EN138Fall 2002

Engineering 138

Design of Civil Engineering StructuresLECTURE 5– Analysis of Tension

Members

EN138Fall 2002

Lecture 5 Outline

• Types of tension members• Limit states for design• Net areas• Staggered holes• Effective net area and shear lag• Block Shear

EN138Fall 2002

• Typical tension members in buildings• Truss diagonals• Columns in uplift• Bracing members• Suspension cables

• Typical shapes used for tension members

Types of Tension Members

EN138Fall 2002

Limit States for Design

• Two key limit states– Yielding in the gross section– Fracture in the net section where holes exist

Tension

“net section”Reduced areaat bolt holes

“gross section”

EN138Fall 2002

Limit States for Design

• Yielding of the “gross” section– Can support load after yield due to strain hardening, but

excessive elongation would result– Nominal strength = Pn = FyAg

– Ultimate strength = Pu = φtPn

– Pu=φtFyAg with φt=0.9

Yield in this entire area causes excessive elongation

Pu

EN138Fall 2002

Limit States for Design

• Fracture in the “net” section – May control if bolt holes are large– Pn=FuAe

– Pu=φ tFuAe with φt=0.75 – NOTE: Yielding of net section not considered since

this area is usually short in length compared to the overall length

Pu

Fracture in reduced section

EN138Fall 2002

Net Areas• Gross cross section area minus any holes, notches, or other

indentations• Standard Bolt holes punched 1/16” larger than actual bolt

to allow for insertion, also assume 1/16” additional size for damage around hole from punching (so add 1/8” to bolt diameter)

3/8” Plate(typ)

3/4” bolt (typ)

Net area = An = (3/8”)*(8”) – 2*(3/4” + 1/8”)*(3/8”) = 2.34 in2

8”

EN138Fall 2002

Effect of Staggered Holes

• Holes are often staggered to increase net areaB

A A

D

C

B

E

g

s

Failure planes:AB

Failure planes:ABE An = Ag – 1 holeABCD An = ???

EN138Fall 2002

Effect of Staggered Holes

• Actual area stressed is combination of tension and shear between holes

• Use empirical formula to estimate net area in tension for staggered holes (per LRFD Section B2)

A

D

C

B

E

g

s

An = Ag – (Ahole + s2/4g)

Add (s2/4g) for each diagonalline in the failure section

EN138Fall 2002

Effect of Staggered Holes

• Example

A

D

C

B

E

2.5”

3”

3”

3”

2.5”

Determine the net area:Ahole = (¾” + 1/8”) = 7/8”

ABCD = 11”-2*(7/8”) = 9.25”

ABCEF = 11”-3*(7/8”) +(32/(4*3)) = 9.125”

ABEF = 11” –2*(7/8”) +(32/(4*6)) = 9.625”

So An = (9.125”)*(1/2”) = 4.56 in2

¾” bolts

F

½” plate

EN138Fall 2002

Effective Net Area

• Entire net area may not be engaged if tensile stress cannot be uniformly transferred between members

“shear lag” transition region,area is not 100% effective intransfer of tensile stresses

PuPu

EN138Fall 2002

Effective Net Area

• Use a reduction factor U to account for nonuniform stress distribution in shear lag region

Ae=AU (LRFD Equation B3-1)A = net area or gross areaU = reduction factor (for bolted or welded conn)

L’

L

P

U factor effectivelyreduces the length Lof a connection to L’

EN138Fall 2002

Effective Net Area

• For bolted members:U = 1- x/L < 0.9 L’

L

x

Pu

x

Centroid of beam

Centroid of lower “Tee”xCentroid of “Tee”

EN138Fall 2002

Effective Net Area

• U factors for design• Permissible U Values for bolted connections

– U=0.90 W,M,S shapes with flange widths not less than 2/3 of the depths, T’s cut from these shapes, provided no fewer than three fasteners per line

– U = 0.85 W,M,S shapes not meeting conditions above, but with at least three fasteners per line

– U=0.75 All members having only two fasteners per line in the direction of stress

EN138Fall 2002

Block Shear

• Tension on one plane and shear on perpendicular plane can cause a “block” of steel to tear out

Tension plane

Shear plane

Tension planeShear plane

Shaded areacan tear outin “block shear”

EN138Fall 2002

Block Shear

• AISC Specification J5.2 for Block Shear:• Total Block Shear Resistance = shear resistance

on shear-failure path + tensile resistance on perpendicular path– Use the ultimate strength in shear (or tension) on the

net section– Use the yield strength in tension (or shear) on the gross

section of the perpendicular sectionShear plane

Tension plane

EN138Fall 2002

Block Shear

( )( )

0.75 0.6

0.75 0.6

where:gross area in shear

gross area in tensionnet area in shearnet area in tension

y gv u ntn

u nv y gt

gv

gt

nv

nt

F A F AR

F A F A

AAAA

φ +=

+

=

=

==

Use largervalue

EN138Fall 2002

Block Shear Example

3”

1 ½”

7”

1 ½”

½” plates7/8” bolts

Standard holes

Determine the ultimate block shear strengthof the connected plates (assume A36 steel)

EN138Fall 2002

Block Shear Example

( )

( )

21 12 2

271 1 1 12 2 8 8 2

212

27 1 18 8 2

Shear path 1-2:4 2.25 in

4 1 1.50 in

Tension path 2-2:7 3.50 in

7 2 3.00 in

gv

nv

gt

nt

A

A

A

A

= × =

= − + × =

= × =

= − × + × =

12

21

7”

3”1 ½”

( )( )

0.75 0.6 36 2 2.25 58 3.00 203 kips

0.75 0.6 58 2 1.5 36 3.50 173 kips

Choosing the larger value, we have:203 kips for block shear

n

n

R

R

φ

φ

× × × + × == × × × + × =

=