--- title: "Frame horizontal deflection at roller support" subject: "Theory of Structures I" chapter: 3 key: "semester_4/theory_of_structures_i:frame-virtual-work-roller-2076" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/frame-virtual-work-roller-2076 site: BCT Engineering --- # Frame horizontal deflection at roller support Theory of Structures I · Chapter 3: Analysis by the Virtual Work Method - Asked in: 2076 Baisakh, 12 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-virtual-work-roller-2076 - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-3 ## Question Calculate horizontal deflection at roller support of the following frame using virtual work method. [Figure: portal frame ABCD; A hinge, D roller; top beam BC (I) of 3 m with 100 kN at 1 m from B, 40 kN horizontal at B; left column AB (2I) of 3 m with 30 kN/m UDL; right column CD (2I).] ## Short answer Horizontal deflection at the roller support is 1880.6/EI (kN·m³) to the right, where EI refers to the beam stiffness; columns have 2I. ## Answer #### Given data Portal frame: A(0,0) hinge, B(0,3), C(3,3), D(3,0) roller. Beam BC: $I$, 3 m, with 100 kN at 1 m from B. Columns AB and CD: $2I$, 3 m. UDL 30 kN/m on AB acting to the right, and 40 kN horizontal at B to the right. Required: horizontal deflection of the roller D by virtual work. #### Real loading: reactions Horizontal: $A_x = 40 + 30\times3 = 130$ kN (to the left). Moments about A (clockwise positive): the horizontal loads give $40\times3 + 90\times1.5 = 255$ kNm clockwise; the 100 kN load gives $100\times1 = 100$ kNm clockwise; the roller reaction $D_y\times3$ acts anticlockwise. $$ D_y = \frac{255 + 100}{3} = 118.33\ \text{kN (up)},\qquad A_y = 100 - 118.33 = -18.33\ \text{kN (i.e. 18.33 kN down)} $$ #### Virtual load Unit horizontal load at D (to the right). The hinge A resists it with $A_x = 1$; the vertical reactions are zero. Hence $m = s$ in AB, $m = 3$ in BC and $m = 3 - s$ in CD (s measured from the start of each member). #### Moments and integrals | Member | $s$ (m) | stiffness | $M$ (kNm) | $m$ | $\int Mm\,ds$ / stiffness | |---|---|---|---|---|---| | AB (A to B) | 0-3 | $2EI$ | $130s - 15s^2$ | $s$ | 433.13 | | BC (B to load) | 0-1 | $EI$ | $255 - 18.33s$ | 3 | 737.50 | | BC (load to C) | 1-3 | $EI$ | $355 - 118.33s$ | 3 | 710.00 | | CD | 0-3 | $2EI$ | 0 | $3 - s$ | 0 | Check: at C, $M = 355 - 118.33\times3 = 0$, as it must be at the roller column (no moment in CD). $$ \Delta_{Dh} = \frac{433.13 + 737.50 + 710.00}{EI} = \frac{1880.6}{EI} $$ **Answer: horizontal deflection of the roller D $= 1880.6/EI$ (kNm$^3$/EI), to the right,** where $EI$ is the stiffness of the beam BC (columns $2EI$). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. To find the horizontal deflection at roller D, the virtual load is: A. Unit horizontal load at D B. Unit vertical load at D C. Unit couple at A D. 100 kN at D Answer: A. A unit load at D in the horizontal direction. (Key point: Virtual load) 2. With a unit horizontal load at the roller D, the virtual moments are: A. m = s in AB, 3 in BC, 3 - s in CD B. Zero everywhere C. m = 3 in AB only D. m = 1 in all Answer: A. The hinge A resists with Ax = 1 and vertical reactions are zero, giving the stated moments. (Key point: Virtual moment m) 3. The vertical reaction at D is: A. 118.33 kN upward B. 100 kN C. 18.33 kN D. 130 kN Answer: A. D_y = (255 + 100)/3 = 118.33 kN. (Key point: Reaction at roller) 4. What check confirms the beam moment expression? A. M at C equals zero, as the roller column carries no moment B. M at C equals 355 C. M at B is zero D. M at A is 255 Answer: A. At C, M = 355 - 118.33 x 3 = 0, consistent with zero moment in column CD. (Key point: Check) 5. Horizontal deflection of D is: A. 1880.6/EI to the right B. 433.13/EI C. 1880.6/EI to the left D. 710/EI Answer: A. 433.13 + 737.50 + 710.00 = 1880.6/EI, positive, so to the right. 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