Stresses in Curved Beams Made of Heterogeneous Materials

The main objective of the present paper is a generalization of some classical results for curved beams made of heterogeneous materials. We consider a beam made of nonhomogeneous, isotropic, linearly elastic material. The elastic parameters depend on the cross-sectional coordinates only. Our investigations include the determination of the normal stress (i.e., the generalization of the Grashof formula), the shearing stress and the curvature change. Interestingly, our newly established formulae have the same structure as the classical ones. We conclude with numerical examples which illustrate the applicability of our results. Stresses in Curved Beams Made of Heterogeneous Materials Publication History: Received: March 30, 2015 Accepted: October 31, 2015 Published: November 05, 2015


Introduction
The majority of the textbooks on strength of materials deal with curved beams made of homogeneous material.Formulae are presented for the calculation of the normal stress, the shearing stress, the change of curvature and the strain energy stored in the curved beam.In this paper our main objective is to generalize these classical results for a curved beam made of heterogeneous isotropic materials By assumption the elastic parameters, i.e., the Young modulus and the Poisson number, depend on the cross sectional coordinates, but are independent of the axial coordinate.Figure 1 shows a part of the beam and the applied orthogonal curvilinear coordinates.Our most important assumptions are as follows: inertia with respect to the axis These concepts have been introduced for a straight beam in paper [1] by Baksa and Ecsedi.
The unit tangent vectors e ξ (s), e n and e ζ (s) of the coordinate lines ξ, n, and ζ are also shown in Figure 1.Let R be the radius of the E-weighted centerline.It is easy to check that e ξ (s), and e ζ (s) satisfy the following relations It is also obvious that the operator ∆ takes the form: We shall assume that (a) the cross section has a translation and a rigid body rotation about the axis , i.e., it remains a plane surface during the deformation, and (b) the deformed centerline remains perpendicular to the cross section.Under these conditions (2) is the displacement field on the cross section, in which u o = u o e ξ + w o e ζ and are the displacement vector and the rotation on the E-weighted centerline, respectively.As is well known is the rigid body body rotation.Consequently from where where at least one factor in the dyadic products denoted, for brevity, by ((...)) is perpendicular to e ξ , we have are the axial strain and the curvature on the E-weighted centerline.

Formulae for the normal stress
Generalization of the Grashof formula.The E-weighted reduced area, first moment and moment of inertia are defined by the following relations: It is clear that the axial force and the bending moment are Here, due to the inequality ,equation is the Hooke law.Upon substitution of the Hook law and then equation (10) into (12) we have As for the bending moment, in a similar way we obtain After solving equation system (13) we have Є 0ξ and κ 0 in terms of the axial force and the bending moment: Let us now substitute these solutions into equation ( 9) so that we can get the axial strain: With the knowledge of the axial strain is the normal stress.In what follows, we attempt to simplify the above formula for the normal stress.First we shall clarify how A eR , Q eR and I eR are related to A e , Q en and I en .Using the power series of fraction R/ (R + ζ) we have because Q en = 0.For the denominator in equation ( 16) the following approximation holds Using this result we can rewrite formula (15): .
, , . ) (10) If we use approximations (17) we can check the following equations Substituting the last two formulae into (20) and the result into the Hooke law we have This formula is the generalization of the Grashof formula for the case of cross sectional inhomogeneity.
A formula for the normal stress assuming pure bending.English textbooks contain a formula for the normal stress under the assumption of pure bending -see for instance equation (4.71) p. 224 in [2].In this subsection it is our aim to generalize the cited equation for heterogenous curved beams.Figure 2 shows the cross section and the geometrical meaning of some notational conventions: ζ o is the coordinate of the neutral axis with radius R, and the radius of an arbitrary point P of the cross section ζ with coordinate is r = R + ζ.Based on equation ( 16) for pure bending, is the formula for the normal stress.We would like to manipulate it into a form similar to the one cited above.
First we shall determine the location of the neutral axis.It follows from condition σ 3 = 0 which needs to be satisfied on the neutral axis that from where is the radius that belongs to the neutral axis.Upon substitution of A eR and Q eR from (11), the radius R of the neutral axis assumes the form For E(η, ζ ) =constant, the above equation coincides with formula (4.66) p. 222 in [2].Now we shall proceed with the determination of the normal stress.Using equation ( 26), the factor in parentheses in equation ( 23 This equation is the generalization of formula (4.71) p. 224 in [2] valid for the homogenous case for curved beams with cross sectional inhomogeneity.

A formula for the shear stress
e eR eR eR eR eR , Our goal in this subsection is to derive a closed form solution for calculating the shear stress.We shall use equilibrium equations for this purpose.This approach results in a relatively simple formula; however, it has the disadvantage that the kinematic equations are not satisfied.The basic idea is well known from the theory of straight beams: we divide a short portion of the beam into two parts and then analyze the equilibrium conditions of one part.
Consider Figure 3 which shows a finite portion of the curved beam with cross sectional inhomogeneity.The left cross section with arc coordinate s B is fixed, coordinate s > s B of the right cross section is regarded as a parameter.We shall use the following assumptions:

3.
The bending moment M and the shear force V are related to each other via equation which is an equilibrium condition.
4. The normal stress σ ξ can be calculated by equation ( 22) for which we assume N = 0 -there is no axial force in the cross section considered.For calculating the shear stresses τ ηξ let us consider the part of the beam with outlines drawn thick in Figure 3 , ( of homogenous material -see pp.358-359 in [3,2002].Now we proceed to determine the strain energy stored in the beam.It is not too difficult to check using equation (45) that the angle change dψ due to the bending moment is Consequently is the work done by the bending moment exerted on an infinitesimal portion of the beam.Let L be the length of the E-weighted centerline.

Numerical examples
Example 1. Figure 5 shows the cross-section of curved beam considered.We assume that the beam is subjected to pure bending M = M e η ; M = 100 Nm.The geometric dimensions are all given in Figure 5.The lower part of the beam is made of steel and the upper part of the beam is made of aluminium.The corresponding material parameters are E 1 = 2:1×10 5 MPa and E 2 = 7 ×10 4 MPa, in that order.Our aim is to depict graphically the normal stress distribution as a function of ζ using the three formulae derived in the previous sections.This allows us to compare the various results.It would also be interesting to check the difference between these formulae regarding the radius of the neutral axis.
First we determine the ordinate z C of the E-weighted centerline in coordinate system γz.Since the E-weighted first moment of the cross section with respect to the axis η should be zero, the following equation holds:

Consequently
With the knowledge of z C one can read off from Figure 5 (51 , , Regarding the E-weighted reduced first moment of the crosssection, equation (11) 2 yields Using the parallel axis theorem, we can determine I eη as Therefore, recalling (11) 3 and utilizing equation (54) we can establish a formula for the E-weighted reduced moment of inertia:

and ζ 2 into equations (52)-(55) we obtain the following numerical values:
With these results we can compute the normal stress 6 using Eqs.(16) -this formula has no simplifications, it is exact under the displacement and stress hypotheses -, (22) -this is a generalization of the Grashof formula -and (31) -this a generalization of the formula that can be found in English textbooks on Strengths of Materials.The computational results are presented in Table 1 and graphically in Figure 6.It is clear that the results obtained by the use of the Grashof formula and by Eq. (31) almost coincide.
As regards Figure 6 the graphs representing the exact solution, the solution obtained by the use of Eq. ( 22) and the solution calculated with Eq. (31) are drawn in blue, red and green.Observe that the red and green curves coincide almost completely.
As for the ordinate of the neutral axis, by setting ,Eqs.( 16), ( 22) and (31) yield -0.8004mm, -0.7771mm and -0.7845mm, respectively.We note that the latter result is exactly the same as the value that can be obtained by using Eq.(26).Example 2. Figure 7 shows a cross-section of the curved beam considered -observe that the beam is the same as in the previous example.Let us assume that the cross-section is subjected to a shear force V=10kN.The shear stress can be calculated using Eqs.(38) and (39).Upon substitution of and (56) 2,5 into (38a) 1   (55) 3 2  We should remark that formula (39) for the shear stress provides an average value -it has been established by using equilibrium conditions.Consequently, the values obtained using this equation are in all probability more accurate if the modulus of elasticity is independent of η, i.e., if it depends on ζ only.

Conclusion
In this paper we have generalized some classical results valid for curved beams made of homogeneous and isotropic material for the case when the curved beam is made of heterogeneous and isotropic material, under the assumption that the elastic parameters, i.e., the Young modulus and the Poisson number, depend on the crosssectional coordinates, but are independent of the axial coordinate.We remark that the Poisson number has played no role in the investigations.
By applying the well known displacement hypotheses we have established (1) three formulae for calculating the normal stress, (2) a formula for the shear stress, and (3) formulae for the change of curvature and the strain energy stored in the beam.
Except the very first formula for the normal stress --see Eq. ( 16) --all the others are generalizations of classical results and each can be applied in paper-and-pencil calculations.
Two numerical examples are presented to illustrate the applicability of the formulae.

Figure 1 :
Figure 1: The unit tangent vectors e ξ (s), e n and e ζ (s) of the coordinate lines ξ, n, and ζ.
) can be rewritten into the form If we use the inequality A eR I eR >> S eR and substitute back the above difference into equation (23), we obtain The last open question is how to transform the fraction R/I eR into a suitable form.The following equation details the transformation step by step: It is worth introducing the notation e = -ζ o .Upon substitution of the result obtained into formula (29) for the normal stress we arrive at its final form:

Figure 2 :
Figure 2: The cross section and the geometrical meaning of some notational conventions.

Page 4 of 8 Figure 3 :
Figure 3: A finite portion of the curved beam with cross sectional inhomogeneity (32) In the present subsection [the radius of curvature] {a point} on the E-weighted centerline before and after deformation are denoted by [R o and R] {P o and P}.The angle formed by the tangent to the E-weighted centerline at Po with the horizontal axis is ψ o .Its change during deformation is ψ oη -the rigid body rotation.The calculation of the curvature change is based on Figure 4 which shows all the quantities mentioned.The infinitesimal arc element ds o on the E-weighted centerline before deformation changes to ds.It is clear that is the axial strain on the E-weighted centerline.Consequently Using the above equation we can manipulate the formula for the curvature change: Here A comparison of equations (10), (42) and (43) yields Substituting ĸ o from (14) and taking into account that in the present case N = 0 and Q eR <<A eR I eR we have that is

Figure 4 :
Figure 4: The calculation of the curvature change is based on which shows all the quantities mentioned.

Figure 5 :
Figure 5: The cross-section of the curved beam considered

Page 7 of 8 Figure 6 :
Figure 6: Graphical representation of Grashof formula computational results.Figure7: Cross-section of the curved beam.

Figure 7 :
Figure 6: Graphical representation of Grashof formula computational results.Figure7: Cross-section of the curved beam.
Kiss L, Szeidl G (2015) Stresses in Curved Beams Made of Heterogeneous Materials.Int J Mech Syst Eng 1: 107.doi: http://dx.doi.org/10.15344/2455-7412/2015/107 Citation: that Before computing the stresses sought we shall set up appropriate formulae for the E-weighted geometrical quantities A eR , Q eR , I eη and I eR .Recalling equation (11)1 we can write we have