Answer to Question #177744 in Differential Equations for SHAIKH SHOAIB

Question

Find the equation of the elastic curve and its maximum deflection for the horizontal, simply supported,

uniform beam of length 2l metres, having uniformly distributed load w kg/metre.

Answer 1

Let 2l = L

and substitute the values in the last

$\bigstar$ Taking coordinate axes x and y as shown, we have for the bending moment at any point x.

Mx = wLx/2 - wx² /2 

$\bigstar$ And deflection equation becomes

EI d² y/dx² = wLx/2 - wx² /2. 

$\bull$ Multiplying both sides by dx and integrating, we obtain 

EI dy/dx = wLx² /4 - wx³ /6 + C1, ------------------ ( 1) 

$\bull$ Where C1 is an integration constant. To evaluate this constant, we note from symmetry that when x = L/2, dy/dx = 0. From this condition, we find 

C1 - wL³ /24 

$\bull$ And equation 1 becomes 

EI dy/dx = -wLx² /4 + wx³ /6 - wL³ /24 ------------------ ( 2) 

$\bull$ Again multiplying both sides by dx and integrating, 

EIy = wLx³ /12 - wx⁴ /24 - wL³ x/24 + C2 

$\bull$ The integration constant C2 is found from the condition that y = 0 when x = 0. 

$\bull$ Thus C2 = 0 and the required equation for the elastic line becomes 

$\bigstar$

$\boxed{y ={ -wx \over{ 24El}} (L^3 -2Lx^2 + x^3 ) }$ {we used ${d^2 y\over {dx^2}} ={ M \over EI }$ }

But L = 2l

so putting the values we get

$\bigstar$

$\boxed{y ={ -wx \over{ 24El}} ((2l)^3 -4lx^2 + x^3 ) }$

$\bigstar$ maximum deflection 

 we set x = L/2 in the equation and obtain 

$\boxed{| δ | ={ 5wL^4 \over384EI }}$

But L = 2l

so putting the values we get

$\bigstar$

$\boxed{| δ | ={ 5w(2l)^4 \over384EI }}$