Cantilever Beams along a UDL:

Consider a cantilever beam loaded with a Udl of w/unit length on total length. Consider a section X-X at a distance x from left end.

Moment, M =- w x(x/2) = - wx²/2  (Hogging BM)

The governing equation for deflection is :

EI (d ² y/ dx² )= M = - w x²/2

Figure

Integrating the Eq. we may get

EI (dy/ dx)        =- Wx³/6 + C₁                       

EIy =- Wx⁴/24 + C₁ x + C₂

The constants C₁ and C₂ can be found from boundary conditions.

The boundary conditions are :

At B, x = l, dy/ dx = 0    ------------ (1)

At B, x = l, y = 0  ------- (2)

Applying the BC (1) into the Eq. we can get

0 = - wl ³ / 6   + C1

∴          C₁  =+  wl ³ / 6

Applying BC (2) into the Eq. we can get

0 =- wl ⁴ /24 +  wl ⁴ /6 + C₂

∴          C₂  =-  wl ⁴ /8

 The slope and deflection equations shall be :

EI (dy/ dx)        =-   w l ³/6 +    w l ³/6

EIy =- + w/24 + wl³x/6 - wl⁴/8

Slope at (x = 0),

θ _A   = + wl³/6 EI

Slope at C (x = l/2) ,

EI θC = -  wl³ /48  +   wl³/ 6   = +7 wl³/48

∴          θ _C     = 7 wl ³ /48 EI

 Deflection at A (x = 0),

y_A  =- wl ⁴/ 8 EI

Deflection at C ( x = l/2) ,

EIy_C  =-(w/ 24 )(l/2) ⁴  + wl³/6 (l/2) -wl⁴/8

= wl ⁴/ (24´16 ) (- 1 + 32 - 48) = -17 wl4/ 384

∴ y_C=- 17 wl ⁴/384 EI