Lifting Lug Design

By:- Syed Jaffry Aco Milic

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DESIGN OF LIFTING LUG INPUT Thickness Diameter of hole Dimension a Dimension e Ultimate steel strength

t= d= a= e= Fu = Fy=

Yeild strength

1.25 1.25 1.125 1.125 58

inches inches inches inches ksi

36

ksi

e

a Geometric 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:

OK

Rule 2: The dimension "e" must be greater than or equal to 0.67 times the hole diameter, d 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 Factor of safety Pw1=Pu/FS

Pu=2.a.t.Fu FS= Pw1=

163.125 kips 5 32.625 kips

Failure 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 1.8 diameter of pin

dpin=

Pw2=0.9.Fy.t.dpin/1.8

Pw2=

0.75 inches 16.875 kips

Failure 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/1.8

Pw3=

22.500 kips

By:- Syed Jaffry Aco Milic

SNC-LAVALIN

Failure 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/1.8d

Pw4=

28.322 kips

Failure 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 inches

Rule 3:

OK

Rule 4: The thickness is greater than or equal to 0.25 times the hole diameter Rule 4:

OK

AISC 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.e A2=2.a.t Compare A1 and A2

A1>=2/3xA2

Reduced dimension 'a'

aeff=

1.40625 in2 2.8125 in2 Not satisfiedreduced 'a' dimension will be used to find tensile capacity 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.

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By:- Syed Jaffry Aco Milic

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4xt>a

aeff

OK

1.125 inches

Requirement 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.a

Reduced dimension 'a'

aeff=

Not satisfiedreduced 'a' dimension will be used to find tensile capacity

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: 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:

By:- Syed Jaffry Aco Milic

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SNC-LAVALIN

W tan b

W tan a b deg a deg

W

W

Load Length of weld along lug thickness Lever arm Length of weld along lug width

W= tw= l= w=

a b tan a tan b fmax (for 3/8 inch weld)

3367.571 1.25 2 3.5

lbs inches inches inches

45 deg 20 deg 1 0.36397 1694 lbs/in

f1

1651.464 lbs

f2

129.0206 lbs

f3

354.4812 lbs

Resultant of f1, f2 and f3

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)

1694 lbs

Difference resultant and fmax

6.82E-13 lbs

Ref. 3

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

Pw6=

3.368 kips

By:- Syed Jaffry Aco Milic

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Lug 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. Load fmax=0.75.Fy/1.8

W= fmax=

Lug width

lw=2.a+d

8283.176 lbs 15 kips

f1

3.5 inches 15000 lbs

Difference between f1 and fmax

0

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

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= Pw7=

3.368 kips

Capacity will be minimum of these

8.283 kips 3.368 kips

CAPACITY

If 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.

By:- Syed Jaffry 6 of 6 SNC-LAVALIN Aco Milic 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.

By:- Syed Jaffry Aco Milic

e

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