A cantilever AB of length l is fixed at A and carries a point load W at the free end B. Using the double integration method with x measured from the fixed end A, the slope dy/dx at any section is (EI = flexural rigidity):
Try it first: pick an option, then read the explanation.
Asked in SSC JE 06/06/2024 (9 am–11 am)
Detailed Explanation
Take a section at distance x from the fixed end. The only load to its right is W at a lever arm (l − x), so the moment is hogging: M = −W(l − x).
\[ EI\frac{d^2y}{dx^2} = -W(l-x) \]
\[ EI\frac{dy}{dx} = -W\left(lx - \frac{x^2}{2}\right) + C \]
At the fixed end x = 0, dy/dx = 0 ⇒ C = 0.
\[ \frac{dy}{dx} = -\frac{W}{2EI}\left(2lx - x^2\right) \]
Check at x = l: \(\theta_B = -\dfrac{Wl^2}{2EI}\), the standard free-end slope.
Why “dy/dx = −W(2lx − x²)/(2EI)” Is Correct
The expression −W(2lx − x²)/(2EI) is zero at the fixed end and equals −Wl²/2EI at the free end, both of which match the known results.
Why the Other Options Are Wrong
- A. dy/dx = −W(2l − x)/(2EI): −W(2l − x)/(2EI) is not zero at x = 0, so it breaks the fixed-end condition.
- B. dy/dx = −W(2lx − x²)/(4EI): The factor 4 is wrong: at x = l it gives Wl²/4EI instead of Wl²/2EI.
- D. dy/dx = −W(2l − x)/(4EI): This is neither zero at the fixed end nor correct at the free end.
Key Point
Cantilever, load at tip, x from fixed end: dy/dx = −W(2lx − x²)/(2EI); θmax = Wl²/2EI.
Key Concepts for Students
- Always check a derived slope against the boundary condition and a known result.
- Deflection: y = −W(3lx² − x³)/(6EI), giving Wl³/3EI at the tip.
- Measuring x from the fixed end makes both constants zero.
Practice more: this is question 5 in Slope and Deflection MCQs – Part 1. Try the whole set and check every answer instantly.