In the double integration method for a simply supported beam carrying a UDL over its entire span, the slope and deflection calculation requires:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 11/10/2023 (9 am–11 am)
Detailed Explanation
Two integrations of EI y'' = M produce two constants. For a symmetrically loaded simply supported beam, the convenient pair of conditions is: zero deflection at a support, and zero slope at mid-span because of symmetry.
\[ EI\frac{d^2y}{dx^2} = \frac{wL}{2}x - \frac{wx^2}{2} \]
Symmetry: \(\frac{dy}{dx} = 0\) at \(x = L/2\) gives \(C_1 = -\dfrac{wL^3}{24}\). Boundary: \(y = 0\) at \(x = 0\) gives \(C_2 = 0\).
Why “the boundary condition as well as the symmetry condition” Is Correct
The book's answer is the boundary condition as well as the symmetry condition: the support condition y = 0 and the symmetry condition dy/dx = 0 at the centre. (Using y = 0 at both supports would also work, but the symmetry condition is the standard, quickest route.) No continuity condition is needed because a single moment equation covers the whole span.
Why the Other Options Are Wrong
- B. the boundary condition as well as the continuity condition: Continuity conditions are needed only when the beam is split into segments with different moment equations (e.g. point loads).
- C. only the symmetry condition: Symmetry alone fixes only C₁; a support condition is still needed for C₂.
- D. only the boundary condition: This works too, but the expected answer for a symmetric beam uses symmetry plus a boundary condition.
Key Point
Symmetric simply supported beam: y = 0 at support (boundary) + dy/dx = 0 at mid-span (symmetry).
Key Concepts for Students
- Continuity conditions match slope and deflection at segment junctions.
- Macaulay's method removes the need for continuity conditions.
- Results: θ = wL³/24EI, δ = 5wL⁴/384EI.
Practice more: this is question 9 in Slope and Deflection MCQs – Part 1. Try the whole set and check every answer instantly.