T-1These are found predominantly as members of plane or space trusses (2D & 3D), as members in transmission towers and as wind bracing (single or double) for single story or high rise steel structures. Among the common shapes used as tension members:Round bar Flat bar Angle Double angle Starred angleCross-section of typical tension members.

T-2The strength of a tension member is controlled by the lowestof the following limiting states:NetArea (An)GrossArea (Ag)TTYielding of the Gross Area (Ag): Fn = Fy AgFailure (Ultimate strength) on the Net Area (An): Fn = Fu AeWhere Ae = effective net Area = UAn U = Reduction Coefficient. C) Block Shear Failure through the end bolt:T

T-3A hole is drilled (or punched) by 1/16 inch greater then the normal diameter of the fastener (rivet or bolt). Hole punching causes some damage to the edges of the hole to the amount of 1/32 inch from each side.Thus the normal hole diameter

T-4What is the net area An for the tension memberas shown in the figure?Solution:TTAg = 4(0.25) = 1.0 sq in.Width to be deducted for holeAn = [Wg (width for hole)] (thickness of plate)Example (T1):

(a) (b)T-5For a group of staggered holes along the tension direction, one must determine the line that produces smallest Net Area.Paths of failureon net sectionEFFECT OF STAGGERED HOLES ON NET AREA :-In the above diagram:p = Pitch or spacing along bolt lines = Stagger Between two adjacent bolt lines (usually s = P/2)g = gage distance transverse to the loading.In case (a) above : An = (Gross width hole dia.) . tIn case (b) above : An = (Gross width hole dia.+ s2/4g) . t

T-6Determine the minimum net area of the plate shown in fig. 3.4.2, assuming in,-diam holes are located as shown:Figure 3.4.2Example 3.4.1Example (T2):

T-7Solution. According to LRFD and ASD-B2, the width used in deducing forholes in the hole diameter plus 1/16 in., and the staggered length correctionIs (s2/4g).1)Path AD (two holes) :2)Path ABD (three holes; two staggers) :3)Path ABC (three holes; two staggers) :(controls)

T-8Angles:When holes are staggered on two legs of an angle, the gage length (g) for use In the (s2/4g) expression is obtained by using length between the centers of the holes measured along the centerline of the angle thickness, i.e., the distance A-B in Fig: 3.4.3. Thus the gage distance g isGage dimension for an angle

T-9Every rolled angle has a standard value for the location of holes (i.e. gage distance ga and gb), depending on the length of the leg and the number of lines of holes. Table shows usual gages for angles as listed in the AISC Manual*.

T-10Determine the net area (An ) for the angle given in figure belowif holes are used?Angle with legs shown *flattened* into one planeExample (T3):9

T-11Solutions. For net area calculation the angle may be visualized as beingflattened into a plate as shown in Figure above.where D is the width to be deducted for the hole.1) Path AC:2) Path ABC:Since the smallest An is 3.75 sq in., that value governs.An = An =

T-12 When some of the cross section (and not all the section) is connected, we need to use effective net area concept :-Ae = U Anwhere, U = Reduction Factor. When all elements of the section are connected, U = 1.0.

When not all elements are connected.i) Transverse Weld Connection:-Ae = UAU = 1.0A = Area of connected part onlye.g. A = 6 x 1/2 = 3 in2ii) Longitudinal Weld Connection :-Ae = Ag UU = 1.0 for L 2 wU = 0.87 for 2w L 1.5 wU = 0.75 for 1.5w L w6GussetplateAngle6x4x1/2TT-13wGussetplateAngle6x4x1/2TLWeld

T-14In bolted connections, the reduction factor (U) is a functionof the eccentricity ( ) in the connection.Thus:-Where:= distance between centroids of elements to the plane of load transferL = Length of the connection in the direction of load.(See Commentary C B 3.1 & C B 3.2)

T-15LFRD Specification for Structural Steel Buildings, December 27, 1999American Institute of Steel Construction

T-16(Commentary P16.1 177 AISC) For bolted or riveted connections the following values for (U) may be used:- W, M or S Shapes with flange width 2/3 depth, and structural tees cut from these shapes, provided connection to the flanges and has 3 fasteners per line in the direction of force, U = 0.90.b) W,M or S Shapes where flanges width < 2/3 depth, and all other shapes, that has no fewer than 3 fasteners per line, U = 0.85c) All members having only two fasteners in the line of stress U = 0.75 For short tension members such as Gusset plates the effective net area equals (An), but must not exceed 0.85 of the gross area (Ag).

Example (T-4)Calculate the Ae values of the following section:- flange width (6.54) > 2/3 x depth (8.0) Three bolts / lineU = 0.90Ag = 8.24 m2An = gross area hole area = 8.24 (2 x 1.0 hole) x web tk 0.285 = 7.68 in2Ae = UAn = 0.9 x 7.68 = 6.912 in2hole dia = 7/8C 9 x 15only 2 bolts / line, U = 0.75Ag = 4.41 m2An = 4.41 (2 x 15/16) 0.285 = 3.875 in2Ae = 0.75 x 3.875 = 2.907 in2(i)(ii)T-17web tk

(iii)L 3 x 3 x 3/8Ag = 2.11 in2An = 2.11 1 x (3/4 + 1/8) x 3/8 = 2.11 -0.328 = 1.782 in2Ae = UAn = 0.852 x 1.782 = 1.518 in2Alternative value of U = 0.85 (3 bolts / line)w 10 x 33T-18

T-19This third mode of failure is limited to thin plates. This failure is a combination of tearing (shear rupture) and of tensile yielding. It is uncommon, but the code provides on extra limit state of (LRFD J 4.3). It is usually checked after design is completed.Failure by tearing outGussetPlateShaded areamay tear outTacbEven as tension members are unlikely to be affected by their stiffness (L/r), it is recommended to limit the maximum slenderness ratio (L/r) for all tension members (except rods) to 300.Max. slenderness = L/rmin 300This is to prevent extra sagging and vibration due to wind.

The general philosophy of LRFD method:For tension members:wheret = resistance reduction factor for tensile membersTn = Nominal strength of the tensile membersTu = Factored load on the tensile members.The design strength tTn is the smaller of:a) Yielding in the gross section;t Tn = t Fy Ag = 0.9 Fy Agb) Fracture of the net section;t Tn = t Fu Ae = 0.75 Fu AeThis is to be followed by check of rupture strength (block shear failure),and limitation of slenderness ratio 300.T-20

Example (T-5):-Find the maximum tensile capacity of a member consisting of 2Ls (6 x 4 x ) can carry for two cases: welded connection, bolted connection 1" dia bolts Fy = 60 ksi Fu = 75 ksi.T-21

Net area = gross area (all sides connected) = 9.50 in2 Yielding Ft = 0.9 Fy Ag = 0.9 x 60 x 9.50 = 513 k Fracture Ft = 0.75 Fu Ae = 0.75 x 75 x 9.5 = 534 k

Thus tension capacity, t Tn = 513 k (yielding controls)(a) welded Connection(b) Bolted ConnectionConsider one LAn Calculation: Wg = gross width = 6 + 4 = 9.5 in.(cont.)T-22

= 6.62 in. (Controls)(fracture controls)T-23Zig-Zag =9

Design is an interactive procedure (trial & error), as we do not have the final connection detail, so the selection is made, connection is detailed, and the member is checked again.Proposed Design Procedure:-Find required (Ag) from factored load .

Find required (Ae) from factored load .

Convert (Ae) to (Ag) by assuming connection detail.

From (ii) & (iii) chose largest (Ag) value

Find required (rmin) to satisfy slenderness

Select a section to satisfy (iv) and (v) above.

Detail the connection for the selected member.

Re-check the member again.T-24

Example (T-6):-A tension member with a length of 5 feet 9 inches must resist a service dead load of 18 kips and a service live load of 52 kips. Select a member with a rectangular cross section. Use A36 steel and assume a connection with one line of 7/8-inch-diameter bolts.Member length = 5.75 ft.T-25

T-26Pu = 1.2 D + 1.6L = 1.2(18) + 1.6(52) = 104.8 kipsBecause Ae = An for this member, the gross area corresponding tothe required net area isTry t = 1 in.Ag = 2.409 + 1(1) = 3.409 in.2

T-27Because 3.409 > 3.235, the required gross area is 3.409 in.2, andRound to the nearest 1/8 inch and try a 1 3 cross section.Check the slenderness ratio:Use a 3 1 bar.

Select a single angle tension member to carry (40 kips DL) and (20 kips LL), member is (15)ft long and will be connected to any one leg by single line of 7/8 diameter bolts. Use A-36 steel.Solution:Step 1) Find Required (Tu):-Tu = 1.2 DL + 1.6 LLTu = 1.4 DL = 1.2 x 40 + 1.6 x 20 or = 1.4 x 40 = 48 + 32 = 80k = 56k Tu = 80k (Controls)T-28Example (T-7):-

Step 2) Find required Ag & Ae:Step 3) Convert (Ae) to (Ag):Since connection to single leg, then use alternative (U) value = 0.85 (more then 3 bolt in a line).For single line 7/8 bolts ; Ag = An + (1)t = 2.16 + t = (Ag)2T-29

Step 4) Find required rmin.Step 5) Select angle:By selecting (t) we get Ag & rmint(Ag)1 (Ag)21/4 2.47 2.413/8 2.47 2.531/2 2.47 2.66select t = 3/8 (Ag)2 = 2.53 in2T-30(Controls)

Selection Ag = 2.67 in2 > 2.53 in2 OKrmin = 0.727 in > 0.6 OKStep 6) Design the bolted connection: (chap. 4).

Step 7) Re-check the section.T-31

Select a pair of MC as shown to carry a factored ultimate load of 490 kips in tension. Assume connection as shown. Steel Fy = 50 ksi, Fu = 65 ksi (A572, grade 50) length = 30 ft.3. Assume that flange thickness ~ 0.5 in and web tk. ~ 0.3 in. (experience !)An = (Ag)2 2 x 1.0 x 0.5 2 x 1.0 x 0.3 = (Ag)2 1.60 (Ag)2 = An + 1.60 = 5.03 + 1.60 = 6.63 in.(controls)T-32U = 1.0 (Well connected)Example (T-8):-

T-33

For built-up members, tie plates are required to make themembers to behave as a single unit. Between tie plates, each member behaves as a single. Therefore, l/r between tie-plates corresponds to that for a single member.For single , rmin = ry ; ry = 1.0 inT-34(N.G.)

Note:Tie-Plates must be used at ends. See Manual for min. sizes. Length of tie-plate 2/3 (dist. between line of connection) = 8"Thickness of tie-plate 1/50 (dist. between line of connection) = 1/2"T-35Therefore one tie-plate at middle must be used.See LFRD D2. (P. 16.1-24)