Respuesta :
To solve this problem, use the superposition method, the total deflection at any point is the sum of the deflections due to the individual loads acting at that point.
a. we need to find the elastic curve equation for segment AB. This can be done by solving for beam deflection due to a point load in the middle of segment AB.
The beam deflection under a point load at distance x from the left support is given by
δ(x) = (w0x^2)/(24EI)
where w0 is the point load, E is the modulus of elasticity of the beam, and I is the moment of inertia of the beam.
Inserting the given value will result in:
δ(x) = (3x^2)/(2425000) = x^2/2000
This is the elastic curve equation for segment AB, as there is a point load in the middle of the segment. For part (b), we need to find the maximum absolute value of deflection between supports A and B. The deflection due to a point load is symmetrical about the midpoint of the segment, so the maximum absolute deflection occurs at the midpoint. Substituting x=L/2=1 into the equation for the elastic curve gives:
δ(x) = (1^2)/2000 = 0.0005m
This is the maximum absolute deflection between supports A and B.
In part (c) we need to find the deflection in the middle of segment AB. Substituting x=L/2=1 into the equation for the elastic curve gives:
δ(x) = (1^2)/2000 = 0.0005m
This is the deflection in the middle of segment AB. In part (d) we need to find the deflection at point C. To do this, solve for the beam deflection due to the evenly distributed load on segment BC and add it to the deflection due to the point load at the middle of segment AB.
b. The deflection of a beam under a uniformly distributed load w per unit length is given by
δ(x) = (wx^3)/(48EI)
Inserting the given value will result in:
δ(x) = (3x^3)/(4825000) = x^3/3200
This is the equation for the elastic curve of segment BC as the load is evenly distributed.
c. To find the deflection at point C, we need to substitute x=L=2 into the elastic curve equation. This will tell you:
δ(x) = (2^3)/3200 = 0.00125m
d. To find the total deflection at point C, the deflection due to the point load at the center of section AB must be added to the deflection due to the evenly distributed load in section BC. Substituting x=L/2=1 into the equation for elastic curve due to point load, we get:
δ(x) = (1^2)/2000 = 0.0005m
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