--- title: "Stiffness method: bent frame member end moments" subject: "Theory of Structures II" chapter: 3 key: "semester_5/theory_of_structures_ii:stiffness-bent-frame-end-moments-inextensible" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/stiffness-bent-frame-end-moments-inextensible site: BCT Engineering --- # Stiffness method: bent frame member end moments Theory of Structures II · Chapter 3: Displacement method - Asked in: 2065 Shrawan, 12 marks - Video: [CH5 Stiffness Matrix (Frame) Part 2/2](https://www.youtube.com/watch?v=PAV07r5Tz7U) — DrM4S4 - 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/stiffness-bent-frame-end-moments-inextensible - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3 ## Question Analyze the bent frame using the stiffness method and find member end moments. Members are inextensible. [Figure: bent frame; column AC (3EI) of 4 m (2 m + 2 m) with 30 kN horizontal at mid-height, A hinged; beam CB (5EI) of 5 m with 20 kN/m UDL, B fixed.] ## Short answer The frame has no sway; the only unknown is rotation θ_C. Using the stiffness method, EIθ_C = 3.0667. Member end moments: M_AC = 0, M_CA = 29.40, M_CB = -29.40, M_BC = 47.80 kN·m. ## Answer #### Assumptions and set-up Column AC is 4 m high ($3EI$) with A hinged; the 30 kN acts at mid-height. Beam CB is 5 m ($5EI$) with 20 kN/m, and B is fixed. The members are inextensible, and B is fixed, so joint C cannot move: **there is no sway**. The only unknown displacement is the rotation $\theta_C$ (kinematic indeterminacy = 1). Clockwise moments positive. #### Fixed-end moments - Column AC (30 kN at mid-height): $PL/8 = 30\times4/8 = 15$ kN·m, so $FEM_{CA} = +15$, $FEM_{AC} = -15$. Because A is hinged, use the modified FEM and stiffness $3EI/L$: $FEM'_{CA} = 15 + \tfrac12\times 15 = 22.5$ kN·m. - Beam CB: $wL^2/12 = 20\times25/12 = 41.667$ kN·m, so $FEM_{CB} = -41.667$, $FEM_{BC} = +41.667$. #### Stiffness equations $$ \begin{aligned} M_{CA} &= \tfrac{3(3EI)}{4}\theta_C + 22.5 = 2.25\,EI\theta_C + 22.5 \\ M_{CB} &= \tfrac{2(5EI)}{5}(2\theta_C) - 41.667 = 4\,EI\theta_C - 41.667 \\ M_{BC} &= \tfrac{2(5EI)}{5}(\theta_C) + 41.667 = 2\,EI\theta_C + 41.667 \\ M_{AC} &= 0 \end{aligned} $$ #### Joint equilibrium at C $$ \begin{aligned} M_{CA} + M_{CB} &= 0 \\ 6.25\,EI\theta_C - 19.167 &= 0 \\ EI\theta_C &= 3.0667 \end{aligned} $$ #### Member end moments (kN·m) | End | Moment | |---|---| | $M_{AC}$ | 0 (hinge) | | $M_{CA}$ | $2.25\times3.0667 + 22.5 = +29.40$ | | $M_{CB}$ | $4\times3.0667 - 41.667 = -29.40$ | | $M_{BC}$ | $2\times3.0667 + 41.667 = +47.80$ | Joint C: $29.40 - 29.40 = 0$. #### Reactions and diagrams - Beam end shears: 46.32 kN at C and 53.68 kN at B. Maximum sagging moment $= -29.40 + 46.32x - 10x^2$ at $x = 46.32/20 = 2.316$ m from C, giving $+24.23$ kN·m. Hogging moments 29.40 at C and 47.80 at B. - Column: moment 0 at A, 15.30 kN·m under the 30 kN load, then reversing to 29.40 kN·m at C (outer face tension at C). - Reactions: A: $H_A = 7.65$ kN (opposing the load), $V_A = 46.32$ kN up. B: $H_B = 22.35$ kN, $V_B = 53.68$ kN up, $M_B = 47.80$ kN·m. Check: $7.65 + 22.35 = 30$ kN; $46.32 + 53.68 = 100 = 20\times5$. **Answer:** $\theta_C = 3.067/EI$ (clockwise); $M_{CA} = 29.40$, $M_{CB} = -29.40$, $M_{BC} = 47.80$ kN·m, $M_{AC} = 0$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is there no sway and only theta_C unknown? A. A is hinged so the frame cannot rotate B. The members are inextensible and B is fixed, so C cannot translate; only rotation of C remains C. The load is symmetrical D. The beam is rigid Answer: B. With B fixed and no axial deformation, C cannot move, leaving theta_C as the single unknown. (Key point: Degrees of freedom) 2. How is the hinge at A handled? A. Use 4EI/L and the original FEM B. Set theta_A = 0 C. Ignore column AC D. Use stiffness 3EI/L and modified FEM_CA = 15 + 7.5 = 22.5 kNm Answer: D. With a hinged far end the stiffness is 3EI/L and the FEM is modified by adding half the far-end FEM. (Key point: Hinged far end) 3. The equation to solve for theta_C is... A. M_CA + M_CB = 0 B. M_CA = M_CB x 2 C. M_BC = 0 D. M_AC = 15 Answer: A. No moment is applied at C, so M_CA + M_CB = 0, giving 6.25 EI theta_C = 19.167. (Key point: Joint equilibrium) 4. Why is M_CB = 4 EI theta_C - 41.667? A. 4EI/L with EI = 5EI, L = 5 gives 4EI, and far end B is fixed with theta_B = 0 B. Because 3EI/L is used C. Because the load is a point load D. Because B rotates Answer: A. M = 2(5EI)/5 (2 theta_C + 0) = 4EI theta_C, then add the UDL FEM -wL^2/12 = -41.667. 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