From the theory of elasticity, if the ends of a horizontal beam (of uniform cros
ID: 3079062 • Letter: F
Question
From the theory of elasticity, if the ends of a horizontal beam (of uniform cross-section and constant density) are supported at the same height in vertical walls, then its vertical displacement y(x) satisfies the Boundary Value Problem where P > 0 is a constant depending on the beam's density and rigidity and L is the distance between supporting walls: Solve the above boundary value problem when L = 4 and P = 24. Show that the maximum displacement occurs at the center of the beam x = L/2 = 2.Explanation / Answer
The differential equation y'''' = -P has y''' = -Px + C1 y'' = -P/2 x^2 + C1 x + C2 y' = -P/6 x^3 + C1/2 x^2 + C2 x + C3 y = -P/24 x^4 + C1/6 x^3 + C2/2 x^3 + C3 x + C4 Now, this is nothing more than a fourth-order polynomial where the leading coefficient is -P/24. The remaining coefficients are arbitrary constants. Examine the form of the boundary conditions, and you'll see why it is advantageous to rewrite the polynomial in the form y = -P/24 x^2(L - x)^2 Note that this polynomial is still a fourth order polynomial where the leading coefficient is -P/24. Now y(0) = 0 (check) y(L) = 0 (check) The derivative is y' = -P/24 [2x(L - x)^2 - 2 x^2 (L-x)] So that y'(0) = 0 (check) y'(L) = 0 (check) The solution satisfies the boundary conditions, and the differential equation, so it is the solution. When L = 4 and P = 24, the solution is y = - x^2(4 - x)^2 The maximum displacement occurs when y' = 0, so solve 0 = - 2x(4 - x)^2 + 2x^2 (4 - x) = -2x(4-x)[ (4-x) - x ] = -2x(4-x)(4 - 2x) The zeros at x=0 and x=4 obviously correspond to the minimum displacement. The zero in the center of the beam, 2x = 4 --> x = 2 is the maximum. You might want to use the second derivative test to confirm it is a maximum, but physical intuition should give it to you too.
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