Prova 1 Resolvida-2022 1

EI dv/dx = -∫ (qx^4/24L + 5qLx^2/24) dx = -9/24L (x^5/5) - 5qL/24 (x^3/3) + C2 = -9x^5/120L - 9Lx^3/72 + C2 Como δ = 0 para x = 0 => C2 = 0 => δ = 1/EI (-9x^5/120L - 9Lx^3/72) Dadas as Em A; x = 0 => { δ = 0, θ = 0 } Em B; x = L => δ = 1/EI (-9L^5/120L - 9L^4/72) = 1/EI (-9L^4/120 - 9L^4/72) θ = 9L^4/24L + 5qL^3/24 = 9L^3/24 + 5L^3/24 para L/2 => δ = 1/EI (-9L^5/120L.32 - 9L^2(3)/72.8) = 1/EI (-9L^4/3840 - 9L^4/576) 2Dadas assunções (2L x 2L) EI dv/dx . EI = -∫ q(2L-x)^3/2L dx = -9/2L ∫ u^3 du = -9/2L (u^4/4) du = -q(2L-x)^4/8L + C1 θ = 0 => x = 2L => C1 = 0 => θ = dv/dx . EI = -q(2L-x)^4/8L + C1 ∫ EI dv/dx = -∫ q(2L-x)^4/8L = +9/8L ∫ u^1 du = -9/8L (u^5/5) + C2 = +9/8L (2L-x)^5 + C2 u = 2L - x du/dx = -1 Como δ(2L) = 0 => C2 = 0 δ = +9/8L (2L-x)^5 para parte C (x = 2L) => { θ = 0, δ = 0 } para x = 3L/2 => θ = +9(2L-3L/2)^4/8L = +9L^4/128L = +9L^3/128 δ = +9/8L (2L-3L/2)^5 = +9/8L (L/2)^5 = +9/8L (L^5/32) = +9.L^4/256 para \hspace{1ex} L \leq x \leq 2L \\ \frac{q}{L} = \frac{y}{2L-x} \\ y = \frac{(2L-x) \cdot q}{2} \\ \\ u = 2L-x \\ \frac{du}{dx} = -1 \\ Esforco cortante \\ \sum F_y = 0 => V - \left( \frac{y \cdot (2L-x)}{2} \right) = 0 \\ V = \frac{(2L-x)^2 \cdot q}{2 L} \\ \\ para \hspace{1ex} x = L => \left( V = \frac{L^2 q}{2 L} = \frac{L q}{2} \right) \\ \\ para \hspace{1ex} x = 2L => V = 0 \\ \\ Momento flator \\ M = \int \left( \frac{(2L-x)^2 \cdot q}{2L} \right) dx \\ = \int \frac{u^2 \cdot q}{2L} du = \frac{-q}{2L} \cdot \frac{u^3}{3} = - q \cdot \frac{(2L-x)^3}{2L} + C \\ \\ Como \hspace{1ex} m(2L) = 0 => C = 0 \\ \\ M = \frac{-q(2L-x)^3}{2L} Esforco cortante \\ Momento flator \\ \\ \frac{L q}{2} \\ \\ \frac{-9 q L^2}{2 L} \\ \\ Inclinacao e deslocamento \hspace{1ex} (0 \leq x \leq L) \\ \frac{d^2 V}{dx^2} = - \frac{M}{EI} \\ EI \int \frac{d^2 V}{dx^2} = - \int \left( \frac{-9 q x^3}{6L} - \frac{5 q L x}{12} \right) dx = \frac{q}{6 \cdot L} \left( \frac{x^4}{4} \right) + \frac{5 q L \cdot x^2}{24} + C_1 \\ \Theta = \frac{dV}{dx} EI = \left( \frac{q x^4}{24 L} + \frac{5 q L x^2}{24} \right) + C_1 => C_1 = 0 => \\ \\ \Theta = \frac{q x^4}{24L} + \frac{5 q L x^2}{24} \\ \\ Como \hspace{1ex} o \hspace{1ex} peso \hspace{1ex} x = 0 :