Design Equations for Steel Tension Members

1. 第二章 拉力桿件

2. 拉力桿件

3. 拉力桿件

4. 拉力桿件

5. 拉力桿件

9. 常見受拉桿件型式

10. 應力-應變曲線 破壞 應力 Fu 極限強度 Fy 降伏強度 破壞 應變硬化 塑性 頸縮 彈性 應變 10

11. 拉力桿件 What is the maximum P? P P LRFD Equation ASD Equation

12. 拉力破壞模式 P P

13. P 1.全斷面降伏 Fy P

14. 1.全斷面降伏 Fy Pn=AgFy

15. 2.淨斷面剪壞 Fu

16. 2.淨斷面剪壞 Fu Fy Fu Pn=AeFu

17. Design of Steel Tension Members Equations for strength of tension members: • For yielding in the gross section: • For fracture in the net section:

18. 1.全斷面降伏 Fy ASD LRFD fPn=0.9AgFy

19. 2.淨斷面剪壞 Ae= An≤ 0.85Ag Fu Fy Fu ASD LRFD fPn=0.75AeFu

20. Net Area(淨面積) • The net area, An, of a member is the sum of the products of the thickness and the net width of each element computed as follows: In computing net area for tension and shear, the width of a bolt hole shall be taken as 1/ 16 in. (2 mm) greater than the nominal dimensionof the hole. 1/16” 1/16” 孔徑放大 鑽孔損失 D

21. 螺栓孔徑使用之大小

22. Determine the net area of the 3/8 × 8-in. plates shown below. The plate is connected at its end with two lines of ¾-in. bolts. 8 in

23. Net Section for Staggered Bolt Holes Recall definition of Net Area, LRFD p. 16.1-10

24. s g

25. Determine the critical net area of the 1/2 -in. thick plates shown below. Using the AISC Spe.(D3.2).The holes are punched for ¾-in. bolts. 21/2 in. 3 in. 3 in. 21/2 in. 3 in.

26. 1/2 -in. thick plates The holes are punched for ¾-in. bolts. CASE 1 21/2 in. 3 in. 3 in. 21/2 in. 3 in.

27. 1/2 -in. thick plates The holes are punched for ¾-in. bolts. CASE 2 21/2 in. 3 in. 3 in. 21/2 in. 3 in.

28. 1/2 -in. thick plates The holes are punched for ¾-in. bolts. CASE 3 21/2 in. 3 in. 3 in. 21/2 in. 3 in.

29. 21/2 in. 3 in. 3 in. 21/2 in. 3 in.

30. For two lines of bolt holes shown below. Determine the pitch that will give a net area DEFG equal to the one along ABC. The holes are punched for ¾-in. bolts. A D 2in E B 2in F 2in G C s in ABC DEFG

31. Determine the net area of the W12×16(Ag=4.71 in.2). The holes are punched for 1-in. bolts.

32. Determine the net area along route ABCDEF for the C15×33.9 (Ag=10.0 in.2). The holes are for ¾-in. bolts.

33. Design Requirements • Ag– Gross cross-sectional area • Ae– Effective net area If tension load is transmitted directly to each of the cross-sectional elements by fasteners or welds: • Ae = An • An = Net cross-sectional area (gross-section minus bolt holes)

34. Design Requirements If tension load transmitted through some but not all of the cross-sectional elements: by fasteners, Ae = AnU by welds, Ae = AgU or Ae = AU

35. Shear Lag 剪力遲滯 • Ae = UAn T T

36. Shear Lag 剪力遲滯 • Ae = UAn T

37. Example of tension transmitted by some but not all of cross-section L –shape with bolts in one leg only Reduction coefficient, Where is the connection eccentricity

41. Pmax = ?

42. LRFD • Pu=  Pn= 0.9Fy Ag , for yielding in the gross section →Pu =  Pn = 0.9Fy Ag = 0.936(95/8) = 182.3 kips • Pu =  Pn = 0.75Fu Ae , for fracture in the net section →Pu =  Pn = 0.75Fu Ae = 0.75583.98 = 173.1 kips • Ae= UAn = 1.0[9 – 3(3/4+1/8)] = 3.98 in.2

43. ASD • Pu= Pn/Ω = Fy Ag /1.67 , for yielding in the gross section →Pu = Pn /Ω = Fy Ag /1.67 = 36(95/8)/1.67 =121.3 kips • Pu = Pn /Ω = Fu Ae /2 , for fracture in the net section →Pu = Pn /Ω = Fu Ae /2 = 583.98/2 = 115.42 kips • Ae= UAn = 1.0[9 – 3(3/4+1/8)] = 3.98 in.2

44. Determine the LRFD tensile design andthe ASD allowable tensile strength of the member (A572 Gr. 50).

45. 6in. 7/8 in. 3/8 in. 6in.

49. 100 80 80 80 100 80 50 50 50 50 20 80 80 80 80 An= Ag=

50. 3.塊剪