The following worksheet calculates the end rotation and deflection of a curved beam resulting from  the application of a force, F, at the tip of the beam.
The approach is to apply Castigliano's 2nd Theorum, decomposing the applied force along orthogonal axes directions and applying a dummy moment at the tip of the beam to solve for tip deflections and rotations, respectively.

William Provancher
December 8, 2004

Forces := {Fx  ->  F * Sin[ϕ],  
        Fy  ->  F * Cos[ϕ]}
geometry1 := {β ->  π/4, ϕ ->  π/
            36}
geometry2 := {xF ->  R * Cos[β],  
        yF ->  R * Sin[β] }
geometry3 := {Emod -> 200 * (10^9),  
        Izz ->  ((π * (d^4))/64)}
geometry4 := { d ->  0.0002,  
        R ->  0.005 }
moment := {Mb ->  
          Md  +  Fy * (xF  -  R * Cos[θ])  +  
            Fx * (yF  -  R * Sin[θ]) }

Formulating the energy stored in the beam.

U  =  (1/(2*Izz*
                Emod)) (∫_(-(π/2))^β)((Mb^2)  /. moment  /. geometry2 ) R \[DifferentialD]θ // 
        Simplify

The beam tip rotation, α, is

α = ((∂_Md) U) /. Md -> 0 // 
        Simplify

In specifying beam geometry {β-> π/4, ϕ-> π/36}

α  /. Forces /. geometry1  // N  // 
      Simplify

Assuming the geometry and material properties to be {Emod -> 200 * (10^9),  
        Izz ->  ((π * (d^4))/64)}  { d-> 0.0002, R-> 0.005 }   results in the following rotation (in Radians) (Force specified in Newtons)

α  /. Forces /. geometry1   /. geometry3  /. 
          geometry4 // N  // Simplify

---------------------------------------------------------------------------------------------------------------------------------------

The beam tip deflection in the x-direction, δx, is

δx  =  ((∂_Fx) U) /. Md -> 0 // 
        Simplify

In specifying beam geometry {β-> π/4, ϕ-> π/36}

δx /. Forces /. geometry1   // N  // 
      Simplify

Assuming the geometry and material properties to be {Emod -> 200 * (10^9),  
        Izz ->  ((π * (d^4))/64)}  { d-> 0.0002, R-> 0.005 }   results in the following rotation (in Radians) (Force specified in Newtons)

δx /. Forces /. geometry1  /. geometry3  /. 
        geometry4 //N

-----------------------------------------------------------------------------------------------------------------------------------

The beam tip deflection in the y-direction, δy, is

δy  =  ((∂_Fy) U) /. Md -> 0

δy  /. Forces /. geometry1   // N  // 
      Simplify

Assuming the geometry and material properties to be {Emod -> 200 * (10^9),  
        Izz ->  ((π * (d^4))/64)}  { d-> 0.0002, R-> 0.005 }   results in the following rotation (in Radians) (Force specified in Newtons)

δy  /. Forces /. geometry1  /. geometry3  /. 
        geometry4 //N

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To check this versus what's shown in Shigley & Mischke's Mechanical Engineering Design (5th Ed) p. 119, make β = π/2 and look at   δy

geometry1 := {β ->  π/2}

U  =  (1/(2*Izz*
                    Emod)) (∫_(-(π/2))^β)((Mb^2)  /. moment  /. 
                        geometry2   /. geometry1  ) 
                  R \[DifferentialD]θ  /. Md -> 0   /. 
          Fx ->  0 // Simplify

δy  =  ((∂_Fy) U)  /. Fx ->  0 /. 
            geometry1 // Simplify  // N

Which matches the formula given by Shigley & Mischke