Design of Tension Members

Moayyad Al Nasra, Ph.D, PE

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Review

Analysis of Tension Member

Summary- Suggested Procedure Step 1: Find the relevant parameters regarding

the tension member, including, length, cross-sectional area, yield stress, ultimate stress, radius of gyration,

Step 2: Check the slenderness ratio l/r 300 (preferred )

Step 3: Find t.Pn based on the gross area t.Pn =0.90(Fy.Ag)

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Step 4: Determine the reduction coefficient U, using the formula for U, and/or AISC

Specifications Table D3.1

U=1-x/L

Step 5: Determine the net area

An= Ag dh.t + [ (s2/4g)] . t

Step 6: Find t.Pn based on fracture effective net section

Ae=U.An t.Pn = 0.75Fu.Ae

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Step 7: Find .Rn based on block shear strength

Rn=0.6FuAnv+UbsFuAnt 0.6FyAgv+UbsFuAnt (AISC Eq. J4-5)

Step 8: Find the lowest value calculated from steps 3, 6 and 7. The lowest value controls.

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Exercise- Analysis of Tension Member

The A572 Grade 50 (Fu=65) tension member shown is connected with three 3/4 in bolts. Calculate the design tensile strength, LRFD, and the allowable tensile strength, ASD, of the member

Shea plane Tension plane

2 4 4-in

10-in

3.5-in

2.5-in

L6X4X1/2

20-ft tension member

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Exercise- Analysis of Tension Member- solution

Step 1: L6X4X1/2 form the steel design manual, Ag=4.75 in

2, rx=1.91-in, ry=1.14-in, x=0.981-in,

Step 2:Check the slenderness ratio, rmin=ryl/r = 20(12)/1.14=210.5 < 300

Step 3:Find t.Pn based on the gross area t.Pn =0.90(Fy.Ag)=0.9(50)(4.75)=231.75 k ASD

Pn=Fy.Ag = 50(4.75)=237.5 k Pn/(=1.67)=237.5/1.67=142.2 k

Step 4: Determine the reduction coefficient U U=1-x/L=1-0.981/(2x4)=0.88

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Exercise- Analysis of Tension Member- solution-contd

Step 5:Determine the net area

An= Ag dh.t + [ (s2/4g)] . t

An=4.75-(3/4+1/8)(1/2)=4.31 in2

Step 6: Find t.Pn based on fracture effective net section

Ae=U.An = 0.88(4.31)=3.79 in2

t.Pn = 0.75Fu.Ae = 0.75(65)(3.79)= 184.9 k

ASD

Pn =Fu.Ae = 65(3.79) =246.4 k

Pn/ =246.4/2.00=132.2 k

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Exercise- Analysis of Tension Member- solution-contd

Step 7:Find .Rn based on block shear strength Rn=0.6FuAnv+UbsFuAnt 0.6FyAgv+UbsFuAnt Agv =(10)(1/2)=5.0 in

2

Anv =[10-(2.5)(3/4+1/8)](1/2)=3.91 in2

Ant = [2.5-(1/2)(3/4+1/8)](1/2)= 1.03 in2

Rn=(0.6)(65)(3.91)+(1.0)(65)(1.03) =219.44k

Rn =(0.6)(50)(5.0)+(1.0)(65)(1.03)=216.95 k

Therefore Rn=216.95 k

. Rn= (0.75)(216.95)= 162.7 k (LRFD)

ASD, Rn/ =216.95/2.00=108.5 k

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Exercise- Analysis of Tension Member- solution-contd

Step 8: The lowest Value

LRFD = lowest of (231.75, 184.9,162.7)

LRFD= 162.7 k

ASD Lowest of (142.2, 132.2, 108.5)

ASD =108.5 k

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Factors affecting the design decision

Safety

Economy

Compactness

Relative dimension

Joint condition

Technical consideration

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Slenderness Ratio (l/r)

Applies basically for members under compression

(providing sufficient stiffness to prevent lateral deflection,

buckling)

Slenderness ratio for members subjected to tension is

limited by AISC steel manual to a max of 300 (in case that

member is subjected to reversed loading, loading during

installation and transportation,)

l=un-braced length laterally

r=radius of gyration=

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Tension Members Design Formulae

Max l/r= 300, min r = l/300.(1)

Pu=t.Fy.Ag Min Ag= Pu /(t.Fy )...(2)

Pu=t.Fu.Ae

Min. Ae= Pu/(t.Fu)..(3)

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Tension Members Design Formulae

Since Ae=U.An

Min. An=Ae/U=Pu/(t.Fu.U)(4)

Also

Min. Ag= Pu/(t.Fu.U) + estimated hole areas .(5)

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Example

Select W10 section of 25 ft length subjected to tensile dead load 90 k, and live load of 80 k. The

member has two lines of bolts in each flange for

-in bolts. (use A572 grade 50 steel)

Use LRFD method

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Solution

Calculate the ultimate, factored load, Pu Pu= 1.4 PD= 1.4(90)= 126 k

Pu= 1.2 PD+1.6PL= 1.2(90)+1.6(80)=236 k Controls

Use Pu=236 k

Compute the minimum Ag required

Min Ag= Pu/(tFy)= 236/(0.90x50)= 5.24 in2

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Or

Min Ag=Pu/(t.Fu.U) + Estimated hole areas

Assume that U = 0.90 from case 7 AISC manual, and assume that flange thickness to

the average of W10s sections ( or pick a flange

thickness of W10 section of area 5.24 in2 or

slightly larger) tf=0.395 in

Min Ag= 236/(0.75x65x0.9) + 4(6/8+1/8)(0.395) =6.76 in2 CONTROLS

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Slenderness ratio criteria l/r = 300 Min r = l/300 = 25x12/300 = 1.0 Select a section of area > 6.76 and r > 1.0 Try W10x26 ( area = 7.61, min r = 1.36, d= 10.33,

bf= 5.77 in., tf= 0.44)

Check the section Pu = tFy.Ag=(0.90)(50)(7.61) = 342.45 k > 236 k ..

OK

U=1- x/L, ( x=1.06 for WT5x13 half of W10x26), L= 2x3 = 6

U=1-(1.06/6)= 0.82

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From Table D3.1 PP.16.1.29 AISC steel design manual 13th edition, U= 0.85 since bf= 5.77 < 2/3(d)=2/3(10.33) = 6.89 Use U=0.85 the larger

An= 7.61-4(6/8+1/8)(0.44) =6.07 in2

Ae= U.An) = 0.85(6.07)= 5.16 in2

Pu= tFu.Ae=0.75(65)(5.16) = 251.5 k > 236 k . OK

Check L/r criteria L/r= 25x12/1.36 = 220 < 300 OK

Check block shear

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Rn=0.6FuAnv+UbsFuAnt 0.6FyAgv+UbsFuAnt Agv=4(8)(.44)=14.08 in

2

Anv=4(8-2.5(3/4+1/8))(0.44)=10.23 in2

Ant=4(1.2-(0.5)(3/4+1/8))(0.44)=1.34 in2

Rn=0.6(65)(10.23) +(1.0)(65)(1.34)=486.07 k

Rn=0.6(50)(14.08)+(1.0)(65)(1.34)=509.5 k

Therefore use Rn=486.07 k

Rn=(0.75)486.07 =364.6> 236 k O.K

Use W10X26

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Design of Tension Members-Suggested Procedure

Step 1: Calculate the ultimate, factored load, Pu

Step 2: Compute the minimum Ag required based on gross area

Min Ag= Pu/(tFy)

Step 3: Assume an appropriate value for U

Step 4:Compute the minimum Ag based on effective area

Min Ag=Pu/(t.Fu.U) + Estimated hole areas

The larger of Ag from step 2 or step 4 will control

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Design of Tension Members-Suggested Procedure, contd

Step 5: Use the Slenderness ratio criteria l/r 300

Min r = l/300

Step 6: Select a section of area > the controlling area in step 4 and r > r-value in step 5

Step 7: Check the section Pu = tFy.Ag > the required Pu otherwise select

larger section

Pu= tFu.Ae (After determining U) > Required Puin step 1 otherwise select larger section

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Design of Tension Members-Suggested Procedure, contd

Step 8 : Check l/r criteria 300 otherwise select larger section

Step 9: Check block shear

Rn=0.6FuAnv+UbsFuAnt 0.6FyAgv+UbsFuAnt .Rn required Pu calculated in step 1, otherwise

adjust connection and/or select larger section

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Design of Rods and Bars

The required area AD= Pu/((0.75).Fu), = 0.75 Example A572 Grade 50 steel rod subjected to tensile dead load

of 12 k and tensile live load of 25 k. Find the diameter of the rod.

Solution Pu= 1.2 (12)+1.6(25)= 54.4 K AD= 54.4 / (0.75x0.75x65)= 1.49 n

2 = d2/4 d= 1.38 in

Use 1 in diameter rod of AD= 1.77 in2

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Exercise : Select the lightest W14 section available to support working tensile loads of

PD= 200 k and PL= 300 k. The member is to be

30 ft long and is assumed to have two lines of

holes for 1-inch bolts in each flange. There

will be at least three holes in each line 4 in. on

center. (Use steel A572 Grade 50). Use LRFD

method (optional block shear criteria

confirmation).

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Solution Pu= 1.2(200)+1.6(300) = 720 k Min Ag= Pu/tFy=720/(0.9x50)= 16 in

2

Assume U=0.90 from AISC Table D3.1 Case 7 and Assume tf= 0.720 in from AISC tables.

Min Ag= Pu/(t..Fu.U) + estimated areas of holes = 720/(0.75x65x0.90) + 4(1+1/8)(0.720) = 19.65 in2

Min r = l/300 = 12x30/300 = 1.20 in. Try W14X68 ( A=20.0 in2, d=14.0 in., bf= 10.0 in., tf= 0.720

in., ry= 2.46 in.)

Check Pu= t.Fy.Ag= 0.90(50)(20.0)= 900 k > 720 k OK

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Exercise: Select a standard threaded rod to resist service loads PD= 15 k and Pl= 18 k,

using A36 steel.

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Solution:

Pu= 1.2x15+1.6x18=46.8 k

AD= Pu/(x0.75xFu) = 46.8/(0.75x0.75x58)=1.434 in2

Use 1 3/8 in. diameter rod with 6 threads per inch (AISC Table 7-18)

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