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This module covers the design requirements, strength limits, and theories related to tension members in steel structures. Topics include yielding, rupture, bearing, and tear-out, as well as stress-strain conditions and ductility considerations.
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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 Tension Module
Strength design requirements: Pu Pn (Pa Pn/Ω)ASD Where varies depending on failure mode. Tension Module
Tensile Strength Strength Limit States: Yielding on Gross Area Rupture on Net Area Block Shear Bearing or Tear-out at Bolts Tension Module
Yielding on Gross Area, Ag Tension Theory
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
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
Yield on Gross Area However, consider how this is affected by the stress-strain conditions. Consider L0=100 inch long tension member. Tension Theory
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
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
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
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
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
Rupture on Effective Net Area, Ae Tension Theory
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
Rupture on Effective Net Area Design typically uses average stress values. This assumption relies on the inherent ductility of steel. Pn Tension Theory
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
Rupture on Effective Net Area This concept describes the Whitmore Section. 30o Pn lw= width of Whitmore Section 30o Tension Theory 42
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
Rupture on Effective Net Area Shear Lag Ae = Effective Net Area An = Net Area Ae ≠ AnDue to Shear Lag Tension Theory 44
Rupture on Effective Net Area Pn l= Length of Connection Tension Theory 45
Rupture on Effective Net Area Pn Rupture Plane l= Length of Connection Tension Theory 46
Rupture on Effective Net Area Pn Distribution of Forces Through Section RupturePlane l= Length of Connection Tension Theory 47
Rupture on Effective Net Area Section Carrying Tension Forces Pn Distribution of Forces Through Section RupturePlane l= Length of Connection Tension Theory 48
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
Block Shear Tension Theory