Keywords

1 Introduction

Composite structures, made of various materials, are widely used in the construction of civil building, industrial and agricultural structures [1, 2]. Composite panels with wooden ribs and covers, made of board materials—are the universal enclosing and supporting structures that can be used as covering, floors, wall envelope; and also as the main load-bearing elements in wooden-frame building. Such elements perform the functions of beams, flooring, lining, can provide thermal protection of the building (panels with the insulation inside the structure), and perform the function of horizontal stiffness diaphragms. There are numerous studies devoted to improving the performance of such panels. A new method is proposed in [3] for joining the outer and inner layers of 3-layer panels, which makes it possible to significantly increase the resistance of joints to longitudinal and transverse shear. The bent strength of prefabricated reinforced concrete wall panels with insulation using various connections between the outer bearing reinforced concrete layers is studied in the article [4]. The research for the ultimate compressive load and analysis of the stability of carbon fiber and glass fiber panels with stiffening ribs «Z»-profile and «L»-profile is presented in [5]. The results of studies of shear strength between layers of modular sandwich panels with a sheath made of high density polyethylene depending on the stamping method are presented in article [6]. The development of wood-based materials and structures base has made it not only technically possible but also economically feasible to use ribbed wall panels and volumetric modules in buildings and structures for various purposes [7,8,9].

The advantages of such panels (Fig. 1) are most evident when it combines load-bearing and enclosing functions, while the main longitudinal ribs take on racks’ role, and the covers take on enclosing layers’ role.

Fig. 1
figure 1

Ribbed wood composite wall panel

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It is possible to increase the load-carrying capacity of the ribs by effectively engaging the covering panels, for which a rigid adhesive bond at the «rib-sheathing» boundary is traditionally used. At the same time, the use of adhesive bonds significantly complicates the technological process of wall panels’ production, which contradicts the main idea of using such structures—simplicity and low production cost. A number of experimental and theoretical studies [10] demonstrated the advisability of taking the sheathing into account when jointing it to the ribs, using semi-rigid mechanical ties. Nails, screws, staples or combined connections based on claw washers are used as mechanical ties. Studies [11] present that taking the participation of the covers due to the screw ties into account gives a significant effect on the stiffness increasing of wall panels during shear when it serve as vertical diaphragms.

However, additional research is needed on the behavior of composite panels, where the ribs are the main load-bearing elements of the frame, and on the influence of the thickness of the covers and the stiffness of the ties on the strength and stability of such panels.

2 Models and Methods

In the engineering design of timber framed ribbed panels it is permissible to consider reduced T- or I-sections where the sheathing acts as the flanges. The normal stresses along the axis of the panel arising in the flanges will be greatest at the ribs and reduce as they move away from the ribs. In the middle of the sheathing the stresses will have a minimum value. Therefore, the reduced width of the sheathing (Fig. 2), rather than the actual width, must be included in the calculation so that the values of the highest longitudinal stresses for the actual and reduced cross-sections are equal [12,13,14].

Fig. 2
figure 2

Schematic of the calculated reduced panel cross-section: a—actual cross-section; b—design cross-section

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According to [15] the possibility of loss of stability in the calculation of centrally-compressed rods is taken into account by the introduction of a bending coefficient, from the condition:

$$ \frac{N}{F \cdot \varphi } \le R_{c} , $$
(1)

where F is the cross-sectional area;

φ is the longitudinal bending coefficient determined according to the formula:

$$ \varphi = \frac{{\sigma_{cr} }}{{R_{c} }}, $$
(2)

where σcr is the value of compressive stresses corresponding to the loss of stability (critical force Ncr).

In order to determine the bending coefficient, it is necessary to determine the fraction of the load that falls on the strut when the critical force Ncr is applied. The system of equations for a three-layer composite element [16,17,18,19,20,21] is presented below:

$$ \left\{ {\begin{array}{*{20}c} {\frac{T^{\prime\prime}}{{\xi_{1} }} = \Delta_{11} T_{1} + \Delta_{12} T_{2} + \Delta_{10} } \\ {\frac{T^{\prime\prime}}{{\xi_{2} }} = \Delta_{21} T_{1} + \Delta_{22} T_{2} + \Delta_{20} } \\ \end{array} } \right., $$
(3)

where Ti is the force at the joints of the composite bar,

Δij, Δiy are coefficients at the unknowns;

Δi0 are free terms;

ξi is the stiffness factor of the shear links in the i-joint, determined by Formula (12).

$$ \begin{aligned} \Delta _{{11}} & = \frac{1}{{E_{1} \cdot A_{1} }} + \frac{1}{{E_{2} \cdot A_{2} }} + \frac{{c_{1}^{2} }}{{\Sigma EI}}; \\ \Delta _{{22}} & = \frac{1}{{E_{2} \cdot A_{2} }} + \frac{{c_{1}^{2} }}{{\Sigma EI}}; \\ \Delta _{{12}} & = \Delta _{{21}} = \frac{{ - 1}}{{E_{2} \cdot A_{2} }} + \frac{{c_{1} c_{2} }}{{\Sigma EI}}; \\ \end{aligned} $$
(4)
$$ \begin{aligned} \Delta_{10} & = \frac{{ - N_{1}^{0} }}{{E_{1} \cdot A_{1} }} + \frac{{N_{2}^{0} }}{{E_{2} \cdot A_{2} }} + \frac{{M^{0} \cdot c_{1} }}{\Sigma EI}, \\ \Delta_{20} & = \frac{{ - N_{2}^{0} }}{{E_{2} \cdot A_{2} }} + \frac{{N_{3}^{0} }}{{E_{3} \cdot A_{3} }} + \frac{{M^{0} \cdot c_{2} }}{\Sigma EI}, \\ \end{aligned} $$
(5)

where E1, A1, E3, A3—modulus of elasticity and cross-section area of outer and inner sheathing respectively; E2, A2—the same, ribs; c1, c2—distance from centre of gravity of outer/inner sheathing to rib centre respectively; N0i—longitudinal load of i-th element; M0—total bending moment from outer load; ΣEI—bending stiffness of a zero shear joint rod.

As we consider the case of central compression and consider the application of external load on the strut with its redistribution onto the sheathing only by the operation of mechanical connections, we assume in further calculations that M0 = \(N_{1}^{0}\) = \(N_{3}^{0}\) = 0.

The system of Eqs. (3) can be represented by two independent equations for the generalized unknown joint forces \(\overline{T}_{12}\):

$$ \overline{T}_{1}^{{\prime\prime}} = \lambda_{1}^{2} \overline{T}_{1} + \overline{R}_{1} ;\quad \overline{T}_{2}^{{\prime\prime}} = \lambda_{2}^{2} \overline{T}_{2} + \overline{R}_{2} $$
(6)

where \(\overline{R}_{1,2}\) are the generalized load terms defined by expressions (7);

\(\lambda_{1,2}\)—the characteristic numbers determined by (8).

$$ \begin{aligned} \overline{R}_{1} & = \sqrt {\xi_{1} } \Delta_{10} \cos \varphi + \sqrt {\xi_{2} } \Delta_{20} \sin \varphi ; \\ \overline{R}_{2} & = - \sqrt {\xi_{1} } \Delta_{10} \sin \varphi + \sqrt {\xi_{2} } \Delta_{20} \cos \varphi , \\ \end{aligned} $$
(7)
$$ \lambda_{1,2} = \sqrt {\frac{1}{2}\left( {\xi_{1} \Delta_{11} + \xi_{2} \Delta_{22} \pm \sqrt {(\xi_{1} \Delta_{11} - \xi_{2} \Delta_{22} )^{2} + 4\Delta_{12}^{2} \xi_{1} \xi_{2} } } \right)} . $$
(8)

The functions of the generalised unknown forces in the joints and \(\overline{T}_{1} \overline{T}_{2}\) are of the form:

$$ \begin{aligned} \overline{T}_{1} & = A_{1} sh\left( {\lambda_{1} x} \right) + B_{1} ch\left( {\lambda_{1} x} \right) + \int\limits_{0}^{x} {R_{1} \left( t \right)sh\left[ {\lambda_{1} \cdot \left( {x - t} \right)} \right]dt;} \\ \overline{T}_{2} & = A_{2} sh\left( {\lambda_{1} x} \right) + B_{2} ch\left( {\lambda_{1} x} \right) + \int\limits_{0}^{x} {R_{2} \left( t \right)sh\left[ {\lambda_{2} \cdot \left( {x - t} \right)} \right]dt.} \\ \end{aligned} $$
(9)

The expressions bellow are used to move from the functions \(\overline{T}_{1}\) and \(\overline{T}_{2}\) to the shear forces T1(x) and T2(x):

$$ \begin{aligned} T_{1} \left( x \right) & = \sqrt {\xi_{1} } \cdot \cos \varphi \cdot \overline{T}_{1} - \sqrt {\xi_{1} } \cdot \sin \varphi \cdot \overline{T}_{2} ; \\ T_{2} \left( x \right) & = \sqrt {\xi_{2} } \cdot \sin \varphi \cdot \overline{T}_{1} - \sqrt {\xi_{2} } \cdot \cos \varphi \cdot \overline{T}_{2} . \\ \end{aligned} $$
(10)

The angle φ is determined by the formula:

$$ \varphi = arctg\left( {\frac{{\lambda_{1}^{2} - \xi \cdot \Delta_{11} }}{{\xi \cdot \Delta_{12} }}} \right). $$
(11)

The joint stiffness coefficients ξi are determined according to the formula:

$$ \xi_{i} = \frac{{n \cdot c_{c,i} }}{{S_{c,i} }}, $$
(12)

where cc,i is the stiffness factor of the joints in the i-th joint, determined experimentally;

Sc,i—spacing of connections; n—amount of longitudinal ribs.

Substituting expressions (4) and (12) into (12) and integrating the right part of the equations, the following equalities are obtained:

$$ \begin{aligned} \overline{T}_{1} \left( x \right) & = A_{1} sh\left( {\lambda_{1} x} \right) + B_{1} sh\left( {\lambda_{1} x} \right) \\ & \quad - \left[ {\frac{{\left( {\sqrt {\xi_{1} } \cdot \Delta_{10} \cdot \cos \varphi + \sqrt {\xi_{2} } \cdot \Delta_{20} \cdot \sin \varphi } \right) \cdot \left( {ch\left( {\lambda_{1} x} \right) - 1} \right)}}{{\lambda_{1}^{2} }}} \right], \\ \end{aligned} $$
(13)
$$ \begin{aligned} \overline{T}_{2} \left( x \right) & = A_{2} sh\left( {\lambda_{2} x} \right) + B_{2} sh\left( {\lambda_{2} x} \right) \\ & \quad - \left[ {\frac{{\left( {\sqrt {\xi_{1} } \cdot \Delta_{10} \cdot \cos \varphi + \sqrt {\xi_{2} } \cdot \Delta_{20} \cdot \sin \varphi } \right) \cdot \left( {ch\left( {\lambda_{1} x} \right) - 1} \right)}}{{\lambda_{2}^{2} }}} \right], \\ \end{aligned} $$
(14)

The arbitrary constants Ai and Bi are determined from the boundary conditions (shear forces at the supports without shear hindrances is equal to 0):

$$ x = 0 \to \overline{{T_{1} }} (0) = \overline{{T_{2} }} (0) = 0;x = l \to \overline{{T_{1} }} (l) = \overline{{T_{2} }} (l) = 0. $$

Substituting boundary conditions into expressions (13) and (14), the following expression is obtained: B1 = B2 = 0,

$$ A_{1} = \frac{{\sqrt {\xi_{1} } \Delta_{10} \cos \varphi \cdot \left( {1 - chl\lambda_{1} } \right) + \sqrt {\xi_{2} } \Delta_{20} \sin \varphi \cdot \left( {1 - chl\lambda_{1} } \right)}}{{\lambda_{1}^{2} \cdot shl\lambda_{1} }}; $$
(15)
$$ A_{2} = \frac{{\sqrt {\xi_{2} } \Delta_{20} \cos \varphi \cdot \left( {1 - chl\lambda_{2} } \right) - \sqrt {\xi_{1} } \Delta_{10} \sin \varphi \cdot \left( {1 - chl\lambda_{1} } \right)}}{{\lambda_{2}^{2} \cdot shl\lambda_{2} }}. $$
(16)

Substituting values A1 and A2 in (13) and (14) and then the values obtained into expressions (10) makes it possible to determine the shear forces Ti in any cross-section along the height of the post. The normal stresses in the ribs are determined by the formula:

$$ \sigma_{r} = \frac{1}{{n_{r} }}\left( {\frac{{ - T_{1} + T_{2} }}{{A_{r} }} \pm \frac{{M_{r} \cdot 0.5 \cdot t_{r} }}{{I_{r} }}} \right), $$
(17)

where Mp is the moment in the rib resulting from the work of the shear bonds, determined by formula (18) (for a symmetrical three-layer panel, Mp = 0); nt is the number of longitudinal ribs.

$$ M_{r} = - \frac{{\left( {T_{1} \cdot c_{1} + T_{2} \cdot c_{2} } \right) \cdot E_{r} \cdot I_{r} }}{\Sigma EI} $$
(18)

To determine the value of Ncr, consider a strut as a composite 3-ply compressed rod, the outer layers of which are sheathing and the inner layers are branches (Fig. 3). The shear bonds in the 2 outside seam are ductility (nails, screws) or rigid (glue) ties. The system of differential equations of a composite compressed rod with account for the longitudinal bend has form:

$$ \left\{ {\begin{array}{*{20}c} {T_{1} ^{\prime\prime}/\xi_{1} = \Delta_{11} T_{1} + \Delta_{12} T_{2} + \Delta_{1y} y + \Delta_{10} \,\,\,\,\,} \\ {T_{2} ^{\prime\prime}/\xi_{2} = \Delta_{21} T_{1} + \Delta_{22} T_{2} + \Delta_{2y} y + \Delta_{20\,} \,\,\,} \\ {\Sigma N^{0} y^{\prime\prime} = \Delta_{y1} T_{1} + \Delta_{y2} T_{2} + \Delta_{yy} y_{2} + \Delta_{y0} } \\ \end{array} } \right., $$
(19)
Fig. 3
figure 3

Dependencies of |Ncr| for a single rib on the stiffness of the bracing cladding to ribs: a—OSB cladding; b—FRC; c—particle board; 1—tob = 12 mm; 2—tob = 15 mm; 3—tob = 18 mm

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Here:

$$ \Delta_{yi} = \Delta_{yi} = \frac{{\Sigma N^{0} }}{\Sigma EI}c_{i} ;\quad \Delta_{yy} = \frac{{\left( {\Sigma N^{0} } \right)^{2} }}{\Sigma EI};\quad \Delta_{y0} = \frac{{\Sigma N^{0} }}{\Sigma EI}M^{0} , $$
(20)

where y is the deflection of the element;

ΣN0—total longitudinal load on the panel.

Since M0 = 0, the coefficients Δi0 and Δy0 are 0, the system is homogeneous. Its solution is Ti = 0, y = 0. However, for some total values of the longitudinal force the homogeneous system has solutions other than zero which correspond to the forms of loss of stability of the compressed composite rod. These values of the total longitudinal force ΣN0 are critical. At articulated end restraint of the rack the boundary conditions are: x = 0 Ti = 0 → y = 0; x = l → Ti = 0, y = 0. The system of differential equations under these boundary conditions has solution [12]:

$$ T_{i} = \alpha_{i} \sin \chi x;\quad y = \alpha_{y} \sin \chi x, $$
(21)

where \(\chi = k\pi /l\), k is a positive integer that takes into account the form of stability loss;

αi, αy are constant coefficients.

The system of equations below is obtained by substituting expressions (21) into (19) and reducing by sinχx:

$$ \left\{ {\begin{array}{*{20}l} {\left( {\Delta_{11} + \frac{{\chi^{2} }}{{\xi_{1} }}} \right)\alpha_{1} + \Delta_{12} \alpha_{2} + \frac{{\Sigma N^{0} }}{\Sigma EI}c_{1} \alpha_{y} = 0} \\ {\Delta_{21} \alpha_{21} + \left( {\Delta_{22} + \frac{{\chi^{2} }}{{\xi_{2} }}} \right)\alpha_{2} + \frac{{\Sigma N^{0} }}{\Sigma EI}c_{2} \alpha_{y} = 0} \\ {\frac{{\Sigma N^{0} }}{\Sigma EI}c_{1} \alpha_{1} + \frac{{\Sigma N^{0} }}{\Sigma EI}c_{2} \alpha_{2} + \left[ {\frac{{\left( {\Sigma N^{0} } \right)^{2} }}{\Sigma EI} + \Sigma N^{0} \chi^{2} } \right]\alpha_{y} = 0} \\ \end{array} } \right.. $$
(22)

The solution of this homogeneous system of ordinary equations in which αi, αy are independent is different from zero only when its determinant is zero. By reducing the system by \( \left( {\Sigma N_{0} } \right)^{2} /\Sigma EI, \) we obtain the equation:

$$ \left| {\begin{array}{*{20}c} {\left( {\Delta_{11} + \chi^{2} /\xi_{1} } \right)} & {\Delta_{12} } & {c_{1} } \\ {\Delta_{21} } & {\left( {\Delta_{22} + \chi^{2} /\xi_{2} } \right)} & {c_{2} } \\ {c_{1} } & {c_{2} } & {\left[ {\Sigma EI + \chi^{2} \left( {\Sigma EI} \right)^{2} /\Sigma N^{0} } \right]} \\ \end{array} } \right| = 0. $$
(23)

Expanding the determinant with respect to ΣN0, the formula for the critical force (Ncr = ΣN0) determination is obtained:

$$ N_{{cr}} = - \frac{{\chi ^{2} \Sigma EI^{2} \cdot \left( {\chi ^{4} + \chi ^{2} \xi _{1} \Delta _{{11}} + \chi ^{2} \xi _{2} \Delta _{{22}} + \xi _{1} \xi _{2} \Delta _{{11}} \Delta _{{22}} - \xi _{1} \xi _{2} \Delta _{{12}} ^{2} } \right)}}{{\left( \begin{aligned} & \chi ^{2} \left( {\chi ^{2} \Sigma EI - c_{1}^{2} \xi _{1} - c_{2}^{2} \xi _{2} + \xi _{1} \Delta _{{11}} \Sigma EI + \xi _{2} \Delta _{{22}} \Sigma EI} \right) \\ & - \xi _{1} \xi _{2} \cdot \left( {c_{2}^{2} \Delta _{{11}} + c_{1}^{2} \Delta _{{22}} - 2c_{{^{1} }} c_{{^{2} }} \Delta _{{12}} - \Delta _{{11}} \Delta _{{22}} \Sigma EI + \Delta _{{12}} ^{2} \Sigma EI} \right) \\ \end{aligned} \right)}}. $$
(24)

The minimum values of the critical force are corresponding to k = 1, which occurs while the rod bends by one half-wave of the sine wave form of longitudinal bending.

3 Results and Discussion

There is a timber-framed wall panel, measuring 1.0 × 3.0 m, with three longitudinal ribs (pine wood). The ribs are hinged at the ends. The cross-section of the ribs is 40 × 100 mm. The covers is connected to the ribs by ductility screw ties. The panel ribs are subjected to longitudinal compressive load transmitted from the upstream panels without eccentricity. There are considered panels with double-sided cladding of 12-, 15- and 18-mm thickness made of OSB/3 oriented chipboard, structural plywood and cement-bonded particleboard. Critical force values and bending coefficients are determined depending on bond stiffness, covers thickness and material. The designed width of the covers according to [22] is 0.9 bp = 0.9 m. The results of calculating the critical force |Ncr| are shown in the graphs in Fig. 3. The values are given for one rib. The critical force at ξ = 0 is determined for the rib without consideration of the sheathing, by the Euler formula.

The presented method makes it possible to determine the value of critical force for the ribs included in the compressed wood-composite wall panels. After determining the values of the critical force |Ncr| according to the Formula (2), it is possible to determine the coefficient of longitudinal bending φ taking into account the participation of the panel covers, and according to the Formula (1) to make a simplified engineering calculation of such structures.

Covers participation together with ribs consideration under compression force can increase the strength and stability of the structure [23, 24]. Depending on the stiffness of the mechanical bonds, the increase in critical force is found to be up to 62% with oriented structural board (OSB) sheathing; up to 69% with plywood (wood-laminated board, glued with the help of urea–formaldehyde composition sheathing) and up to 64% with cement bonded particle board (CBPB). It is evident from the graphs in Fig. 3 that an increase in shear ties stiffness gives a significantly greater increase in |Ncr| values than an increase in sheath thickness. Thus, an increase in connection stiffness from 0 to 10,000 kN/m2 results in increase of |Ncr| values by 52–69%, while an increase of cover thickness by 1.5 times (from 12 to 18 mm)— by 6–8%. The use of plywood cladding gives the greatest effect of increasing |Ncr|, up to 4.5% compared with OSB, up to 2.6% with CBPB.

4 Conclusions

  1. 1.

    The presented mathematical model makes it possible to estimate the stress–strain state of timber-framed wall panels and determine the value of the critical force, taking into account the covers, involved by connection to the ribs using ductility mechanical ties.

  2. 2.

    The two-sided covering of the ribs can significantly increase the critical force value |Ncr|. The greatest increase in the critical force |Ncr| is observed when the mechanical shear ties fixing the sheathing to the ribs are rigid. Increasing the thickness of the sheathing as well as the material used (OSB, plywood, CBPB) has slight effect.

  3. 3.

    In order to increase the strength and stability of timber-wall panels, it is first necessary to reduce the shear resistance of the mechanical ties at the «rib-cover» seams, which can be achieved both by using more shear-resistant ties and by reducing the spacing of the ties.