--- title: "Frame rotation and vertical deflection at free end" subject: "Theory of Structures I" chapter: 3 key: "semester_4/theory_of_structures_i:frame-rotation-vertical-free-end-2073m" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/frame-rotation-vertical-free-end-2073m site: BCT Engineering --- # Frame rotation and vertical deflection at free end Theory of Structures I · Chapter 3: Analysis by the Virtual Work Method - Asked in: 2073 Magh, 10 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/frame-rotation-vertical-free-end-2073m - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-3 ## Question Determine the rotation and vertical deflection at free end. [Figure: frame fixed at the base of the left column (2EI, 3 m + 1 m high, 50 kN horizontal at 1 m above base); top beam of 6 m (1.5EI) with 20 kN/m UDL; right column EI hanging 2 m with 30 kN horizontal load at its free end.] ## Short answer Rotation at the free end is 1512.5/EI rad clockwise; vertical deflection is 7275/EI downward. Virtual work with unit couple and unit vertical load applied at the free end yields the results. ## Answer #### Given data Frame fixed at the base A. Left column: 4 m high ($2EI$), with 50 kN horizontal (to the right) 1 m above the base. Beam: 6 m ($1.5EI$) with UDL 20 kN/m (down). Right column: $EI$, hanging 2 m from the beam end, with 30 kN horizontal (to the left) at its free end. Required: rotation and vertical deflection at the free end D. ```text 20 kN/m (1.5EI, 6 m) B ==================== C | | EI, 2 m | D <-- 30 kN | 2EI, 3 m | |--> 50 kN (1 m above A) | 1 m A (fixed) ``` #### Virtual systems - **Rotation:** unit couple (clockwise) at D. - **Vertical deflection:** unit load (downward) at D. The frame is a cantilever, so $M$ and $m$ are found from the free end D without reactions. $s$ is measured from D along DC, from C along CB, and from B down BA. All moments below are the **clockwise** moments, about the section, of the forces between D and the section; $M$ and $m$ use the same convention. #### Bending moments (kNm) | Member | $s$ (m) | stiffness | $M$ | $m_\theta$ | $m_v$ | |---|---|---|---|---|---| | DC | 0-2 | $EI$ | $30s$ | 1 | 0 | | CB | 0-6 | $1.5EI$ | $60 + 10s^2$ | 1 | $s$ | | BA (upper 3 m) | 0-3 | $2EI$ | $420 - 30s$ | 1 | 6 | | BA (lower 1 m) | 3-4 | $2EI$ | $270 + 20s$ | 1 | 6 | Every load (the 30 kN to the left at D, the UDL and the 50 kN to the right on the column) gives a clockwise moment, so $M$ has the same sign in every member. #### Integrals $\int Mm\,ds$ / stiffness | Member | Rotation | Vertical | |---|---|---| | DC | 60.0 | 0 | | CB | 720.0 | 2880.0 | | BA (3 m) | 562.5 | 3375.0 | | BA (1 m) | 170.0 | 1020.0 | | **Total** | **1512.5** | **7275.0** | Both totals are positive, so the free end moves in the sense of the virtual loads: $$ \theta_D = \frac{1512.5}{EI}\ \text{(clockwise)},\qquad \Delta_{Dv} = \frac{7275}{EI}\ \text{(downward)} $$ **Answer: rotation at the free end $= 1512.5/EI$ rad (clockwise); vertical deflection $= 7275/EI$ m (downward), with $EI$ in kNm$^2$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which virtual loads are used to find the rotation and the vertical deflection at the free end D? A. A unit vertical load for both B. A unit couple for both C. A unit couple at D for rotation and a unit vertical load at D for deflection D. A unit horizontal load for rotation and a unit couple for deflection Answer: C. The virtual action must match the displacement wanted: couple for rotation, force for deflection. (Key point: Virtual systems) 2. Why can M and m be found without calculating the support reactions? A. The supports are hinged B. The frame is a cantilever, so moments can be taken from the free end D C. The frame is symmetric D. The loads are small Answer: B. With a fixed base and free end D, the section moment is found from the loads between the section and the free end. (Key point: Reactions not needed) 3. What is m for the rotation case in every member? A. It equals s B. It equals 1, because a unit couple gives the same moment everywhere C. It equals 6 D. It equals 0 in the beam Answer: B. A unit couple applied at the free end of a cantilever frame produces a constant moment of 1 in all members. (Key point: Virtual moment for rotation) 4. Why is m = 6 in column BA for the vertical deflection? A. The unit load is 6 kN B. The stiffness of BA is 6EI C. The unit vertical load at D acts at a horizontal distance of 6 m (the beam length) from the column D. The column is 6 m high Answer: C. The moment of a unit vertical load at D about any section in BA is its lever arm, the 6 m horizontal span of the beam. 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