2022
DOI: 10.1016/j.jcsr.2021.107066
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Seismic performance of steel chevron braced frames designed according to Japanese practice

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Cited by 7 publications
(32 citation statements)
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“…Weak beam mechanism: β0goodbreak=κ+x·r02+x·r0$$ {\beta}_0=\frac{\kappa +x\cdotp {r}_0}{2+x\cdotp {r}_0} $$ The above equations are consistent with Seki et al 15 . In these equations, x=Nu/Ny$$ x={N}_u/{N}_y $$ is the compressive‐to‐tensile strength ratio of the brace.…”
Section: Plastic Mechanism and Plastic Strengthsupporting
confidence: 75%
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“…Weak beam mechanism: β0goodbreak=κ+x·r02+x·r0$$ {\beta}_0=\frac{\kappa +x\cdotp {r}_0}{2+x\cdotp {r}_0} $$ The above equations are consistent with Seki et al 15 . In these equations, x=Nu/Ny$$ x={N}_u/{N}_y $$ is the compressive‐to‐tensile strength ratio of the brace.…”
Section: Plastic Mechanism and Plastic Strengthsupporting
confidence: 75%
“…Furthermore, κ$$ \kappa $$, 1.0κ2.0$$ 1.0\le \kappa \le 2.0 $$, is the assumed limit for the maximum possible unbalanced vertical force over the plastic resistance of the beam, Vb0=4Mp/l$$ {V}_{b0}=4{M}_p/l $$, when subject to reverse end moments Mp$$ {M}_p $$. The factor κ$$ \kappa $$ addresses the beam end moments that act in combination with the unbalanced vertical force 3,4,15 . The maximum value0.25emκ=2.0$$ \kappa =2.0 $$ corresponds to the case when both end moments Mp$$ {M}_p $$ act in the direction to oppose the unbalanced vertical force.…”
Section: Plastic Mechanism and Plastic Strengthmentioning
confidence: 99%
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