Title 16 Double Integration Method.pdf ✅

16 Double Integration Method SOLUTIONS SITUATION 1. Solving for the moment equation using the cutting plane shown, EIy ′′ = wL 2 x − wx ( x 2 ) EIy ′′ = wL 2 x − 1 2 wx 2 Solving for the general solution, EIy ′ = wL 4 x 2 − wx 3 6 + C1 EIy = wLx 3 12 − wx 4 24 + C1x + C2 From the support conditions, at x = 0, then y = 0. Also, at x = L, y = 0. From the first support condition, EI(0) = wL(0) 3 12 − w(0) 4 24 + C1 (0) + C2 → C2 = 0 From the second support condition, EI(0) = wL(L) 3 12 − w(L) 4 24 + C1L + 0 C1 = − wL 3 24 Slope equation: θ = y ′ = 1 EI ( wLx 2 4 − wx 3 6 − wL 3 24 ) Deflection equation:


θ = y ′ = 1 EI (5x 2 − 12⟨x − 3.5⟩ 2 − 595 12 ) θleft = 1 EI [5(0) 2 − 12⟨0 − 3.5⟩ 2 − 595 12 ] θleft = − 595 12EI or − 49.5833 EI ▣ 5. The rotation at the right support is the value of θ at x = 6. θ = y ′ = 1 EI (5x 2 − 12⟨x − 3.5⟩ 2 − 595 12 ) θright = 1 EI [5(6) 2 − 12⟨6 − 3.5⟩ 2 − 595 12 ] θright = 665 12EI or 55.4167 EI ▣ 6. The deflection under the load is the value of δ at x = 3.5. δ = y = 1 EI ( 5 3 x 3 − 4⟨x − 3.5⟩ 3 − 595 12 x) δP = 1 EI [ 5 3 (3.5) 3 − 4⟨3.5 − 3.5⟩ 3 − 595 12 (3.5)] δP = − 1225 12EI or − 102.0833 EI ▣ 7. Locate the maximum deflection. The maximum deflection in the beam is when θ = 0. Assuming that 0 ≤ x ≤ 3.5, θ = y ′ = 1 EI (5x 2 − 12⟨x − 3.5⟩ 2 − 595 12 ) 0 = 1 EI (5x 2 − 595 12 ) → x = 3.149074 m Since 3.149 m < 3.5 m, then the assumption is correct. The maximum deflection is the value of δ at the computed x = 3.149074. δ = y = 1 EI ( 5 3 x 3 − 4⟨x − 3.5⟩ 3 − 595 12 x) δmaximum = 1 EI [ 5 3 (3.149) 3 − 595 12 (3.149)] δmaximum = − 104.0944 EI