid-span. #### Shear force diagram (kN) - Columns: constant $10$ kN (horizontal), acting inwards at the base. - Beam: $+60$ kN at B, falling linearly through zero at mid-span to $-60$ kN at C. #### Normal thrust diagram (kN) - Columns: $-60$ (compression) throughout. - Beam: $-10$ (compression) throughout. **Answer: $H = 10$ kN, $V = 60$ kN and $M = 13.33$ kNm at each base; $M_B = M_C = -26.67$ kNm; mid-span $M = +33.33$ kNm.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the degree of indeterminacy of a portal frame fixed at both bases? A. 1 B. 2 C. 3 D. 6 Answer: C. r = 6 reactions, so D_s = 6 - 3 = 3; H, V and M at D are the redundants. (Key point: Indeterminacy) 2. Which matrix equation is solved? A. [f]{X} = {Δ0} B. [f]{X} = -{Δ0} C. {X} = [f]{Δ0} D. [f] = {X}{Δ0} Answer: B. Displacements at the released support must vanish: [f]{X} + {Δ0} = 0. (Key point: Compatibility form) 3. What reactions at each base come from the symmetry and the solution? A. V = 120 kN, H = 0 B. V = 60 kN, H = 10 kN inwards, M = 13.33 kNm C. V = 30 kN, H = 60 kN D. V = 60 kN, H = 26.67 kN Answer: B. Total load 30 x 4 = 120 kN shared equally gives 60 kN each; solving gives H = 10 kN and M = 13.33 kNm. (Key point: Symmetry results) 4. What is the bending moment at B and C and at beam mid-span? A. -26.67 kNm and +33.33 kNm B. +13.33 kNm and -26.67 kNm C. -40 kNm and +60 kNm D. 0 and 120 kNm Answer: A. M_B = 13.33 - 10 x 4 = -26.67 kNm; mid-span M = -26.67 + 60x2 - 15x4 = +33.33 kNm. 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