Subassemblage tests and design of steel channels assembled buckling-restrained braces

Abstract

This study investigates the load-carrying behavior of a proposed all-steel buckling-restrained brace (BRB) under seismic loading. The proposed steel channels assembled BRB (SCA-BRB) consists of a steel core plate encased by four steel channels assembled by high-strength bolts. Self-weight of an SCA-BRB is relatively small since the use of concrete is eliminated, leading to easier transportation and erection of the BRB. Firstly, by simplifying the restraining system of an SCA-BRB into a 2-chord battened column, the reduction factor of the elastic buckling load of the restraining system considering the discrete connection of bolts was derived. On this basis, the formulas in predicting ultimate resistance of an SCA-BRB were deduced along with the formulas of the lower limit of restraining ratio, and the upper limit of bolt spacing. Then, subassemblage test results of four specimens were reported. All the test specimens maintained stable load-carrying capacity during the loading process, and satisfactory energy-dissipating ability was achieved. Finite element (FE) models were established, and they well predicted the hysteretic responses of the test specimens. A total of 25 additional FE models was analyzed to perform a parametric study, verifying the effect of the member length, number of bolts, steel grade, gap size and geometrical imperfection on the hysteretic response and ultimate resistance of BRB. Finally, a design procedure of the SCA-BRB was proposed based on theoretical formulas, which have been validated by tests and FE results.

Appendix

A detailed design procedure of an SCA-BRB is provided as shown below.

• 1. Provided design parameters

The BRB length is $l = 6.0\;{\text{m}}$; the length of the restraining system is $l_{0} = 5.38\;{\text{m}}$; the required ultimate resistance is $P_{\hbox{max} } = 1000\;{\text{kN}}$; the gap size is $g = 1.5\;{\text{mm}}$; the initial imperfection is $i = l/1000 = 6.0\;{\text{mm}}$; Q235 steel is used in the steel core and Q345 steel is used in the restraining system; 10.9 grade high-strength bolts is used with a shear strength of $310{\text{MPa}}$.

• 2. Determine the cross-sectional dimension of the steel core

Based on the required ultimate resistance of BRB, the required yield load of the core can be calculated as

$$ P_{y} = \frac{{P_{\hbox{max} } }}{\eta } = \frac{1000}{1.7} = 588\;{\text{kN}} $$

The Q235 steel is adopted in the steel core, thus the required cross-sectional area of the core can be calculated as

$$ A_{c} = \frac{{P_{y} }}{{f_{y,c} }} = \frac{{588 \times 10^{3} }}{235} = 2502\;{\text{mm}}^{2} $$

The cross-sectional dimension of the core is determined as $t_{c} \times b_{c} = 16 \times 157\;{\text{mm}}$, where the cross-sectional area is $A_{c} = 2512\;{\text{mm}}^{2}$ and the yield load is $P_{y} = 590\;{\text{kN}}$.

• 3. Determine the dimension of the steel channels

Let $\zeta = \zeta_{\hbox{min} } = 2.0$, according to Eq. (22), the required plastic moment capacity of the restraining system can be calculated as

$$ W_{p,0} = \frac{{\zeta \eta P_{y} \left( {i + g} \right)}}{{\left( {\zeta - \eta } \right)f_{y,r} }} = \frac{{3.4 \cdot 590 \times 10^{3} \cdot \left( {1.5 + 6.0} \right)}}{0.3 \cdot 345} = 1.454 \times 10^{5} \;{\text{mm}}^{3} $$

According to Eq. (21), the required moment of inertia of the restraining system can be calculated as

$$ I_{0} = \frac{{\zeta \cdot P_{y} l_{0}^{2} }}{{\omega \cdot \pi^{2} E}} = \frac{{2.0 \cdot 590 \times 10^{3} \cdot 5380^{2} }}{{0.9 \cdot \pi^{2} \cdot 2.06 \times 10^{5} }} = 1.867 \times 10^{7} \;{\text{mm}}^{4} $$

The cross-sectional dimension of the L–R channels is determined as $160 \times 63 \times 4.0 \times 5.5\;{\text{mm}}$ and that of the T–B channels is determined as $180 \times 68 \times 4.5 \times 6.0\;{\text{mm}}$, thus the plastic moment capacity of the restraining system is $W_{p,0} = 2.692 \times 10^{5} \;{\text{mm}}^{3}$ and the moment of inertia of the restraining system is $I_{0} = 1.886 \times 10^{7} \;{\text{mm}}^{4}$.

• 4. Determine the bolt spacing and number of bolts

The plastic moment capacity of a single L–R channel can be calculated as

$$ M_{p,1} = W_{p,1} f_{y,r} = 1.593 \times 10^{4} \cdot 345 = 5.496 \times 10^{6} \;{\text{N}}\;{\text{mm}} $$

The buckling wavelength of the core can be calculated as

$$ l_{w} = \sqrt {\frac{{4\pi^{2} E_{t} I_{c} }}{{P_{y} }}} = \sqrt {\frac{{4\pi^{2} \cdot 0.05 \cdot 2.06 \times 10^{5} \cdot 5.359 \times 10^{4} }}{{590 \times 10^{3} }}} = 122\;{\text{mm}} $$

The upper limit of the bolt spacing can be calculated as

$$ l_{1,\hbox{max} } = \frac{{l_{w} }}{g} \cdot \frac{{M_{p,1} }}{{\eta P_{y} }} = \frac{122}{1.5} \cdot \frac{{5.496 \times 10^{6} }}{{1.7 \cdot 590 \times 10^{3} }} = 446\;{\text{mm}} $$

The number of bolted sections is determined as $N_{b} = 14$ and correspondingly the bolt spacing is $l_{1} = 414\;{\text{mm}}$. Since 4 bolts are installed in each of the bolted sections, a total of 56 high-strength bolts are used in this BRB member.

• 5. Check the ultimate resistance of the BRB member

Based on the dimension of the steel channels and the bolt spacing, the slenderness ratio values can be calculated as

$$ \lambda_{0} = \frac{{l_{0} }}{{\sqrt {I_{0} /A_{0} } }} = \frac{5380}{{\sqrt {1.886 \times 10^{7} /5722} }} = 93.7 $$

$$ \lambda_{1} = \frac{{l_{1} }}{{\sqrt {I_{1} /A_{1} } }} = \frac{414}{{\sqrt {5.089 \times 10^{5} /1289} }} = 20.8 $$

According to Eq. (11), the reduction factor can be calculated as

$$ \omega = \left( {1 + \frac{{\pi^{2} }}{12}\frac{{\lambda_{1}^{2} }}{{\lambda_{0}^{2} }}} \right)^{ - 1} = \left( {1 + \frac{{\pi^{2} }}{12}\frac{{20.8^{2} }}{{93.7^{2} }}} \right)^{ - 1} = 0.961 $$

According to Eq. (14), the global elastoplastic buckling load can be calculated as

$$ \begin{aligned} P_{\hbox{max} ,g} & = \frac{{M_{p,0} }}{{i + g + M_{p,0} /\omega P_{cr,0} }} \\ & = \frac{{2.692 \times 10^{5} \cdot 345}}{{6.0 + 1.5 + 2.692 \times 10^{5} \cdot 345/\left( {0.961 \cdot 1.325 \times 10^{6} } \right)}} \cdot \frac{1}{1000} = 1154.6\;{\text{kN}} > P_{\hbox{max} } = 1000\;{\text{kN}} \\ \end{aligned} $$

According to Eq. (19), the local elastoplastic buckling load can be calculated as

$$ \begin{aligned} P_{\hbox{max} ,l} & = \frac{{l_{w} }}{{l_{1} }} \cdot \frac{{M_{p,1} }}{g} \\ & = \frac{122}{414} \cdot \frac{{5.496 \times 10^{6} }}{1.5} \cdot \frac{1}{1000} = 1079.7\;{\text{kN}} > P_{\hbox{max} } = 1000\;{\text{kN}} \\ \end{aligned} $$

• 6. Determine the bolt size

According to Eq. (16), the extrusion force can be calculated as

$$ F_{e} = \frac{4g}{{l_{w} }} \cdot P_{\hbox{max} ,l} = \frac{4 \cdot 1.5}{122} \cdot 1079.7 = 53.1\;{\text{kN}} $$

The total extrusion force along the BRB length can be calculated as

$$ F_{e} \cdot \frac{{l_{0} }}{{l_{w} }} = 53.1 \cdot \frac{5380}{122} = 2341.6\;{\text{kN}} $$

The required cross-sectional area of each high-strength bolts can be calculated as

$$ \frac{{2341.6\;{\text{kN}}}}{{2N_{b} \cdot 310\;{\text{MPa}}}} = 269.8\;{\text{mm}}^{2} $$

The bolt size is determined to be M22 with an efficient cross-sectional area of $3 0 3. 4\;{\text{mm}}^{2}$.

• 7. Finite element validation

As tabulated in Table 4, Model 30 was constructed based on the geometrical parameters provided above in the example of practical design. Eigenvalue buckling analysis was firstly performed, and the elastic buckling load of a pin-ended SCA-BRB was obtained as $1280\;{\text{kN}}$. Since the elastic buckling load of the core plate is negligible compared with that of the restraining system, the obtained elastic buckling load can be approximately regarded as the elastic buckling load of the restraining system, i.e. $P_{cr,r} = 1280\;{\text{kN}}$. Accordingly, the reduction factor can be calculated as $\omega = 0.966$, which is slightly greater than the theoretical result ($\omega = 0.961$). Therefore, even the difference between the elastic buckling loads of the whole member and the restraining system is taken into consideration, the theoretical formula of the reduction factor [Eq. (11)] is satisfactorily accurate.

On the other hand, the hysteretic behavior of the practical design example was further investigated. As can be seen in Fig. 18 that Model 30 maintained stable during the elastoplastic FE analysis, validating that the recommended design procedure would lead to a safe and reliable design of an SCA-BRB.

Fig. 18

Hysteretic behavior of Model 30