Determination of Tensile Elastic Parameters from Brazilian Tensile Test: Theory and Experiments

Abstract

Elastic properties of rocks, including Young’s modulus and Poisson’s ratio, are usually obtained from the uniaxial compressive strength (UCS) test, while tensile strength is obtained from direct or indirect (Brazilian) tensile test. In this paper, we developed an analytical model as well as an improved test procedure, called modified Brazilian tensile (MBT) test, which can simultaneously measure the tensile strength, Young’s modulus and Poisson’s ratio in a single test by attaching two strain gauges at the center of each end face of the specimen. To validate the proposed model, both the UCS and modified Brazilian tensile tests were conducted using two types of aluminum alloy (AL 6061 and AL 7075). Results showed that the Young’s modulus obtained from UCS tests and MBT tests are within the range of experimental error. Poisson’s ratios determined from the two test methods were consistent. The proposed model was further validated with Colton sandstones. The Young’s modulus and the Poisson’s ratio obtained from the UCS test and the MBT test agreed well. We also conducted numerical modeling for further comparison, and the results also verified the reasonableness of the proposed model.

Appendices

Appendix A: Derivation of Mathematical Model for Elastic Modulus and Poisson’s Ratio in Brazilian Tensile Test with Different Geometries of Strain Gauges

See Fig. 15.

The solution for an isotropic Brazilian disc subjected to concentrated loads is (Muskhelishvili 1958):

$${\sigma _x}=\frac{{2P}}{{\pi L}}\left\{ {\frac{{\cos {\theta _1}{{\sin }^2}{\theta _1}}}{{{r_1}}}+\frac{{\cos {\theta _2}{{\sin }^2}{\theta _2}}}{{{r_2}}} - \frac{1}{D}} \right\},$$

(30)

$${\sigma _y}= - \frac{{2P}}{{\pi L}}\left\{ {\frac{{{{\cos }^3}{\theta _1}}}{{{r_1}}}+\frac{{{{\cos }^3}{\theta _2}}}{{{r_2}}} - \frac{1}{D}} \right\},$$

(31)

$${\tau _{xy}}=\frac{{2P}}{{\pi L}}\left\{ {\frac{{{{\cos }^2}{\theta _1}\sin {\theta _1}}}{{{r_1}}}+\frac{{{{\cos }^2}{\theta _2}\sin {\theta _2}}}{{{r_2}}}} \right\},$$

(32)

where ${\sigma _x}$ and ${\sigma _y}$ are normal stress along x- and y-directions, respectively, in MPa; ${\tau _{xy}}$ is the shear stress, in MPa; P is the load, in N; D is the diameter of the tested disc, in mm; L is the length of the disc, in mm; ${\theta _1}$ and ${\theta _2}$ are positive when the point A is at the right of load P, and they are negative when A is at left of load P. ${r_1}$ and ${r_2}$ are the distances from point A to the loading points B and C. According to the triangle relations:

$$r_{2}^{2}=r_{1}^{2}+{D^2} - 2{r_1}D\cos {\theta _1},$$

(33)

$$\cos {\theta _2}=\frac{{{D^2}+r_{2}^{2} - r_{1}^{2}}}{{2{r_2}D}}=\frac{{D - {r_1}\cos {\theta _1}}}{{{r_2}}},$$

(34)

$$\sin {\theta _2}=\sqrt {1 - {{\cos }^2}{\theta _2}} =\frac{{{r_1}\sin {\theta _1}}}{{{r_2}}}.$$

(35)

According to right triangle regulation, the coordinate of point A (x, y) is:

$$x={r_1}\sin {\theta _1},$$

(36)

$$y=\frac{D}{2} - {r_1}\cos {\theta _1},$$

(37)

$${r_1}=\sqrt {{{\left( {\frac{D}{2} - y} \right)}^2}+{x^2}} .$$

(38)

Then:

$$\sin {\theta _1}=\frac{x}{{\sqrt {{{\left( {\frac{D}{2} - y} \right)}^2}+{x^2}} }},$$

(39)

$$\cos {\theta _1}=\frac{{2/D - y}}{{\sqrt {{{\left( {\frac{D}{2} - y} \right)}^2}+{x^2}} }}.$$

(40)

Put (38), (39) and (40) into (33), (34) and (35):

$$r_{2}^{2}={\left( {\frac{D}{2}+y} \right)^2}+{x^2},$$

(41)

$$\cos {\theta _2}=\frac{{D/2+y}}{{\sqrt {{{\left( {\frac{D}{2}+y} \right)}^2}+{x^2}} }},$$

(42)

$$\sin {\theta _2}=\frac{x}{{\sqrt {{{\left( {\frac{D}{2}+y} \right)}^2}+{x^2}} }}.$$

(43)

Put Eqs. (38)–(43) into Eqs. (30), (31) and (32), we get the analytical solution of stress state for the isotropic rock disc:

$${\sigma _x}=\frac{{2P}}{{\pi L}}\left\{ {\frac{{\left( {\left( {\frac{D}{2}} \right) - y} \right){x^2}}}{{{{\left( {{{\left( {\left( {\frac{D}{2}} \right) - y} \right)}^2}+{x^2}} \right)}^2}}}+\frac{{\left( {\left( {\frac{D}{2}} \right)+y} \right){x^2}}}{{{{\left( {{{\left( {\left( {\frac{D}{2}} \right)+y} \right)}^2}+{x^2}} \right)}^2}}} - \frac{1}{D}} \right\},$$

(44)

$${\sigma _y}= - \frac{{2P}}{{\pi L}}\left\{ {\frac{{{{\left( {\left( {\frac{D}{2}} \right) - y} \right)}^3}}}{{{{\left( {{{\left( {\left( {\frac{D}{2}} \right) - y} \right)}^2}+{x^2}} \right)}^2}}}+\frac{{{{\left( {\left( {\frac{D}{2}} \right)+y} \right)}^3}}}{{{{\left( {{{\left( {\left( {\frac{D}{2}} \right)+y} \right)}^2}+{x^2}} \right)}^2}}} - \frac{1}{D}} \right\},$$

(45)

$${\tau _{xy}}=\frac{{2P}}{{\pi L}}\left\{ {\frac{{{{\left( {\left( {\frac{D}{2}} \right) - y} \right)}^2}x}}{{{{\left( {{{\left( {\left( {\frac{D}{2}} \right) - y} \right)}^2}+{x^2}} \right)}^2}}} - \frac{{{{\left( {\left( {\frac{D}{2}} \right)+y} \right)}^2}x}}{{{{\left( {{{\left( {\left( {\frac{D}{2}} \right)+y} \right)}^2}+{x^2}} \right)}^2}}}} \right\}.$$

(46)

Two strain gauges, measuring strains in x- and y-directions, are pasted at the center of both sides of the Brazilian disc (Fig. 16). The geometry of both strain gauges are not necessary identical. For general, we assume that two stain gauges’ geometry are not the same. For the strain gauge used to measure tensile strain the half-width is d₁ and the half-length is l₁, see Fig. 16a. For the strain gauge used to measure compressive strain the half-width is d₂ and the half-length is l₂, see Fig. 16b. The width and length of strain gauge maybe varied by strain gauge manufactures.

For strain gauge #1 with half-length of ${l_1}$ and half-width of ${d_1}$, the average stress along x-axial and y-axial are:

$$\overline {{{\sigma _{x1}}}} =\frac{1}{{{d_1}}}\mathop \int \limits_{0}^{{{d_1}}} \frac{{2P}}{{\pi L}}\left\{ {\frac{{\left( {\left( {\frac{D}{2}} \right) - y} \right){d_1}^{2}}}{{{{\left( {{{\left( {\left( {\frac{D}{2}} \right) - y} \right)}^2}+{d_1}^{2}} \right)}^2}}}+\frac{{\left( {\left( {\frac{D}{2}} \right)+y} \right){d_1}^{2}}}{{{{\left( {{{\left( {\left( {\frac{D}{2}} \right)+y} \right)}^2}+{d_1}^{2}} \right)}^2}}} - \frac{1}{D}} \right\}{\text{d}}y,$$

(47)

$$\overline {{{\sigma _{y1}}}} = - \frac{1}{{{l_1}}}\mathop \int \limits_{0}^{{{l_1}}} \frac{{2P}}{{\pi L}}\left\{ {\frac{{{{\left( {\left( {\frac{D}{2}} \right) - {l_1}} \right)}^3}}}{{{{\left( {{{\left( {\left( {\frac{D}{2}} \right) - {l_1}} \right)}^2}+{x^2}} \right)}^2}}}+\frac{{{{\left( {\left( {\frac{D}{2}} \right)+{l_1}} \right)}^3}}}{{{{\left( {{{\left( {\left( {\frac{D}{2}} \right)+{l_1}} \right)}^2}+{x^2}} \right)}^2}}} - \frac{1}{D}} \right\}{\text{d}}x.$$

(48)

Similarly, for strain gauge #2 which is attached to the other end of disc and has half-length of ${l_2}$ and half-width of ${d_2}$, the average stress along x-axial and y-axial are:

$$\overline {{{\sigma _{x2}}}} =\frac{1}{{{l_2}}}\mathop \int \limits_{0}^{{{l_2}}} \frac{{2P}}{{\pi L}}\left\{ {\frac{{\left( {\left( {\frac{D}{2}} \right) - y} \right){l_2}^{2}}}{{{{\left( {{{\left( {\left( {\frac{D}{2}} \right) - y} \right)}^2}+{l_2}^{2}} \right)}^2}}}+\frac{{\left( {\left( {\frac{D}{2}} \right)+y} \right){l_2}^{2}}}{{{{\left( {{{\left( {\left( {\frac{D}{2}} \right)+y} \right)}^2}+{l_2}^{2}} \right)}^2}}} - \frac{1}{D}} \right\}{\text{d}}y,$$

(49)

$$\overline {{{\sigma _{y2}}}} = - \frac{1}{{{d_2}}}\mathop \int \limits_{0}^{{{d_2}}} \frac{{2P}}{{\pi L}}\left\{ {\frac{{{{\left( {\left( {\frac{D}{2}} \right) - {d_2}} \right)}^3}}}{{{{\left( {{{\left( {\left( {\frac{D}{2}} \right) - {d_2}} \right)}^2}+{x^2}} \right)}^2}}}+\frac{{{{\left( {\left( {\frac{D}{2}} \right)+{d_2}} \right)}^3}}}{{{{\left( {{{\left( {\left( {\frac{D}{2}} \right)+{d_2}} \right)}^2}+{x^2}} \right)}^2}}} - \frac{1}{D}} \right\}{\text{d}}x.$$

(50)

Letting,

$${A_{x1}}=\left( {\frac{D}{2}} \right) - {d_1},$$

(51)

$${B_{x1}}=\left( {\frac{D}{2}} \right)+{d_1},$$

(52)

$${A_{y1}}=\left( {\frac{D}{2}} \right) - {l_1},$$

(53)

$${B_{y1}}=\left( {\frac{D}{2}} \right)+{l_1},$$

(54)

$${A_{x2}}=\left( {\frac{D}{2}} \right) - {l_2},$$

(55)

$${B_{x2}}=\left( {\frac{D}{2}} \right)+{l_2},$$

(56)

$${A_{y2}}=\left( {\frac{D}{2}} \right) - {d_2},$$

(57)

$${B_{y2}}=\left( {\frac{D}{2}} \right)+{d_2}.$$

(58)

And integrating Eqs. (47)–(50) gives:

$$\overline {{{\sigma _{x1}}}} =\frac{P}{{\pi L}}\left[ {{d_1}\left( {\frac{1}{{A_{{x1}}^{2}+d_{1}^{2}}} - \frac{1}{{B_{{x1}}^{2}+d_{1}^{2}}}} \right) - \frac{2}{D}} \right],$$

(59)

$$\overline {{{\sigma _{y1}}}} = - \frac{{2P}}{{\pi {l_1}L}}\left\{ {\frac{{{A_{y1}}^{3}}}{2}\left[ {\frac{1}{{{A_{y1}}^{3}}}\arctan \frac{{{l_1}}}{{{A_{y1}}}}+\frac{{{l_1}}}{{{A_{y1}}^{2}({l_1}^{2}+{A_{y1}}^{2})}}} \right]+\frac{{{B_{y1}}^{3}}}{2}\left[ {\frac{1}{{{B_{y1}}^{3}}}\arctan \frac{{{l_1}}}{{{B_{y1}}}}+\frac{{{l_1}}}{{{B_{y1}}^{2}({l_1}^{2}+{B_{y1}}^{2})}}} \right] - \frac{{{l_1}}}{D}} \right\},$$

(60)

$$\overline {{{\sigma _{x2}}}} =\frac{P}{{\pi L}}\left[ {{l_2}\left( {\frac{1}{{A_{{x2}}^{2}+l_{2}^{2}}} - \frac{1}{{B_{{x2}}^{2}+l_{2}^{2}}}} \right) - \frac{2}{D}} \right],$$

(61)

$$\overline {{{\sigma _{y2}}}} = - \frac{{2P}}{{\pi {d_2}L}}\left\{ {\frac{{{A_{y2}}^{3}}}{2}\left[ {\frac{1}{{{A_{y2}}^{3}}}\arctan \frac{{{d_2}}}{{{A_{y2}}}}+\frac{{{d_2}}}{{{A_{y2}}^{2}({d_2}^{2}+{A_{y2}}^{2})}}} \right]+\frac{{{B_{y2}}^{3}}}{2}\left[ {\frac{1}{{{B_{y2}}^{3}}}\arctan \frac{{{d_2}}}{{{B_{y2}}}}+\frac{{{d_2}}}{{{B_{y2}}^{2}({d_2}^{2}+{B_{y2}}^{2})}}} \right] - \frac{{{d_2}}}{D}} \right\}.$$

(62)

For a given specimen and strain gauges, disc diameter$~D$, disc length $L$ and strain gauge geometry parameters ${l_1}$, ${d_1}$, ${l_2}$ and ${d_2}$are constants. Grou** those constants to simplify Eqs. (59–62) and letting:

$${F_1}=\frac{1}{{\pi L}}\left[ {{d_1}\left( {\frac{1}{{A_{{x1}}^{2}+d_{1}^{2}}} - \frac{1}{{B_{{x1}}^{2}+d_{1}^{2}}}} \right) - \frac{2}{D}} \right],$$

(63)

$${G_1}= - \frac{1}{{\pi {l_1}L}}\left\{ {{A_{y1}}^{3}\left[ {\frac{1}{{{A_{y1}}^{3}}}\arctan \frac{{{l_1}}}{{{A_{y1}}}}+\frac{{{l_1}}}{{{A_{y1}}^{2}({l_1}^{2}+{A_{y1}}^{2})}}} \right]+{B_{y1}}^{3}\left[ {\frac{1}{{{B_{y1}}^{3}}}\arctan \frac{{{l_1}}}{{{B_{y1}}}}+\frac{{{l_1}}}{{{B_{y1}}^{2}({l_1}^{2}+{B_{y1}}^{2})}}} \right] - \frac{{2{l_1}}}{D}} \right\},$$

(64)

$${F_2}=\frac{1}{{\pi L}}\left[ {{l_2}\left( {\frac{1}{{A_{{x2}}^{2}+l_{2}^{2}}} - \frac{1}{{B_{{x2}}^{2}+l_{2}^{2}}}} \right) - \frac{2}{D}} \right],$$

(65)

$${G_2}= - \frac{1}{{\pi {d_2}L}}\left\{ {{A_{y2}}^{3}\left[ {\frac{1}{{{A_{y2}}^{3}}}\arctan \frac{{{d_2}}}{{{A_{y2}}}}+\frac{{{d_2}}}{{{A_{y2}}^{2}({d_2}^{2}+{A_{y2}}^{2})}}} \right]+{B_{y2}}^{3}\left[ {\frac{1}{{{B_{y2}}^{3}}}\arctan \frac{{{d_2}}}{{{B_{y2}}}}+\frac{{{d_2}}}{{{B_{y2}}^{2}({d_2}^{2}+{B_{y2}}^{2})}}} \right] - \frac{{2{d_2}}}{D}} \right\}.$$

(66)

We can rewrite Eqs. (49)–(62) as:

$$\overline {{{\sigma _{x1}}}} =P{F_1},$$

(67)

$$\overline {{{\sigma _{y1}}}} =P{G_1}$$

(68)

$$\overline {{{\sigma _{x2}}}} =P{F_2},$$

(69)

$$\overline {{{\sigma _{y2}}}} =P{G_2}.$$

(70)

According to linearly elasticity theory, the tensile strain (${\varepsilon _x}$) and compressive strain (${\varepsilon _y}$) are generated by both $\overline {{{\sigma _x}}}$ and $~\overline {{{\sigma _y}}}$.

$${\varepsilon _x}=\frac{1}{{{E_{\text{t}}}}}\left( {\overline {{{\sigma _{x1}}}} - {{\nu}_{\text{t}}}\overline {{{\sigma _{y1}}}} } \right),$$

(71)

$${\varepsilon _y}=\frac{1}{{{E_{\text{t}}}}}\left( { - {{\nu}_{\text{t}}}\overline {{{\sigma _{x2}}}} +\overline {{{\sigma _{y2}}}} } \right),$$

(72)

where ${\varepsilon _x}$ is strain in x-direction; ${\varepsilon _y}$ is strain in y direction; ${E_{\text{t}}}$ is the tensile elastic modulus; and ${{\nu}_{\text{t}}}$ is the tensile Poisson’s ratio.

Substituting Eqs. (67)–(70) into Eqs. (71) and (72) gives:

$${\varepsilon _x}=\frac{P}{{{E_{\text{t}}}}}\left( {{F_1} - {{\nu}_{\text{t}}}{{\text{G}}_1}} \right),$$

(73)

$${\varepsilon _y}=\frac{P}{{{E_{\text{t}}}}}\left( { - {{\nu}_{\text{t}}}{F_2}+{{\text{G}}_2}} \right).$$

(74)

From Eqs. (73) and (74), we can obtain the ratio of the tensile strain (${\varepsilon _x}$) to compressive strain (${\varepsilon _y}$) ratio:

$$\frac{{{\varepsilon _x}}}{{{\varepsilon _y}}}=\frac{{{F_1} - {{\nu}_{\text{t}}}{{\text{G}}_1}}}{{ - {{\nu}_{\text{t}}}{F_2}+{{\text{G}}_2}}}.$$

(75)

Then tensile Poisson’s ratio can be expressed as:

$${{\nu}_{\text{t}}}=\frac{{{F_1} - \frac{{{\varepsilon _x}}}{{{\varepsilon _y}}}{G_2}}}{{{G_1} - \frac{{{\varepsilon _x}}}{{{\varepsilon _y}}}{F_2}}}.$$

(76)

Exchanging ${\varepsilon _x}$ and ${E_{\text{t}}}$ in Eq. (71), we get tensile elastic modulus formula related to $P,~{\varepsilon _x}$ and ${{\nu}_{\text{t}}}$:

$${E_{\text{t}}}=\frac{P}{{{\varepsilon _x}}}\left( {{F_1} - {{\nu}_{\text{t}}}{{\text{G}}_1}} \right).$$

(77)

Also, exchanging ${\varepsilon _y}$ and ${E_{\text{t}}}$ in Eq. (72), we get another tensile elastic modulus formula related to $P,~{\varepsilon _y}$ and ${{\nu}_{\text{t}}}$:

$${E_{\text{t}}}=\frac{P}{{{\varepsilon _y}}}\left( { - {{\nu}_{\text{t}}}{F_2}+{{\text{G}}_2}} \right).$$

(78)

As can be seen from Eqs. (76, 77) (and/or Eq. 78), both Poisson’s ratio and Young’s modulus are function of load (P) and radial and axial strains (${\varepsilon _x}$ and ${\varepsilon _y}$), and ${\varepsilon _x}$ and ${\varepsilon _y}$ are stress dependent. To eliminate stress dependent elastic constants, we use an alternative method based on the superposition principle. Since the elastic properties are obtained from the linear part of stress–strain curve, the tensile Poisson’s ratio and Young’s modulus can also be written as:

$${{\nu}_{\text{t}}}=\frac{{{F_1} - \frac{{\Delta {\varepsilon _x}}}{{\Delta {\varepsilon _y}}}{G_1}}}{{{G_1} - \frac{{\Delta {\varepsilon _x}}}{{\Delta {\varepsilon _y}}}{F_2}}},$$

(79)

$${E_{\text{t}}}=\frac{{\Delta P}}{{\Delta {\varepsilon _x}}}\left( {{F_1} - {{\nu}_{\text{t}}}{{\text{G}}_1}} \right),$$

(80)

$${E_{\text{t}}}=\frac{{\Delta P}}{{\Delta {\varepsilon _y}}}\left( { - {{\nu}_{\text{t}}}{{\text{F}}_2}+{{\text{G}}_2}} \right).$$

(81)

In this way, we can determine tensile Poisson’s ratio and tensile Young’s modulus from a single Brazilian tensile test.

Note The stress that used to calculate the tensile Young’s Modulus and Poisson’s ratio is the average stress in the strain gauge measured area. If the strain gauge measured area is small enough, the stress is uniform distributed in this area and the average stress can be used to calculate the tensile Young’s modulus and Poisson’s ratio (Figs. 8, 9). However, if the strain gauge measure area is so large that exceed the stress uniform distributed section, the calculation is questionable. Thus, the strain gauge dimension (length/width) must be optimized considering the dimension of the test sample.

Appendix B: Detailed Integration of Average Tensile Stress and Average Compressive Stress

• 1. Integration of average tensile stress

  The average stress along x-axis is:

  $$\overline {{{\sigma _x}}} =\frac{1}{l}\mathop \int \limits_{0}^{l} \frac{{2P}}{{\pi L}}\left\{ {\frac{{\left( {\left( {\frac{D}{2}} \right) - y} \right){l^2}}}{{{{\left( {{{\left( {\left( {\frac{D}{2}} \right) - y} \right)}^2}+{l^2}} \right)}^2}}}+\frac{{\left( {\left( {\frac{D}{2}} \right)+y} \right){l^2}}}{{{{\left( {{{\left( {\left( {\frac{D}{2}} \right)+y} \right)}^2}+{l^2}} \right)}^2}}} - \frac{1}{D}} \right\}{\text{d}}y,$$

  (82)
  $$\overline {{{\sigma _x}}} =\frac{{2P}}{{\pi Ll}}\left\{ {\mathop \int \limits_{0}^{l} \left[ {\frac{{\left( {\left( {\frac{D}{2}} \right) - y} \right){l^2}}}{{{{\left( {{{\left( {\left( {\frac{D}{2}} \right) - y} \right)}^2}+{l^2}} \right)}^2}}}} \right]dy+\mathop \int \limits_{0}^{l} \left[ {\frac{{\left( {\left( {\frac{D}{2}} \right)+y} \right){l^2}}}{{{{\left( {{{\left( {\left( {\frac{D}{2}} \right)+y} \right)}^2}+{l^2}} \right)}^2}}}} \right]{\text{d}}y - \mathop \int \limits_{0}^{l} \frac{1}{D}{\text{d}}y} \right\}.$$

  (83)

  We integrate Eq. (83) separately:

  • 1.1
    $$\mathop \int \limits_{0}^{l} \left[ {\frac{{\left( {\left( {\frac{D}{2}} \right) - y} \right){l^2}}}{{{{\left( {{{\left( {\left( {\frac{D}{2}} \right) - y} \right)}^2}+{l^2}} \right)}^2}}}} \right]{\text{d}}y.$$

    Let $u=\left( {\frac{D}{2}} \right) - y,$

    $$\mathop \int \limits_{0}^{l} \left[ {\frac{{\left( {\left( {\frac{D}{2}} \right) - y} \right){l^2}}}{{{{({{(\left( {\frac{D}{2}} \right) - y)}^2}+{l^2})}^2}}}} \right]{\text{d}}y= - \mathop \int \limits_{{\frac{D}{2}}}^{{\frac{D}{2} - l}} \left[ {\frac{{u{l^2}}}{{{{({u^2}+{l^2})}^2}}}} \right]{\text{d}}u= - \frac{{{l^2}}}{2}\mathop \int \limits_{{\frac{D}{2}}}^{{\frac{D}{2} - l}} \left[ {\frac{1}{{{{({u^2}+{l^2})}^2}}}} \right]{\text{d}}{u^2}= - \frac{{{l^2}}}{2}\mathop \int \limits_{{\frac{D}{2}}}^{{\frac{D}{2} - l}} \left[ {\frac{1}{{{{({u^2}+{l^2})}^2}}}} \right]{\text{d}}({u^2}+{l^2})=\frac{{{l^2}}}{2}\left[ {\frac{1}{{{u^2}+{l^2}}}} \right]_{{\frac{D}{2}}}^{{\frac{D}{2} - l}}=\frac{{{l^2}}}{2}\left[ {\frac{1}{{{{\left( {\frac{D}{2} - l} \right)}^2}+{l^2}}} - \frac{1}{{{{\left( {\frac{D}{2}} \right)}^2}+{l^2}}}} \right]$$

    (84)
  • 1.2
    $$~\mathop \int \limits_{0}^{l} \left[ {\frac{{\left( {\left( {\frac{D}{2}} \right)+y} \right){l^2}}}{{{{\left( {{{\left( {\left( {\frac{D}{2}} \right)+y} \right)}^2}+{l^2}} \right)}^2}}}} \right]{\text{d}}y.$$

    Let $v=\left( {\frac{D}{2}} \right)+y$

    $$\begin{gathered} \mathop \int \limits_{0}^{l} \left[ {\frac{{\left( {\left( {\frac{D}{2}} \right)+y} \right){l^2}}}{{{{\left( {{{\left( {\left( {\frac{D}{2}} \right)+y} \right)}^2}+{l^2}} \right)}^2}}}} \right]{\text{d}}y={l^2}\mathop \int \limits_{{\frac{D}{2}}}^{{\frac{D}{2}+l}} \left[ {\frac{v}{{{{({v^2}+{l^2})}^2}}}} \right]{\text{d}}v=\frac{{{l^2}}}{2}\mathop \int \limits_{{\frac{D}{2}}}^{{\frac{D}{2}+l}} \left[ {\frac{1}{{{{({v^2}+{l^2})}^2}}}} \right]{\text{d}}{v^2}=\frac{{{l^2}}}{2}\mathop \int \limits_{{\frac{D}{2}}}^{{\frac{D}{2}+l}} \left[ {\frac{1}{{{{({v^2}+{l^2})}^2}}}} \right]{\text{d}}({v^2}+{l^2})=\frac{{{l^2}}}{2}\left[ { - \frac{1}{{{v^2}+{l^2}}}} \right]_{{\frac{D}{2}}}^{{\frac{D}{2}+l}}= - \frac{{{l^2}}}{2}\left[ {\frac{1}{{{{\left( {\frac{D}{2}+l} \right)}^2}+{l^2}}} - \frac{1}{{{{\left( {\frac{D}{2}} \right)}^2}+{l^2}}}} \right] \end{gathered}$$

    (85)
  • 1.3
    $$- \mathop \int \limits_{0}^{l} \frac{1}{D}{\text{d}}y= - \frac{l}{D}.$$

    Then put the results of 1.1, 1.2 and 1.3 into the following equation:

    $$\begin{aligned} \overline {{{\sigma _x}}} & =\frac{{2P}}{{\pi Ll}}\left\{ {\mathop \int \limits_{0}^{l} \left[ {\frac{{\left( {\left( {\frac{D}{2}} \right) - y} \right){l^2}}}{{{{\left( {{{\left( {\left( {\frac{D}{2}} \right) - y} \right)}^2}+{l^2}} \right)}^2}}}} \right]{\text{d}}y+\mathop \int \limits_{0}^{l} \left[ {\frac{{\left( {\left( {\frac{D}{2}} \right)+y} \right){l^2}}}{{{{({{(\left( {\frac{D}{2}} \right)+y)}^2}+{l^2})}^2}}}} \right]{\text{d}}y - \mathop \int \limits_{0}^{l} \frac{1}{D}{\text{d}}y} \right\} \\ & =\frac{{2P}}{{\pi Ll}}\left\{ {\frac{{{l^2}}}{2}\left[ {\frac{1}{{{{\left( {\frac{D}{2} - l} \right)}^2}+{l^2}}} - \frac{1}{{{{\left( {\frac{D}{2}} \right)}^2}+{l^2}}}} \right] - \frac{{{l^2}}}{2}\left[ {\frac{1}{{{{\left( {\frac{D}{2}+l} \right)}^2}+{l^2}}} - \frac{1}{{{{\left( {\frac{D}{2}} \right)}^2}+{l^2}}}} \right] - \frac{l}{D}} \right\} \\ & =\frac{P}{{\pi Ll}}\left\{ {{l^2}\left[ {\frac{1}{{{{\left( {\frac{D}{2} - l} \right)}^2}+{l^2}}} - \frac{1}{{{{\left( {\frac{D}{2}+l} \right)}^2}+{l^2}}}} \right] - \frac{{2l}}{D}} \right\}. \\ \end{aligned}$$

    (86)

    Let

    $$A=\left( {\frac{D}{2}} \right) - l,$$

    (87)
    $$B=\left( {\frac{D}{2}} \right)+l.$$

    (88)

    Then

    $$\overline {{{\sigma _x}}} =\frac{P}{{\pi Ll}}\left\{ {{l^2}\left[ {\frac{1}{{{A^2}+{l^2}}} - \frac{1}{{{B^2}+{l^2}}}} \right] - \frac{{2l}}{D}} \right\},$$

    (89)
    $$\overline {{{\sigma _x}}} =\frac{P}{{\pi L}}\left\{ {l\left[ {\frac{1}{{{A^2}+{l^2}}} - \frac{1}{{{B^2}+{l^2}}}} \right] - \frac{2}{D}} \right\}.$$

    (90)

• 2. Integration of average compressive stress

  The average stress along y-axis is:

  $$\overline {{{\sigma _y}}} = - \frac{1}{l}\mathop \int \limits_{0}^{l} \frac{{2P}}{{\pi L}}\left\{ {\frac{{{{\left( {\left( {\frac{D}{2}} \right) - l} \right)}^3}}}{{{{\left( {{{\left( {\left( {\frac{D}{2}} \right) - l} \right)}^2}+{x^2}} \right)}^2}}}+\frac{{{{\left( {\left( {\frac{D}{2}} \right)+l} \right)}^3}}}{{{{\left( {{{\left( {\left( {\frac{D}{2}} \right)+l} \right)}^2}+{x^2}} \right)}^2}}} - \frac{1}{D}} \right\}{\text{d}}x,$$

  (91)
  $$\overline {{{\sigma _y}}} = - \frac{{2P}}{{\pi lL}}\mathop \int \limits_{0}^{l} \left\{ {\frac{{{{\left( {\left( {\frac{D}{2}} \right) - l} \right)}^3}}}{{{{\left( {{{\left( {\left( {\frac{D}{2}} \right) - l} \right)}^2}+{x^2}} \right)}^2}}}+\frac{{{{\left( {\left( {\frac{D}{2}} \right)+l} \right)}^3}}}{{{{\left( {{{\left( {\left( {\frac{D}{2}} \right)+l} \right)}^2}+{x^2}} \right)}^2}}} - \frac{1}{D}} \right\}{\text{d}}x.$$

  (92)

  Put Eq. (87) and Eq. (88) into Eq. (92), get

  $$\overline {{{\sigma _y}}} = - \frac{{2P}}{{\pi lL}}\mathop \int \limits_{0}^{l} \left\{ {\frac{{{A^3}}}{{({{({A^2}+{x^2})}^2}}}+\frac{{{B^3}}}{{{{({B^2}+{x^2})}^2}}} - \frac{1}{D}} \right\}{\text{d}}x,$$

  (93)
  $$\overline {{{\sigma _y}}} = - \frac{{2P}}{{\pi lL}}\left\{ {\mathop \int \limits_{0}^{l} \left[ {\frac{{{A^3}}}{{({{({A^2}+{x^2})}^2}}}} \right]{\text{d}}x+\mathop \int \limits_{0}^{l} \left[ {\frac{{{B^3}}}{{({{({B^2}+{x^2})}^2}}}} \right]{\text{d}}x - \mathop \int \limits_{0}^{l} \frac{1}{D}{\text{d}}x} \right\}.$$

  (94)

  We integrate Eq. (94) separately:

  • 2.1
    $$\mathop \int \limits_{0}^{l} \left[ {\frac{{{A^3}}}{{({{({A^2}+{x^2})}^2}}}} \right]{\text{d}}x.$$

    Let $\frac{x}{A}=v$, then

    $$\mathop \int \limits_{0}^{l} \left[ {\frac{{{A^3}}}{{({{({A^2}+{x^2})}^2}}}} \right]{\text{d}}x=\mathop \int \limits_{0}^{{\frac{l}{A}}} \frac{1}{{{{(1+{v^2})}^2}}}{\text{d}}v.$$

    (95)

    Let

    $$v=\tan t.$$

    (96)

    Then ${\text{d}}v={\sec ^2}t{\text{d}}t,$

    $$\mathop \int \limits_{0}^{{\frac{l}{A}}} \frac{1}{{{{(1+{v^2})}^2}}}{\text{d}}v=\mathop \int \limits_{0}^{{\arctan \frac{l}{A}}} \frac{1}{{{{\sec }^4}t}}{\sec ^2}t{\text{d}}t=\mathop \int \limits_{0}^{{\arctan \frac{l}{A}}} {\cos ^2}t{\text{d}}t=\mathop \int \limits_{0}^{{\arctan \frac{l}{A}}} \left( {\frac{{~\cos 2t+1}}{2}} \right){\text{d}}t=\frac{1}{2}[\sin t\cos t]_{0}^{{\arctan \frac{l}{A}}}+\frac{1}{2}\arctan \frac{l}{A},$$

    (97)
    $$\sin t\cos t=\frac{{\sin t{{\cos }^2}t}}{{\cos t}}=\tan t\frac{1}{{{{\sec }^2}t}}=\frac{{\tan t}}{{{{\tan }^2}t+1}}.$$

    (98)

    Put Eq. (96) into Eq. (97).

    Then

    $$\sin t\cos t=\frac{v}{{{v^2}+1}}.$$

    (99)

    Put Eq. (99) into Eq. (97):

    $$\mathop \int \limits_{0}^{{\frac{l}{A}}} \frac{1}{{{{(1+{v^2})}^2}}}{\text{d}}v=\frac{1}{2}[\sin t\cos t]_{0}^{{\arctan \frac{l}{A}}}+\frac{1}{2}\arctan \frac{l}{A}=\frac{1}{2}\left[ {\frac{v}{{{v^2}+1}}} \right]_{0}^{{\frac{l}{A}}}+\frac{1}{2}\arctan \frac{l}{A}=\frac{1}{2}\left( {\frac{{lA}}{{{l^2}+{A^2}}}} \right)+\frac{1}{2}\arctan \frac{l}{A}=\frac{{{A^3}}}{2}\left[ {\frac{l}{{{A^2}({l^2}+{A^2})}}+\frac{1}{{{A^3}}}\arctan \frac{l}{A}} \right].$$

    (100)

  • 2.2 Similarly, we get the second part of Eq. (94):

    $$\mathop \int \limits_{0}^{l} \left[ {\frac{{B^{3} }}{{((B^{2} + x^{2} )^{2} }}} \right]{\text{d}}x = \frac{{B^{3} }}{2}\left[ {\frac{l}{{B^{2} (l^{2} + B^{2} )}} + \frac{1}{{B^{3} }}\arctan \frac{l}{B}} \right].$$
  • 2.3
    $$- \,\mathop \int \limits_{0}^{l} \frac{1}{D}{\text{d}}x= - \frac{l}{D}.$$

    Then put the integration results of 2.1, 2.2 and 2.3 into Eq. (94):

    $$\begin{aligned} \overline {{{\sigma _y}}} & = - \frac{{2P}}{{\pi lL}}\left\{ {\mathop \int \limits_{0}^{l} \left[ {\frac{{{A^3}}}{{({{({A^2}+{x^2})}^2}}}} \right]{\text{d}}x+\mathop \int \limits_{0}^{l} \left[ {\frac{{{B^3}}}{{({{({B^2}+{x^2})}^2}}}} \right]{\text{d}}x - \mathop \int \limits_{0}^{l} \frac{1}{D}{\text{d}}x} \right\} \\ ~ & = - \frac{{2P}}{{\pi lL}}\left\{ {\frac{{{A^3}}}{2}\left[ {\frac{l}{{{A^2}({l^2}+{A^2})}}+\frac{1}{{{A^3}}}\arctan \frac{l}{A}} \right]+\frac{{{B^3}}}{2}\left[ {\frac{l}{{{B^2}({l^2}+{B^2})}}+\frac{1}{{{B^3}}}\arctan \frac{l}{B}} \right] - \frac{l}{D}} \right\}. \\ \end{aligned}$$

    (101)