Derivation of some fundamental formulae of strength of materials by energy method

Abstract

In this paper, three fundamental formulae of strength of materials are derived by the application of the theorem of minimum of strain energy. In the first example the torsion of non-homogeneous circular bar is considered. The second example deals with the in-plane bending of non-homogeneous curved beam. The third one is concerned with the pure bending of non-homogeneous elastic prismatic bars.

Keywords

Torsion circular bar curved beam bending non-homogeneous

Introduction

Energy methods are useful in solving many complicated problems of structural mechanics through two major approaches: approximate solutions of boundary value problem by Ritz method and formulation of governing equations of finite element analysis.^1–3 Another important application of energy methods is the derivation of approximate equations for a given boundary value problem of elasticity. In this paper, the theorem of minimum of strain energy is used to derive the formula of stress distribution as a function of cross-sectional coordinates and the applied load for torsion of circular bar, for in-plane bending of curved beam and for non-symmetrical bending of prismatic beam. In all cases, the structural members are non-homogeneous. Theorem of the minimum of strain energy, which will be used here, can be formulated as:^4,5 The strain energy U of an elastic body in equilibrium under the action of prescribed forces in an absolute minimum for the true stresses on the set of all values of the functional U determined by the statically admissible stresses. The statically admissible stresses satisfy all the equations of equilibrium.^4,5

Torsion of circular bar

In this section the Saint-Venant torsion of non-homogeneous circular bar is considered. The shear modulus depends on the radial coordinate r (Figure 1). The length of the beam is L and the applied torque is denoted by T. We assume that according to the Saint-Venant's theory of torsion of circular bar

σ_{r} = σ_{ϕ} = σ_{z} = τ_{rz} = τ_{r ϕ} = 0 τ_{ϕ z} = τ (r)

(1)

Figure 1.

Saint-Venant torsion of circular bar.

In equation (1), $σ_{r}, σ_{ϕ}, σ_{z}$ are the normal stresses and $τ_{r ϕ}, τ_{rz}, τ_{ϕ z}$ are the shearing stresses. Equation (1) refers to the cylindrical coordinate system $Or ϕ z$ (Figure 1).

Here, we must note, the stress field satisfies the equation of mechanical equilibrium and traction boundary condition which are in this problem formulated as^3,4

\frac{\partial τ_{rz}}{\partial r} + \frac{τ_{rz}}{r} + \frac{1}{r} \frac{\partial τ_{ϕ z}}{\partial ϕ} = 0 in A

(2)

τ_{rz} = 0 on \partial A

(3)

where A is the cross section of the circular bar and

\partial A

is the boundary curve of A. The strain energy per unit length of the twisted beam is expressed in terms of

τ = τ (r)

U [τ (r)] = \frac{1}{2} \int_{0}^{2 π} \int_{0}^{R} \frac{[τ (r)] 2}{G (r)} r d r d ϕ = π \int_{0}^{R} \frac{[τ (r)] 2}{G (r)} r d r

(4)

In equation (4) $G = G (r)$ is the shear modulus. The connection between the applied torque T and shearing stress function $τ = τ (r)$ is as follows

T = K [τ (r)] K [τ (r)] = 2 π \int_{0}^{2 π} r 2 τ (r) d r

(5)

In order that we get the distribution of shearing stress as a function of radial coordinate r we look for the minimum of the strain energy per unit length of bar under the subsidiary condition 5. The method of Lagrange multiplier will be used.^1,5 We consider the next functional

F [τ (r), λ] = π \int_{0}^{R} \frac{[τ (r)] 2}{G (r)} r d r - λ (2 π \int_{0}^{R} r 2 τ (r) d r - T)

(6)

where λ is the Lagrange multiplier. From the minimum condition of

U [τ (r)]

under the constraint given by equation (5) it follows that^1,5 the first variation of

F [τ (r), λ]

with respect to

τ (r)

and λ vanishes, that is

δ F = 2 π \int_{0}^{R} [\frac{τ (r)}{G (r)} r - λ r 2] δ τ d r - δ λ [2 π \int_{0}^{R} r 2 τ (r) d r - T] = 0

(7)

Since $δ τ (r)$ and $δ λ$ are arbitrary we obtain from equation (7)

τ (r) = λ r G (r)

(8)

T = 2 π λ \int_{0}^{R} r 3 G (r) d r

(9)

Let

S = 2 π \int_{0}^{R} r 3 G (r) d r

(10)

be. Combination of equations (8) and (9) with equation (10) gives

τ (r) = \frac{T}{S} r G (r)

(11)

since

λ = \frac{T}{S}

(12)

The strain energy per unit length of circular bar in terms of the applied torque is as follows

U = \frac{T 2}{2 S}

(13)

The validity of formula (13) follows from equations (4) and (11). Equation (13) shows that S is the torsional rigidity of the non-homogeneous circular bar. By the use of Prandtl formulation of Saint-Venant torsion, equations (10) and (11) were derived by Horgan and Chan.⁶

It is known that the rate of twist ϑ can be computed as^1,5,6

ϑ = \frac{T}{S}

(14)

Comparison of equation (12) with equation (14) shows that $λ = ϑ$ , that is the mechanical meaning of the Lagrange multiplier λ is the rate of twist.

In-plane bending of non-homogeneous curved beam

Figure 2 shows the non-homogeneous curved beam which has uniform curvature. Its material is linearly elastic, isotropic and non-homogeneous. The applied bending moment is denoted by $M_{0}$ . The plane $z = 0$ is the plane of symmetry for the geometrical and material properties. The Young's modulus E depends on the cross-sectional coordinate r and z such that

E (r, z) = E (r, - z) (r, z) \in A

(15)

Figure 2.

In-plane bending of curved beam.

The cross section A of the curved beam is shown in Figure 2. In the strength of materials the equilibrium equations are formulated under the next assumptions for in-plane bending

\frac{d M}{d ϕ} = 0 that is M = M_{0} - α \leq ϕ \leq α

(16)

$σ_{r}, σ_{z}, τ_{r ϕ}, τ_{rz}, τ_{ϕ z}$ are neglected in comparison with $σ_{ϕ} = σ_{ϕ} (r, z)$

K_{1} [σ_{ϕ} (r, z)] = \int_{A} σ_{ϕ} (r, z) d A = 0

(17)

K_{2} [σ_{ϕ} (r, z)] = \int_{A} r σ_{ϕ} (r, z) d A - M_{0} = 0

(18)

The strain energy per unit polar angle $Δ ϕ = 1$ (Figure 2) can be computed as^1,5

U [σ_{ϕ} (r, z)] = \frac{1}{2} \int_{A} \frac{[σ_{ϕ} (r, z)] 2}{E (r, z)} r d A

(19)

In order that we get the distribution of normal stress as a function of cross-sectional coordinates $r, z$ we will use the theorem of minimum of strain energy. We look for the minimum of $U [σ_{ϕ} (r, z)]$ under the subsidiary conditions (17) and (18). The above defined minimum problem is solved by the applications of Lagrange multipliers. At first we define a new functional

F [σ_{ϕ} (r, z), λ_{1}, λ_{2}] = U [σ_{ϕ} (r, z)] - λ_{1} K_{1} [σ_{ϕ} (r, z)] - λ_{2} K_{2} [σ_{ϕ} (r, z)]

(20)

The stationarity condition of $F [σ_{ϕ} (r, z), λ_{1}, λ_{2}]$ leads to solution of the formulated minimum problem

δ F = \int_{A} [\frac{σ_{ϕ} (r, z)}{E (r, z)} r - λ_{1} - λ_{2} r] δ σ_{ϕ} d A - δ λ_{1} \int_{A} σ_{ϕ} d A - δ λ_{2} (M_{0} - \int_{A} r σ_{ϕ} d A) = 0

(21)

Since $δ σ_{ϕ} (r, z)$ , $δ λ_{1}$ and $δ λ_{2}$ are arbitrary from variational equation (21) it follows that

σ_{ϕ} (r, z) = E (r, z) [\frac{λ_{1}}{r} + λ_{2}]

(22)

\int_{A} σ_{ϕ} (r, z) d A = 0 \int_{A} r σ_{ϕ} (r, z) d A = M_{0}

(23)

In the next we define some cross-sectional properties that are $E_{0}, R, r_{c}$ and e⁷

E_{0} = \frac{1}{A} \int_{A} E (r, z) d A \frac{E_{0} A}{R} = \int_{A} \frac{E (r, z)}{r} d A

(24)

E_{0} A r_{c} = \int_{A} r E (r, z) d A e = r_{c} - R

(25)

By the use of cross-sectional properties introduced by equations (24) and (25) from equations (22) and (23) we obtain the formula of normal stress $σ_{ϕ}$ in terms of $M_{0}$ as

σ_{ϕ} (r, z) = \frac{E (r, z)}{E_{0} A e} (1 - \frac{R}{r}) M_{0}

(26)

since

λ_{1} = - \frac{M_{0} R}{E_{0} A e} λ_{2} = \frac{M_{0}}{E_{0} A e}

(27)

Substitution for $N = 0$ in formula (A.1) of paper by Ecsedi and Dluhi⁷ we get the same result for $σ_{ϕ}$ as given by equation (26).

Pure bending of non-homogeneous bar

Figure 3 shows the non-homogeneous bar which is subjected to bending moment $M$ . The axis z is the E-weighted center line of the bar, which connects the E-weighted centres of cross-section. The Young's modulus E depends on the cross-sectional coordinates x and y, that is we have⁸

\int_{A} E (x, y) R d A = 0 R = x e_{x} + y e_{y}

(28)

Figure 3.

Pure bending of non-homogeneous bar.

In the framework of strength of materials the equilibrium stress field is characterized by the equations in this case

σ_{x} = σ_{y} = τ_{xy} = τ_{xz} = τ_{yz} = 0, σ_{z} = σ_{z} (x, y)

(29)

K_{1} [σ_{z} (x, y)] = \int_{A} σ_{z} (x, y) d A = 0

(30)

K_{2} [σ_{z} (x, y)] = \int_{A} R σ_{z} (x, y) d A - e_{z} \times M = 0

(31)

Equations (29)–(31) refer to the coordinate system $Oxyz$ (Figure 3) and $σ_{x}, σ_{y}, σ_{z}$ are the normal stresses and $τ_{xy}, τ_{xz}, τ_{yz}$ are the shearing stresses and the cross between two vectors in equation (31) denotes their vectorial product. It is obvious that the bending moment vector does not depend on the axial coordinate z. The strain energy per unit length is as follows

U [σ_{z} (x, y)] = \frac{1}{2} \int_{A} \frac{[σ_{z} (x, y)] 2}{2 E (x, y)} d A

(32)

Our aim is to search the minimum of U under the subsidiary conditions 30 and 31. The method of Lagrange multipliers will be used. We define a new functional which contains the constraints given by equations (30) and (31)

F [σ_{z} (x, y), λ_{1}, λ_{2}] = U [σ_{z} (x, y)] - λ_{1} K_{1} [σ_{z} (x, y)] - λ_{2} \cdot K_{2} [σ_{z} (x, y)]

(33)

In equation (31) the scalar product of two vectors is denoted by dot. The necessary condition of minimum is formulated by the variational equation

δ F = \int_{A} (\frac{σ_{z} (x, y)}{E (x, y)} - λ_{1} - λ_{2} \cdot R) δ σ_{z} - δ λ_{1} \int_{A} σ_{z} (x, y) d A - δ λ_{2} \cdot [\int_{A} R σ_{z} (x, y) d A - e_{z} \times M] = 0

(34)

Since $δ σ_{z} (x, y)$ and $δ λ_{1}, δ λ_{2}$ are arbitrary we obtain from equation (34)

σ_{z} (x, y) = E (x, y) λ_{1} - E (x, y) λ_{2} \cdot R

(35)

\int_{A} σ_{z} (x, y) d A = 0, \int_{A} R σ_{z} (x, y) d A = M

(36)

Substitution of equation (35) into equation (36₁) gives

λ_{1} = 0

(37)

that is we have

σ_{z} (x, y) = E (x, y) λ_{2} \cdot R

(38)

Let $λ_{2} = λ_{2} m$ be, where $| m | = 1$ . Equation of neutral axis of normal stress, which is given by the equation $σ_{z} = 0$ , can be written as

m \cdot R = m_{x} x + m_{y} y = 0

(39)

The neutral axis (zero line of stresses) goes through the E-weighted centre of cross section (Figure 4). The direction of the neutral axis n is given by the unit vector $n = n_{x} e_{x} + n_{y} e_{y}$ . From equation (36₂) and 38 it follows that

e_{z} \times M = λ_{2} \int_{A} E (x, y) R (R \cdot m) d A = λ_{2} (\int_{A} E (x, y) R \circ R d A) \cdot m = λ_{2} J \cdot m

(40)

Figure 4.

Illustration of the neutral axis.

In equation (40) circle between the two vectors denotes their dyadic product and $J$ is the Euler tensor, whose representation by sum of dyadic products is as follows

e_{y} \circ e_{x}) + J_{x} e_{y} \circ e_{y}

(41)

J_{x} = \int_{A} E y 2 d A J_{y} = \int_{A} E x 2 d A J_{xy} = \int_{A} Exy d A

(42)

By a simple calculation from equation (40) we get

M = λ_{2} J \cdot n

(43)

since

n = m \times e_{z}

. Equation (43) gives a possibility to determine n. It is evident

n = \frac{J - 1 \cdot M}{| J - 1 \cdot M |}

(44)

From equation (43) we have

n \cdot M = M_{n} = λ_{2} n \cdot J \cdot n = λ_{2} J_{n}

(45)

that is

λ_{2} = \frac{M_{n}}{J_{n}}

(46)

Let $η = m \cdot R$ be (Figure 4). The combination of equation (38) with equation (46) gives the bending formula

σ_{z} (x, y) = E (x, y) \frac{M_{n}}{J_{n}} η

(47)

according to paper Baksa and Ecsedi.⁸

A recommendation for the method presented to use in teaching of elasticity

We recommend to introduce the presented method of the derivation of considered formulae of strength of materials in the study of first-year graduate level course in elasticity, where the students get to know on the essential variational methods of the linearized theory of elasticity such as principle of minimum of potential energy, principle of minimum of complementary energy, principle of minimum of strain energy etc. Traditionally taught energy methods are primarily applied to look for the approximate solutions of a given boundary value problem of elasticity by means of Ritz method. In this paper we give examples how the theorem of the principle of minimum of strain energy can be used to obtain the well-known formulae of mechanics of solids. This type of application of energy methods does not exceed the level of a first-year graduate course in elasticity.

Conclusions

The paper illustrates the applications of energy methods by three examples in the field of strength of materials. Examples refer to non-homogeneous structural members. By the use of principle of minimum of strain energy the stress distributions are determined for torsion of circular bar, for in-plane bending of curved beam and for pure bending of prismatic beam. The presented approaches do not follow the direct methods used in textbooks of strength of materials and elasticity.

Footnotes

Acknowledgements

This research was (partially) carried out in the framework of the Center of Excellence of Innovative Engineering Design and Technologies at the University of Miskolc.

Conflict of interest

None declared.

Funding

This research received no specific grant from any funding agency in the public, commercial, or not-for-profit sectors.

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