Page Comparison

...

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--$$ \normalsize A_b = \dfrac%7B \pi d%5e2%7D %7B4%7D = \dfrac%7B \pi 16%5e2%7D %7B4%7D = 201.062\; \mathrm%7Bmm%5e2%7D $$
Fnv	Image Removed		Rn	Image Removed	Mathinline body --uriencoded--$$ \normalsize F_%7Bnv%7D=0.450F_%7Bub%7D=0.450 \times 800 = 360\; \mathrm%7BN/mm%7D $$
Rn	Mathinline body --uriencoded--$$ \normalsize R_n = F_%7Bn%7D \times A_b = R_%7Bnv%7D = n ( m F_%7Bnv%7D A_b) $$ Mathinline body --uriencoded--$$ \normalsize R_n = 4 \times (1 \times 360 \times 201.062 \times 10%5e%7B-3%7D )%7D = 289.53 \; \; \mathrm%7BkN%7D $$	AISC 360-16 J3-1
R n / Ω Image Removed	Mathinline body --uriencoded--$$ \normalsize R_n/ \Omega = 289.53 / 2 = 144.765 \; \; \mathrm%7BkN%7D $$

...

Bearing strength limit states of the connection part that are “shear tear out” and “ovalization of bolt hole” for both end and inner bolts are checked according to AISC 360-16.

dh	16+2=18 mm
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
Lc,spacing Image Removed	Mathinline body --uriencoded--$$ \normalsize L_%7Bc,spacing%7D = s - d_h = 60 - 18 =42\; \mathrm%7Bmm%7D $$
Rn-spacing Image Removed	Mathinline body --uriencoded--\begin%7Baligned%7D\normalsize R_%7Bn%7D=\mathrm%7Bmin%7D\left[\begin%7Bmatrix%7D 1.2 ( 42) ( 7.1 )(362.846 \times 10%5e%7B-3%7D) \\ 2.4 ( 16 ) ( 7.1 )(362.846 \times 10%5e%7B-3%7D) \end%7Bmatrix%7D\right]\end%7Baligned%7D = 98.926 \; \mathrm%7BkN%7D
Rn / Ω Image Removed	Mathinline body --uriencoded--$$ \normalsize R_n/ \Omega = 395.704 / 2 = 197.852 \; \; \mathrm%7BkN%7D $$

...

Bearing strength limit states of the connection part that are “shear tear out” and “ovalization of bolt hole” for both end and inner bolts are checked according to AISC 360-16.

dh	16+2=18 mm
Lc,edge Image Removed	Mathinline body --uriencoded--$$ \normalsize L_%7Bc,edge%7D = L_e - 0.5d_h = 45 -0.5 \times 18 = 36\; \mathrm%7Bmm%7D $$
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_%7Bn%7D=\mathrm%7Bmin%7D\left[\begin%7Bmatrix%7D 1.2 ( 36 ) ( 12 )(362.846 \times 10%5e%7B-3%7D) \\ 2.4 ( 16 ) (12 )(362.846 \times 10%5e%7B-3%7D) \end%7Bmatrix%7D\right]\end%7Baligned%7D = 167.199\; \mathrm%7BkN%7D
Lc,spacing Image Removed	Mathinline body --uriencoded--$$ \normalsize L_%7Bc,spacing%7D = s - d_h = 60 - 18 =42 $$
Rn-spacing Image Removed	Mathinline body --uriencoded--\begin%7Baligned%7D\normalsize R_%7Bn%7D=\mathrm%7Bmin%7D\left[\begin%7Bmatrix%7D 1.2 ( 42 ) ( 12 )(362.846 \times 10%5e%7B-3%7D) \\ 2.4 ( 16 ) (12 )(362.846 \times 10%5e%7B-3%7D) \end%7Bmatrix%7D\right]\end%7Baligned%7D = 167.199 \; \mathrm%7BkN%7D
Rn Image Removed	Mathinline body --uriencoded--\normalsize R_%7Bn%7D = n_e \times R_%7Bn,edge%7D + n_s \times R_%7Bn,spacing%7D Mathinline body --uriencoded--$$ \normalsize R_%7Bn%7D = 4 \times 167.199 = 668.797 \; \mathrm%7BkN%7D $$
Rn / Ω Image Removed	Mathinline body --uriencoded--$$ \normalsize R_n/ \Omega = 668.797 / 2 = 334.399 \; \; \mathrm%7BkN%7D $$

Beam Shear Yield

In the case of the block shear limit state, the gross area yielding of the tensile plane The shear strength of connecting elements in shear is the minimum value obtained according to the limit states of shear yielding and shear rupture. Shear yielding is checked according to AISC 360-16.

Ag Image Removed	Mathinline body --uriencoded--$$ \normalsize A_%7Bg%7D = L_pt_p = ( 240 - 30.7) \times 6.2 = 1297.66 \; \; \mathrm%7Bmm%5e2%7D $$
Fy	235.359 N/mm2
Rn Image Removed	Mathinline body --uriencoded--\normalsize R_n = 0.6F_%7By%7D A_g Mathinline body --uriencoded--$$ \normalsize R_n = 0.6 \times 235.359 \times 10%5e%7B-3%7D \times 1297.66 = 183.25 \; \; \mathrm%7BkN%7D $$	AISC 360-16 J4-3
Rn / Ω Image Removed	Mathinline body --uriencoded--$$ \normalsize R_n/ \Omega = 183.25 / 1.5 = 122.166 \; \; \mathrm%7BkN%7D $$

Plate Shear Yield

The gross area yielding of the tensile plane shear strength of connecting elements in shear is the minimum value obtained according to the limit states of shear yielding and shear rupture. Shear yielding is checked according to AISC 360-16.

Ag Image Removed	Mathinline body --uriencoded--$$ \normalsize A_%7Bg%7D = L_pt_p = 2 \times 150 \times 12 = 3600 \; \; \mathrm%7Bmm%5e2%7D $$
Fy	235.359 N/mm2
Rn	Mathinline body --uriencoded--$$ \normalsize R_n Image Removed = 0.6F_%7By%7D A_g $$ Mathinline body --uriencoded--$$ \normalsize R_n = 0.6 \times 235.359 \times 10%5e%7B-3%7D \times 3600 = 508.16 \; \; \mathrm%7BkN%7D $$	AISC 360-16 J4-3
Rn / Ω Image Removed	Mathinline body --uriencoded--$$ \normalsize R_n/ \Omega = 508.16 / 1.5 = 338.77 \; \; \mathrm%7BkN%7D $$

Plate Shear Rupture

The net area rupture of the tensile plane of the connection part shear strength of connecting elements in shear is the minimum value obtained according to the limit states of shear yielding and shear rupture. Shear rupture is checked according to AISC 360-16.

Anv Image Removed	Mathinline body --uriencoded--$$ \normalsize A_%7Bnv%7D = t_p(d_b-n_bd_e) = 2 \times 12 \times (150 - 2 \times 20) = 2640 \; \; \mathrm%7Bmm%5e2%7D $$
Fu	362.846 N/mm2
Rn Image Removed	Mathinline body --uriencoded--$$ \normalsize R_n = 0.6F_%7Bu%7D A_%7Bnv%7D $$ Mathinline body --uriencoded--$$ \normalsize R_n = 0.6 \times 362.846 \times 10%5e%7B-3%7D \times 2640 = 574.75 \; \; \mathrm%7BkN%7D $$	AISC 360-16 J4-4
Rn / Ω Image Removed	Mathinline body --uriencoded--$$ \normalsize R_n/ \Omega = 574.75 / 2 = 287.374 \; \; \mathrm%7BkN%7D $$

...

The block shear limit state is checked according to AISC 360-16. All block shear modes combined with tensile failure on one plane and shear failure on a perpendicular plane are checked according to AISC 360-16.

Ag Image Removed	Mathinline body --uriencoded--\normalsize A_%7Bg%7D = (60 + 45) \times 12 \times 2 = 2520 \; \; \mathrm%7Bmm%5e2%7D
Anv Image Removed	Mathinline body --uriencoded--\normalsize A_%7Bnv%7D = ((60 + 45)-1.5 \times 20) \times 12 \times 2 = 1800 \; \; \mathrm%7Bmm%5e2%7D
Ant Image Removed	Mathinline body --uriencoded--\normalsize A_%7Bnt%7D = 2 \times 12 \times (40-0.5 \times 20) = 720 \; \; \mathrm%7Bmm%5e2%7D
Fy	235.359 N/mm2
Fu	362.846 N/mm2
Ubs	1.0
Image Removed	Mathinline body --uriencoded--$$ \normalsize U_%7Bbs%7DF_%7Bu%7DA_%7Bnt%7D=1 \times 362.846 \times 10%5e%7B-3%7D \times 720 = 261. 25 $$ Mathinline body --uriencoded--$$ \normalsize 0.6 F_%7Bu%7D A_%7Bnv%7D = 0.6 \times 362.846 \times 10%5e%7B-3%7D \times 1800 = 391.874 $$ Mathinline body --uriencoded--$$ \normalsize 0.6 F_%7By%7D A_%7Bg%7D = 0.6 \times 235.359 \times 10%5e%7B-3%7D \times 2520 = 355.863 $$
Rn Image Removed	Mathinline body --uriencoded--\begin%7Baligned%7D \normalsize R_n = \mathrm%7Bmin%7D \left[\begin%7Bmatrix%7D 0.6F_uA_%7Bnv%7D \\0.6F_yA_%7Bg%7D \end%7Bmatrix%7D\right] \end%7Baligned%7D + U_%7Bbs%7DF_uA_%7Bnt%7D Mathinline body --uriencoded--$$ \normalsize R_%7Bn%7D =355.863 + 261.25 = 617.113 \; \mathrm%7BkN%7D $$	AISC 360-16 J4-5
Rn / Ω Image Removed	Mathinline body --uriencoded--$$ \normalsize R_n/ \Omega = 617.113 / 2 = 308.556 \; \; \mathrm%7BkN%7D $$

Welding Strength

Fe	480000 kN480 N/mmm2
w	7.07 mm
Fu	362.846  N/mm2
t	6.2 mm
l	135.856 mm
Rnw	Image Removed
RnBM	Image Removed
Rn	Image Removed	Mathinline body --uriencoded--$$ \normalsize R_%7Bnw%7D = 0.6 \times F_e \times 2 \times 0.707 \times w \times l $$ Mathinline body --uriencoded--$$ \normalsize R_%7Bnw%7D = 0.6 \times 480 \times 2 \times 0.707 \times 7.072 \times 135.856 \times 10%5e%7B-3%7D = 391.258 \; \mathrm%7BkN%7D $$
RnBM	Mathinline body --uriencoded--$$ \normalsize R_%7BnBM%7D = 0.6 F_u t l = 0.6 \times 362.846 \times 6.2 \times 135.856 \times 10%5e%7B-3%7D = 183.376\; \mathrm%7BkN%7D $$
Rn	Mathinline body --uriencoded--$$ \normalsize R_n = min (R_%7Bnw%7D,R_%7BnBM%7D) = 183.376\; \mathrm%7BkN%7D $$
Rn / Ω Image Removed	Mathinline body --uriencoded--$$ \normalsize R_n/ \Omega = 183.376 / 2 = 91.688 \; \; \mathrm%7BkN%7D $$

...