LRFD TensionMembers Sample

7/30/2019 LRFD TensionMembers Sample

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Lecture 32 - Page 1 of 11

Lecture 32 Steel Design Tension members and beams

Structural steel is the strongest material commercially available for theconstruction of buildings and other structures. It is widely used for theconstruction of the tallest and widest buildings and other structures. Local labor

and availability make structural steel the material of choice for many applications.The design, fabrication and erection of structural steel for buildings within the US(and much of the world) is dictated by the American Institute of SteelConstruction, AISC http://www.aisc.org

IBC Struc tural Steel Design

Steel construction information is found in Chapter 22 of the IBC. In particular,Section 2205 states:

The design, fabrication and erection of structural steel for buildings andother structures shall be in accordance with the AISC.


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AISC Methods of Design

There are two methods of structural design of structural steel:

1) Allowable Stress Design (ASD)

The ASD method has been around for many decades and is still in use today.As with wood and other materials that also use this method, it is based on thepremise that the allowable stress must be greater than the actual anticipatedstress. This is the method that will be used in the design of s tructuralsteel members for our d iscuss ion, and is based on the 9th edition of theAISC Manual of Steel Construction ASD, 1989.

2) Load and Resistance Factor Design (LRFD)

The LRFD method is relatively new the first edition of LRFD Manual of SteelConstruction was in 1986 (the 3rd edition issued in 2002). It is based on astatistical approach of applying load factors to the anticipated service loadsand selecting members such that the ultimate strength of the membersexceeds the factored loads. These load factors are listed in the IBC Section1605.2.1. The method is a bit more complicated and is not as widely used(yet) as the ASD method. However, design by the LRFD typically yieldslighter-weight members and is therefore more efficient. This method isgaining increased support by architects and engineers and will most likelyreplace the ASD method within the next decade. It is the basis of s teeldesign in the AECT 460 Structural Steel Design.

ASD Manual 9t ed. (1989) LRFD Manual 3r ed. (2002)


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Tension Member Design

The allowable tensile stress, Ft, shall not exceed 60% of the yield stress on thegross area nor 50% of the ultimate stress on the net area. Or, re-writing, theallowable tensile load on a steel member, Pallow:

where:Fy = yield stress for steel, KSI

= 36 KSI for ASTM A36 steel= 50 KSI for ASTM A572 & A992 grade 50 steels

Fu = ultimate stress for steel, KSI= 58 KSI for ASTM A36 steel= 65 KSI for ASTM A572 & A992 grade 50 steels

Ag = gross cross-sectional area of member, in2

Anet = net area of member accounting for reduction of area for bolt holes= Ag [(No. of bolts)(bolt dia. x material thickness)]

0.60(Fy)(Ag)yielding on gross area

or

0.50(Fu)(Anet)fracture on net area

Pallow < smaller

Gross area of angle Net area of angle

2 bolt holes


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Example 1GIVEN: A steel A36 C10x20 channel carries a tensile load = 120 kips. It isfastened to the structure with 3 diameter bolts.REQUIRED:

1) Determine if the channel is acceptable based upon yielding on the gross

area.2) Determine if the channel is acceptable based upon fracture on the netarea.

Step 1 Determine Pallow for yielding on gross area:

Pallow = 0.60(Fy)(Ag)

= 0.60(36 KSI)(5.88 in2)

Pallow = 127 kips

Since Pallow = 127 kips > 120 kips channel is acceptable

120 kips

Channel with bolt holes

Section cut thru bolt holes


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Step 2 Determine Pallow for fracture on net area:

Pallow = 0.50(Fu)(Anet)

= 0.50(Fu)(Ag [(No. of bolts)(bolt dia. x material thickness)])

= 0.50(58 KSI)(5.88 in2 [(3 bolts)( x 0.379 in)])

Pallow = 146 kips

Since Pallow = 146 kips > 120 kips channel is acceptableExample 2GIVEN: A steel-framed building is stabilized against wind force with adiagonal tension brace as shown below. The bracing member is to be 2 equal leg back-to-back angles.

REQUIRED: Design the lightest weight back-to-back equal leg anglesusing A572 steel. The angles are to be welded to the steel gusset plate,so there is no need to investigate fracture on the net area.

web thickness


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Step 1 Determine required area based on yielding on gross area:

Pallow = 0.60(Fy)(Ag)

Rearranging to solve for Ag:

Ag >)(60.0 yF

P

>)50(60.0

86

KSI

Kips

Ag > 2.87 in2

Step 2 Determine size of single angle:

From the analysis, Ag

= 2.87 in2 for 2 angles

Area for 1 angle = (2.87 in2)= 1.44 in2

Possibilities:

Angle Size: Area (in2): Weight (lbs per foot):

L6x6x5/16 3.65 12.4L5x5x5/16 3.03 10.3L4x4x1/4 1.94 6.6

L3 x 3 x 1.69 5.8L3x3x 1.44 4.9

Use 3x3x back-to-back angles Ag = 2.88 in2 > 2.87 in2


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Beam Design

Steel beams are designed on the basis ofbending stress. The allowablebending stress, Fb, is based on the following:

Width of compression flange Distance between lateral supports of compression flange

Orientation of beam

Type of loading, i.e., single-curvature or double-curvature

Recalling that a typical wide-flange beam under normal loading conditions hascompression acting in the top flange and tension acting in the bottom flange asshown below:

The compression in the flange of a beam behaves much the same ascompression in a column, i.e., unless it is adequately laterally suppor ted, thebeam is likely to buckle. This buckling in a beam is referred to as lateraltorsional buckling. The beam will twist out of plane and fail, similar to a yardstick. Increasing the distance between lateral support dramatically reduces themoment capacity of the beam.

If a graph of the moment capacity of a beam were plotted against the distancebetween lateral supports, it would take the following shape:

Momentcapacity

Distance between lateral supports


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Most beams have adequate lateral support of the compression flange, usually inthe form of flooring, roofing or by other beams framing into it. Assuming this isthe case, the allowable bending stress, Fb is defined as:

Fb = 0.66Fy > fb

where:Fb = allowable bending stress, KSI

Fy = yield stress, KSI= 50 KSI for ASTM A992 steel (new wide-flange beams)= 50 KSI for ASTM A572 steel (high strength angles, channels)= 36 KSI for ASTM A36 steel (low strength angles, channels, plates)= 46 KSI for ASTM A500 steel (tube steel referred to as Hollow Structural

Section, HSS)

fb = actual bending stress

=xS

Mmax

The design of steel beams involves selecting the section modulus by setting theactual bending stress equal to the allowable bending stress. Therefore:

Required Sx =bF

ftM )/"12(max


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Example 3GIVEN: A simply-supported beam (continuously laterally braced) isloaded as shown below. Assume the weight of the beam is alreadyincluded.REQUIRED:

1) Design the lightest weight wide-flange beam using A992 steel.2) Design the lightest weight back-to-back channel beam using A36 steel.3) Design the lightest weight square tube Hollow Structural Section (HSS)

beam using A500 steel.

Step 1 Determine the maximum moment, Mmax, on the beam:

Mmax =8

2wL

=8

)'20)(1750( 2PLF

= 87500 ft-lb

= 87.5 kip-ft

Step 2 Design lightest weight wide-flange beam using A992 steel:

Required Sx =bF

ftM )/"12(max

=)50(66.0

)/"12(5.87

KSI

ftftkip

= 31.8 in3

20-0

1750 PLF


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Possibilities:

Beam size: Sx (in3) Weight (lbs per foot):

W8x40 35.3 40W10x30 32.4 30

W12x26 33.4 26W14x26 35.3 26W16x26 38.4 26W18x35 57.6 46

Use W16x26 since Sx = 38.4 in3 > 31.8 in3 total wt. = 26 PLF

Step 3 Design lightest weight back-to-back channels using A36 steel:

Required Sx =bF

ftM )/"12(max

=)36(66.0

)/"12(5.87

KSI

ftftkip

= 44.2 in3 for 2 channels

Required Sx for 1 channel = (44.2 in3)

= 22.1 in3

Possibilities:

Channel size: Sx (in3) Weight (lbs per foot):

C12x25 24.1 25C15x33.9 42.0 33.9

Use ][ 12x25 since Sx = 2(24.1 in3) > 44.2 in3 total wt. = 50 PLF


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Step 4 Design lightest weight square tube HSS using A500 steel:

Required Sx =bF

ftM )/"12(max

=)46(66.0

)/"12(5.87

KSI

ftftkip

= 34.6 in3

Possibilities:

Tube size: Sx (in3) Weight (lbs per foot):

9x9x0.50 42.9 55.6610x10x0.3125 36.7 40.35

12x12x0.25 44.1 39.4314x14x0.3125 74.6 57.36

Use 12x12x0.25 HSS since Sx = 36.7 in3 > 34.6 in3 total wt. = 39.43 PLF

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