--- title: "Force method: frame, SFD and BMD" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:force-method-frame-5knm-20kn-sfd" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/force-method-frame-5knm-20kn-sfd site: BCT Engineering --- # Force method: frame, SFD and BMD Theory of Structures II · Chapter 2: Force method - Asked in: 2070 Chaitra, 15 marks - Video: [Force Method for Frames - Structural Analysis](https://www.youtube.com/watch?v=pVX1AYEEqsE) — Civil Engineering with Tanya J. Laird - 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-5knm-20kn-sfd - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Draw shear force and bending moment diagrams for the frame given below. Use the force method. [Figure: frame; beam B-C-D with 5 kN/m UDL on BC (5 m) and a 2 m overhang CD with 20 kN at D; 20 kN horizontal at B; columns EI: AB (A fixed) of height 4 m, middle column CE hinged at E, right side with a 2 m dimension marked.] ## Short answer Using force method with redundants at hinge E, horizontal reaction H_E = 1.93 kN left, vertical V_E = 47.63 kN up. Bending moments: M_A = 44.35 kNm, M_B = 27.93 kNm, M_C = -47.72 kNm, M_D = 0; zero moment in BC at 2.86 m from B. Shear forces: AB = 18.07 kN, BC at B = -2.63 kN, at C = -27.63 kN, CD = 20 kN, CE = 1.93 kN. ## Answer Reading of the figure: left column AB (4 m, A fixed); beam B-C-D with BC = 5 m carrying 5 kN/m and a 2 m overhang CD with 20 kN at D; a middle column CE (4 m) hinged at E, below C; 20 kN horizontal at B (towards the right). $EI$ is constant for all members. $r = 3 + 2 = 5$, so $D_s = 2$. Release the hinge at E and take $X_1$ = horizontal reaction ($\rightarrow$), $X_2$ = vertical reaction ($\uparrow$) at E as redundants. The primary structure is fixed at A. ```text 20 kN-> B ============ C ====== D 5 kN/m on BC | | v 20 kN | 4 m | 4 m | o E (hinge) A (fixed) |<--- 5 m --->|<- 2 ->| ``` #### Unit and load moments (+ tension on the inner/lower face) $M_0$ is the moment in the primary structure under the real loads (20 kN at B, 5 kN/m on BC, 20 kN at D), with the column CE unstressed; $m_1$ and $m_2$ are the moments due to unit forces at E. Their values at the key sections are given in the table below. $$ \begin{aligned} f_{11} &= \frac{122.667}{EI} \\ f_{12} &= \frac{90.000}{EI} \\ f_{22} &= \frac{141.667}{EI} \\ \Delta_{10} &= \frac{-4050.0}{EI} \\ \Delta_{20} &= \frac{-6574.0}{EI} \end{aligned} $$ $$ \begin{aligned} 122.667\,X_1 + 90.000\,X_2 &= 4050.0 \\ 90.000\,X_1 + 141.667\,X_2 &= 6574.0 \end{aligned} $$ $$ X_1 = H_E = -1.93\ \text{kN},\qquad X_2 = V_E = 47.63\ \text{kN} $$ #### Final bending moments (kNm) | Section | $M_0$ | $m_1$ | $m_2$ | $M=M_0+\sum X m$ | |---|---|---|---|---| | A | -282.50 | 0.00 | 5.00 | -44.35 | | B (column top) | -202.50 | 4.00 | 5.00 | 27.93 | | mid BC | -105.63 | 4.00 | 2.50 | 5.73 | | C (beam, left) | -40.00 | 4.00 | 0.00 | -47.72 | | C (overhang) | -40.00 | 0.00 | 0.00 | -40.00 | | D | 0.00 | 0.00 | 0.00 | 0.00 | | C (column) | 0.00 | 4.00 | 0.00 | -7.72 | | E | 0.00 | 0.00 | 0.00 | 0.00 | At the joint C the member moments balance: -47.72 = -40.00 + (-7.72). #### Shear force (kN) | Member | SF | |---|---| | AB | +18.07 | | BC (at B / at C) | -2.63 / -27.63 | | CD (overhang) | +20.00 | | CE | +1.93 | #### Reactions - E (hinge): $H_E = 1.93$ kN (towards the left), $V_E = 47.63$ kN ($\uparrow$) - A (fixed): $H_A = 18.07$ kN (towards the left), $V_A = 2.63$ kN (downward), $M_A = 44.35$ kNm Check: $\Sigma V$: $-2.63 + 47.63 = 5\times5 + 20 = 45$ kN; $\Sigma H$: $20 - 18.07 - 1.93 = 0$. In BC the moment falls from $+27.93$ kNm at B through zero at 2.86 m from B to -47.72 kNm at C. **Answer: BMD ordinates: $M_A = 44.35$ kNm, $M_B = 27.93$ kNm, $M_C = -47.72$ kNm (beam side), $M_D = 0$ and the BM in BC changes sign at 2.86 m from B; the SFD values are in the table above.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The frame has D_s = 2. What redundants are chosen? A. Moments at A and B B. Horizontal and vertical reactions at hinge E C. Shear and moment at C D. Vertical reactions at A and D Answer: B. Releasing the hinge E leaves a cantilever fixed at A; X1 and X2 are the E reactions. (Key point: Redundants) 2. What is the form of the equations for X1 and X2? A. Two simultaneous equations f11X1+f12X2 = -Δ10 and f21X1+f22X2 = -Δ20 (zero displacement at E) B. X1 and X2 are found independently C. Equilibrium of the whole frame alone D. Only one equation is needed Answer: A. Displacements at E in both directions must be zero, giving two coupled equations with f12 = f21. (Key point: Compatibility equations) 3. How are final bending moments obtained? A. M = M0 only B. M = X1 m1 + X2 m2 only C. M = M0 + X1 m1 + X2 m2 D. M = M0 - m1 - m2 Answer: C. Load moments in the primary structure plus the redundants multiplied by unit moments. (Key point: Superposition) 4. At joint C the member-end moments -47.72 = -40.00 + (-7.72) show which check? A. Shear check B. Joint moment equilibrium C. Axial force check D. Deflection check Answer: B. Moments of the members meeting at a joint without applied couple must balance. 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