User Input

Load Case 1 - Anchor Additional Tension From Moment Caused by V_y

Shear key shear force

 = from user load input

 = 175.00

[kips]

Refer to sketch below , shear key's shear reaction takes moment to base plate center line and this moment will cause additional tensile force on anchors

Base plate & grout thickness

 = 2.000

[in]

 = 1.000

[in]

Shear key depth

 = 8.000

[in]

Shear key shear V to base plate center moment arm distance

 = 0.5( d_sk - g) + g + 0.5 t_p

 = 5.500

[in]

Moment by shear key shear

 = V_u d_ms

 = 80.21

[kip-ft]

Anchor out-out spacing - in shear direction

 = from user input

 = 10.000

[in]

Column depth - in shear direction

 = sect W18X86

 = 18.400

[in]

Exterior anchhor moment arm

 = d + 0.5( s₁ - d )

 = 14.200

[in]

Anchor number along exterior anchhor

 = from user input

 = 2

Single anchor tension from moment caused by shear key

 = 

 = 33.89

[kips]

Load Case 2 - Anchor Additional Tension From Moment Caused by V_y

Shear key shear force

 = from user load input

 = 175.00

[kips]

Refer to sketch below , shear key's shear reaction takes moment to base plate center line and this moment will cause additional tensile force on anchors

Base plate & grout thickness

 = 2.000

[in]

 = 1.000

[in]

Shear key depth

 = 8.000

[in]

Shear key shear V to base plate center moment arm distance

 = 0.5( d_sk - g) + g + 0.5 t_p

 = 5.500

[in]

Moment by shear key shear

 = V_u d_ms

 = 80.21

[kip-ft]

Anchor out-out spacing - in shear direction

 = from user input

 = 10.000

[in]

Column depth - in shear direction

 = sect W18X86

 = 18.400

[in]

Exterior anchhor moment arm

 = d + 0.5( s₁ - d )

 = 14.200

[in]

Anchor number along exterior anchhor

 = from user input

 = 2

Single anchor tension from moment caused by shear key

 = 

 = 33.89

[kips]

Check min anchor dimensions as per PIP STE05121 Application of ASCE Anchorage Design for Petrochemical Facilities - 2018   Table 1 as shown below.

This check is NOT a code requirement.   User can turn this check On/Off by changing setting at Anchor Bolt --> Anchor Bolt - Config & Setting --> Check min anchor spacing and edge distance as per PIP STE05121 Table 1

Anchor rod grade and dia

 = A320 Gr L7

 = 1¹⁄₂

[in]

Anchor edge distance

 = 13.000

[in]

 = 10.000

[in]

 = 13.000

[in]

 = 10.000

[in]

Min anchor edge distance required

 = from PIP STE05121 Table 1 below

 = 9.000

[in]

PIP STE05121 Table 1

Min anchor edge distance

 = min(c₁ , c₂ , c₃ , c₄ )

 = 10.000

[in]

 ≥ c_min

OK

Min anchor spacing required

 = from PIP STE05121 Table 1 below

 = 6.000

[in]

PIP STE05121 Table 1

Anchor bolt pattern

 = from user input

 = 4A

Min anchor spacing

 = from user input

 = 10.000

[in]

 ≥ s_min

OK

Min anchor embedment required

 = from PIP STE05121 Table 1 below

 = 18.000

[in]

PIP STE05121 Table 1

Min anchor embedment depth

 = from user input

 = 20.000

[in]

 ≥ h_min

OK

Table 1 from PIP STE05121 Application of ASCE Anchorage Design for Petrochemical Facilities - 2018

When anchor reinforcement or supplementary reinforcement is provided, the check on min edge distance
is not required. Only min anchor spacing is checked as per ACI 318-19 Table 17.9.2(a)

ACI 318-19 17.9.1

ACI 318-19 17.9.1

Anchor rod grade and dia

 = A320 Gr L7

 = 1¹⁄₂

[in]

Min anchor spacing required

 = 6 x d_a

 = 9.000

[in]

ACI 318-19 Table 17.9.2(a)

Anchor bolt pattern

 = from user input

 = 4A

Min anchor spacing

 = from user input

 = 10.000

[in]

 ≥ s_min

OK

ACI 318-19 Table 17.9.2(a)

There is no tensile load applied to anchor/anchor group, either no tensile load from user load input or it's small moment
case and tension caused by moment is offset by axial compression load, so Anchor Tension Resistance and Tension - Shear
Interaction checks are Not Applicable

There is no shear load from user load input or shear key is used and all shear is taken by shear key, so
Anchor Shear Resistance and Tension - Shear Interaction checks are Not Applicable

Seismic - Tension

    Not Applicable

ACI 318-19 17.10.5.1

There is no tensile load applied to anchor/anchor group, either there is no tensile load from user load input or
tension caused by moment is offset by axial compression load, so Seismic Tension check is NOT required

Seismic - Shear

    Not Applicable

ACI 318-19 17.10.6.1

There is no shear load applied to anchor/anchor group, so Seismic Shear check is NOT required

Check min anchor dimensions as per PIP STE05121 Application of ASCE Anchorage Design for Petrochemical Facilities - 2018   Table 1 as shown below.

This check is NOT a code requirement.   User can turn this check On/Off by changing setting at Anchor Bolt --> Anchor Bolt - Config & Setting --> Check min anchor spacing and edge distance as per PIP STE05121 Table 1

Anchor rod grade and dia

 = A320 Gr L7

 = 1¹⁄₂

[in]

Anchor edge distance

 = 13.000

[in]

 = 10.000

[in]

 = 13.000

[in]

 = 10.000

[in]

Min anchor edge distance required

 = from PIP STE05121 Table 1 below

 = 9.000

[in]

PIP STE05121 Table 1

Min anchor edge distance

 = min(c₁ , c₂ , c₃ , c₄ )

 = 10.000

[in]

 ≥ c_min

OK

Min anchor spacing required

 = from PIP STE05121 Table 1 below

 = 6.000

[in]

PIP STE05121 Table 1

Anchor bolt pattern

 = from user input

 = 4A

Min anchor spacing

 = from user input

 = 10.000

[in]

 ≥ s_min

OK

Min anchor embedment required

 = from PIP STE05121 Table 1 below

 = 18.000

[in]

PIP STE05121 Table 1

Min anchor embedment depth

 = from user input

 = 20.000

[in]

 ≥ h_min

OK

Table 1 from PIP STE05121 Application of ASCE Anchorage Design for Petrochemical Facilities - 2018

When anchor reinforcement or supplementary reinforcement is provided, the check on min edge distance
is not required. Only min anchor spacing is checked as per ACI 318-19 Table 17.9.2(a)

ACI 318-19 17.9.1

ACI 318-19 17.9.1

Anchor rod grade and dia

 = A320 Gr L7

 = 1¹⁄₂

[in]

Min anchor spacing required

 = 6 x d_a

 = 9.000

[in]

ACI 318-19 Table 17.9.2(a)

Anchor bolt pattern

 = from user input

 = 4A

Min anchor spacing

 = from user input

 = 10.000

[in]

 ≥ s_min

OK

ACI 318-19 Table 17.9.2(a)

Anchor rod effective section area

 = 1.41

[in²]

 = 125.0

[ksi]

Anchor rod steel strength in tension

 = A_se f_uta

 = 176.25

[kips]

ACI 318-19 17.6.1.2

Anchor group axial tensile force

 = from user load input

 = -85.00

[kips]

in tension

No of anchors in the group

 = 

 = 4

Refer to Anchor Forces Calculation section above , shear key's shear reaction takes moment to base plate center line and this moment will cause additional tensile force on anchors

Single anchor tension from moment caused by shear key reaction force

 = from Anchor Forces Calculation above

 = 33.89

[kips]

in tension

Single anchor tensile force

 = T_sk - P / n_t

 = 55.14

[kips]

Strength reduction factor

 = 0.75

ACI 318-19 17.5.3(a)

 = 0.75 x 176.25

 = 132.19

[kips]

 = 0.42

 > T

OK

Concrete & rebar strength

 = 4.5

[ksi]

 = 60.0

[ksi]

Ver rebar no & area

 = #8

 = 0.79

[in²]

 = 1"

 = 1.0

 = 1.0

ACI 318-19 Table 25.4.3.2

 = 1.0

 = 0.90

Hook rebar development length

 = max( d1.5b , 8d_b , 6" )

 = 14.636

[in]

ACI 318-19 25.4.3.1

Actual rebar development length

 = h_ef - top cover (2") - d_ar tan35

 = 15.199

[in]

 = 0.53

 > l_min

OK

ACI 318-19 25.4.3.1

Anchor group tensile load

 = from user load input

 = 85.00

[kips]

No of ver rebar effective to resist anchor tension

 = from user input

 = 4.0

Rebar resistance factor

 = 0.75

ACI 318-19 17.5.3

Anchor reinft breakout resistance

 = φ_s n_v A_s f_y

 = 142.01

[kips]

ACI 318-19 17.5.2.1 (a)

 = 0.60

 > N_u

OK

Anchor head net bearing area & conc strength

 = 3.12

[in²]

 = 4.5

[ksi]

Single bolt pullout resistance

 = 8 A_brg f_c

 = 112.25

[kips]

ACI 318-19 17.6.3.2.2a

Pullout cracking factor

 = for cracked concrete

 = 1.00

ACI 318-19 17.6.3.3.1(b)

Anchor group axial tensile force

 = from user load input

 = -85.00

[kips]

in tension

No of anchors in the group

 = 

 = 4

Refer to Anchor Forces Calculation section above , shear key's shear reaction takes moment to base plate center line and this moment will cause additional tensile force on anchors

Single anchor tension from moment caused by shear key reaction force

 = from Anchor Forces Calculation above

 = 33.89

[kips]

in tension

Single anchor tensile force

 = T_sk - P / n_t

 = 55.14

[kips]

Strength reduction factor

 = 0.70

 pullout strength is always Condition B

ACI 318-19 17.5.3(c)

 = φ_tc Ψ_cP N_p

 = 78.57

[kips]

Seismic design strength reduction

 = x 1.0   not applicable

 = 78.57

[kips]

ACI 318-19 17.10.5.4(c)

 = 0.70

 > T

OK

Anchor edge distance

 = 13.000

[in]

 = 10.000

[in]

 = 13.000

[in]

 = 10.000

[in]

Anchor out-out spacing

 = 10.000

[in]

 = 10.000

[in]

Side Edges Along X-X Axis - Width Edges

Anchor edge distance in Y direction

 = min (c₁ , c₃ )

 = 13.000

[in]

Anchor embedment depth

 = from user input

 = 20.000

[in]

Side blowout check is required on this edge or not

 = check if h_ef > 2.5 c_a1

 = False

ACI 318-19 17.6.4.1

Side blowout check is NOT required

ACI 318-19 17.6.4.1

Side Edges Along Y-Y Axis - Depth Edges

Anchor edge distance in X direction

 = min (c₂ , c₄ )

 = 10.000

[in]

Anchor embedment depth

 = from user input

 = 20.000

[in]

Side blowout check is required on this edge or not

 = check if h_ef > 2.5 c_a2

 = False

ACI 318-19 17.6.4.1

Side blowout check is NOT required

ACI 318-19 17.6.4.1

Anchor group governing tensile resistance is the minimum value of the resistance values in
the following limit states

No of anchors in anchor group
resisting tension

 = from Anchor Forces Calculation above

 = 4

Anchor rod tensile resistance

 = 4 x 132.19

 = 528.75

[kips]

Anchor concrete breakout resistance

 = from anchor reinft tensile breakout

 = 142.01

[kips]

resistance calc above

Anchor pullout resistance

 = 4 x 78.57

 = 314.29

[kips]

Anchor side blowout resistance

 = from anchor side blowout calc above

 = N/A

Anchor group governing tensile resistance

 = minimum of above values

 = 142.01

[kips]

There is no shear load from user load input or shear key is used and all shear is taken by shear key, so
Anchor Shear Resistance and Tension - Shear Interaction checks are Not Applicable

Seismic - Tension

    Not Applicable

ACI 318-19 17.10.5.1

Seismic SDC < C or E <= 0.2U , additional seismic requirements in ACI 318-19 17.10.5.3 is NOT required

ACI 318-19 17.10.5.3

Seismic - Shear

    Not Applicable

ACI 318-19 17.10.6.1

There is no shear load applied to anchor/anchor group, so Seismic Shear check is NOT required

Check if h_ef / h_sl ≥ 2.5

 = h_ef / h_sl

 = 2.86

ACI 318-19

 ≥ 2.5

OK

17.11.1.1.8 (a)

Check if h_ef / c_sl ≥ 2.5

 = h_ef / c_sl

 = 4.00

 ≥ 2.5

OK

17.11.1.1.8 (b)

Concrete Breakout Toward Free Edge

User needs to provide hor anchor reinforcement such as hor hair pin or enclosed stirrup around the shear key
pocket to resist concrete breakout. When the non-enclosed rebar such as hor hair pin is used , user must
ensure that rebar has adequate development length l_d or l_dh on both sides of failure plane as shown in
ACI 318-19 Fig. R17.5.2.1b(i)

17.5.2.1
Fig. R17.5.2.1b(i)

* For tie reinforcement, only the top most 2 or 3 layers of ties (2" from TOC and 2x3" after) are effective

* Assume 100% of hor. tie bars can develop full yield strength as per user's choice in Anchor Reinforcement input

Total number of hor tie bar

 = n_leg (leg) x n_lay (layer)

 = 18

Shear load on anchor group

 = from user load input

 = 175.00

[kips]

Rebar resistance factor

 = 0.75

17.5.3 (a)

Hor rebar area & strength

 = 0.31

[in²]

 = 60.0

[ksi]

Single tie bar tension resistance

 = φ_s f_yh A_s

 = 13.95

[kips]

Total tie bar tension resistance

 = n x T_r

 = 251.10

[kips]

17.5.2.1 (b)

 = 0.70

 > V_u

OK

Shear lug sect HSS8.625X0.625

 = 8.625

[in]

 = 0.581

[in]

Shear lug embed depth

 = from user input

 = 8.000

[in]

Grout thickness

 = from user input

 = 1.000

[in]

Conc compressive strength

 = from user input

 = 4.5

[ksi]

Compressive load in anchor group

 = from user load input

 = 545.00

[kips]

Base plate area

 = 17.000 x 21.000

 = 357.00

[in²]

Shear lug bearing factor

 = 1 + 4 ≤ 2.0

 = 2.00

ACI 318-19 17.11.2.2.1c

Shear in Y direction

 = from user input

 = 175.00

[kips]

Shear lug bearing area

 = OD (d - g )

 = 60.38

[in²]

Shear lug bearing strength

 = 1.7 f_c A_ef,sl Ψ_b,sl

 = 923.74

[kips]

ACI 318-19 17.11.2.1

Resistance factor-LRFD

 = 0.65

ACI 318-19 17.11.1.1.4

 = 

 = 600.43

[kips]

 = 0.29

 > V_y

OK

Shear key sect HSS8.625X0.625

 = 8.625

[in]

 = 0.581

[in]

 = 37.66

[in³]

 = 37.66

[in³]

Shear key steel yield strength

 = 46.0

[ksi]

Shear key embed depth

 = from user input

 = 8.000

[in]

Grout thickness

 = from user input

 = 1.000

[in]

Shear in Y direction

 = from user input

 = 175.00

[kips]

Moment caused by shear

 = V_y [ g + 0.5(d - g) ]

 = 65.63

[kip-ft]

Shear key flexural capacity

 = F_y Z_s

 = 144.36

[kip-ft]

Resistance factor-LRFD

 = 0.90

 = 

 = 129.93

[kip-ft]

 = 0.51

 > M_s

OK

Shear key sect HSS8.625X0.625

 = 8.625

[in]

 = 0.581

[in]

Shear key embed depth

 = from user input

 = 8.000

[in]

Grout thickness

 = from user input

 = 1.000

[in]

Shear key steel yield strength

 = 46.0

[ksi]

Shear in Y direction

 = from user input

 = 175.00

[kips]

CHS sect wall center line radius

 = 0.5 ( OD - t )

 = 4.022

[in]

CHS sect shear area

 = π r t

 = 7.34

[in²]

Shear key shear strength

 = 0.6 F_y A_w

 = 202.62

[kips]

Resistance factor-LRFD

 = 0.90

 = 

 = 182.36

[kips]

 = 0.96

 > V_y

OK

Check if h_ef / h_sl ≥ 2.5

 = h_ef / h_sl

 = 2.86

ACI 318-19

 ≥ 2.5

OK

17.11.1.1.8 (a)

Check if h_ef / c_sl ≥ 2.5

 = h_ef / c_sl

 = 4.00

 ≥ 2.5

OK

17.11.1.1.8 (b)

Concrete Breakout Toward Free Edge

User needs to provide hor anchor reinforcement such as hor hair pin or enclosed stirrup around the shear key
pocket to resist concrete breakout. When the non-enclosed rebar such as hor hair pin is used , user must
ensure that rebar has adequate development length l_d or l_dh on both sides of failure plane as shown in
ACI 318-19 Fig. R17.5.2.1b(i)

17.5.2.1
Fig. R17.5.2.1b(i)

* For tie reinforcement, only the top most 2 or 3 layers of ties (2" from TOC and 2x3" after) are effective

* Assume 100% of hor. tie bars can develop full yield strength as per user's choice in Anchor Reinforcement input

Total number of hor tie bar

 = n_leg (leg) x n_lay (layer)

 = 18

Shear load on anchor group

 = from user load input

 = 175.00

[kips]

Rebar resistance factor

 = 0.75

17.5.3 (a)

Hor rebar area & strength

 = 0.31

[in²]

 = 60.0

[ksi]

Single tie bar tension resistance

 = φ_s f_yh A_s

 = 13.95

[kips]

Total tie bar tension resistance

 = n x T_r

 = 251.10

[kips]

17.5.2.1 (b)

 = 0.70

 > V_u

OK

Shear lug sect HSS8.625X0.625

 = 8.625

[in]

 = 0.581

[in]

Shear lug embed depth

 = from user input

 = 8.000

[in]

Grout thickness

 = from user input

 = 1.000

[in]

Conc compressive strength

 = from user input

 = 4.5

[ksi]

Anchor rod effective section area

 = 1.41

[in²]

 = 125.0

[ksi]

Anchor rod steel strength in tension

 = A_se f_uta

 = 176.25

[kips]

ACI 318-19 17.6.1.2

No of anchors in tension

 = 

 = 4

Tensile load in anchor group

 = from user load input

 = -85.00

[kips]

Shear lug bearing factor

 = 1 + ≤ 1.0

 = 0.88

ACI 318-19 17.11.2.2.1a

Shear in Y direction

 = from user input

 = 175.00

[kips]

Shear lug bearing area

 = OD (d - g )

 = 60.38

[in²]

Shear lug bearing strength

 = 1.7 f_c A_ef,sl Ψ_b,sl

 = 406.18

[kips]

ACI 318-19 17.11.2.1

Resistance factor-LRFD

 = 0.65

ACI 318-19 17.11.1.1.4

 = 

 = 264.02

[kips]

 = 0.66

 > V_y

OK

Shear key sect HSS8.625X0.625

 = 8.625

[in]

 = 0.581

[in]

 = 37.66

[in³]

 = 37.66

[in³]

Shear key steel yield strength

 = 46.0

[ksi]

Shear key embed depth

 = from user input

 = 8.000

[in]

Grout thickness

 = from user input

 = 1.000

[in]

Shear in Y direction

 = from user input

 = 175.00

[kips]

Moment caused by shear

 = V_y [ g + 0.5(d - g) ]

 = 65.63

[kip-ft]

Shear key flexural capacity

 = F_y Z_s

 = 144.36

[kip-ft]

Resistance factor-LRFD

 = 0.90

 = 

 = 129.93

[kip-ft]

 = 0.51

 > M_s

OK

Shear key sect HSS8.625X0.625

 = 8.625

[in]

 = 0.581

[in]

Shear key embed depth

 = from user input

 = 8.000

[in]

Grout thickness

 = from user input

 = 1.000

[in]

Shear key steel yield strength

 = 46.0

[ksi]

Shear in Y direction

 = from user input

 = 175.00

[kips]

CHS sect wall center line radius

 = 0.5 ( OD - t )

 = 4.022

[in]

CHS sect shear area

 = π r t

 = 7.34

[in²]

Shear key shear strength

 = 0.6 F_y A_w

 = 202.62

[kips]

Resistance factor-LRFD

 = 0.90

 = 

 = 182.36

[kips]

 = 0.96

 > V_y

OK

Suggested minimum base plate thickness for rigidity

 = max ( m/4 , n/4 )

 = 1.015

[in]

 = 0.51

 < t_p

OK

Factored compression force

 = from user input

 = 545.00

[kips]

ACI 318-19

 = min ( √A₂ / A₁ , 2 )

 = 1.739

Table 14.5.6.1

Concrete strength & strength reduction factor

 = 4.5

[ksi]

 = 0.65

Table 21.2.1 (d)

Pedestal bearing strength

 = φ_c k 0.85 f_c A₁

 = 1543.8

[kips]

Table 14.5.6.1

 = 0.35

 > P_u

OK

Factored forces on base plate

 = 545.00

[kips]

 = 0.00

[kip-ft]

Column sect W18X86

 = 18.400

[in]

 = 11.100

[in]

ACI 318-19

Concrete strength & strength reduction factor

 = 4.5

[ksi]

 = 0.65

Table 21.2.1 (d)

 = 

 = 0.331

3.1.2 on Page 16

 = min( , 1 )

 = 0.633

3.1.2 on Page 16

 = λ

 = 2.263

[in]

3.1.3 on Page 17

 = max( m , n , λ n' )

 = 4.060

[in]

Base plate strength & strength reduction factor

 = 50.0

[ksi]

 = 0.90

Base plate thickness

 = from user input

 = 2.000

[in]

Base plate min. thickness required

 = L [ ]0.5

 = 1.058

[in]

3.1.2 on Page 16

 = 0.53

 < t_p

OK

Thinner part joined thickness

 = 

 = 0.770

[in]

Min fillet weld size allowed

 = 

 = 0.313

[in]

AISC 15^th  Table J2.4

Fillet weld size provided

 = 

 = 0.313

[in]

 ≥ w_min

OK

Fillet weld size provided

 = 

 = 0.313

[in]

Min fillet weld length allowed

 = 4 x w

 = 1.250

[in]

AISC 15^th  J2.2b

Min fillet weld length

 = 0.5 b_fc - k_1c

 = 4.487

[in]

 ≥ L_min

OK

Thinner part joined thickness

 = 

 = 0.480

[in]

Min fillet weld size allowed

 = 

 = 0.188

[in]

AISC 15^th  Table J2.4

Fillet weld size provided

 = 

 = 0.313

[in]

 ≥ w_min

OK

Fillet weld size provided

 = 

 = 0.313

[in]

Min fillet weld length allowed

 = 4 x w

 = 1.250

[in]

AISC 15^th  J2.2b

Min fillet weld length

 = d_b - 2 k_b

 = 15.150

[in]

 ≥ L_min

OK

Column section W18X86

 = 18.400

[in]

 = 11.100

[in]

 = 0.770

[in]

 = 1.063

[in]

Forces on W shape flange weld

 = 545.00

[kips]  (C)

 = 0.00

[kips]

Flange fillet weld length-double fillet

 = ( b_fc + b_fc - 2k_1c )/2 as double fillet

 = 10.037

[in]

Column flange to base plate fillet weld size

 = from user input

 = ⁵⁄₁₆

[in]

Weld shear stress-dbl fillet by V_x

 = V_x / 2L

 = 0.00

[kip/in]

Weld stress combined - max

 = f_v

 = 0.000

[kip/in]

AISC 15^th  Eq 8-11

Weld stress load angle

 = 

 = 0.0

[°]

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

 = 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

 = 18.56

[kip/in]

AISC 15^th  Eq 8-1

Base metal - column flange

 = 0.770

[in]

 = 65.0

[ksi]

Base metal - column flange 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

 = 30.03

[kip/in]

AISC 15^th  Eq J4-4

Double fillet linear shear strength

 = min ( R_n-w , R_n-b )

 = 18.559

[kip/in]

AISC 15^th  Eq 9-2

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq 8-1

 = 

 = 13.919

[kip/in]

 = 0.00

 > f_max

OK

The moment M_x-x , M_y-y and V_x are all taken care by W shape flange weld and the column axial force is compressive force, so W shape web to base plate weld only takes shear in strong axis direction V_y

Forces on column web weld

 = 545.00

[kips]  (C)

 = 175.00

[kips]

Column section W18X86

 = 18.400

[in]

 = 1.625

[in]

 = 0.480

[in]

Fillet weld length - double fillet

 = d_c - 2 k_c

 = 15.150

[in]

Column flange to base plate fillet weld size

 = from user input

 = ⁵⁄₁₆

[in]

Weld stress from shear force

 = V_y / L

 = 11.551

[kip/in]

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

 = 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

 = 18.56

[kip/in]

AISC 15^th  Eq 8-1

Base metal - column web

 = 0.480

[in]

 = 65.0

[ksi]

Base metal - column web 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

 = 18.72

[kip/in]

AISC 15^th  Eq J4-4

Double fillet linear shear strength

 = min ( R_n-w , R_n-b )

 = 18.559

[kip/in]

AISC 15^th  Eq 9-2

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq 8-1

 = 

 = 13.919

[kip/in]

 = 0.83

 > f_v

OK

Suggested minimum base plate thickness for rigidity

 = max ( m/4 , n/4 )

 = 1.015

[in]

 = 0.51

 < t_p

OK

Factored tension force

 = from user input

 = 85.00

[kips]

No of anchors in tension

 = anchor bolt pattern - 4A

 = 4

Anchor rod effective section area

 = 1.41

[in²]

 = 125.0

[ksi]

Strength reduction factor

 = 0.75

ACI 318-19 17.5.3(a)

Anchor rod tensile resistance

 = φ_ts n_t A_se f_uta

 = 528.75

[kips]

ACI 318-19 17.6.1.2

 = 0.16

 > T_u

OK

Base plate thickness

 = from user input

 = 2.000

[in]

Bending to Column Web

Moment lever arm

 = f₁ - 0.5 t_w

 = 4.760

[in]

Example 4.5 on Page 34

Moment to column flange

 = T_b x a

 = 8.43

[kip-ft]

Effective plate width

 = 2 x a

 = 9.520

[in]

Example 4.5 on Page 35

Base plate strength & strength reduction factor

 = 50.0

[ksi]

 = 0.90

Base plate required thickness

 = ( )0.5

 = 0.972

[in]

Example 4.5 on Page 35

 = 0.49

 < t_p

OK

Thinner part joined thickness

 = 

 = 0.770

[in]

Min fillet weld size allowed

 = 

 = 0.313

[in]

AISC 15^th  Table J2.4

Fillet weld size provided

 = 

 = 0.313

[in]

 ≥ w_min

OK

Fillet weld size provided

 = 

 = 0.313

[in]

Min fillet weld length allowed

 = 4 x w

 = 1.250

[in]

AISC 15^th  J2.2b

Min fillet weld length

 = 0.5 b_fc - k_1c

 = 4.487

[in]

 ≥ L_min

OK

Thinner part joined thickness

 = 

 = 0.480

[in]

Min fillet weld size allowed

 = 

 = 0.188

[in]

AISC 15^th  Table J2.4

Fillet weld size provided

 = 

 = 0.313

[in]

 ≥ w_min

OK

Fillet weld size provided

 = 

 = 0.313

[in]

Min fillet weld length allowed

 = 4 x w

 = 1.250

[in]

AISC 15^th  J2.2b

Min fillet weld length

 = d_b - 2 k_b

 = 15.150

[in]

 ≥ L_min

OK

Column section W18X86

 = 18.400

[in]

 = 11.100

[in]

 = 0.770

[in]

 = 1.063

[in]

Forces on W shape flange weld

 = -85.00

[kips]  (T)

 = 0.00

[kips]

Flange fillet weld length-double fillet

 = ( b_fc + b_fc - 2k_1c )/2 as double fillet

 = 10.037

[in]

Column flange to base plate fillet weld size

 = from user input

 = ⁵⁄₁₆

[in]

W shape and flange area

 = 25.30

[in²]

 = 8.55

[in²]

Tensile axial force taken by w shape one side flange

 = P_t A_f / A

 = -28.72

[kips]

Weld tensile stress by P_ft

 = P_ft / L

 = -2.86

[kip/in]

Weld shear stress-dbl fillet by V_x

 = V_x / 2L

 = 0.00

[kip/in]

Weld stress combined - max

 = ( f2t + f2v )0.5

 = 2.861

[kip/in]

AISC 15^th  Eq 8-11

Weld stress load angle

 = tan^-1 ( )

 = 90.0

[°]

Fillet weld leg size

 = ⁵⁄₁₆

[in]

 = 90.0

[°]

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 )

 = 1.50

AISC 15^th  Page 8-9

Fillet weld shear strength

 = 0.6 (C₁ x 70 ksi) 0.707 w n C₂

 = 27.84

[kip/in]

AISC 15^th  Eq 8-1

Base metal - column flange

 = 0.770

[in]

 = 65.0

[ksi]

Base metal - column flange is in tension, tensile rupture as per AISC 15^th  Eq J4-2 is checked

AISC 15^th  J2.4

Base metal tensile rupture

 = F_u t

 = 50.05

[kip/in]

AISC 15^th  Eq J4-2

Double fillet linear shear strength

 = min ( R_n-w , R_n-b )

 = 27.838

[kip/in]

AISC 15^th  Eq 9-2

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq 8-1

 = 

 = 20.879

[kip/in]

 = 0.14

 > f_max

OK

The moment M_x-x , M_y-y and V_x are all taken care by W shape flange weld , so W shape web to base plate weld only takes shear in strong axis direction V_y and part of axial tensile load P_w

Forces on column web weld

 = 85.00

[kips]  (T)

 = 175.00

[kips]

Column section W18X86

 = 18.400

[in]

 = 1.625

[in]

 = 0.480

[in]

Fillet weld length - double fillet

 = d_c - 2 k_c

 = 15.150

[in]

W shape and web area

 = 25.30

[in²]

 = 7.27

[in²]

Tensile axial force taken by w shape web

 = P_t A_w / A

 = 24.43

[kips]

Column web to base plate fillet weld size

 = from user input

 = ⁵⁄₁₆

[in]

Weld tensile stress from axial load

 = P_w / L

 = 1.61

[kip/in]

Weld shear stress from shear load

 = V_y / L

 = 11.55

[kip/in]

Weld stress combined - max

 = ( f2a + f2v )0.5

 = 11.663

[kip/in]

AISC 15^th  Eq 8-11

Weld stress load angle

 = tan^-1 ( )

 = 7.9

[°]

Fillet weld leg size

 = ⁵⁄₁₆

[in]

 = 7.9

[°]

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 )

 = 1.03

AISC 15^th  Page 8-9

Fillet weld shear strength

 = 0.6 (C₁ x 70 ksi) 0.707 w n C₂

 = 19.04

[kip/in]

AISC 15^th  Eq 8-1

Base metal - column web

 = 0.480

[in]

 = 65.0

[ksi]

Base metal - column web is in tension, tensile rupture as per AISC 15^th  Eq J4-2 is checked

AISC 15^th  J2.4

Base metal tensile rupture

 = F_u t

 = 31.20

[kip/in]

AISC 15^th  Eq J4-2

Double fillet linear shear strength

 = min ( R_n-w , R_n-b )

 = 19.036

[kip/in]

AISC 15^th  Eq 9-2

Resistance factor-LRFD

 = 0.75

AISC 15^th  Eq 8-1

 = 

 = 14.277

[kip/in]

 = 0.82

 > f_max

OK