Page Comparison

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Bearing strength limit states of the connection plate that 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	Rn-edge	Image Removed Mathinline
Rn	Image Removed	AISC 360-16 J3-6a
body --uriencoded--$$ \normalsize L_%7Bc,edge%7D = L_e - 0.5d_h = 55 - 0.5 \times 27 =41.5 $$
Rn Image Removed	Mathinline
ΦRn	Image Removed

Required

Available

Ratio

Control

0.031 kN

126.392 kN

0.00024

√

Web Plate Shear Yield

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.

A_g

Image Removed

 

F_y

235 N/mm²

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	Mathinline body --uriencoded--\begin%7Baligned%7D \normalsize R_n = \mathrm%7Bmin%7D \left[\begin%7Bmatrix%7D 1.2(41.5)(9.4)(360 \times 10%5e%7B-3%7D) \\2.4(24)(9.4)(360 \times 10%5e%7B-3%7D) \end%7Bmatrix%7D\right] \end%7Baligned%7D = 168.52
Rn

Image Removed

AISC 360-16 J4-3

ΦR_n

Image Removed

Mathinline body --uriencoded--$$ \normalsize R_%7Bn%7D = n_eR_%7Bn,edge%7D =1 \times 168.52 =168.52 $$
ΦRn	Mathinline body $$ \normalsize \varphi R_n = 0.75 \times 168.52 = 126.39 $$

Web Plate Shear

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Yield

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 rupture yielding is checked according to AISC 360-16.

A

_nv

g

Image Removed

Mathinline

 

A_e

Image Removed

body --uriencoded--$$ \normalsize A_%7Bg%7D = h_p \times t_ p = 2 \times 325 \times 10 =6500 $$
F

_u

y

360

235 N/mm2
Rn	Mathinline

Image Removed

AISC 360-16 J4-4

ΦR_n

Image Removed

 

Required

Available

Ratio

Control

0.075 kN

771.12 kN

0.0001

√

Beam Shear Rupture

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Web Plate Shear Rupture

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 rupture is checked according to AISC 360-16.

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A_nv

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A_e

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AISC 360-16.

Beam Shear Rupture

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 rupture is checked according to AISC 360-16.

Web Plate Block Shear

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	Mathinline body --uriencoded--$$ \normalsize A_%7Bg%7D = 2\times 10 \times ( 325 -49.5 ) =5510 $$
Anv	Mathinline body --uriencoded--$$ \normalsize A_%7Bnv%7D = 5510 - 2 \times ( 2.5 \times 29 \times 10 ) =4060 $$
Ant	Mathinline body --uriencoded--$$ \normalsize A_%7Bnt%7D = 10 \times 2 \times ( 55 - 0.5 \times 29 ) = 810 $$
Fy	235 N/mm2
Fu	360 N/mm2

R_n

Image Removed

AISC 360-16 J4-4

ΦR_n

Image Removed

 

Required

Available

Ratio

Control

0.075 kN

552.76 kN

0.0001

√

Web Plate Block Shear

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
Anv	Image Removed
Ant	Image Removed
Fy	235 N/mm2
Fu	360 N/mm2
Ubs	1.0
	Image Removed
Rn	Image Removed Ubs	1.0
	Mathinline body --uriencoded--$$ \normalsize U_%7Bbs%7D F_u A_%7Bnt%7D = 1 \times 360 \times 10%5e%7B-3%7D \times 810 = 291.6 $$ Mathinline body --uriencoded--$$ \normalsize 0.6 F_%7Bu%7D A_%7Bnv%7D = 0.6 \times 360 \times 10%5e%7B-3%7D \times 4060 = 876.96 $$ Mathinline body --uriencoded--$$ \normalsize 0.6 F_%7By%7D A_%7Bg%7D = 0.6 \times 235 \times 10%5e%7B-3%7D \times 5510 = 776.91 $$
Rn	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 = 776.91 + 291.6 =1068.51	AISC 360-16 J4-5
ΦRn Image Removed	Mathinline body $$ \normalsize \varphi R_n = 0.75 \times 1068.51 = 801.38 $$

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