Sheet1DESIGN OF LIFTING LUGINPUTThicknesst=1.25inchesDiameter of
holed=1.25inchesDimension aa=1.125inchesDimension
ee=1.125inchesUltimate steel strengthFu=58ksiYeild
strengthFy=36ksiGeometric Guidelines:There are some geometric
guidelines to be considered as recommended in Ref 1. They will be
called Rule 1 and Rule 2.Rule 1: The dimension "a" must be greater
than or equal to half the hole diameter, d.Rule 1:OK0Rule 2: The
dimension "e" must be greater than or equal to 0.67 times the hole
diameter, dRule 2:OK0Evaluation based on Failure Mode:Failure Mode
1:This failure mode involves tension failure on both sides of the
hole.Ultimate tensile loadPu=2.a.t.Fu163.125kipsFactor of
safetyFS=5Pw1=Pu/FSPw1=32.625kipsFailure Mode 2:This Failure mode
involves bearing failure at the pin/lifting lug interface. Often
the pin diameter is much less than the hole diameter. Let us assume
a pin diameter 1/2" less than the hole diameter. Using a bearing
stress of 0.9Fy, and a "factor" of 1.8diameter of
pindpin=0.75inchesPw2=0.9.Fy.t.dpin/1.8Pw2=16.875kipsFailure Mode
3:This Failure mode involves shear failure as the pin tries to push
out a block of steel through the edge of the lug plate. The shear
area is twice the cross-sectional area beyond the hole for the
pin.Pw3=2x0.4.Fy.e.t/1.8Pw3=22.500kipsFailure Mode 4:This failure
mode involves tensile failure as the pin tries to push out of a
block of steel through the edge of the lug plate. Assume a block of
steel 0.8d in length.Pw4=1.67x0.67Fy.e2.t/1.8dPw4=28.322kipsFailure
Mode 5:This failure mode involves the out-of-plane buckling failure
of the lug. Per Ref. 1, this failure is prevented by ensuring a
minimum thickness of lug of 0.5 inches and 0.25 times the hloe
diameter d. These are refered to as Rule 3 and Rule 4 here.Rule 3:
The thickness of lug is greater than or equal to 0.5 inchesRule
3:OK0Rule 4: The thickness is greater than or equal to 0.25 times
the hole diameterRule 4:OK0AISC Code Checks per Section D3.2:The
D3.2 section of AISC code has three separate geometry checks that
can be applied to the lifting lug. If these requirements are not
met, a smaller value for "a" should be used for the calculation of
tensile capacity.Requirement 1:This requirement states that the
minimum net area beyond the pin hole, parallel to the axis of the
member (A1), shall not be less than 2/3 of the net area across the
pin hole (A2).A1=t.e1.40625in2A2=2.a.t2.8125in2Compare A1 and
A2A1>=2/3xA2Not satisfied- reduced 'a' dimension will be used to
find tensile capacityReduced dimension
'a'aeff=0.84375inchesRequirement 2:This requirement states that the
distance transverse to the axis of a pin-connected plate from the
edge of the pin hole to the edge of the member, that is dimension
'a' shall not exceed 4 times the thickness at the pin
hole.4xt>aOKaeff1.125inchesRequirement 3:This requirement states
that the diameter of the pin hole shall not be less than 1.25 times
distance from the edge of pin hole to the edge of plate
'a'.d>1.25.aNot satisfied- reduced 'a' dimension will be used to
find tensile capacityReduced dimension 'a'aeff=1inchesTensile
capacity based on these 3 requirementsUse minimum
aeffaeff0.84375inchesPw5=2.aeffx0.45.Fy.t/1.8Pw518.984kipsWeld
between Lug and Base Plate:This is typically the weakest link in
the overhead lifting lug, due to off-set loading. In general, the
lug is rarely directly over the item to be rigged. Conservatively,
let us assume that the off-set is a maximum of 45 degrees in the
plane of the lug and 20 degress normal to the plane of the lug. The
additional loads due to off-set can be determined by statics to be
as follows:LoadW=3367.5710704021lbsLength of weld along lug
thicknesstw=1.25inchesLever arml=2inchesLength of weld along lug
widthw=3.5inchesa45degb20degtan a1tan b0.3639702343fmax (for 3/8
inch weld)1694lbs/inFrom Table 3.24 of Steel Handbook, for 3/8
inche weld factored shear resistance is 5710 lbs/in. Divided by a
factor of safety of 1.8 we get 1694 lbs/in. (Ref.
4)f11651.4638325082lbsf2129.0205927792lbsf3354.4811653055lbsResultant
of f1, f2 and f31694lbsDifference resultant and fmax0lbsRef. 3In
order to find Pw6, the difference between the resultant and fmax
should be zero. To get this, go to Tools menu and click on Goal
Seek. You will get the following window. Fill in as shown below and
click OKPw6=3.368kipsLug Base Material:The analysis is similar to
the weld above except that there is no interaction between tension
and shear. The capacity is based on the maximum tensile stress at
the base of the
lug.LoadW=8283.1760177168lbsfmax=0.75.Fy/1.8fmax=15kipsLug
widthlw=2.a+d3.5inchesf115000lbsDifference between f1 and fmax0In
order to find Pw7, the difference between f1 and fmax should be
zero. To get this, go to Tools menu and click on Goal Seek. You
will get the following window. Fill in as shown below and click
OKPw78.283kipsCONCLUSION:Pw1=32.625kipsPw2=16.875kipsPw3=22.500kipsPw4=28.322kipsPw5=18.984kipsPw6=3.368kipsPw7=8.283kipsCapacity
will be minimum of these3.368kipsCAPACITYIf additional capacity is
desired, the angles a and b can be restricted as needed to increase
the capcity of the lug.References:1. David T. Ricker, "Design and
Construction of Lifting Beams", Engineering Journal, 4th Quarter,
1991.2. AISC Manual of steel Construction (ASD), 9th edition,
1989.3. Omer Blodgett, "Design of Welded Structures", 1966.4. CISC
Handbook of Steel Construction, 1997.Notes:1. As discussed in Ref.
1, using a factor of 1.8 on AISC allowables results in a factor of
safety of 5 for A36 steel. This is in line with ASME B30.20 which
required a design factor of 3 on yield strength and ANSI N14.6
which requires a design factor of 3 on yield strength and 5 on
ultimate strength. This is also in line with the load ratings for
other components of the lifting assembly such as slings, shackles,
etc.
&L&"Arial,Bold"By:- Syed Jaffry Aco
Milic&C&"Arial,Bold"SNC-LAVALIN&R&P of 6WW tan aa
degWW tan bb degae
Sheet2
Sheet3
