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

$\bigstar$

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