# Design of Tension Members

## Presentation on theme: "Design of Tension Members"— Presentation transcript:

Design of Tension Members
Structural Elements Subjected to Axial Tensile Forces Trusses Bracing for Buildings and Bridges Cables in Suspension and Cable-Stayed Bridges

Design of Tension Members Tables for the Design
LAST TIME Design of Tension Members Tables for the Design Threaded Rods and Cables

Design of Tension Members LAST TIME
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 LRFD max LRFD min ASD etc max ASD min

Design of Tension Members LAST TIME
Determine required Area LRFD To prevent yielding To avoid fracture Yielding controls if

Design of Tension Members LAST TIME
Determine required Area ASD To prevent yielding To avoid fracture Yielding controls if

LRFD - Example LAST TIME
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

LRFD - Example LAST TIME
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

LRFD - Example LAST TIME
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

LRFD - Example LAST TIME
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

Step 1: Required Strength
ASD - Example LAST TIME 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

Step 3: Plate Selection based on Ag - Same as LRFD
ASD - Example LAST TIME 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 - Same as LRFD Try thickness t = 1 in Choose PL 1 X 3-1/2 See Manual pp1-8 for availability of plate products

OK ASD - Example LAST TIME Step 4: Check Effective Area
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

LRFD - Example LAST TIME
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 LAST TIME
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 3-1/2 1-1/8 7/8 3/4 5/8 g1 2-1/4 g2

Example LAST TIME 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 LAST TIME
Step 1: Required Strength Step 2: Required Areas

Angle - Example LAST TIME
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 LAST TIME
Step 4: Check Effective Area Length of connection not known 4 – bolts in direction of load U=0.8 NG

Angle - Example LAST TIME
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 LAST TIME
Step 4: Check Effective Area Length of connection not known 4 – bolts in direction of load U=0.8 NG

Angle - Example LAST TIME
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 LAST TIME
Step 4: Check Effective Area Iterative Process 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 A36

Step 1: Required Strength Step 2: Choose L based on Pu
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

NG Angle - Example Step 3: Check Effective Area
Length of connection not known 4 – bolts in direction of load U=0.8 NG

Angle - Example Shape did not work because table values are for Ae/Ag=0.75 In this problem Ae/Ag=3.1/4.75 = 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

OK Angle - Example Step 5: Check Effective Area
Length of connection not known 4 – bolts in direction of load U=0.8 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 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 20ft 180(2.5)=450 lb 180(5)=900 lb ……
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 20ft 130 lb 130 lb …… PURLINS M8x6.5
Purlin Load = 6.5(20) = 130 lb 20ft 130 lb 130 lb ……

Step 1 – Load Analysis 20ft 400(2.5)=1000 lb 400(5)=2000 lb …… SNOW
Snow Load = 20(20) = 400 lb/ft 20ft 400(2.5)=1000 lb 400(5)=2000 lb ……

Interior Joint 0.1(900+130+2000)=303 lb
Step 1 – Load Analysis Dead Load of Truss Assume 10% of all other loads End Joint 0.1(9(20)(20) )=158 lb Interior Joint 0.1( )=303 lb 158 lb 303 lb ……

Step 1 – Load Analysis = 738 lb = 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 3 – Required Areas

Step 4: T Selection based on Ag
Choose MT5x A=1.10 in2 See Manual pp1-68

Step 5 Check Effective Area
NG

Step 6 TRY NEXT LARGER Choose MT6X5 A=1.46 in2 See Manual pp1-68

Step 7 Check Effective Area
OK

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