Analysis Of Tension Members

Engineering 138 Design of Civil Engineering Structures LECTURE 5– Analysis of Tension Members

EN138 Fall 2002

Lecture 5 Outline • • • • • •

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

EN138 Fall 2002

Types of Tension Members • Typical tension members in buildings • • • •

Truss diagonals Columns in uplift Bracing members Suspension cables

• Typical shapes used for tension members

EN138 Fall 2002

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

Tension

“gross section”

“net section” Reduced area at bolt holes

EN138 Fall 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

Pu

Yield in this entire area causes excessive elongation

EN138 Fall 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 EN138 Fall 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) 3/4” bolt (typ) 3/8” Plate 8” (typ)

Net area = An = (3/8”)*(8”) – 2*(3/4” + 1/8”)*(3/8”) = 2.34 in2 EN138 Fall 2002

Effect of Staggered Holes • Holes are often staggered to increase net area B

E D C

g

B A Failure planes: AB

A

s

Failure planes: ABE An = Ag – 1 hole ABCD An = ??? EN138 Fall 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) E D

An = Ag – (Ahole + s2/4g) C

g

B A

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

s EN138 Fall 2002

Effect of Staggered Holes • Example

¾” bolts

A 2.5”

B

3”

C E D

F 3”

½” plate

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 EN138 Fall 2002

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

Pu

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

Pu

EN138 Fall 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 area U = reduction factor (for bolted or welded conn) L’

L

P U factor effectively reduces the length L of a connection to L’ EN138 Fall 2002

Effective Net Area • For bolted members: U = 1- x/L < 0.9

L’

Pu

x L

Centroid of “Tee”

x

Centroid of beam Centroid of lower “Tee”

x EN138 Fall 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

EN138 Fall 2002

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

Shaded area can tear out in “block shear”

Tension plane Shear plane

Tension plane

EN138 Fall 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 section Shear plane

Tension plane

EN138 Fall 2002

Block Shear 0.75 ( 0.6 Fy Agv + Fu Ant )  φ Rn =  0.75 ( 0.6 Fu Anv + Fy Agt ) where:

Use larger value

Agv = gross area in shear Agt = gross area in tension Anv = net area in shear Ant = net area in tension

EN138 Fall 2002

Block Shear Example ½” plates

7/8” bolts

7”

Standard holes

3” 1 ½”

1 ½”

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

EN138 Fall 2002

Block Shear Example 1

Shear path 1-2:

2

Agv = 4 12 × 12 = 2.25 in 2 7”

1

2

Anv =  4 12 − 1 12 ( 87 + 81 )  × 12 = 1.50 in 2 Tension path 2-2: Agt = 7 × 12 = 3.50 in 2

1 ½”

3”

Ant =  7 − 2 × ( 87 + 81 )  × 12 = 3.00 in 2

0.75 ( 0.6 × 36 × 2 × 2.25 + 58 × 3.00 ) = 203 kips φ Rn =  0.75 ( 0.6 × 58 × 2 × 1.5 + 36 × 3.50 ) = 173 kips Choosing the larger value, we have: φ Rn = 203 kips for block shear

EN138 Fall 2002