Brace sect HSS6.875X0.500

 = A500 Gr.C Round

 = 46.0

[ksi]

Ratio of expected Fy to specified min Fy

 = 1.30

AISC 341-16  Table A3.1

 = 9.360

[in²]

 = 2.272

[in]

 = 29000

[ksi]

Brace member length & effective length factor K

 = 144.0

[in]

 = 1.00

Brace expected yield strength in tension

 = R_y F_y A_g

 = 559.7

[kips]

AISC 341-16 F2.6c (1)

Right Brace seismic design force in tension

 = P_et

 = -559.7

[kips]

AISC 341-16 F2.6c (1)

Member length L & effective length factor K

 = 144.0

[in]

 = 1.00

Member radius of gyration & elastic modulus

 = 2.272

[in]

 = 29000

[ksi]

Member slenderness ratio

 = K x L  / r

 = 63.37

Elastic buckling stress

 = 

 = 71.27

[ksi]

AISC 15^th  Eq E3-4

when     ≤   4.71 ( ) 0.5 = 103.72

AISC 15^th  E3

Critical stress

 = 0.658 ( R_y F_y / F_e )  R_y F_y

 = 42.09

[ksi]

AISC 15^th  Eq E3-2

Brace expected yield strength in compression

 = min ( R_y F_y A_g , 1.14 F_cr A_g )

 = 449.1

[kips]

AISC 341-16 F2.3

Brace force in compression

 = from user input in load section

 = 0.0

[kips]

Right Brace seismic design force in compression

 = P_ec

 = 449.1

[kips]

AISC 341-16 F2.6c (2)

Right Brace seismic design force in compression - post-buckling

 = 0.3 x P_ec

 = 134.7

[kips]

AISC 341-16 F2.3 (ii)

Brace sect HSS6.875X0.500

 = A500 Gr.C Round

 = 46.0

[ksi]

Ratio of expected Fy to specified min Fy

 = 1.30

AISC 341-16  Table A3.1

 = 9.360

[in²]

 = 2.272

[in]

 = 29000

[ksi]

Brace member length & effective length factor K

 = 144.0

[in]

 = 1.00

Brace expected yield strength in tension

 = R_y F_y A_g

 = 559.7

[kips]

AISC 341-16 F2.6c (1)

Left Brace seismic design force in tension

 = P_et

 = -559.7

[kips]

AISC 341-16 F2.6c (1)

Member length L & effective length factor K

 = 144.0

[in]

 = 1.00

Member radius of gyration & elastic modulus

 = 2.272

[in]

 = 29000

[ksi]

Member slenderness ratio

 = K x L  / r

 = 63.37

Elastic buckling stress

 = 

 = 71.27

[ksi]

AISC 15^th  Eq E3-4

when     ≤   4.71 ( ) 0.5 = 103.72

AISC 15^th  E3

Critical stress

 = 0.658 ( R_y F_y / F_e )  R_y F_y

 = 42.09

[ksi]

AISC 15^th  Eq E3-2

Brace expected yield strength in compression

 = min ( R_y F_y A_g , 1.14 F_cr A_g )

 = 449.1

[kips]

AISC 341-16 F2.3

Brace force in compression

 = from user input in load section

 = 0.0

[kips]

Left Brace seismic design force in compression

 = P_ec

 = 449.1

[kips]

AISC 341-16 F2.6c (2)

Left Brace seismic design force in compression - post-buckling

 = 0.3 x P_ec

 = 134.7

[kips]

AISC 341-16 F2.3 (ii)

Refer to AISC 341-16 F2.3(i), LC1 & LC2 are the load cases in which all braces are assumed to resist forces corresponding to their expected strength in tension P_{s_t}   or in compression P_{s_ci}

F2.3(ii), LC3 & LC4 are the load cases in which all braces are assumed to resist forces corresponding to their
expected strength in tension P_{s_t}   and all braces in compression are assumed to resist their expected
compressive post-buckling strength P_{s_cii}

LC1       Right Brace   P_{s_t} =-559.7 kips (T)                   Left Brace   P_{s_ci} =449.1 kips (C)

AISC 341-16 F2.3(i)

LC2       Right Brace   P_{s_ci} =449.1 kips (C)                   Left Brace   P_{s_t} =-559.7 kips (T)

LC3       Right Brace   P_{s_t} =-559.7 kips (T)                   Left Brace   P_{s_cii} =134.7 kips (C)       post-buckling

AISC 341-16 F2.3(ii)

LC4       Right Brace   P_{s_cii} =134.7 kips (C)                   Left Brace   P_{s_t} =-559.7 kips (T)       post-buckling

Refer to AISC DG29  Fig. 4-5 ~ Fig. 4-7 for all charts and definitions of variables and symbols shown in calculation below

Right brace axial force

 = from seismic brace force calc

 = -559.7

[kips]

in tension

Right brace to hor line angle

 = from user input

 = 45.0

[°]

 = -395.8

[kips]

 = -395.8

[kips]

Left brace axial force

 = from seismic brace force calc

 = 449.1

[kips]

in compression

Left brace to hor line angle

 = from user input

 = 45.0

[°]

 = 317.6

[kips]

 = 317.6

[kips]

 = 27.000

[in]

 = 27.000

[in]

 = L₁ + L₂

 = 54.000

[in]

 = ( L₂ - L₁ ) /2

 = 0.000

[in]

 = 11.050

[in]

 = 31.294

[in]

 = -364.44

[kip-ft]

 = -292.42

[kip-ft]

AISC DG29  Fig. 4-6

Shear

 = H₁ - H₂

 = -713.3

[kips]

Axial

 = V₁ + V₂

 = -78.2

[kips]

in tension

Moment

 = M₁ + M₂

 = -656.86

[kip-ft]

AISC DG29  Fig. 4-7

Shear

 = ( V₁ + V₂ ) + - V₁

 = 64.7

[kips]

Axial

 = ( H₁ - H₂ ) x -1 + H₁

 = -39.1

[kips]

in tension

Moment

 =  (V₁+V₂ ) + (H₁-H₂ ) +

 = 43.00

[kip-ft]

- V₁ Δ - H₁ (e + )

Refer to AISC DG29  Fig. 4-5 ~ Fig. 4-7 for all charts and definitions of variables and symbols shown in calculation below

Right brace axial force

 = from seismic brace force calc

 = 449.1

[kips]

in compression

Right brace to hor line angle

 = from user input

 = 45.0

[°]

 = 317.6

[kips]

 = 317.6

[kips]

Left brace axial force

 = from seismic brace force calc

 = -559.7

[kips]

in tension

Left brace to hor line angle

 = from user input

 = 45.0

[°]

 = -395.8

[kips]

 = -395.8

[kips]

 = 27.000

[in]

 = 27.000

[in]

 = L₁ + L₂

 = 54.000

[in]

 = ( L₂ - L₁ ) /2

 = 0.000

[in]

 = 11.050

[in]

 = 31.294

[in]

 = 292.42

[kip-ft]

 = 364.44

[kip-ft]

AISC DG29  Fig. 4-6

Shear

 = H₁ - H₂

 = 713.3

[kips]

Axial

 = V₁ + V₂

 = -78.2

[kips]

in tension

Moment

 = M₁ + M₂

 = 656.86

[kip-ft]

AISC DG29  Fig. 4-7

Shear

 = ( V₁ + V₂ ) + - V₁

 = -64.7

[kips]

Axial

 = ( H₁ - H₂ ) x -1 + H₁

 = -39.1

[kips]

in tension

Moment

 =  (V₁+V₂ ) + (H₁-H₂ ) +

 = 43.00

[kip-ft]

- V₁ Δ - H₁ (e + )

Check HSS section limiting width-to-thickness ratio for HSS wall in compression as Highly Ductile section
per AISC Seismic Design Manual 3rd Ed Table 1-D

AISC SDM 3^rd  Table 1-D

CHS sect HSS6.875X0.500

 = 6.875

[in]

 = 0.465

[in]

HSS sect HSS6.875X0.500

 = A500 Gr.C Round

 = 46.0

[ksi]

 = 29000

[ksi]

Ratio of expected Fy to specified min Fy

 = 1.30

AISC 341-16  Table A3.1

CHS width-to-thickness ratio limit

 = 0.053

 = 25.70

AISC SDM 3^rd  Table 1-D

CHS width-to-thickness ratio actual

 = D/t

 = 14.78

 ≤ λ_hd

OK

CHS sect HSS6.875X0.500

 = 6.875

[in]

 = 0.465

[in]

 = 9.360

[in²]

Gusset plate thickness

 = from user input

 = 0.750

[in]

HSS cut slot width

 = t_gp + 1/8"

 = 0.875

[in]

HSS brace net area

 = A_g - 2 w t

 = 8.546

[in²]

Reinforcing plate

 = 1.250

[in]

 = 1.250

[in]

Reinforcing plate area

 = w_r x t_r

 = 1.563

[in²]

CHS 1/2 net area A₁ = 0.5A_nb

 = 4.273

[in²]

 = 2.040

[in]

Reinforce plate

 = 1.563

[in²]

 = 4.063

[in]

Dist to centroid of comb sect

 = 

 = 2.582

[in]

Length of connection

 = 

 = 26.000

[in]

Shear lag factor

 = 1 - x / L

 = 0.901

AISC 15^th  Table D3.1

Total net area

 = A_nb + 2 x A_r

 = 11.671

[in²]

Total effective net area

 = U A_n

 = 10.512

[in²]

The brace effective net area shall not be less than the brace gross area

AISC 341-16 F2.5b (3)

HSS sect HSS6.875X0.500

 = brace gross area

 = 9.360

[in²]

Total brace effective net area

 = U A_n

 = 10.512

[in²]

 ≥ A_g

OK

AISC 341-16 F2.5b (3)

The specified minimum yield strength of the reinforce plate shall be at least the specified
minimum yield strength of the brace

AISC 341-16 F2.5b (3)(i)

HSS sect HSS6.875X0.500

 = A500 Gr.C Round

 = 46.0

[ksi]

Reinforce plate

 = A992

 = 50.0

[ksi]

 ≥ F_y

OK

AISC 341-16 F2.5b (3)(i)

Thinner part joined thickness

 = 

 = 0.465

[in]

Min fillet weld size allowed

 = 

 = 0.188

[in]

AISC 15^th  Table J2.4

Fillet weld size provided

 = 

 = 0.250

[in]

 ≥ w_min

OK

Fillet weld size provided

 = 

 = 0.250

[in]

Min fillet weld length allowed

 = 4 x w

 = 1.000

[in]

AISC 15^th  J2.2b

Min fillet weld length

 = 

 = 26.000

[in]

 ≥ L_min

OK

HSS sect HSS6.875X0.500 wall thick

 = 0.465

[in]

 = 46.0

[ksi]

HSS brace wall-gusset overlap length

 = 26.000

[in]

Ratio of expected Fy to specified min Fy

 = 1.30

AISC 341-16  Table A3.1

Beam axial load

 = from seismic load calc

 = 559.7

[kips]

HSS brace wall-gusset shear yielding

 = 0.6 R_y F_y t L x 4 walls

 = 1735.2

[kips]

AISC 15^th  Eq J4-3

Resistance factor-LRFD

 = 1.00

AISC 15^th  Eq J4-3

 = 

 = 1735.2

[kips]

 = 0.32

 > P_u

OK

HSS sect HSS6.875X0.500 wall thick

 = 0.465

[in]

 = 62.0

[ksi]

HSS brace wall-gusset overlap length

 = 26.000

[in]

Ratio of expected Fu to specified min Fu

 = 1.20

AISC 341-16  Table A3.1

Beam axial load

 = from seismic load calc

 = 559.7

[kips]

HSS brace wall-gusset shear rupture

 = 0.6 R_t F_u t L x 4 walls

 = 2158.8

[kips]

AISC 15^th  Eq J4-4

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq J4-4

 = 

 = 1619.1

[kips]

 = 0.35

 > P_u

OK

Plate thickness

 = 0.750

[in]

Plate strength

 = 50.0

[ksi]

 = 65.0

[ksi]

C shape weld group size

 = 26.000

[in]

 = 6.875

[in]

Gross area subject to shear

 = b t_p x 2

 = 39.000

[in²]

Net area subject to shear

 = A_gvb

 = 39.000

[in²]

Net area subject to tension

 = d t_p

 = 5.156

[in²]

Block shear strength required

 = 

 = 559.7

[kips]

Uniform tension stress factor

 = 1.00

AISC 15^th  Fig C-J4.2

Bolt shear resistance provided

 = min (0.6F_u A_nv , 0.6F_y A_gv ) +

 = 1505.2

[kips]

AISC 15^th  Eq J4-5

U_bs F_u A_nt

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq J4-5

 = 

 = 1128.9

[kips]

 = 0.50

 > V_u

OK

Plate size

 = 29.411

[in]

 = 0.750

[in]

Plate yield strength

 = 50.0

[ksi]

Plate gross area in shear

 = b_p t_p

 = 22.058

[in²]

Tensile force required

 = 

 = 559.7

[kips]

Plate tensile yielding strength

 = F_y A_g

 = 1102.9

[kips]

AISC 15^th  Eq J4-1

Resistance factor-LRFD

 = 0.90

AISC 15^th  Eq J4-1

 = 

 = 992.6

[kips]

 = 0.56

 > P_u

OK

Plate size

 = 29.411

[in]

 = 0.750

[in]

Plate tensile strength

 = 65.0

[ksi]

Plate net area in tension

 = b_p t_p

 = 22.058

[in²]

Tensile force required

 = 

 = 559.7

[kips]

Plate tensile rupture strength

 = F_u A_nt

 = 1433.8

[kips]

AISC 15^th  Eq J4-2

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq J4-2

 = 

 = 1075.3

[kips]

AISC 15^th  Eq J4-2

 = 0.52

 > P_u

OK

Fillet weld leg size

 = ¹⁄₄

[in]

 = 0.0

[°]

Electrode strength

 = 70.0

[ksi]

 = 1.00

AISC 15^th  Table 8-3

Number of weld line

 = 2   for double fillet

Load angle coefficient

 = ( 1 + 0.5 sin^1.5 θ )

 = 1.00

AISC 15^th  Page 8-9

Fillet weld shear strength

 = 0.6 (C₁ x 70 ksi) 0.707 w n C₂

 = 14.847

[kip/in]

AISC 15^th  Eq 8-1

Base metal - brace

 = 0.750

[in]

 = 65.0

[ksi]

Base metal - brace is in shear, shear rupture as per AISC 15^th  Eq J4-4 is checked

AISC 15^th  J2.4

Base metal shear rupture

 = 0.6 F_u t

 = 29.250

[kip/in]

AISC 15^th  Eq J4-4

Double fillet linear shear strength

 = min ( R_n-w , R_n-b )

 = 14.847

[kip/in]

AISC 15^th  Eq 9-2

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq 8-1

 = 

 = 11.135

[kip/in]

Shear resistance required

 = 

 = 559.7

[kips]

Fillet weld leg size

 = ¹⁄₄

[in]

 = 26.000

[in]

when L/w = 104.0 >100 , weld length reduction applied

AISC 15^th  J2.2b

Weld length reduction factor

 = 1.2 - 0.002 (L/w) ≤ 1.0

 = 0.99

AISC 15^th  Eq J2-1

Weld length used for design

 = β x L

 = 51.584

[in]

Fillet weld length - double fillet

 = 

 = 51.584

[in]

Shear resistance provided

 = φ R_n x L

 = 574.4

[kips]

 = 0.97

 > V_u

OK

Reinforcing plate

 = 1.250

[in]

 = 1.250

[in]

 = 50.0

[ksi]

Ratio of expected Fy to specified min Fy

 = 1.10

AISC 341-16  Table A3.1

Reinforcing plate area

 = w_r x t_r

 = 1.563

[in²]

Required strength of weld

 = R_y F_y A_r

 = 85.9

[kips]

Longitudinal weld length

 = reinforce plate length

 = 8.000

[in]

Transverse weld length

 = reinforce plate width

 = 1.250

[in]

Total weld length - single fillet weld

 = 2 x L_L + L_T

 = 17.250

[in]

AISC 15^th  Eq J2-10a

 = 0.85 x 2 x L_L + 1.5 x L_T

 = 15.475

[in]

AISC 15^th  Eq J2-10b

 = max ( L₁ , L₂ )

 = 17.250

[in]

AISC 15^th  J2.4 (c)

Fillet weld leg size

 = ¹⁄₄

[in]

 = 0.0

[°]

Electrode strength

 = 70.0

[ksi]

 = 1.00

AISC 15^th  Table 8-3

Number of weld line

 = 1   for single fillet

Load angle coefficient

 = ( 1 + 0.5 sin^1.5 θ )

 = 1.00

AISC 15^th  Page 8-9

Fillet weld shear strength

 = 0.6 (C₁ x 70 ksi) 0.707 w n C₂

 = 7.424

[kip/in]

AISC 15^th  Eq 8-1

Base metal - reinforce plate

 = 1.250

[in]

 = 65.0

[ksi]

Base metal - reinforce plate is in shear, shear rupture as per AISC 15^th  Eq J4-4 is checked

AISC 15^th  J2.4

Base metal shear rupture

 = 0.6 F_u t

 = 48.750

[kip/in]

AISC 15^th  Eq J4-4

Single fillet linear shear strength

 = min ( R_n-w , R_n-b )

 = 7.424

[kip/in]

AISC 15^th  Eq 9-2

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq 8-1

 = 

 = 5.568

[kip/in]

Shear resistance required

 = 

 = 85.9

[kips]

Fillet weld length - single fillet

 = 

 = 17.250

[in]

Shear resistance provided

 = φ R_n x L

 = 96.0

[kips]

 = 0.89

 > P_u

OK

Plate size

 = 29.411

[in]

 = 0.750

[in]

 = 50.0

[ksi]

 = 29000

[ksi]

Plate gross area in compression

 = b_p t_p

 = 22.058

[in²]

Plate radius of gyration

 = t_p / √12

 = 0.217

[in]

Plate effective length factor

 = 

 = 0.65

Plate unbraced length

 = 

 = 12.778

[in]

Plate slenderness

 = 0.65 x L_u  / r

 = 38.36

when     >  25 , use Chapter E

AISC 15^th  J4.4 (b)

Elastic buckling stress

 = 

 = 194.49

[ksi]

AISC 15^th  Eq E3-4

when     ≤ 4.71 ( ) 0.5 = 113.43

AISC 15^th  E3 (a)

Critical stress

 = 0.658 ( F_y / F_e )   F_y

 = 44.90

[ksi]

AISC 15^th  Eq E3-2

Plate compression required

 = P_c = 449.1

 = 449.1

[kips]

Plate compression provided

 = F_cr x A_g

 = 990.4

[kips]

AISC 15^th  Eq E3-1

Resistance factor-LRFD

 = 0.90

AISC 15^th  E1

 = 

 = 891.4

[kips]

 = 0.50

 > P_u

OK

Fillet weld leg size

 = ¹⁄₄

[in]

 = 0.0

[°]

Electrode strength

 = 70.0

[ksi]

 = 1.00

AISC 15^th  Table 8-3

Number of weld line

 = 2   for double fillet

Load angle coefficient

 = ( 1 + 0.5 sin^1.5 θ )

 = 1.00

AISC 15^th  Page 8-9

Fillet weld shear strength

 = 0.6 (C₁ x 70 ksi) 0.707 w n C₂

 = 14.847

[kip/in]

AISC 15^th  Eq 8-1

Base metal - brace

 = 0.750

[in]

 = 65.0

[ksi]

Base metal - brace is in shear, shear rupture as per AISC 15^th  Eq J4-4 is checked

AISC 15^th  J2.4

Base metal shear rupture

 = 0.6 F_u t

 = 29.250

[kip/in]

AISC 15^th  Eq J4-4

Double fillet linear shear strength

 = min ( R_n-w , R_n-b )

 = 14.847

[kip/in]

AISC 15^th  Eq 9-2

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq 8-1

 = 

 = 11.135

[kip/in]

Shear resistance required

 = 

 = 449.1

[kips]

Fillet weld leg size

 = ¹⁄₄

[in]

 = 26.000

[in]

when L/w = 104.0 >100 , weld length reduction applied

AISC 15^th  J2.2b

Weld length reduction factor

 = 1.2 - 0.002 (L/w) ≤ 1.0

 = 0.99

AISC 15^th  Eq J2-1

Weld length used for design

 = β x L

 = 51.584

[in]

Fillet weld length - double fillet

 = 

 = 51.584

[in]

Shear resistance provided

 = φ R_n x L

 = 574.4

[kips]

 = 0.78

 > V_u

OK

Reinforcing plate

 = 1.250

[in]

 = 1.250

[in]

 = 50.0

[ksi]

Ratio of expected Fy to specified min Fy

 = 1.10

AISC 341-16  Table A3.1

Reinforcing plate area

 = w_r x t_r

 = 1.563

[in²]

Required strength of weld

 = R_y F_y A_r

 = 85.9

[kips]

Longitudinal weld length

 = reinforce plate length

 = 8.000

[in]

Transverse weld length

 = reinforce plate width

 = 1.250

[in]

Total weld length - single fillet weld

 = 2 x L_L + L_T

 = 17.250

[in]

AISC 15^th  Eq J2-10a

 = 0.85 x 2 x L_L + 1.5 x L_T

 = 15.475

[in]

AISC 15^th  Eq J2-10b

 = max ( L₁ , L₂ )

 = 17.250

[in]

AISC 15^th  J2.4 (c)

Fillet weld leg size

 = ¹⁄₄

[in]

 = 0.0

[°]

Electrode strength

 = 70.0

[ksi]

 = 1.00

AISC 15^th  Table 8-3

Number of weld line

 = 1   for single fillet

Load angle coefficient

 = ( 1 + 0.5 sin^1.5 θ )

 = 1.00

AISC 15^th  Page 8-9

Fillet weld shear strength

 = 0.6 (C₁ x 70 ksi) 0.707 w n C₂

 = 7.424

[kip/in]

AISC 15^th  Eq 8-1

Base metal - reinforce plate

 = 1.250

[in]

 = 65.0

[ksi]

Base metal - reinforce plate is in shear, shear rupture as per AISC 15^th  Eq J4-4 is checked

AISC 15^th  J2.4

Base metal shear rupture

 = 0.6 F_u t

 = 48.750

[kip/in]

AISC 15^th  Eq J4-4

Single fillet linear shear strength

 = min ( R_n-w , R_n-b )

 = 7.424

[kip/in]

AISC 15^th  Eq 9-2

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq 8-1

 = 

 = 5.568

[kip/in]

Shear resistance required

 = 

 = 85.9

[kips]

Fillet weld length - single fillet

 = 

 = 17.250

[in]

Shear resistance provided

 = φ R_n x L

 = 96.0

[kips]

 = 0.89

 > P_u

OK

Check HSS section limiting width-to-thickness ratio for HSS wall in compression as Highly Ductile section
per AISC Seismic Design Manual 3rd Ed Table 1-D

AISC SDM 3^rd  Table 1-D

CHS sect HSS6.875X0.500

 = 6.875

[in]

 = 0.465

[in]

HSS sect HSS6.875X0.500

 = A500 Gr.C Round

 = 46.0

[ksi]

 = 29000

[ksi]

Ratio of expected Fy to specified min Fy

 = 1.30

AISC 341-16  Table A3.1

CHS width-to-thickness ratio limit

 = 0.053

 = 25.70

AISC SDM 3^rd  Table 1-D

CHS width-to-thickness ratio actual

 = D/t

 = 14.78

 ≤ λ_hd

OK

CHS sect HSS6.875X0.500

 = 6.875

[in]

 = 0.465

[in]

 = 9.360

[in²]

Gusset plate thickness

 = from user input

 = 0.750

[in]

HSS cut slot width

 = t_gp + 1/8"

 = 0.875

[in]

HSS brace net area

 = A_g - 2 w t

 = 8.546

[in²]

Reinforcing plate

 = 1.250

[in]

 = 1.250

[in]

Reinforcing plate area

 = w_r x t_r

 = 1.563

[in²]

CHS 1/2 net area A₁ = 0.5A_nb

 = 4.273

[in²]

 = 2.040

[in]

Reinforce plate

 = 1.563

[in²]

 = 4.063

[in]

Dist to centroid of comb sect

 = 

 = 2.582

[in]

Length of connection

 = 

 = 26.000

[in]

Shear lag factor

 = 1 - x / L

 = 0.901

AISC 15^th  Table D3.1

Total net area

 = A_nb + 2 x A_r

 = 11.671

[in²]

Total effective net area

 = U A_n

 = 10.512

[in²]

The brace effective net area shall not be less than the brace gross area

AISC 341-16 F2.5b (3)

HSS sect HSS6.875X0.500

 = brace gross area

 = 9.360

[in²]

Total brace effective net area

 = U A_n

 = 10.512

[in²]

 ≥ A_g

OK

AISC 341-16 F2.5b (3)

The specified minimum yield strength of the reinforce plate shall be at least the specified
minimum yield strength of the brace

AISC 341-16 F2.5b (3)(i)

HSS sect HSS6.875X0.500

 = A500 Gr.C Round

 = 46.0

[ksi]

Reinforce plate

 = A992

 = 50.0

[ksi]

 ≥ F_y

OK

AISC 341-16 F2.5b (3)(i)

Thinner part joined thickness

 = 

 = 0.465

[in]

Min fillet weld size allowed

 = 

 = 0.188

[in]

AISC 15^th  Table J2.4

Fillet weld size provided

 = 

 = 0.250

[in]

 ≥ w_min

OK

Fillet weld size provided

 = 

 = 0.250

[in]

Min fillet weld length allowed

 = 4 x w

 = 1.000

[in]

AISC 15^th  J2.2b

Min fillet weld length

 = 

 = 26.000

[in]

 ≥ L_min

OK

Plate size

 = 29.411

[in]

 = 0.750

[in]

 = 50.0

[ksi]

 = 29000

[ksi]

Plate gross area in compression

 = b_p t_p

 = 22.058

[in²]

Plate radius of gyration

 = t_p / √12

 = 0.217

[in]

Plate effective length factor

 = 

 = 0.65

Plate unbraced length

 = 

 = 12.778

[in]

Plate slenderness

 = 0.65 x L_u  / r

 = 38.36

when     >  25 , use Chapter E

AISC 15^th  J4.4 (b)

Elastic buckling stress

 = 

 = 194.49

[ksi]

AISC 15^th  Eq E3-4

when     ≤ 4.71 ( ) 0.5 = 113.43

AISC 15^th  E3 (a)

Critical stress

 = 0.658 ( F_y / F_e )   F_y

 = 44.90

[ksi]

AISC 15^th  Eq E3-2

Plate compression required

 = P_c = 449.1

 = 449.1

[kips]

Plate compression provided

 = F_cr x A_g

 = 990.4

[kips]

AISC 15^th  Eq E3-1

Resistance factor-LRFD

 = 0.90

AISC 15^th  E1

 = 

 = 891.4

[kips]

 = 0.50

 > P_u

OK

Fillet weld leg size

 = ¹⁄₄

[in]

 = 0.0

[°]

Electrode strength

 = 70.0

[ksi]

 = 1.00

AISC 15^th  Table 8-3

Number of weld line

 = 2   for double fillet

Load angle coefficient

 = ( 1 + 0.5 sin^1.5 θ )

 = 1.00

AISC 15^th  Page 8-9

Fillet weld shear strength

 = 0.6 (C₁ x 70 ksi) 0.707 w n C₂

 = 14.847

[kip/in]

AISC 15^th  Eq 8-1

Base metal - brace

 = 0.750

[in]

 = 65.0

[ksi]

Base metal - brace is in shear, shear rupture as per AISC 15^th  Eq J4-4 is checked

AISC 15^th  J2.4

Base metal shear rupture

 = 0.6 F_u t

 = 29.250

[kip/in]

AISC 15^th  Eq J4-4

Double fillet linear shear strength

 = min ( R_n-w , R_n-b )

 = 14.847

[kip/in]

AISC 15^th  Eq 9-2

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq 8-1

 = 

 = 11.135

[kip/in]

Shear resistance required

 = 

 = 449.1

[kips]

Fillet weld leg size

 = ¹⁄₄

[in]

 = 26.000

[in]

when L/w = 104.0 >100 , weld length reduction applied

AISC 15^th  J2.2b

Weld length reduction factor

 = 1.2 - 0.002 (L/w) ≤ 1.0

 = 0.99

AISC 15^th  Eq J2-1

Weld length used for design

 = β x L

 = 51.584

[in]

Fillet weld length - double fillet

 = 

 = 51.584

[in]

Shear resistance provided

 = φ R_n x L

 = 574.4

[kips]

 = 0.78

 > V_u

OK

Reinforcing plate

 = 1.250

[in]

 = 1.250

[in]

 = 50.0

[ksi]

Ratio of expected Fy to specified min Fy

 = 1.10

AISC 341-16  Table A3.1

Reinforcing plate area

 = w_r x t_r

 = 1.563

[in²]

Required strength of weld

 = R_y F_y A_r

 = 85.9

[kips]

Longitudinal weld length

 = reinforce plate length

 = 8.000

[in]

Transverse weld length

 = reinforce plate width

 = 1.250

[in]

Total weld length - single fillet weld

 = 2 x L_L + L_T

 = 17.250

[in]

AISC 15^th  Eq J2-10a

 = 0.85 x 2 x L_L + 1.5 x L_T

 = 15.475

[in]

AISC 15^th  Eq J2-10b

 = max ( L₁ , L₂ )

 = 17.250

[in]

AISC 15^th  J2.4 (c)

Fillet weld leg size

 = ¹⁄₄

[in]

 = 0.0

[°]

Electrode strength

 = 70.0

[ksi]

 = 1.00

AISC 15^th  Table 8-3

Number of weld line

 = 1   for single fillet

Load angle coefficient

 = ( 1 + 0.5 sin^1.5 θ )

 = 1.00

AISC 15^th  Page 8-9

Fillet weld shear strength

 = 0.6 (C₁ x 70 ksi) 0.707 w n C₂

 = 7.424

[kip/in]

AISC 15^th  Eq 8-1

Base metal - reinforce plate

 = 1.250

[in]

 = 65.0

[ksi]

Base metal - reinforce plate is in shear, shear rupture as per AISC 15^th  Eq J4-4 is checked

AISC 15^th  J2.4

Base metal shear rupture

 = 0.6 F_u t

 = 48.750

[kip/in]

AISC 15^th  Eq J4-4

Single fillet linear shear strength

 = min ( R_n-w , R_n-b )

 = 7.424

[kip/in]

AISC 15^th  Eq 9-2

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq 8-1

 = 

 = 5.568

[kip/in]

Shear resistance required

 = 

 = 85.9

[kips]

Fillet weld length - single fillet

 = 

 = 17.250

[in]

Shear resistance provided

 = φ R_n x L

 = 96.0

[kips]

 = 0.89

 > P_u

OK

HSS sect HSS6.875X0.500 wall thick

 = 0.465

[in]

 = 46.0

[ksi]

HSS brace wall-gusset overlap length

 = 26.000

[in]

Ratio of expected Fy to specified min Fy

 = 1.30

AISC 341-16  Table A3.1

Beam axial load

 = from seismic load calc

 = 559.7

[kips]

HSS brace wall-gusset shear yielding

 = 0.6 R_y F_y t L x 4 walls

 = 1735.2

[kips]

AISC 15^th  Eq J4-3

Resistance factor-LRFD

 = 1.00

AISC 15^th  Eq J4-3

 = 

 = 1735.2

[kips]

 = 0.32

 > P_u

OK

HSS sect HSS6.875X0.500 wall thick

 = 0.465

[in]

 = 62.0

[ksi]

HSS brace wall-gusset overlap length

 = 26.000

[in]

Ratio of expected Fu to specified min Fu

 = 1.20

AISC 341-16  Table A3.1

Beam axial load

 = from seismic load calc

 = 559.7

[kips]

HSS brace wall-gusset shear rupture

 = 0.6 R_t F_u t L x 4 walls

 = 2158.8

[kips]

AISC 15^th  Eq J4-4

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq J4-4

 = 

 = 1619.1

[kips]

 = 0.35

 > P_u

OK

Plate thickness

 = 0.750

[in]

Plate strength

 = 50.0

[ksi]

 = 65.0

[ksi]

C shape weld group size

 = 26.000

[in]

 = 6.875

[in]

Gross area subject to shear

 = b t_p x 2

 = 39.000

[in²]

Net area subject to shear

 = A_gvb

 = 39.000

[in²]

Net area subject to tension

 = d t_p

 = 5.156

[in²]

Block shear strength required

 = 

 = 559.7

[kips]

Uniform tension stress factor

 = 1.00

AISC 15^th  Fig C-J4.2

Bolt shear resistance provided

 = min (0.6F_u A_nv , 0.6F_y A_gv ) +

 = 1505.2

[kips]

AISC 15^th  Eq J4-5

U_bs F_u A_nt

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq J4-5

 = 

 = 1128.9

[kips]

 = 0.50

 > V_u

OK

Plate size

 = 29.411

[in]

 = 0.750

[in]

Plate yield strength

 = 50.0

[ksi]

Plate gross area in shear

 = b_p t_p

 = 22.058

[in²]

Tensile force required

 = 

 = 559.7

[kips]

Plate tensile yielding strength

 = F_y A_g

 = 1102.9

[kips]

AISC 15^th  Eq J4-1

Resistance factor-LRFD

 = 0.90

AISC 15^th  Eq J4-1

 = 

 = 992.6

[kips]

 = 0.56

 > P_u

OK

Plate size

 = 29.411

[in]

 = 0.750

[in]

Plate tensile strength

 = 65.0

[ksi]

Plate net area in tension

 = b_p t_p

 = 22.058

[in²]

Tensile force required

 = 

 = 559.7

[kips]

Plate tensile rupture strength

 = F_u A_nt

 = 1433.8

[kips]

AISC 15^th  Eq J4-2

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq J4-2

 = 

 = 1075.3

[kips]

AISC 15^th  Eq J4-2

 = 0.52

 > P_u

OK

Fillet weld leg size

 = ¹⁄₄

[in]

 = 0.0

[°]

Electrode strength

 = 70.0

[ksi]

 = 1.00

AISC 15^th  Table 8-3

Number of weld line

 = 2   for double fillet

Load angle coefficient

 = ( 1 + 0.5 sin^1.5 θ )

 = 1.00

AISC 15^th  Page 8-9

Fillet weld shear strength

 = 0.6 (C₁ x 70 ksi) 0.707 w n C₂

 = 14.847

[kip/in]

AISC 15^th  Eq 8-1

Base metal - brace

 = 0.750

[in]

 = 65.0

[ksi]

Base metal - brace is in shear, shear rupture as per AISC 15^th  Eq J4-4 is checked

AISC 15^th  J2.4

Base metal shear rupture

 = 0.6 F_u t

 = 29.250

[kip/in]

AISC 15^th  Eq J4-4

Double fillet linear shear strength

 = min ( R_n-w , R_n-b )

 = 14.847

[kip/in]

AISC 15^th  Eq 9-2

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq 8-1

 = 

 = 11.135

[kip/in]

Shear resistance required

 = 

 = 559.7

[kips]

Fillet weld leg size

 = ¹⁄₄

[in]

 = 26.000

[in]

when L/w = 104.0 >100 , weld length reduction applied

AISC 15^th  J2.2b

Weld length reduction factor

 = 1.2 - 0.002 (L/w) ≤ 1.0

 = 0.99

AISC 15^th  Eq J2-1

Weld length used for design

 = β x L

 = 51.584

[in]

Fillet weld length - double fillet

 = 

 = 51.584

[in]

Shear resistance provided

 = φ R_n x L

 = 574.4

[kips]

 = 0.97

 > V_u

OK

Reinforcing plate

 = 1.250

[in]

 = 1.250

[in]

 = 50.0

[ksi]

Ratio of expected Fy to specified min Fy

 = 1.10

AISC 341-16  Table A3.1

Reinforcing plate area

 = w_r x t_r

 = 1.563

[in²]

Required strength of weld

 = R_y F_y A_r

 = 85.9

[kips]

Longitudinal weld length

 = reinforce plate length

 = 8.000

[in]

Transverse weld length

 = reinforce plate width

 = 1.250

[in]

Total weld length - single fillet weld

 = 2 x L_L + L_T

 = 17.250

[in]

AISC 15^th  Eq J2-10a

 = 0.85 x 2 x L_L + 1.5 x L_T

 = 15.475

[in]

AISC 15^th  Eq J2-10b

 = max ( L₁ , L₂ )

 = 17.250

[in]

AISC 15^th  J2.4 (c)

Fillet weld leg size

 = ¹⁄₄

[in]

 = 0.0

[°]

Electrode strength

 = 70.0

[ksi]

 = 1.00

AISC 15^th  Table 8-3

Number of weld line

 = 1   for single fillet

Load angle coefficient

 = ( 1 + 0.5 sin^1.5 θ )

 = 1.00

AISC 15^th  Page 8-9

Fillet weld shear strength

 = 0.6 (C₁ x 70 ksi) 0.707 w n C₂

 = 7.424

[kip/in]

AISC 15^th  Eq 8-1

Base metal - reinforce plate

 = 1.250

[in]

 = 65.0

[ksi]

Base metal - reinforce plate is in shear, shear rupture as per AISC 15^th  Eq J4-4 is checked

AISC 15^th  J2.4

Base metal shear rupture

 = 0.6 F_u t

 = 48.750

[kip/in]

AISC 15^th  Eq J4-4

Single fillet linear shear strength

 = min ( R_n-w , R_n-b )

 = 7.424

[kip/in]

AISC 15^th  Eq 9-2

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq 8-1

 = 

 = 5.568

[kip/in]

Shear resistance required

 = 

 = 85.9

[kips]

Fillet weld length - single fillet

 = 

 = 17.250

[in]

Shear resistance provided

 = φ R_n x L

 = 96.0

[kips]

 = 0.89

 > P_u

OK

Thinner part joined thickness

 = 

 = 0.750

[in]

Min fillet weld size allowed

 = 

 = 0.250

[in]

AISC 15^th  Table J2.4

Fillet weld size provided

 = 

 = 0.438

[in]

 ≥ w_min

OK

Fillet weld size provided

 = 

 = 0.438

[in]

Min fillet weld length allowed

 = 4 x w

 = 1.750

[in]

AISC 15^th  J2.2b

Min fillet weld length

 = 

 = 84.686

[in]

 ≥ L_min

OK

Plate size

 = 54.000

[in]

 = 0.750

[in]

Plate yield strength

 = 50.0

[ksi]

Plate gross area in shear

 = b_p t_p

 = 40.500

[in²]

Shear force required

 = 

 = 713.3

[kips]

Plate shear yielding strength

 = 0.6 F_y A_gv

 = 1215.0

[kips]

AISC 15^th  Eq J4-3

Resistance factor-LRFD

 = 1.00

AISC 15^th  Eq J4-3

 = 

 = 1215.0

[kips]

 = 0.59

 > V_u

OK

Plate size

 = 54.000

[in]

 = 0.750

[in]

Plate tensile strength

 = 65.0

[ksi]

Plate net area in shear

 = b_p t_p

 = 40.500

[in²]

Shear force in demand

 = 

 = 713.3

[kips]

Plate shear rupture strength

 = 0.6 F_u A_nv

 = 1579.5

[kips]

AISC 15^th  Eq J4-4

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq J4-4

 = 

 = 1184.6

[kips]

 = 0.60

 > V_u

OK

Gusset Edge Equivalent Normal Force

Refer to AISC DG29  Fig. B-1 for formula below to calculate gusset edge equivalent normal force

Gusset edge axial force

 = 

 = -78.2

[kips]

Gusset edge moment force

 = 

 = 656.86

[kip-ft]

Gusset edge interface length

 = 

 = 54.000

[in]

Gusset edge equivalent normal force

 = N -

 = -662.1

[kips]

AISC DG29  Fig B-1

Plate size

 = 54.000

[in]

 = 0.750

[in]

Plate yield strength

 = 50.0

[ksi]

Plate gross area in shear

 = b_p t_p

 = 40.500

[in²]

Tensile force required

 = 

 = 662.1

[kips]

Plate tensile yielding strength

 = F_y A_g

 = 2025.0

[kips]

AISC 15^th  Eq J4-1

Resistance factor-LRFD

 = 0.90

AISC 15^th  Eq J4-1

 = 

 = 1822.5

[kips]

 = 0.36

 > P_u

OK

Gusset Edge Equivalent Normal Force

Refer to AISC DG29  Fig. B-1 for formula below to calculate gusset edge equivalent normal force

Gusset edge axial force

 = 

 = -78.2

[kips]

Gusset edge moment force

 = 

 = 656.86

[kip-ft]

Gusset edge interface length

 = 

 = 54.000

[in]

Gusset edge equivalent normal force

 = N -

 = -662.1

[kips]

AISC DG29  Fig B-1

Plate size

 = 54.000

[in]

 = 0.750

[in]

Plate tensile strength

 = 65.0

[ksi]

Plate net area in tension

 = b_p t_p

 = 40.500

[in²]

Tensile force required

 = 

 = 662.1

[kips]

Plate tensile rupture strength

 = F_u A_nt

 = 2632.5

[kips]

AISC 15^th  Eq J4-2

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq J4-2

 = 

 = 1974.4

[kips]

AISC 15^th  Eq J4-2

 = 0.34

 > P_u

OK

Gusset plate

 = 54.000

[in]

 = 0.750

[in]

 = 50.0

[ksi]

Shear plate - gross area

 = b_p x t_p

 = 40.500

[in²]

Shear plate - plastic modulus

 = ( b_p x t2p ) / 4

 = 546.75

[in³]

Flexural strength available

 = φ F_y Z_p     φ=0.90

 = 2050.31

[kip-ft]

Flexural strength required

 = from gusset interface forces calc

 = 656.86

[kip-ft]

Axial strength available

 = from axial tensile yield check

 = 1822.5

[kips]

Axial strength required

 = from gusset interface forces calc

 = -78.2

[kips]

Shear strength available

 = from shear yielding check

 = 1215.0

[kips]

Shear strength required

 = from gusset interface forces calc

 = 713.3

[kips]

Flexural yield interaction

 = ( )2 + ( + )2

 = 0.48

AISC 15^th  Eq 10-5

 < 1.0

OK

Gusset plate

 = 54.000

[in]

 = 0.750

[in]

 = 65.0

[ksi]

Net area of plate

 = b_p x t_p

 = 40.500

[in²]

Plastic modulus of net section

 = ( b_p x t2p ) / 4

 = 546.75

[in³]

Flexural strength available

 = φ F_u Z_net     φ=0.75

 = 2221.17

[kip-ft]

Flexural strength required

 = from gusset interface forces calc

 = 656.86

[kip-ft]

Axial strength available

 = from axial tensile rupture check

 = 1974.4

[kips]

Axial strength required

 = from gusset interface forces calc

 = -78.2

[kips]

Shear strength available

 = from shear rupture check

 = 1184.6

[kips]

Shear strength required

 = from gusset interface forces calc

 = 713.3

[kips]

Flexural rupture interaction

 = ( )2 + ( + )2

 = 0.48

AISC 15^th  Eq 10-5

 < 1.0

OK

Plate size

 = 31.294

[in]

 = 0.750

[in]

Plate yield strength

 = 50.0

[ksi]

Plate gross area in shear

 = b_p t_p

 = 23.471

[in²]

Shear force required

 = 

 = 64.7

[kips]

Plate shear yielding strength

 = 0.6 F_y A_gv

 = 704.1

[kips]

AISC 15^th  Eq J4-3

Resistance factor-LRFD

 = 1.00

AISC 15^th  Eq J4-3

 = 

 = 704.1

[kips]

 = 0.09

 > V_u

OK

Plate size

 = 31.294

[in]

 = 0.750

[in]

Plate tensile strength

 = 65.0

[ksi]

Plate net area in shear

 = b_p t_p

 = 23.471

[in²]

Shear force in demand

 = 

 = 64.7

[kips]

Plate shear rupture strength

 = 0.6 F_u A_nv

 = 915.3

[kips]

AISC 15^th  Eq J4-4

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq J4-4

 = 

 = 686.5

[kips]

 = 0.09

 > V_u

OK

Gusset Edge Equivalent Normal Force

Refer to AISC DG29  Fig. B-1 for formula below to calculate gusset edge equivalent normal force

Gusset edge axial force

 = 

 = -39.1

[kips]

Gusset edge moment force

 = 

 = 43.00

[kip-ft]

Gusset edge interface length

 = 

 = 31.294

[in]

Gusset edge equivalent normal force

 = N -

 = -105.1

[kips]

AISC DG29  Fig B-1

Plate size

 = 31.294

[in]

 = 0.750

[in]

Plate yield strength

 = 50.0

[ksi]

Plate gross area in shear

 = b_p t_p

 = 23.471

[in²]

Tensile force required

 = 

 = 105.1

[kips]

Plate tensile yielding strength

 = F_y A_g

 = 1173.5

[kips]

AISC 15^th  Eq J4-1

Resistance factor-LRFD

 = 0.90

AISC 15^th  Eq J4-1

 = 

 = 1056.2

[kips]

 = 0.10

 > P_u

OK

Gusset Edge Equivalent Normal Force

Refer to AISC DG29  Fig. B-1 for formula below to calculate gusset edge equivalent normal force

Gusset edge axial force

 = 

 = -39.1

[kips]

Gusset edge moment force

 = 

 = 43.00

[kip-ft]

Gusset edge interface length

 = 

 = 31.294

[in]

Gusset edge equivalent normal force

 = N -

 = -105.1

[kips]

AISC DG29  Fig B-1

Plate size

 = 31.294

[in]

 = 0.750

[in]

Plate tensile strength

 = 65.0

[ksi]

Plate net area in tension

 = b_p t_p

 = 23.471

[in²]

Tensile force required

 = 

 = 105.1

[kips]

Plate tensile rupture strength

 = F_u A_nt

 = 1525.6

[kips]

AISC 15^th  Eq J4-2

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq J4-2

 = 

 = 1144.2

[kips]

AISC 15^th  Eq J4-2

 = 0.09

 > P_u

OK

Gusset plate

 = 31.294

[in]

 = 0.750

[in]

 = 50.0

[ksi]

Shear plate - gross area

 = b_p x t_p

 = 23.471

[in²]

Shear plate - plastic modulus

 = ( b_p x t2p ) / 4

 = 183.62

[in³]

Flexural strength available

 = φ F_y Z_p     φ=0.90

 = 688.58

[kip-ft]

Flexural strength required

 = from gusset interface forces calc

 = 43.00

[kip-ft]

Axial strength available

 = from axial tensile yield check

 = 1056.2

[kips]

Axial strength required

 = from gusset interface forces calc

 = -39.1

[kips]

Shear strength available

 = from shear yielding check

 = 704.1

[kips]

Shear strength required

 = from gusset interface forces calc

 = 64.7

[kips]

Flexural yield interaction

 = ( )2 + ( + )2

 = 0.02

AISC 15^th  Eq 10-5

 < 1.0

OK

Gusset plate

 = 31.294

[in]

 = 0.750

[in]

 = 65.0

[ksi]

Net area of plate

 = b_p x t_p

 = 23.471

[in²]

Plastic modulus of net section

 = ( b_p x t2p ) / 4

 = 183.62

[in³]

Flexural strength available

 = φ F_u Z_net     φ=0.75

 = 745.96

[kip-ft]

Flexural strength required

 = from gusset interface forces calc

 = 43.00

[kip-ft]

Axial strength available

 = from axial tensile rupture check

 = 1144.2

[kips]

Axial strength required

 = from gusset interface forces calc

 = -39.1

[kips]

Shear strength available

 = from shear rupture check

 = 686.5

[kips]

Shear strength required

 = from gusset interface forces calc

 = 64.7

[kips]

Flexural rupture interaction

 = ( )2 + ( + )2

 = 0.02

AISC 15^th  Eq 10-5

 < 1.0

OK

 = 713.3

[kips]

 = -78.2

[kips]   in tension

 = 656.86

[kip-ft]

Gusset-beam fillet weld length

 = 

 = 54.000

[in]

Weld stress from axial force

 = N / L_w

 = -1.448

[kip/in]

in tension

Weld stress from shear force

 = V / L_w

 = 13.209

[kip/in]

Weld stress from moment force

 = 

 = 16.219

[kip/in]

Weld stress combined - max

 = [ (f_a - f_b )2 + f2v ]0.5

 = 22.059

[kip/in]

AISC 15^th  Eq 8-11

Weld resultant load angle

 = tan^-1 [( f_b - f_a ) / f_v ]

 = 53.2

[°]

Fillet weld leg size

 = ⁷⁄₁₆

[in]

 = 53.2

[°]

Electrode strength

 = 70.0

[ksi]

 = 1.00

AISC 15^th  Table 8-3

Number of weld line

 = 2   for double fillet

Load angle coefficient

 = ( 1 + 0.5 sin^1.5 θ )

 = 1.36

AISC 15^th  Page 8-9

Fillet weld shear strength

 = 0.6 (C₁ x 70 ksi) 0.707 w n C₂

 = 35.293

[kip/in]

AISC 15^th  Eq 8-1

Base metal - gusset plate

 = 0.750

[in]

 = 65.0

[ksi]

Base metal - gusset plate is in shear, shear rupture as per AISC 15^th  Eq J4-4 is checked

AISC 15^th  J2.4

Base metal shear rupture

 = 0.6 F_u t

 = 29.250

[kip/in]

AISC 15^th  Eq J4-4

Double fillet linear shear strength

 = min ( R_n-w , R_n-b )

 = 29.250

[kip/in]

AISC 15^th  Eq 9-2

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq 8-1

 = 

 = 21.938

[kip/in]

When gusset plate is directly welded to beam or column, apply 1.25 ductility factor
to allow adequate force redistribution in the weld group

AISC 15^th  Page 13-11

Weld strength used for design after applying ductility factor

 = φ R_n x ( 1/1.25 )

 = 17.550

[kip/in]

 = 1.26

 < f_max

NG

Gusset Edge Equivalent Normal Force

Refer to AISC DG29  Fig. B-1 for formula below to calculate gusset edge equivalent normal force

Gusset edge axial force

 = 

 = -78.2

[kips]

Gusset edge moment force

 = 

 = 656.86

[kip-ft]

Gusset edge interface length

 = 

 = 54.000

[in]

Gusset edge equivalent normal force

 = N -

 = -662.1

[kips]

AISC DG29  Fig B-1

Concentrated force from gusset

 = 

 = 662.1

[kips]

Beam section

 = 22.100

[in]

 = 1.150

[in]

 = 0.720

[in]

 = 1.650

[in]

 = 50.0

[ksi]

Length of bearing

 = gusset-beam weld length

 = 54.000

[in]

Beam web local yielding strength

 = F_y t_w ( 5 k + l_b )

 = 2241.0

[kips]

AISC 15^th  Eq J10-2

Resistance factor-LRFD

 = 1.00

 = 

 = 2241.0

[kips]

 = 0.30

 > P_u

OK

Gusset Edge Equivalent Normal Force

Refer to AISC DG29  Fig. B-1 for formula below to calculate gusset edge equivalent normal force

Gusset edge axial force

 = 

 = -78.2

[kips]

Gusset edge moment force

 = 

 = 656.86

[kip-ft]

Gusset edge interface length

 = 

 = 54.000

[in]

Gusset edge equivalent normal force

 = N -

 = -662.1

[kips]

AISC DG29  Fig B-1

Concentrated force from gusset

 = 

 = 662.1

[kips]

Beam section

 = 22.100

[in]

 = 1.150

[in]

 = 0.720

[in]

 = 1.650

[in]

 = 50.0

[ksi]

 = 29000

[ksi]

Length of bearing

 = gusset-beam weld length

 = 54.000

[in]

Beam web local crippling strength

 = 0.8 t2w [1+3( )^1.5 ] x

 = 2923.0

[kips]

AISC 15^th  Eq J10-4

( )^0.5

Resistance factor-LRFD

 = 0.75

AISC 15^th  J10.3

 = 

 = 2192.3

[kips]

 = 0.30

 > P_u

OK

Beam Web Effective Length for Transmitting Shear

Refer to AISC Design Example v14.2 Page IIC-70 for formula below to calculate beam web effective length
in transmitting shear along Sect a-a

Beam sect W21X147

 = 22.100

[in]

 = 12.500

[in]

 = 1.150

[in]

 = 0.720

[in]

 = 1.650

[in]

 = 50.0

[ksi]

Gusset edge interface length

 = 

 = 54.000

[in]

 = 0.90

 = 1.00

Beam web effective length for transmitting shear

 = L + 5k +

 = 122.146

[in]

Gusset edge shear (Sect a-a)

 = 

 = 713.3

[kips]

Beam web shear strength

 = 0.6 F_y t_w L_eff

 = 2638.4

[kips]

AISC 15^th  Eq J4-3

Resistance factor-LRFD

 = 1.00

AISC 15^th  Eq J4-3

 = 

 = 2638.4

[kips]

 = 0.27

 > V_u

OK

Beam sect W21X147

 = 22.100

[in]

 = 0.720

[in]

Right brace axial force

 = from user input

 = -559.7

[kips]

in tension

Right brace to hor line angle

 = from user input

 = 45.0

[°]

Right brace force ver component

 = P₁ sin θ₁

 = -395.8

[kips]

Gusset edge shear (Sect b-b)

 = 

 = 64.7

[kips]

Transfer force from chev brace on the other side of beam or column

 = from user input

 = -46.0

[kips]

in compression

Beam web transverse shear

 = V₁ + V' - A_b

 = 285.1

[kips]

Beam web shear strength

 = 0.6 F_y d t_w C_v

 = 477.4

[kips]

AISC 15^th  Eq G2-1

 = 1.00

AISC 15^th  Eq G2-2

Resistance factor-LRFD

 = 1.00

AISC 15^th  Eq G2-1

 = 

 = 477.4

[kips]

 = 0.60

 > V_u

OK

Plate size

 = 54.000

[in]

 = 0.750

[in]

Plate yield strength

 = 50.0

[ksi]

Plate gross area in shear

 = b_p t_p

 = 40.500

[in²]

Shear force required

 = 

 = 713.3

[kips]

Plate shear yielding strength

 = 0.6 F_y A_gv

 = 1215.0

[kips]

AISC 15^th  Eq J4-3

Resistance factor-LRFD

 = 1.00

AISC 15^th  Eq J4-3

 = 

 = 1215.0

[kips]

 = 0.59

 > V_u

OK

Plate size

 = 54.000

[in]

 = 0.750

[in]

Plate tensile strength

 = 65.0

[ksi]

Plate net area in shear

 = b_p t_p

 = 40.500

[in²]

Shear force in demand

 = 

 = 713.3

[kips]

Plate shear rupture strength

 = 0.6 F_u A_nv

 = 1579.5

[kips]

AISC 15^th  Eq J4-4

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq J4-4

 = 

 = 1184.6

[kips]

 = 0.60

 > V_u

OK

Gusset Edge Equivalent Normal Force

Refer to AISC DG29  Fig. B-1 for formula below to calculate gusset edge equivalent normal force

Gusset edge axial force

 = 

 = -78.2

[kips]

Gusset edge moment force

 = 

 = 656.86

[kip-ft]

Gusset edge interface length

 = 

 = 54.000

[in]

Gusset edge equivalent normal force

 = N -

 = -662.1

[kips]

AISC DG29  Fig B-1

Plate size

 = 54.000

[in]

 = 0.750

[in]

Plate yield strength

 = 50.0

[ksi]

Plate gross area in shear

 = b_p t_p

 = 40.500

[in²]

Tensile force required

 = 

 = 662.1

[kips]

Plate tensile yielding strength

 = F_y A_g

 = 2025.0

[kips]

AISC 15^th  Eq J4-1

Resistance factor-LRFD

 = 0.90

AISC 15^th  Eq J4-1

 = 

 = 1822.5

[kips]

 = 0.36

 > P_u

OK

Gusset Edge Equivalent Normal Force

Refer to AISC DG29  Fig. B-1 for formula below to calculate gusset edge equivalent normal force

Gusset edge axial force

 = 

 = -78.2

[kips]

Gusset edge moment force

 = 

 = 656.86

[kip-ft]

Gusset edge interface length

 = 

 = 54.000

[in]

Gusset edge equivalent normal force

 = N -

 = -662.1

[kips]

AISC DG29  Fig B-1

Plate size

 = 54.000

[in]

 = 0.750

[in]

Plate tensile strength

 = 65.0

[ksi]

Plate net area in tension

 = b_p t_p

 = 40.500

[in²]

Tensile force required

 = 

 = 662.1

[kips]

Plate tensile rupture strength

 = F_u A_nt

 = 2632.5

[kips]

AISC 15^th  Eq J4-2

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq J4-2

 = 

 = 1974.4

[kips]

AISC 15^th  Eq J4-2

 = 0.34

 > P_u

OK

Gusset plate

 = 54.000

[in]

 = 0.750

[in]

 = 50.0

[ksi]

Shear plate - gross area

 = b_p x t_p

 = 40.500

[in²]

Shear plate - plastic modulus

 = ( b_p x t2p ) / 4

 = 546.75

[in³]

Flexural strength available

 = φ F_y Z_p     φ=0.90

 = 2050.31

[kip-ft]

Flexural strength required

 = from gusset interface forces calc

 = 656.86

[kip-ft]

Axial strength available

 = from axial tensile yield check

 = 1822.5

[kips]

Axial strength required

 = from gusset interface forces calc

 = -78.2

[kips]

Shear strength available

 = from shear yielding check

 = 1215.0

[kips]

Shear strength required

 = from gusset interface forces calc

 = 713.3

[kips]

Flexural yield interaction

 = ( )2 + ( + )2

 = 0.48

AISC 15^th  Eq 10-5

 < 1.0

OK

Gusset plate

 = 54.000

[in]

 = 0.750

[in]

 = 65.0

[ksi]

Net area of plate

 = b_p x t_p

 = 40.500

[in²]

Plastic modulus of net section