DESIGN OF LIFTING LUG INPUT
Thickness t= 1.25 inches
Diameter of hole d= 1.25 inches
Dimension a a= 1.125 inches
Dimension e e= 1.125 inches
Ultimate steel strength Fu= 58 ksi
Yeild strength Fy= 36 ksi
Geometric Guidelines:
Rule 1: OK
Rule 2: OK
Evaluation based on Failure Mode: Failure Mode 1:
This failure mode involves tension failure on both sides of the hole. Ultimate tensile load Pu=2.a.t.Fu 163.125 kips
Factor of safety FS= 5
Pw1=Pu/FS Pw1= 32.625 kips
Failure Mode 2:
diameter of pin dpin= 0.75 inches
Pw2=0.9.Fy.t.dpin/1.8 Pw2= 16.875 kips
Failure Mode 3:
Pw3=2x0.4.Fy.e.t/1.8 Pw3= 22.500 kips
There are some geometric guidelines to be considered as recommended in Ref 1. They will be called Rule 1 and Rule 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 1.8
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.
Rule 1: The dimension "a" must be greater than or equal to half the hole diameter, d.
Rule 2: The dimension "e" must be greater than or equal to 0.67 times the hole diameter, d
a
Failure Mode 4:
Pw4=1.67x0.67Fy.e2.t/1.8d Pw4= 28.322 kips
Failure Mode 5:
Rule 3: OK
Rule 4: OK
AISC Code Checks per Section D3.2:
Requirement 1:
A1=t.e 1.40625 in2 A2=2.a.t 2.8125 in2 Compare A1 and A2 A1>=2/3xA2
Reduced dimension 'a' aeff= 0.84375 inches
Requirement 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.
Rule 4: The thickness is greater than or equal to 0.25 times the hole diameter
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.
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).
Not satisfied- reduced 'a' dimension will be used to find tensile
capacity
Rule 3: The thickness of lug is greater than or equal to 0.5 inches 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.
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.
4xt>a
aeff 1.125 inches
Requirement 3:
d>1.25.a
Reduced dimension 'a' aeff= 1 inches
Tensile capacity based on these 3 requirements
Use minimum aeff aeff 0.84375 inches
Pw5=2.aeffx0.45.Fy.t/1.8 Pw5 18.984 kips
Weld between Lug and Base Plate:
OK
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'.
Not satisfied- reduced 'a' dimension will be used to find tensile
capacity
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:
Load W= 3367.571 lbs Length of weld along lug thickness tw= 1.25 inches
Lever arm l= 2 inches
Length of weld along lug width w= 3.5 inches
a 45 deg
b 20 deg
tan a 1
tan b 0.36397
fmax (for 3/8 inch weld) 1694 lbs/in
f1 1651.464 lbs
f2 129.0206 lbs
f3 354.4812 lbs
Resultant of f1, f2 and f3 1694 lbs
Difference resultant and fmax 6.82E-13 lbs Ref. 3
Pw6= 3.368 kips
In 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 OK
From 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) W W tan a a deg W W tan b b deg
Lug Base Material:
Load W= 8283.176 lbs
fmax=0.75.Fy/1.8 fmax= 15 kips
Lug width lw=2.a+d 3.5 inches
f1 15000 lbs
Difference between f1 and fmax 0
Pw7 8.283 kips CONCLUSION: Pw1= 32.625 kips Pw2= 16.875 kips Pw3= 22.500 kips Pw4= 28.322 kips Pw5= 18.984 kips Pw6= 3.368 kips Pw7= 8.283 kips
Capacity will be minimum of these 3.368 kips CAPACITY
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:
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.
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.
In 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 OK
If additional capacity is desired, the angles a and b can be restricted as needed to increase the capcity of the lug.
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.
