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The required plate thickness is determined according to AISC Steel Design Guide 1 Eq.3.3.15a.
fc | 25000 kN/m2 | |
A1  Image Removed | | Mathinline |
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| body | --uriencoded--$$ \normalsize A_%7B1%7D = 900 \times 500 \times 10%5e%7B-6%7D = 0.45 $$ |
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| body | --uriencoded--$$ \normalsize A_%7B2%7D = 900 \times 500 \times 10%5e%7B-6%7D = 0.45 $$ |
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qmax  Image Removed | | Mathinline |
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| body | --uriencoded--$$ \normalsize q_%7Bmax%7D = \varphi_c f_%7Bp(max)%7D B = 0.65 \times 21250 \times 0.5 = 6906.25 $$ |
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| body | --uriencoded--$$ \normalsize f_%7Bp(max)%7D = 0.85 f_c \sqrt %7B \dfrac%7BA_2%7D%7BA_1%7D %7D \leq 1.7f_c $$ |
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| body | --uriencoded--$$ \normalsize f_%7Bp(max)%7D = 0.85 \times 25000 \sqrt %7B \dfrac%7B0.45%7D%7B0.45%7D %7D = 21250 \leq 1.7 \times 25000 =42500 $$ |
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e  Image Removed | | Mathinline |
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| body | --uriencoded--$$ \normalsize e = \dfrac%7BM%7D%7BP%7D = \dfrac%7B365.17%7D%7B739.22%7D =493.994 $$ |
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ecrit  Image Removed | | Mathinline |
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| body | --uriencoded--$$ \normalsize e_%7Bcrit%7D = \dfrac%7BN%7D%7B2%7D - \dfrac%7BP%7D%7B2q_%7Bmax%7D%7D = \dfrac%7B900%7D%7B2%7D - \dfrac%7B739.22%7D%7B2 \times 6906.25%7D =396.482 $$ |
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Y  Image Removed | | Mathinline |
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| body | --uriencoded--$$ \normalsize Y = \left ( f + \dfrac%7BN%7D%7B2%7D \right ) \mp \sqrt%7B \left ( f +\dfrac%7BN%7D%7B2%7D \right )%5e2 - \dfrac%7B2P (e + f)%7D %7Bq_%7Bmax%7D%7D %7D $$ |
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Y  Image Removed | | Mathinline |
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| body | --uriencoded--$$ \small Y = \left ( 405 + \dfrac%7B900 %7D%7B2%7D \right ) \mp \sqrt%7B \left ( 405 +\dfrac%7B900%7D%7B2%7D \right )%5e2 - \dfrac%7B2 \times 739.22 (493.994 + 405)%7D %7B6906.25%7D %7D $$ |
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| Mathinline |
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| body | $$ \small Y =121.123 $$ |
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f  Image Removed | | Mathinline |
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| body | --uriencoded--$$ \normalsize f = \dfrac%7BN%7D%7B2%7D - l_e = \dfrac%7B900%7D%7B2%7D - 45 = 405 $$ |
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le | 45 mm | |
l | max(m,n)=141.25 | |
m  Image Removed | | Mathinline |
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| body | --uriencoded--$$ \normalsize m = \dfrac%7BN - 0.95h%7D%7B2%7D = \dfrac%7B900 -0.95 \times 650%7D%7B2%7D =141.25 $$ |
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n  Image Removed | | Mathinline |
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| body | --uriencoded--$$ \normalsize n = \dfrac%7BB - 0.8b%7D%7B2%7D = \dfrac%7B500 -0.8 \times 300%7D%7B2%7D =130 $$ |
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h | 650 mm | |
b | 300 mm | |
N | 900 mm | |
B | 500 mm | |
M | 365.17 kNm | |
P | 739.22 kN | |
y | 355 N/mm2 | |
treq  Image Removed | | Mathinline |
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| body | --uriencoded--$$ \normalsize t_%7Breq%7D = 2 \sqrt %7B \dfrac%7B \varphi_c f_%7Bp(max) %7D Y( l - \dfrac%7BY%7D%7B2%7D ) %7D%7B\varphi F_y %7D %7D =41.11 $$ |
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| AISC DG-1-2nd 3.3.15.a |
Required | Available | Ratio | Control |
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41.11 mm | 42 mm | 0.979 | √ |
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The required base plate thickness for tension is determined according to AISC Steel Design Guide 1 Eq.3.4.7a.
fc | 25000 kN/m2 | |
A1 Image Removed | | Mathinline |
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| body | --uriencoded--$$ \normalsize A_%7B1%7D = 900 \times 500 \times 10%5e%7B-6%7D = 0.45 $$ |
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A2 Image Removed | | Mathinline |
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| body | --uriencoded--$$ \normalsize A_%7B2%7D = 900 \times 500 \times 10%5e%7B-6%7D = 0.45 $$ |
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qmax Image Removed | | Mathinline |
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| body | --uriencoded--$$ \normalsize q_%7Bmax%7D = \varphi_c f_%7Bp(max)%7D B = 0.65 \times 21250 \times 0.5 = 6906.25 $$ |
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e Image Removed | | Mathinline |
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| body | --uriencoded--$$ \normalsize e = \dfrac%7BM%7D%7BP%7D = \dfrac%7B365.17%7D%7B739.22%7D =493.994 $$ |
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ecrit Image Removed | | Mathinline |
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| body | --uriencoded--$$ \normalsize e_%7Bcrit%7D = \dfrac%7BN%7D%7B2%7D - \dfrac%7BP%7D%7B2q_%7Bmax%7D%7D = \dfrac%7B900%7D%7B2%7D - \dfrac%7B739.22%7D%7B2 \times 6906.25%7D =396.482 $$ |
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T  Image Removed | | Mathinline |
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| body | --uriencoded--$$ \normalsize T = q_%7Bmax%7D Y - P = 6906.25 \times 121.123 - 739.22 = 97.286 $$ |
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Y Image Removed | | Mathinline |
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| body | --uriencoded--$$ \normalsize Y = \left ( f + \dfrac%7BN%7D%7B2%7D \right ) \mp \sqrt%7B \left ( f +\dfrac%7BN%7D%7B2%7D \right )%5e2 - \dfrac%7B2P (e + f)%7D %7Bq_%7Bmax%7D%7D %7D $$ |
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Y Image Removed | | Mathinline |
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| body | --uriencoded--$$ \small Y = \left ( 405 + \dfrac%7B900 %7D%7B2%7D \right ) \mp \sqrt%7B \left ( 405 +\dfrac%7B900%7D%7B2%7D \right )%5e2 - \dfrac%7B2 \times 739.22 (493.994 + 405)%7D %7B6906.25%7D %7D $$ |
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| Mathinline |
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| body | $$ \small Y =121.123 $$ |
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f Image Removed | | Mathinline |
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| body | --uriencoded--$$ \normalsize f = \dfrac%7BN%7D%7B2%7D - l_e = \dfrac%7B900%7D%7B2%7D - 45 = 405 $$ |
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x  Image Removed | | Mathinline |
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| body | $$ \normalsize x = m - l_e = 141.25 - 45 = 96.25 $$ |
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le | 45 mm | |
m Image Removed | | Mathinline |
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| body | --uriencoded--$$ \normalsize m = \dfrac%7BN - 0.95h%7D%7B2%7D = \dfrac%7B900 -0.95 \times 650%7D%7B2%7D =141.25 $$ |
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h | 650 mm | |
N | 900 mm | |
B | 500 mm | |
M | 365.17 kNm | |
P | 739.22 kN | |
Fy | 355 N/mm2 | |
treq  Image Removed | | Mathinline |
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| body | --uriencoded--$$ \normalsize t_%7Breq%7D = 2 \sqrt %7B \dfrac%7B Tx %7D%7B\varphi F_y B %7D %7D = 2 \sqrt %7B \dfrac%7B 97.288 \times 96.25 %7D%7B 0.9 \times 355 \times 500 %7D %7D =15.312 $$ |
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| AISC DG-1-2nd 3.4.7a |
Required | Available | Ratio | Control |
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15.312 mm | 42 mm | 0.365 | √ |
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The limit state of the anchor rod tension rupture is checked according to AISC 360-16.
Ab  Image Removed | | Mathinline |
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| body | --uriencoded--$$ \normalsize A_b = \dfrac%7B \pi d%5e2%7D %7B4%7D = \dfrac%7B \pi 24%5e2%7D %7B4%7D = 452.389 $$ |
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Fn | 750000 kN/m2 | |
Rn  Image Removed | | Mathinline |
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| body | --uriencoded--$$ \normalsize R_n = F_n A_b = 750 \times 452.389 \times 10%5e%7B-3%7D =339.292 $$ |
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| AISC 360-16 J3-1 |
ΦRn  Image Removed | | Mathinline |
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| body | --uriencoded--$$ \normalsize \varphi R_n = 0.75 \times 339.292 = 254.469 \; \; \mathrm%7BkN%7D $$ |
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Required | Available | Ratio | Control |
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28.779 kN | 254.469 kN | 0.113 | √ |
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Ψ | 1.0 | ACI 318M-08 D.5.3.6 |
Np |  | ACI 318M-08 (D-15) |
Abgr | 7401.592 mm2 | |
fc | 25000 kN/m2 | |
Rn |  | ACI 318M-08 (D-14) |
ΦRn |  | |
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