Civil Engineering Archive: Questions from May 09, 2022
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The general solution of the Partial differential equation uze + 2xy + y = 18e24+ is Ou=F(-2+ y) +G(-2+ y) +2e2x+y Ou= F(-2+ y) + G (-+ + y) + 2e-+ Ou=F(-2+ y) + 2G (-1 + y) +1842-- Ou=F(-2+y)+ G(-x +1 answer -
3. If A=0.04 m2, E=2.19x106 Ton/m2, L=3.00 m and k=2,000 Ton/m, determine the Equivalent Stiffness of the system. 4. If EI=320 Ton m2, L=4.00 m and k=60 Ton/m, determine the Equivalent Stiffness of th
3. Si A=0.04 m2, E=2.19x106 Ton/m2, L=3.00 m y k=2,000 Ton/m, determine la Rigidez Equivalente del sistema. A, E L h 5 m р 4. Si EI=320 Ton m?, L=4.00 m y k=60 Ton/m, determine la Rigidez Equivalen1 answer -
5. Determine the static and dynamic Degrees of Freedom (DOF) for each case. The members are axially non-deformable. 6. Determine the static and dynamic Degrees of Freedom (DOF) for each case. Conside
5. Determine los GDL estáticos y dinámicos para cada caso. Los miembros son indeformables axialmente. El, = 0 EI m m EL ETC ΕΙ. EI Determine los GDL estáticos y dinámicos para cada caso. Consi1 answer -
A blimp is supported by the three cables shown in Figure P4-18. The net vertical thrust developed by the airship is 8,000N, the wind drag is equal to 1,000N, the wind is "horizontal and perpendicular
4-18 Un dirigible es sostenido por los tres cables representados en la Figura P4-18. El empuje vertical neto que desarrolla el diri- gible es 8.000N, el arrastre del viento es igual a 1.000N, éste es1 answer