Cantilever Beams along a Central Point Load:

Consider a cantilever beam loaded with a point load 'W' at the centre 'C'. Consider a section X-X at a distance x from left end.

Moment, M = - W [x - l /2]                (Hogging BM)

The governing equation for deflection is  following:

EI d ² y / dx²  = M = - W [x - l/2 ]

Figure

Integrating the Eq. (105), we can get

EI dy/dx= -W/2[ x-l/2]² + C₁           

EIy =-  W/6 [ x- l/2]³ + C₁ x+ C₂

 The constants C₁ and C₂ can be found from the 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. (106), we may get

0 = -(W/6)(l - (l/2)³ + Wl²/8 ´l + C₂

∴          C₂  =+ 5Wl³/48

Applying BC (2) into the Eq. (107), we may get

0 =- (w/6) (l-(l/2) ³+ Wl²/8 ´ C₂

∴          C₂  = - 5 W l³ /48

The slope & deflection equations shall be :

 EI( dy/dx )=- (W/2)[x-(l/2)]² + Wl²/8

EIy = (w/6)[x-(l/2)]³+Wl²/8 - 5Wl³/48

Slope at (x = 0),

θ_A = + W l ² / 8 EI

Slope at C (x = l /2) ,

 θc = + W l ² / 8 EI

Deflection at A (x = 0),

y _A  =-5 W l³ / 48 EI

Deflection at C (x = l /2),

EIy       =+ (W l ²/8).(l/2) - 5Wl³/48 = - Wl³/24

∴          y _C        =- W l ₂/24 EI

Cantilever Beam along a Central Point Load