Cantilever Beams along a Triangular Load:

Let a cantilever beam loaded with a triangular load as illustrated in Figure

Figure

Consider a section X-X at a distance x from A as illustrated in Figure

Intensity of loading = w/l . x

Moment,  M =-(( ½) × (w/l) x  × x)(x/3) = - w x3/6l

The governing equation for deflection is following :

EI d ² y/ dx²      = M = - w x³/6 l

Integrating the eq, we may get

EI (dy/ dx)     = - w x⁴/24 l + C₁

EIy=- w x⁵/120 l + C₁ x + C₂

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

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

At B, x = l, y = 0                                ------------- (2)

Applying the boundary condition (1) into the Eq. (150), we may get

0 =       - wl ³ / 24        + C₁

 ∴         C₁ =+ wl ³/  24

Applying the boundary condition (2) into the Eq. (151), we may get

0 =- wl ⁴ /120  + wl ⁴ /24+ C2

∴          C₂  =+  wl³ /24

 The slope and deflection equations shall be :

EI (dy/ dx) =- w x⁴/24 l + w l ³/24

EIy =-  w x⁵/120 l  + (w l ³ /24 )x - w l ³ / 30

Slope at A, (x = 0),

θ_A = + wl³ /24 EI

Slope at C (x = l /2),

EI θ_C = - (w / 24 l )(l/2) ⁴ +  wl ³ /24

= (wl³/24)[-(1/16)+1] =  + 15 w l ³ / (24 × 16)

∴          θ _C = +  15 wl³   /384 EI

Deflection at A, (x = 0),

y _A  =- wl ₄ /30 EI

Deflection at C, ( x = l/2) ,

EI yC  =- wl ⁴/120    + wl ⁴/48 - wl ⁴/  30

=- wl ⁴/480 [4 + 10 - 16] =- wl ⁴/240

∴ y_C = wl ⁴ /240 EI