--- title: "Cantilever: slope and deflection under load P1" subject: "Strength of Materials" chapter: 7 key: "semester_3/strength_of_materials:cantilever-slope-deflection-under-p1" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/cantilever-slope-deflection-under-p1 site: BCT Engineering --- # Cantilever: slope and deflection under load P1 Strength of Materials · Chapter 7: Theory of Flexure - Asked in: 2069 Chaitra, 5 marks - Video: [Superposition Method - Slopes and Deflections](https://www.youtube.com/watch?v=PHwg9Q64FVs) — Concrete Dream - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/cantilever-slope-deflection-under-p1 - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-7 ## Question Find the slope and deflection under the load $P_1$. [Figure: cantilever fixed at the left end; $P_1 = 20\ \text{kN}$ at 3 m from the fixed end; $P_2 = 30\ \text{kN}$ at the free end 2 m beyond; $E = 2\times10^5\ \text{MPa}$, $I = 2\times10^8\ \text{mm}^4$.] ## Short answer Using superposition with EI=4×10¹³ Nmm², the slope under the 20 kN load is 0.0101 rad and the deflection is 18.0 mm downward. ## Answer **Given:** $P_1 = 20$ kN at $a = 3$ m from the fixed end; $P_2 = 30$ kN at the free end ($L = 5$ m). $E = 2\times10^5$ N/mm², $I = 2\times10^8$ mm⁴. $$ EI = 2\times10^5\times2\times10^8 = 4\times10^{13}\ \text{N mm}^2 = 4\times10^4\ \text{kNm}^2 $$ Use superposition. Take $x$ from the fixed end; the moment due to $P_2$ at the free end is $M = P_2(L - x)$. #### Effect of $P_1$ (at its own position) $$ \theta_1 = \frac{P_1 a^2}{2EI} = \frac{20\times10^3\times3000^2}{2\times4\times10^{13}} = 0.00225\ \text{rad} $$ $$ y_1 = \frac{P_1 a^3}{3EI} = \frac{20\times10^3\times3000^3}{3\times4\times10^{13}} = 4.5\ \text{mm} $$ #### Effect of $P_2$ (at free end) at the position of $P_1$ Integrate $EI\,y'' = P_2(L-x)$ with $y = y' = 0$ at the fixed end, then put $x = 3$ m: $$ EI\,\theta_2 = P_2\left[Lx - \frac{x^2}{2}\right]_{x=3} = 30\,(15 - 4.5) = 315\ \text{kNm}^2 \Rightarrow \theta_2 = 0.007875\ \text{rad} $$ $$ EI\,y_2 = P_2\left[\frac{Lx^2}{2} - \frac{x^3}{6}\right]_{x=3} = 30\,(22.5 - 4.5) = 540\ \text{kNm}^3 \Rightarrow y_2 = 13.5\ \text{mm} $$ #### Total under $P_1$ $$ \theta = 0.00225 + 0.007875 = 0.0101\ \text{rad} $$ $$ y = 4.5 + 13.5 = 18.0\ \text{mm} $$ **Answer: slope under $P_1$ = 0.0101 rad (about 0.58°); deflection under $P_1$ = 18.0 mm downward.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the slope and deflection under P1 found in this answer? A. Add the effect of P1 at its own position and the effect of P2 at the position of P1 B. Use only P2 because it is the larger load C. Use only P1 because the question asks about P1 D. Average the effects of P1 and P2 Answer: A. Both loads are on the beam, so by superposition the effects of each at the point under P1 are added. (Key point: Superposition) 2. What is the deflection under P1 due to P1 alone (a = 3 m)? A. P1 a^3/(8EI) = 1.7 mm B. P1 a^3/(3EI) = 4.5 mm C. P1 a^2/(2EI) = 0.00225 mm D. P1 a^3/(48EI) = 0.28 mm Answer: B. A cantilever with a point load at distance a deflects P1 a^3/3EI at the load, which gives 4.5 mm. (Key point: Effect of P1) 3. What bending moment expression is integrated for the effect of P2 at the free end? A. M = P2 x, with x from the fixed end B. M = P2 L only, constant along the beam C. M = P2(L - x), with x from the fixed end D. M = P2 L^2/2 Answer: C. The moment at a section is P2 times its distance from the free end, L - x. (Key point: Moment due to P2) 4. What are the total slope and deflection under P1? A. 0.00225 rad and 4.5 mm B. 0.007875 rad and 13.5 mm C. 0.0101 rad and 4.5 mm D. 0.0101 rad and 18.0 mm Answer: D. Slope 0.00225 + 0.007875 = 0.0101 rad; deflection 4.5 + 13.5 = 18.0 mm. 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