Govt. Exams
Entrance Exams
For column fixed at both ends, effective length L_e = 0.5L (K=0.5), giving maximum buckling resistance.
I = bd³/12, A = bd, r = √(I/A) = √(d²/12) = d/(2√3) for axis parallel to base.
For a cantilever beam with a uniformly distributed load, the slope at the free end is found by integrating the bending moment equation twice using the differential equation of the elastic curve.
For a cantilever beam fixed at one end with UDL w acting downward, consider a section at distance x from the free end.
The slope is obtained from integrating the bending moment using the standard relation where EI(d²y/dx²) = M(x).
Integrating once with respect to x to find slope θ:
At the fixed end (x = L), the slope is zero: θ(L) = 0, so C₁ = 0.
At the free end (x = 0), substitute to find the slope:
The slope at the free end is (B) wL³/6EI
Maximum 3 members can be cut in a 2D truss to maintain a determinate section with 3 equilibrium equations available.
For a simply supported beam with central point load, maximum bending moment occurs at the center (L/2) and equals PL/4.
Superposition principle applies only to linear elastic systems with small displacements where strain-displacement and stress-strain relationships are linear.
A hinged/pin support prevents translation in both horizontal and vertical directions (2 restraints), thus has 2 unknown reactions (Hx and Vy). Rotation is free.
In method of sections for trusses, we can cut maximum 3 members (one truss has 3 equilibrium equations: ΣFx=0, ΣFy=0, ΣM=0) to find member forces.
Flexibility method removes redundant reactions/members equal to degree of indeterminacy to create a primary determinate structure, then applies compatibility equations.
Castigliano's second theorem: ∂U/∂P = deflection at point of load application. U is strain energy and P is the applied load.
