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 Φt.Pn based on the gross area

– Φt.Pn =0.90(Fy.Ag)

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• Step 4: Determine the shear lag factor U, using

AISC Specifications Table D3., pp. 16.1-28

• Step 5: Determine the net area

– An= Ag – Σ dh.t + [Σ (s2/4g)] . t

• Step 6: Find Φt.Pn based on fracture effective

net section

– Ae=U.An

– Φt.Pn = 0.75Fu.Ae

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• Step 7: Find Φ.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 in2, rx=1.91-in, ry=1.14-in, x=0.981-in, …

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

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

– ASD • Pn=Fy.Ag = 50(4.75)=237.5 k

• Pn/(Ω=1.67)=237.5/1.67=142.2 k

• Step 4: Determine the shear lag factor U – U=1-x/L=1-0.981/(2x4)=0.88

– Or U=0.60 form Table D3.1 Case 8 AISC 14th ed. Pp.16.1.228

– Use the larger U, U= 0.88

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Exercise- Analysis of Tension Member- solution-cont’d

• Step 5:Determine the net area

– An= Ag – Σ dh.t + [Σ (s2/4g)] . t

– An=4.75-(3/4+1/8)(1/2)=4.31 in2

• Step 6: Find Φt.Pn based on fracture effective

net section

– Ae=U.An = 0.88(4.31)=3.79 in2

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

– ASD

• Pn =Fu.Ae = 65(3.79) =246.4 k

• Pn/Ω =246.4/2.00=132.2 k

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Exercise- Analysis of Tension Member- solution-cont’d

• Step 7:Find Φ.Rn based on block shear strength

– Rn=0.6FuAnv+UbsFuAnt ≤ 0.6FyAgv+UbsFuAnt

– Agv =(10)(1/2)=5.0 in2

– 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

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

– ASD, Rn/Ω =216.95/2.00=108.5 k

•

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Exercise- Analysis of Tension Member- solution-cont’d

• 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=Φt.Fy.Ag

• Min Ag= Pu /(Φt.Fy )……….……..(2)

• Pu=Φt.Fu.Ae

• Min. Ae= Pu/(Φt.Fu)………………..(3)

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Tension Members Design Formulae

• Since Ae=U.An

• Min. An=Ae/U=Pu/(Φt.Fu.U)………(4)

• Also

• Min. Ag= Pu/(Φt.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

¾-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/(ΦtFy)= 236/(0.90x50)= 5.24 in2

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• Or

• Min Ag=Pu/(Φt.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 = ΦtFy.Ag=(0.90)(50)(7.61) = 342.45 k > 236 k

….. OK

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• From Table D3.1 PP.16.1.28 AISC steel design manual 14th 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= ΦtFu.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 in2

• 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

• ΦRn=(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/(ΦtFy)

• Step 3: Assume an appropriate value for U

• Step 4:Compute the minimum Ag based on effective area

– Min Ag=Pu/(Φt.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, cont’d

• 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 = ΦtFy.Ag > the required Pu otherwise select larger section

– Pu= ΦtFu.Ae (After determining U) > Required Pu in step 1 otherwise select larger section

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Design of Tension Members-Suggested Procedure, cont’d

• Step 8 : Check l/r criteria ≤ 300 otherwise

select larger section

• Step 9: Check block shear

– Rn=0.6FuAnv+UbsFuAnt ≤ 0.6FyAgv+UbsFuAnt

– Φ.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/(Φ(0.75).Fu), Φ= 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 n2 = πd2/4 d= 1.38 in

• Use 1 ½ 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/ΦtFy=720/(0.9x50)= 16 in2

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

• Min Ag= Pu/(Φt..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= Φt.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/(Φx0.75xFu) =

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

• Use 1 3/8 in. diameter rod with 6 threads per

inch (AISC Table 7-17, 14th ed. Pp.7-81)

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