Effect of a longitudinal crack on the flexural performance of bamboo culms

Abstract

Splitting parallel to the culm fibers is common in full-culm bamboo structural members, even early in a structure’s lifespan. Currently, there is insufficient knowledge on the effect of splitting on member performance, which induces significant uncertainties in bamboo member engineering design. This is a potential threat to the safety of existing and future full-culm bamboo structures. This study investigates analytically the effect of a longitudinal crack on the stiffness of an originally intact bamboo culm in flexure. The study develops analytical expressions that describe stiffness loss in two flexure cases (a three-point bending and a four-point bending test) and verifies them with available experimental results and numerical simulations. Main cause of the stiffness loss is torsion-induced deflections, with secondary cause being shear deformations. Importantly, stiffness loss solely depends on two dimensionless parameters: the shape factor (radius-to-thickness ratio) and a factor that is a function of material properties and ratio of shear span length to culm diameter. Additionally, the study proves analytically that friction at the load application points mitigates torsion-induced deflections. This has important implications for bamboo structure design and testing standards, indicating that the manner in which loads are transferred on beams affects the apparent beam stiffness when a crack appears.

Appendices

A Schematic representation of the torsion problem

See Figs. 11 and 12.

Fig. 11

Open cross section and shear center (a), and total torque diagrams for the three-point (b) and four-point (c) bending test

Fig. 12

Boundary conditions to solve the mixed torsion and warping problem for a the three-point and b the four-point bending test

B Torsion in the three-point bending test

Fig. 13

Normalized angle of twist $\theta$ along the beam for the three-point and four-point bending test ($P_\mathrm{{tot}}$ is the total load for each test and l is the total span length). The plot assumes the same span length for both tests (normalized shear span length $n_{3p}=1.5n_{4p}=10$), $E_\parallel /G_\parallel =4.5$ and shape factor $\phi$ = 5 (Eq. (22))

In the three-point bending test, and considering half of the beam because of symmetry, the load at end k (Fig. 12a) is P/2. The torque acting at the end k, because of the load is $T_0=0.5P\cdot 2R=PR$ (Fig. 12a). Applying the boundary conditions, the resulting solution for angle of twist $\theta _{3p}$ is (Fig. 13):

$$\begin{aligned} E_\parallel {J_\mathrm{{w}} }\theta _{3p} \left( z \right) =\frac{{PR}}{{{\lambda ^2}}}\left[ z{ - \frac{{\sinh \left( {\lambda z} \right) }}{{\lambda \cosh \left( {\lambda {L_{3p}}} \right) }}} \right] , \end{aligned}$$

(B1)

where $0 \le {z} \le L_{3p}$ (Fig. 12a). Figure 13 shows that the angle of twist in the three-point bending test is larger than in the four-point bending test, assuming the same total load and total span length. The expressions for Saint-Venant torque $T_{\mathrm{{s}},3p}$ and warping bimoment $B_{\mathrm{{w}},3p}$ ensue by substituting Eq. (B1) into Eqs. (10) and (11):

$$\begin{aligned} {T_{\mathrm{{s}},3p}}\left( z \right)= & {} PR\left[ {1 - \frac{{\cosh \left( {\lambda z} \right) }}{{\cosh \left( {\lambda L_{3p}} \right) }}} \right] , \end{aligned}$$

(B2)

$$\begin{aligned} {B_{\mathrm{{w}},3p}}\left( z \right)= & {} - PR\frac{{\sinh \left( {\lambda z} \right) }}{{\lambda \cosh \left( {\lambda L_{3p}} \right) }}, \end{aligned}$$

(B3)

where P is the load at the middle of the culm and $L_{3p}$ is the shear span length of the three-point bending test, equal to half the total span. Note that, although the applied boundary conditions assume that the culm does not twist at the supports, Eqs. (B2) and (B3) are valid regardless, as long as the rest of the boundary conditions remain the same. Whether the culm twists at the supports or not, only affects the resulting angle of twist $\theta$ (Eq. B1).

Subsequently, the deflection due to torsion for the three-point bending test ($\delta _{t,3p}$) occurs by substituting Eqs. (B2) and (B3), into Eq. (17), taking into account Eqs. (12) and (13), and calculating the integrals for $z_m=L_{3p}$. $\overline{T}_s$ and $\overline{B}_w$ in Eq. (17) (Saint-Venant torque and warping bimoment, because of the virtual load) occur by setting P = 1 in Eqs. (B2), (B3). Thus,

$$\begin{aligned} \delta _{t,3p}=\frac{{6Pn{\phi ^3}}}{{\pi {G_\parallel }R}} \cdot \left[ {1 - \frac{{\tanh \left( {\lambda L_{3p}} \right) }}{{\lambda L_{3p}}}} \right] , \end{aligned}$$

(B4)

where $\phi$ is the dimensionless shape factor of the culm (Eq. 22) and n is the dimensionless shear span length ($n=L_{3p}/\left( 2R\right)$). Subsequently, combining Eq. (B4) with Eq. (1),

$$\begin{aligned} \frac{{{\delta _{\mathrm{{t}},3p}}}}{{{\delta _{b,3p}}}} = \frac{{9}}{{2}}{\Theta }{\phi ^2}\left[ {1 - \frac{{\tanh \left( {\lambda L_{3p}} \right) }}{{\lambda L_{3p}}}} \right] , \end{aligned}$$

(B5)

where $\Theta$ is the shear-torsion deflection factor, as in Eq. (6). The deflection ratio of Eq. (B5) solely depends on the dimensionless terms $\phi$ and $\Theta$ (see also Eq. 25). Finally, the total deflection of the culm $\delta _{\mathrm{{tot}},3p}$, during the three-point bending test, when there is a longitudinal crack at the side of the culm, occurs as a function of the deflection due to the bending moment $\delta _{b,3p}$ from Eqs. (1), (5) and (B5):

$$\begin{aligned} {\delta _{\mathrm{{tot}},3p}} = {\delta _{b,3p}} + {\delta _{s,3p}} + {\delta _{t,3p}}=\left\{ {1 + \frac{{{3 }}}{{c_{i,3p}}}\Theta \cdot \left\{ {\frac{1}{{{k_\mathrm{{s}}}}} + 12{\phi ^2}\left[ {1 - \frac{{\tanh \left( {\lambda L_{3p}} \right) }}{{\lambda L_{3p}}}} \right] } \right\} } \right\} {\delta _{b,3p}}, \end{aligned}$$

(B6)

where $c_{i,3p}=8$.