✂️ 전단 설계, 기준 6종 비교 (조항, 식, 원문, 계산)
RCKDS 14 20 22콘크리트구조 전단, 비틀림 설계기준 (2022)국토교통부, KCSC 무료RCKDS 14 20 10콘크리트구조 해석, 설계 원칙 (2021)국토교통부, KCSC 무료, 참조FRPKDS 14 20 68GFRP 보강근 콘크리트구조 설계기준 (2024)국토교통부, KCSC 무료FRPKDS 24 50 05GFRP 보강근 콘크리트교 설계기준 (2024)국토교통부, KCSC 무료참고KCS 24 50 05GFRP 보강근 콘크리트교 시공 시방 (2024)국토교통부, KCSC 무료, 시공 시방서참고KDS 14 20 68 부록GFRP 보강근 재료, 품질 시방 역할 (2024)KDS 14 20 68 : 2024 발췌, 개정안에서 KCS 이관 예정참고KEC 잠정지침 2022도로공사 GFRP 잠정 설계, 시공지침 (2022)한국도로공사, 내부 실무지침(보유 자료, 비공개)참고KEC 일위대가 2024도로공사 GFRP 일위대가 (2024)한국도로공사, 내부 실무지침(보유 자료, 비공개)FRP 해외ACI 440.1R-15FRP 보강근 콘크리트 설계, 시공 지침 (2015)ACI, 저작권(로컬 열람), 파일 없음(공식 링크)FRP 해외ACI 440.11-22GFRP 보강근 콘크리트 구조 설계코드 (2022)ACI, 저작권(로컬 열람), 스캔본, 파일 없음(공식 링크)FRP 해외AASHTO-2018GFRP 보강 콘크리트교 설계지침 2판 (2018)AASHTO, 저작권(로컬 열람), 스캔본, 파일 없음(공식 링크)
⚠ 어떤 원문이 열리나 (기준마다 다름)
KDS, KCS (국토교통부 고시) → 누구나 열람. 저작권법 제7조 비보호 저작물이며 국가건설기준센터에서 무료로 받을 수 있음.
ACI, AASHTO (해외 기준) → 저작권 자료라 이 PC에 있는 사본으로만 열람. 링크가 안 열리면 발행처 공식 미리보기, 상점으로 연결됨.
도로공사 지침 → 발주처 내부 자료라 비공개. 조항 번호와 내용만 표에 옮겨 적었다.
원문 PDF는 웨일, 크롬에서 열면 해당 쪽과 위치까지 자동으로 이동함.
ACI, AASHTO (해외 기준) → 저작권 자료라 이 PC에 있는 사본으로만 열람. 링크가 안 열리면 발행처 공식 미리보기, 상점으로 연결됨.
도로공사 지침 → 발주처 내부 자료라 비공개. 조항 번호와 내용만 표에 옮겨 적었다.
원문 PDF는 웨일, 크롬에서 열면 해당 쪽과 위치까지 자동으로 이동함.
📊 기준 6종 비교 차트 (같은 예제에서 기준별 값)
노란 테두리 = 지금 고른 FRP 기준붉은 점선 = 모든 기준에 같은 한계붉은 짧은 선 + 숫자 = 그 기준만의 한계
약칭 KDS 14 = KDS 14 20 68 (건축, 일반), KDS 24 = KDS 24 50 05 (교량), ACI 15 = ACI 440.1R-15, ACI 22 = ACI 440.11-22, AASHTO = AASHTO GFRP 보강 콘크리트교 설계지침 2판 (2018)
| 항목 | RC, KDS 14 20 22 | KDS 14 20 68 | KDS 24 50 05 | ACI 440.1R-15 | ACI 440.11-22 | AASHTO GFRP 2018 |
|---|---|---|---|---|---|---|
| 예제 결과 $V_u = 150$ kN | $\phi V_n = \boxed{\mathbf{409.6}\ \text{kN}}$ $V_u/\phi V_n = \mathbf{0.37}\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}$ $s\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}$ | $\phi V_n = \boxed{\mathbf{221.7}\ \text{kN}}$ $V_u/\phi V_n = \mathbf{0.68}\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}$ $s\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}$ | $\phi V_n = \boxed{\mathbf{187.8}\ \text{kN}}$ $V_u/\phi V_n = \mathbf{0.80}\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}$ $s\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}$ | $\phi V_n = \boxed{\mathbf{189.8}\ \text{kN}}$ $V_u/\phi V_n = \mathbf{0.79}\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}$ $s\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}$ | $\phi V_n = \boxed{\mathbf{223.6}\ \text{kN}}$ $V_u/\phi V_n = \mathbf{0.67}\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}$ $s\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}$ | $\phi V_n = \boxed{\mathbf{173.9}\ \text{kN}}$ $V_u/\phi V_n = \mathbf{0.86}\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}$ $s\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}$ |
| 설계 원칙 $\phi$ | 4.1.1 (4.1-1) $$ V_u \le \phi V_n,\quad V_n = V_c + V_s $$ $\phi_v=0.75$ $$\begin{aligned}\phi V_n &= 0.75\,(V_c + V_s) = 0.75\,(185.9 + 360.3) = \boxed{\mathbf{409.6}\ \text{kN}} \\ \frac{V_u}{\phi V_n} &= \frac{150}{409.6} = \mathbf{0.37}\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$ | $$ \phi V_n \ge V_u,\quad V_n = V_c + V_f $$ $\phi_v=0.75$ $$\begin{aligned}\phi V_n &= 0.75\,(V_c + V_f) = 0.75\,(92.9 + 202.7) = \boxed{\mathbf{221.7}\ \text{kN}} \\ \frac{V_u}{\phi V_n} &= \frac{150}{221.7} = \mathbf{0.68}\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$ | 6.1, 6.2.1, 4.5 (6.1-1)(6.2-1) $$ V_r=\phi V_n,\quad V_n = V_c + V_f $$ $\phi_v=0.75$ $$\begin{aligned}\phi V_n &= 0.75\,(V_c + V_f) = 0.75\,(88.3 + 162.1) = \boxed{\mathbf{187.8}\ \text{kN}} \\ \frac{V_u}{\phi V_n} &= \frac{150}{187.8} = \mathbf{0.80}\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$ | $$ \phi V_n \ge V_u,\quad V_n = V_c + V_f $$ $\phi=0.75$ ↗ ACI 318 준용 (원문 미보유) $$\begin{aligned}\phi V_n &= 0.75\,(V_c + V_f) = 0.75\,(90.9 + 162.1) = \boxed{\mathbf{189.8}\ \text{kN}} \\ \frac{V_u}{\phi V_n} &= \frac{150}{189.8} = \mathbf{0.79}\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$ | 22.5.1.1, 21.2 (22.5.1.1) $$ V_n = V_c + V_f $$ $\phi=0.75$ (표 21.2.1) 단면 상한 (22.5.1.2) $V_u \le \phi\,0.2 f'_c b_w d$ $$\begin{aligned}\phi V_n &= 0.75\,(V_c + V_f) = 0.75\,(95.5 + 202.7) = \boxed{\mathbf{223.6}\ \text{kN}} \\ \frac{V_u}{\phi V_n} &= \frac{150}{223.6} = \mathbf{0.67}\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}} \\ V_u &\le \phi\,0.2 f_{ck} b_w d = 0.75\times0.2\times30\times400\times509/10^3 = 916\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$ | 2.7.3.3, 2.5.5.2 (2.7.3.3-1) $$ V_n = V_c + V_f $$ $\phi=0.75$ $$\begin{aligned}\phi V_n &= 0.75\,(V_c + V_f) = 0.75\,(85.9 + 145.9) = \boxed{\mathbf{173.9}\ \text{kN}} \\ \frac{V_u}{\phi V_n} &= \frac{150}{173.9} = \mathbf{0.86}\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$ |
| 콘크리트 전단강도 $V_c$ | 4.2.1 (4.2-1)(4.2-2) $$ V_c=\frac{1}{6}\left(1+\frac{N_u}{14A_g}\right)\lambda\sqrt{f_{ck}}\,b_w d $$ 축력 없으면 $(4.2\text{-}1)$ $$\begin{aligned}V_c &= \tfrac{1}{6}\Big(1+\tfrac{N_u}{14A_g}\Big)\lambda\sqrt{f_{ck}}\,b_w d \\ &= \tfrac{1}{6}\Big(1+\tfrac{0}{14\times240{,}000}\Big)(1.0)\sqrt{30}\times400\times509 \\ &= \boxed{\mathbf{185.9}\ \text{kN}}\end{aligned}$$ | 4.3.2 (4.3-1)–(4.3-3) $$ V_c=\frac{1}{6}(1+k_p)\,k_s\,\beta_v\sqrt{f_{ck}}\,b_w d $$ $k_p=N_u/14A_g$ $k_s=(0.3/d)^{0.25}$ (0.75–1.1 최소보강 만족 시 1) $\beta_v=0.5$ 독자식 $$\begin{aligned}k_p &= N_u/(14A_g) = 0/(14\times240{,}000) = 0.000 \\ k_s &= (0.3/d)^{0.25} = (0.3/0.509)^{0.25} = 0.88 \\ &\to [0.75,\,1.1] \to 0.88\ \to\ 1.00\ (\text{최소보강 만족}) \\ V_c &= \tfrac{1}{6}(1+0.000)(1.00)(\beta_v\,0.5)\sqrt{30}\times400\times509 = \boxed{\mathbf{92.9}\ \text{kN}}\end{aligned}$$ | 6.2.1(2) (6.2-2) $$ V_c=\frac{2}{5}\sqrt{f_{ck}}\,b_w(kd) $$ $k$ = 균열단면 중립축비 (4.3-4) $$\begin{aligned}n_f &= E_f/E_c = 45{,}000/27{,}537 = 1.63 \\ \rho_f &= A_f/(b_w d) = 3{,}040/(400\times509) = 0.0149 \\ k &= \sqrt{2\rho_f n_f+(\rho_f n_f)^2}-\rho_f n_f = 0.198 \\ kd &= 0.198\times509 = 100.7\ \text{mm} \\ V_c &= \tfrac{2}{5}\sqrt{30}\times400\times100.7 = \boxed{\mathbf{88.3}\ \text{kN}}\end{aligned}$$ | 8.2 (8.2a)(8.2b) $$ V_c=\frac{2}{5}\sqrt{f'_c}\,b_w c,\quad k=\sqrt{2\rho_f n_f+(\rho_f n_f)^2}-\rho_f n_f $$ $c=kd$ (Tureyen & Frosch) $$\begin{aligned}n_f &= E_f/E_c = 45{,}000/25{,}743 = 1.75 \\ \rho_f &= A_f/(b_w d) = 3{,}040/(400\times509) = 0.0149 \\ k &= \sqrt{2\rho_f n_f+(\rho_f n_f)^2}-\rho_f n_f = 0.204 \\ kd &= 0.204\times509 = 103.8\ \text{mm} \\ V_c &= \tfrac{2}{5}\sqrt{30}\times400\times103.8 = \boxed{\mathbf{90.9}\ \text{kN}}\end{aligned}$$ | 표 22.5.5.1, 22.5.5.1.3 (a)(b) $$ V_c=\max\!\left(5\lambda_s k_{cr},\ 0.8\lambda_s\right)\sqrt{f'_c}\,b_w d\ \text{[psi]} \;\to\; \max(0.42\lambda_s k_{cr},\,0.066\lambda_s)\sqrt{f_{ck}}\,b_w d $$ 크기효과 $\lambda_s=\sqrt{2/(1+0.004d)}\le1$ (최소보강 미만 시) 하한 $k_{cr}\ge0.16$ 상당 $$\begin{aligned}n_f &= E_f/E_c = 45{,}000/25{,}743 = 1.75 \\ \rho_f &= A_f/(b_w d) = 3{,}040/(400\times509) = 0.0149 \\ k &= \sqrt{2\rho_f n_f+(\rho_f n_f)^2}-\rho_f n_f = 0.204 \\ \lambda_s &= 1.00\ (A_{fv}\ge A_{fv,min}) \\ V_c &= \max(0.42k_{cr},\,0.066)\,\lambda_s\sqrt{f_{ck}}\,b_w d \\ &= \max(0.086,\,0.066)\times1.00\times\sqrt{30}\times400\times509 = \boxed{\mathbf{95.5}\ \text{kN}}\end{aligned}$$ | 2.7.3.4, 2.7.3.6.1 (2.7.3.4-1) $$ V_c=0.0316\,\beta\sqrt{f'_c}\,b_v d_v\ \text{[ksi]},\quad \beta=5.0k,\ \theta=45^\circ $$ 간이법 → SI $\tfrac{2}{5}k\sqrt{f_{ck}}\,b_v d_v$ $$\begin{aligned}n_f &= E_f/E_c = 45{,}000/25{,}743 = 1.75 \\ \rho_f &= A_f/(b_w d) = 3{,}040/(400\times509) = 0.0149 \\ k &= \sqrt{2\rho_f n_f+(\rho_f n_f)^2}-\rho_f n_f = 0.204 \\ d_v &= \max(0.9d,\,0.72h) = \max(458,\,432) = 458\ \text{mm} \\ V_c &= 0.42\,k\sqrt{f_{ck}}\,b_v d_v = 0.42\times0.204\times\sqrt{30}\times400\times458 \\ &= \boxed{\mathbf{85.9}\ \text{kN}}\end{aligned}$$ |
| 스터럽 전단강도 $V_s$, $V_f$, 상한 | 4.3.4(2), (8) (4.3-3) $$ V_s=\frac{A_v f_{yt} d}{s}\le\frac{2}{3}\lambda\sqrt{f_{ck}}\,b_w d $$ $$\begin{aligned}A_v &= 2\times\pi\times13^2/4 = 265\ \text{mm}^2 \\ V_s &= \frac{A_v f_{yt} d}{s} = \frac{265\times400\times509}{150} = \boxed{\mathbf{360.3}\ \text{kN}} \\ V_{s,max} &= \tfrac{2}{3}\sqrt{30}\times400\times509 = 743.4\ \ge V_s\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$ | 4.3.3.4(1), (4) (4.3-7)(4.3-10) $$ V_f=\frac{A_{fv} f_{ft} d}{s}\le 0.25\,\xi_v f_{ck} b_w z,\quad \xi_v=0.6(1-f_{ck}/250),\ z=0.85d $$ $$\begin{aligned}A_v &= 2\times\pi\times13^2/4 = 265\ \text{mm}^2 \\ V_f &= \frac{A_{fv} f_{ft} d}{s} = \frac{265\times225\times509}{150} = \boxed{\mathbf{202.7}\ \text{kN}} \\ V_{f,max} &= 0.25\,\xi_v f_{ck} b_w z = 0.25\times0.528\times30\times400\times433 = 685.3\ \ge V_f\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$ | 6.2.1(4) (6.2-3) $$ V_f=\frac{A_{fv} f_{fv} d}{s} $$ $$\begin{aligned}A_v &= 2\times\pi\times13^2/4 = 265\ \text{mm}^2 \\ V_f &= \frac{A_{fv} f_{ft} d}{s} = \frac{265\times180\times509}{150} = \boxed{\mathbf{162.1}\ \text{kN}} \\ V_{f,max} &= 0.66\sqrt{30}\times400\times509 = 736.0\ \ge V_f\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$ | 8.2, 8.2.3 (8.2c) $$ V_f=\frac{A_{fv} f_{fv} d}{s},\quad V_f\le 8\sqrt{f'_c}\,b_w d\ [\text{psi}]=0.66\sqrt{f_{ck}}\,b_w d $$ 8.2.3: ACI 318 상한 8√f′c 권장 (복부 압괴 0.18–0.3f′c 대신) $$\begin{aligned}A_v &= 2\times\pi\times13^2/4 = 265\ \text{mm}^2 \\ V_f &= \frac{A_{fv} f_{ft} d}{s} = \frac{265\times180\times509}{150} = \boxed{\mathbf{162.1}\ \text{kN}} \\ V_{f,max} &= 0.66\sqrt{30}\times400\times509 = 736.0\ \ge V_f\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$ | 22.5.8.1, 22.5.8.5.3 (22.5.8.1)(22.5.8.5.3) $$ V_f\ge\frac{V_u}{\phi}-V_c,\quad V_f=\frac{A_{fv} f_{ft} d}{s} $$ 상한은 22.5.1.2 단면 제한으로 $$\begin{aligned}A_v &= 2\times\pi\times13^2/4 = 265\ \text{mm}^2 \\ V_f &= \frac{A_{fv} f_{ft} d}{s} = \frac{265\times225\times509}{150} = \boxed{\mathbf{202.7}\ \text{kN}} \\ V_{f,max} &= 0.2 f_{ck} b_w d - V_c = 0.2\times30\times400\times509/10^3 - 95.5 = 1{,}126.1\ \ge V_f\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$ | 2.7.3.5 (2.7.3.5-1)(2.7.2.5-1) $$ V_f=\frac{A_{fv} f_{fv} d_v\cot\theta}{s}\le0.25\sqrt{f'_c}\,b_v d_v\ [\text{ksi}]=0.66\sqrt{f_{ck}}\,b_v d_v $$ $\theta=45^\circ$ → $\cot\theta=1$ $$\begin{aligned}A_v &= 2\times\pi\times13^2/4 = 265\ \text{mm}^2 \\ V_f &= \frac{A_{fv} f_{ft} d}{s} = \frac{265\times180\times458}{150} = \boxed{\mathbf{145.9}\ \text{kN}} \\ V_{f,max} &= 0.66\sqrt{30}\times400\times458 = 662.4\ \ge V_f\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$ |
| 스터럽 설계응력 | $$ f_{yt}\le 500\ \text{MPa} $$ $$\begin{aligned}f_{yt} &= 400\ \text{MPa} \le 500\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$ | 4.3.3.4(2) (4.3-8) $$ f_{ft}=0.005E_f\le f_{fb} $$ $f_{fb}$ = 굽힘부 설계기준인장강도 (앱: $(0.05r_b/d_b+0.3)f_{fu}$ 근사 직접 입력 가능) $$\begin{aligned}f_{fb} &= (0.05\,r_b/d_b + 0.3)\,f_{fu} \\ &= (0.05\times3+0.3)\times850 = 382\ \text{MPa} \\ 0.005E_f &= 0.005\times45{,}000 = 225\ \text{MPa} \\ f_{ft} &= \min(225,\ 382) = \boxed{\mathbf{225}\ \text{MPa}}\end{aligned}$$ | 6.2.1(5) (6.2-4) $$ f_{fv}=0.004E_f\le f_{fb} $$ $f_{fb}$ (3.2-6) $=(0.05r_b/d_b+0.3)f_{fu}$ $$\begin{aligned}f_{fb} &= (0.05\,r_b/d_b + 0.3)\,f_{fu} \\ &= (0.05\times3+0.3)\times800 = 360\ \text{MPa} \\ 0.004E_f &= 0.004\times45{,}000 = 180\ \text{MPa} \\ f_{ft} &= \min(180,\ 360) = \boxed{\mathbf{180}\ \text{MPa}}\end{aligned}$$ | 8.2, 6.2.1 (8.2d)(6.2.1) $$ f_{fv}=0.004E_f\le f_{fb},\quad f_{fb}=\left(0.05\frac{r_b}{d_b}+0.3\right)f_{fu} $$ $$\begin{aligned}f_{fb} &= (0.05\,r_b/d_b + 0.3)\,f_{fu} \\ &= (0.05\times3+0.3)\times800 = 360\ \text{MPa} \\ 0.004E_f &= 0.004\times45{,}000 = 180\ \text{MPa} \\ f_{ft} &= \min(180,\ 360) = \boxed{\mathbf{180}\ \text{MPa}}\end{aligned}$$ | $$ f_{ft}\le\min(f_{fb},\ 0.005E_f),\quad f_{fb}=C_E f_{fb}^* $$ 0.005 (0.004 아님) 2026-08-28 원문 대조로 수정 $$\begin{aligned}f_{fb} &= (0.05\,r_b/d_b + 0.3)\,f_{fu} \\ &= (0.05\times3+0.3)\times850 = 382\ \text{MPa} \\ 0.005E_f &= 0.005\times45{,}000 = 225\ \text{MPa} \\ f_{ft} &= \min(225,\ 382) = \boxed{\mathbf{225}\ \text{MPa}}\end{aligned}$$ | 2.7.3.5 (2.7.3.5-2)(2.7.3.5-3) $$ f_{fv}=0.004E_f\le f_{fb},\quad f_{fb}=\left(0.05\frac{r_b}{d_b}+0.3\right)f_{fd} $$ $$\begin{aligned}f_{fb} &= (0.05\,r_b/d_b + 0.3)\,f_{fu} \\ &= (0.05\times3+0.3)\times800 = 360\ \text{MPa} \\ 0.004E_f &= 0.004\times45{,}000 = 180\ \text{MPa} \\ f_{ft} &= \min(180,\ 360) = \boxed{\mathbf{180}\ \text{MPa}}\end{aligned}$$ |
| 소요 스터럽량 | 4.3.4 (4.3-3) 역산 $$ \frac{A_v}{s}=\frac{V_u-\phi V_c}{\phi f_{yt} d} $$ $$\begin{aligned}\phi V_c &= 0.75\times185.9 = 139.4 < V_u = 150\ \to\ \text{스터럽 필요} \\ \frac{A_v}{s} &= \frac{V_u-\phi V_c}{\phi f d} = \frac{(150-139.4)\times10^3}{0.75\times400\times509} \\ &= \boxed{\mathbf{0.069}\ \text{mm²/mm}} \\ s_{req} &= A_v\big/\tfrac{A_v}{s} = 265/0.069 = 3{,}823\ \text{mm}\end{aligned}$$ | 4.3.3.4(3) (4.3-9) $$ \frac{A_{fv}}{s}=\frac{V_u-\phi V_c}{\phi f_{ft} d} $$ $$\begin{aligned}\phi V_c &= 0.75\times92.9 = 69.7 < V_u = 150\ \to\ \text{스터럽 필요} \\ \frac{A_v}{s} &= \frac{V_u-\phi V_c}{\phi f d} = \frac{(150-69.7)\times10^3}{0.75\times225\times509} \\ &= \boxed{\mathbf{0.935}\ \text{mm²/mm}} \\ s_{req} &= A_v\big/\tfrac{A_v}{s} = 265/0.935 = 284\ \text{mm}\end{aligned}$$ | 6.2.1(6) (6.2-5) $$ \frac{A_{fv}}{s}=\frac{V_u-\phi V_c}{\phi f_{fv} d} $$ $$\begin{aligned}\phi V_c &= 0.75\times88.3 = 66.2 < V_u = 150\ \to\ \text{스터럽 필요} \\ \frac{A_v}{s} &= \frac{V_u-\phi V_c}{\phi f d} = \frac{(150-66.2)\times10^3}{0.75\times180\times509} \\ &= \boxed{\mathbf{1.220}\ \text{mm²/mm}} \\ s_{req} &= A_v\big/\tfrac{A_v}{s} = 265/1.220 = 218\ \text{mm}\end{aligned}$$ | 8.2 (8.2e) $$ \frac{A_{fv}}{s}=\frac{V_u-\phi V_c}{\phi f_{fv} d} $$ $$\begin{aligned}\phi V_c &= 0.75\times90.9 = 68.2 < V_u = 150\ \to\ \text{스터럽 필요} \\ \frac{A_v}{s} &= \frac{V_u-\phi V_c}{\phi f d} = \frac{(150-68.2)\times10^3}{0.75\times180\times509} \\ &= \boxed{\mathbf{1.190}\ \text{mm²/mm}} \\ s_{req} &= A_v\big/\tfrac{A_v}{s} = 265/1.190 = 223\ \text{mm}\end{aligned}$$ | 22.5.8.1, R22.5.8.5 (22.5.8.1)(R22.5.8.5) $$ \frac{A_{fv}}{s}=\frac{V_u-\phi V_c}{\phi f_{ft} d} $$ $$\begin{aligned}\phi V_c &= 0.75\times95.5 = 71.6 < V_u = 150\ \to\ \text{스터럽 필요} \\ \frac{A_v}{s} &= \frac{V_u-\phi V_c}{\phi f d} = \frac{(150-71.6)\times10^3}{0.75\times225\times509} \\ &= \boxed{\mathbf{0.913}\ \text{mm²/mm}} \\ s_{req} &= A_v\big/\tfrac{A_v}{s} = 265/0.913 = 291\ \text{mm}\end{aligned}$$ | 2.7.3.5 (2.7.3.5-1) 역산 $$ \frac{A_{fv}}{s}=\frac{V_u/\phi-V_c}{f_{fv} d_v} $$ $$\begin{aligned}\phi V_c &= 0.75\times85.9 = 64.5 < V_u = 150\ \to\ \text{스터럽 필요} \\ \frac{A_v}{s} &= \frac{V_u-\phi V_c}{\phi f d} = \frac{(150-64.5)\times10^3}{0.75\times180\times458} \\ &= \boxed{\mathbf{1.383}\ \text{mm²/mm}} \\ s_{req} &= A_v\big/\tfrac{A_v}{s} = 265/1.383 = 192\ \text{mm}\end{aligned}$$ |
| 최소 전단보강 $A_{v,min}$ | 4.3.3(3) (4.3-1) $$ A_{v,min}=0.0625\sqrt{f_{ck}}\frac{b_w s}{f_{yt}}\ \ge\ 0.35\frac{b_w s}{f_{yt}} $$ $V_u > \tfrac{1}{2}\phi V_c$ 구간 $$\begin{aligned}A_{v,min} &= \max(0.0625\sqrt{30},\,0.35)\,\frac{b_w s}{f_{yt}} \\ &= \max(0.342,\,0.35)\times\frac{400\times150}{400} = \boxed{\mathbf{52}\ \text{mm²}} \\ A_v &= 265 \ge 52\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$ | 4.3.3.3 (4.3-4)(4.3-5) $$ A_{fv,min}=\max\!\left(0.0625\sqrt{f_{ck}},\ 0.35\right)\frac{b_w s}{f_{ft}} $$ $V_u \ge 0.5\phi V_c$ 구간 $$\begin{aligned}A_{fv,min} &= \max(0.0625\sqrt{30},\,0.35)\,\frac{b_w s}{f_{ft}} \\ &= 0.350\times\frac{400\times150}{225} = \boxed{\mathbf{93}\ \text{mm²}} \\ A_{fv} &= 265 \ge 93\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$ | 6.2.2 (6.2-8) $$ A_{fv,min}=0.35\frac{b_w s}{f_{fv}} $$ $V_u > \phi V_c/2$ 구간 $$\begin{aligned}A_{fv,min} &= 0.35\,\frac{b_w s}{f_{ft}} \\ &= 0.350\times\frac{400\times150}{180} = \boxed{\mathbf{117}\ \text{mm²}} \\ A_{fv} &= 265 \ge 117\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$ | 8.2.2 (8.2.2) $$ A_{fv,min}=0.35\frac{b_w s}{f_{fv}} $$ [50 $b_w s/f_{fv}$ psi] $$\begin{aligned}A_{fv,min} &= 0.35\,\frac{b_w s}{f_{ft}} \\ &= 0.350\times\frac{400\times150}{180} = \boxed{\mathbf{117}\ \text{mm²}} \\ A_{fv} &= 265 \ge 117\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$ | 9.6.3.1, 9.6.3.4 (a)(b) $$ A_{fv,min}=\max\!\left(0.75\sqrt{f'_c},\ 50\right)\frac{b_w s}{f_{ft}}\ \text{[psi]}\ \to\ \max(0.062\sqrt{f_{ck}},\,0.35)\frac{b_w s}{f_{ft}} $$ $V_u\ge\phi\,2.5k_{cr}\sqrt{f'_c}b_w\,d$ (= ½φV_c) 구간 $$\begin{aligned}A_{fv,min} &= \max(0.062\sqrt{30},\,0.35)\,\frac{b_w s}{f_{ft}} \\ &= 0.350\times\frac{400\times150}{225} = \boxed{\mathbf{93}\ \text{mm²}} \\ A_{fv} &= 265 \ge 93\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$ | 2.7.2.4 (2.7.2.4-1) $$ A_{fv}\ge0.05\frac{b_v s}{f_{fv}}\ \text{[ksi]}\ \to\ 0.35\frac{b_v s}{f_{fv}} $$ 0.05 ksi = 0.345 MPa 2026-08-28 원문 대조로 수정(이전 0.083√$f_{ck}$ 오기) $$\begin{aligned}A_{fv,min} &= 0.35\,\frac{b_w s}{f_{ft}} \\ &= 0.350\times\frac{400\times150}{180} = \boxed{\mathbf{117}\ \text{mm²}} \\ A_{fv} &= 265 \ge 117\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$ |
| 간격 제한 $s_{max}$ | $$ s\le d/2,\ 600\ \text{mm} $$ $V_s>\tfrac{1}{3}\lambda\sqrt{f_{ck}}b_w\,d$ 이면 절반 $$\begin{aligned}s_{max} &= \min(d/2,\,600) = \min(254,\,600) = 254\ \text{mm} \\ V_s &= 360.3 \le \tfrac{1}{3}\sqrt{f_{ck}}\,b_w d = 371.7\ \to\ \text{절반 규정 미적용} \\ s_{allow} &= \min(s_{req},\,s_{min},\,s_{max}) = \min(3{,}823,\,758,\,254) \\ &= \boxed{\mathbf{254}\ \text{mm}} \ \ge s = 150\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$ | $$ s\le d/2,\ 600\ \text{mm} $$ KDS 14 20 22 4.3.2 준용 (단 4.3.2(2) 제외) $$\begin{aligned}s_{max} &= \min(d/2,\,600) = \min(254,\,600) = 254\ \text{mm} \\ V_f &= 202.7 \le \tfrac{1}{3}\sqrt{f_{ck}}\,b_w d = 371.7\ \to\ \text{절반 규정 미적용} \\ s_{allow} &= \min(s_{req},\,s_{min},\,s_{max}) = \min(284,\,427,\,254) \\ &= \boxed{\mathbf{254}\ \text{mm}} \ \ge s = 150\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$ | $$ s\le d/2,\ 600\ \text{mm} $$ 6.3(2) $r_b/d_b\ge3$ (3) 90° 갈고리 꼬리 $12d_b$ $$\begin{aligned}s_{max} &= \min(d/2,\,600) = \min(254,\,600) = 254\ \text{mm} \\ V_f &= 162.1 \le \tfrac{1}{3}\sqrt{f_{ck}}\,b_w d = 371.7\ \to\ \text{절반 규정 미적용} \\ s_{allow} &= \min(s_{req},\,s_{min},\,s_{max}) = \min(218,\,341,\,254) \\ &= \boxed{\mathbf{218}\ \text{mm}} \ \ge s = 150\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$ | $$ s\le d/2,\ 600\ \text{mm (24 in.)} $$ 절반 규정은 ACI 318 준용(앱: $V_f>\tfrac{1}{3}\sqrt{f_{ck}}b_w\,d$ 이면 d/4 300) $$\begin{aligned}s_{max} &= \min(d/2,\,600) = \min(254,\,600) = 254\ \text{mm} \\ V_f &= 162.1 \le \tfrac{1}{3}\sqrt{f_{ck}}\,b_w d = 371.7\ \to\ \text{절반 규정 미적용} \\ s_{allow} &= \min(s_{req},\,s_{min},\,s_{max}) = \min(223,\,341,\,254) \\ &= \boxed{\mathbf{223}\ \text{mm}} \ \ge s = 150\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$ | 9.7.6.2.2 표 9.7.6.2.2 $$ V_f\le4\sqrt{f'_c}b_w\,d:\ s\le d/2,\ 24\ \text{in};\qquad V_f>4\sqrt{f'_c}b_w\,d:\ s\le d/4,\ 12\ \text{in} $$ SI: $4\sqrt{f'_c}=0.33\sqrt{f_{ck}}$ 600/300 mm $$\begin{aligned}s_{max} &= \min(d/2,\,600) = \min(254,\,600) = 254\ \text{mm} \\ V_f &= 202.7 \le \tfrac{1}{3}\sqrt{f_{ck}}\,b_w d = 371.7\ \to\ \text{절반 규정 미적용} \\ s_{allow} &= \min(s_{req},\,s_{min},\,s_{max}) = \min(291,\,427,\,254) \\ &= \boxed{\mathbf{254}\ \text{mm}} \ \ge s = 150\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$ | $$ s\le0.5d,\ 600\ \text{mm (24 in.)} $$ 절반 규정 없음 2026-08-28 원문 대조로 수정(이전 0.8d_v/0.4d_v 오기) $$\begin{aligned}s_{max} &= \min(0.5d,\,600) = \min(254,\,600) = 254\ \text{mm} \\ s_{allow} &= \min(s_{req},\,s_{min},\,s_{max}) = \min(192,\,341,\,254) \\ &= \boxed{\mathbf{192}\ \text{mm}} \ \ge s = 150\ \ \color{#0a8a2a}{\checkmark\ \textbf{OK}}\end{aligned}$$ |
| 환경감소계수 $C_E$ (참고) | 해당 없음 RC 해당 없음 | $$ C_E=0.85 $$ 노출환경 무관 (GFRP) $$\begin{aligned}f_{fu} &= C_E\,f_{fu}^* = 0.85\times1000 = \boxed{\mathbf{850}\ \text{MPa}}\end{aligned}$$ | $$ C_E=0.8\ /\ 0.7 $$ 비노출 / 노출 (GFRP) $$\begin{aligned}f_{fu} &= C_E\,f_{fu}^* = 0.80\times1000 = \boxed{\mathbf{800}\ \text{MPa}}\end{aligned}$$ | $$ C_E=\begin{cases}\text{GFRP }0.8/0.7\\ \text{CFRP }1.0/0.9\\ \text{AFRP }0.9/0.8\end{cases} $$ 섬유 종류 × 노출 $$\begin{aligned}f_{fu} &= C_E\,f_{fu}^* = 0.80\times1000 = \boxed{\mathbf{800}\ \text{MPa}}\end{aligned}$$ | $$ C_E=0.85 $$ 노출 무관 ASTM D7957 내구성 요건 $$\begin{aligned}f_{fu} &= C_E\,f_{fu}^* = 0.85\times1000 = \boxed{\mathbf{850}\ \text{MPa}}\end{aligned}$$ | $$ C_E=0.8\ /\ 0.7 $$ 비노출 / 노출 (GFRP) $$\begin{aligned}f_{fu} &= C_E\,f_{fu}^* = 0.80\times1000 = \boxed{\mathbf{800}\ \text{MPa}}\end{aligned}$$ |
| 이 앱의 검증 근거 | 손계산 대조 ($V_{c}$ 185.9, φV_n 409.7, $A_{v}$/s 0.069, s 255) ✔ 2차 독립구현 무작위 300케이스 × 6항목 대조 일치 (verify_shear_independent.py, 0.2%) | 공식 예제 없음 → (4.3-1) 손계산, 화면값 손검산 (verify_engine.py) ✔ 2차 독립구현 무작위 300케이스 × 6항목 대조 일치 (verify_shear_independent.py, 0.2%) | 공식 예제 없음 → 식 대조 (6.2-2)–(6.2-8) ✔ 2차 독립구현 무작위 300케이스 × 6항목 대조 일치 (verify_shear_independent.py, 0.2%) | 설계예제 8M (SI) 대조: k, φV_c, $f_{fb}$, $f_{fv}$, $A_{fv}$/s, 간격 3종 2% 이내 통과 ✔ 2차 독립구현 무작위 300케이스 × 6항목 대조 일치 (verify_shear_independent.py, 0.2%) | 스캔 원문 눈대조 (21.2, 20.2.2.3, 2.2.6, 표 22.5.5.1, 22.5.8.1, 8.5.3, 9.6.3.1, 3.4, 9.7.6.2.2) → 식 반영. SI 계수(0.42, 0.066, 0.062)는 psi→MPa 환산 ✔ 2차 독립구현 무작위 300케이스 × 6항목 대조 일치 (verify_shear_independent.py, 0.2%) | 스캔 원문 눈대조 (2.4.2.1, 2.5.5.2, 2.7.2.4, 2.5, 2.6, 2.8, 2.7.3.3–3.6.1) → 간이법(β=5k, θ=45°) 반영. 일반법(MCFT 2.7.3.6.2) 미구현 ✔ 2차 독립구현 무작위 300케이스 × 6항목 대조 일치 (verify_shear_independent.py, 0.2%) |