The exact value 8 =w()is obtained from eqn.9 as L 24EI2(-2 384EI We observe that 61 6 0.8 384 i. e the approximate solution underpredicts the maximum deflection by 20% However, if we consider the following approximation with 3 degrees of freedom (note it also satisfies the essential boundary conditions, eqn. 11) ar)=CiI(L-r)+C2. (L-r)+c3I(L-r) and require that II(c1, C2, c3) be a minimum aII 4cEⅠL+2c2EⅠL2+2c3EⅠL3 Lg0 0 6 2a1B2+42ED3+43E1+0=0 2clEIL+4c2EIL+<CEILs L5 q0 whose solution is El If you replace this values in eqn. 14 and evaluate the deflection at the midpoint of the beam you obtain the exact solution !!� The exact value δ = w( L 2 ) is obtained from eqn.9 as: δ = − q0 L L − L� � L2 + LL − �L�2� = − 5 q0L4 24EI 2 2 2 2 384 EI We observe that: 1 δ1 96 4 = = = 0.8 δ 5 5 384 i.e. the approximate solution underpredicts the maximum deflection by 20%. However, if we consider the following approximation with 3 degrees of freedom (note it also satisfies the essential boundary conditions, eqn.11): w3(x) = c1x(L − x) + c2x2 (L − x) + c3x(L − x) 2 (14) and require that Π(c1, c2, c3) be a minimum: ∂Π ∂Π ∂Π = 0, = 0, = 0, ∂c1 ∂c2 ∂c3 i.e.: 4 c1 EI L + 2 c2 EI L2 + 2 c3 EI L3 + L3 q0 = 0 6 2 c1 EI L2 + 4 c2 EI L3 + 4 c3 EI L4 + L4 q0 = 0 12 2 c1 EI L3 + 4 c2 EI L4 + 24 c3 EI L5 + L5 q0 = 0 5 20 whose solution is: − (L2 q0) − (L q0) q0 c1 → , c2 → , c3 → 24 EI 24 EI 24 EI If you replace this values in eqn. 14 and evaluate the deflection at the midpoint of the beam you obtain the exact solution !!! 4