The basic differential equation of the elastic curve for a simply supported, uniformly loaded beam is given as y ΕΙ t dx² wLx 2 where E = 200× 10° Pa is the modulus of elasticity, w = 15000× sin² (x) N/m, L = 3 m, and I = 3x 10-4 m4 is the moment of inertia. The boundary conditions are y(0) = y(L) = 0. Solve for the deflection of the beam using the finite-difference approach (Ax = 0.1 m). Plot the deflection in mm versus the length in m. W wx² L

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
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The basic differential equation of the elastic curve for a simply supported, uniformly loaded
beam is
given as
wLx
wx²
1|²y = 10LX-2
ΕΙ
where E = 200× 10° Pa is the modulus of elasticity, w = 15000x sin² (x) N/m, L = 3 m,
and I = 3x 10-4 mª is the moment of inertia. The boundary conditions are y(0) = y(L) =
0. Solve for the deflection of the beam using the finite-difference approach (Ax = 0.1 m).
Plot the deflection in mm versus the length in m.
t
W
L
OXO
Transcribed Image Text:The basic differential equation of the elastic curve for a simply supported, uniformly loaded beam is given as wLx wx² 1|²y = 10LX-2 ΕΙ where E = 200× 10° Pa is the modulus of elasticity, w = 15000x sin² (x) N/m, L = 3 m, and I = 3x 10-4 mª is the moment of inertia. The boundary conditions are y(0) = y(L) = 0. Solve for the deflection of the beam using the finite-difference approach (Ax = 0.1 m). Plot the deflection in mm versus the length in m. t W L OXO
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