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The bending of an elastic rod is described by the elastic curve passing through centres of gravity of rod's cross-sections. At small bendings the equation of this curve takes the form N(x)=EI(d^2y)/(dx^2), where N(x) is the bending moment of the elastic forces in the cross-section corresponding to the x coordinate, E is Young's modulus, I is the moment of inertia of the cross-section relative to the axis pasing through the neutral layer (I=intz^2dS, figure) Suppose one end of a steel rod of a square cross-section with side a is embedded into a wall, the protruding section being of length l (figure). Assuming the mass of the rod to be negligible, find the shape of the elastic curve and the deflection of the rod lambda, if its end A experiences (a) the bending moment of the couple N_0, , (b) a force F oriented along the y axis. |
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Answer» Solution :We use the equation given above and use the result that when y is small `1/R~~(d^2y)/(DX^2)`. Thus, `(d^2y)/(dx^2)=(N(x))/(EI)` (a) Here `N(x)=N_0` is a constant. Then integration gives, `(dy)/(dx)=(N_0x)/(EI)+C_1` But `((dy)/(dx))=0` for `x=0`, so `C_1=0`. Integrating again, `y=(N_0x^2)/(2EI)` where we have used `y=0` for `x=0` to set the constant of integration at zero. This is the equation of a parabola. The sage of the free end is `LAMBDA=y(x=l)=(N_0l^2)/(2El)` (b) In this case `N(x)=F(l-x)` because the load F at the EXTREMELY is balanced by a similar FORCE at F directed upward and they constitute a couple. Then `(d^2y)/(dx^2)=(F(l-x))/(EI)`. Integrating, `(dy)/(dx)=(F(lx-x^//2))/(EI)+C_1` As before `C_1=0`. Integrating again, using `y=0` for `x=0` `y=(F((lx^2)/(2)-(x^3)/(6)))/(EI)` here `lambda=(Fl^3)/(3El)` Here for a square cross section `I=underset(-a//2)overset(a//2)intz^2adz=a^4//12`. |
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