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Design Equations for Steel Tension Members

Explore the design equations for steel tension members, including gross and net areas, bolt holes, shear lag, and example calculations. Learn about LRFD and ASD approaches for determining tensile strength.

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Design Equations for Steel Tension Members

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  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

  6. 常見受拉桿件型式

  7. 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

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

  9. 拉力破壞模式 P P

  10. P 1.全斷面降伏 Fy P

  11. 1.全斷面降伏 Fy Pn=AgFy

  12. 2.淨斷面剪壞 Fu

  13. 2.淨斷面剪壞 Fu Fy Fu Pn=AeFu

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

  15. 1.全斷面降伏 Fy ASD LRFD fPn=0.9AgFy

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

  17. 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

  18. 螺栓孔徑使用之大小

  19. 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

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

  21. s g

  22. 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.

  23. 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.

  24. 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.

  25. 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.

  26. 21/2 in. 3 in. 3 in. 21/2 in. 3 in.

  27. 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

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

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

  30. 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)

  31. 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

  32. Shear Lag 剪力遲滯 • Ae = UAn T T

  33. Shear Lag 剪力遲滯 • Ae = UAn T

  34. 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

  35. Pmax = ?

  36. 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

  37. 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

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

  39. 6in. 7/8 in. 3/8 in. 6in.

  40. 100 80 80 80 100 80 50 50 50 50 20 80 80 80 80 An= Ag=

  41. 3.塊剪

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