Using finite element method, determine the reaction moments at the fixed ends of the beam as shown. The beam is divided into two elements. At node 2, a downward force and an ACW moment are applied. Take the modulus of elasticity to be 150 GPa and the moment of inertia of the cross section as 0.001 m^4 . 35 kN-m 5 m 5 m 1 2 3 50 kN a) Reaction moments at nodes 1 and 3 = 71.25 kN-m ACW and 53.75 kN-m CW respectively. b) Reaction moments at nodes 1 and 3 = 17.5 kN-m ACW and 17.5 kN-m ACW respectively. c) Reaction moments at nodes 1 and 3 = 42.5 kN-m ACW and 25 kN-m CW respectively. d) Reaction moments at nodes 1 and 3 = 105 kN-m ACW and 70 kN-m CW respectively.
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The stiffness matrix for a beam element in bending is given by: K = (E*I/L) * [12 -6L; -6L 4L^2] where E is the modulus of elasticity, I is the moment of inertia, and L is the length of the element. For element 1 (from node 1 to node 2), L = 5 m. So, K1 = Show more…
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