Design of Tension Members Structural Elements Subjected to

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

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 Required Strength LRFD max (Nominal Resistance) LRFD min

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, A 36 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, A 36 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 pp 1 -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, A 36 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, A 36 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, A 36 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 Leg 8 7 6 5 4 31/2 3 2 -1/2 2 1 -3/4 1 -1/2 1 -3/8 1 -1/4 1 g 4 -1/2 4 3 -1/2 3 2 -1/2 2 1 -3/4 1 -3/8 1 -1/8 1 7/8 3/4 5/8 g 1 3 2 -1/2 2 -1/4 2 g 2 3 3 2 -1/2 1 -3/4

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 A 36

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 L 6 x 4 x 1/2 A=4. 75, rmin=0. 864 See Manual pp 1 -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 L 5 x 3 -1/2 x 5/8 A=4. 92, rmin=0. 746 See Manual pp 1 -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 L 8 x 4 x 1/2 A=5. 75, rmin=0. 863 See Manual pp 1 -42

Angle - Example Step 4: Check Effective Area Length of connection not known 4 – bolts in direction of load U=0. 8 OK

TABLES FOR DESIGN OF TENSION MEMBERS

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 A 36

Example – Using Tables Step 1: Required Strength Step 2: Choose L based on Pu Choose L 6 x 4 x 1/2 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 L 8 x 4 x 1/2 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

Tension Members in Roof Trussed Pin Hinge Supporting walls: reinforced concrete, concrete block, brick or combination

Tension Members in Roof Trussed

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 longitudinal welds at the flange. Use A 992 steel and the following load data (wind is not considered) Purlins M 8 x 6. 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 180(2. 5)=450 lb …… 20 ft 180(5)=900 lb

Step 1 – Load Analysis PURLINS M 8 x 6. 5 Purlin Load = 6. 5(20) = 130 lb …… 20 ft 130 lb

Step 1 – Load Analysis SNOW Snow Load = 20(20) = 400 lb/ft 20 ft 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)+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 S 2000 lb

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 3 – Required Areas

Step 4: T Selection based on Ag Choose MT 5 x 3. 75 A=1. 10 in 2 See Manual pp 1 -68

Step 5 Check Effective Area NG

Step 6 TRY NEXT LARGER Choose MT 6 X 5 A=1. 46 in 2 See Manual pp 1 -68

Step 7 Check Effective Area OK

Step 8 – Check Slenderness Assume bracing points at panel points OK