Entrance Exams
Govt. 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).