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## Design of Tension Members

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**Design of Tension Members**Structural Elements Subjected to Axial Tensile Forces Trusses Bracing for Buildings and Bridges Cables in Suspension and Cable-Stayed Bridges**LAST TIME**• Design of Tension Members • Tables for the Design • Threaded Rods and Cables**LRFD**max LRFD min Design of Tension Members • Objective • Find a member with adequate gross and net areas • Find a member that satisfies L/r<300 • Does not apply to cables and rods Available Strength (Nominal Resistance) Required Strength**Design of Tension Members**Determine required Area To prevent yielding To avoid fracture Yielding controls if**Example**Tension member with a length 5’-9” resists D=18 kips and L=52 kips Select a member with rectangular cross section, A36 steel and one line 7/8” bolts Step 1: Required Strength Step 2: Required Areas**Example**Tension member with a length 5’-9” resists D=18 kips and L=52 kips Select a member with rectangular cross section, A36 steel and one line 7/8” bolts Step 3: Plate Selection based on Ag Try thickness t = 1 in Choose PL 1 X 3-1/2 See Manual pp1-8 for availability of plate products**Example**Tension member with a length 5’-9” resists D=18 kips and L=52 kips Select a member with rectangular cross section, A36 steel and one line 7/8” bolts Step 4: Check Effective Area OK**Example**Tension member with a length 5’-9” resists D=18 kips and L=52 kips Select a member with rectangular cross section, A36 steel and one line 7/8” bolts Step 4: Check Slenderness OK**LRFD - Example**Tension member with a length 5’-9” resists D=18 kips and L=52 kips Select a member with rectangular cross section, A36 steel and one line 7/8” bolts Step 4: Check Slenderness OK**Angles as Tension Members**• Must have enough room for bolts (if bolted connection) • Space is a problem if 2 lines of bolts in a leg • Usual fabrication practice – standard hole location Manual pp 1-46**Example**• Select and unequal-leg angle tension member 15 feet long to resist a service dead load of 35 kips and a service live load of 70 kips. Use A36**Angle - Example**Step 1: Required Strength Step 2: Required Areas**Angle - Example**Step 3: Angle Selection based on Ag Two lines of bolts, therefore min. length of one leg = 5 in see table Choose L6x4x1/2 A=4.75, rmin=0.864 See Manual pp1-42**Angle - Example**Step 4: Check Effective Area Length of connection not known 4 – bolts in direction of load U=0.85 NG**Angle - Example**Step 3: Angle Selection based on Ag – TRY NEXT LARGER Two lines of bolts, therefore min. length of one leg = 5 in see table Choose L5 x 3-1/2 x 5/8 A=4.92, rmin=0.746 See Manual pp1-42**Angle - Example**Step 4: Check Effective Area Length of connection not known 4 – bolts in direction of load U=0.8 NG**Angle - Example**Step 3: Angle Selection based on Ag – TRY NEXT LARGER Two lines of bolts, therefore min. length of one leg = 5 in see table Choose L8 x 4 x 1/2 A=5.75, rmin=0.863 See Manual pp1-42**Angle - Example**Step 4: Check Effective Area Length of connection not known 4 – bolts in direction of load U=0.8 OK**Example**• Select and unequal-leg angle tension member 15 feet long to resist a service dead load of 35 kips and a service live load of 70 kips. Use A36**Example – Using Tables**Step 1: Required Strength Step 2: Choose L based on Pu Choose L6x4x1/2 A=4.75, rmin=0.980 See Manual pp 5-15**Angle - Example**Step 3: Check Effective Area Length of connection not known 4 – bolts in direction of load U=0.85 NG**Angle - Example**Shape did not work because table values are for Ae/Ag=0.75 In this problem Ae/Ag=3.29/4.75 = 0.693 Enter table with adjusted Pu as**Example – Using Tables**Step 4: Choose L based on ADJUSTED Pu Choose L8x4x1/2 A=5.75, rmin=0.863 See Manual pp 5-14**Angle - Example**Step 5: Check Effective Area Length of connection not known 4 – bolts in direction of load U=0.85 OK**Tension Members in Roof Trusses**• Main supporting elements of roof systems where long spans are required • Used when the cost and weight of a beam would be prohibitive • Often used in industrial or mill buildings**Pin**Hinge Tension Members in Roof Trussed Supporting walls: reinforced concrete, concrete block, brick or combination**Tension Members in Roof Trusses**Sag Rods are designed to provide lateral support to purlins and carry the component of the load parallel to the roof Located at mid-point, third points, or more frequently**Tension Members in Roof Trusses**Bottom Chord in tension Top Chord in compression Web members: some in compression some in tension Wind loads may alternate force in some members**Tension Members in Roof Trusses**Chord Members are designed as continuous Joint rigidity introduces small moments that are usually ignored Bending caused by loads applied directly on members must be taken into account**Tension Members in Roof Trusses**Working Lines Intersect at the Working Point in each joint • Bolted Truss: Working Lines are the bolt lines • Welded Truss: Working Lines are the centroidal axes of the welds • For analysis: Member length from working point to working point**Tension Members in Roof Trusses**Bolted trusses Double Angles for chords Double Angles for web members Single Gusset plate**Tension Members in Roof Trusses**Welded trusses Structural Tee shapes are used in chords Angles are used in web members Angles are usually welded to the stem of the Tee**Tension Members in Roof Trusses**Welded trusses Structural Tee shapes are used in chords Angles are used in web members Angles are usually welded to the stem of the Tee**Example**Select a structural Tee for the bottom chord of the Warren roof truss. Trusses are welded and spaced at 20 feet. Assume bottom chord connection is made with 9-inch long longitudinal welds at the flange. Use A992 steel and the following load data (wind is not considered) Purlins M8x6.5 Snow 20 psf horizontal projection Metal Deck 2 psf Roofing 4 psf Insulation 3 psf**Step 1 – Load Analysis**DEAD (excluding purlins) Deck 2 psf Roof 4 psf Insulation 3 psf Total 9 psf Total Dead Load = 9(20) = 180 lb/ft 20ft 180(2.5)=450 lb 180(5)=900 lb ……**Step 1 – Load Analysis**PURLINS M8x6.5 Purlin Load = 6.5(20) = 130 lb 20ft 130 lb 130 lb ……**Step 1 – Load Analysis**SNOW Snow Load = 20(20) = 400 lb/ft 20ft 400(2.5)=1000 lb 400(5)=2000 lb ……**Step 1 – Load Analysis**Dead Load of Truss Assume 10% of all other loads End Joint 0.1(9(20)(20)+130+1000)=158 lb Interior Joint 0.1(900+130+2000)=303 lb 158 lb 303 lb ……**Step 1 – Load Analysis**450+130+158 = 738 lb 900+130+303 = 1333 lb …… D 1000 lb 2000 lb S**Step 2 – Required Force**1.2(0.74) + 1.6(1) = 2.48 kips 1.2(1.33)+1.6(2)= 4.8 kips ……**Step 2 – Required Force**Method of Sections**Step 4: T Selection based on Ag**Choose MT5x3.75 A=1.10 in2 See Manual pp1-68**Step 6 TRY NEXT LARGER**Choose MT6X5 A=1.46 in2 See Manual pp1-68**Step 8 – Check Slenderness**Assume bracing points at panel points OK