·

Engenharia Civil ·

Análise Estrutural

Send your question to AI and receive an answer instantly

Ask Question

Recommended for you

Preview text

TABELA 3.2d\nMOMENTOS DE ENGASTAMENTO PERFEITO\nCARREGAMENTO\nM_BA = - \\frac{P}{120 l^2} c^2 (40e^2 - 45c e + 12c^2)\nM_CD = \\frac{P}{20 l^2} c^2 (5c e - 4c^2)\nM_DC = - \\frac{P}{30 l^2} c^2 (10l^2 - 15c e + 6c^2)\nM_EF = \\frac{P}{30 l^2} c^2 (5l^2 - 3c^2)\nM_BA = - \\frac{P}{120 l^2} c^2 (20e^2 - 15c e + 3c^2)\nM_CD = \\frac{P}{60 l^2} c^2 (5c e - 3c^2)\nM_DC = - \\frac{P}{60 l^2} c^2 (10a e + 3c^2)\nM_EF = \\frac{P}{120 l^2} c^2 (10l^2 - 3c^2)\nM_BA = - \\frac{P}{8 l^2} (l^3 - 2a^2 l^2 + a^3)\nM_CD = \\frac{P}{12 l} (l^3 - 2a^2 l^2 + a^3)\nM_DC = - \\frac{P}{12 l} (l^3 - 2a^2 l^2 + a^3)\nM_EF = \\frac{P}{8 l} (l^3 - 2a^2 l^2 + a^3)\nExtraído de SCHREYER (1965). Convenção de GRINTER.\nRevista e adaptada por Libânio M. Pinheiro, Bruna Catoia e Thiago Catoia. Barra monengastada\nK = \\frac{3EI}{L}\nBarra blengastada\nK = \\frac{4EI}{L}\nAnálise de Estruturas\nk = \\frac{3EI}{L}\nk = \\frac{4EI}{L}\n(k)\nk = \\frac{3EI}{L^2} (c)\n(k)\nk = \\frac{3EI}{L^2}\n(k)\n{k}\n{k}\n\\{ \nE_0 + k_1 d_1 + k_2 d_2 = 0 \nE_0 + k_1 d_1 + k_2 d_2 = 0 \nM_a = M_b' * K_1 d_1 \nM_bc = M_b' * K_{k_1 d_1} + K'_{k_2 d_2} \n\\}