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&

LIFTING LUG DESIGN

CALCULATION REPORT

Page 1 of 6

1 A. CHECK ON LUG STRENGTH

2 (1) LUG DIMENSIONS

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11 Material :

12 SS400

13 Fy = 235 MPa

14 Fu = 400 MPa

15

16 The geometric properties are as follow :

17 e = 85 mm

18 T1 = t = 20 mm

19 D = w = 180 mm D/wnew = 210 mm

20 d = 60 mm

21

22 Check dimensional requirements using AISC 360-05 Specification Section D5.2

23 1. be= 2t + 16 mm Check dimensional req. after update D/w :

24 = 56 mm < 60 mm Controls 1. be= 2t + 16 mm

25 2. a > 1.33be= 56 mm < 75 mm Controls

26 60 mm > 74 mm NOT OK 2. a > 1.33be

27 3. w > 2be + d 75 mm > 74 mm OK

28 180 mm > 172 mm OK 3. w > 2be + d

29 4. c > a 210 mm > 172 mm OK

30 60 mm > 60 mm OK 4. c > a

31 75 mm > 75 mm OK

32 # Recommendation : Changed dimension D/w to 210 mm instead of 180 mm and changed dimension center

33 line pin hole to edge side to 105 mm instead of 90 mm.

34 (2) DESIGN LOAD

35 Max. Load W = kg

36 Impact Load Factor f = 2

37 Design Load Wd = kg

38 Number of lug used N = 4

39 Design Load per Lug Fv = kg

40 b = 30 deg1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

11000

22000

5500

Page 2 of 6

1 P = Fv/cos b = kg

2 Fh = P*sin b = kg

3

4 q = 5 deg

5H = Fv*tan q = 481 kg

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7Resultan force, R = SQRT (Fv

2 + Fh

2 + H

2)

8 R = kg

9= N

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11 (3) STRENGTH CHECK

12 a) Tensile Rupture

13 Calculate the available tensile rupture strength on the effective net area

14 Pn = Fu (2tbe) (Spec. Eq. D5-1)

15= N

16 From AISC Specification Section D5.1, the available tensile rupture strength is :

17 Pn , where Wt = 2.00 for ASD

18 Wt

19 Pn = N > N --> OK

20

21 b) Shear Rupture

22 Asf = 2t(a + d/2)

23 = mm2

24 Pn = 0.6FuAsf (Spec. Eq. D5-2)

25= N

26 From AISC Specification Section D5.1, the available shear rupture strength is :

27 Pn , where Wsf = 2.00 for ASD

28 Wsf

29 Pn = N > N --> OK

30

31 c) Bearing `

32 Apb = td

33 = mm2

34 Rn = 1.8FyApb (Spec. Eq. J7-1)

35 = N

36 From AISC Specification Section J7, the available bearing strength is :

37 Pn , where W = 2.00 for ASD

38 W

39 Pn = N > N --> OK

40

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

896000

448000 62459

3600

864000

432000

3175

6369.1

62459

6351

62459

1200

507600

253800 62459

Page 3 of 6

1

2d) Tensile Yielding

3Ag = wt

4= mm

2

5Pn = FyAg (Spec. Eq. D2-1)

6= N

7 From AISC Specification Section D2, the available tensile yielding strength is :

8 Pn , where Wt = 1.67 for ASD

9 Wt

10Pn = N > N --> OK

11

12e) Flexural Yielding

13M = Re

14= N.mm

15

16From AISC Manual Table 17-27, the geometric properties are as follow :

17Sx = tw

2

186

19= mm

3

20Zx = tw

2

214

22= mm

3

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24e-1 Calculate Nominal Flexural Strength, Mn

25Check limit from AISC Specification Section F11.1

26Lbd < 0.08E , where : Lb = 85 mm

27 t2 Fy E = MPa

2845 < 68 ,therefore, the yielding limit state applies

29

30Mn = Mp (Spec. Eq. F11-1)

31= FyZ < 1.6My

321.6My = 1.6FySx

33= N.mm

34Mp = FyZx

35= N.mm < N.mm --> OK

36So, use Mn = Mp = N.mm

37

38e-2 Lateral-Torsional Buckling (AISC Specification Section F11.2)

39As per previously calculated Lbd < 0.08E , therefore, the lateral torsional buckling limit state does

40 not apply. t2 Fy

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4200

987000

591018 62459

6E+07

5E+07 6E+07

5E+07

5309024

147000

220500

199948

Page 4 of 6

1

2From AISC Specification Section F1, the available flexural strength is :

3 Mn , where Wb = 1.67 for ASD

4 Wb

5Mn = N.mm > N.mm --> OK

6

7Conclusion : The available strength is governed by the bearing strength limit state.

8

9 (4) CHECK ON WELDMENT STRENGTH

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25Pa = N

26a. The maximum fillet weld size that can be used notation per AISC Specification Section J2.2b

27Because the thickness material more than 6 mm, the maximum filled weld size :

28wfmax = t - 2 mm (Spec. J2.2b (b) )

29 = 18 mm

30While, from AISC Specification Table J2.4, the minimum size of fillet weld, based on a material thickness

31 of 20 mm is 8 mm. (Spec. Table J2.4)

32

33 b. Length of Weld Required

34 # The nominal weld strength per mm used = 14 mm weld.

35 Rn = FnwAwe (Spec. Eq. J2-3, J2-4, J2-5)

36 = 0.6FEXXAwe , where FEXX = 483 MPa

37 = N/mm

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3E+07 5E+06

62459

2866.7

Page 5 of 6

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2 PaW , where W = 2.00 for ASD

3 Rn

4 Min. Length of weld = 44 mm, but the actual length of weld (lw) = 300 mm of weld on each side.

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6 # From AISC Specification Section J2.2b, for longtudinal filled weld used alone in end connections of flat-bar

7 tension members, the length of each fillet weld shall be not less than the perpendicular distance between them.

8 lw > w

9 300 mm > 210 mm --> OK

10

11Conclusion : Weld size use 14 mm, and length of weld use 300 mm of weld on each side.

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