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 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
Required Strength
1.4D
LRFD max 1.2D  1.6L
ASD max
D
DL
etc
(Nominal Resistance)
LRFD
ASD
min
min
0.9 Fy Ag
0.75Fu Ae
0.6 Fy Ag
0.5Fu Ae
Design of Tension Members LAST TIME
Determine required Area
LRFD
To prevent yielding
Pu
Pu  0.90Fy Ag  Ag 
0.9 Fy
To avoid fracture
Pu
Pu  0.75Fu Ae  Ae 
0.75Fy
Yielding controls if
0.90 Fy Ag  0.75Fu Ae
Fy
Ae
 1.2
Ag
Fu
Design of Tension Members LAST TIME
• Determine required Area ASD
To prevent yielding
Pa
Pa
 Fy Ag  Ag 
0.6
0.6Fy
To avoid fracture
Pa
Pa
 Fu Ae  Ae 
0.5
0.5Fu
Yielding controls if
Fy
Ae
 1.2
Ag
Fu
Fy Ag
Fu Ae

1.67
2
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
1.4 D  1.418  25.2kips
Pu  max 
 104.8kips
1.2 D  1.6 L  1.218  1.652   104.8kips
Step 2: Required Areas
Pu
104.8
Ag ,req 

 3.235 in 2
t Fy 0.936
Pu
104.8
Ae,req 

 2.409 in 2
t Fu 0.7558
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
Ag ,req  3.235 in 2  wreq (1)  wreq  3.235 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
Ae  UAn  1Ag  Ahole  

 7 1 
11x 3.5       2.5 in 2  Ae,req  2.409
 8 8 

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
wt 3 3.51


 0.2917 in 4
12
12
3
I min
A  3.51  3.5 in 2
I  Ar  r 
2
I min
 0.2887 in 2
A
L
5.75
max 
 239 in 2  300 in 2
r 0.2887
OK
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
1D  18  18kips
Pa  max 
 70.0kips
 D  L  18  52   70kips
Step 2: Required Areas
Pa
70.0
Ag ,req 

 3.24 in 2
0.6Fy 0.636
Pa
70.0
Ae ,req 

 2.414 in 2
0.5Fu 0.558
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
Ag ,req  3.241 in 2  wreq (1)  wreq  3.241 in
Choose PL 1 X 3-1/2
See Manual pp1-8 for availability of plate products
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 4: Check Effective Area
Ae  UAn  1Ag  Ahole  

 7 1 
11x 3.5       2.5 in 2  Ae,req  2.414
 8 8 

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
wt 3 3.51


 0.2917 in 4
12
12
3
I min
A  3.51  3.5 in 2
I  Ar  r 
2
I min
 0.2887 in 2
A
L
5.75
max 
 239 in 2  300 in 2
r 0.2887
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
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 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
1.4 D  35  49kips
Pu  max 
 154kips
1.2 D  1.6 L  1.235  1.670  154kips
Step 2: Required Areas
Pu
154
Ag ,req 

 4.75 in 2
0.9 Fy 0.936
Pu
154
Ae ,req 

 3.54 in 2
0.75Fu 0.558
rreq
L
15(12)


 0.6 in
300u
300
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
Ag ,req
Pu
154


 4.75 in 2
0.9 Fy 0.936
Choose L6x4x1/2 A=4.75, rmin=0.864
See Manual pp1-42
Angle - Example LAST TIME
Step 4: Check Effective Area
 3 1  1 
An  Ag  Ahole  4.75  2     3.875in 2
 4 8  2 
Length of connection not known
4 – bolts in direction of load U=0.8
Ae  UAn  0.83.875  3.10in 2  Ae,req  3.54
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
Ag ,req
Pu
154


 4.75 in 2
0.9 Fy 0.936
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
 3 1  5 
An  Ag  Ahole  4.92  2     3.826in 2
 4 8  8 
Length of connection not known
4 – bolts in direction of load U=0.8
Ae  UAn  0.83.826  3.06in 2  Ae,req  3.54
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
Ag ,req
Pu
154


 4.75 in 2
0.9 Fy 0.936
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
 3 1  5 
An  Ag  Ahole  5.75  2     4.875in 2
 4 8  8 
Length of connection not known
4 – bolts in direction of load U=0.8
Ae  UAn  0.84.875  3.9in 2  Ae,req  3.54
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
Example – Using Tables
Step 1: Required Strength
1.4 D  35  49kips
Pu  max 
 154kips
1.2 D  1.6 L  1.235  1.670  154kips
Step 2: Choose L based on Pu
Choose L6x4x1/2
A=4.75, rmin=0.980
yielding : t Pn  154 kips
rupture : t Pn  155 kips
See Manual pp 5-15
Angle - Example
Step 3: Check Effective Area
 3 1  5 
An  Ag  Ahole  4.75  2     3.875in 2
 4 8  8 
Length of connection not known
4 – bolts in direction of load U=0.8
Ae  UAn  0.83.875  3.10in 2
t Pn  t Fu Ae  0.75(58)( 3.10)  135kips  Pu  154kips
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 = 0.6526
Enter table with adjusted Pu as
0.75
 Pu
actual ratio
Example – Using Tables
Step 4: Choose L based on ADJUSTED Pu
0.75
Pu 
154  177kips
0.6525
Choose L8x4x1/2
A=5.75, rmin=0.863
yielding : t Pn  186 kips
rupture : t Pn  187 kips
See Manual pp 5-14
Angle - Example
Step 5: Check Effective Area
 3 1  1 
An  Ag  Ahole  5.75  2     4.875in 2
 4 8  2 
Length of connection not known
4 – bolts in direction of load U=0.8
Ae  UAn  0.84.875  3.90in 2
t Pn  t Fu Ae  0.75(58)( 3.90)  170kips  Pu  154kips
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 9inch 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
180(2.5)=450 lb
……
20ft
180(5)=900 lb
Step 1 – Load Analysis
PURLINS M8x6.5
Purlin Load = 6.5(20) = 130 lb
130 lb
……
20ft
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
S
2000 lb
Step 2 – Required Force
Pu  1.2 D  1.6S
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
M
E
 0  FIJ  48.04 kips
Step 3 – Required Areas
Ag ,req
Pu
FIJ
48.04



 1.07 in 2
0.9 Fy 0.9 Fy 0.950
Ae,req
Pu
FIJ
48.04



 0.985 in 2
0.75Fu 0.75Fu 0.7565
Step 4: T Selection based on Ag
Choose MT5x3.75 A=1.10 in2
See Manual pp1-68
Step 5 Check Effective Area
x
 1.51 
U  1  1 
  0.8322
L
 9 
Ae  UAg  0.83221  0.915 in 2  Ae,req  0.985 in 2
NG
Step 6 TRY NEXT LARGER
Choose MT6X5 A=1.46 in2
See Manual pp1-68
Step 7 Check Effective Area
x
 1.51 
U  1  1 
  0.7933
L
 9 
Ae  UAg  0.79331.46  1.16 in 2  Ae,req  0.985 in 2
OK
Step 8 – Check Slenderness
Assume bracing points at panel points
L 5(12)

 101  300
r 0.594
OK