--- title: "Overhang beam deflection and slope at C" subject: "Theory of Structures I" chapter: 3 key: "semester_4/theory_of_structures_i:beam-vw-overhang-c-10kn" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/beam-vw-overhang-c-10kn site: BCT Engineering --- # Overhang beam deflection and slope at C Theory of Structures I · Chapter 3: Analysis by the Virtual Work Method - Asked in: 2072 Asoj, 12 marks - Video: [Moment-Area Method - Step by Step Explanation and Solved Example - Deflection and Slope](https://www.youtube.com/watch?v=jkwPzvWsqIY) — AF Math & Engineering - 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/beam-vw-overhang-c-10kn - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-3 ## Question Determine the deflection and slope at C in the overhanging beam shown in figure below by using virtual work (unit load) method. Take $EI = 100000\ \text{kNm}^2$. [Figure: beam ABC, A hinge, B roller at 10 m from A, overhang BC 5 m with 10 kN downward at C.] ## Short answer Virtual work with unit load and unit couple. Real moments M=5x (AB), 10s (BC). Integrate Mm/EI: deflection = 1250/100000 = 0.0125 m (12.5 mm downward); slope = 291.67/100000 = 2.917x10^-3 rad clockwise. ## Answer #### Given data Beam ABC: A hinge, B roller at $AB = 10$ m, overhang $BC = 5$ m, with $P = 10$ kN downward at C. $EI = 100\,000$ kNm$^2$. ```text A B C ^---------------^-------+ 10 kN 10 m 5 m ``` #### Real loading $$ R_A\times10 = -10\times5 \;\Rightarrow\; R_A = -5\ \text{kN (down)},\quad R_B = 15\ \text{kN} $$ Hogging moments: $M = 5x$ in AB ($x$ from A), $M = 10s$ in BC ($s$ from C). $M_B = 50$ kNm. #### Deflection at C: unit vertical load at C Virtual moments (hogging): $m = 0.5x$ in AB, $m = s$ in BC. $$ \Delta_C = \frac{1}{EI}\left[\int_0^{10} 5x\,(0.5x)\,dx + \int_0^5 10s\cdot s\,ds\right] = \frac{1}{EI}\left[\frac{2.5\times1000}{3} + \frac{10\times125}{3}\right] $$ $$ \Delta_C = \frac{833.33 + 416.67}{EI} = \frac{1250}{EI} = \frac{1250}{100\,000} = 0.0125\ \text{m} $$ #### Slope at C: unit couple at C Virtual moments: $m = x/10$ in AB, $m = 1$ in BC (same sense as $M$). $$ \theta_C = \frac{1}{EI}\left[\int_0^{10} 5x\,\frac{x}{10}dx + \int_0^5 10s\cdot1\,ds\right] = \frac{1}{EI}\left[166.67 + 125\right] = \frac{291.67}{EI} $$ $$ \theta_C = \frac{291.67}{100\,000} = 2.917\times10^{-3}\ \text{rad (clockwise)} $$ **Answer: deflection at C $= 12.5$ mm downward; slope at C $= 2.917\times10^{-3}$ rad (clockwise).** (Check: $\Delta_C = \dfrac{Pa^2(L+a)}{3EI} = 12.5$ mm, $\theta_C = \dfrac{Pa(2L+3a)}{6EI}$.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With A hinge, B roller 10 m away, and 10 kN down at C (5 m beyond B), what is R_A? A. 15 kN downward B. 5 kN downward C. 10 kN upward D. 5 kN upward Answer: B. Moments about B give R_A x 10 = -10 x 5, so R_A = -5 kN (down) and R_B = 15 kN. (Key point: Reaction at A) 2. For a unit vertical load at C, what are the virtual moments m? A. m = x/10 in AB and m = 1 in BC B. m = 0 in AB and m = s in BC C. m = 5x in AB and m = 10s in BC D. m = 0.5x in AB and m = s in BC (s from C) Answer: D. The unit load gives R_A = 0.5 (down), so m = 0.5x in AB, and m = s in the overhang. (Key point: Virtual moments for deflection) 3. What is the deflection at C? A. 2.917 mm downward B. 12.5 mm downward C. 12.5 mm upward D. 1250 mm upward Answer: B. Delta_C = (833.33 + 416.67)/EI = 1250/100000 = 0.0125 m downward. (Key point: Deflection result) 4. What is applied at C to find the slope there, and what is m in the overhang BC? A. A unit couple, m = 1 in BC and m = x/10 in AB B. A unit horizontal load, m = 1 in BC C. A unit couple, m = 0 in BC D. A unit vertical load, m = s Answer: A. A unit couple at C gives a constant moment of 1 on BC and a linear moment x/10 on AB; the result is theta_C = 291.67/EI = 2.917e-3 rad. 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