15 questions. Tap an option to check it instantly, then read the solution.
Question 31Hard
A cantilever 2 m long and 10 cm deep tapers in plan from a width of 24 cm at the fixed end to zero at its free end, and carries a point load of 5 kN at the free end. If E = 0.20 × 10⁶ N/mm², the deflection of the free end is:
b ∝ x and M = Wx, so M/I is constant: uniform curvature 1/R = WL/(EI₀). δ = L²/(2R) = 6WL³/(Eb₀d³) = 6 × 5000 × 2000³/(2 × 10⁵ × 240 × 100³) = 5 mm.
Question 32Medium
A simply supported beam of length L carries two equal and opposite (unlike) couples M at its two ends, so that the beam is in pure bending. If EI is constant, the central deflection is:
Constant M: EIy'' = −M. With y = 0 at ends and y' = 0 at centre, EIy = −Mx²/2 + MLx/2 ⇒ δ = ML²/8EI.
Question 33Medium
A cantilever 2 m long, fixed at its left end, carries a downward point load of 9 kN at its right (free) end. Find the magnitude and direction of the couple to be applied at the free end so that the vertical deflection there is zero.
WL³/3EI = ML²/2EI ⇒ M = 2WL/3 = 2 × 9 × 2/3 = 12 kN·m. It must bend the beam upward (sagging), i.e. act anticlockwise at the right end. (Book key: clockwise.)
Question 34Easy
What is the slope at the end of a simply supported beam of span 2 m carrying a load of 5 kg per unit length over the entire span?
δᵢ = ∂U/∂Pᵢ. In the usual modern numbering (followed by the book's key) this is Castigliano's second theorem; Castigliano's first theorem is Pᵢ = ∂U/∂δᵢ.
All three are true: σ = My/I is zero at the N.A.; τ = VAȳ/Ib is zero at the outer fibres; with τ = 0 there, the bending stress at the outer fibre is a principal stress.
Question 45Easy
The ratio of the maximum deflection of a simply supported beam with a central load W to that of a cantilever of the same length with a load W at its free end is: