shearlug example

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TECHNICAL CORRECTION PIP STE05121October 2006 Anchor Bolt Design Guide

Process Industry Practices Page 19 of 24

will develop friction even when the column or vertical vessel is in uplift. This

downward load can be considered in calculating frictional resistance. Care shall

be taken to assure that the downward load that produces frictional resistance

occurs simultaneously with the shear load. In resisting horizontal loads, the

friction resistance attributable to downward force from overturning moment may be used.

The frictional resistance can also be used in combination with shear lugs to resist

the factored shear load. The frictional resistance should not be used in

combination with the shear resistance of anchors unless a mechanism exists to

keep the base plate from slipping before the anchors can resist the load (such as

welding the washer to the base plate).

Note: If the design requires welding the washer to the base plate, plain

washers or steel plate (rather than hardened washers) must be

specified to ensure that a good weld can be produced.

8.2 Calculating Resisting Friction ForceThe resisting friction force, Vf , may be computed as follows:

Vf = P

P = normal compression force

= coefficient of friction

The materials used and the embedment depth of the base plate determine the

value of the coefficient of friction. (Refer to Figure F for a pictorial

representation.)

a. = 0.90 for concrete placed against as-rolled steel with the contact plane a

full plate thickness below the concrete surface. b. = 0.70 for concrete or grout placed against as-rolled steel with the

contact plate coincidental with the concrete surface.

c. = 0.55 for grouted conditions with the contact plane between grout and

as-rolled steel above the concrete surface.

9. Shear Lug Design

Normally, friction and the shear capacity of the anchors used in a foundation adequately

resist column base shear forces. In some cases, however, the engineer may find the shear

force too great and may be required to transfer the excess shear force to the foundation

by another means. If the total factored shear loads are transmitted through shear lugs orfriction, the anchor bolts need not be designed for shear.

A shear lug (a plate or pipe stub section, welded perpendicularly to the bottom of the

base plate) allows for complete transfer of the force through the shear lug, thus taking the

shear load off of the anchors. The bearing on the shear lug is applied only on the portion

of the lug adjacent to the concrete. Therefore, the engineer should disregard the portion

of the lug immersed in the top layer of grout and uniformly distribute the bearing load

through the remaining height.

Shear Lug Verification Example

PIP STE05121 Anchor Bolt Design Guide

PIP - Oct 2006


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PIP STE05121 TECHNICAL CORRECTION Anchor Bolt Design Guide October 2006

Page 20 of 24 Process Industry Practices

The shear lug should be designed for the applied shear portion not resisted by friction

between the base plate and concrete foundation. Grout must completely surround the lug

plate or pipe section and must entirely fill the slot created in the concrete. When using a

pipe section, a hole approximately 2 inches in diameter should be drilled through the

base plate into the pipe section to allow grout placement and inspection to assure that

grout is filling the entire pipe section.

9.1 Calculating Shear Load Applied to Shear Lug

The applied shear load, Vapp, used to design the shear lug should be computed as

follows:

Vapp = Vua - Vf

9.2 Design Procedure for Shear Lug Plate

Design of a shear lug plate follows (for an example calculation, see Appendix

Example 3, this Practice):

a. Calculate the required bearing area for the shear lug:

Areq = Vapp / (0.85 * * fc’) = 0.65

b. Determine the shear lug dimensions, assuming that bearing occurs only on

the portion of the lug below the grout level. Assume a value of W, the lug

width, on the basis of the known base plate size to find H, the total height

of the lug, including the grout thickness, G:

H = (Areq /W) + G

c. Calculate the factored cantilever end moment acting on a unit length of the

shear lug:

Mu = (Vapp/W) * (G + (H-G)/2)

d. With the value for the moment, the lug thickness can be found. The shear

lug should not be thicker than the base plate:

t = [(4 * Mu)/(0.9*f ya)]0.5

e. Design weld between plate section and base plate.

f. Calculate the breakout strength of the shear lug in shear. The method

shown as follows is from ACI 349-01, Appendix B, section B.11:

Vcb = AVc*4* *[f c’]0.5

where

AVc = the projected area of the failure half-truncated pyramid defined

by projecting a 45-degree plane from the bearing edges of the

shear lug to the free edge. The bearing area of the shear lug

shall be excluded from the projected area.

= concrete strength reduction factor = 0.85


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Verification Example

Javier Encinas, PE

Shear Lug Verification

Page # ___

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ASDIP Steel 3.2.5 STEEL BASE PLATE DESIGN www.asdipsoft.com

GEOMETRY&olumn Section '''''''''''''''''

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ColumnSection.................

Width Length

Column..........

Plate..............

Concrete

Support

RodOffset.....

ThicknessofGrout............

Wp1

Wp2

Lp1

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in

in

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!erticalLoadP................

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&esign'ccentricit(e.......

&esign'ccentricit()s*L+,

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kft

kip

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PlateSteelStrength/(....

PierConcreteStrengthf0c

ksi

ksi

"earingstress 22.+13.-413.-56-.1ksi

"earingstrength 6-.748.-4 63.3ksi 9C)1-.13.1

:nderstrengthfactor;6-., 9C)


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'ccentricit( -.-412+22.6-.-in *L+,613.-+,62.8in

#aBDearingstress 622.

13.-413.-

,4-.-412

[email protected]

#inDearingstress6 622.

13.-413.-

,4-.-412

[email protected]

"earingatcriticalsection -.18.2+13.-4-.1-.156-.1ksi

#omentduetoDearing

#D6-.148.2@+24-.1-.1548.2@+86-.,kin+in

Platethickness 6 6-.2Ein 9)SC&G=18.1.2

"


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Rod#aterialSpecification...... /138, :se35Rodsf(a68,.-ksifuta67.-ksi

9nchorRodSi%e.... 1Fdiam.B12.-inemD. 9se6-.,[email protected]@

9C)&.

Totaltensionforceu6-.-kip =oftensionrods6- Tensionforceperrodui6-.-kip

Steelstrengthofanchorsintension 9C)&..1

Steelstrength-.,-,47.-68.1kip 9C)'H.&25

:nderstrengthfactor;6-.E 9C)&.3.8

Steelstrengthratio6 6-.-

-.E48.16-.-- 9C)&.3.1.1*1.-O

ConcreteDreakoutstrengthofanchorsintension 9C)&..2

oReinforcing"arsProIided

'ffectiIeemDedment 1E.-+1.611.,Ein 9C)&..2.8

9nchorgrouparea

9nc61E.?,.547.-?7.-?,.56E,.-in@ 9C)&..2.1

Singleanchorarea


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Javier Encinas, PE


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ConcretesidefaceDloJoutstrengthofanchorsintension 9C)&..3

SidefaceDloJout sDg6.9.'mDed*2.Ca₁12.-*2.4,.61,.85 9C)&..3.1

Tension&esignRatio6 6 -.-- *1.-O 9C)&.3.1.1

9C)&.

ShearresistedD(ShearLug?/riction

Totalshearforce!u6 3-.-kip CompressionforceC622.kip /rictioncoeff.6 -.2-

/rictionstrength -.E422.4-.2-68.3kip

/rictionstrengthratio6 63-.-

-.E-48.361.-- *61.-O

ShearlugJidthWl612.-in Shearlugheight$l68.-in Shearlugthicknesstl6 1.-in

SteelstrengthofluginfleBure

Lugmoment 8,.,41.-?8.-1.-5+256E8.8kin

LugfleBuralstrength 12.-48,41.-@+361-7.-kin 9)SC/.11

:nderstrengthfactor;6-.1.-?24-.2-+85412.-A6.2kip

ResultantperlugJidthR6 6 618-.7kip

Weldstress -.,4E-41?-.56,8.-ksi 9)SC'H.K2.5

Weldstrength ,8.-4-.E-E4-.2-42412.-62,E.8kip

:nderstrengthfactor;6-.E 9)SCK8D.3

Weldstrengthratio6 6

18-.7

-.E42,E.8 6-., 9)SC"8.8*1.-O

ConcreteDearingstrengthofluginshear

LugDearingarea 8.-1.-5412.--623.-in@

LugDearingstrength 1.848.-423.-6


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ConcreteDreakoutstrengthofluginshear 9C)83

shearlug example

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