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)

Top 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]

Top Brace seismic design force in compression

 = P_ec

 = 449.1

[kips]

AISC 341-16 F2.6c (2)

Top Brace seismic design force in compression - post-buckling

 = 0.3 x P_ec

 = 134.7

[kips]

AISC 341-16 F2.3 (ii)

Brace sect HSS7.500X0.500

 = A500 Gr.C Round

 = 46.0

[ksi]

Ratio of expected Fy to specified min Fy

 = 1.30

AISC 341-16  Table A3.1

 = 10.300

[in²]

 = 2.493

[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

 = 615.9

[kips]

AISC 341-16 F2.6c (1)

Bot Brace seismic design force in tension

 = P_et

 = -615.9

[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

 = 2.493

[in]

 = 29000

[ksi]

Member slenderness ratio

 = K x L  / r

 = 57.77

Elastic buckling stress

 = 

 = 85.76

[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

 = 44.66

[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 )

 = 524.4

[kips]

AISC 341-16 F2.3

Brace force in compression

 = from user input in load section

 = 0.0

[kips]

Bot Brace seismic design force in compression

 = P_ec

 = 524.4

[kips]

AISC 341-16 F2.6c (2)

Bot Brace seismic design force in compression - post-buckling

 = 0.3 x P_ec

 = 157.3

[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       Top Brace   P_{s_t} =-559.7 kips (T)                   Bot Brace   P_{s_ci} =524.4 kips (C)

AISC 341-16 F2.3(i)

LC2       Top Brace   P_{s_ci} =449.1 kips (C)                   Bot Brace   P_{s_t} =-615.9 kips (T)

LC3       Top Brace   P_{s_t} =-559.7 kips (T)                   Bot Brace   P_{s_cii} =157.3 kips (C)       post-buckling

AISC 341-16 F2.3(ii)

LC4       Top Brace   P_{s_cii} =134.7 kips (C)                   Bot Brace   P_{s_t} =-615.9 kips (T)       post-buckling

Top and bottom brace force

 = -559.7

[kips] (T)

 = 524.4

[kips] (C)

Beam end shear & transfer force

 = 19.9

[kips]

 = -46.3

[kips]

Refer to AISC 15^th  Page 13-4 and Fig. 13-2 for all charts and definitions of variables and symbols shown in calculation below

 = 12.350

[in]

 = 6.450

[in]

 = 14.000

[in]

 = 9.625

[in]

 = 45.0

[°]

 = e_b tanθ - e_c

 = 5.900

[in]

AISC 15^th  Eq. 13-16

 = tan2 θ + ()2

 = 3.116

AISC 15^th  Eq. 13-24

 = α ( tan θ + )

 = 34.364

AISC 15^th  Eq. 13-23

 = [ K' tan θ + K ()2 ] / D

 = 14.000

[in]

AISC 15^th  Eq. 13-21

 = ( K' - K tan θ ) / D

 = 8.100

[in]

AISC 15^th  Eq. 13-22

 = [ ( e_b + β ) 2 + ( e_c + α ) 2 ]0.5

 = 28.921

[in]

AISC 15^th  Eq. 13-6

Brace axial force

 = from seismic brace force calc

 = -559.7

[kips]

in tension

Shear force

 = ( β / r ) P_u

 = -156.8

[kips]

AISC 15^th  Eq. 13-2

Axial force

 = ( e_c / r ) P_u

 = -124.8

[kips]

AISC 15^th  Eq. 13-3

Moment

 = H_c ( β - β )

 = 15.86

[kip-ft]

AISC 15^th  Eq. 13-19

Shear force

 = ( α / r ) P_u

 = -270.9

[kips]

AISC 15^th  Eq. 13-5

Axial force

 = ( e_b / r ) P_u

 = -239.0

[kips]

AISC 15^th  Eq. 13-4

Moment

 = V_b ( α - α )

 = 0.00

[kip-ft]

AISC 15^th  Eq. 13-17

Refer to AISC 15^th  Page 13-4 and Fig. 13-2 for all charts and definitions of variables and symbols shown in calculation below

 = 12.350

[in]

 = 6.450

[in]

 = 14.750

[in]

 = 10.375

[in]

 = 45.0

[°]

 = e_b tanθ - e_c

 = 5.900

[in]

AISC 15^th  Eq. 13-16

 = tan2 θ + ()2

 = 3.021

AISC 15^th  Eq. 13-24

 = α ( tan θ + )

 = 35.720

AISC 15^th  Eq. 13-23

 = [ K' tan θ + K ()2 ] / D

 = 14.750

[in]

AISC 15^th  Eq. 13-21

 = ( K' - K tan θ ) / D

 = 8.850

[in]

AISC 15^th  Eq. 13-22

 = [ ( e_b + β ) 2 + ( e_c + α ) 2 ]0.5

 = 29.981

[in]

AISC 15^th  Eq. 13-6

Brace axial force

 = from seismic brace force calc

 = 524.4

[kips]

in compression

Shear force

 = ( β / r ) P_u

 = 154.8

[kips]

AISC 15^th  Eq. 13-2

Axial force

 = ( e_c / r ) P_u

 = 112.8

[kips]

AISC 15^th  Eq. 13-3

Moment

 = H_c ( β - β )

 = -14.34

[kip-ft]

AISC 15^th  Eq. 13-19

Shear force

 = ( α / r ) P_u

 = 258.0

[kips]

AISC 15^th  Eq. 13-5

Axial force

 = ( e_b / r ) P_u

 = 216.0

[kips]

AISC 15^th  Eq. 13-4

Moment

 = V_b ( α - α )

 = 0.00

[kip-ft]

AISC 15^th  Eq. 13-17

Beam end shear reaction

 = from user input

 = 19.9

[kips]

Top brace gusset-beam axial force

 = 

 = -239.0

[kips]

AISC 15^th  Eq. 13-4

Bot brace gusset-beam axial force

 = 

 = 216.0

[kips]

AISC 15^th  Eq. 13-4

Beam to column shear force

 = R_b + V_b-top - V_b-bot

 = -435.1

[kips]

AISC 15^th  Page 13-4

Top brace gusset-column axial force

 = 

 = -124.8

[kips]

AISC 15^th  Eq. 13-3

Bot brace gusset-column axial force

 = 

 = 112.8

[kips]

AISC 15^th  Eq. 13-3

Transfer force from adjacent bay

 = from user input

 = -46.3

[kips]

Beam to column axial force

 = ( H_c-top + H_c-bot ) x -1 - A_b

 = 58.3

[kips]

AISC 15^th  Page 13-4

This force is not for use in connection calc. It's output here for user input connection forces equilibrium check only.

P_bm - Beam member axial force is different from P_b-c - Beam to column interface axial force as shown above.

P_bm - Beam member axial force is from structural analysis output and cannot be used directly in beam end to column connection design as this force is interrupted by brace gusset to beam interface force before beam end reaching the column. This force is actually not needed from user's input for beam end to column connection design.

P_b-c - Beam to column interface axial force is calculated from user's input of brace axial forces and trasnfer force using uniform force method. This force is used in the beam end to column connection design.

P_bm - Beam member axial force is not needed for the beam end to column connection design and is calculated here for verification purpose only. If it matches the structural analysis output, that means equilibrium is reached and user's input of brace axial forces and trasnfer force are correct.

Top brace axial force

 = from seismic brace force calc

 = -559.7

[kips]

in tension

Top brace to ver line angle

 = from user input

 = 45.0

[°]

Top brace gusset-column axial force

 = from calc shown above

 = -124.8

[kips]

AISC 15^th  Eq. 13-3

Bot brace axial force

 = from seismic brace force calc

 = 524.4

[kips]

in compression

Bot brace to ver line angle

 = from user input

 = 45.0

[°]

Bot brace gusset-column axial force

 = from calc shown above

 = 112.8

[kips]

AISC 15^th  Eq. 13-3

Beam to column interface axial force

 = from calc shown above

 = 58.3

[kips]

AISC 15^th  Page 13-4

Beam member axial force

 = (H_ct - P_t sinθ_t ) + (H_cb - P_b sinθ_b )

 = 71.3

[kips]

in compression

+ P_b-c

Top and bottom brace force

 = 449.1

[kips] (C)

 = -615.9

[kips] (T)

Beam end shear & transfer force

 = 19.9

[kips]

 = -46.3

[kips]

Refer to AISC 15^th  Page 13-4 and Fig. 13-2 for all charts and definitions of variables and symbols shown in calculation below

 = 12.350

[in]

 = 6.450

[in]

 = 14.000

[in]

 = 9.625

[in]

 = 45.0

[°]

 = e_b tanθ - e_c

 = 5.900

[in]

AISC 15^th  Eq. 13-16

 = tan2 θ + ()2

 = 3.116

AISC 15^th  Eq. 13-24

 = α ( tan θ + )

 = 34.364

AISC 15^th  Eq. 13-23

 = [ K' tan θ + K ()2 ] / D

 = 14.000

[in]

AISC 15^th  Eq. 13-21

 = ( K' - K tan θ ) / D

 = 8.100

[in]

AISC 15^th  Eq. 13-22

 = [ ( e_b + β ) 2 + ( e_c + α ) 2 ]0.5

 = 28.921

[in]

AISC 15^th  Eq. 13-6

Brace axial force

 = from seismic brace force calc

 = 449.1

[kips]

in compression

Shear force

 = ( β / r ) P_u

 = 125.8

[kips]

AISC 15^th  Eq. 13-2

Axial force

 = ( e_c / r ) P_u

 = 100.2

[kips]

AISC 15^th  Eq. 13-3

Moment

 = H_c ( β - β )

 = -12.73

[kip-ft]

AISC 15^th  Eq. 13-19

Shear force

 = ( α / r ) P_u

 = 217.4

[kips]

AISC 15^th  Eq. 13-5

Axial force

 = ( e_b / r ) P_u

 = 191.8

[kips]

AISC 15^th  Eq. 13-4

Moment

 = V_b ( α - α )

 = 0.00

[kip-ft]

AISC 15^th  Eq. 13-17

Refer to AISC 15^th  Page 13-4 and Fig. 13-2 for all charts and definitions of variables and symbols shown in calculation below

 = 12.350

[in]

 = 6.450

[in]

 = 14.750

[in]

 = 10.375

[in]

 = 45.0

[°]

 = e_b tanθ - e_c

 = 5.900

[in]

AISC 15^th  Eq. 13-16

 = tan2 θ + ()2

 = 3.021

AISC 15^th  Eq. 13-24

 = α ( tan θ + )

 = 35.720

AISC 15^th  Eq. 13-23

 = [ K' tan θ + K ()2 ] / D

 = 14.750

[in]

AISC 15^th  Eq. 13-21

 = ( K' - K tan θ ) / D

 = 8.850

[in]

AISC 15^th  Eq. 13-22

 = [ ( e_b + β ) 2 + ( e_c + α ) 2 ]0.5

 = 29.981

[in]

AISC 15^th  Eq. 13-6

Brace axial force

 = from seismic brace force calc

 = -615.9

[kips]

in tension

Shear force

 = ( β / r ) P_u

 = -181.8

[kips]

AISC 15^th  Eq. 13-2

Axial force

 = ( e_c / r ) P_u

 = -132.5

[kips]

AISC 15^th  Eq. 13-3

Moment

 = H_c ( β - β )

 = 16.84

[kip-ft]

AISC 15^th  Eq. 13-19

Shear force

 = ( α / r ) P_u

 = -303.0

[kips]

AISC 15^th  Eq. 13-5

Axial force

 = ( e_b / r ) P_u

 = -253.7

[kips]

AISC 15^th  Eq. 13-4

Moment

 = V_b ( α - α )

 = 0.00

[kip-ft]

AISC 15^th  Eq. 13-17

Beam end shear reaction

 = from user input

 = 19.9

[kips]

Top brace gusset-beam axial force

 = 

 = 191.8

[kips]

AISC 15^th  Eq. 13-4

Bot brace gusset-beam axial force

 = 

 = -253.7

[kips]

AISC 15^th  Eq. 13-4

Beam to column shear force

 = R_b + V_b-top - V_b-bot

 = 465.4

[kips]

AISC 15^th  Page 13-4

Top brace gusset-column axial force

 = 

 = 100.2

[kips]

AISC 15^th  Eq. 13-3

Bot brace gusset-column axial force

 = 

 = -132.5

[kips]

AISC 15^th  Eq. 13-3

Transfer force from adjacent bay

 = from user input

 = -46.3

[kips]

Beam to column axial force

 = ( H_c-top + H_c-bot ) x -1 - A_b

 = 78.6

[kips]

AISC 15^th  Page 13-4

This force is not for use in connection calc. It's output here for user input connection forces equilibrium check only.

P_bm - Beam member axial force is different from P_b-c - Beam to column interface axial force as shown above.

P_bm - Beam member axial force is from structural analysis output and cannot be used directly in beam end to column connection design as this force is interrupted by brace gusset to beam interface force before beam end reaching the column. This force is actually not needed from user's input for beam end to column connection design.

P_b-c - Beam to column interface axial force is calculated from user's input of brace axial forces and trasnfer force using uniform force method. This force is used in the beam end to column connection design.

P_bm - Beam member axial force is not needed for the beam end to column connection design and is calculated here for verification purpose only. If it matches the structural analysis output, that means equilibrium is reached and user's input of brace axial forces and trasnfer force are correct.

Top brace axial force

 = from seismic brace force calc

 = 449.1

[kips]

in compression

Top brace to ver line angle

 = from user input

 = 45.0

[°]

Top brace gusset-column axial force

 = from calc shown above

 = 100.2

[kips]

AISC 15^th  Eq. 13-3

Bot brace axial force

 = from seismic brace force calc

 = -615.9

[kips]

in tension

Bot brace to ver line angle

 = from user input

 = 45.0

[°]

Bot brace gusset-column axial force

 = from calc shown above

 = -132.5

[kips]

AISC 15^th  Eq. 13-3

Beam to column interface axial force

 = from calc shown above

 = 78.6

[kips]

AISC 15^th  Page 13-4

Beam member axial force

 = (H_ct - P_t sinθ_t ) + (H_cb - P_b sinθ_b )

 = 71.6

[kips]

in compression

+ P_b-c

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

 = 28.541

[in]

 = 0.750

[in]

Plate yield strength

 = 50.0

[ksi]

Plate gross area in shear

 = b_p t_p

 = 21.406

[in²]

Tensile force required

 = 

 = 559.7

[kips]

Plate tensile yielding strength

 = F_y A_g

 = 1070.3

[kips]

AISC 15^th  Eq J4-1

Resistance factor-LRFD

 = 0.90

AISC 15^th  Eq J4-1

 = 

 = 963.3

[kips]

 = 0.58

 > P_u

OK

Plate size

 = 28.541

[in]

 = 0.750

[in]

Plate tensile strength

 = 65.0

[ksi]

Plate net area in tension

 = b_p t_p

 = 21.406

[in²]

Tensile force required

 = 

 = 559.7

[kips]

Plate tensile rupture strength

 = F_u A_nt

 = 1391.4

[kips]

AISC 15^th  Eq J4-2

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq J4-2

 = 

 = 1043.5

[kips]

AISC 15^th  Eq J4-2

 = 0.54

 > 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

 = 12.000

[in]

Transverse weld length

 = reinforce plate width

 = 1.250

[in]

Total weld length - single fillet weld

 = 2 x L_L + L_T

 = 25.250

[in]

AISC 15^th  Eq J2-10a

 = 0.85 x 2 x L_L + 1.5 x L_T

 = 22.275

[in]

AISC 15^th  Eq J2-10b

 = max ( L₁ , L₂ )

 = 25.250

[in]

AISC 15^th  J2.4 (c)

Fillet weld leg size

 = ⁵⁄₁₆

[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

 = 9.279

[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 )

 = 9.279

[kip/in]

AISC 15^th  Eq 9-2

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq 8-1

 = 

 = 6.960

[kip/in]

Shear resistance required

 = 

 = 85.9

[kips]

Fillet weld length - single fillet

 = 

 = 25.250

[in]

Shear resistance provided

 = φ R_n x L

 = 175.7

[kips]

 = 0.49

 > P_u

OK

Plate size

 = 28.541

[in]

 = 0.750

[in]

 = 50.0

[ksi]

 = 29000

[ksi]

Plate gross area in compression

 = b_p t_p

 = 21.406

[in²]

Plate radius of gyration

 = t_p / √12

 = 0.217

[in]

Plate effective length factor

 = 

 = 0.60

Plate unbraced length

 = 

 = 16.420

[in]

Plate slenderness

 = 0.60 x L_u  / r

 = 45.50

when     >  25 , use Chapter E

AISC 15^th  J4.4 (b)

Elastic buckling stress

 = 

 = 138.23

[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

 = 42.98

[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

 = 919.9

[kips]

AISC 15^th  Eq E3-1

Resistance factor-LRFD

 = 0.90

AISC 15^th  E1

 = 

 = 827.9

[kips]

 = 0.54

 > 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

 = 12.000

[in]

Transverse weld length

 = reinforce plate width

 = 1.250

[in]

Total weld length - single fillet weld

 = 2 x L_L + L_T

 = 25.250

[in]

AISC 15^th  Eq J2-10a

 = 0.85 x 2 x L_L + 1.5 x L_T

 = 22.275

[in]

AISC 15^th  Eq J2-10b

 = max ( L₁ , L₂ )

 = 25.250

[in]

AISC 15^th  J2.4 (c)

Fillet weld leg size

 = ⁵⁄₁₆

[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

 = 9.279

[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 )

 = 9.279

[kip/in]

AISC 15^th  Eq 9-2

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq 8-1

 = 

 = 6.960

[kip/in]

Shear resistance required

 = 

 = 85.9

[kips]

Fillet weld length - single fillet

 = 

 = 25.250

[in]

Shear resistance provided

 = φ R_n x L

 = 175.7

[kips]

 = 0.49

 > 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

 = 

 = 17.250

[in]

 ≥ L_min

OK

Plate size

 = 17.250

[in]

 = 0.750

[in]

Plate yield strength

 = 50.0

[ksi]

Plate gross area in shear

 = b_p t_p

 = 12.938

[in²]

Shear force required

 = 

 = 156.8

[kips]

Plate shear yielding strength

 = 0.6 F_y A_gv

 = 388.1

[kips]

AISC 15^th  Eq J4-3

Resistance factor-LRFD

 = 1.00

AISC 15^th  Eq J4-3

 = 

 = 388.1

[kips]

 = 0.40

 > V_u

OK

Plate size

 = 17.250

[in]

 = 0.750

[in]

Plate tensile strength

 = 65.0

[ksi]

Plate net area in shear

 = b_p t_p

 = 12.938

[in²]

Shear force in demand

 = 

 = 156.8

[kips]

Plate shear rupture strength

 = 0.6 F_u A_nv

 = 504.6

[kips]

AISC 15^th  Eq J4-4

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq J4-4

 = 

 = 378.4

[kips]

 = 0.41

 > V_u

OK

Plate size

 = 17.250

[in]

 = 0.750

[in]

Plate yield strength

 = 50.0

[ksi]

Plate gross area in shear

 = b_p t_p

 = 12.938

[in²]

Tensile force required

 = 

 = 124.8

[kips]

Plate tensile yielding strength

 = F_y A_g

 = 646.9

[kips]

AISC 15^th  Eq J4-1

Resistance factor-LRFD

 = 0.90

AISC 15^th  Eq J4-1

 = 

 = 582.2

[kips]

 = 0.21

 > P_u

OK

Plate size

 = 17.250

[in]

 = 0.750

[in]

Plate tensile strength

 = 65.0

[ksi]

Plate net area in tension

 = b_p t_p

 = 12.938

[in²]

Tensile force required

 = 

 = 124.8

[kips]

Plate tensile rupture strength

 = F_u A_nt

 = 840.9

[kips]

AISC 15^th  Eq J4-2

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq J4-2

 = 

 = 630.7

[kips]

AISC 15^th  Eq J4-2

 = 0.20

 > P_u

OK

Gusset plate

 = 17.250

[in]

 = 0.750

[in]

 = 50.0

[ksi]

Shear plate - gross area

 = b_p x t_p

 = 12.938

[in²]

Shear plate - plastic modulus

 = ( b_p x t2p ) / 4

 = 55.79

[in³]

Flexural strength available

 = φ F_y Z_p     φ=0.90

 = 209.22

[kip-ft]

Flexural strength required

 = from gusset interface forces calc

 = 15.86

[kip-ft]

Axial strength available

 = from axial tensile yield check

 = 582.2

[kips]

Axial strength required

 = from gusset interface forces calc

 = -124.8

[kips]

Shear strength available

 = from shear yielding check

 = 388.1

[kips]

Shear strength required

 = from gusset interface forces calc

 = 156.8

[kips]

Flexural yield interaction

 = ( )2 + ( + )2

 = 0.25

AISC 15^th  Eq 10-5

 < 1.0

OK

Gusset plate

 = 17.250

[in]

 = 0.750

[in]

 = 65.0

[ksi]

Net area of plate

 = b_p x t_p

 = 12.938

[in²]

Plastic modulus of net section

 = ( b_p x t2p ) / 4

 = 55.79

[in³]

Flexural strength available

 = φ F_u Z_net     φ=0.75

 = 226.66

[kip-ft]

Flexural strength required

 = from gusset interface forces calc

 = 15.86

[kip-ft]

Axial strength available

 = from axial tensile rupture check

 = 630.7

[kips]

Axial strength required

 = from gusset interface forces calc

 = -124.8

[kips]

Shear strength available

 = from shear rupture check

 = 378.4

[kips]

Shear strength required

 = from gusset interface forces calc

 = 156.8

[kips]

Flexural rupture interaction

 = ( )2 + ( + )2

 = 0.24

AISC 15^th  Eq 10-5

 < 1.0

OK

 = 156.8

[kips]

 = -124.8

[kips]   in tension

 = 15.86

[kip-ft]

Gusset-column fillet weld length

 = from user input

 = 17.250

[in]

Weld stress from axial force

 = H_c / L_wc

 = -7.235

[kip/in]

in tension

Weld stress from shear force

 = V_c / L_wc

 = 9.090

[kip/in]

Weld stress from moment force

 = 

 = 3.838

[kip/in]

Weld stress combined - max

 = [ (f_a - f_b )2 + f2v ]0.5

 = 14.326

[kip/in]

AISC 15^th  Eq 8-11

Weld resultant load angle

 = tan^-1 [( f_b - f_a ) / f_v ]

 = 50.6

[°]

Fillet weld leg size

 = ⁷⁄₁₆

[in]

 = 50.6

[°]

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 )

 = 1.34

AISC 15^th  Page 8-9

Fillet weld shear strength

 = 0.6 (C₁ x 70 ksi) 0.707 w n C₂

 = 34.810

[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]

 = 0.82

 > f_max

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

 = 

 = -124.8

[kips]

Gusset edge moment force

 = 

 = 15.86

[kip-ft]

Gusset edge interface length

 = 

 = 17.250

[in]

Gusset edge equivalent normal force

 = N -

 = -168.9

[kips]

AISC DG29  Fig B-1

Concentrated force from gusset

 = 

 = 168.9

[kips]

Column section

 = 12.900

[in]

 = 0.990

[in]

 = 0.610

[in]

 = 1.590

[in]

 = 50.0

[ksi]

Length of bearing

 = Gusset/Column interface length

 = 17.250

[in]

Column web local yielding strength

 = F_y t_w ( 5 k + l_b )

 = 768.6

[kips]

AISC 15^th  Eq J10-2

Resistance factor-LRFD

 = 1.00

 = 

 = 768.6

[kips]

 = 0.22

 > P_u

OK

W sect W12X106

 = 12.900

[in]

 = 0.610

[in]

 = 50.0

[ksi]

Gusset to column axial force

 = from gusset interface force calc

 = 124.8

[kips]

Column shear strength

 = 0.6 F_y d t_w

 = 236.1

[kips]

AISC 15^th  Eq J4-3

Resistance factor-LRFD

 = 1.00

AISC 15^th  Eq J4-3

 = 

 = 236.1

[kips]

 = 0.53

 > V_u

OK

Plate size

 = 17.250

[in]

 = 0.750

[in]

Plate yield strength

 = 50.0

[ksi]

Plate gross area in shear

 = b_p t_p

 = 12.938

[in²]

Shear force required

 = 

 = 125.8

[kips]

Plate shear yielding strength

 = 0.6 F_y A_gv

 = 388.1

[kips]

AISC 15^th  Eq J4-3

Resistance factor-LRFD

 = 1.00

AISC 15^th  Eq J4-3

 = 

 = 388.1

[kips]

 = 0.32

 > V_u

OK

Plate size

 = 17.250

[in]

 = 0.750

[in]

Plate tensile strength

 = 65.0

[ksi]

Plate net area in shear

 = b_p t_p

 = 12.938

[in²]

Shear force in demand

 = 

 = 125.8

[kips]

Plate shear rupture strength

 = 0.6 F_u A_nv

 = 504.6

[kips]

AISC 15^th  Eq J4-4

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq J4-4

 = 

 = 378.4

[kips]

 = 0.33

 > V_u

OK

Plate size

 = 17.250

[in]

 = 0.750

[in]

Plate yield strength

 = 50.0

[ksi]

Plate gross area in shear

 = b_p t_p

 = 12.938

[in²]

Tensile force required

 = 

 = 100.2

[kips]

Plate tensile yielding strength

 = F_y A_g

 = 646.9

[kips]

AISC 15^th  Eq J4-1

Resistance factor-LRFD

 = 0.90

AISC 15^th  Eq J4-1

 = 

 = 582.2

[kips]

 = 0.17

 > P_u

OK

Gusset plate

 = 17.250

[in]

 = 0.750

[in]

 = 50.0

[ksi]

Shear plate - gross area

 = b_p x t_p

 = 12.938

[in²]

Shear plate - plastic modulus

 = ( b_p x t2p ) / 4

 = 55.79

[in³]

Flexural strength available

 = φ F_y Z_p     φ=0.90

 = 209.22

[kip-ft]

Flexural strength required

 = from gusset interface forces calc

 = 12.73

[kip-ft]

Axial strength available

 = from axial tensile yield check

 = 582.2

[kips]

Axial strength required

 = from gusset interface forces calc

 = 100.2

[kips]

Shear strength available

 = from shear yielding check

 = 388.1

[kips]

Shear strength required

 = from gusset interface forces calc

 = 125.8

[kips]

Flexural yield interaction

 = ( )2 + ( + )2

 = 0.12

AISC 15^th  Eq 10-5

 < 1.0

OK

Gusset plate

 = 17.250

[in]

 = 0.750

[in]

 = 65.0

[ksi]

Net area of plate

 = b_p x t_p

 = 12.938

[in²]

Plastic modulus of net section

 = ( b_p x t2p ) / 4

 = 55.79

[in³]

Flexural strength available

 = φ F_u Z_net     φ=0.75

 = 226.66

[kip-ft]

Flexural strength required

 = from gusset interface forces calc

 = 12.73

[kip-ft]

Shear strength available

 = from shear rupture check

 = 378.4

[kips]

Shear strength required

 = from gusset interface forces calc

 = 125.8

[kips]

Flexural rupture interaction

 = ( )2 + ( )2

 = 0.11

AISC 15^th  Eq 10-5

 < 1.0

OK

 = 125.8

[kips]

 = 100.2

[kips]   in compression

 = 12.73

[kip-ft]

Gusset-column fillet weld length

 = from user input

 = 17.250

[in]

Weld stress from axial force

 = H_c / L_wc

 = 5.809

[kip/in]

in compression

Weld stress from shear force

 = V_c / L_wc

 = 7.293

[kip/in]

Weld stress from moment force

 = 

 = 3.080

[kip/in]

Weld stress combined - max

 = f_v

 = 7.293

[kip/in]

AISC 15^th  Eq 8-11

Weld resultant load angle

 = weld only has shear component

 = 0.0

[°]

Fillet weld leg size

 = ⁷⁄₁₆

[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

 = 25.982

[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 )

 = 25.982

[kip/in]

AISC 15^th  Eq 9-2

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq 8-1

 = 

 = 19.487

[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 )

 = 15.589

[kip/in]

 = 0.47

 > f_max

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

 = 

 = 100.2

[kips]

Gusset edge moment force

 = 

 = 12.73

[kip-ft]

Gusset edge interface length

 = 

 = 17.250

[in]

Gusset edge equivalent normal force

 = N +

 = 135.6

[kips]

AISC DG29  Fig B-1

Concentrated force from gusset

 = 

 = 135.6

[kips]

Column section

 = 12.900

[in]

 = 0.990

[in]

 = 0.610

[in]

 = 1.590

[in]

 = 50.0

[ksi]

Length of bearing

 = Gusset/Column interface length

 = 17.250

[in]

Column web local yielding strength

 = F_y t_w ( 5 k + l_b )

 = 768.6

[kips]

AISC 15^th  Eq J10-2

Resistance factor-LRFD

 = 1.00

 = 

 = 768.6

[kips]

 = 0.18

 > 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

 = 

 = 100.2

[kips]

Gusset edge moment force

 = 

 = 12.73

[kip-ft]

Gusset edge interface length

 = 

 = 17.250

[in]

Gusset edge equivalent normal force

 = N +

 = 135.6

[kips]

AISC DG29  Fig B-1

Concentrated force from gusset

 = 

 = 135.6

[kips]

Column section

 = 12.900

[in]

 = 0.990

[in]

 = 0.610

[in]

 = 1.590

[in]

 = 50.0

[ksi]

 = 29000

[ksi]

Length of bearing

 = Gusset/Column interface length

 = 17.250

[in]

when l_N  ≥ d/2 , use Eq J10-4

AISC 15^th  Eq J10-4

Column web local crippling strength

 = 0.8 t2w [1+3( )^1.5 ] x

 = 1342.7

[kips]

AISC 15^th  Eq J10-4

( )^0.5

Resistance factor-LRFD

 = 0.75

AISC 15^th  J10.3

 = 

 = 1007.0

[kips]

 = 0.13

 > P_u

OK

W sect W12X106

 = 12.900

[in]

 = 0.610

[in]

 = 50.0

[ksi]

Gusset to column axial force

 = from gusset interface force calc

 = 100.2

[kips]

Column shear strength

 = 0.6 F_y d t_w

 = 236.1

[kips]

AISC 15^th  Eq J4-3

Resistance factor-LRFD

 = 1.00

AISC 15^th  Eq J4-3

 = 

 = 236.1

[kips]

 = 0.42

 > V_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

 = 

 = 26.000

[in]

 ≥ L_min

OK

Plate size

 = 26.000

[in]

 = 0.750

[in]

Plate yield strength

 = 50.0

[ksi]

Plate gross area in shear

 = b_p t_p

 = 19.500

[in²]

Shear force required

 = 

 = 270.9

[kips]

Plate shear yielding strength

 = 0.6 F_y A_gv

 = 585.0

[kips]

AISC 15^th  Eq J4-3

Resistance factor-LRFD

 = 1.00

AISC 15^th  Eq J4-3

 = 

 = 585.0

[kips]

 = 0.46

 > V_u

OK

Plate size

 = 26.000

[in]

 = 0.750

[in]

Plate tensile strength

 = 65.0

[ksi]

Plate net area in shear

 = b_p t_p

 = 19.500

[in²]

Shear force in demand

 = 

 = 270.9

[kips]

Plate shear rupture strength

 = 0.6 F_u A_nv

 = 760.5

[kips]

AISC 15^th  Eq J4-4

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq J4-4

 = 

 = 570.4

[kips]

 = 0.47

 > V_u

OK

Plate size

 = 26.000

[in]

 = 0.750

[in]

Plate yield strength

 = 50.0

[ksi]

Plate gross area in shear

 = b_p t_p

 = 19.500

[in²]

Tensile force required

 = 

 = 239.0

[kips]

Plate tensile yielding strength

 = F_y A_g

 = 975.0

[kips]

AISC 15^th  Eq J4-1

Resistance factor-LRFD

 = 0.90

AISC 15^th  Eq J4-1

 = 

 = 877.5

[kips]

 = 0.27

 > P_u

OK

Plate size

 = 26.000

[in]

 = 0.750

[in]

Plate tensile strength

 = 65.0

[ksi]

Plate net area in tension

 = b_p t_p

 = 19.500

[in²]

Tensile force required

 = 

 = 239.0

[kips]

Plate tensile rupture strength

 = F_u A_nt

 = 1267.5

[kips]

AISC 15^th  Eq J4-2

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq J4-2

 = 

 = 950.6

[kips]

AISC 15^th  Eq J4-2

 = 0.25

 > P_u

OK

Gusset plate

 = 26.000

[in]

 = 0.750

[in]

 = 50.0

[ksi]

Shear plate - gross area

 = b_p x t_p

 = 19.500

[in²]

Shear plate - plastic modulus

 = ( b_p x t2p ) / 4

 = 126.75

[in³]

Flexural strength available

 = φ F_y Z_p     φ=0.90

 = 475.31

[kip-ft]

Flexural strength required

 = from gusset interface forces calc

 = 0.00

[kip-ft]

Axial strength available

 = from axial tensile yield check

 = 877.5

[kips]

Axial strength required

 = from gusset interface forces calc

 = -239.0

[kips]

Shear strength available

 = from shear yielding check

 = 585.0

[kips]

Shear strength required

 = from gusset interface forces calc

 = 270.9

[kips]

Flexural yield interaction

 = ( )2 + ( + )2

 = 0.29

AISC 15^th  Eq 10-5

 < 1.0

OK

Gusset plate

 = 26.000

[in]

 = 0.750

[in]

 = 65.0

[ksi]

Net area of plate

 = b_p x t_p

 = 19.500

[in²]

Plastic modulus of net section

 = ( b_p x t2p ) / 4

 = 126.75

[in³]

Flexural strength available

 = φ F_u Z_net     φ=0.75

 = 514.92

[kip-ft]

Flexural strength required

 = from gusset interface forces calc

 = 0.00

[kip-ft]

Axial strength available

 = from axial tensile rupture check

 = 950.6

[kips]

Axial strength required

 = from gusset interface forces calc

 = -239.0

[kips]

Shear strength available

 = from shear rupture check

 = 570.4

[kips]

Shear strength required

 = from gusset interface forces calc

 = 270.9

[kips]

Flexural rupture interaction

 = ( )2 + ( + )2

 = 0.29

AISC 15^th  Eq 10-5

 < 1.0

OK

 = 270.9

[kips]

 = -239.0

[kips]   in tension

 = 0.00

[kip-ft]

Gusset-beam fillet weld length

 = 

 = 26.000

[in]

Weld stress from axial force

 = V_b / L_wb

 = -9.192

[kip/in]

in tension

Weld stress from shear force

 = H_b / L_wb

 = 10.419

[kip/in]

Weld stress from moment force

 = 

 = 0.000

[kip/in]

Weld stress combined - max

 = [ (f_a - f_b )2 + f2v ]0.5

 = 13.895

[kip/in]

AISC 15^th  Eq 8-11

Weld resultant load angle

 = tan^-1 [( f_b - f_a ) / f_v ]

 = 41.4

[°]

Fillet weld leg size

 = ⁷⁄₁₆

[in]

 = 41.4

[°]

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 )

 = 1.27

AISC 15^th  Page 8-9

Fillet weld shear strength

 = 0.6 (C₁ x 70 ksi) 0.707 w n C₂

 = 32.973

[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]

 = 0.79

 > f_max

OK

Concentrated force from gusset

 = 

 = 239.0

[kips]

Beam section

 = 24.700

[in]

 = 1.090

[in]

 = 0.650

[in]

 = 1.590

[in]

 = 50.0

[ksi]

Length of bearing

 = Gusset/Beam interface length

 = 26.000

[in]

Gusset plate corner clip

 = from user input

 = 1.000

[in]

Distance from normal force applied point to member end

 = 0.5 l_b + clip

 = 14.000

[in]

when l_N  ≤ d , use AISC 15^th  Eq J10-3

AISC 15^th  Eq J10-3

Beam web local yielding strength

 = F_y t_w ( 2.5 k + l_b )

 = 974.2

[kips]

AISC 15^th  Eq J10-3

Resistance factor-LRFD

 = 1.00

 = 

 = 974.2

[kips]

 = 0.25

 > P_u

OK

Plate size

 = 26.000

[in]

 = 0.750

[in]

Plate yield strength

 = 50.0

[ksi]

Plate gross area in shear

 = b_p t_p

 = 19.500

[in²]

Shear force required

 = 

 = 217.4

[kips]

Plate shear yielding strength

 = 0.6 F_y A_gv

 = 585.0

[kips]

AISC 15^th  Eq J4-3

Resistance factor-LRFD

 = 1.00

AISC 15^th  Eq J4-3

 = 

 = 585.0

[kips]

 = 0.37

 > V_u

OK

Plate size

 = 26.000

[in]

 = 0.750

[in]

Plate tensile strength

 = 65.0

[ksi]

Plate net area in shear

 = b_p t_p

 = 19.500

[in²]

Shear force in demand

 = 

 = 217.4

[kips]

Plate shear rupture strength

 = 0.6 F_u A_nv

 = 760.5

[kips]

AISC 15^th  Eq J4-4

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq J4-4

 = 

 = 570.4

[kips]

 = 0.38

 > V_u

OK

Plate size

 = 26.000

[in]

 = 0.750

[in]

Plate yield strength

 = 50.0

[ksi]

Plate gross area in shear

 = b_p t_p

 = 19.500

[in²]

Tensile force required

 = 

 = 191.8

[kips]

Plate tensile yielding strength

 = F_y A_g

 = 975.0

[kips]

AISC 15^th  Eq J4-1

Resistance factor-LRFD

 = 0.90

AISC 15^th  Eq J4-1

 = 

 = 877.5

[kips]

 = 0.22

 > P_u

OK

Gusset plate

 = 26.000

[in]

 = 0.750

[in]

 = 50.0

[ksi]

Shear plate - gross area

 = b_p x t_p

 = 19.500

[in²]

Shear plate - plastic modulus

 = ( b_p x t2p ) / 4

 = 126.75

[in³]

Flexural strength available

 = φ F_y Z_p     φ=0.90

 = 475.31

[kip-ft]

Flexural strength required

 = from gusset interface forces calc

 = 0.00

[kip-ft]

Axial strength available

 = from axial tensile yield check

 = 877.5

[kips]

Axial strength required

 = from gusset interface forces calc

 = 191.8

[kips]

Shear strength available

 = from shear yielding check

 = 585.0

[kips]

Shear strength required

 = from gusset interface forces calc

 = 217.4

[kips]

Flexural yield interaction

 = ( )2 + ( + )2

 = 0.19

AISC 15^th  Eq 10-5

 < 1.0

OK

Gusset plate

 = 26.000

[in]

 = 0.750

[in]

 = 65.0

[ksi]

Net area of plate

 = b_p x t_p

 = 19.500

[in²]

Plastic modulus of net section

 = ( b_p x t2p ) / 4

 = 126.75

[in³]

Flexural strength available

 = φ F_u Z_net     φ=0.75

 = 514.92

[kip-ft]

Flexural strength required

 = from gusset interface forces calc

 = 0.00

[kip-ft]

Shear strength available

 = from shear rupture check

 = 570.4

[kips]

Shear strength required

 = from gusset interface forces calc

 = 217.4

[kips]

Flexural rupture interaction

 = ( )2 + ( )2

 = 0.15

AISC 15^th  Eq 10-5

 < 1.0

OK