92
18-1
K = 0.5일 때 pq 다이어그램에서 응력경로 기울기 구하기(세로가 1이라고 할 때 가로가 몇인지...)
σ h = 0.5 σ v {\displaystyle \sigma _{h}=0.5\sigma _{v}}
p = σ v + σ h 2 = 1.5 σ v 2 = 0.75 σ v {\displaystyle {\begin{aligned}p&={\frac {\sigma _{v}+\sigma _{h}}{2}}={\frac {1.5\sigma _{v}}{2}}\\&=0.75\sigma _{v}\\\end{aligned}}}
q = σ v − σ h 2 = 0.5 σ v 2 = 0.25 σ v {\displaystyle {\begin{aligned}q&={\frac {\sigma _{v}-\sigma _{h}}{2}}={\frac {0.5\sigma _{v}}{2}}\\&=0.25\sigma _{v}\\\end{aligned}}}
K = q p = 0.25 0.75 = 1 3 {\displaystyle K={\frac {q}{p}}={\frac {0.25}{0.75}}={\frac {1}{3}}}
