--- title: "Castigliano: slope at A, deflection at D" subject: "Theory of Structures II" chapter: 2 key: "semester_5/theory_of_structures_ii:castigliano-slope-a-deflection-d" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/castigliano-slope-a-deflection-d site: BCT Engineering --- # Castigliano: slope at A, deflection at D Theory of Structures II · Chapter 2: Force method - Asked in: 2073 Shrawan, 10 marks - Video: [CASTIGLIANO'S THEOREM in Just Over 10 Minutes!](https://www.youtube.com/watch?v=JH6iOhOe8h0) — Less Boring Lectures - 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/castigliano-slope-a-deflection-d - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2 ## Question Determine slope at A and deflection at D of the beam shown in figure below using Castigliano's theorem. [Figure: beam A-B-C-D with 30 kN at B (4 m from A), 10 kN at D; BC = 6 m, CD = 2 m overhang; A hinged, C roller.] ## Short answer A dummy couple at A and treating D’s load as variable P, compute partial derivatives of M. Integration yields slope at A = 158.67/EI rad clockwise, deflection at D = 176/EI upward due to overhang uplift. ## Answer Beam ABCD: A hinged, C roller (AC = 10 m), loads 30 kN at B (4 m from A) and 10 kN at the end D of the 2 m overhang CD. EI is constant. Apply a dummy couple $M'$ (clockwise) at A, and treat the load at D as a variable $P$ (equal to 10 kN at the end). ```text 30 kN 10 kN v v A o------B---------------------C o--D hinge roller |<- 4 ->|<------- 6 ------->|<- 2 ->| ``` #### Reactions (with $M'$ and $P$) Moments about A: $$ R_C = \frac{30(4) + 12P + M'}{10},\qquad R_A = 30 + P - R_C $$ At $M' = 0$, $P = 10$: $R_C = 24$ kN, $R_A = 16$ kN. #### Bending moments ($x$ from A for AC; $s$ from D for the overhang) $$ \begin{aligned} AB\ (0 \le x \le 4):\ & M = M' + R_Ax, \\ BC\ (4 \le x \le 10):\ & M = M' + R_Ax - 30(x-4), \\ DC\ (0 \le s \le 2):\ & M = -Ps \end{aligned} $$ so $R_A = 30 + P - R_C = 18 - 0.2P - 0.1M'$, with $\dfrac{\partial R_A}{\partial M'} = -0.1$ and $\dfrac{\partial R_A}{\partial P} = -0.2$. | Range | $\partial M/\partial M'$ | $\partial M/\partial P$ | |---|---|---| | AB, BC | $1 - 0.1x$ | $-0.2x$ | | DC | 0 | $-s$ | At $M' = 0$ and $P = 10$: $R_A = 16$ kN, $M = 16x$ in AB, $M = 120 - 14x$ in BC, $M = -10s$ in the overhang. #### Slope at A $$ EI\,\theta_A = \int_0^4 16x\,(1-0.1x)\,dx + \int_4^{10}(120-14x)(1-0.1x)\,dx = 93.87 + 64.80 $$ $$ \theta_A = \frac{158.67}{EI}\ \text{rad (clockwise, the positive sense of } M'\text{)} $$ #### Deflection at D $$ EI\,\delta_D = \int_0^4 16x(-0.2x)\,dx + \int_4^{10}(120-14x)(-0.2x)\,dx + \int_0^2(-10s)(-s)\,ds $$ $$ EI\,\delta_D = -68.27 - 134.40 + 26.67 = -176.00 $$ The negative sign shows that D moves opposite to the 10 kN load. $$ \delta_D = \frac{176}{EI}\ \text{upward} $$ (The 30 kN load at B tilts the beam at C so that the overhang rises; this outweighs the downward bending of the overhang under its own 10 kN load, which is only $160/EI$.) **Answer: slope at A $= \dfrac{158.67}{EI}$ rad (clockwise); deflection at D $= \dfrac{176}{EI}$ (upward).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the slope at A found by Castigliano's theorem when there is no couple at A? A. Differentiate U with respect to the 30 kN load B. Apply a dummy couple M' at A, differentiate U with respect to M', then set M' = 0 C. Differentiate with respect to R_C D. Integrate M only Answer: B. Slope = ∂U/∂M' evaluated with M' = 0; a dummy couple is needed because none acts at A. (Key point: Dummy load) 2. Why is the 10 kN load at D treated as a variable P? A. To find deflection at D as ∂U/∂P, then substitute P = 10 B. To remove it C. Because it is a couple D. Because it equals R_A Answer: A. Deflection at the load point is ∂U/∂P, taken after substituting P = 10. (Key point: Load as variable) 3. Why does R_A depend on M' and P so that ∂M/∂M' = 1 - 0.1x? A. Because R_A is constant B. Because R_A = 18 - 0.2P - 0.1M' changes with them, so the derivative includes R_A terms C. Because the overhang is rigid D. Because EI changes Answer: B. Equilibrium gives R_A in terms of M' and P; differentiating M = M' + R_A x gives 1 + x dR_A/dM' = 1 - 0.1x. (Key point: Reaction dependence) 4. What is the slope at A? A. 176/EI B. 93.87/EI C. 64.80/EI D. 158.67/EI clockwise Answer: D. EI θ_A = 93.87 + 64.80 = 158.67. (Key point: Slope result) 5. What is the deflection at D and its direction? A. 176/EI upward B. 176/EI downward C. 160/EI downward D. 68.27/EI upward Answer: A. EI δ_D = -68.27 - 134.40 + 26.67 = -176; the negative sign means opposite to the load, i.e. upward. 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