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Design of Steel Structures (3rd Civil) New Cairo Academy 2008-2009 3. DESIGN OF TENSION MEMBERS 3.1 INTRODUCTION Tension members are encounters in most steel structures. They occur as principal structural members in bridges and roof trusses, in truss structures such as transmission and microwave towers and wind bracing systems in multi-story buildings. These members can be defines as “ Members which carry only axial tensile forces”. Some examples of these members are shown in figure (3.1) and their cross-sections are shown in figure (3.2). Figure (3.1) Samples of Steel Structures Containing Tension Members Prof. Ahmed Abdel-Salam Ismail El-Serwi 1 Design of Steel Structures (3rd Civil) New Cairo Academy 2008-2009 Figure (3.2) Samples of Cross-Sections of Tension Members 3.2 TENSION MEMBER RESISTANCE: The maximum resistance of a tension member “Tr” is calculated using the following equation: Tr = Anet . Ft (3-1) where Anet is the effective net area of the cross section and Ft is the allowable stress of the steel in tension which is defined by the Egyptian Code of Practice, ECP 2001, as: Ft = 0.58 Fy (3-2) 3.2.1 Allowable Tensile Stresses for Steel The allowable tensile strength for steel is calculated using equation (3-2) in which, Fy is the nominal yield stress of steel. Table (3-1) presents the nominal yield stress and the nominal ultimate stress for different steel grades used in structural design. Prof. Ahmed Abdel-Salam Ismail El-Serwi 2 Design of Steel Structures (3rd Civil) New Cairo Academy 2008-2009 Table (3-1) Nominal Yield Stresses and Ultimate Stresses for Different Steel Grades Nominal Values of Yield Stress Fy and Ultimate Strength Fu Thickness “t” Grade t ≤ 40 mm 40 mm ≤ t ≤ 100mm of Steel Fy Fu Fy Fu t/cm2 t/cm2 t/cm2 t/cm2 St 37 2.40 3.60 2.15 3.40 St 44 2.80 4.40 2.55 4.10 St 52 3.60 5.20 3.35 4.90 According to table (3-1) and equation (3-2) , Ft can be taken from table (3-2): Table (3-2) Allowable Tensile Stresses for Different Steel Grades Grade of Steel Ft (t/cm2) t ≤ 40 mm 40 mm ≤ t ≤ 100mm St 37 1.40 1.30 St 44 1.60 1.50 St 52 2.10 2.0 The above values for the allowable stress, Ft, may be increased by 20% if secondary stresses are considered (case of loading II). 3.2.2 Effective Net Area “Anet” Whenever tension member is to be fastened by means of bolts or rivets, holes must be provided at the connection. As a result, the member cross-sectional area at the connection is reduced and the strength of the member may also be reduced depending on the size and location of the holes. The effective area in resisting the tensile stress of the cross-section is called the effective net area Anet. The tension member resistance must be calculated based on the effective net area of the cross-section Anet which can be calculated as follows: i- For Lined up Holes: Whenever there is only one hole or multiple holes lined up ransverse to the loading direction, there is only one potential line may exist. The controlling failure line is that which passes through the bolt hole/ or holes (line AB in Figure). Prof. Ahmed Abdel-Salam Ismail El-Serwi 3 Design of Steel Structures (3rd Civil) New Cairo Academy 2008-2009 Anet = Agross – n (d+ 2 mm) t (3-3) Where n = number of bolts in the cross section perpendicular to the direction of loading. t = the thickness of plate element at which the hole lies. d = bolt dimeter It is noted that the hole diameter exceeds the bolt diameter by 2 mm. ii- For Staggered Holes: Whenever there is more thaone hole and the holes are not lined up transverse to the loading direction, more than one potential line may exist. The controlling failure line is that which gives the minimum area. In the opposite figure, there are two lines of staggered holes. The failure line may be through one hole (section A-B), or it might be along a diagonal path (section A-C). Anet (Path A-B) = Agross – (d+2mm) . t (3-4) Anet (Path A-C) = Agross – 2. (d+2mm) . t +( s2/4g)xt (3-5) The net area Anet is the minimum of the two values. 3.3 MEMBERS USED IN TRUSSES Prof. Ahmed Abdel-Salam Ismail El-Serwi 4 Design of Steel Structures (3rd Civil) New Cairo Academy 2008-2009 Unsymmetric Sections (Single Angles) Used in Vertical and diagonal members of truss Minimum radius of gyration is iv = 0.2 a Symmetric Sections (Star-Shaped) Used in Vertical members of truss at intersection with long. Bracing, and long tension members Minimum radius of gyration is iu = 0.385 a Symmetric Sections (two angles back-to-back) Used in chord members of truss and web members of high values of force. Minimum radius of gyration is ix = 0.3 a Maximum radius of gyration is iy = 0.45 a 3.4 APPLICATION ON TRUSS MEMBERS WITH BOLTED CONNECTIONS Truss members are considered symmetric sections when the center of gravity of the members coincides with the center of gravity of the connecting gusset plate. 3.4.1 Symmetric Sections: Sections composed of two angles back-to-back and two angles star shaped (where there is no eccentricity at the location of the C.G. of the member and the C.G. of the gusset plate) are considered symmetric sections. Net Area = Gross Area – Area of Bolt’s holes Anet = Agross - AHoles Where the diameter of the bolt hole = diameter of the bolt + 2 mm for drilled bolts. 3.4.2 Unsymmetric Sections (single angle) In single angle truss members where there is eccentricity between the C.G. of the member Prof. Ahmed Abdel-Salam Ismail El-Serwi 5 Design of Steel Structures (3rd Civil) New Cairo Academy 2008-2009 and the C.G. of the gusset plate at the location of the connection (the system shown in figure), the net area of the member is calculated as follows: 3.A1 Anet= A1 + A2 ( ) (3-6) 3 A1 + A2 Where A1 is the net area of the connected leg and A2 is the area of the unconnected leg. 3.4.3 Unsymmetric sections (double angles). The two angles are considered unsymmetric when they lie on one side of the gusset plate as shown in figure. To allow for the eccentricity of the connection (the system shown in figure), the net area of the member is calculated as follows: 5.A1 Anet = A1 + A2 ( ) (3-7) 5 A1 + A2 Where A1 is the net area of the connected leg and A2 is the area of the unconnected leg. Prof. Ahmed Abdel-Salam Ismail El-Serwi 6 Design of Steel Structures (3rd Civil) New Cairo Academy 2008-2009 3.5 APPLICATION ON TRUSS MEMBERS WITH WELDED CONNECTIONS 3.5.1 Symmetric Sections: Anet = Agross 3.5.2 Unsymmetric sections ( e.g. single angle ): 3.A1 Anet = A1 + A2 ( ) 3 A1 + A2 Where A1 is the net area of the connected leg and A2 is the area of the unconnected leg (there are no bolt holes to be deducted from areas A1 and A2). 3.5.3Unsymmetric Sections (Double Angles): Prof. Ahmed Abdel-Salam Ismail El-Serwi 7 Design of Steel Structures (3rd Civil) New Cairo Academy 2008-2009 3.A1 Anet = A1 + A2 ( ) 3 A1 + A2 Where A1 is the net area of the connected leg and A2 is the area of the unconnected leg (there are no bolt holes to be deducted from areas A1 and A2). 3.6 STIFFNESS REQUIREMENTS FOR TENSION MEMBERS: To avoid sag, vibration and slack of tension member the ECP, 2001defined: 3.6.1 The Maximum Slenderness Ratio The maximum slenderness ratio of a tension member (excluding wires) : 1. Tension member in λ= leff / r Buildings 300 Bridges: Roadway Bridges 180 Railway Bridges 160 Vertical hungers 300 Bracing system 200 3.6.2 The length/depth ratio (Recommended Values) Another stiffness requirements for tension members (excluding wires) is: Prof. Ahmed Abdel-Salam Ismail El-Serwi 8 Design of Steel Structures (3rd Civil) New Cairo Academy 2008-2009 Tension member in l/d Buildings 60 Roadway Bridges 35 Railway Bridges 30 3.7 DESIGN PROCEDURES: Step1: Estimate the cross section: Force Bolted connections – Symmetric Sections Arequired = 0.85 Ft Force Bolted connections – Unsymmetric Sections Arequired = 0.85 x 0.85 x Ft Force Welded connections – Symmetric Sections Arequired = Ft Force Welded connections – Unsymmetric Sections Arequired = 0.85 x Ft Then choose a suitable cross-section Step2. Check of the Chosen Member: Force 1-Check of the actual Stress fact f act = ≤ 0.58 Fy Anet l 2-Check of the length top depth ratio ≤ 60 (for diagonals and chord members only) d 3-Check member stiffness λ max ≤ 300 l b −in l λ max is the maximum of λin = and λ out = b −out in case of two angles back-to back; rx ry lb − max and λ max = in case of two angles star-shaped ru −1L Prof. Ahmed Abdel-Salam Ismail El-Serwi 9 Design of Steel Structures (3rd Civil) New Cairo Academy 2008-2009 lb − max and λ max = in case of single angle sections. rv −1L 4-minimum angle leg “a” shall be ≥ 3 d + t Where d is the bolt diameter and t is the angle leg thickness. Example 1: Design the lower chord tension member “A” shown in the Figure. The force in the member is 30t (Case II) and the bolts used in the connections are 16 non-pretensioned bolts. Data: Force = 30 t (Case II ) Length = 3.00 m Lx= 3.0 m and Ly = 9.0 m Estimation of the cross section: Force Areq = 0.85 Ft Areq = 30 / ( 0.85 x 1.4 x 1.20 ) = 21.0 cm2 Choose 2 angles back-to-back for the lower chord. A of 1L = 21.0/2= 10.50 cm2 Knowing that for 2 angles back-to-back with equal legs: rx=0.3 a and ry = 0.45 a λx = Lx / (rx =0.3a)=300/0.3 a= 300 a req=3.33 cm λy = Ly / (ry=0.45 a)= 900/(0.45x300) = 6.67 cm For construction : a-t > 3d = 4.8 cm. Prof. Ahmed Abdel-Salam Ismail El-Serwi 10 Design of Steel Structures (3rd Civil) New Cairo Academy 2008-2009 From the tables choose 2Ls 80x80x8 Check: 1. Strength: Anet = Agross - AHoles Anet = 2 ( 12.3 - (1.6+0.2) * 0.8 ) = 21.72 cm2. Fact = 30 / 21.72 = 1.38 t/cm2 < 1.4x1.2 ( case II ) =1.68 t/cm2 Safe 2. Stiffness: l/d = 300 / 8 = 37.5 < 60 OK lx/rx = 300 / 2.42 = 124 < 300 OK Iy = 2 ( Iy (1L) + A(1L) • { ey + 0.5 • tg. pl. }2 ) = 2 ( 72.3 + 12.3 ( 2.26+0.5x1.0 )2 ) = 332 cm4 A = 2 x 12.3 = 24.6 cm2 Iy 332 ry = = =3.67 A2 L 24.6 Ly/ry = 900 / 3.67 = 245 < 300 OK 3. Construction requirement: a = 8.0 cm > 3 d + t = 3x1.6+0.9 = 5.7 cm OK Example 2: Design the same member of Example 1 with welded connections. Estimation of the cross section: Areq = 30 / (1.4 x 1.2) = 17.86 cm2 Choose 2 angles back-to-back for the lower chord. A of 1L = 17.86/2 = 8.92 cm2 For 2 angles back-to-back with equal legs: rx=0.3 a and ry=0.45 a rx = Lx / (rx=0.3 a)=300/(0.3x300) = 3.33 cm ry = Ly / (ry=0.45 a) = 900/(0.45x300) = 6.67 cm From the tables choose 2Ls 70x70x7 Check: 1. Strength: Anet = Agross = 2 x 9.4= 18.8 cm2. Prof. Ahmed Abdel-Salam Ismail El-Serwi 11 Design of Steel Structures (3rd Civil) New Cairo Academy 2008-2009 fca = 30 / 18.8 = 1.595 t/cm2< 1.4x1.2 ( case II ) =1.68 t/cm2. Safe 2. Stiffness: l/d = 300/7 = 42.85 < 60 O.K. Safe l x 300 = = 142.8 < 300 O.K. rx 2 .1 t gusset 2 1 I y = 2[ I y (1L ) + A1L (e y + ) ] = 2[42.4 + 9.4(1.97 + ) 2 ] = 199.5 cm4 2 2 2 A = 2x9.4 = 18.8 cm Iy 199.5 ry = = 3.26 A 18.8 ly 900 = = 276 < 300 O.K. ry 3.26 Example 3: Design the same tension member of Example 1 using unequal angles. Use 16 non- pretensioned bolts in all the truss connections and 10 mm gusset plates. Data: Force = + 30 t (Case II), Member length = 3.0 m, Lin = 3.0 m, Lout=9.0 m Estimation of the cross section: Areq = 30 / ( 0.85 x 1.4 x 1.2 ) = 21.0 cm2. Choose 2 angles back-to-back for the lower chord. Area of 1L = 21.0/2= 10.50 cm2 Choose 2 Ls back-to-back 65x100x7 (choose unequal angle and use the longer leg to add stiffness for the member around the Y-Y axis as shown in the Figure. Check: 1. Strength: Anet = 2 ( 11.2 - 1.8 x 0.7 ) = 19.88 cm2. • Fact = 30 / 19.88 = 1.51 t/cm2 < 1.4x1.20 t/cm2. Safe Prof. Ahmed Abdel-Salam Ismail El-Serwi 12 Design of Steel Structures (3rd Civil) New Cairo Academy 2008-2009 2. Stiffness: • l/d = 300 / 6.5 = 46.2 < 60 OK • lx/ry = 300 / 1.84 = 163 < 300 OK Iy = 2 [ 113 + 11.2 { (1.0/2 + 3.23 }2] = 537.7 cm4 537.7 ry = = 4.899 2 x11.2 ly/ry = 900 / 4.899 = 184 < 300 OK 3. Construction requirement: a = 6.5 cm > 3d+t = 3x1.6+0.7 = 5.5 cm OK Example 4 Design a diagonal member in a truss if the tensile force in the member is 6 t (Case I) and the member length is 3.6 m. The bolts used in all the truss connections are M16 non- pretensioned bolts grade 4.6, with 10 mm thick gusset plates. Data: Force = + 6t (Case I) Length = 3.6 m. Estimation of the member cross section: Assume the section is single angle (unsymmetric section) Areq = 6.0 / ( 0.85 x 0.85 x 1.4 ) = 5.93 cm2 For 1 L: rv=0.2 a L / (rv=0.2 a)=300 a = 360/(0.2x300) = 6.0 cm for construction : a > 3d+ t a-t > 4.8 cm. To satisfy the above 3 requirements: choose 1L 60x60x6 Check: 1. Strength: Prof. Ahmed Abdel-Salam Ismail El-Serwi 13 Design of Steel Structures (3rd Civil) New Cairo Academy 2008-2009 A1 = 6.0x0.6 - (1.6+0.2)x0.6 = 2.52 cm2 A2 = (6.0-0.6)x0.6 = 3.24 cm2 Anet =A1 + A2{ 3A1 / (3A1 + A2 )} = 2.52 + 3.24 {3 x 2.52 / ( 3 x 2.52 + 3.24 )} = 4.79 cm2 • Fact = 6 / 4.79 = 1.25 t/cm2 < 0.58 fy = 1.40 t/cm2 Safe 2. Stiffness • l/d = 360 / 6.0 = 60 OK • l/rv = 360 / 1.17 = 308> 300 unsafe Use L 65x65x7 and recheck stiffness only: l/rv = 360 / 1.26 = 286 < 300 OK Example 5 Design the shown tension member if the length of the member is 6.0 m and it carries a force of 6 t. The bolts used in the connections are M16 non-pretensioned bolts with 10 mm gusset plates. Estimation of the member cross section: Areq = 6/ ( 0.85 x 1.4 ) = 5.04 cm2 (Assuming that the cross section will be symmetric about the guest plate: Star-shaped cross-section) Prof. Ahmed Abdel-Salam Ismail El-Serwi 14 Design of Steel Structures (3rd Civil) New Cairo Academy 2008-2009 A of one angle = 5.04/2 = 2.5 cm2 For star-shaped angle: ru=0.385a a = L / 0.385 a < 300 a > 600/0.385x300 = 5.2 cm L/d < 60 , d> 600/60 = 10 cm (d=2.a + tG) Construction requirements: a-t > 4.8 cm. From table choose 2 Ls star-shaped 55x55x5. Check: 1.Strength: Anet = 2 ( 5.32 - (1.6+0.2)x0.5) = 8.84 cm2 • Fact = 6/8.84 = 0.67 t/cm2< 0.58 fy t/cm2 safe. 2. Stiffness: l/d = 600 / ( 2x5.5+1.0 ) = 50 > 60 OK l/ru = 600 /2.09 = 287 < 300 OK Example 6: Design the lower chord tension member “A” shown in the Figure using circular hollow section (tube). The force in the member is 30t (Case II) and the member is connected to the truss members by using weld. Data: Force = 30 t (Case II ) Length = 3.00 m Lx= 3.0 m and Ly = 9.0 m Prof. Ahmed Abdel-Salam Ismail El-Serwi 15 Design of Steel Structures (3rd Civil) New Cairo Academy 2008-2009 Estimation of the cross section: Force D=133 mm Areq = Ft Areq = 30 / (1.4 x 1.20 ) = 17.86 cm2 Choose CHS 133x5 t=5 mm A = 20.1 cm2 rx= ry = 4.53 cm Check: 2. Strength: Agross= 20.1 cm2 fact = 30 / 20.1 = 1.49 t/cm2 < 1.4x1.2 ( case II ) =1.68 t/cm2 Safe 2. Stiffness: l/d = 300 / 13.3 = 22.55 < 60 OK lx/rx = 300 / 4.53 = 66.23 < 300 OK Prof. Ahmed Abdel-Salam Ismail El-Serwi 16

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