Analysis of Tension Members
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Transcript of Analysis of Tension Members
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
× × × + × == × × × + × =
=