--- title: "Force method: reactions at D, frame with 110 kN" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:force-method-frame-110kn-30knm-d" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/force-method-frame-110kn-30knm-d site: BCT Engineering --- # Force method: reactions at D, frame with 110 kN Theory of Structures II · Chapter 2: Force method - Asked in: 2070 Asar, 10 marks - Video: [Indeterminate Frame by Force Method](https://www.youtube.com/watch?v=Z7BR7BvpoLU) — STATICS THE EASY WAY - 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-110kn-30knm-d - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Determine horizontal and vertical reactions at support D of the frame shown in figure below using the force method. [Figure: portal frame ABCD; beam BC (2EI) of 5 m with 30 kN/m UDL; columns AB and CD (EI) 4 m high; 110 kN horizontal at B; A fixed, D hinged.] ## Short answer Using force method with hinge D released, horizontal reaction H_D = 33.72 kN (left), vertical reaction V_D = 126.32 kN (up). Fixed end A: H_A = 76.28 kN left, V_A = 23.68 kN up, M_A = 183.40 kNm anticlockwise. ## 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 110 kN horizontal load acts at B (towards C), A is fixed and D is hinged. $D_s = 3 + 2 - 3 = 2$. Release the hinge at D and take $X_1$ = horizontal reaction ($\rightarrow$) and $X_2$ = vertical reaction ($\uparrow$). The primary structure is the cantilever fixed at A. ```text 110 kN-> B ============ C 30 kN/m | | | 4 m | 4 m | | ///A o D (hinge) ``` #### Moment diagrams in the primary structure (+ tension on the inner face) - $M_0$: in the beam, $-375$ kNm at B (UDL resultant $150\times2.5$) falling parabolically to 0 at C; in the column AB it varies linearly from $-375$ at B to $-(375 + 110\times4) = -815$ kNm at A. - $m_1$ (unit horizontal force at D): CD from 0 to 4, beam constant 4, column AB 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) $$ \begin{aligned} f_{11} &= \frac{82.667}{EI} \\ f_{12} &= \frac{65.000}{EI} \\ f_{22} &= \frac{120.833}{EI} \\ \Delta_{10} &= \frac{-5423.3}{EI} \\ \Delta_{20} &= \frac{-13071.9}{EI} \end{aligned} $$ #### Compatibility $$ \begin{aligned} 82.667\,X_1 + 65.000\,X_2 &= 5423.3 \\ 65.000\,X_1 + 120.833\,X_2 &= 13071.9 \end{aligned} $$ $$ X_1 = H_D = -33.72\ \text{kN},\qquad X_2 = V_D = 126.32\ \text{kN} $$ #### Final bending moments (kNm) | Section | $M_0$ | $m_1$ | $m_2$ | $M=M_0+\sum X m$ | |---|---|---|---|---| | A | -815.00 | 0.00 | 5.00 | -183.40 | | B (column) | -375.00 | 4.00 | 5.00 | 121.72 | | mid BC | -93.75 | 4.00 | 2.50 | 87.17 | | C (beam) | 0.00 | 4.00 | 0.00 | -134.88 | | D | 0.00 | 0.00 | 0.00 | 0.00 | Maximum sagging moment in BC: 131.07 kNm at 0.79 m from B. #### Reactions - D (hinge): $H_D = 33.72$ kN (towards the left), $V_D = 126.32$ kN ($\uparrow$) - A (fixed): $H_A = 76.28$ kN (towards the left), $V_A = 23.68$ kN ($\uparrow$), $M_A = 183.40$ kNm (anticlockwise) Check: $\Sigma H$: $110 - 76.28 - 33.72 = 0$; $\Sigma V$: $23.68 + 126.32 = 30\times5 = 150$ kN. **Answer: at the hinged support D, the horizontal reaction is $33.72$ kN (towards the left) and the vertical reaction is $126.32$ kN (upward).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With A fixed and D hinged, D_s = 2. The primary structure is: A. The cantilever fixed at A, with the hinge at D released B. Simply supported frame C. Fixed-fixed frame D. Three-hinged frame Answer: A. Releasing D leaves a cantilever frame; the horizontal and vertical reactions at D are redundants. (Key point: Redundants and primary structure) 2. The equations 82.667X1 + 65X2 = 5423.3 and 65X1 + 120.833X2 = 13071.9 come from: A. Moment equilibrium at A B. Zero horizontal and vertical displacement at D C. Shear at B D. Zero rotation at A Answer: B. Δi0 + Σ f_ij Xj = 0 with Δ10, Δ20 negative gives these. (Key point: Compatibility equations) 3. X1 = -33.72 kN means: A. H_D acts towards the right B. H_D acts towards the left with 33.72 kN C. H_D is zero D. The solution is invalid Answer: B. X1 was assumed rightward; negative means 33.72 kN to the left. (Key point: Sign of X1) 4. Which check confirms the horizontal reactions? A. 110 - 76.28 - 33.72 = 0 B. 110 + 76.28 = 33.72 C. 30 x 5 = 110 D. M_A = M_B Answer: A. ΣH = 0: the 110 kN load is balanced by H_A and H_D. 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