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TENSION MEMBER Design

Teaching Modules for Steel Instruction. TENSION MEMBER Design. Developed by Scott Civjan University of Massachusetts, Amherst. P. TENSION MEMBER: Structural member subjected to tensile axial load. P. Strength design requirements: P u   P n ( P a  P n / Ω )ASD

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TENSION MEMBER Design

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  1. Teaching Modules for Steel Instruction TENSION MEMBER Design Developed by Scott Civjan University of Massachusetts, Amherst

  2. P TENSION MEMBER: Structural member subjected to tensile axial load. P Tension Module

  3. Strength design requirements: Pu  Pn (Pa  Pn/Ω)ASD Where  varies depending on failure mode. Tension Module

  4. Tensile Strength Strength Limit States: Yielding on Gross Area Rupture on Net Area Block Shear Bearing or Tear-out at Bolts Tension Module

  5. Yielding on Gross Area, Ag Tension Theory

  6. Yield on Gross Area When a member is loaded the strength is limited by the yielding of the entire cross section. P=FyA L0 eyL0 D Tension Theory

  7. Yield on Gross Area When a member is loaded the strength is limited by the yielding of the entire cross section. P P=FyA D L P eyL0 D Tension Theory

  8. Yield on Gross Area However, consider how this is affected by the stress-strain conditions. Consider L0=100 inch long tension member. Tension Theory

  9. Yield on Gross Area Fu Stress Esh Fy E eu .1 to .2 ey .001 to .002 esh .01 to .03 eu .1 to .2 er .2 to .3 Strain Tension Theory 9

  10. Yield on Gross Area Fu Stress Esh Fy Δy = 0.0015(100) = 0.15” E eu .1 to .2 ey .001 to .002 esh .01 to .03 eu .1 to .2 er .2 to .3 Strain Tension Theory 10

  11. Yield on Gross Area Fu Stress Esh Fy Δsh = 0.02(100) = 2” Δy = 0.0015(100) = 0.15” E eu .1 to .2 ey .001 to .002 esh .01 to .03 eu .1 to .2 er .2 to .3 Strain Tension Theory 11

  12. Yield on Gross Area Δu = 0.15(100) = 15” Fu Stress Esh Fy Δsh = 0.02(100) = 2” Δy = 0.0015(100) = 0.15” E eu .1 to .2 ey .001 to .002 esh .01 to .03 eu .1 to .2 er .2 to .3 Strain Tension Theory 12

  13. Yield on Gross Area Consider L0 = 100 inch long tension member. ΔYield = approx. 0.00172(100) = 0.172” ΔOnset of Strain Hardening = approx. 0.02(100) = 2” ΔPeak Load = approx. 0.15(100) = 15” Excessive deformations defines “Failure” for tension member yielding. Limit to FyAg. Tension Theory

  14. Rupture on Effective Net Area, Ae Tension Theory

  15. Rupture on Effective Net Area If holes are included in the cross section less area resists the tension force. Bolt holes are larger than the bolt diameter. In addition, processes of punching holes can damage the steel around the perimeter. Tension Theory

  16. Rupture on Effective Net Area Design typically uses average stress values. This assumption relies on the inherent ductility of steel. Pn Tension Theory

  17. Rupture on Effective Net Area Design typically uses average stress values. This assumption relies on the inherent ductility of steel. Initial stresses will typically include stress concentrations due to higher strains at these locations. Pn Tension Theory 17

  18. Rupture on Effective Net Area Design typically uses average stress values. This assumption relies on the inherent ductility of steel. Highest strain locations yield, then elongate along plastic plateau while adjacent stresses increase with additional strain. Pn Tension Theory 18

  19. Rupture on Effective Net Area Design typically uses average stress values. This assumption relies on the inherent ductility of steel. Pn Eventually at very high strains the ductility of steel results in full yielding of the cross section. Tension Theory 19

  20. Rupture on Effective Net Area Design typically uses average stress values. This assumption relies on the inherent ductility of steel. Pn Therefore average stresses are typically used in design. Tension Theory 20

  21. Rupture on Effective Net Area Similarly, bolts and surrounding material will yield prior to rupture due to the inherent ductility of steel. Therefore assume each bolt transfers equal force . Pn Tension Theory 21

  22. Rupture on Effective Net Area Shear Lag affects members where: Only a portion of the cross section is connected, Connection does not have sufficient length. Tension Theory

  23. Rupture on Effective Net Area The plate will fail in the line with the highest force (for similar number of bolts in each line). Each bolt line shown transfers 1/3 of the total force. Net area reduced by hole area Pn Cross Section 3 1 Bolt line 2 Tension Theory 23

  24. Rupture on Effective Net Area The plate will fail in the line with the highest force (for similar number of bolts in each line). Each bolt line shown transfers 1/3 of the total force. Pn Pn Net area reduced by hole area Pn Cross Section 3 1 Bolt line 2 Tension Theory 24

  25. Rupture on Effective Net Area The plate will fail in the line with the highest force (for similar number of bolts in each line). Each bolt line shown transfers 1/3 of the total force. Pn/6 Pn 2/3Pn Pn/6 Net area reduced by hole area Pn Cross Section 3 1 Bolt line 2 Tension Theory 25

  26. Rupture on Effective Net Area The plate will fail in the line with the highest force (for similar number of bolts in each line). Each bolt line shown transfers 1/3 of the total force. Pn/6 Pn/6 Pn 1/3Pn Pn/6 Pn/6 Pu Pu Net area reduced by hole area Pn Cross Section 3 1 Bolt line 2 Tension Theory 26

  27. Rupture on Effective Net Area Pn/6 Pn/6 Pn/6 0 Pn/6 Pn/6 Pn/6 The plate will fail in the line with the highest force (for similar number of bolts in each line). Each bolt line shown transfers 1/3 of the total force. Pn Net area reduced by hole area Pn Cross Section 3 1 Bolt line 2 Tension Theory 27

  28. Rupture on Effective Net Area The plate will fail in the line with the highest force (for similar number of bolts in each line). Each bolt line shown transfers 1/3 of the total force. Bolt line 1 resists Pn in the plate. Bolt line 2 resists 2/3Pn in the plate. Bolt line 3 resists 1/3Pnin the plate. Force in plate Net area reduced by hole area Pn 1/3 Pn 2/3 Pn 0 Pn Cross Section 3 1 Bolt line 2 Tension Theory

  29. Rupture on Effective Net Area Consider how this is affected by the stress-strain conditions. Consider L0=1 inch diameter holes. 1 inch Pn Tension Theory

  30. Rupture on Effective Net Area Fu Stress Esh Fy E eu .1 to .2 ey .001 to .002 esh .01 to .03 eu .1 to .2 er .2 to .3 Strain Tension Theory 30

  31. Rupture on Effective Net Area Fu Stress Esh Fy = 50 ksi Δ = 0.0017(1) = 0.0017” E eu .1 to .2 ey .001 to .002 esh .01 to .03 eu .1 to .2 er .2 to .3 Strain Tension Theory 31

  32. Rupture on Effective Net Area Fu Stress Esh Fy Δ = 0.02(1) = 0.02” Δ = 0.0015(1) = 0.0015” E eu .1 to .2 ey .001 to .002 esh .01 to .03 eu .1 to .2 er .2 to .3 Strain Tension Theory 32

  33. Rupture on Effective Net Area Δu = 0.15(1) = 0.15” Fu Stress Esh Fy Δsh = 0.02(1) = 0.02” Δy = 0.0017(1) = 0.0017” E eu .1 to .2 ey .001 to .002 esh .01 to .03 eu .1 to .2 er .2 to .3 Strain Tension Theory 33

  34. Rupture on Effective Net Area Consider L0=1 inch hole diameter. ΔYield = approx. 0.00172(1) = 0.00172” ΔOnset of Strain Hardening = approx. 0.02(1) = 0.02” ΔPeak Load = approx. 0.15(1) = 0.15” Failure at net area can achieve Fu so long as ductility is available. Tension Theory 34

  35. Rupture on Effective Net Area For a plate with a typical bolt pattern the rupture plane is shown. Yield on Ag would occur along the length of the member. Both failure modes depend on cross-sectional areas. Rupture failure across section at lead bolts. Pn Yield failure (elongation) occurs along the length of the member. Tension Theory

  36. Rupture on Effective Net Area What if holes are not in a line perpendicular to the load? Need to include additional length/area of failure plane due to non-perpendicular path. Pn g s Additional strength depends on: Geometric length increase Combination of tension and shear stresses Combined effect makes a direct calculation difficult. Tension Theory

  37. Rupture on Effective Net Area Boundary of force transfer into the plate from each bolt. Pn As the force is transferred from each bolt it spreads through the tension member. This is sometimes called the “flow of forces” Note that the forces from the left 4 bolts act on the full cross section at the failure plane (bolt line nearest load application). Tension Theory

  38. Rupture on Effective Net Area Now consider a much wider plate. Pn At the rupture plane (right bolts) forces have not engaged the entire plate. Tension Theory

  39. Rupture on Effective Net Area Now consider a much wider plate. Rupture Plane Pn At rupture plane (right bolts) forces have not engaged the entire plate. Tension Theory 39

  40. Rupture on Effective Net Area Now consider a much wider plate. Rupture Plane Portion of member carrying no tension. Pn At the rupture plane (right bolts) forces have not engaged the entire plate. Tension Theory 40

  41. Rupture on Effective Net Area Now consider a much wider plate. Rupture Plane Portion of member carrying no tension. Pn Effective length of rupture plane At the rupture plane (right bolts) forces have not engaged the entire plate. Tension Theory 41

  42. Rupture on Effective Net Area This concept describes the Whitmore Section. 30o Pn lw= width of Whitmore Section 30o Tension Theory 42

  43. Rupture on Effective Net Area Shear Lag Accounts for distance required for stresses to distribute from connectors into the full cross section. Largest influence when Only a portion of the cross section is connected. Connection does not have sufficient length. Tension Theory

  44. Rupture on Effective Net Area Shear Lag Ae = Effective Net Area An = Net Area Ae ≠ AnDue to Shear Lag Tension Theory 44

  45. Rupture on Effective Net Area Pn l= Length of Connection Tension Theory 45

  46. Rupture on Effective Net Area Pn Rupture Plane l= Length of Connection Tension Theory 46

  47. Rupture on Effective Net Area Pn Distribution of Forces Through Section RupturePlane l= Length of Connection Tension Theory 47

  48. Rupture on Effective Net Area Section Carrying Tension Forces Pn Distribution of Forces Through Section RupturePlane l= Length of Connection Tension Theory 48

  49. Rupture on Effective Net Area Pn Area not Effective in Tension Due to Shear Lag Shear lag less influential when l is long, or if outstanding leg has minimal area or eccentricity Effective Net Area in Tension Tension Theory 49

  50. Block Shear Tension Theory

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