--- title: "Force method: frame reactions at hinged support" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:force-method-frame-50knm-hinged-d" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/force-method-frame-50knm-hinged-d site: BCT Engineering --- # Force method: frame reactions at hinged support Theory of Structures II · Chapter 2: Force method - Asked in: 2078 Kartik, 10 marks - Video: [Force Method for Indeterminate Structures - Intro to Structural Analysis](https://www.youtube.com/watch?v=kwjt1H3Ycjw) — Structures with Prof. H - 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-50knm-hinged-d - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Determine reactions at the hinged support of the frame loaded as shown in the figure using the force method. [Figure: frame ABCD, beam BC of span 10 m with I = 2I carrying 50 kN/m UDL; column AB 5 m long with 1.5I, A fixed; column CD 3 m long with I, D hinged.] ## Answer Reading of the figure: A (0, 0) fixed; B (0, 5) with column AB 5 m of stiffness $1.5EI$; beam BC of span 10 m with stiffness $2EI$ carrying 50 kN/m; column CD 3 m of stiffness $EI$ with D hinged. As the columns are of different lengths, D is taken 2 m above the level of A. Here $EI$ stands for $E\times I$ of the lightest member. The frame has $r = 3 + 2 = 5$, so $D_s = 2$. Release the hinge at D and take the reactions at D as redundants: $X_1$ = horizontal ($\rightarrow$ positive) and $X_2$ = vertical ($\uparrow$ positive). The primary structure is a cantilever fixed at A. ```text B ================ C 50 kN/m | | | 5 m | 3 m | o D (hinge) A/// ``` #### Moment diagrams (sign: + tension on the inner face of the frame) - $M_0$ (50 kN/m on the cantilever): $M_B = M_{A} = -50(10)^2/2 = -2500$ kNm, falling parabolically to 0 at C. - $m_1$ (unit horizontal force at D): CD varies 0 to 3, beam constant 3, column from 3 at B down to $-2$ at A. - $m_2$ (unit vertical force at D): beam varies 10 to 0 from B to C, column constant 10. #### Flexibility coefficients and load terms $$ \begin{aligned} f_{11} &= \frac{61.778}{EI} \\ f_{12} &= \frac{91.667}{EI} \\ f_{22} &= \frac{500.000}{EI} \\ \Delta_{10} &= \frac{-16666.7}{EI} \\ \Delta_{20} &= \frac{-114583.3}{EI} \end{aligned} $$ #### Compatibility ($\Delta_{D,x} = \Delta_{D,y} = 0$) $$ \begin{aligned} 61.778\,X_1 + 91.667\,X_2 &= 16666.7 \\ 91.667\,X_1 + 500.000\,X_2 &= 114583.3 \end{aligned} $$ Solving: $$ X_1 = -96.51\ \text{kN},\qquad X_2 = 246.86\ \text{kN} $$ #### Final bending moments $M = M_0 + X_1 m_1 + X_2 m_2$ (kNm) | Section | $M_0$ | $m_1$ | $m_2$ | $M=M_0+\sum X m$ | |---|---|---|---|---| | A | -2500.00 | -2.00 | 10.00 | 161.62 | | B (column) | -2500.00 | 3.00 | 10.00 | -320.93 | | B (beam) | -2500.00 | 3.00 | 10.00 | -320.93 | | mid BC | -625.00 | 3.00 | 5.00 | 319.77 | | C (beam) | 0.00 | 3.00 | 0.00 | -289.53 | | C (column) | 0.00 | 3.00 | 0.00 | -289.53 | | D | 0.00 | 0.00 | 0.00 | 0.00 | The maximum sagging moment in BC is 319.87 kNm at 5.07 m from B. #### Reactions - Hinge D: $H_D = 96.51$ kN ($\leftarrow$), $V_D = 246.86$ kN ($\uparrow$) - Fixed support A: $H_A = 96.51$ kN ($\rightarrow$), $V_A = 253.14$ kN ($\uparrow$), $M_A = 161.62$ kNm (clockwise; the moment at A puts the inner face of the column in tension) Check: $\Sigma V = 253.14 + 246.86 = 500.00$ kN $= 50 \times 10 = 500$ kN; $\Sigma H = 96.51 - 96.51 = 0$. Moments about A: the UDL gives $2500$ kNm clockwise, while $V_D$ ($246.86\times10 = 2468.6$) and $H_D$ ($96.51\times2 = 193.0$) give $2661.6$ kNm anticlockwise, so the support couple must be $161.62$ kNm clockwise. **Answer: at the hinged support D, $H_D = 96.51$ kN (towards the left) and $V_D = 246.86$ kN (upward).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How many redundants does the frame have and what are taken as redundants? A. One: the vertical reaction at D only B. Two (r = 5, Ds = 2): the horizontal and vertical reactions at the hinge D C. Three: all reactions at the fixed end A D. Two: the moments M_B and M_C at the corners Answer: B. r = 3 + 2 = 5 gives Ds = 2. Releasing the hinge D leaves a cantilever fixed at A, with X1 (horizontal) and X2 (vertical) as redundants. (Key point: Degree and redundants) 2. What conditions give the two equations for X1 and X2? A. The slopes at A and D are zero B. The sum of horizontal forces and moments is zero C. Horizontal and vertical displacements at the hinge D are both zero D. The deflection at mid BC and at C is zero Answer: C. D is a fixed hinge, so both its displacements vanish; this gives two simultaneous flexibility equations. (Key point: Compatibility) 3. What is M0 at B and A in the cantilever primary structure? A. -2500 kNm, from 50 x 10^2 / 2 (hogging) B. -500 kNm, the total load C. +2500 kNm sagging D. -1250 kNm, from wL^2/4 Answer: A. The 50 kN/m UDL acts over 10 m on the cantilever, giving 50(10)^2/2 = 2500 kNm hogging at the corner. (Key point: Primary-structure moment) 4. X1 came out as -96.51 kN with right (->) taken positive. What does this mean? A. The reaction at D acts to the right with magnitude 96.51 kN B. The solution is wrong and must be redone C. The reaction at D is vertical, not horizontal D. The horizontal reaction at D acts to the left with magnitude 96.51 kN Answer: D. A negative result means the actual direction is opposite to the assumed one, so H_D = 96.51 kN towards the left. 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