Page 196 - Shigley's Mechanical Engineering Design

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Deflection and Stiffness 171
Combining the four terms gives the total strain energy

M dθ F Rdθ MF θ dθ CF Rdθ
2 2 2
r
θ
U = + − + (4–37)
2AeE 2AE AE 2AG
The deflection produced by the force F can now be found. It is
∂U M ∂M F θ R ∂F θ
δ = = dθ + dθ
∂F AeE ∂F AE ∂F
1 ∂(MF θ ) CF r R ∂F r

− dθ + dθ (4–38)
AE ∂F AG ∂F
This equation is general and may be applied to any section of a thick-walled circular
curved beam with application of appropriate limits of integration.
For the specific curved beam in Fig. 4–12b, the integrals are evaluated from 0 to π.
Also, for this case we find
∂M
M = FR sin θ = R sin θ
∂F
∂F θ
F θ = F sin θ = sin θ
∂F
∂(MF θ )
2 2 2
MF θ = F R sin θ = 2FR sin θ
∂F
∂F r
F r = F cos θ = cos θ
∂F
Substituting these into Eq. (4–38) and factoring yields

FR 2 π 2 FR π 2 2FR π 2
δ = sin θ dθ + sin θ dθ − sin θ dθ
AeE 0 AE 0 AE 0
CF R π 2
+ cos θ dθ
AG 0
π FR 2 π FR π FR πCF R π FR 2 π FR πCF R (4–39)
= + − + = − +
2AeE 2AE AE 2AG 2AeE 2AE 2AG
Because the first term contains the square of the radius, the second two terms will be
small if the frame has a large radius.
For curved sections in which the radius is significantly larger than the thickness, say
R/h > 10, the effect of the eccentricity is negligible, so that the strain energies can be
approximated directly from Eqs. (4–17), (4–23), and (4–25) with a substitution of Rdθ
for dx. Further, as R increases, the contributions to deflection from the normal force
and tangential force becomes negligibly small compared to the bending component.
Therefore, an approximate result can be obtained for a thin circular curved member as

2
. M
U = Rdθ R/h > 10 (4–40)
2EI

∂U . 1 ∂M
δ = = M Rdθ R/h > 10 (4–41)
∂F EI ∂F