Question

MA Review Launch the simulation, then answer the question Slope and Displacement by Integration Part A The easac curve changes shape based on how the beam is supported and the amount of load. The elastic curve can be determined by double Integration of the internal moment expression of the beam and applying is boundary conditions In this activity, you wil observe the folowing when a load is applied to the two beams 1. Elasoc (detecton) curve of the beam 2. Change in elastic curve What is the order of the elastic curve (i.e., the polynomial in terms of the location coordinate x) for the simply supported beam configuration in this animation? O Second Cantilever Beam P O Fourth First O Third Simply Supported Beam Submit Request Answer Provide Feedback Next > Drag Slider to change the load

Answer 1

P Ra Rb a b L

From figure we can see that

$\\\\R_A+R_B=P \\\\M_A=0 \\\\LR_B=aP \\\\R_B=\frac{a}{L}P \\\\R_A=P-\frac{a}{L}P=P\frac{L-a}{L} \\\\R_A=\frac{b}{L}P$

3 Ra

MEnd = 0 M = Rat b P- L

P M Ra a *

$\\\\\sum M_{End}=0 \\\\M=R_ax-P(x-a)=P\frac{b}{L}x-P(x-a) \\\\M=P(\frac{b}{L}x-x)+Pa=\frac{Px}{L}(b-L)+Pa \\\\M=Pa-\frac{Pax}{L}$

$M=Pa(1-\frac{x}{L})$

since for both

$\\\\M \propto x \\\\M=EI \frac{d^2y}{dx^2} \\\\\frac{d^2y}{dx^2}=\frac{M}{EI} \\\\y =\int \int \frac{M}{EI} dx dx \\\\y = \int \int \frac{Pa}{EI}(1-\frac{x}{L}) dx dx \\\\y =\frac{Pa}{EI}(\frac{x^2}{2}-\frac{x^3}{6L})$

So the answer is third order polynomial

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