For a cantilever AB (fixed at A) of length L carrying a point load P at the free end, the slope equation from the double integration method is EI dy/dx = −P(Lx − x²/2), with x measured from the fixed end. The maximum deflection is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 06/06/2024 (9 am–11 am)
Detailed Explanation
Integrating the slope equation once more gives the deflection equation. The fixed end supplies the second boundary condition (y = 0 at x = 0), and the maximum deflection occurs at the free end.
\[ EIy = -P\left(\frac{Lx^2}{2} - \frac{x^3}{6}\right) + C_2,\qquad y(0) = 0 \Rightarrow C_2 = 0 \]
\[ EIy_B = -P\left(\frac{L^3}{2} - \frac{L^3}{6}\right) = -\frac{PL^3}{3} \Rightarrow y_B = -\frac{PL^3}{3EI} \]
Why “−PL³/(3EI)” Is Correct
The free-end deflection is −PL³/(3EI) (negative = downward). Notice the dimensions: a deflection must have L³ with P, while a slope has L².
Why the Other Options Are Wrong
- A. −PL²/(6EI): PL²/6EI has the dimensions of a slope, not a deflection.
- C. −PL²/(3EI): PL²/3EI also has slope dimensions.
- D. −PL³/(6EI): PL³/6EI is half the correct value (missing the Lx²/2 term).
Key Point
Cantilever with tip load: δmax = PL³/3EI at the free end.
Key Concepts for Students
- Dimension check: P·L²/EI → slope (radians); P·L³/EI → deflection (length).
- Maximum deflection of a cantilever is always at the free end for downward loads.
- Sign convention: downward deflection negative when y is taken upward.
Practice more: this is question 6 in Slope and Deflection MCQs – Part 1. Try the whole set and check every answer instantly.