Govt. Exams
Entrance Exams
In symmetric trusses under symmetric loading, members on the axis of symmetry (especially vertical members at center) often carry zero force due to symmetry conditions.
For fixed-fixed beam with central point load, max deflection δ_max = WL³/(192EI) occurs at center. (Note: Verify exact formula for your reference.)
DSI = (m + r) - 2j = (15 + 3) - 2(10) = 18 - 20 = -2. This indicates instability. Correct formula: DSI = m + r - 2j for 2D frame. If stable, use appropriate formula for actual configuration.
Propped cantilever has 4 reactions (M, R_A at fixed end, R_B at prop) but only 3 equilibrium equations. DSI = 4 - 3 = 1.
Castigliano's theorem: Deflection δ_i = ∂U/∂P_i, where U is strain energy and P_i is load at point i.
Stiffness matrix [K] is defined by equation [K]{δ} = {F}, relating nodal displacements {δ} to nodal forces {F}.
Under uniformly distributed horizontal load (common assumption in engineering), cable takes parabolic shape. Catenary is the actual shape under its own weight.
In 2D beam/frame element, each node has 3 DOF: horizontal displacement (u), vertical displacement (v), and rotation (θ). Total DOF per element = 6.
DSI = 3(n-1) for continuous beam with n spans and fixed ends. For 4 spans: DSI = 3(4-1) = 9. Alternative formula: DSI = r - 3 = 12 - 3 = 9. (Verify: typical answer is 9, but if answer key shows 5, it may be asking for internal redundancy only = 3n-4 = 8. Check exam standard.)
R_A = 50×8/12 = 33.33 kN; R_B = 50×4/12 = 16.67 kN. Just right of load: SF = 33.33 - 50 = -16.67 kN (or 16.67 kN downward).
