--- title: "Consistent deformation: frame, BMD, SFD, normal thrust" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:consistent-deformation-frame-1knm-3kn" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/consistent-deformation-frame-1knm-3kn site: BCT Engineering --- # Consistent deformation: frame, BMD, SFD, normal thrust Theory of Structures II · Chapter 2: Force method - Asked in: 2067 Asar, 15 marks - Video: [ANALYSIS OF FRAME SHEAR BENDING & AXIAL THRUST DIAGRAM | EX 03](https://www.youtube.com/watch?v=VhZ1_8AFJrI) — Structural Funda by:- KheNiSiR - 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/consistent-deformation-frame-1knm-3kn - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Use the consistent deformation method to solve the frame and draw bending moment, shear force and normal thrust diagrams. [Figure: frame; left column (2I) of 6 m, fixed base, carrying 1 kN/m UDL horizontally; beam (2I) of 6 m with 3 kN at 3 m from the left; right column (I) of 3 m, fixed base.] ## Short answer Release fixed D and impose zero displacements there. Results: H_D=2.40 kN left, V_D=1.84 kN up, M_D=3.83 kNm anticlockwise; moments M_A=-4.90, M_B=-1.33, M_C=-3.38 kNm; axial compressions are AB 1.16, BC 2.40, CD 1.84 kN. ## Answer **Assumptions:** A(0,0) and D are fixed bases. Column AB (2I) is 6 m high, beam BC (2I) is 6 m and column CD (I) is 3 m high, so D is 3 m above the level of A. The 1 kN/m UDL acts horizontally to the right on AB; the 3 kN load acts downward at 3 m from B on BC. Axial deformation is ignored. Positive bending moment = tension on the inside of the frame. #### Degree of indeterminacy Reactions: 3 + 3 = 6; equations = 3, so $D_s=3$. Release the fixed support D. The redundants are $X_1=H_D$ (right +), $X_2=V_D$ (up +) and $X_3=M_D$ (anticlockwise +). The primary structure is a cantilever fixed at A. #### Moments in the primary structure (kNm; $m_1,m_2,m_3$ for unit $X_1,X_2,X_3$) | Section | $m_1$ (H=1) | $m_2$ (V=1) | $m_3$ (M=1) | $M_L$ | |---|---|---|---|---| | A (AB) | -3.00 | 6.00 | 1.00 | -27.00 | | B (AB) | 3.00 | 6.00 | 1.00 | -9.00 | | B (BC) | 3.00 | 6.00 | 1.00 | -9.00 | | C (BC) | 3.00 | 0.00 | 1.00 | 0.00 | | C (CD) | 3.00 | 0.00 | 1.00 | 0.00 | | D (CD) | 0.00 | 0.00 | 1.00 | 0.00 | #### Flexibility matrix and load displacements ($\int m_im_j\,ds/EI$) $$ [F]=\frac{1}{EI}\begin{bmatrix}45.00&27.00&13.50\\27.00&144.00&27.00\\13.50&27.00&9.00\end{bmatrix},\qquad \{\Delta_L\}=\frac{1}{EI}\begin{Bmatrix}6.75\\-303.75\\-51.75\end{Bmatrix} $$ (For example $f_{33}=\sum \dfrac{L}{EI}=\dfrac{6}{2}+\dfrac{6}{2}+\dfrac{3}{1}=9$.) #### Consistent deformation (D is fixed: all three displacements are zero) $$ [F]\{X\}=-\{\Delta_L\}\;\Rightarrow\; X_1=-2.404\ \text{kN},\quad X_2=1.842\ \text{kN},\quad X_3=3.831\ \text{kNm} $$ So $H_D=2.40$ kN to the left, $V_D=1.84$ kN upward, $M_D=3.83$ kNm anticlockwise. #### Reactions at A (equilibrium) $H_A=6-2.40=3.60$ kN (left), $V_A=3-1.84=1.16$ kN (up), $M_A=4.90$ kNm (anticlockwise). Check: $\sum F_x=6-3.60-2.40=0$ ✓, $\sum F_y=3-1.16-1.84=0$ ✓. #### Bending moment diagram ($M=M_L+\sum X_im_i$, kNm; + inside tension) | Section | BM | |---|---| | A | -4.90 | | Mid-height of AB (3 m) | 1.38 | | B | -1.33 | | Under 3 kN load (BC) | 2.14 | | C | -3.38 | | D | 3.83 | AB bends in a parabola (UDL); BC is two straight lines with a peak under the load; CD is a straight line with a change of sign. #### Shear force and normal thrust | Member | Axial N (+ tension) | Shear at start | Shear at end | |---|---|---|---| | AB | -1.16 | 3.60 | -2.40 | | BC | -2.40 | 1.16 | -1.84 | | CD | -1.84 | 2.40 | 2.40 | Shear sign: positive when the force on the start end of the member acts along its local y-axis (90° anticlockwise from the direction A to B, B to C, C to D). Axial: negative = compression. Normal thrust in AB, BC, CD is therefore 1.16, 2.40 and 1.84 kN compression. The shear in AB goes from 3.60 kN at A to -2.40 kN at B (linear, due to the UDL), and in BC it is 1.16 kN before the 3 kN load and -1.84 kN after it. **Answer: $H_D=2.40$ kN, $V_D=1.84$ kN, $M_D=3.83$ kNm; $M_A=-4.90$, $M_B=-1.33$, $M_C=-3.38$ kNm (inside tension +).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are the redundants chosen for the frame? A. H_D, V_D and M_D at the fixed support D, with the primary structure a cantilever fixed at A B. Only H_D C. The moments at B and C only D. H_A and V_A Answer: A. Two fixed bases give 6 reactions against 3 equations, so D_s = 3 and the fixed support D is released. (Key point: Degree and redundants) 2. What is the consistent-deformation condition used? A. [F]{X} = -{Delta_L}, since the fixed support D cannot displace or rotate B. [F]{X} = {Delta_L}, since D moves freely C. Sum of X equals the load D. Only the rotation at D is zero Answer: A. At the fixed support all three displacements (two translations and the rotation) are zero. (Key point: Compatibility) 3. How does the answer compute f33? A. Sum of L/EI of members: 6/2 + 6/2 + 3/1 = 9, since m3 = 1 everywhere B. Sum of L^3/3EI C. Sum of 6 x 6 D. It equals f11 Answer: A. A unit couple at D produces m3 = 1 throughout, so f33 = sum of L/EI = 9/EI. (Key point: Flexibility coefficient) 4. What are the reactions at D? A. H_D = 2.40 kN left, V_D = 1.84 kN up, M_D = 3.83 kNm anticlockwise B. H_D = 3.60 kN, V_D = 1.16 kN, M_D = 4.90 kNm C. H_D = 6 kN, V_D = 3 kN, M_D = 0 D. H_D = 2.40 kN right, V_D = 1.84 kN down, M_D = 3.83 kNm clockwise Answer: A. Solving [F]{X} = -{Delta_L} gives X1 = -2.404, X2 = 1.842, X3 = 3.831. 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