--- title: "Force method: frame with 120 k load" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:force-method-frame-120k-sfd-bmd" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/force-method-frame-120k-sfd-bmd site: BCT Engineering --- # Force method: frame with 120 k load Theory of Structures II · Chapter 2: Force method - Asked in: 2074 Chaitra, 10 marks - Video: [FORCE METHOD (2 redundancies): Analysis of Statically Indeterminate Structures I Structural Analysis](https://www.youtube.com/watch?v=D4q0fM0kye8) — Study with Alfred Zimo - 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-120k-sfd-bmd - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Analyze the structure given below using the force method. Draw shear force and bending moment diagrams. [Figure: frame a-d-b-c; horizontal member a-b of 24 ft (12 ft + 12 ft, 2I) with 120 kip downward load at d (mid-span); vertical member b-c of 18 ft (3I); a hinged, c hinged.] ## Short answer Using Ha as redundant, Ha = 20.00 kip, Va = 45.00 kip, Vc = 75.00 kip. Maximum bending moments: 540.0 kip-ft sagging at midspan, –360.0 kip-ft hogging at corner b. ## Answer Reading of the figure: beam a-b is 24 ft (12 ft + 12 ft) with stiffness $2EI$, carrying 120 kip at d (mid-span); column b-c is 18 ft high with stiffness $3EI$; a and c are hinged. Units: kip and ft. $r = 4$, so $D_s = 1$. Take the horizontal reaction $H_a$ at a as the redundant (a becomes a roller). ```text 120 k a o----d----b ^ | H_a | 18 ft | o c |<- 12 ->|<- 12 ->| ``` #### Primary structure (a on a roller) Reactions: $V_a = V_c = 60$ kip. | Section | $M_0$ (kip-ft) | $m$ (unit $H_a$) | |---|---|---| | a | 0 | 0 | | d | 720 | -9 | | b | 0 | -18 | | c | 0 | 0 | ($M_0 = 60\times12 = 720$ at d. A unit horizontal force at a causes a vertical reaction pair of $18/24 = 0.75$, so $m_d = -0.75\times12 = -9$ and $m_b = -18$ (hogging).) #### Coefficients (unit-load method; beam stiffness $2EI$, column stiffness $3EI$) $$ \begin{aligned} f_{11} &= \frac{1}{2EI}\left[\frac{12}{3}(81) + \frac{12}{3}(81 + 162 + 324)\right] + \frac{1}{3EI}\left[\frac{18}{3}(324)\right] = \frac{162 + 1134 + 648}{EI} = \frac{1944}{EI} \\ \Delta_{10} &= \frac{1}{2EI}\left[\frac{12}{3}(720)(-9) + \frac{12}{6}\{2(720)(-9) + 720(-18)\}\right] = \frac{-12960 - 25920}{EI} = \frac{-38880}{EI} \end{aligned} $$ #### Compatibility (a does not move horizontally) $$ f_{11}H_a + \Delta_{10} = 0 \ \Rightarrow\ H_a = \frac{38880}{1944} = 20.00\ \text{kip} $$ #### Reactions - a: $H_a = 20.00$ kip ($\rightarrow$), $V_a = 45.00$ kip ($\uparrow$) - c: $H_c = 20.00$ kip ($\leftarrow$), $V_c = 75.00$ kip ($\uparrow$) (The unit force on the primary structure gives a vertical reaction at a of $0.75$ per unit $H$, so $V_a = 60 - 0.75\times20 = 45$ kip and $V_c = 60 + 15 = 75$ kip.) #### Bending moment $M = M_0 + H_a m$ (kip-ft) | Point | BM | Tension side | |---|---|---| | a | 0 | | | d | 540.0 | bottom | | b | -360.0 | outside (top of beam, outer face of column) | | c | 0 | | #### Shear force (kip) | Member | SF | |---|---| | a-d | +45.0 | | d-b | -75.0 | | b-c | +20.0 (constant) | #### Axial force (kip) Beam: $-20.0$ (compression); column: $-75.0$ (compression). **Answer: $H_a = H_c = 20.00$ kip; $V_a = 45.00$ kip, $V_c = 75.00$ kip; $M_d = 540.0$ kip-ft (sagging), $M_b = -360.0$ kip-ft (hogging).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which redundant is chosen for this frame with both a and c hinged? A. Moment at b B. Horizontal reaction H_a (a becomes a roller) C. Vertical reaction at c D. Shear in column Answer: B. r = 4 so D_s = 1; releasing the horizontal restraint at a gives a determinate primary structure. (Key point: Redundant) 2. What are M0 and m at d in the primary structure? A. M0 = 720, m = -9 B. M0 = 0, m = -18 C. M0 = 720, m = -18 D. M0 = 360, m = 9 Answer: A. M0 = 60 x 12 = 720; unit H_a produces vertical reactions 0.75, so m_d = -0.75 x 12 = -9. (Key point: M0 and m) 3. What is H_a? A. 38.88 kip B. 60 kip C. 20.00 kip D. 45 kip Answer: C. H_a = 38880/1944 = 20 kip from f11 H_a + Δ10 = 0. (Key point: Thrust) 4. Why do the vertical reactions change from 60 kip each to 45 and 75 kip? A. The column weight is added B. Because H_a produces a vertical couple of 0.75 per unit that adds 15 kip at c and removes 15 at a C. Because of axial shortening D. Because the load is not at mid-span Answer: B. The horizontal reaction at a, acting at 18 ft height difference, shifts 0.75 x 20 = 15 kip: Va = 45, Vc = 75. 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