2.1 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:

A eR   =  ∫ A R R + ζ  E( η,ζ )dA,  Q eR  =   ∫ A R R + ζ E( η,ζ )ζdA I eR  =  ∫ A R R + ζ E( η,ζ ) ζ 2 dA . (11)

It is clear that the axial force and the bending moment are

N =  ∫ A σ ξ dA , M =   ∫ A ζ σ ξ dA . (12)

Here, due to the inequality σ ξ >> σ η , σ ξ , σ ξ  = E (η,ζ)  ε ζ ,equation is the Hooke law. Upon substitution of the Hook law and then equation (10) into (12) we have

N =  ε οξ ∫ A R R + ζ E( η,ζ )dA +  κ ο   ∫ A R R + ζ E( η,ζ ) ζdA =  ε οξ A eR  +  κ ο Q eR (13a)

As for the bending moment, in a similar way we obtain

M =  ε οξ   ∫ A R R + ζ E 2 ( η,ζ )ζdA +  κ ο   ∫ A R R + ζ E( η,ζ )  ζ 2 dA =  ε οξ Q eR  +  κ ο I eR  . (13b)

After solving equation system (13) we have Є_0ξ and κ₀ in terms of the axial force and the bending moment:

ε οξ  =  1 Q 2 eR  −  A eR I eR ( M Q eR − N I eR ),  κ ο =  1 Q 2 eR −  A eR I eR ( N Q eR −M A eR ) (14)

Let us now substitute these solutions into equation (9) so that we can get the axial strain:

ε ξ  =  R R + ζ   1 A eR I eR −  Q 2 eR [ ( I eR −ζ Q eR ) N−M ( Q eR − ζ A eR ) ] . (15)

With the knowledge of the axial strain

σ ξ = E( η,ζ ) ε ξ = E( η,ζ ) R R + ζ   1 A eR I eR −  Q 2 eR [ ( I eR  − ζ Q eR )N−M( Q eR − ζ A eR ) ] (16)

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

A eR =  ∫ A ( 1− ζ R + ζ 2 R 2 − ... )  E( η,ζ )dA ≅  ∫ A E( η,ζ ) dA ︸ A e − 1 r ∫ A E( η,ζ ) ζdA ︸ Q eη =

=  A e − Q eη R =  A e , (17a)

Q eR =  ∫ A ( 1−  ζ R + ζ 2 R 2 −... )  ζE( η,ζ )dA≅   ≅  ∫ A ζE( η,ζ ) dA ︸ Q eη −  1 R ∫ A ζ 2 E( η,ζ ) dA ︸ I eη = Q eη − I eη R =− I eR R (17b)

and

I eR =  ∫ A ( 1− ζ R + ζ 2 R 2 −... )   ζ 2 E( η,ζ )dA ≅  I eη (17c)

because Q_en = 0. For the denominator in equation (16) the following approximation holds

A eR I eR ( 1− Q eR 2 I eR A eR ) ≃  A eR I eR (1− 1 R 2 I 2 eη A eR I eR ) ≃  A eR I eR ( 1− 1 R 2 I eη 2 A eR )≃ A eR I eR (18)

since

1>> 1 R 2 I eη A eR . (19)

Using this result we can rewrite formula (15):

ε ξ ≈ R R+ζ 1 A eR I eR [ ( I eR −ζ Q eR )N−M( Q eR −ζ A eR ) ]=  =( 1−ζ Q eR I eR )  R R+ζ N A eR +( − Q eR A eR I eR + ζ I eR ) R R+ζ M. (20)

If we use approximations (17) we can check the following equations

Q eR I eR ≃− 1 R I eη I eR ≃− 1 R ,    1 R R R+ζ M A eR ≃ M R A eR . (21)

Substituting the last two formulae into (20) and the result into the Hooke law we have

σ ξ =E( η,ζ )( N A eR + M R A eR + M I eR R R+ζ ζ ). (22)

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,

σ ξ =E( η,ζ ) R R+ζ 1 A eR I eR − Q 2 eR ( ζ A eR − Q eR )M (23)

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 σ₃ = 0 which needs to be satisfied on the neutral axis that

Q eR = ζ ο A eR . (24)

Consequently

Q eR =( R+ ζ ο −R ) A eR =( R ¯ −R ) A eR = R ¯ A eR −R A eR  , (25)

from where

R ¯ = Q eR A eR +R (26)

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

R ¯ =R+ ∫ A E( η,ζ )ζ R r dA ∫ A R r E( η,ζ )dA = ∫ A E( η,ζ )ζ R r dA+ R 2 ∫ A E( η,ζ ) r dA R ∫ A E( η,ζ ) r dA =

            = 1 R ∫ A E( η,ζ )ζ R r dA+R ∫ A E( η,ζ ) r dA ∫ A E( η,ζ ) r dA = ∫ A E( η,ζ )dA ∫ A E( η,ζ ) r dA

or equivalently

R ¯ = ∫ A E( η,ζ )dA ∫ A E( η,ζ ) r dA (27)

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) can be rewritten into the form

ζ A eR − Q eR =( r−R ) A eR +R A eR − R ¯ A eR =(r− R ¯ ) A eR . (28)

If we use the inequality A eR I eR    >>   S eR 2 and substitute back the above difference into equation (23), we obtain

σ ξ =E( η,ζ ) 1 r M R I eR ( r− R ¯ ) (29)

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:

I eR = ∫ A E( η,ζ ) R r ζ 2 dA= ∫ A E( η,ζ ) R R+ζ−R r ζdA=

   = ∫ A E( η,ζ ) R r−R r ζdA=R ∫ A E( η,ζ ) ζdA− ∫ A E( η,ζ ) R 2 1 r ( r−R )dA=

R Q eη ︸ 0 − ∫ A E( η,ζ ) R 2 1 r ( r−R )dA= − R 2 ∫ A E ( η,ζ )dA+ R 3 ∫ A E( η,ζ ) r dA=

=− R 2 A e + R 3 ∫ A E( η,ζ ) r dA= A e ( R 3 R ¯ − R 2 )=

= A e R 2 ( R R ¯ −1 )= A e R 2 ( R− R ¯ R ) ︷ − ζ ο =− A e R R ¯ 2 ζ ο . (30)

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:

σ ξ =E( η,ζ ) M r r− R ¯ A e e . (31)

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.

2.2 A formula for the shear stress

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:

1. The shear stresses τ_ξ = τ_ηξe_η + τ_ζξe_ζ on the line ζ = ζ = constant intersect each other in one point which coincides with the intersection point of the tangents to the contour of the cross section at ζ = ζ = constant. Consequently, τ_ηξ = -τ_ηξ(-η), which means that τ_ηξ(η) is an odd function of η.
2. The shear stress τ_ηξ is constant if ζ = constant. The bending moment M and the shear force V are related to each other via equation

   
   dM ds =−V (32)

   which is an equilibrium condition.
3. 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. It is bounded by the marked endfaces A B ' ,  A ' , the cylinder with radius ρ O  +  ζ ^ is a part of the lateral surface. By assumption the lateral surface is unloaded.

The equilibrium equation for the part of the beam considered is of the form

∫ A ' ( σ ξ e ξ + τ ξ ) dA− ∫ A B ' ( σ ξ e ξ + τ ξ ) dA−                            − ∫ s B s R+ ζ ^ R υ ( ζ ^ ) τ ξζ ( ζ ^ ) e ξ ( ξ )dξ=0. (33)

If we take into account that the shear stress - τ ξ ζ ( ζ ^ ) e ξ (s) is constant on the cylindrical surface with radius R +  ζ ^ , and the fact that

R+ ζ ^ R υ( ζ ^ )dξ=dA

is the surface element, then it follows that the last integral in (33) is the resultant of the shear stresses.

Let us differentiate equation (33) with respect to s. Then (a) substitute the derivatives of the unit vectors e_ξ and e_ζ with respect to s; (b) take into account that (1) the integral over A B ' is constant therefore its derivative is zero; (2) τ_ηξ is an odd function of η therefore its integral is zero; (3) the derivative of an integral with respect to the upper limit is the integrand. Then

∫ A ' d σ ξ ds e ξ dA− ∫ A ' σ ξ R e ζ dA+ d ds ∫ A ' τ ηξ e η dA+ ︸ =0             + ∫ A ' ( d τ ζξ R e ζ + τ ζξ R e ξ ) dA− R+ ζ ^ R υ( ζ ^ ) τ ξζ ( ζ ^ ) e ζ ( s )=0.

If we now dot multiply throughout by e_ξ we obtain

∫ A ' d σ ξ ds dA+ ∫ A τ ζξ R dA− R+ ζ ^ R υ( ζ ^ ) τ ξζ ( ζ ^ )=0. (34)

Let e_max be the distance of the top of the cross section from the point Ce. This is always less than R. The area A' can be given as a product υ( ζ ^ )h( ζ ^ ) where h( ζ ^ ) is less than e_max. Consequently

∫ A ' τ ζξ ( η,ζ ) R dA≃ 1 R h( ζ ^ )υ( ζ ^ ) τ ξζ ( ζ ^ ),    h( ζ ^ ) R <<1

is an upper limit of the second integral in (34). Really if we take into account that the shear stress is taken on the line ζ ^ (instead of being taken at inner points of A') we can readily check the validity of the previous statement. On the basis of this estimation the second term in (34) can be neglected if we compare it to the third term. Omitting this term results in the following equation

∫ A ' d σ ξ ds dA= R+ ζ ^ R υ( ζ ^ ) τ ξζ ( ζ ^ )  (35)

for the calculation of the shear stress τ ξζ ( ζ ^ ) .After rearranging we obtain the following average value

τ ξζ ( ζ ^ )= R R+ ζ ^ 1 υ( ζ ^ ) ∫ A ' d σ ξ ds dA . (36)

Upon substitution of the normal stress from (22) – N = 0 by assumption – we have

τ ξζ ( ζ ^ )= R R+ ζ ^ 1 υ( ζ ^ ) ∫ A ' d ds [ E( η,ζ )( M R A eR + M I eR R R+ζ ) ] dA. (37)

A further transformation yields

τ ξζ ( ζ ^ )= R R+ ζ ^ 1 υ( ζ ^ ) dM ds ∫ A ' ( E( η,ζ ) R A eR + E I eR R R+ζ ζ ) dA=

= R R+ ζ ^ 1 υ( ζ ^ ) dM ds 1 I eR ( R I eR R 2 A eR ∫ A ' E( η,ζ )dA+ ∫ A ' R R+ζ ζE( η,ζ )dA ).

Introducing the notations

α e = I eR R 2 A eR ;   Q eη ' = ∫ A ' E( η,ζ ) R R+ζ ζdA (38a)

and

∫ A ' E( η,ζ ) dA= A e ' (38b)

and substituting (32) we obtain

τ ξζ ( ζ ^ )=− R R+ ζ ^ V I eR υ( ζ ^ ) ( R α e A e ' + Q eη ' ) (39)

which is a formula for the average value of the shear stress. This result is a generalization of a classical formula valid for curved beams made of homogenous material – see pp. 358-359 in [3, 2002].

2.3 Curvature change, strain energy

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 P_o 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

ε οξ = ds−d s ο d s ο (40)

is the axial strain on the E-weighted centerline. Consequently

d s ο = ds 1+ ε οξ . (41)

Using the above equation we can manipulate the formula for the curvature change:

1 R − 1 R ο = d( ψ ο + ψ οη ) ds − d ψ ο d s ο = d( ψ ο + ψ οη ) ds − d ψ ο ds ( 1+ ε οξ )= d ψ οη ds − ε οξ d ψ ο ds . (42)

Here

ε οξ d ψ ο ds = ε οξ d ψ ο d s ο ( 1+ ε ο ξ )= ε οξ d ψ ο d s ο = ε οξ 1 R ο . (43)

A comparison of equations (10), (42) and (43) yields

1 R − 1 R ο = κ ο − ε οξ 1 R . (44)

Substituting ĸ_o from (14) and taking into account that in the present case _N = 0 and Q 2 eR << A eR I eR we have

1 R − 1 R ο =− M Q 2 eR − A eR I eR ( A eR + Q eR R ︸ ≈0 )≃ M A eR A eR I eR − Q 2 eR ≃ M A eR A eR I eR = M I eR

that is

1 R − 1 R ο = M I eR . (45)

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

dψ= ds R − d s ο R ο ≃ ds R − ds R ο = M I eR ds (46)

Consequently

dU= 1 2 Mdψ= 1 2 M 2 I eR ds (47)

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. After integration

U= 1 2 ∫ L M 2 I eR ds (48)

2.4 Numerical examples

Example 1. Figure 5 shows the cross-section of the 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 E₁ = 2:1×10⁵ MPa and E₂ = 7 ×10⁴ 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:

Q eη = Q ey − z C A e =0. (49)

Consequently

z C = Q ey A e = E 1 b 1 2 A 1 + E 2 ( b 1 + b 1 2 ) A 2 E 1 A 1 + E 2 A 2 =12mm. (50)

With the knowledge of z_C one can read off from Figure 5 that

ζ 1 − =− z C =−12mm,  ζ k =4mm,  ζ 2 + =20mm. (51)

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

A eR = ∫ L R+ζ−ζ R+ζ E( η,ζ )adζ= ∫ A E( η,ζ ) dA− E 1 a ︸ A e ∫ ζ − 1 ζ + 1 ζ R+ζ dζ − E 2 a ∫ ζ − 1 ζ + 1 ζ R+ζ dζ=

= A e − E 1 a[ ζ−Rln( 1+ ζ R ) ] | ζ 1 − ζ κ − E 2 a[ ζ−Rln( 1+ ζ R ) ]| ζ κ ζ + 2 =

= A 1 E 1 + A 2 E 2 +a[ ( E 2 − E 1 ) ζ κ + E 1 ζ 1 − − E 2 ζ 2 + ]+

+a[ ( E 2 − E 1 )Rln( 1+ ζκ R )− E 1 Rln( 1− ζ 1 − R )+ E 2 Rln( 1+ ζ 2 + R ) ]. (52)

Regarding the E-weighted reduced first moment of the crosssection, equation (11)₂ yields

Q eR = ∫ R+ζ−ζ R+ζ E( η,ζ )ζadζ= Q eη ︸ =0 − ∫ ζ 2 E( η,ζ ) R+ζ adζ=− ∫ ζ 2 E( η,ζ ) R+ζ adζ=

=− E 1 a[ 1 2 ( ζ κ ) 2 − ζ κ R+ R 2 ln( R+ ζ κ )− 1 2 ( ζ 1 − ) 2 + ζ 1 − R− R 2 ln( R+ ζ 1 − ) ]−  − E 2 a[ 1 2 ( ζ 2 + ) 2 − ζ 2 + R+ R 2 ln( R+ ζ 2 + )− 1 2 ( ζ κ ) 2 + ζ κ R− R 2 ln( R+ ζ κ ) ] (53)

Using the parallel axis theorem, we can determine I_eη as

I eη = ∫ A E( η,ζ ) ζ 2 dA= E 1 ∫ A 1 ζ 2 dA + E 2 ∫ A 2 ζ 2 dA= = E 1 [ a b 1 3 12 + ( z C − b 1 2 ) 2 a b 1 ]+ E 2 [ a b 2 3 12 + ( b 1 − z C + b 2 2 ) 2 a b 2 ]. (54)

Therefore, recalling (11)₃ and utilizing equation (54) we can establish a formula for the E-weighted reduced moment of inertia:

I eR= ∫ A 1 ∪ A 2 R R+ζ ζ 2 E( η,ζ )dA= ∫ A 1 ∪ A 2 R+ζ−ζ R+ζ ζ 2 E( η,ζ )dA= I e η − − ∫ A 1 ∪ A 2 ζ 3 R+ζ E( η,ζ )dA= I eη +a( E 2 − E 1 )( ζ κ R 2 − 1 2 ( ζ κ ) 2 R− R 3 ln( R+ ζ κ )+ 1 3 ( ζ κ ) 3 )− − E 1 a( − 1 3 ( ζ 1 − ) 3 + 1 2 ( ζ 1 − ) 2 R− ζ 1 − R 2 + R 3 ln( R+ ζ 1 − ) )− −a E 2 ( + 1 3 ( ζ 2 + ) 3 − 1 2 ( ζ 2 + ) 2 R+ ζ 2 + R 2 − R 3 ln( R+ ζ 2 + ) ) . (55)

Substituting now a, b₁, b₂, R, A₁, E₁, A₂, E₂, ζ_k, ζ^-₁ and ζ⁺₂ into equations (52)-(55) we obtain the following numerical values:

A e =1.4336× 10 8 N,  A eR =1.4477× 10 8 N,  Q eR =−1.1588× 10 8 Nmm, I eη =9.9396× 10 9 Nm m 2 ,  I eR =9.5024× 10 9 Nm m 2 . (56)

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 σ ζ (ζ)=0 ,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 we have

α= I eR R 2 A eR = 9.5024× 10 9 82 2 ×1.447731712× 10 8 =9.7615× 10 −3 .

If ζ ^ ∈[ 4,20 ] A e ' = ∫ A ' E 2 dA=70000×32× ∫ ζ ^ 20 dζ=44 800 000 −2240 000 ζ ^ N.

We get in the same way that

Q eη ' = ∫ A ' E 2 R R+ζ ζdA=70000×32× ∫ ζ ^ 20 82 82+ζ dζ= =2.24× 10 6 ×( 1640−6724ln2−6724ln3−6724ln17−82 ζ ^ +6724ln(82+ ζ ^ ) )= =2.24× 10 6 ×( −29458.3172−82 ζ ^ +6724ln(82+ ζ ^ ) )Nmm.

With α, A e ' ( ζ ^ ) and Q eη ' ( ζ ^ ) we can determine the shear stress from (39):

τ ξζ ( ζ ^ )=− R R+ ζ ^ V I eR υ( ζ ^ ) ( R α e A e ' + Q eη ' )= =− 82 82+ ζ ^ × 50000 9.5024× 10 9 ×32 [ 82×9.7615× 10 −3 ×(44  800  000−2240  000 ζ ^ )+2.24× 10 6 ×( −29458.3172−82 ζ ^ +6724ln( 82+ ζ ^ ) ) ]= = 3.7150 82+ ζ ^ ×( −2.0195× 10 3 ζ ^ −7.2099× 10 5 +1.64× 10 5 ln( 82+ ζ ^ ) )N/m m 2 . if ζ∈[ −12,4 ] then A e ' = ∫ A ' E 1 dA=210000×32× ∫ −12 ζ ^ dζ=6720  000 ζ ^ +80 640 000N, Q eη ' = ∫ A ' E 1 R R+ζ ζdA=210000×32 ∫ −12 ζ ^ 82ζ 82+ζ dζ= =551 040 000 ζ ^ −45 185 280 000ln( 82+ ζ ^ )+1.985 81927× 10 11 Nmm. (57)

Consequently

τ ξζ ( ζ ^ )=− R R+ ζ ^ V I eR υ( ζ ^ ) ( R α e A e ' + Q eη ' )= =− 82 82+ ζ ^ × 50000 9.5024× 10 9 ×32 [ 82×9.7615× 10 −3 ×(6720  000−80 640 000)+551 040 000 ζ ^ −45 185 280 000ln( 82+ ζ ^ )+1.985  81927× 10 11 ]= = 3.7150 82+ ζ ^ ×( −2.0195× 10 3 ζ ^ −7.2099× 10 5 +1.64× 10 5 ln( 82+ ζ ^ ) )N/m m 2 . (58)

The values in Table 2 are computed using equations (57) and (58).

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.