Page Comparison

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Strength Controls

Base Plate Thickness (

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Compression)

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 body --uriencoded--$$ \normalsize A_%7B1%7D = 900 \times 500 \times 10%5e%7B-6%7D = 0.45 \; \mathrm%7Bm%5e2%7D $$
A2 Image Removed	Mathinline body --uriencoded--$$ \normalsize A_%7B2%7D = 900 \times 500 \times 10%5e%7B-6%7D = 0.45\; \mathrm%7Bm%5e2%7D $$
qmax Image Removed	Mathinline body --uriencoded--$$ \normalsize q_%7Bmax%7D = \varphi_c f_%7Bp(max)%7D B = 0.65 \times 21250 \times 0.5 = 6906.25 \; \mathrm%7BkN/m%7D $$
fp(max) Image Removed	Mathinline body --uriencoded--$$ \normalsize f_%7Bp(max)%7D = 0.85 f_c \sqrt %7B \dfrac%7BA_2%7D%7BA_1%7D %7D \leq 1.7f_c $$
fp(max) Image Removed	Mathinline body --uriencoded--$$ \normalsize f_%7Bp(max)%7D = 0.85 \times 25000 \sqrt %7B \dfrac%7B0.45%7D%7B0.45%7D %7D = 21250 \leq 1.7 \times 25000 =42500 \; \mathrm%7BkN/m%5e2%7D $$
e Image Removed	Mathinline body --uriencoded--$$ \normalsize e = \dfrac%7BM%7D%7BP%7D = \dfrac%7B365.17%7D%7B739.22%7D \times 10%5e3 =493.994 \; \mathrm%7Bmm%7D $$
ecrit Image Removed	Mathinline body --uriencoded--$$ \normalsize e_%7Bcrit%7D = \dfrac%7BN%7D%7B2%7D - \dfrac%7BP%7D%7B2q_%7Bmax%7D%7D = \dfrac%7B900%7D%7B2%7D - \dfrac%7B739.22%7D%7B2 \times 6906.25 \times 10%5e%7B-3%7D%7D =396.482 \; \mathrm%7Bmm%7D $$
Y Image Removed	Mathinline body --uriencoded--$$ \normalsize Y = \left ( f + \dfrac%7BN%7D%7B2%7D \right ) \mp \sqrt%7B \left ( f +\dfrac%7BN%7D%7B2%7D \right )%5e2 - \dfrac%7B2P (e + f)%7D %7Bq_%7Bmax%7D%7D %7D $$
Y Image Removed	Mathinline body --uriencoded--$$ \small Y = \left ( 405 + \dfrac%7B900 %7D%7B2%7D \right ) \mp \sqrt%7B \left ( 405 +\dfrac%7B900%7D%7B2%7D \right )%5e2 - \dfrac%7B2 \times 739.22 (493.994 + 405)%7D %7B6906.25%7D %7D $$ Mathinline body --uriencoded--$$ \small Y =121.123\; \mathrm%7Bmm%7D $$
f Image Removed	Mathinline body --uriencoded--$$ \normalsize f = \dfrac%7BN%7D%7B2%7D - l_e = \dfrac%7B900%7D%7B2%7D - 45 = 405 \; \mathrm%7Bmm%7D $$
le	45 mm
l	max(m,n)=141.25
m Image Removed	Mathinline body --uriencoded--$$ \normalsize m = \dfrac%7BN - 0.95h%7D%7B2%7D = \dfrac%7B900 -0.95 \times 650%7D%7B2%7D =141.25\; \mathrm%7Bmm%7D $$
n Image Removed	Mathinline body --uriencoded--$$ \normalsize n = \dfrac%7BB - 0.8b%7D%7B2%7D = \dfrac%7B500 -0.8 \times 300%7D%7B2%7D =130\; \mathrm%7Bmm%7D $$
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 body --uriencoded--$$ \normalsize t_%7Breq%7D = 2 \sqrt %7B \dfrac%7B \varphi_c f_%7Bp(max) %7D Y( l - \dfrac%7BY%7D%7B2%7D ) %7D%7B\varphi F_y %7D %7D =41.11\; \mathrm%7Bmm%7D $$	AISC DG-1-2nd 3.3.15.a

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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 body --uriencoded--$$ \normalsize A_%7B1%7D = 900 \times 500 \times 10%5e%7B-6%7D = 0.45 \; \mathrm%7Bm%5e2%7D $$
A2 Image Removed	Mathinline body --uriencoded--$$ \normalsize A_%7B2%7D = 900 \times 500 \times 10%5e%7B-6%7D = 0.45 \; \mathrm%7Bm%5e2%7D $$
qmax Image Removed	Mathinline body --uriencoded--$$ \normalsize q_%7Bmax%7D = \varphi_c f_%7Bp(max)%7D B = 0.65 \times 21250 \times 0.5 = 6906.25 \; \mathrm%7BkN/m%7D $$
e Image Removed	Mathinline body --uriencoded--$$ \normalsize e = \dfrac%7BM%7D%7BP%7D = \dfrac%7B365.17%7D%7B739.22%7D \times 10%5e3 =493.994\; \mathrm%7Bmm%7D $$
ecrit Image Removed	Mathinline body --uriencoded--$$ \normalsize e_%7Bcrit%7D = \dfrac%7BN%7D%7B2%7D - \dfrac%7BP%7D%7B2q_%7Bmax%7D%7D = \dfrac%7B900%7D%7B2%7D - \dfrac%7B739.22%7D%7B2 \times 6906.25 \times 10%5e%7B-3%7D %7D =396.482\; \mathrm%7Bmm%7D $$
T Image Removed	Mathinline body --uriencoded--$$ \normalsize T = q_%7Bmax%7D Y - P = 6906.25 \times 121.123 \times 10%5e%7B-3%7D - 739.22 = 97.286 \; \mathrm%7BkN%7D $$
Y	Image Removed		Y	Image Removed Mathinline body --uriencoded--$$ \normalsize Y = \left ( f + \dfrac%7BN%7D%7B2%7D \right ) \mp \sqrt%7B \left ( f +\dfrac%7BN%7D%7B2%7D \right )%5e2 - \dfrac%7B2P (e + f)%7D %7Bq_%7Bmax%7D%7D %7D $$
Y	Mathinline body --uriencoded--$$ \small Y = \left ( 405 + \dfrac%7B900 %7D%7B2%7D \right ) \mp \sqrt%7B \left ( 405 +\dfrac%7B900%7D%7B2%7D \right )%5e2 - \dfrac%7B2 \times 739.22 (493.994 + 405)%7D %7B6906.25%7D %7D $$ Mathinline body --uriencoded--$$ \small Y =121.123\; \mathrm%7Bmm%7D $$
f Image Removed	Mathinline body --uriencoded--$$ \normalsize f = \dfrac%7BN%7D%7B2%7D - l_e = \dfrac%7B900%7D%7B2%7D - 45 = 405\; \mathrm%7Bmm%7D $$
x Image Removed	Mathinline body --uriencoded--$$ \normalsize x = m - l_e = 141.25 - 45 = 96.25\; \mathrm%7Bmm%7D $$
le	45 mm
m Image Removed	Mathinline body --uriencoded--$$ \normalsize m = \dfrac%7BN - 0.95h%7D%7B2%7D = \dfrac%7B900 -0.95 \times 650%7D%7B2%7D =141.25\; \mathrm%7Bmm%7D $$
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 body --uriencoded--$$ \normalsize t_%7Breq%7D = 2 \sqrt %7B \dfrac%7B Tx %7D%7B\varphi F_y B %7D %7D = 2 \sqrt %7B \dfrac%7B 97.288 \times 96.25 \times 10%5e3 %7D%7B 0.9 \times 355 \times 500 %7D %7D =15.312\; \mathrm%7Bmm%7D $$	AISC DG-1-2nd 3.4.7a

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The limit state of the anchor rod tension rupture is checked according to AISC 360-16.

Ab Image Removed	Mathinline body --uriencoded--$$ \normalsize A_b = \dfrac%7B \pi d%5e2%7D %7B4%7D = \dfrac%7B \pi 24%5e2%7D %7B4%7D = 452.389\; \mathrm%7Bmm%5e2%7D $$
Fn	750000 kN/m2
Rn Image Removed	Mathinline body --uriencoded--$$ \normalsize R_n = F_n A_b = 750 \times 452.389 \times 10%5e%7B-3%7D =339.292 \; \mathrm%7BkN%7D $$	AISC 360-16 J3-1
ΦRn Image Removed	Mathinline body --uriencoded--$$ \normalsize \varphi R_n = 0.75 \times 339.292 = 254.469 \; \; \mathrm%7BkN%7D $$

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Ψ	1.0	ACI 318M-08 D.5.3.6
Np Image Removed	Mathinline body --uriencoded--$$ \normalsize N_p = 8A_%7Bbgr%7D f_c = 8 \times 7401.592 \times 25000 \times 10%5e%7B-6%7D $$ Mathinline body --uriencoded--$$ \normalsize N_p = 1480.318\; \mathrm%7BkN%7D $$	ACI 318M-08 (D-15)
Abgr	7401.592 mm2
fc	25000 kN/m2
Rn Image Removed	Mathinline body --uriencoded--$$ \normalsize R_n = \psi N_p = 1.0 \times 1480.318 \; \mathrm%7BkN%7D = 1480.318 $$	ACI 318M-08 (D-14)
ΦRn Image Removed	Mathinline body --uriencoded--$$ \normalsize \varphi R_n = 0.70 \times 1480.318 = 1036.222 \; \mathrm%7BkN%7D $$

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The calculation is made using the Elastic method, one of the methods selected in the steel analysis settings tab. In this check, the operation is performed on half of the symmetry axis and is calculated to form a force pair with the required force.

Ab Image Removed	Mathinline body --uriencoded--$$ \normalsize A_b = \dfrac%7B \pi d%5e2%7D %7B4%7D = \dfrac%7B \pi 24%5e2%7D %7B4%7D = 452.389 \; \mathrm%7Bmm%5e2%7D $$
Fn Image Removed	Mathinline body --uriencoded--$$ \normalsize F_n = 0.450 F_%7Bub%7D = 0.450 \times 1000 =450 \; \mathrm%7BN/mm%5e2%7D $$
Rn Image Removed	Mathinline body --uriencoded--$$ \normalsize R_n = F_n A_b = 450 \times 452.389 \times 10%5e%7B-3%7D =203.575\; \mathrm%7BkN%7D $$	AISC 360-16 J3-1
ΦRn	Image Removed Mathinline body --uriencoded--$$ \normalsize \varphi R_n = 0.75 \times 203.575 = 152.681 \; \mathrm%7BkN%7D $$

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The bearing strength limit states of the base plate, which are “shear tear out” and “ovalization of bolt hole” for both end and inner bolts, are checked according to AISC 360-16.

dh	24+3=27 mm
Lc,edge Image Removed	Mathinline body --uriencoded--$$ \normalsize L_%7Bc,edge%7D = L_e - 0.5d_h = 45 - 0.5 \times 27 =31.5 \; \mathrm%7Bmm%7D $$
Rn Image Removed	Mathinline body --uriencoded--\begin%7Baligned%7D \normalsize R_n = \mathrm%7Bmin%7D \left[\begin%7Bmatrix%7D 1.2L_c \times t \times F_u \\2.4d \times t \times F_u \end%7Bmatrix%7D\right] \end%7Baligned%7D	AISC 360-16 J3-6a
Rn-edge Image Removed	Mathinline body --uriencoded--\begin%7Baligned%7D \normalsize R_n = \mathrm%7Bmin%7D \left[\begin%7Bmatrix%7D 1.2(31.5)(42)(470 \times 10%5e%7B-3%7D) \\2.4(24)(42)(470 \times 10%5e%7B-3%7D) \end%7Bmatrix%7D\right] \end%7Baligned%7D =746.17\; \mathrm%7BkN%7D
Rn Image Removed	Mathinline body --uriencoded--$$ \normalsize R_%7Bn%7D = n_eR_%7Bn,edge%7D =1 \times 746.17 =746.17 \; \mathrm%7BkN%7D $$
ΦRn Image Removed	Mathinline body --uriencoded--$$ \normalsize \varphi R_n = 0.75 \times 746.17 = 559.628 \; \mathrm%7BkN%7D $$

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