SS Beams with an End Couple:

∑ F_y  = 0 so that R_A  + R_B  = 0

Taking moments around A,

R_B  × l = M ₀

∴          R_B  = +   M₀ / l  (↑)

Figure: SS Beam with End Couple

From Equation. (66) & (67),

R _A   =-   M ₀ / l (↓)

Let a section X-X at a distance from x from A,

 M = R _A  . x = - (M ₀ / l ) x

The governing equation for deflection is :

EI (d  y/dx) = (M₀/2l )x²  +C₁

Integrating the Eq. (69), we will get

EI dy /dx=- (M 0/2l ) x²  + C₁

EIy =- M ₀ /6l x³   + C₁ x + C₂

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

at A, x = 0, y = 0               --------- (1)

at B, x = 0,  y = 0               ----------- (2)

By applying the boundary condition (1) into the Eq. (71), C₂ = 0.

By applying the BC (2) into the Equation (71),

0 =- M 0 l² /6  + C _l  l                ∴ C₁  = M ₀ l/6

The slope & deflection equation shall be :

EI dy/ dx =- M ₀ x² /2 l + M ₀ l /6

EIy =- M ₀ /6 l x³ + M₀lx/6

Slope at A, (x = 0)

θ _A = + M ₀ l/6EI

Slope at B, (x = l)

EI θ _B  = - M ₀ l/2 + M ₀ l /6= - M ₀ l/3

∴   θ_B = - M ₀ l / 3 EI

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

EI θ _C  = - M ₀  ( l/2) ²+ M ₀ l /6 = - M ₀ l/8 + M ₀ l/6

∴          θ_C = + M ₀ l /24 EI

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

EIy =- M ₀ /6l(  l /2)³ + (M ₀ l/6) ( l /2)

=-   M₀  l ² /48  +  M₀  l ²/12 = M ₀ l ² / 16

∴          y_C   =+ M₀  l ²/16 EI

Figure