Standard Deformations of Nonlinear Elastic Structural Elements with Power-Law Constitutive Model

Abstract

In this paper, the case of the power dependence between strain and stress is studied, along with the way in which this dependence modifies the calculation methodologies and the results that are obtained in classic cases of stress. The main cases studied are compression (squashing), tension (pulling), bending, shear (cutting), and torsion (twisting). Simple relationships are thus obtained for a wide class of materials that fall into this category. They can be useful to designers because they provide information on mechanical structures in a short time with good precision.

1. Introduction

The theory of elasticity (and implicitly the strength of materials) has as its fundamental hypothesis the linear relationship between strain and stress. Calculations based on this assumption give good results for steel and some other materials (usually materials often used in engineering practice). However, although the calculations are built on this assumption, for many materials this linear relationship does not work, and better or worse approximations are obtained depending on the real dependence between strain and stress. These materials include cast iron, copper, brass, lead, rubber, and some plastic or composite materials that are widely used in current technology. In general, the characteristic strain–stress curves of these materials that do not obey Hooke’s law are well approximated with the help of power laws. The power strain–stress curve shows how the stress decreases/increases following a power law as the strain increases. This curve represents the stress–strain variation law when we subject the material to tension. All forces acting on a material can be considered as combinations of just a few basic types: compression (squashing), tension (pulling), bending, shear (cutting), and torsion (twisting). Practically, we have two basic forces: compression and tension. Shear and bending are combinations of compression and tension, and torsion (circular bending) is a combination of compression, tension, shear, and bending. When a new parameter, time, needs to be added to the theory of viscoelasticity, nonlinear constitutive models are commonly used. Nonlinear viscoelastic materials have been modeled for a variety of foods, soft biological tissues, and polymers. In [1], all of these models were presented unitarily utilizing classical conclusions from continuous media mechanics. When examining the behavior of a fibrous boron/aluminum composite with a metallic matrix, the law of elastoplastic homogenization was helpful [2]. The integral equation derived from this composite’s micromechanical study was solved using Fourier analysis. In [3], it was suggested to analyze nonlinear systems in order to offer probabilistic information for structural analysis. A new notion was introduced, the concept of the variability response function, to capture the effect of the stochastic spectral characteristics of uncertain parameters. Micromechanics is the method generally used to study the behavior of nonlinear viscous materials. A simplified model of porous materials with nonlinear components was presented in [4]. The proposed scheme stands out due to its simplicity and the study of a simple “quadratic” shape for the deformation speed. The dynamics of the deformations of structures essentially depend on the constitutive law, and if it is nonlinear, each case must be analyzed separately. A study proposing a scheme that automatically modifies the Jacobian depending on the considered constitutive law was provided in [5]. An engineering application was analyzed to illustrate the proposed method. A new formulation of the constitutive laws in the case of nonlinear isotropic materials was proposed in [6]. The existence of a differentiable convex isotropic potential was assumed in this modeling. Conclusions useful to researchers were presented for cases where nonlinear elasticity problems are approached in this way. An interesting approach to a hyperbolic system that was obtained in an elastodynamic analysis of a 1D system was presented in [7]. Analytical solutions for different constitutive laws were presented. Large deformations are generally the result of nonlinear constitutive laws. A useful model for the study of such cases in the case of post-buckling analysis of trusses was presented in [8]. The obtained solutions were compared with the exact solutions of the equilibrium equations. The paper presented a modified method with finite elements for the nonlinear analysis of beam structures located in a 2D plane, using a particular model with finite elements obtained in [9]. The beam model was Timoshenko. This method is relatively simple to implement in common CAE 2020 software that is available for structural analysis. Recent results in this field were presented in [10,11,12,13,14,15,16]. The development of research in the nonlinear field was boosted by the appearance of composite materials that, in general, have nonlinear behavior. As an example, polymer composites reinforced with fibers have nonlinear behavior and show hyperelastic characteristics. For the study of such materials, micromechanical models that operate according to the hypothesis of the existence of a hyperelastic constitutive law represent the main path followed [17]. Arbitrary complete three-dimensional constitutive equations are used for the study of plates and shells. Models in which more parameters are used, thus allowing the study of more complex phenomena, were offered in [18]. The considered constitutive laws determine the method that is used to analyze the nonlinear, anisotropic, and asymmetric inelastic mechanical behavior of composites [19,20]. Numerical examples confirm obtained results. A modified constitutive model, which assumes an elaborate approach, is used to study large nonlinear stress–strain evolutions [21]. The use of this model was suggested by experimental results, and based on the experimental results, it was proven to be a powerful tool for studying the mechanical behaviors of polymers. The importance of constitutive laws in the study of such problems was highlighted in [22,23,24]. The Finite Element Method has proven to be the most suitable numerical tool for addressing such problems. Numerous works have addressed the problem of using FEM to solve problems in which the behavior of a material is nonlinear. We mention only a few of them [25,26,27,28,29,30,31,32]. Experiments performed to validate theoretical results and comparisons with results from the literature have demonstrated the efficiency and versatility of the proposed numerical procedure. Particular attention was given to the study of human tissue. Thus, blood flow through blood vessels and the nonlinear behavior of blood vessel tissue have been intensively studied [33,34,35]. A model for studying skeletal muscles was presented in [36], and the soft tissue of the human body was analyzed in [37]. Engineering applications in which nonlinear materials are used can be found in [38,39,40,41,42,43,44,45,46,47], and theoretical developments related to this problem can be found in [48,49,50,51,52,53,54,55,56,57,58].

In conclusion, there are numerous practical situations that require the study of nonlinear materials and a rich body of work dealing with different types of nonlinearities. This paper deals with a particular class of such materials, those in which the strain–stress law has a power form. In this paper, a unitary study is carried out for these materials, obtaining results that can be used in practice with ease. In this way, engineers are offered results that can be obtained simply in a relatively short time, which is important in a real project.

2. Models and Methods: Power-Law Strain–Stress Diagram

The characteristic curves of materials with power-law behavior can be expressed with the help of some shape strain–stress laws:

$$\epsilon ={\left({\displaystyle \frac{\sigma }{{\sigma }_0}}\right)}^n\mathrm{and}\gamma ={\left({\displaystyle \frac{\tau }{{\tau }_0}}\right)}^n,$$

(1)

where n is real, $\epsilon$ is the normal strain, $\gamma$ is the shear strain, $\sigma$ is the normal stress, $\tau$ is the shear stress, and ${\sigma }_0$ and ${\tau }_0$ are material constants with the physical dimension of the Young’s modulus [21]. Their values and the exponent n can be determined by laboratory tests for traction/compression, bending, or twisting.

Figure 1 presents characteristic curves for different values of n. In particular, the curve for n = 1 corresponds to materials obeying Hooke’s law. The curve for n = $\infty$ corresponds to the plastic stress in Prandtl’s schematization, and n = 0 corresponds to a completely rigid material.

Experimental determination of the characteristics n and ${\sigma }_0$ can be performed through a tensile test in which at least two measurements are made. Thus, $\epsilon$ and $\sigma$ are measured in two states of loading, (1) and (2), denoted by ${\sigma }_1$ and ${\sigma }_2$, and ${\epsilon }_1$ and ${\epsilon }_2$, respectively. Based on the relationships

$${\epsilon }_1={\left({\displaystyle \frac{{\sigma }_1}{{\sigma }_0}}\right)}^n\mathrm{and}{\epsilon }_2={\left({\displaystyle \frac{{\sigma }_2}{{\sigma }_0}}\right)}^n,$$

(2)

the following equations can be obtained through simple calculations:

$$n={\displaystyle \frac{\log \frac{{\epsilon }_1}{{\epsilon }_2}}{\log \frac{{\sigma }_1}{{\sigma }_2}}}\mathrm{and}{\sigma }_0={\displaystyle \frac{{\sigma }_1}{{{\epsilon }_1}^{\frac{1}{n}}}}.$$

(3)

Similarly, for a torsion test where measurements are performed in at least two distinct states, one obtains

$$n={\displaystyle \frac{\log \frac{{\gamma }_1}{{\gamma }_2}}{\log \frac{{\tau }_1}{{\tau }_2}}}\mathrm{and}{\tau }_0={\displaystyle \frac{{\tau }_1}{{{\gamma }_1}^{\frac{1}{n}}}}.$$

(4)

Obviously, more measurements can be made. Then, the least squares method can be used to obtain better estimates for these two quantities.

This type of problem, in which the relationship between strain and stress follows a power law, has been identified and mentioned for a long time [21], but it has not been the subject of systematic and unitary studies. For example, we present results obtained through measurements in a university laboratory for three different materials (Figure 2, Figure 3 and Figure 4), based on which n and ${\sigma }_0$ were calculated in each case.

In what follows, the particularities implied by the existence of a power-law relationship between stress and strain will be studied.

3. Results: Basic Loading Cases

3.1. Tension: Compression

In the following analysis, some cases that are often encountered in practice will be studied.

(a)

Assumptions related to linear material resistance are kept, including Bernoulli’s hypothesis. Based on this hypothesis, $\epsilon =ct$ for one arbitrary section at every point of the section. Therefore, for the points of this section, the stress values are the same:

$$\sigma ={\sigma }_0{\left(ct\right)}^{{\displaystyle \frac{1}{n}}}.$$

(5)

If the normal stress in the section is denoted by N, it can be calculated based on the equilibrium of the stresses acting on the section:

$$N={\displaystyle {\int }_A\sigma \text{}dA}=\sigma A.$$

(6)

This results in

$$\sigma ={\displaystyle \frac{N}{A}},$$

(7)

(as in linear elasticity). If a bar has the length l and the constant section A, the lengthening/shortening of the bar can be calculated as follows:

$$\Delta l=\epsilon \cdot l={\left({\displaystyle \frac{\sigma }{{\sigma }_0}}\right)}^{n}l={\left({\displaystyle \frac{N}{{\sigma }_{0}A}}\right)}^{n}l;$$

(8)

(b)

In what follows, the case of a bar with an inhomogeneous section is considered. We take the case of a bar made of three different materials marked 1, 2, and 3 (Figure 5). By taking a section and denoting it with $N_1$, $N_2$, and $N_3$, the normal forces on the three materials from the balance equation are deduced:

$$N=N_1+N_2+N_3.$$

(9)

The three materials working together in tension/compression have the same deformation and result:

$$\Delta l_1=\Delta l_2=\Delta l_3,$$

(10)

or

$${\left({\displaystyle \frac{N_1}{{\sigma }_{01}{A}_1}}\right)}^{n_1}={\left({\displaystyle \frac{N_2}{{\sigma }_{02}{A}_2}}\right)}^{n_2}={\left({\displaystyle \frac{N_3}{{\sigma }_{03}{A}_3}}\right)}^{n_3}.$$

(11)

Substituting the values of $N_2$ and $N_3$ extracted from relation (11) in Equation (10), we obtain

$$N=N_1+{\displaystyle \frac{{\sigma }_{02}{A}_2}{{\left({\sigma }_{01}{A}_1\right)}^{\frac{n_1}{n_2}}}}{N}_1^{{\displaystyle \frac{n_1}{n_2}}}+{\displaystyle \frac{{\sigma }_{03}{A}_3}{{\left({\sigma }_{01}{A}_1\right)}^{\frac{n_1}{n_3}}}}{N}_1^{{\displaystyle \frac{n_1}{n_3}}}.$$

(12)

Based on Equation (12), $N_1$ can be taken out using a numerical method for solving the equation. Then, it is simple to obtain $N_2$ and $N_3$ using Equation (11).

(c)

If a uniformly heated bar is prevented from expanding, it will be subjected to compression (Figure 6). The compression force (N) is calculated as follows. Noting the temperature difference with $\Delta t$, linear expansion is expressed as

$$\Delta l=\alpha l\Delta t,$$

(13)

where $\alpha$ is the coefficient of linear expansion. Using Equation (8), we obtain

$${\left({\displaystyle \frac{N}{{\sigma }_{0}A}}\right)}^n=\alpha l\Delta t,$$

(14)

wherefrom we can obtain

$$N={\sigma }_{0}A{\left(\alpha \Delta t\right)}^{{\displaystyle \frac{1}{n}}}.$$

(15)

The normal stress is

$$\sigma ={\displaystyle \frac{N}{A}}={\sigma }_0{\left(\alpha \Delta t\right)}^{{\displaystyle \frac{1}{n}}};$$

(16)

3.2. Torsion of Bars with Circular Sections

The assumptions of linear resistance are kept. The radius of the section is denoted with R, and the radius where the element dA is located is denoted with ρ. The stress ($\tau$) acts on the section due to the torsional moment (M_r):

$$M_r={\displaystyle {\int }_A\tau \rho dA}.$$

(17)

Based on geometric considerations, the sliding angle is deduced using Figure 7.

$$\gamma =\rho \theta ,$$

(18)

wherefrom we can obtain

$$\tau ={\tau }_0{\gamma }^{{\displaystyle \frac{1}{n}}}.$$

(19)

Substituting Equations (18) and (19) in Equation (17) we obtain

$$M_r={\tau }_0{\theta }^{{\displaystyle \frac{1}{n}}}{\displaystyle {\int }_A{\rho }^{\frac{n+1}{n}}dA}={\tau }_0{\theta }^{{\displaystyle \frac{1}{n}}}{I}_r,$$

(20)

where it is noted that

$$I_r={\displaystyle {\int }_A{\rho }^{\frac{n+1}{n}}dA},$$

(21)

which represents the generalized polar moment of the cross-section. This results in the following equations after simple calculations:

$$\tau ={\displaystyle \frac{M_r{\rho }^{\frac{1}{n}}}{I_r}},$$

(22)

$$\varphi =l\theta ={\left({\displaystyle \frac{M_r}{{\tau }_0{I}_r}}\right)}^{n}l,$$

(23)

and

$${\tau }_{\max }={\displaystyle \frac{M_r}{W_r}},$$

(24)

where it is noted that

$$W_r={\displaystyle \frac{M_r}{R^{\frac{1}{n}}}}.$$

(25)

This represents the generalized polar strength modulus.

For the circular section, the values for I_r and W_r are

$$I_r={\displaystyle {\int }_A{\rho }^{\frac{n+1}{n}}dA}={\displaystyle {\int }_0^R{\rho }^{\frac{n+1}{n}}2\pi \rho d\rho }=2\pi {\displaystyle {\int }_0^R{\rho }^{\frac{2n+1}{n}}d\rho }={\displaystyle \frac{2\pi n}{3n+1}}{R}^{{\displaystyle \frac{3n+1}{n}}};$$

(26)

$$W_r={\displaystyle \frac{2\pi n}{3n+1}}{R}^3.$$

(27)

For the tubular sections of radii r and R, in an analogous way, we obtain

$$I_r={\displaystyle \frac{2\pi n}{3n+1}}\left(R^{{\displaystyle \frac{3n+1}{n}}}-r^{{\displaystyle \frac{3n+1}{n}}}\right);W_r={\displaystyle \frac{2\pi n}{\left(3n+1\right)R^{\frac{1}{n}}}}\left(R^{{\displaystyle \frac{3n+1}{n}}}-r^{{\displaystyle \frac{3n+1}{n}}}\right).$$

(28)

If the tube wall is very thin and the thickness of the tube wall is denoted with t, it results in the following equation:

$$r=R-t.$$

(29)

In this case, it can be written successively as follows:

$$\begin{array}{l}{\displaystyle \frac{\left(R^{\frac{3n+1}{n}}-r^{\frac{3n+1}{n}}\right)}{R^{\frac{1}{n}}}}=R^3-{\left({\displaystyle \frac{r}{R}}\right)}^{{\displaystyle \frac{1}{n}}}{r}^3=R^3-{\left(1-{\displaystyle \frac{t}{R}}\right)}^{{\displaystyle \frac{1}{n}}}{R}^3{\left(1-{\displaystyle \frac{t}{R}}\right)}^3\cong \\ R^3-R^3\left(1-{\displaystyle \frac{t}{R}}\right)\left(1-{\displaystyle \frac{t}{nR}}\right)\cong {\displaystyle \frac{3n+1}{n}}{R}^{2}t,\end{array}$$

Then,

$$W_r=2\pi R^{2}t,$$

(30)

3.3. Calculation of a Helical Spring

The deformation of a spring (f) with $n_s$ spirals is calculated, starting with the angular deformation of an element of the spiral with a length of ds (Figure 8).

$$d\varphi ={\displaystyle \frac{M_r^{n}ds}{{\tau }_0^n{I}_r^n}},$$

(31)

where

$$M_r=PR.$$

(32)

The meanings of the notations can be seen in Figure 8.

Based on this relationship,

$$df=Rd\varphi ,$$

(33)

which results in the total deformation of the spring:

$$f=R{\displaystyle \int d\varphi }=R{\displaystyle \frac{M_r^{n}2\pi n_{s}R}{{\tau }_0^n{I}_r^n}}=R{\displaystyle \frac{2\pi n_s{P}^n{R}^{n+2}}{{\tau }_0^n{I}_r^n}},$$

(34)

where I_r is given by relation (26):

$$I_r={\displaystyle \frac{\pi n{d}^{\frac{3n+1}{n}}}{2^{\frac{2n+1}{n}(3n+1)}}}.$$

(35)

By substituting the above displacement, we obtain

$$f={\displaystyle \frac{8(3n+1)\sqrt[n]{2}{P}^n{R}^{n+2}{n}_s}{n{\tau }_0^n{d}^{\frac{3n+1}{n}}}}.$$

(36)

3.4. Bending

(a)

Bending is treated using the assumptions of linear elasticity. Taking into account Bernoulli’s hypothesis, the elongation of a fiber located at a distance (y) from the neutral axis is deduced. It is found (Figure 9) that

$$\epsilon ={\displaystyle \frac{y}{\rho }},$$

(37)

where $\rho$ is the radius of the curvature of the average fiber near the considered section.

The stress is

$$\sigma ={\sigma }_0{\left({\displaystyle \frac{y}{\rho }}\right)}^{{\displaystyle \frac{1}{n}}}.$$

(38)

Based on the equilibrium equation (Figure 9), it follows that

$$M_i={\displaystyle {\int }_A\sigma ydA}={\displaystyle \frac{{\sigma }_0}{{\rho }^{\frac{1}{n}}}}{\displaystyle {\int }_A{y}^{\frac{n+1}{n}}dA}={\displaystyle \frac{{\sigma }_0{I}_z}{{\rho }^{\frac{1}{n}}}},$$

(39)

where it is noted that

$$I_z={\displaystyle {\int }_A{y}^{\frac{n+1}{n}}dA}.$$

(40)

This represents the generalized axial moment of the cross-section. Based on Equation (25), this results in

$${\displaystyle \frac{1}{\rho }}={\left({\displaystyle \frac{M_i}{{\sigma }_0{I}_z}}\right)}^n.$$

(41)

Introducing this result to Equation (38) generates the following stress equation:

$$\sigma ={\displaystyle \frac{M_i{y}^{\frac{1}{n}}}{I_z}}.$$

(42)

The maximum stress is

$$\sigma ={\displaystyle \frac{M_i}{W_z}},$$

(43)

where it is noted that

$$W_z={\displaystyle \frac{I_z}{y^{\frac{1}{n}}}}.$$

(44)

This represents the generalized axial strength modulus.

If the other equilibrium equations for the section are also written:

$${\displaystyle {\int }_A\sigma dA}=0\mathrm{and}{\displaystyle {\int }_A\sigma zdA}=0,$$

(45)

and their values are replaced by relation (38), we obtain

$${\displaystyle \frac{{\sigma }_0}{{\rho }^{\frac{1}{n}}}}{\displaystyle {\int }_A{y}^{\frac{1}{n}}dA}=0\mathrm{and}{\displaystyle \frac{{\sigma }_0}{{\rho }^{\frac{1}{n}}}}{\displaystyle {\int }_A{y}^{\frac{1}{n}}zdA}=0.$$

(46)

From the first relationship, it can be deduced that, in general, the neutral axis does not pass through the center of gravity. The position of the neutral axis is determined using the following equation:

$${\displaystyle {\int }_{A_1}{y}^{\frac{1}{n}}dA}={\displaystyle {\int }_{A_2}{y}^{\frac{1}{n}}dA},$$

(47)

where A₁ and A₂ are the areas of the section on either side of the neutral axis. In the case of symmetrical sections, the neutral axis passes through the center of gravity.

From the second relation (46) it follows that the y-axis must be an axis of symmetry of the section.

(b)

$I_z$ and $W_z$ are calculated below for some sections that are often used in practice:

• A rectangular section (Figure 10a) with a surface element:

  $$dA=bdy.$$

  (48)
  From this we obtain

  $$I_z=2b{\displaystyle {\int }_0^{\frac{h}{2}}{y}^{\frac{n+1}{n}}}dy={\displaystyle \frac{2nb}{2n+1}}{\left({\displaystyle \frac{h}{2}}\right)}^{{\displaystyle \frac{2n+1}{n}}};$$

  (49)

  $$W_z={\displaystyle \frac{nb{h}^2}{2\left(2n+1\right)}},$$

  (50)

  $${\sigma }_{\max }={\displaystyle \frac{2\left(2n+1\right)}{nb{h}^2}}{M}_i,$$

  (51)
• Section I (Figure 10b):

  $$I_z=2\left[b{\displaystyle {\int }_0^{\frac{h}{2}}{y}^{\frac{n+1}{n}}}dy+B{\displaystyle {\int }_{\frac{h}{2}}^{\frac{H}{2}}{y}^{\frac{n+1}{n}}}dy\right]={\displaystyle \frac{2n}{2n+1}}\left[B{\left({\displaystyle \frac{H}{2}}\right)}^{{\displaystyle \frac{2n+1}{n}}}-\left(B-b\right){\left({\displaystyle \frac{h}{2}}\right)}^{{\displaystyle \frac{2n+1}{n}}}\right];$$

  (52)

  $$W_z={\displaystyle \frac{I_z}{{\left(\frac{H}{2}\right)}^{\frac{1}{n}}}}={\displaystyle \frac{n}{2n+1}}\left[B{H}^2-\left(B-b\right)h^2{\left({\displaystyle \frac{h}{H}}\right)}^{{\displaystyle \frac{1}{n}}}\right];$$

  (53)

Circular section (Figure 10c):

$$I_z=2{\displaystyle {\int }_0^R{y}^{\frac{n+1}{n}}2z}dy,$$

(54)

where

$$z=R\sin \alpha ;y=R\cos \alpha ;dy=-R\sin \alpha d\alpha .$$

(55)

Substituting these values in Equation (54) results in

$$I_z=4R^{{\displaystyle \frac{3n+1}{n}}}{\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^2\alpha {\cos }^{\frac{n+1}{n}}\alpha }d\alpha .$$

(56)

This integral can only be calculated analytically for whole values of powers. For non-integer values, numerical calculation can be applied. Particular values are calculated below:

• n = 1: $I_z={\displaystyle \frac{\pi }{4}}{R}^4$;
• n = 1/2: $I_z={\displaystyle \frac{8}{15}}{R}^5$;
• n = 1/3: $I_z={\displaystyle \frac{\pi }{8}}{R}^6$;
• n = $\infty$: $I_z={\displaystyle \frac{4}{3}}{R}^3$.

$W_z$ can also be determined and can be expressed as follows:

$$W_z={\displaystyle \frac{I_z}{R^{\frac{1}{n}}}}=4R^3{\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^2\alpha {\cos }^{\frac{n+1}{n}}}\alpha d\alpha .$$

(57)

The neutral axis (z) is determined for the triangular section (Figure 11). Its position is marked with an x.

For A₁,

$${\displaystyle {\int }_{A_1}{y}^{\frac{1}{n}}}dA=\frac{b}{h}{\displaystyle {\int }_0^{h-x}{y}^{\frac{1}{n}}}\left(h-x-y\right)dy=\frac{n^2}{\left(n+1\right)\left(2n+1\right)}{\displaystyle \frac{b}{h}}{\left(h-x\right)}^{{\displaystyle \frac{2n+1}{n}}}.$$

(58)

For A₂,

$${\displaystyle {\int }_{A_2}{y}^{\frac{1}{n}}}dA=\frac{b}{h}{\displaystyle {\int }_0^x{y}^{\frac{1}{n}}}\left(h-x+y\right)dy=\frac{n}{\left(n+1\right)\left(2n+1\right)}{\displaystyle \frac{b}{h}}{\left(x\right)}^{{\displaystyle \frac{n+1}{n}}}\left[\left(2n+1\right)h-nx\right].$$

(59)

The neutral axis results from Equation (47):

$$n{(h-x)}^{{\displaystyle \frac{2n+1}{n}}}=x^{{\displaystyle \frac{n+1}{n}}}\left[\left(2n+1\right)h-nx\right].$$

(60)

In the case of n = 0.5, the equation becomes

$${\left(h-x\right)}^4=x^3\left(4h-x\right),$$

(61)

which, by changing the variable t = x/h, becomes

$$2t^4-8t^3+6t^2-4t+1=0,$$

(62)

which gives t = 0.359 as a convenient solution. Therefore,

x = 0.359 h.

(63)

The neutral axis is displaced from the center of gravity by

(0.359 − 0.333) h = 0.026 h.

(64)

3.5. Inclined Bending

This involves a compound load that can no longer be solved by superposing effects as for linear elasticity. α denotes the inclination of the plane of forces with respect to the y-axis, which is considered a symmetry axis of the section, and β denotes the inclination of the neutral axis with respect to the z-axis (Figure 12). Bernoulli’s hypothesis can be expressed by the following relation:

$$\epsilon =K\eta$$

(65)

where K is a constant to be determined and

$$\eta =z\sin \beta +y\cos \beta .$$

(66)

Equilibrium equations:

$${\displaystyle \underset{A}{\int }\sigma zdA}=M_i\sin \alpha ;$$

(67)

$${\displaystyle \underset{A}{\int }\sigma ydA}=M_i\cos \alpha ,$$

(68)

Simple transformations lead to the following relations:

$${\sigma }_0{K}^{{\displaystyle \frac{1}{n}}}{\displaystyle \underset{A}{\int }{\left(z\sin \beta +y\cos \beta \right)}^{\frac{1}{n}}zdA}=M_i\sin \alpha ;$$

(69)

$${\sigma }_0{K}^{{\displaystyle \frac{1}{n}}}{\displaystyle \underset{A}{\int }{\left(z\sin \beta +y\cos \beta \right)}^{\frac{1}{n}}ydA}=M_i\cos \alpha ,$$

(70)

K and $\beta$ can also be determined using this system of equations. Further,

$${\sigma }_{\max }={\sigma }_0{K}^{{\displaystyle \frac{1}{n}}}{\eta }_{\max }^{{\displaystyle \frac{1}{n}}}.$$

(71)

As the system of equations shown above is difficult to solve, solutions can be found that lead more easily to results in particular cases. This shows how effects can be calculated in the case of a rectangular section (Figure 13).

The equilibrium equations are written using the projections on the neutral axis:

$$2{\displaystyle \underset{A^{\prime }}{\int }\sigma \eta dA}+2{\displaystyle \underset{A^{″}}{\int }\sigma \eta dA}=M_i\cos (\beta -\alpha ),$$

(72)

$$2{\displaystyle \underset{A^{\prime }}{\int }\sigma \nu dA}+2{\displaystyle \underset{A^{″}}{\int }\sigma \nu dA}=M_i\sin (\beta -\alpha ),$$

(73)

where

(a)

In the region A’,

$$dA={\displaystyle \frac{b}{\cos \beta }}d\eta ,$$

(74)

$\eta$ varies between 0 and ${\eta }_0$:

$${\eta }_0={\displaystyle \frac{h}{2}}\cos \beta -{\displaystyle \frac{b}{2}}\sin \beta ,$$

(75)

and

$$\nu =\eta tg\beta .$$

(76)

(b)

In the region A”,

$$dA={\displaystyle \frac{b}{\cos \beta }}{\displaystyle \frac{{\eta }_{\max }-\eta }{{\eta }_{\max }-{\eta }_0}}d\eta ,$$

(77)

$\eta$ varies between ${\eta }_0$ and ${\eta }_{\max }$:

$${\eta }_{\max }={\displaystyle \frac{h}{2}}\cos \beta +{\displaystyle \frac{b}{2}}\sin \beta ,$$

(78)

and

$$\nu ={\eta }_0\left(tg\beta +ctg2\beta \right)-\eta ctg2\beta .$$

(79)

Equations (72) and (73) become

$${\displaystyle \frac{2b}{\cos \beta }}\left({\displaystyle {\int }_0^{{\eta }_0}\sigma \eta d\eta }+{\displaystyle {\int }_{{\eta }_0}^{{\eta }_{\max }}\sigma \frac{{\eta }_{\max }-\eta }{{\eta }_{\max }-{\eta }_0}\eta d\eta }\right)=M_i\cos (\beta -\alpha );$$

(80)

$${\displaystyle \frac{2b}{\cos \beta }}\left({\displaystyle {\int }_0^{{\eta }_0}\sigma tg\beta \eta d\eta }+{\displaystyle {\int }_{{\eta }_0}^{{\eta }_{\max }}\sigma \frac{\left({\eta }_0-\eta \cos 2\beta \right)\left({\eta }_{\max }-\eta \right)}{\sin 2\beta \left({\eta }_{\max }-{\eta }_0\right)}\eta d\eta }\right)=M_i\cos (\beta -\alpha ).$$

(81)

Replacing $\sigma$ with the value results in

$$\sigma ={\sigma }_0{\epsilon }^{{\displaystyle \frac{1}{n}}}={\sigma }_0{K}^{{\displaystyle \frac{1}{n}}}{\eta }^{{\displaystyle \frac{1}{n}}}.$$

(82)

This results in

$${\displaystyle {\int }_0^{{\eta }_0}{\eta }^{\frac{n+1}{n}}d\eta }+{\displaystyle \frac{1}{{\eta }_{\max }-{\eta }_0}}{\displaystyle {\int }_{{\eta }_0}^{{\eta }_{\max }}\left({\eta }_{\max }-\eta \right){\eta }^{\frac{n+1}{n}}d\eta }=M_i{\displaystyle \frac{\cos (\beta -\alpha )\cos \beta }{2{\sigma }_0{K}^{\frac{1}{n}}b}}$$

(83)

$$tg\beta {\displaystyle {\int }_0^{{\eta }_0}{\eta }^{\frac{n+1}{n}}d\eta }+{\displaystyle \frac{1}{\left({\eta }_{\max }-{\eta }_0\right)\sin \beta }}{\displaystyle {\int }_{{\eta }_0}^{{\eta }_{\max }}\left({\eta }_0-\eta \cos 2\beta \right)\left({\eta }_{\max }-\eta \right){\eta }^{\frac{n+1}{n}}d\eta }=M_i{\displaystyle \frac{\sin (\beta -\alpha )\cos \beta }{2{\sigma }_0{K}^{\frac{1}{n}}b}}$$

(84)

After integration,

$${\displaystyle \frac{n}{2n+1}}{\eta }_0^{{\displaystyle \frac{2n+1}{n}}}+{\displaystyle \frac{1}{{\eta }_{\max }-{\eta }_0}}\left[{\displaystyle \frac{n}{2n+1}}{\eta }_{\max }\left({\eta }_{\max }^{{\displaystyle \frac{2n+1}{n}}}-{\eta }_0^{{\displaystyle \frac{2n+1}{n}}}\right)-{\displaystyle \frac{n}{3n+1}}\left({\eta }_{\max }^{{\displaystyle \frac{3n+1}{n}}}-{\eta }_0^{{\displaystyle \frac{3n+1}{n}}}\right)\right]=M_i{\displaystyle \frac{\cos (\beta -\alpha )\cos \beta }{2{\sigma }_0{K}^{\frac{1}{n}}b}},$$

(85)

$$\begin{array}{l}{\displaystyle \frac{n}{2n+1}}{\eta }_0^{{\displaystyle \frac{2n+1}{n}}}tg\beta +{\displaystyle \frac{1}{\left({\eta }_{\max }-{\eta }_0\right)\sin 2\beta }}[{\displaystyle \frac{n}{n+1}}{\eta }_0{\eta }_{\max }\left({\eta }_{\max }^{{\displaystyle \frac{n+1}{n}}}-{\eta }_0^{{\displaystyle \frac{n+1}{n}}}\right)-{\displaystyle \frac{n\left({\eta }_0+{\eta }_{\max }\cos 2\beta \right)}{2n+1}}\left({\eta }_{\max }^{{\displaystyle \frac{2n+1}{n}}}-{\eta }_0^{{\displaystyle \frac{2n+1}{n}}}\right)+\\ {\displaystyle \frac{n}{3n+1}}\cos 2\beta \left({\eta }_{\max }^{{\displaystyle \frac{3n+1}{n}}}-{\eta }_0^{{\displaystyle \frac{3n+1}{n}}}\right)]=M_i{\displaystyle \frac{\sin (\beta -\alpha )\cos \beta }{2{\sigma }_0{K}^{\frac{1}{n}}b}}.\end{array}$$

(86)

It is denoted that

$$E_1\left(\beta \right)K^{{\displaystyle \frac{1}{n}}}={\displaystyle \frac{M_i\cos \left(\beta -\alpha \right)\cos \beta }{2{\sigma }_{0}b{K}^{\frac{1}{n}}}},$$

(87)

and

$$E_2\left(\beta \right)={\displaystyle \frac{M_i\sin \left(\beta -\alpha \right)\cos \beta }{2{\sigma }_{0}b{K}^{\frac{1}{n}}{E}_1\left(\beta \right)}}.$$

(88)

By dividing, we obtain

$$tg\left(\beta -\alpha \right)={\displaystyle \frac{E_2\left(\beta \right)}{E_1\left(\beta \right)}}.$$

(89)

Knowing the angle $\beta$, we proceed to the calculation of the coefficient K:

$$K^{{\displaystyle \frac{1}{n}}}={\displaystyle \frac{M_i\cos \left(\beta -\alpha \right)\cos \beta }{2{\sigma }_{0}b{E}_1\left(\beta \right)}}.$$

(90)

Now, the following equation can be calculated:

$${\sigma }_{\max }={\sigma }_0{K}^{{\displaystyle \frac{1}{n}}}{\left({\displaystyle \frac{h}{2}}\cos \beta +{\displaystyle \frac{b}{2}}\sin \beta \right)}^{{\displaystyle \frac{1}{n}}}.$$

(91)

3.6. Bending with Traction

Let a beam be subjected in a section to the bending moment $M_i$ and the axial force N. The section shows symmetry with respect to the y-axis located in the plane of the bending forces (Figure 14). The neutral axis is displaced with respect to the center of gravity by the distance $y_0$. Bernoulli’s hypothesis is expressed by

$$\epsilon =Kt,$$

(92)

where t is the distance from the section to the neutral axis and K is a constant to be determined. The stress is

$$\sigma ={\sigma }_0{\epsilon }^{{\displaystyle \frac{1}{n}}}={\sigma }_0{K}^{{\displaystyle \frac{1}{n}}}{t}^{{\displaystyle \frac{1}{n}}}.$$

(93)

The equilibrium equations are

$${\displaystyle {\int }_A\sigma }dA=N;$$

(94)

$${\displaystyle {\int }_A\sigma t}dA=M_i+N{y}_0,$$

(95)

which can also be written as

$${\sigma }_0{K}^{{\displaystyle \frac{1}{n}}}\left[{\displaystyle {\int }_{A^{″}}{t}^{\frac{1}{n}}}dA-{\displaystyle {\int }_{A^{\prime }}{t}^{\frac{1}{n}}}dA\right]=N;$$

(96)

$${\sigma }_0{K}^{{\displaystyle \frac{1}{n}}}\left[{\displaystyle {\int }_{A^{″}}{t}^{\frac{n+1}{n}}}dA+{\displaystyle {\int }_{A^{\prime }}{t}^{\frac{n+1}{n}}}dA\right]=M_i+N{y}_0.$$

(97)

K and $y_0$ are also determined from this system of equations. The maximum stress is

$${\sigma }_{\max }={\sigma }_0{K}^{{\displaystyle \frac{1}{n}}}{\left(h^n+y_0\right)}^{{\displaystyle \frac{1}{n}}}.$$

(98)

For a rectangular elemental section with dimensions of b, h, dA = bdt and the system of equations becomes

$${\left({\displaystyle \frac{h}{2}}+y_0\right)}^{{\displaystyle \frac{n+1}{n}}}-{\left({\displaystyle \frac{h}{2}}-y_0\right)}^{{\displaystyle \frac{n+1}{n}}}={\displaystyle \frac{n+1}{n}}\cdot {\displaystyle \frac{N}{b{\sigma }_0{K}^{\frac{1}{n}}}},$$

(99)

$${\left({\displaystyle \frac{h}{2}}+y_0\right)}^{{\displaystyle \frac{2n+1}{n}}}+{\left({\displaystyle \frac{h}{2}}-y_0\right)}^{{\displaystyle \frac{2n+1}{n}}}={\displaystyle \frac{2n+1}{n}}\cdot {\displaystyle \frac{M+N\frac{h}{2}v}{b{\sigma }_0{K}^{\frac{1}{n}}}}.$$

(100)

Introducing the variable $v={\displaystyle \frac{2y_0}{h}}$ yields

$${\left(1+v\right)}^{{\displaystyle \frac{n+1}{n}}}-{\left(1-v\right)}^{{\displaystyle \frac{n+1}{n}}}={\displaystyle \frac{n+1}{n}}{\left({\displaystyle \frac{2}{h}}\right)}^{{\displaystyle \frac{n+1}{n}}}{\displaystyle \frac{N}{b{\sigma }_0{K}^{\frac{1}{n}}}},$$

(101)

$${\left(1+v\right)}^{{\displaystyle \frac{2n+1}{n}}}+{\left(1-v\right)}^{{\displaystyle \frac{2n+1}{n}}}={\displaystyle \frac{2n+1}{n}}{\left({\displaystyle \frac{2}{h}}\right)}^{{\displaystyle \frac{2n+1}{n}}}{\displaystyle \frac{M+N\frac{h}{2}v}{b{\sigma }_0{K}^{\frac{1}{n}}}}.$$

(102)

By dividing the two relations, an equation in v is obtained:

$${\displaystyle \frac{{\left(1+v\right)}^{\frac{2n+1}{n}}+{\left(1-v\right)}^{\frac{2n+1}{n}}}{{\left(1+v\right)}^{\frac{n+1}{n}}-{\left(1-v\right)}^{\frac{n+1}{n}}}}-{\displaystyle \frac{2n+1}{n+1}}v={\displaystyle \frac{2n+1}{n+1}}\cdot {\displaystyle \frac{2}{h}}\cdot {\displaystyle \frac{M}{N}}.$$

(103)

If bending is predominant ($0\le v\le 1$), there will generally be both traction and compression effects.

The maximum stress is

$${\sigma }_{\max }={\sigma }_0{K}^{{\displaystyle \frac{1}{n}}}{\left({\displaystyle \frac{h}{2}}+y_0\right)}^{{\displaystyle \frac{1}{n}}}.$$

(104)

3.7. Shear

The problem of shear with bending is studied. The assumptions used by Juravski [58] are kept (Figure 15). The equilibrium of the beam element between sections I and II and between the section through a plane parallel to the neutral plane distanced from it by y gives

$${\displaystyle \underset{A_y}{\int }\left(\sigma +d\sigma \right)}dA-{\displaystyle \underset{A_{y1}}{\int }\sigma }dA-{\tau }_{xy}dx=0.$$

(105)

Introducing the stress values gives

$$\sigma ={\displaystyle \frac{M_i{y}^{\frac{1}{n}}}{I_z}}\mathrm{and}\sigma +d\sigma ={\displaystyle \frac{\left(M_i+d{M}_i\right)y^{\frac{1}{n}}}{I_z}},$$

(106)

which results in

$${\displaystyle {\int }_{A_y}\frac{d{M}_i{y}^{\frac{1}{n}}}{I_z}dA}={\tau }_{xy}bdx.$$

(107)

It is noted that

$${\displaystyle {\int }_{A_y}{y}^{\frac{1}{n}}dA}=S_z,$$

(108)

which results in

$${\tau }_{xy}={\displaystyle \frac{T{S}_z}{b{I}_z}},$$

(109)

In the following equation, the tangential stress distribution is calculated in the case of a bar in a rectangular section (Figure 16):

$$S_z=b{\displaystyle {\int }_{A_y}{y}^{\frac{1}{n}}dA}={\displaystyle \frac{bn}{n+1}}\left[{\left({\displaystyle \frac{h}{2}}\right)}^{{\displaystyle \frac{n+1}{n}}}-y^{{\displaystyle \frac{n+1}{n}}}\right],$$

(110)

With the value of I_z from Equation (49), it is determined using Equation (109) that

$${\tau }_{xy}={\displaystyle \frac{\left(2n+1\right)T}{2(n+1)b}}{\left({\displaystyle \frac{2}{h}}\right)}^{{\displaystyle \frac{2n+1}{n}}}\left[{\left({\displaystyle \frac{h}{2}}\right)}^{{\displaystyle \frac{n+1}{n}}}-y^{{\displaystyle \frac{n+1}{n}}}\right].$$

(111)

The maximum value for y = 0 is

$${\tau }_{\max }={\displaystyle \frac{\left(2n+1\right)T}{2(n+1)A}}.$$

(112)

4. Conclusions

In the case of the nonlinearity of the dependence between stresses and deformations, new problems arose, some very complicated, when we proposed determining the state of strain and stress in an elastic body. The behavior and distribution of stress and strain depend on the constitutive law for the respective material. In this work, the main cases of force loading of simple bars were analyzed where the dependence between strain and stress is a power law. The main differences that appeared compared to the classical treatment, where dependence is linear, were highlighted. The principle of overlapping effects no longer worked, and as a result, calculation became seriously complicated. For some cases encountered more often in practice, calculations were performed within this work. It is obvious that in these cases the main benefit came from the numerical calculations for the encountered situations. The cases presented in this study can constitute a solid basis for further research in this field. The power law was studied, but this analysis can be developed for other situations with other variation laws. In the context of the appearance and use of new and composite materials, the study of such problems becomes important for engineering applications. There are therefore possibilities for the development of research in the future. A wealth of research addressing various forms of nonlinearities and a multitude of real-world scenarios necessitate the study of nonlinear materials. One class of these materials—those whose strain–stress variation follows a power law—was the subject of this paper. A unitary investigation was conducted for these materials in this paper, yielding conclusions that are easily applicable in reality. This gives engineers solutions that are easy to achieve in a short amount of time, which is crucial for a real project.