chapter 4 - tension members

7/26/2019 Chapter 4 - Tension Members

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Dr. J. BERLIN P. JUANZON CE, MBA, MSCM



Tension members are axially loadedmembers stressed in tension and are usedin steel structures in various forms. Theyare used in trusses as web and chordmembers, hanger and sag rods, diagonalbracing for lateral stability, and lap splices

such as in a moment connections.



The basic design check for a tension

member is to provide enough cross-sectional

area to resist the applied tensile force.



For members subjected to tension, the two basicmodes of failure are tensile yielding and tensilerupture.

Tensile yielding occurs when the stress on the

gross area of the section is large enough to causeexcessive deformation.The expression for tensile yielding on the gross

area is

Pn = ϕF y A g Eqn 4.1where

Φ = 0.90,Fy = Minimum yield stress, and

Ag = Gross area of the tension member.



Tensile rupture occurs when the stress on theeffective area of the section is large enough to cause themember to fracture, which usually occurs across a lineof bolts where the tension member is weakest.

The expression for tensile rupture on the effectivearea is

Pn = ϕF u Ae Eqn 4.2

whereϕ = 0.75

Fu = Minimum tensile stress, and

Ae = Effective area of the tension member.



The design strength of a tension member is the

smaller of the two expressions indicated inequations 4.1 and 4.2.

The gross area, Ag, of a tension member issimply the total cross-sectional area of the member

in question.The effective area, Ae, of a tension member is

described as follows:

Ae = An*U Where: An = Net Area

U = Shear lag factorNote that for a tension member that is connected by

welds, the net area equals the gross area(i.e., An= Ag).



The net area of a tension member with fastenersthat are in line is the difference between the grosscross-sectional area and the area of the bolt holes:

An = A g - Aholes

Where

Aholes = n(db + 1 ⁄8)t

n = number of bolt holes along the failure plane,

db = bolt diameter,t = material thickness



Tension members with in-line fasteners

The actual hole size used in the design calculations will be 1 ⁄ 16 in. + 1 ⁄ 16

in. = 1 ⁄ 8 in. for most bolted connections in tension



The shear lag factor (U ) accounts for the non-uniform stress distribution when some of the

elements of a tension member are not directly

connected, such as a single angle or WT member



Table D3.1 of the AISCM gives the value for theshear lag factor, U , for various connectionconfigurations. With the exception of plates andround hollow structural sections (HSS) memberswith a single concentric gusset plate andlongitudinal welds, the shear lag factor is

U = 1 -

Where: x = Distance from the centroid of the

connected part to the connection plane

L = Connection length



For the bolted tension member

shown, determine the shear lag factor, U ;

the net area, An; and the effective area,

Ae.



For the welded tension member

shown in Figure 4-7, determine the shear lag factor, U ; the net area, An; and the

effective area, Ae.



Determine if the channel is adequate for the

applied tension load shown. The channel is ASTM A36;it is connected with four 5 ⁄ 8-in. diameter bolts. Neglect

block shear.



From equation 4.1 based on gross area:P n = ϕF y Ag

Pn = (0.90)(36)(3.37) = 109kips > Pu = 75kips –SAFE

From equation 4.2 based on net area:

P n = ϕF u Ae

Pn = (0.75)(58)(32.61) = 113kips > Pu = 75kips –SAFE



Tension members with zigzag fasteners



For tension members with a series of holes

in a diagonal or zigzag pattern, which might beused when bolt spacing is limited there may

exist several possible planes of failure that

need to be investigated.

For a failure plane where one or more of thefailure planes is at an angle, then the following

term is added to the net width of the member

for each diagonal portion that is present along

the failure plane:



Determine the maximum factored load that

can be applied in tension to the angle shown. Theangle is ASTM A36; it is connected with four 3 ⁄ 4-in. diameter bolts. Neglect block shear.



From equation based on gross area:P n = ϕF y Ag

Pn = (0.90)(36)(3.61) = 116kips

From equation based on net area:

P n = ϕF u Ae

Pn = (0.75)(58)(2.39) = 104kips

Smaller value controls : use Pn = 104kips



Tension member ruptures in both shear andtension



Block Shear Failure



The nominal strength based on shear yielding is

Rn = 0.60F y A gvThe nominal strength based on shear rupture is

Rn = 0.60F u Anvwhere

Fy = Minimum yield stress

Fu = Minimum tensile stress

Agv = Gross area subject to shear

Anv = Net area subject to shear



To determine the design strength inshear yielding and shear rupture, the nominal

strength Rn, is multiplied by a -factor of 1.0and 0.75, respectively, when the shear doesnot occur simultaneously with tension

stress.



Combining the available tension and shear strength yields the

expression for the available block shear strength:

Pn = ϕ (0.60F u Anv + U bsF u Ant) ≤ϕ(0.60F

y A

gv+ U

bsF

u A

nt)

Where: ϕ= 0.75

Fu = Minimum tensile stress,Fy = Minimum yield stress,

Agv = Gross area subjected to shear, Ant = Net area subjected to tension , Anv = Net area subjected to shear,Ubs = 1.0 for uniform tension stress and

0.50 for non-uniform tension stress.



The U bs term in equation is a reduction factorthat accounts for a non-uniform stressdistribution. Section C-J4.3 of the AISCM gives

examples of connections with uniform andnon-uniform tension stress distribution, but themost common case is to have a uniform stressdistribution and, therefore, Ubs = 1.0 for most

cases.



For the connection shown in Problem 3.0,

determine if the channel and gusset plate areadequate for the applied tension load consideringblock shear. Assume that the width of the plate issuch that block shear along the failure plane

shown in Figure below controls the design of theplate.

Fy = 36ksi

Fu = 58ksi



Using equation for Block Shear:

Pu = ϕ (0.60F u Anv + U bsF u Ant) ≤ ϕ(0.60F y A gv + U bsF u Ant)

For Plate:

Pu = 0.75((0.60)(58)(3.28) + (1.0)(58)(1.21))

≤ 0.75((0.60)(36)(4.12) + (1.0)(58)(1.21))

= 138kips > 119kips Smaller value controlsPu = 119kips > 75kips (Applied Load) Safe



For Channel :

Pu = 0.75((0.60)(58)(1.92) + (1.0)(58)(0.715))

≤ 0.75((0.60)(36)(2.42) + (1.0)(58)(0.715))

= 81.20 kips > 70.3kips Smaller value controls

Pu = 70.3 kips < 75kips (Applied Load) Fail



The design of a tension member can besummarized as follows:

1. Determine the minimum gross area from the

tensile yielding failure mode equation:



2. Determine the minimum net area from thetensile fracture failure mode equation:

where the net area is found from equation:



3. Use the larger Ag value from the above

equations, and select a trial member size basedon the larger value of Ag.

4. For tension members, AISC specificationSection D1 suggests that the slenderness ratio KL

/ r min should be ≤ 300 to prevent flapping orflutter of the member,

where:K = Effective length factor (usually assumed to be

1.0 for tension members),L = Unbraced length of the tension member, and

r min = Smallest radius of gyration of themember.

Th ll t di f ti f ll d



The smallest radius of gyration for rolledsections can be obtained from part 1 of the

AISCM . For other sections, such as plates, the

radius of gyration can be calculated from

Where:

Imin is the smallest moment of inertia.

If the above equation cannot be satisfied (i.e., the

member is too slender), the member should be pre-

tensioned. Allow for 5% to 10% pretension force in the design

of the member.



5. Using equation for Block Shear , determinethe block shear capacity of the selected tensionmember.

If Pn (block shear) is greater than Pu, themember is adequate.

If Pn (block shear) is less than Pu, increasethe member size and repeat step 5 until

Pn (block shear) ≥ Pu.Where: Pn = Nominal Force

Pu = Ultimate Applied Force

chapter 4 - tension members

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