### Bearing capacity

Presently, the mainstream theory for calculating the bearing performance of a bolted joint in timber engineering is the EYM, which assumes that the external force undertaken by the joint is the yield force *P*_{y} when the timber reaches the dowel bearing yield strength *f*_{e,y} or the bolt reaches the yield moment *M*_{y}. Scholars [6, 17] also used the dowel bearing ultimate strength of wood *f*_{e,u} and the ultimate moment of the bolt *M*_{u} to calculate the bearing capacity of the bolted joint *P*_{p} by the yield theory. This section presents the following assumptions: (1) the force–displacement relations for the dowel bearing behavior of wood are ideal elastoplasticity; and (2) when the full length of the timber or the timber segment between two plastic hinges reaches the dowel bearing ultimate strength of the wood, the load is the bearing capacity of the bolted joint. The parameters include the bolt diameter *d*, the timber specimen thickness *l*, the ultimate bending moment for the bolt section *M*_{u}, and the dowel bearing ultimate strength of the wood *f*_{e,u}.

The free body diagrams in the ultimate state for three failure patterns are given in Fig. 10.

The equation of equilibrium for Pattern I is as follows:

$$\frac{{P_{{\text{p}}} }}{2} = f_{{\text{e,u}}} \cdot d \cdot \frac{l}{2}.$$

(2)

The equations of equilibrium for Pattern III are as follows:

$$\frac{{P_{{\text{p}}} }}{2} + f_{{\text{e,u}}} \cdot d \cdot a = f_{{\text{e,u}}} \cdot d(\frac{1}{2}l - a),$$

(3)

$$M_{{\text{u}}} + f_{{\text{e,u}}} \cdot d \cdot a(\frac{1}{2}l - \frac{1}{2}a) = \frac{1}{2}f_{{\text{e,u}}} \cdot d(\frac{1}{2}l - a)^{2} .$$

(4)

The equations of equilibrium for Pattern IV are as follows:

$$\frac{{P_{{\text{p}}} }}{2} = f_{{\text{e,u}}} \cdot d \cdot b,$$

(5)

$$2M_{{\text{u}}} = \frac{1}{2}f_{{\text{e,u}}} \cdot d \cdot b^{2} .$$

(6)

Solving Eqs. (2)–(6), we can express the bearing capacity of the joint as follows:

$$\left\{ \begin{gathered} P_{{\text{p}}} = f_{{\text{e,u}}} \cdot d \cdot l{\kern 1pt} {\kern 1pt} \left( {{\text{Pattern}}{\kern 1pt} {\kern 1pt} {\text{I}}} \right){\kern 1pt} {\kern 1pt} \hfill \\ P_{{\text{p}}} = f_{{\text{e,u}}} \cdot d \cdot l\left( {\sqrt {2 + \frac{{16M_{{\text{u}}} }}{{f_{{\text{e,u}}} \cdot d \cdot l^{2} }}} - 1} \right){\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \left( {{\text{Pattern}}{\kern 1pt} {\kern 1pt} {\text{III}}} \right) \hfill \\ P_{{\text{p}}} = 4\sqrt {M_{{\text{u}}} \cdot f_{{\text{e,u}}} \cdot d} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \left( {{\text{Pattern}}{\kern 1pt} {\kern 1pt} {\text{IV}}} \right) \hfill \\ \end{gathered} \right..$$

(7)

### Elastic stiffness

The elastic stiffness is analyzed by supposing that (1) the contact relation between the bolt and the timber is a Euler–Bernoulli beam acting on an elastic foundation [18, 19]; (2) the contact stiffness of the points on the elastic foundation is taken as the dowel bearing stiffness of the wood; and (3) the constraint influence of the nut and washer is not considered. The parameters include the modulus of elasticity of the steel bolt *E* and the dowel bearing stiffness of the wood *k*_{s}.

Because single-bolted joints are bilaterally symmetric (Fig. 11a), only the right portion is analyzed. The positive directions of the *x* and *w* axes and the position of the origin are shown in Fig. 11b. The free body diagram for a bolt microelement at any *x* is shown in Fig. 11c.

From the force balance in the *w*-axis direction, we have

$$V - k_{{\text{s}}} w(x){\text{d}}x - (V + {\text{d}}V) = 0.$$

(8)

Hence,

$$\frac{{{\text{d}}V}}{{{\text{d}}x}} + k_{{\text{s}}} w(x) = 0.$$

(9)

The relations for the shear force, bending moment, and curvature of the Euler–Bernoulli beam are as follows:

$$V = \frac{{{\text{d}}M}}{{{\text{d}}x}},$$

(10)

$$M = {EI}\frac{{{\text{d}}{}^{2}w(x)}}{{{\text{d}}x^{2} }}.$$

(11)

From formulas (9)–(11), we have the following:

$$\frac{{{\text{d}}{}^{4}w(x)}}{{{\text{d}}x^{4} }} + \frac{{k_{{\text{s}}} }}{{{EI}}}w(x) = 0.$$

(12)

With \(l_{{\text{c}}} = \sqrt[4]{{\frac{{4{EI}}}{{k_{{\text{s}}} }}}}\) and \(\xi = \frac{x}{{l_{{\text{c}}} }}\), we obtain the following equation:

$$\frac{1}{{l_{{\text{c}}}^{4} }}\frac{{{\text{d}}{}^{4}w(\xi )}}{{{\text{d}}\xi^{4} }} + \frac{4}{{l_{{\text{c}}}^{4} }}w(\xi ) = 0.$$

(13)

Hence,

$$\frac{{{\text{d}}{}^{4}w(\xi )}}{{{\text{d}}\xi^{4} }} + 4w(\xi ) = 0.$$

(14)

This homogeneous differential equation is then solved, and the general solution is expressed below:

$$w(\xi ) = A_{1} e^{\xi } (\cos \xi + i\sin \xi ) + A_{2} e^{\xi } (\cos \xi - i\sin \xi ){\kern 1pt} {\kern 1pt} {\kern 1pt} + A_{3} e^{ - \xi } (\cos \xi + i\sin \xi ) + A_{4} e^{ - \xi } (\cos \xi - i\sin \xi ).$$

(15)

In formula (15), *A*_{1}, *A*_{2}, *A*_{3,} and *A*_{4} are constants. With \(B_{1} = A_{1} + A_{2}\), \(B_{2} = A_{1} i - A_{2} i\), \(B_{3} = A_{3} + A_{4}\) and \(B_{4} = A_{3} i - A_{4} i\), we obtain the following formula:

$$w(\xi ) = B_{1} e^{\xi } \cos \xi + B_{2} e^{\xi } \sin \xi + B_{3} e^{ - \xi } \cos \xi + B_{4} e^{ - \xi } \sin \xi .$$

(16)

As *x* (namely, \(\xi\)) increases, the deflection *w* gradually approaches 0, while in formula (16), the terms containing \(e^{\xi } \cos \xi\) and \(e^{\xi } \sin \xi\) diverge as *x* increases; thus, \(B_{1} = B_{2} = 0\).

With \(g_{1} = e^{ - \xi } \cos \xi\), \(g_{2} = e^{ - \xi } \sin \xi\), \(g_{3} = g_{1} + g_{2}\), and \(g_{4} = g_{1} - g_{2}\), we have \(\frac{{{\text{d}}g_{1} }}{{{\text{d}}\xi }} = - g_{3}\), \(\frac{{{\text{d}}g_{2} }}{{{\text{d}}\xi }} = g_{4}\), \(\frac{{{\text{d}}g_{3} }}{{{\text{d}}\xi }} = - 2g_{2}\), and \(\frac{{{\text{d}}g_{4} }}{{{\text{d}}\xi }} = - 2g_{1}\). Therefore, the deflection *w*, angle of rotation \(\theta\), bending moment *M,* and shear force *V* at any *x* are expressed as follows:

$$w(\xi ) = B_{3} g_{1} + B_{4} g_{2} ,$$

(17)

$$\theta (\xi ) = \frac{{{\text{d}}w(x)}}{{{\text{d}}x}} = \frac{1}{{l_{{\text{c}}} }}( - B_{3} g_{3} + B_{4} g_{4} ),$$

(18)

$$M(\xi ) = {EI}\frac{{{\text{d}}{}^{2}w(x)}}{{{\text{d}}x^{2} }} = \frac{{2{EI}}}{{l_{{\text{c}}}^{2} }}(B_{3} g_{2} - B_{4} g_{1} ),$$

(19)

$$V(\xi ) = \frac{{{\text{d}}M(x)}}{{{\text{d}}x}} = \frac{{2{EI}}}{{l_{{\text{c}}}^{3} }}(B_{3} g_{4} + B_{4} g_{3} ).$$

(20)

At \(x = 0\) (namely, \(\xi = 0\)), the shear force \(V_{0}\) and bending moment \(M_{0}\) (Fig. 12b) can be substituted into formulas (19) and (20) to obtain the following formulas:

$$B_{3} = \frac{{l_{{\text{c}}} }}{{2{EI}}}(l_{{\text{c}}} V_{0} + M_{0} ),$$

(21)

$$B_{4} = - \frac{{l_{{\text{c}}} }}{{2{EI}}}M_{0} .$$

(22)

Thus, formulas (17)–(20) can be expressed as follows:

$$w = \frac{{l_{{\text{c}}}^{2} }}{{2{EI}}}(l_{{\text{c}}} V_{0} g_{1} + M_{0} g_{4} ),$$

(23)

$$\theta = - \frac{{l_{{\text{c}}} }}{{2{EI}}}(l_{{\text{c}}} V_{0} g_{3} + 2M_{0} g_{1} ),$$

(24)

$$M = l_{{\text{c}}} V_{0} g_{2} + M_{0} g_{3} ,$$

(25)

$$V = V_{0} g_{4} - \frac{2}{{l_{{\text{c}}} }}M_{0} g_{2} .$$

(26)

The elastic stiffness *k*_{e} of the bolted joint is described as the proportional relation of the external force *P* with the displacement \(\Delta\) of the midpoint of the bolt. \(\Delta\) is composed of the deflection of the bolt on the timber foundation and the deflection of the bolt on the steel plate foundation. Because the deflection of the bolt on the steel plate foundation is far smaller than the deflection of the bolt on the timber foundation and may thus be negligible, \(\Delta\) may be approximately considered to be \(\Delta \approx w_{0}\).

At \(x = 0\)(namely, \(\xi = 0\)), the shear force \(V_{0} = \frac{1}{2}P\) is plugged into formula (23), and thus,

$$\Delta = w_{0} = \frac{{l_{{\text{c}}}^{2} }}{{2{EI}}}\left( {l_{{\text{c}}} \frac{P}{2} + M_{0} } \right).$$

(27)

Assuming that

$$\alpha = \frac{{M_{0} }}{{l_{{\text{c}}} P}},$$

(28)

where \(\alpha\) is a constant, the following formula can be obtained:

$$P = \frac{1}{(1 + 2\alpha )}\frac{{4{EI}}}{{l_{{\text{c}}}^{3} }}\Delta .$$

(29)

Let

$$\beta = \frac{1}{(1 + 2\alpha )},$$

(30)

where \(\beta\) is a constant. Substituting \(l_{{\text{c}}} = \sqrt[4]{{\frac{{4{EI}}}{{k_{{\text{s}}} }}}}\) into formula (29) yields

$$P = \beta \left( {k_{{\text{s}}} \sqrt[4]{{\frac{{4{EI}}}{{k_{{\text{s}}} }}}}} \right)\Delta .$$

(31)

Thus, the elastic stiffness *k*_{e} of the bolted joint satisfies the following formula:

$$k_{{\text{e}}} = \frac{P}{\Delta } = \beta \left( {k_{{\text{s}}} \sqrt[4]{{\frac{{4{EI}}}{{k_{{\text{s}}} }}}}} \right),$$

(32)

where *k*_{s} and *EI* are determined by the material properties and geometries of the timber and the bolt. \(\beta\) is discussed as follows: (1) When the rotation of the bolt midpoint (*x* = 0) is fully constrained (\(\theta_{0} = 0\)), we substitute \(\theta_{0} = 0\) into formulas (24), (28), and (30) and solve them to obtain \(\beta = 2\); (2) when the rotation of the bolt midpoint (*x* = 0) is not fully constrained (\(M_{0} = 0\)), we substitute \(M_{0} = 0\) into formulas (28) and (30) and solve them to obtain \(\beta = 1\); thus, \(\beta\) is in the range of 1 to 2. In the elastic stage of loading of the bolted joint with a slotted-in steel plate, the rotation at the midpoint of the bolt is small, and \(\beta\) can be taken as approximately 2.