design of tension members

Design of Tension Members

Structural Elements Subjected to Axial Tensile Forces

Cables in Suspension and Cable-Stayed Bridges

Trusses

Bracing for Buildings and Bridges

LAST TIME

• Design of Tension Members

• Tables for the Design

• Threaded Rods and Cables


• 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

Required StrengthAvailable Strength(Nominal Resistance)

eu AF75.0gy AF9.0

LRFD minLD 6.12.1 D4.1

LRFD max


Determine required Area

To prevent yielding y

uggyu F

PAAFP

9.090.0

To avoid fracture y

ueeuu F

PAAFP

75.075.0

Yielding controls if eugy AFAF 75.090.0

u

y

g

e

F

F

A

A2.1

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

kips

kipsLD

kipsDPu 8.104

8.104526.1182.16.12.1

2.25184.14.1max

Step 1: Required Strength

2

, 235.3369.0

8.104in

F

PA

yt

ureqg

Step 2: Required Areas

2

, 409.25875.0

8.104in

F

PA

ut

ureqe

Example



Step 3: Plate Selection based on Ag

inwwinA reqreqreqg 235.3)1(235.3 2,

Try thickness t = 1 in

Choose PL 1 X 3-1/2See Manual pp1-8 for availability of plate products

Example



Step 4: Check Effective Area

409.25.28

1

8

75.3x11

1

,2

reqe

holegne

Ain

AAUAAOK

Example



Step 4: Check Slenderness

433

min in 2917.012

15.3

12

wtI

2in 5.315.3 A

2min2 in 2887.0A

IrArI

22 in 300in 2392887.0

75.5max

r

LOK

LRFD - Example



Step 4: Check Slenderness

433

min in 2917.012

15.3

12

wtI

2in 5.315.3 A

2min2 in 2887.0A

IrArI

22 in 300in 2392887.0

75.5max

r

LOK

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 locationManual 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 7/8 3/4 5/8

g1 3 2-1/2 2-1/4 2

g2 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 A36

Angle - Example

kips

kipsLD

kipsDPu 154

154706.1352.16.12.1

49354.1max


2

, 75.4369.0

154

9.0in

F

PA

y

ureqg

Step 2: Required Areas

2

, 54.3585.0

154

75.0in

F

PA

u

ureqe

inL

ru

req 6.0300

)12(15

300

Angle - Example

Step 3: Angle Selection based on Ag

Two lines of bolts, therefore min. length of one leg = 5 insee table

Choose L6x4x1/2 A=4.75, rmin=0.864See Manual pp1-42

2

, 75.4369.0

154

9.0in

F

PA

y

ureqg

Angle - Example


54.329.3875.385.0 ,2 reqene AinUAA NG

Length of connection not known4 – bolts in direction of load U=0.85

2875.32

1

8

1

4

3275.4 inAAA holegn

Angle - Example

Step 3: Angle Selection based on Ag – TRY NEXT LARGER


Choose L5 x 3-1/2 x 5/8 A=4.92, rmin=0.746See Manual pp1-42

2

, 75.4369.0

154

9.0in

F

PA

y

ureqg

Angle - Example


54.3252.3826.385.0 ,2 reqene AinUAA NG


2826.38

5

8

1

4

3292.4 inAAA holegn

Angle - Example

Step 3: Angle Selection based on Ag – TRY NEXT LARGER


Choose L8 x 4 x 1/2 A=5.75, rmin=0.863See Manual pp1-42

2

, 75.4369.0

154

9.0in

F

PA

y

ureqg

Angle - Example


54.314.4875.485.0 ,2 reqene AinUAA OK


2875.48

5

8

1

4

3275.5 inAAA holegn

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

Example – Using Tables

kips

kipsLD

kipsDPu 154

154706.1352.16.12.1

49354.1max


Step 2: Choose L based on Pu

Choose L6x4x1/2 A=4.75, rmin=0.980

kipsPyielding nt 154 : kipsPrupture nt 155 :

See Manual pp 5-15

Angle - Example


229.3875.385.0 inUAA ne

NG


2875.38

5

8

1

4

3275.4 inAAA holegn

kipsPkipsAFP ueutnt 1543.143)29.3)(58(75.0

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

uPratioactual

75.0

Example – Using Tables

Step 4: Choose L based on ADJUSTED Pu

Choose L8x4x1/2 A=5.75, rmin=0.863

kipsPyielding nt 186 : kipsPrupture nt 187 :

See Manual pp 5-14

kipsPu 16715469.0

75.0

Angle - Example


214.4875.485.0 inUAA ne

OK


2875.42

1

8

1

4

3275.5 inAAA holegn

kipsPkipsAFP ueutnt 154180)14.4)(58(75.0

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

PinHinge

Supporting walls: reinforced concrete, concrete block, brick or combination

Tension Members in Roof Trussed


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


Bottom Chord in tensionTop Chord in compressionWeb members: some in compression some in tension

Wind loads may alternate force in some members


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


• 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

Working Lines Intersect at the Working Point in each joint


Bolted trusses

Double Angles for chords

Double Angles for web members

Single Gusset plate


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(5)=900 lb

……

180(2.5)=450 lb


PURLINS M8x6.5

Purlin Load = 6.5(20) = 130 lb 20ft

……

130 lb 130 lb


SNOWSnow Load = 20(20) = 400 lb/ft

20ft

……

400(2.5)=1000 lb 400(5)=2000 lb


Dead Load of TrussAssume 10% of all other loads

……

End Joint 0.1(9(20)(20)+130+1000)=158 lb

158 lb 303 lb

Interior Joint 0.1(900+130+2000)=303 lb


……

450+130+158 = 738 lb 900+130+303 = 1333 lb

D

S

1000 lb 2000 lb

Step 2 – Required Force

……

1.2(0.74) + 1.6(1) =

2.48 kips1.2(1.33)+1.6(2)=

4.8 kips

SDPu 6.12.1

Step 2 – Required Force

Method of Sections

kips 04.480 IJE FM

Step 3 – Required Areas

2

, 07.1509.0

04.48

9.09.0in

F

F

F

PA

y

IJ

y

ureqg

2

, 985.06575.0

04.48

75.075.0in

F

F

F

PA

u

IJ

u

ureqe

Step 4: T Selection based on Ag

Choose MT5x3.75 A=1.10 in2

See Manual pp1-68

Step 5 Check Effective Area

2,

2 in 985.0in 915.018322.0 reqege AUAA NG

8322.09

51.111

L

xU

Step 6 TRY NEXT LARGER

Choose MT6X5 A=1.46 in2 See Manual pp1-68

Step 7 Check Effective Area

2,

2 in 985.0in 16.146.17933.0 reqege AUAA OK

7933.09

51.111

L

xU

Step 8 – Check Slenderness

Assume bracing points at panel points

300101594.0

)12(5

r

LOK

design of tension members

Documents

angle selection

areasexampletension

known4 bolts

a36 steel

rectangular cross section

resists d

largertwo lines of bolts

tension membersmust