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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