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Monther Dwaikat Assistant Professor Department of Building Engineering

Monther Dwaikat Assistant Professor Department of Building Engineering An-Najah National University. 68402: Structural Design of Buildings II 61420: Design of Steel Structures 62323: Architectural Structures II. Tension Member Design. Table of Contents. Typical Tension Members

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Monther Dwaikat Assistant Professor Department of Building Engineering

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  1. Monther Dwaikat Assistant Professor Department of Building Engineering An-Najah National University 68402: Structural Design of Buildings II61420: Design of Steel Structures62323: Architectural Structures II Tension Member Design 68402

  2. Table of Contents • Typical Tension Members • Introductory Concepts • Design Strength • Effective and Net Areas • Staggered Bolted Connections • Block Shear • Design of Tension Members • Slenderness Requirements 68402

  3. Tension Members • Applications • In bridge, roof and floor trusses, bracing systems, towers, and tie rods • Consist of angles, channels, tees, plates, W or S shapes, or combinations 68402

  4. Typical Tension Members • Tension chord in a truss 68402

  5. Typical Tension Members • Cables • Ties 68402

  6. Tension Members • Commonly Used Sections: • W/H shapes • Square and Rectangular or round HSS • Tees and Double Tees • Angles and double angles • Channel sections • Cables 68402

  7. Introductory Concepts • Stress: The stress in the column cross-section can be calculated as f - assumed to be uniform over the entire cross-section. P - the magnitude of load A - the cross-sectional area normal to the load • The stress in a tension member is uniform throughout the cross-section except: • near the point of application of load • at the cross-section with holes for bolts or other discontinuities, etc. 68402

  8. Average Stress distribution Gross Area Net Area Design Strength 68402

  9. Introductory Concepts • For example, consider an 200 x 10 mm. bar connected to a gusset plate & loaded in tension as shown below in Fig. 2.1 20 mm hole diameter 200 x 10 mm plate Fig. 2.1 Example of tension member 68402

  10. Introductory Concepts • Area of bar at section a – a = 200 x 10 = 2000 mm2 • Area of bar at section b – b = (200 – 2 x 20 ) x 10 = 1600 mm2 • Therefore, by definition the reduced area of section b – b will be subjected to higher stresses • However, the reduced area & therefore the higher stresses will be localized around section b – b. • The unreduced area of the member is called its gross area = Ag • The reduced area of the member is called its net area = An 68402

  11. Steel Stress-strain Behavior • The stress-strain behavior of steel is shown below in Fig. 2.2 E = 200 GPa Fy = 248 MPa Fig. 2.2 Stress-strain behavior of steel 68402

  12. Steel Stress-strain Behavior • In Fig. 2.2: • E - the elastic modulus = 200 GPa. • Fy ­ the yield stress Fu - the ultimate stress • y­ is the yield strain u­ the ultimate strain • Deformations are caused by the strain . Fig. 2.2 indicates that the structural deflections will be small as long as the material is elastic (f < Fy) • Deformations due to the strain  will be large after the steel reaches its yield stress Fy. 68402

  13. Design Strength • We usually determine the strength “capacity” of any structural element based on possible scenarios of failure! • Possible failures of a tension member include • Yield of the element • Fracture of element • The stress of axially loaded elements can be determined as • The stress is therefore a function of the cross sectional area thus the presence of holes will change the stress. • Bolted connections reduce the area of the cross section. 68402

  14. Design Strength • A tension member can fail by reaching one of two limit states: • excessive deformation • fracture • Excessive deformation can occur due to the yielding of the gross section (for example section a-a from Fig. 2.1) along the length of the member • Fracture of the net section can occur if the stress at the net section (for example section b-b in Fig. 2.1) reaches the ultimate stress Fu. • The objective of design is to prevent these failure before reaching the ultimate loads on the structure (Obvious). • This is also the load & resistance factor design approach for designing steel structures 68402

  15. Load & Resistance Factor Design • The load & resistance factor design approach is recommended by AISC for designing steel structures. It can be understood as follows: • Step I. Determine the ultimate loads acting on the structure • The values of D, L, W, etc. are nominal loads (not maximum or ultimate) • During its design life, a structure can be subjected to some maximum or ultimate loads caused by combinations of D, L, or W loading. • The ultimate load on the structure can be calculated using factored load combinations. The most relevant of these load combinations are given below: 68402

  16. Load & Resistance Factor Design • 1.4 D • 1.2 D + 1.6 L + 0.5 (Lr or S) • 1.2 D + 1.6 (Lr or S) + (0.5 L or 0.8 W) • 1.2 D + 1.6 W + 0.5 L + 0.5 (Lr or S) • 0.9 D + 1.6 W • Step II. Conduct linear elastic structural analysis • Determine the design forces (Pu­, Vu, & Mu) for each structural member 68402

  17. Load & Resistance Factor Design • Step III. Design the members • The failure (design) strength of the designed member must be greater than the corresponding design forces calculated in Step II: Rn - the calculated failure strength of the member  - the resistance factor used to account for the reliability of the material behavior & equations for Rn Qi - the nominal load i - the load factor used to account for the variability in loading & to estimate the ultimate loading 68402

  18. Design Strength of Tension Members • Yielding of the gross section will occur when the stress f reaches Fy. • Therefore, nominal yield strength = Pn = Ag Fy • Factored yield strength = t Pn t = 0.9 for tension yielding limit state 68402

  19. Design Strength of Tension Members • Facture of the net section will occur after the stress on the net section area reaches the ultimate stress Fu • Therefore, nominal fracture strength = Pn = Ae Fu • Where, Ae is the effective net area, which may be equal to the net area or smaller. • The topic of Ae will be addressed later. • Factored fracture strength = t Ae Fu where: t = 0.75 for tension fracture limit state 68402

  20. t dbolt dhole Net Area • We calculate the net area by deducting the width of the “bolts + some tolerance around the bolt” • Use a tolerance of 1.6 mm above the diameter hole which is typically 1.6 mm larger than the bolt diameter • Rule b 68402

  21. Design Strength • Tensile strength of a section is governed by two limit states: • Yield of gross area (excessive deformation) • Fracture of net area • Thus the design strength is one of the following Load Effect YIELD FRACTURE • The difference in the f factor for the two limit states represent the • Seriousness of the fracture limit state • The reliability index (probability of failure) assumed with each limit state 68402

  22. Important Notes • Why is fracture (& not yielding) the relevant limit state at the net section? Yielding will occur first in the net section. However, the deformations induced by yielding will be localized around the net section. These localized deformations will not cause excessive deformations in the complete tension member. Hence, yielding at the net section will not be a failure limit state. • Why is the resistance factor (t) smaller for fracture than for yielding? The smaller resistance factor for fracture (t = 0.75 as compared to t = 0.90 for yielding) reflects the more serious nature & consequences of reaching the fracture limit state. 68402

  23. Important Notes • What is the design strength of the tension member? The design strength of the tension member will be the lesser value of the strength for the two limit states (gross section yielding & net section fracture). • Where are the Fy & Fu values for different steel materials? The yield & ultimate stress values for different steel materials are dependent on type of steel. 68402

  24. Ex. 2.1 – Tensile Strength • A 125 x 10 mm bar of A572 (Fy = 344 MPa) steel is used as a tension member. It is connected to a gusset plate with six 20 mm. diameter bolts as shown below. Assume that the effective net area Ae equals the actual net area An & compute the tensile design strength of the member. 20 mm hole diameter 200 x 10 mm plate A572 (Fy = 344 MPa) 68402

  25. Ex. 2.1 – Tensile Strength • Gross section area = Ag = 125 x 10 = 1250 mm2 • Net section area (An) • Bolt diameter = db = 20 mm. • Nominal hole diameter = dh = 20 + 1.6 = 21.6 mm • Hole diameter for calculating net area = 21.6 + 1.6 = 23.2 mm • Net section area = An = (125 – 2 x (23.2)) x 10 = 786 mm2 • Gross yielding design strength = t Pn = t Fy Ag • Gross yielding design strength = 0.9 x 344 x 1250/1000 = 387 kN 68402

  26. Ex. 2.1 – Tensile Strength • Fracture design strength = t Pn = t Fu Ae • Assume Ae = An (only for this problem) • Fracture design strength = 0.75 x 448 x 786/1000 = 264 kN • Design strength of the member in tension = smaller of 264 kN & 387 kN • Therefore, design strength = 264 kN (net section fracture controls). 68402

  27. Effective Net Area • The connection has a significant influence on the performance of a tension member. A connection almost always weakens the member & a measure of its influence is called joint efficiency. • Joint efficiency is a function of: • material ductility • fastener spacing • stress concentration at holes • fabrication procedure • shear lag. • All factors contribute to reducing the effectiveness but shear lag is the most important. 68402

  28. Effective Net Area • Shear lag occurs when the tension force is not transferred simultaneously to all elements of the cross-section. This will occur when some elements of the cross-section are not connected. • For example, see the figure below, where only one leg of an angle is bolted to the gusset plate. 68402

  29. Effective Net Area • A consequence of this partial connection is that the connected element becomes overloaded & the unconnected part is not fully stressed. • Lengthening the connection region will reduce this effect • Research indicates that shear lag can be accounted for by using a reduced or effective net area Ae • Shear lag affects both bolted & welded connections. Therefore, the effective net area concept applied to both types of connections. • For bolted connection, the effective net area is Ae = U An • For welded connection, the effective net area is Ae = U Ag 68402

  30. Effective Net Area • The way the tension member is connected affects its efficiency because of the “Shear Lag” phenomenon • Shear lag occurs when the force is transmitted to the section through part of the section (not the whole section) • To account for this stress concentration in stress, an area reduction factor “U” is used Bolted Connections Welded Connections Over stressed Under stressed 68402

  31. Effective Net Area • Where, the reduction factor U is given by: U = 1- ≤ 0.9 (4.7) - the distance from the centroid of the connected area to the plane of the connection L - the length of the connection. • If the member has two symmetrically located planes of connection, is measured from the centroid of the nearest one – half of the area. 68402

  32. Bolted Connections Welded Connections Effective Net Area • Increasing the connection length reduces the shear lag effect • Some special cases govern bolted and welded connections 68402

  33. Effective Net Area • The distance L is defined as the length of the connection in the direction of load. • For bolted connections, L is measured from the center of the bolt at one end to the center of the bolt at the other end. • For welded connections, it is measured from one end of the connection to other. • If there are weld segments of different length in the direction of load, L is the length of the longest segment. 68402

  34. U for Bolted Connections OR • Two major groups of bolted connections • Connections with at least three bolts per line • W,M and S shapes and T cut from them connected in flange with • All other shapes • Connections with only two bolts per line 68402

  35. Ex. 2.2 – Design Strength • Determine the effective net area & the corresponding design strength for the single angle tension member in the figure below. The tension member is an L 4 x 4 x 3/8 made from A36 steel. It is connected to a gusset plate with 15 mm diameter bolts, as shown in Figure below. The spacing between the bolts is 75 mm center-to-center. L 4 x 4 x 3/8 db = 15 mm 68402

  36. Ex. 2.2 – Design Strength • Gross area of angle = Ag = 1850 mm2 T = 9.5 mm • Net section area = An • Bolt diameter = 15 mm. • Hole diameter for calculating net area = 15 +3.2 = 18.2 mm. • Net section area = Ag – 18.2 x 9.5 = 1850 –172.9 = 1677.1 mm2 • is the distance from the centroid of the area connected to the plane of connection • For this case is equal to the distance of centroid of the angle from the edge. • This value is given in the section property table. • = 28.7 mm. 68402

  37. Ex. 2.2 – Design Strength • L is the length of the connection, which for this case will be equal to 2 x 75 = 150 mm. • Effective net area = Ae = 0.809 x 1677.1 in2 = 1357 mm2 • Gross yielding design strength = t Ag Fy = 0.9 x 1850 x 248/1000 = 412.9 kN • Net section fracture = t Ae Fu = 0.75 x 1357 x 400/1000 = 407.1 kN • Design strength = 407.1 kN (net section fracture governs) • (Lower of the two values) 68402

  38. Ex. 2.3 – Design Strength • Determine the design strength of an ASTM A992 W8 x 24 with four lines if 20 mm diameter bolts in standard holes, two per flange, as shown in the Figure below. Assume the holes are located at the member end & the connection length is 225 mm. Also calculate at what length this tension member would cease to satisfy the slenderness limitation in LRFD specification. db = 20 mm 75 mm 75 mm 75 mm 68402

  39. Ex. 2.3 – Design Strength • For ASTM A992 material: Fy = 344 MPa; & Fu = 448 MPa • For the W8 x 24 section: • Ag = 4570 mm2 d = 201 mm. • tw = 6.2 mm. bf = 165 mm. • tf = 10.2 mm. ry = 40.9 mm. • Gross yielding design strength = t Pn = t Ag Fy = 0.90 x 4570 x 344/1000 = 1414.9 kN • Net section fracture strength = t Pn = t Ae Fu = 0.75 x Ae x 448 • Ae = U An - for bolted connection • An = Ag – (no. of holes) x (diameter of hole) x (thickness of flange) • An = 4570 – 4 x (20+3.2) x 10.2. • An = 3623 mm2 68402

  40. Ex. 2.3 – Design Strength • What is for this situation? • is the distance from the edge of the flange to the centroid of the half (T) section 68402

  41. Special Cases for Welded Connections • If some elements of the cross-section are not connected, then Ae will be less than An • For a rectangular bar or plate Ae will be equal to An • However, if the connection is by longitudinal welds at the ends as shown in the figure below, then Ae = UAg • Where, U = 1.0 for L ≥ 2w U = 0.87 for 1.5 w ≤ L < 2 w U = 0.75 for w ≤ L < 1.5 w • L = length of the pair of welds ≥ w • w = distance between the welds or width of plate/bar 68402

  42. Ex. 2.3 – Design Strength • The calculated value is not accurate due to the deviations in the geometry • But, U ≤ 0.90. Therefore, assume U = 0.90 68402

  43. Ex. 2.3 – Design Strength • Net section fracture strength = t Ae Fu = 0.75 x 0.9 x 3623 x 448/1000 = 1095.6 kN • The design strength of the member is controlled by net section fracture = 1095.6 kN • According to LRFD specification, the maximum unsupported length of the member is limited to 300 ry = 300 x 40.9 = 12270 mm = 12.27 m. 68402

  44. Special Cases for Welded Connections 68402

  45. Special Cases for Welded Connections • For any member connected by transverse welds alone, Ae = area of the connected element of the cross-section 68402

  46. W L U for Welded Connections OR • Two major groups of welded connections • General case • W,M and S shapes and T cut from them connected in flange with • All other shapes • Special case for plates welded at their ends • Any member with transverse welds all around ONLY 68402

  47. Ex. 2.4 – Tension Design Strength • Consider the welded single angle L 6x 6 x ½ tension member made from A36 steel shown below. Calculate the tension design strength. 42.4 mm 140 mm 68402

  48. Ex. 2.4 – Tension Design Strength • Ag = 3720 mm2 • An = 3720 mm2 - because it is a welded connection • Ae = U An • = 42.4 mm for this welded connection • L = 152 mm for this welded connection • Gross yielding design strength = tFy Ag = 0.9 x 248 x 3720/1000 = 830 kN • Net section fracture strength = tFu Ae = 0.75 x 400 x 0.72 x 3720/1000 = 803 kN • Design strength = 803 kN (net section fracture governs) 68402

  49. Design of Tension Members • The design of a tension member involves finding the lightest steel section (angle, wide-flange, or channel section) with design strength (Pn) greater than or equal to the maximum factored design tension load (Pu) acting on it. •  Pn ≥ Pu • Pu is determined by structural analysis for factored load combinations •  Pnis the design strength based on the gross section yielding, net section fracture & block shear rupture limit states. 68402

  50. Design of Tension Members • For net section fracture limit state, Pn = 0.75 x Ae x Fu • Therefore, 0.75 x Ae x Fu ≥ Pu • Therefore, Ae ≥ • But, Ae = U An • U & An - depend on the end connection. • Thus, designing the tension member goes hand-in-hand with designing the end connection, which we have not covered so far. 68402

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