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Bending of Curved Members

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Details

This lesson considers the stresses caused by the application of equal and opposite couples to members that are initially curved. The lesson will be limited to curved members of uniform cross section possessing a plane of symmetry in which the bending couples are applied, and it will be assumed that all stresses remain below the proportional limit.

If the initial curvature of the member is small, i.e., if its radius of curvature is large compared to the depth of its cross section, a good approximation can be obtained for the distribution of stresses by assuming the member to be straight and using the formulas derived in the lessons: Deformations in a Symmetric Member in Pure Bending and Stresses and Deformations in the Elastic Range. However, when the radius of curvature and the dimensions of the cross section of the member are of the same order of magnitude, a different method of analysis must be used.

Consider the curved member of uniform cross section shown in the following figure:



Its transverse section is symmetric with respect to the y axis (the middle figure) and, in its unstressed state, its upper and lower surfaces intersect the vertical xy plane along arcs of circle AB and FG centered at C.

Equal and opposite couples M and M' are now applied in the plane of symmmetry of the member (as shown in the right figure). A reasoning similar to that of the lesson Deformations in a Symmetric Member in Pure Bending would show that any transverse plane section containing C will remain plane, and that the various arcs of circle indicated in the above figure at left, will be transformed into circular and concentric arcs with a center C' different from C. More specifically, if the couples M and M' are directed as shown, the curvature of the various arcs of circle will increase; that is A'C' < AC. It is also noted that the couples M and M' will cause the length of the upper surface of the member to decrease (A'B' < AB) and the length of the lower surface to increase (F'G' > FG). It is concluded that a neutral surface must exist in the member, the length of which remains constant. The intersection of the neutral surface with the xy plane has been represented in the above figure at left by the arc DE of radius R, and in the above figure at right by the arc D'E' of radius R'. Denoting by θ and θ' the central angles corresponding respectively to DE and D'E', the fact that the length of the neutral surface remains constant can be expressed by:

(Eq1)

Rθ = R'θ'

Considering now the arc of circle JK located at a distance y above the neutral surface, and denoting respectively by r and r' the radius of this arc before and after the bending couples have been applied, the deformation of JK can be expressed as:

(Eq2)

δ = r'θ' − rθ

Observing from the above figure that:

(Eq3)

r = R − y

and:

(Eq4)

r' = R' − y

Substituting Eq3 and Eq4 into Eq2:

δ = (R' − y)θ' − (R − y)θ

or, recalling Eq1 and setting θ' − θ = Δθ:

(Eq5)

δ = −yΔθ

The normal strain ε_x in the elements of JK is obtained by dividing the deformation δ by the original length rθ of arc JK. It is written:

(Eq6)

εx = δ rθ = −  yΔθ rθ

or, recalling Eq3:

(Eq7)

εx =  − Δθ θ y R − y

The relation obtained shows that, while each transverse section remains plane, the normal strain ε_x does not vary linearly with the distance y from the neutral surface.

The normal stress σ_x can now be obtained from Hooke's law Eq2, σ_x = Eε_x, by substituting for ε_x from Eq7. The following is then:

(Eq8)

σx =  − EΔθ θ y R − y

or, alternatively, recalling Eq3:

(Eq9)

σx =  − EΔθ θ R − r r

Eq8 shows that, like ε_x, the normal stress σ_x does not vary linearly with the distance y from the neutral surface. Plotting σ_x versus y, an arc of hyperbola is obtained as shown:



In order to determine the location of the neutral surface in the member and the value of the coefficient E Δθ/θ used in Eq7 and Eq8, it is recalled that the elementary forces acting on any transverse section must be statically equivalent to the bending couple M. Expressing, as was done in the lesson Symmetric Member in Pure Bending for a straight member, that the sum of the elementary forces acting on the section must be zero, and that the sum of their moments about the transverse z-axis must be equal to the bending moment M, the following equations are written:

(Eq10)

∫ σxdA = 0

and:

(Eq11)

∫ (−yσxdA) = M

Substituting for σ_x from Eq