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Analysis of Tension Members

Moayyad Al Nasra, Ph.D, PE

(c) Al Nasra

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Analysis of Tension Members

Moayyad Al Nasra, Ph.D, PE

(c) Al Nasra 2

Analysis of Tension Members

• Types of tension members:

– L- section,

– round bars,

– flat bars,

– double-angle,

– T-section,

– I-section,

– built-up

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Strength of tension members AISC 14th edition page 16.1-26

a.) Gross-section

Pn= Fy.Ag AISC D2-1

Pu=Φt.Fy.Ag

Φt=0.90

b.) Net-section

Pn=Fu.Ae AISC D2-2

Pu=Φt.Pn=Φt.Fu.Ae

Φt=0.75

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The design strength of a tension member

Φt.Pn, is to be the smaller of the above equations

Where:

Pn = nominal tensile force

Pu = ultimate tensile strength

Fy = yield stress

Fu = ultimate stress

Ae = effective area

Ag = gross area

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Net Area

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• Net area = gross area – areas of holes

• Holes are usually punched 1/16 inch larger

than the diameter of the bolt. Also the

punching of the hole is assumed to damage or

even destroy 1/16 inch more of the

surrounding metal. Therefore the area of the

hole is 1/8 inch in diameter larger than the bolt

diameter.

• Example 2-1:

– Net Area=An=(3/8)(6)-2[3/4+1/8](3/8)=1.59 in2

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Effective area

• To account for the non-uniform distribution of stresses at the connection sections, a reduction factor will be introduced. The further the section from the connection the more uniform the stress becomes

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• (concentration of stresses around the holes). Therefore AISC introduced a reduction

• factor, U, so that – Ae=AnU AISC D3-1 (AISC 14th pp. 16.1-27)

• Where U = shear lag factor determined from

table D3.1 AISC 14th pp.16.1-28 – U=1-x/L for tension members see case 2 table D3.1

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• (table D3.1 AISC 14th edition steel design

manual shows that for W, M, S section use

U=0.9 for bf≥2/3 d, and use U=0.85 for bf<2/3

d)

• Where:

– L= length of a connection

– X= distance measured from the plane of the

connection to the centroid of the area of the whole

section. Can be obtained from the AISC manual.

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• Example: Determine the LRFD tensile

strength of a W10X45 with two lines of ¾

inch bolts in each flange using A572 grade 50

steel, with Fy=50 ksi, and Fu = 65 ksi, and the

AISC specification. There are assumed to be at

least three bolts in each line 4 inches on center,

and the bolts are not staggered with respect to

each other.

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• From AISC manual W10x45 (Ag=13.3 in2, d=10.10 in., bf=8.02 in., tf=0.620 in.), Also

• WT5x22.5, x=0.907 in. ( half of W10x45)

• Pu= Φt. Fy.Ag=(0.90)(50)(13.3)=598.5 k

• An= 13.3-4(3/4+1/8)(0.620)=11.13 in2

– But, bf=8.02”>(2/3)d=(2/3)10.1=6.73, from case 7 AISC 14th manual table 3.1, U=0.9 pp. 16.1-28

• Use U=0.9

• Ae=U.An=((0.90)(11.13)=10.02 in2

• Pu=Φt.Fu.Ae=(0.75)(65)(10.02)=488.5 K

• Therefore the LRFD tensile strength of the section is 488.5 K

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Bolted splice plates

• For bolted splice plates

• Ae=An≤0.85 Ag

• Example 2-2

• Same as example 2-1

• Ag=(3/8)(6)=2.25 in2

• 0.85Ag=0.85(2.25)=1.91 in2

• Ae=An=1.59 in2 < 0.85Ag=1.91 in2

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Effect of Staggered Holes

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• The joint will fail at the weakest section

• To compute the net width of a tension member

along a zig-zag section:

– Net width=gross width-diameter of holes along the

zig-zag section + S2/(4g)

•

• Example Determine the critical net area of ¼ “

thick plate. The holes are punched for ¾” bolts

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• Solution: Possible sections are abc, and abde.

• Hole diameters to be subtracted =3/4+1/8=7/8 in.

• Net width:

– Abc: 7-7/8 = 6.125 in.

– Abde: 7-2(7/8)+42/(4x3)=6.58 in.

• Therefore the section abc controls, net width=6.125

• Net area=An=6.125(1/4)=1.53 in2

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• Example: Design bolts configuration by

optimizing the failure mechanism, using the

previous example.

• Change the value of, S, to make the net width

of abc, the same as the net width of abde.

• 6.125=7-2(7/8)+S2/(4x3) then solve for

S=3.24 inch

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Net Area

• In General

– bn=b – Σ dh + Σ (s2/4g)

b dh

s

g

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Net Area

• Where

– bn = net width

– b = gross width

– dh = width of hole ( diameter)

– s= pitch spacing

– g = gage spacing

For members of uniform thickness, t, then

An = bn.t

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Net Area

• For members of non-uniform thickness, the net

area can be calculated as follows:

• An= Ag – Σ dh.t + [Σ (s2/4g)] . t

• The critical net area is the net area having the

least value. It is obtained by checking all

possible failure paths

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Net Area, Example

• Compute the net area of the 7/8 X 12 plate

shown. The holes are for 3/4 –in bolts

3 in

3 in

3 in

3 in

1.5-in

¾ - in. bolts 7/8 X 12 –in plate

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Net Area, solution

3 in

3 in

3 in

3 in

1.5-in

A

B

C

D

E

=s

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Net Area, Solution

• Net width

– ABCD = 12 – (2)(3/4+1/8)=10.25 in.

– ABECD = 12 – (3)(3/4+1/8) + (2)[1.52/(4x3)]=9.75

in

– The 9.75 is the lowest and it governs

Net area = (9.75)(0.875)= 8.53 in2

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Effective Net Areas

• When a member other than a flat plate or bar is loaded in axial tension until failure occurs across its net section, its actual tensile failure stress will be probably be less than the coupon tensile strength of the steel, unless all of the various elements which make up the section are counted so that stress is transferred uniformly across the section

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• In the transition region the stress in the connected part of the member may very well exceed Fy, and go into the strain – hardening range. Unless the load is reduced, the member may fracture prematurely. The farther we move from the connection, the more uniform the stress becomes. In the transition region, the shear transfer has “lagged” and the phenomenon is referred to as “shear lag”

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• In such a situation, the flow of tensile stress between the full member cross section and the smaller connected cross section is not 100 percent effective. As a result, the AISC Specification (D3) states that the effective net area, Ae, of such a member is to be determined by multiplying as area, A, ( which is the net or gross area or directly connected area) by a reduction factor U. The use of a factor such as U accounts for the non-uniform stress distribution, in a simple manner.

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• Ae= An( U )

• The value of the reduction coefficient, U, is affected by the cross section of the member and by the length of its connection.

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The area effective in resisting tension can be appreciably increased by shortening the width of the unconnected leg and lengthening the width of the connected leg, as shown

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• One measure of the effectiveness of a member such as an angle connected by one leg is the distance, X, measured from the plane of the connection to the centroid of the area of the whole section. The smaller the value of X, the larger is the effective area of the member, and thus the larger is the member’s design strength.

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Bolted Members • Should a tension load be transmitted by bolts, A

equals the net area An of the member and U is computed as follows:

• Ae= An( U )

• Table D3.1 of the AISC 14th ed. Specification provides a detailed list of shear lag factors U for different situations pp. 16.1.28.

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Example

• Determine the LRFD tensile strength and the ASD allowable tensile design strength for a W12X30 with two lines of ¾ -in diameter bolts in each flange using A572 Grade 50 steel, with Fy

= 50 Ksi, and Fu= 65 ksi, and the AISC Specification. There are assumed to be at least three bolts in each line 4 in. on center, and the bolts are not staggered with respect to each other

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• Solution

• Using W12x30 (Ag= 8.79 in2, d= 12.3 in, bf= 6.52, tf= 0.440 in)

• Nominal or available tensile strength= Pn= FyAg = (50 ksi)(8.79 in2)=439.5 k

• A.) Gross section yielding

• LRFD with φt=0.9, φt Pn=0.9(439.5)=395.55 k

• ASD: Ωt= 1.67, Pn/ Ωt=438.5/1.67=263.2 k

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• B.)Tensile rupture strength

• An = 8.79- (4)(3/4+1/8)(0.44)=7.25 in2

• Referring to tables in Manual for one half of W12X30 (WT6X15)m we find

• bf=6.52 < (2/3) d= (2/3)(12.3)=8.2 in

• U from Table D3.1 case 7 is 0.85

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• Ae= UAn =0.85(7.25)=6.16 in2

• Pn= FuAe=(65)(6.16)=400.56 k

• LRFD: φt=0.75, φt Pn=0.75(400.56)=300.4 k

• ASD: Ωt= 2.00, Pn/ Ωt=400.56/2=200.3 k

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Example • Determine the LRFD design strength and the

ASD allowable strength for an A36 (Fy=36 ksi, Fu= 58 ksi) L6X6X3/8 in that is connected at its ends with one line of four 7/8 in diameter bolts in standard holes 3 in on center in one leg of the angle

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• Solution:

• Using an L6x6X3/8 (Ag=4.38 in2, y=x=1.62 in) nominal or available strength of L

• Pn= FyAg= (36 ksi)(4.38 in2)= 157.7 k

• A.) Gross section yielding

• LRFD: with φt=0.9, φt Pn=0.9(157.7 k) =141.9k

• ASD: Ωt= 1.67, Pn/ Ωt=157.7/1.67=94.4 k

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• B.) Tensile rupture strength

• An = 4.38 in2-(1)(7/8+1/8)(3/8)=4.00 in2

• Length of connection = L= 3(3 in) = 9 in.

• U=1-x/L=1-1.62/9=0.82

• From case 8 AISC 14th manual table D3.1 for 4 or more fasteners in the direction of loading, U= 0.80. Use calculated U=0.82 (if U is calculated per case 2 Table D3.1, the larger value is permitted to be used)

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• Ae= AnU=(4.00 in2)(0.82)=3.28 in2

• Pn=FuAe=(58 ksi)(3.28 in2)= 190.2 k

• LRFD: φt=0.75, φt Pn=0.75(190.2)=142.6 k

• ASD: : Ωt= 2.00, Pn/ Ωt=190.2 / 2 =95.1 k

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Welded Members • When tension loads are transferred by welds, the rules

from AISC Table D3.1, that are to be determine values for A and U ( Ae as for bolted connections = AU) are follows:

• 1.) Should the load be transmitted only by longitudinal welds to other a plate member, or by longitudinal welds in combination with transverse welds, A is to equal the gross area of the member Ag.

• 2.) Should a tension load be transmitted only by transverse welds, A is to equal the area of the directly connected elements and U is equal 1.0, and An = area of the directly connected element ( Case 3 in Table D3.1)

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• 3.) Tests have shown that when flat plates or bars connected by longitudinal fillet welds are used as tension members, they may fail prematurely by shear lag at the corners if the welds are too far apart. Therefore, the AISC Specification states that when such situation are encountered, the length of the welds may not be less than the width of the plates or bars. The letter A represents the area of the plate, and UA is the effective net area. For such situation the values of U to be used ( AISC 14th ed. Pp.16.1.28 Specification Table D3.1, Case 4) are as follows

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L

w

L ≥ 2 W use U= 1.0 2W > L ≥ 1.5 W use U= 0.87 1.5W > L ≥ W use U= 0.75

Example • The 1x8 in plate shown in connected to 1x12 in plate with

longitudinal filler welds to transfer a tensile load. Determine the LRFD tensile design strength and the ASD allowable tensile strength of the member if Fy=50 ksi and Fu = 65 ksi

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Longitudinal fillet welds L=10 in

W=8 in

PL1X8 PL 1x12

• Solution : Considering the nominal or available tensile strength of the smaller PL 1x8

• Pn= FyAg= (50 ksi)(1x8) = 400 k

• A.) Gross section yielding

• LRFD: with φt=0.9, φt Pn=0.9(400 k)= 360 k

• ASD: : Ωt= 1.67, Pn/ Ωt=400/1.67 = 239.5 k

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• B.) Tensile rupture strength

• 1.5 W = 1.5 (8) 12 in > L = 10 > w = 8 in

• U= 0.75 from Table D3.1 Case 4

• Ae= AU= (8 in2)(0.75)=6 in2

• Pn= Fu Ae= (65 ksi)(6 in2)= 390 k

• LRFD: φt=0.75, φt Pn=0.75(390)=292.5 k

• ASD: Ωt= 2.00, Pn/ Ωt=390/2.00=195 k

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Example • Compute the LRFD design strength and ASD allowable strength

of the angle shown. It is welded on the ends and sides of the 8-in leg only. Fy = 50 ksi and Fu = 70 ksi

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8 in

6 in

L8x6x3/4 X=1.56 in, A=9.9 in2

• Solution: Nominal or available tensile strength of L = Pn= FyAg= (50Ksi)(9.94 in2)= 497 k

• A.) Gross section yielding

• LRFD: with φt=0.9, φt Pn=0.9(497)=447.3 k

• ASD: : Ωt= 1.67, Pn/ Ωt=497/1.67= 297.6

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• B.) Tensile rupture strength ( As only one leg of L is connected, a reduced effective area needs to be computed)

• U=1-x/L=1-1.56/6= 0.74

• Ae= AgU=(9.94)(0.74)= 7.36 in2

• Pn= FuAe= (70)(7.36)=515.2 k

• LRFD: φt=0.75, φt Pn=0.75(515.2)=386.4 k

• ASD: Ωt= 2.00, Pn/ Ωt=515.2/2.0=257.6 k

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Slenderness Ratio

• The AISC steel design manual specification D1 lists a preferred (but not required) maximum slenderness ration (SR) of 300. Rods and wires are excluded from this recommendation,

• SR=l/r

• l= un-braced length

• r= radius of gyration = sqrt(I/A)

• I= moment of inertia

• A= cross-sectional area

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Block Shear

• The formulae used by LRFD (Φt.Pn) and ASD

(Pn/Ωt) to calculate the allowable strengths of

tension members are not always the controlling

criteria. The allowable strength in tension may

be controlled by block shear strength, where

the failure may occur along a path involving

tension in one plane and shear on a

perpendicular plane. So it is possible for a

block of steel to tear out.

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Block Shear The cross-hatched parts may tear out

Bolted angle

Bolted W

Section

Shear plane Tension plane

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Block Shear • The AISC specifications (J4.3) states that the

block shear design strength of a particular member is to be determined by

– Computing the tensile fracture strength on the net section in one direction and adding to that value the shear yield strength on the gross area on the perpendicular segment

– Computing the shear fracture strength on the gross area subject to tension and adding it to the tensile yield strength on the net area subject to shear on the perpendicular segment. The expression to apply is the one with the larger rupture term

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Block Shear

• The AISC specification (J4.3) states that the available strength, Rn, for the block shear rupture design strength is as follows:

• Rn=0.6FuAnv+UbsFuAnt ≤ 0.6FyAgv+UbsFuAnt (AISC 14th ed. Eq. J4-5,pp. 16.1-129 )

• Φ =0.75 (LRFD), Ω = 2.00 (ASD)

• In which

– Anv =net area subjected to shear, in2 (mm2)

– Ant = net area subjected to tension, in2 (mm2)

– Agv = gross area subjected to shear, in2 (mm2)

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Shear Block • To account for the fact that stress distribution may

not be uniform on the tensile plane for some connections, AISC introduced a reduction factor, Ubs. Should the tensile stress distribution be uniform, Ubs. = 1.0 according to AISC specification (J4.3) (i.e. gusset plates, single-row beam connection,…). For non-uniform stress tensile stresses, Ubs. =0.5 (i.e. multiple-row beam end connection,…)

• Should the block shear strength of a connection be insufficient, it may be increased by increasing the edge distance and/or the bolt spacing.

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Block shear, Example

Determine the block shearing strength

(LRFD, ASD), W12X45 (A242 Grade

50 steel and 7/8-in bolts

2-in 3-in 3-in

5.5-in

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Block shear, Solution

Using a W12X45 (tf=0.575-in, bf=8.05-in)

Agv=(4)(8)(0.575)=18.4 in2 (2 each flange)

Anv=(4)[8-2.5(7/8+1/8)](0.575)=12.65 in2 (2.5 bolts each of the 4

sides)

Ant=(4)[1.275-(.5)(7/8+1/8)](0.575)=1.78 in2 (1/2 bolt in tension)

8-in

1.275-in

5.5-in

1.275-in

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• Ubs = 1.0 uniforn tensile stress

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

• Rn= 0.6(70)(12.65)+1.0(70)(1.78) = 655.9 k

• <0.6(50)(18.4)+1.0(70)(1.78)=676.6 k

• Therefore Rn =655.9 k

• LRFD

– Φ. Rn = (0.75)(655.9)=491.9k

ADS

Rn/Ω=655.9/2=327.9k

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60

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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Built-up Tension Member

• Section D4 and J3.5 (AISC 14th ed. Pp. 16.1-122) of the AISC Specification provides a set of definite rules describing how the different parts of built-up tension members are connected together.

• 1.) When a tension member is built up from element in continuous contact with each other, such as plate and a shape, or two plates, the longitudinal spacing of connectors between those elements must not exceed 24 times the thickness of the thinner plate or 12 in if the member is to be painted, or if it is not to be painted and not to be subjected to corrosive conditions.

• 2.) Should the member consist of unpainted weathering steel elements in continuous contact and be subject to atmospheric corrosion, the maximum permissible connectors spacings are 14 times the thickness of the thinner plate, or 7 in.

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• 3.) Should a tension member be built up from two or more shapes separated by intermitted fillers, the shapes preferably should be connected to each other at intervals such that the slenderness ratio of the individual shapes between the fasteners does not exceed 300.

• 4.) The distance from the center of any bolts to the nearest edge of the connected part under consideration may not be larger than 12 times the thickness of the connected part, or 6 in.

• 5.) For elements in continuous contact with each other, the spacing of connectors are given in Sections J3.3 through J3.5 of the AISC Specification

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Example

• The two C12X30s shown in the figure below have been selected to support a dead tensile working load of 120 k and a 240 k live working load. The member is 30 ft long, consisting of A36 steel, and has one line of three 7/8 bolt in each channel flange 3 in. on center. Using the AISC Specification, determine whether the member is satisfactory and design the necessary tie plate. Assume centers of holes are 1.75 in from the backs of the channels.

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• Solution: Using C12X30s ( Ag = 8.81 in2 each, tf = 0.501 in, Ix = 162 in4 each, Iy = 5.12 in4 each, y-axis 0.0674 from back of C, ry= 0.762 in)

• Loads to be resisted:

– LRFD: Pn = 1.2(120)+(1.6)(240) = 528 k

– ASD: Pn= 120+240 = 360 k

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• A.) Gross section yielding

• Pn = FyAg=(36)(2x8.81)=634.2 k

– LRFD: with φt=0.9, φt Pn=0.9(634.2)=570.9 k

– > 528 k OK

– ASD: Ωt= 1.67, Pn/ Ωt=634.2/1.67= 379.8 k

– >360 k OK

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• B.) Tensile rupture strength

• An= 2(8.81-(2)(7/8+1/8)(0.501))=15.62 in2

• U=1-x/L=1-0.674/(2x3)=0.89

• Pn=FuUAn=(58)(15.62)(0.89)=806.3 k

– LRFD: φt=0.75, φt Pn=0.75(806.3)=604.7 k

– >528 k OK

– ASD: Ωt= 2.00, Pn/ Ωt=806.3/2=403.1 k

– > 360 k OK

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• Slenderness ratio

• Ix= 2(162)=324 in4

• Iy= 2(5.12)+2(8.81)(5.326)^2=510 in4

• Rx=sqrt(324/17.62)=4.29 in <ry=sqrt(510/17.62)=5.38 in

• Lx/rx=(12x30)/4.29= 83.9 < 300 OK

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• Design of tie plates (AISC Specification D4) • Distance between lines of bolts = 12.00 – 2(1 ¾) = 8.50 in • Minimum length of tie plates = (2/3)8.5)= 5.67 in (say 6 in) • Minimum thickness of tie plate = ( 1/50 ) (8.50) = 0.17 in (

say 3/16) • Minimum width of tie plates = 8.5+2(1 ½)= 11.5 in (say 12

in) • Maximum preferable spacing of tie plates • Least r one C = 0.762 in • Maximum preferable L/r = 300= 12L/0.762 • L=19.05 ft • Use 3/16x6x1 ft 0 in tie plate 15 ft 0 in on center

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Exercise

• A tension member is to consist of four equal leg angles, arranged as shown in the accompanying illustration to support the service loads, PD = 180 k and PL = 320 k. The member is assumes to be 24 ft long and is to have one line of three ¾ in bolts in each leg. Design the member with 50 ksi steel, Fu = 65 ksi, and including the necessary tie plate. Neglect block shear.

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105

Compression Members

Moayyad Al Nasra, Ph.D, PE

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Failure Mechanism

• Flexural bucking, large L/r

• Local buckling, non-compact

• Torsional bucking, non-uniform

• Yielding, very short column

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Bucking depends on

• l/r

• Connection

• Eccentricity

• Material imperfection

• Initial crookedness

• Residual Stresses

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Residual Stresses

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Euler Formula

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• PCr=π2E.I/L2

………………………………………………(1)

• I=Ar2 ………………..……………………(2)

• P/A=Fe= π2E/(L/r)2 ......................................(3)

• Where

• Fe= Critical stress = Critical force/area

• l/r = Slenderness ratio < 200 for columns

• r= radius of gyration =

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• Example : Find the bucking load of W10X30

steel column if

– L= 18 ft

– L= 9 ft

– L= 4.5 ft

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Solution • W10X30 (A=8.84 in2, rx= 4.38 in., ry= 1.37 in.)

• Min r=ry=1.37 in. • L/r=(18x12)/1.37= 157.7 • Fe= π2(29,000 ksi)/(157.7)2 = 11.5 ksi, • Pcr= 11.5(8.84) = 101.8 k

• L/r= (9x12)/1.37=78.8 • Fe= π2(29,000 ksi)/(78.8)2= 46.1 ksi • Pcr=46.1(8.84) = 407.1 k

• L/r= 4.5x12/1.37=39.4 • Fe= π2(29,000 ksi)/(39.4)2= 184.2 ksi • Pcr=184.2(8.84)=1628.6 k

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Effective Length

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• The effective length is the length between the

two inflection points ( points of zero moments)

• K= effective length factor, depends on the joint

condition. (See AISC steel design manual 14th

edition pp. 16.1.511, Table C-A-7.1)

• In practice there is no perfect hinge nor perfect

fixed joint, AISC introduced design values of

K.

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LRFD Column Formulas Compressive Strength for Flexural Buckling of members without Slender Elements

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• Pn=Fcr.Ag (AISC 14th ed., E3-1, pp. 16.1-33) • Pu=ΦPn=ΦcFcrAg, • Φc =0.90 • The flexural bucking stress, Fcr, is determined as follows:

– When Kl/r≤ 4.71 .(sqrt(E/Fy)), (or Fy/Fe≥2.25) – Fcr=[0.658(Fy/Fe)]Fy …………….AISC E3-2 PP.16.1.33) – When Kl/r >4.71 .(sqrt(E/Fy)), (or Fy/Fe<2.25) – Fcr=0.877 Fe …………………….AISC E3-3, PP.16.1.33

• • Fe= π2E/(Kl/r)2 • Pn= nominal axial load • Pu= Design factored load • ΦcFcr=design stress for axial member (Table 4.22, pp.4.322,

AISC 14th edition)

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• Example: Find the design strength of 12 –ft hinged W10x30 column. Use Fy=50 ksi

• Solution • W10x30 ( A=8.84 in2, rx= 4.38 in., ry= 1.37 in.) • K=1.0 for hinged columns • Min r = ry=1.37 • Kl/r=1(12x12)/1.37=105.1 • Kl/r=105.1 < 4.71 (sqrt(29,000/50)) =113 • Fcr=[0.658(Fy/Fe)]Fy • • Fe= π2E/(Kl/r)2= π2(29,000 ksi)/(105.1)2= 25.91 ksi • Fy/Fe=50/25.91= 1.93 • Fcr= [0.6581.93](50)=22.29 ksi • Pu=0.9(8.84)(22.29)=177.38 K • • OR • From the design Tables , AISC 14th edition, PP. 4.324 and for Kl/r =105.1 • ΦcFcr=20.1 ksi • Pu= ΦcFcrA=20.1 (8.84)=177.68

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OR From the design Tables , AISC 14th edition, PP. 4.324 and for Kl/r =105.1 ΦcFcr=20.1 ksi Pu= ΦcFcrA=20.1 (8.84)=177.68

Available Critical Stress for Compression Members, AISC Table 4-22, Fy=50 ksi, pp.4-324

Kl/r ASD, ksi LRFD, ksi

Fcr/Ωc ΦcFcr

104 13.6 20.4

105 13.4 20.1

106 13.2 19.8

119 (c) Al Nasra

• Exercise

• Determine the LRFD design strength for the

15 –ft hinged W10X45 column. Use LRFD

specifications and a steel with Fy=50 ksi.

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• Solution

• W10x45 (A=13.3, rx=4.32, ry=2.01)

• Kl/r= (1.0)(15x12)/2.01=89.55

• From Table 4.22 – pp.4.318

• ΦcFcr=25.05 ksi

• Pu= ΦcFcrA=25.05(13.3)=333.2k

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Example

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Solution

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124

• C15x50 ( A=14.7 in2, Ix=404 in4, Iy=11.0 in4

• A=(3/4)(18)+(2)(14.7)=13.5+29.4=42.9 in2

• Y=[(13.5)(0.375)+(2)(14.7)(8.25)]/42.9=5.77 in

• Ix=(2)(404)+2(14.7)(2.48)2+(13.5)(5/393)2+(1/12)(18)(3/4)3=1382 in4

• Iy=(1/12)(3/4)(18)3+2[11.0+(14.7)(4.799)2]=1063.6 in4

• ry=sqrt(1063.6/42.9)=4.98 in

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125

• KL/r=(12)(20)/4.98 = 48.19

• Referring to AISC Table 4-22

– LRFD : φcFcr=37.94 ksi, φcPn=(37.94)(42.9)=1628k

– ASD: Fcr/Ωc =25.26 ksi, Pn/Ωc =(25.26)(42.9)=1084 k

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126

• Exercise

• Determine the LRFD design strength for the 15

–ft hinged W10X45 column. Use LRFD

specifications and a steel with Fy=50 ksi.

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127

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128

• Solution

• W10x45 (A=13.3, rx=4.32, ry=2.01)

• Kl/r= (1.0)(15x12)/2.01=89.55

• From Table 4.22 – pp.4.318

• ΦcFcr=25.05 ksi

• Pu= ΦcFcrA=25.05(13.3)=333.2k

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129

130

Design of Columns, LRFD – Axially Loaded

Moayyad Al Nasra, PhD, PE

131 (c) Al Nasra

Trial and Error Procedure:

• Calculate the factored axial load, Pu, – Pu= 1.4 PD – Pu= 1.2 PD + 1.6 PL – ….

• Assume (KL/r) ( ~ 50 for common columns) • Find ΦcFcr , form Table 4.22 corresponding to the assumed KL/r • Calculate required area Arequired=Pu/ ΦcFcr • Select a member and calculate the larger of (KL/r)x, or (KL/r)y • Find, ΦcFcr, from table 4.22 (AISC 14th edition), corresponding to the

calculated (KL/r) in step 5 • Calculate ΦcPn=Pu= ΦcFcrA and compare with the required Pu

– If Pu calculated is slightly larger than Pu required The section is satisfactory – If Pu calculated is less than the Pu required select larger section and go to

step 5.

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• Example : Select the lightest available W14 section to support the axial compression loads PD =150 k and PL 200 k if KL 14 ft and A572 steel grade 50 is used.

• • Solution : • Pu= 1.2(150)+1.6(200) = 500 k • Assume KL/r = 50 • ΦcFcr from AISC Table 4.22 = 37.5 • Arequired= Pu/( ΦcFcr) =500/37.5 =13.33 in2 • Try W14 X 48 ( A=14.1 in2, rx=5.85 in., ry = 1.91 in.) • (KL/r)y= 12(14)/1.91=87.96 • ΦcFcr= 25.52 ksi from AISC Table 4.22, pp. 4-324 • ΦcPn= 25.52(14.1)= 359.8 k <500 k ………………..N.G. • Try W14X61 (A=17.9 in2, ry= 2.45 in.) • (KL/r)y= 12x14/2.45 = 68.57 • ΦcFcr= 31.93 ksi • ΦcPn= 31.93(17.9)= 571.5 k > 500 k …………………………… O.K. • Subsequent check of W14x53 shows it will not do • Use W14X61

133 (c) Al Nasra

Available Critical Stress for Compression Members, AISC 14th Ed. Table 4-22, Fy=50 ksi

Kl/r ASD, ksi LRFD, ksi

Fcr/Ωc ΦcFcr

65 22.0 33.0

70 20.9 31.4

75 19.8 29.8

80 18.8 28.2

85 17.7 26.5

90 16.6 24.9

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LRFD Design Tables

• Procedure

– Calculate Pu, required

– Calculate KL , for the weak axis

– Select a section with Pu calculated > Pu required

135 (c) Al Nasra

• Example same as the previous one

– Pu= 500 k

– KL = 14 ft

– Select W14x61 ( AISC 14th ed. page 4-16)

Available Strength in Axial Compression, Kips, W- shapes, AISC 14th ed. Table 4-1, Fy=50 ksi, PP. 4-16

Shape W14x

Wt/ft 74 68 61 53

Design, LRFD ΦcPn ΦcPn ΦcPn ΦcPn

KL, ft 13 735 671 599 433

14 701 640 571 401

15 667 608 543 369

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Example • Select the lightest available W12 section, using

both the LRFD and ASD methods for the following conditions: Fy=50 ksi, PD=250 k , PL=400 k, KxLx=26 ft and KyLy = 13 ft

– By Trial and error

– By Using AISC tables

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• Solution: Using trial and error to select a section, using the LRFT expressions, and then checking the section with both the LRFD, and ASD methods.

• LRFD – Pn= (1.2)(250)+(1.6)(400)=940 k – Assume KL/r= 50 – φcFcr==37.5 ksi (AISC Table 4-22, pp 4-323) – A, required = 900/37.5 24 in2

– Try W12X87 (A=25 in2, rx=5.38 in, ry=3.07 in) – (kl/r)x=(12)(26)/5.38=57.99 Governs – (KL/r)y=(12)(13)/3.07=50.81 – φcFcr= 35.2 ksi (Table 4-22) – φcPn=(35.2)*25.6)=901 K<940 k N.G – Try W12x96, φcPn=994 k> 940 OK

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138

• ASD – P=250+400 650 k

– Assume KL/r= 50

– Fcr/Ωc=24.9 ksi (AISC Table 4-22, pp 4-323)

– A, required= 650/24.9 = 26.10 in2

– Try W12x87 (A= 25.6 in2, rx=5.38 in, ry=3.07 in)

– (KL/r)x=(12)(26)/5.38=57.99 Governs

– (KL/r)y=(12)(13)/3.07=50.81

– Fcr/Ωc=23.4 ksi (Table 4-22)

– Pn/Ωc=(23.4)(25.6)=599 k< 650 k N.G.

– Try W12x96, Pn/Ωc=662 k >650 k O.K.

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139

Axially Loaded Compression Members

Sidesway inhibited: there is something present other than just columns and girders to prevent sidesway.

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140

• Sideway Uninhibited: Resistance to horizontal translation is supplied only by the bending strength and stiffness of the girders and beams of the frame in question.

• Rotational Stiffness = ratio of the sum of the column stiffness to the girder stiffness – G=Σ(column stiffness)/ Σ(Girder stiffness) =

Σ(I/L)c/Σ(I/L)g

– For pinned column G=infinite. It is recommended that G be made equal to 10 where such non-rigid supports are used.

– For rigid connection, G=0, but from a practical standpoint use G=1.0

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141

K Factor • To determine K value,( AISC 14th ed. pp 16.1-

512, and pp 16.1.-513)

– Select the appropriate alignment chart (sidesway inhibited, or sidesway uninhibited)

– Compute G at each end of the column and label the values GA and GB as desired.

– Draw a straight line on the chart between the GA and GB values, and read K where the line hits the center K scale.

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142

Alignment Charts – Sidesway Inhibited

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GA GB K

0.0

50

0.5

0.5

0.7

1.0

0.0

0.5

50

Alignment Charts – Sidesway Uninhibited

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GA GB K

0.0

100

4.0

0.5

2.0

20.0

0.0

4.0

100

Stiffness Reduction Factor • The Alignment Charts were prepared with the

assumption of elastic failure. The chart K values are too conservative and should be corrected by introducing a reduction factor- Stiffness Reduction Factor, SRF, τa=Fcr,inelastic/Fcr,elastic≈(Pu/A)/Fcr,elastic and ≈ (Pa/A)/Fcr,elastic (Table 4.21 AISC 14th Manual, pp 4-321)

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145

Inelastic Buckling • A direct design method for considering inelastic buckling is

presented in the manual. It involves the following steps: – Calculate Pu or Pa and select a trail size – Calculate Pu/Ag or Pa/Ag and pick the SRF τa from Table 4.21 AISC

Manual, pp 4-321. If the Pu/A, or Pa/A is less than the values given in the table, the column is in the elastic range and no reduction needs to be made.

– The value of G, elastic is computed and multiplied by the SRF and K is picked from the chart

– The effective slenderness ratio KL/r is computed and φcFcr or Fcr/Ω is obtained from the manual and multiplied by the column area to obtain Pu or Pa. If this value is appreciably different from the value computed in step 1, another trial column size is attempted and the four steps are repeated.

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146

Example

• Determine the effective length factors for columns EF, FG and KL of the frame shown, assuming that the frame is subject to sidesway and that all of the assumption on which the alignment charts were developed are met.

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148

• K Factors

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149

Column GA GB K

EF 10.00 (2)(69.42)/183.33= 0.76

1.84

FG (2)(69.42)/183.33= 0.76

(69.42+49.75)/183.3= 0.65

1.26

KL (61.67+39.58)/83.3=1.22

39.58/27.33=1.45 1.44

Example • Select a W14 section for column AB in the frame

shown if PD= 250 K, PL = 500 k, and Fy= 50 ksi, and only in-plane behavior is considered. Furthermore, assume that the column immediately above and below AB are approximately the same size as AB, and also that all the other assumption on which the alignment charts were developed are met. What if inelastic behavior is considered.

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150

• AB

• W21x68 W21x68 14 ft

• W21x68 W21x68 14 ft

• 28 ft 28 ft 28 ft 14 ft

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151

Solution • LRFD, Pu=(1.2)(250)+(1.6)(500)=1100 k • ASD, Pa =250+500 = 750 k • A.) Elastic Design

– Beams are W21X68 ( Ix=1480 in4) – Assume KL= 14 ft for columns – Try W14x99 ( A= 29.1 in2, Ix= 110 in4, rx/ry= 1.66) – GA=GB= (2)(110/14)/[(2)(1480/28)]=1.50 – K= 1.48 from sidesway uninhibited charts – Equivalent KyLy=(KxLx)/(rx/ry)=1.48x14/1.66=12.48 ft – LRFD: use W14x99, φcPn=1160 k >1100 k OK (note that 1160 is

an interpolation from Table 4-1, AISC 14th ed. pp. 4-15, 1150 and 1170)

– ASD: Use W14x99, Pn/Ωc = 774 k > 750 k OK (774k is an interpolation from Table 4-1 pp 4-15, 767 and 781)

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152

• B.) Inelastic Design

– Try W14x90 (A=26.5 in2, Ix= 999 in4, rx/ry=1.66)

– LRFD: Pu/A= 1100/26.5=41.51 ksi

– SRF from AISC 14th ed. Table 4-21, pp 4-321 = 0.564 (interpolated between 0.538 and 0.590)

– GA=GB=[(2)(999/14)]/[(2)(1480/28)](0.564)=0.761

– K= from sideway uninhibited charts = 1.42

– KxLx=(1.42)(14)=20

– Equivalent KyLy=KxLx/(rx/ry)=20/1.66=12 ft

– Use W14x99 (1170>1100)

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• ASD: – Pa/A=750/26.5=28.30 ksi

– SRF from AISC Table 4-21, pp 4-317=0.302

– GA=GB=[(2)(999/14)]/[(2)(1480/28](0.302) =0.408

– K from Sidesway uninhibited charts = 1.24

– KxLx=(1.24)(14)=KxLx/(rx/ry)=20/1.66=12 ft

– Use 14x99 (780 > 750, AISC 14th ed page 4-15)

– Note Stiffness reduction factor, changed from the previous edition of AISC, that made ASD and LRFD results converge

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154

Design of Base Plates, Concentrically Loaded Columns

155 (c) Al Nasra

• A1=BN = area of the base plate

• Pp=P/A1= Pressure on the concrete foundation

• Usually the area of the concrete foundation “A2” is larger than the area of the base plate A1.

• LRFD allows reduction in the design value of A1 if A2>A1 by a factor

– Sqrt(A2/A1)≤2 for less than full area of concrete support

•

156 (c) Al Nasra

Base Plate Design

• Nominal bearing strength Pp is determined as

– Pp= (0.85 fc’A1) for full area of concrete support

– Pu= ΦcPp=Φc(0.85 fc’A1)

– LRFD equation J8-1, AISC 14th ed. Pp 16.1-132

– Φc=0.65 for bearing on concrete (LRFD)

– Ωc=2.31 (ASD)

(c) Al Nasra 157

Base Plate Design

• Nominal bearing strength Pp is determined as

– Pp= (0.85 fc’A1) sqrt(A2/A1) )≤1.7 fc

’A1

– for less than full area of concrete support

– Sqrt(A2/A1)≤2

– Pu= ΦcPp

– LRFD equation J8-2, AISC 14th ed. Pp 16.1-132

– Φc=0.65 for bearing on concrete (LRFD)

– Ωc=2.31 (ASD)

(c) Al Nasra 158

Base Plate Design

• A1= Pu/ (ΦcPp )

• Where

• A1 should not be less than the dimensions of the column

• A1≥ bfd

(c) Al Nasra 159

Base Plate Design

• For efficient dimensions, N and B can be selected such that

• N=√A1+0.5(0.95 d -0.80bf)

• And B=A1/N

(c) Al Nasra 160

Base Plate Thickness

• Plate thickness, t

• tmin=(l)(sqrt(2Pu/(0.9FyBN)) LRFD

• tmin=(l)(sqrt(3.33Pu/(FyBN)) ASD – AISC 14th ed, eqns 14-7a, and 14-7b, pp. 14-6

• Where, – l=max(m,n,λn’)

• m=(N-0.95d)/2

• n= (B-0.8bf)/2

• n’ = sqrt(dbf)/4

(c) Al Nasra 161

Base Plate Thickness

(c) Al Nasra 162

𝛌 =𝟐 𝐗

𝟏+ 𝟏 − 𝐗≤ 𝟏

AISC 14th edition, 14-5, pp. 14-5

X can be obtain from for following (AISC 14th ed. 14-6a

and 14-6b pp. 14-5)

For LRFD

𝐗 = 𝟒𝐝𝐛𝐟

(𝐝+ 𝐛𝐟)𝟐

𝐏𝐮

∅𝐜𝐏𝐩

And for ASD

𝐗 = 𝟒𝐝𝐛𝐟

(𝐝+ 𝐛𝐟)𝟐 𝛀𝐜𝐏𝐮

𝐏𝐩

• Example

• Design a base plate with A36 steel for W12x106 column with a dead working load of 100 k and a working live load of 420 k. The concrete 28-day strength is 4 ksi. Footing size is 12 ft by 12 ft

•

• Solution

• Using W12x106 ( d= 12.9 in., bf= 12.2 in.)

• Pu= 1.2(100)+1.6(420)=792 k

• A2=12x12(12x12)=20,736 in2

163 (c) Al Nasra

Example (cont’d)

• Note that A2 is many times larger than hat will be the base plate area, less than full concrete area will provide support, and sqrt (A2/A1) can be taken as 2 ( max is 2).

• Pp= (0.85 fc’A1) sqrt(A2/A1) )≤1.7 fc

’A1

• Pu= ΦcPp=(0.65)(0.85(4)(2) )A1=792

• Solve for A1, A1=179.2 in2

• Base plate must be larger than the column dimensions=12.9(12.2)= 157.38 in2 OK

• N=sqrt(179.2)+0.5(0.95(12.9)-0.8(12.2))=14.6 in

• Use N=15 in

(c) Al Nasra 164

Example,cont’d

• B=A1/N=179.2/15=11.9 in use 13 in

• Use 15x13 plate

(c) Al Nasra 165

Checking the bearing strength of the concrete

• ΦcPp=Φc(0.85)fcA1(sqrt(A2/A1))

• =(0.65)(0.85)(4.0)(13x15)(2)=862 k

• >792 O.K.

• Pp=862/0.65=1326 k

166 (c) Al Nasra

Thickness Design

• Computing m,n,n’

• m= (N-0.95d)/2=(15-0.95(12.9))/2=1.37 in

• n= (B-0.8bf)/2=(13-0.8(12.2))/2=1.62 in

• n’ = sqrt(dbf)4= sqrt(12.9x12.2)/4 = 3.14 in

• X=((4x12.9x12.2)/(12.9+12.2)2[792/(0.65x1326)]

• X=0.92

• l= max [1.37,1.12,(0.92x3.14=2.88)]=2.88

• tmin=(l)(sqrt(2Pu/(0.9FyBN)) LRFD

167 (c) Al Nasra

Thickness Design, cont’d

• t = 2.88 (sqrt(2x792/(0.9x50x15x13))=1.22 in

• Use 1.25 inch

• Use plate of 13”x15”x1.25” as minimum dimsions

(c) Al Nasra 168

169

Introduction to Beam Design

Moayyad Al Nasra, PhD,PE

(c) Al Nasra 170

• Beams are usually said to be members that support transverse loads.

• Types of beams

– Joists

– Rafters

– Lintels

– Spandrel

– Stringers

– Floor beams

– Girders

(c) Al Nasra 171

Bending Stresses • The flexural formula maw be written as follows:

• fb= MC/I=M/S

(c) Al Nasra 172

• fb Fy Fy Fy • Variation in bending stresses due to increasing moment about x -axis

(c) Al Nasra 173

Elastic Design

• C • 2/3 d • T=(1/2)FY(d/2)b=Fybd/4=C • Fy

• The elastic design theory can be summarized as follows; the maximum load that a structure could support was assumed to equal the load that the first caused a stress somewhere in the structure to equal to the yield stress of the material

(c) Al Nasra 174

Elastic Section Modulus • The resisting moment equals T or C times the

lever arm between them, as follows:

• My=(Fybd/4)(2/3)d=Fybd2/6

• The elastic section modulus can again be seen to equal to S=bd2/6

(c) Al Nasra 175

Plastic Design

•

• d C

• T=Fy(d/2)b=C

• b

• Mp=Mn=T(d/2)=C(d/2)=Fy(bd/2)(d/2)= Fy(bd2/4)

• The plastic section modulus = Z=(bd2/4)

• Mp/My=Z/S=1.5

(c) Al Nasra 176

Example

• Find the values of S and Z, and the shape factor about the horizontal x axis. The

thickness is 1 in, total depth = 18 in, the flange width = 12 in

(c) Al Nasra 177

Solution • Elastic calculations

– A=(2)(12)(1)+(16)(1)=40 in2

– I=(1/12)(12)(18)3-(1/12)(11)(16)3=2077.3 in4

– S=2077.8/9.00=230.8 in3

– My=FyS=(50)(230.8)/12=962 ft-k

(c) Al Nasra 178

• Plastic calculations

– Z=(2)(12)(1)(8.5)+(2)(1)(8)(4)=268 in3 (area by dist. between local centroid and global centoid)

– Mn=FyZ=(50)(268)/12=117 ft-k

– Shape factor

• S.F.=268/230.8=1.16

(c) Al Nasra 179

Theory of Plastic Analysis • Those part of the structure that have been

stressed to the yield stress cannot resist additional stresses, but instead, will yield the amount required to permit the extra load or stresses to be transferred to other parts of the structure where the stresses are below the yield stress, and thus in the elastic range and able to resist increased stress.

(c) Al Nasra 180

Virtual Work Method

• wn k/ft W16x77

• 30 ft

• a aL/2 a

• 2a

• L/2

•

(c) Al Nasra 181

• For W16x77, Zx=150 in3

• The external = internal work

• External work= WnL times the average deflection. The average deflection = ½ the deflection at the center plastic hinge= (1/2 x a X L/2).

• The external work= work absorbed by the hinges or the sum of Mn at each plastic hinge times the angle through which it works.

(c) Al Nasra

182

• Mn(a+2a+a)=WnL(1/2 X a X L/2)

• Mn=wnL2/16

• Wn=16Mn/L2

• Mn = FyZ=(50)(150)/12=625 ft k

• Wn= (16)(625)/302=11.11 k/ft

(c) Al Nasra 183

Location of Plastic Hinge

• wn k/ft

• L

• a 0.586aL 1.414a

• 2.414 a

• 0.586 L

•

(c) Al Nasra

184

Continuous Beam - Example

• Wn k.ft W27x178 (Zx=570 in3)

• 20 ‘ 30 ‘ 20’

• 1.4 a a a a a 1.4 a

• Mn1 Mn2 Mn3

(c) Al Nasra 185

• Mn,1 (3.414 a)=(1/2)(wn)(20)(11.72a)

• Mn,1= 34.33 wn, wn= Mn1/34.33

• Mn,2=(4a)= 30wn(1/2)(15 a)

• Mn,2= 56.25 wn, wn= Mn,2 /56.25

• Mn,3= 3.414 a= (1/2 wn) (20) (11.72 a)

• Mn,3 = 34.33 wn, wn = Mn,3 /34.33

• Using W27x178, the max is in the 2nd span

• Mn = FyZ=(50)(570)/12= 2375 ft k

• Wn= Mn/56.25= 2375/56.25= 42.22 k ft

(c) Al Nasra 186

187

Design of Beams for Moment

Moayyad Al Nasra, PhD, PE

188 (c) Al Nasra

Design of Beams for Moment Full-Lateral Support of the Top Flange

• Procedure

• Calculate the max. Factored moment, Mu

• Calculate the required plastic modulus, “Z”

• Zrequired= Mu/ΦbFy, Φb=0.90

• Select the lightest section from AISC tables

• Check Compactness

189 (c) Al Nasra

190 (c) Al Nasra

Compactness

• A compact section is a section that is capable of developing a fully plastic stress distribution before buckling

• A non-compact section is a section in which the yield stress can be reached in some, but not all, of its compression elements before buckling occurs.

191 (c) Al Nasra

• Example

• Select the lightest W beam to support the following loads

• 4 in. concrete slab

• 20 psf dead load

• 40 psf live load

• Fy= 50 ksi

192 (c) Al Nasra

193 (c) Al Nasra

• Solution • Dead load = 20psf(10-ft)+(4/12 ft)(150 pcf)(10-ft) = 700

lb/ft • Assume beam self weight = 50 lb/ft • Total dead load =750 lb/ft • Live load = 40 (10) = 400 lb/ft • Wu= 1.2(750)+1.6(400) =1540 lb/ft • Mu= wul2/8 = (1540 lb/ft)(20ft)2/8 = 77000 lb.ft =77 k.ft • Zx,required= Mu/ΦbFy = (77 k.ft)(12 in./ft)/(0.90(50)) = 20.53 in3 • Select a section using AISC steel design manual, 14th

edition, LRFD, PP. 1-27 , pick a section with Zx= 21.67 > Zx,

required= 20.53 in3, select W10X19 • PP. 3-77, under beam for 20-ft length Pick W10x19, Load =

32.4k > 1.54 k/ft(20-ft)=30.8 k • PP. 3-132 under beam , locate Mu = 77 k.ft, and unbraced

length lb=0, read up and to the right, read W10x19.

194 (c) Al Nasra

PP. 1-24 , pick a section with Zx= 21.67 > Zx, required= 20.53 in3, select W10X19

W shapes, Dimensions, Table 1-1, AISC, 14th edition, PP. 1-27

Shape Axis X-X

I, in4 S, in3 r, in. Z, in3

W10x26 144 27.9 4.35 31.3

W10x22 118 23.2 4.27 26.0

W10x19 96.3 18.8 4.14 21.6

W10x17 81.9 16.2 4.05 18.7

195 (c) Al Nasra

under beam for 20-ft length Pick W10x19, Load = 32.4 > 1.54(20)=30.8 k

Maximum Total Uniform Loads, Kips, W shapes, Table

3-6, AISC 14 ed., pp. 3-77, Fy=50 ksi,

Shape W10x

22 19 17

Design ASD LRFD ASD LRFD ASD LRFD

Span, ft 19 27.3 41.1 22.7 34.1 19.6 29.5

20 25.9 39.0 21.6 32.4 18.7 28.1

21 24.7 37.1 20.5 30.9 17.8 26.7

196 (c) Al Nasra

PP. 3-132 AISC 14th ed. under beam , locate Mu = 77 k.ft, and unbraced length lb=0, read up and to the right, read W10x19.

Available Moment vs. Unbraced length, W shaped, Table 3-10, AISC, Fy=50 ksi, Cb=1,

PP. 3-132

72

78

84

2 4 6 8

W12x19

W10x19

W12x16

Un-braced length, ft

ΦMn,

Kip-ft

197 (c) Al Nasra

• Exercise

• Select the most economical section, with Fy = 50 ksi, assuming full lateral support for the compression flange. D=1.2 k/ft, L=3.0 k/ft, beam total length = 36 ft.

198 (c) Al Nasra

199 (c) Al Nasra

• Solution

• Assume the beam self weight = 90 lb/ft

• Wu=1.2(1.29)+1.6(3.0)=6.348 k/ft

• Mu=(6.348)(36)2/8= 1028.4 ft.k

• From AISC Table 3-6, pp. 3-46, for load of 6.348(36)=228.5 pick W30x90 of 236>228.5k

• Or from design charts, AISC pp.3-113 Mu=1028 K.ft, lb=0 Use W30x90

200 (c) Al Nasra

201

Beam – Column Design

Moayyad Al Nasra, PhD, PE

(c) Al Nasra 202

Bending and Axial Force • Examples

– Eccentrically loaded columns

– Crooked columns

– Columns subjected to wind and other lateral loadings

– Columns in rigid frames are subjected moments

– Centrifugal effect of traffic on curved bridges

– Many others

(c) Al Nasra 203

Members Subjected to Bending and axial Tension

• For Pr/Pc ≥ 0.2 • Pr/Pc+(8/9) [(Mrx/Mcx)+(Mry/Mcy) ≤ 1.0 • AISC 14th ed. eqn H1-1a, pp. 16.1-73 • And For Pr/Pc < 0.2 • Pr/2Pc + [(Mrx/Mcx)+(Mry/Mcy) ≤ 1.0 • AISC 14th ed. eqn H1-1b, pp. 16.1-733

(c) Al Nasra 204

• In which

• Pr= Required tensile strength

• Pc = Design tensile strength (φc Pn) or allowable tensile strength (Pn/Ωc)

• Mr= required flexural strength

• Mc= design flexural strength (φb Mn) or allowable flexural strength (Mn/φb)

(c) Al Nasra 205

Example • A W14 X 30 tension member with no holes,

consisting of 50 ksi steel, is subjected to the service loads PD = 80 k, and PL = 100 k, and to the service moments MD= 10 k-ft and ML= 12 k-ft. Is the member satisfactory if Lb = 4.0 ft and if Cb = 1.0 ?

(c) Al Nasra 206

Note

– Lb= laterally un-braced length

– Cb= lateral-torsional buckling modification factor for non-uniform moment diagram

– Cb=12.5Mmax/(2.5Mmax+3MA+4MB+3MC)

• AISC 14th ed. Eqn. F1-1, pp. 16.1-46

– Mmax=absolute max moment, MA, MB, and MC= moment at quarter point, centerline and three-quarter point of the un-braced segment, k.in

(c) Al Nasra 207

Note cont’d

• Cb=1.0 for cantilever or overhangs where the free end is un-braced

• For doubly symmetric members with no transverse loading between brace points – Cb=1.0, for the case of equal end moments of

opposite sign (uniform moment)

– Cb=2.27, for the case of equal end moments of the same sign (reverse curvature bending)

– Cb=1.67, when one end moment equal zero.

(c) Al Nasra 208

Solution - LRFD • Using W 14x30 ( Ag=8.85 in2,)

• Lp=limiting lateral un-braced length = 1.76rysqrt(E/Fy)=5.26 ft > 4 ft, AISC F2-5, pp. 16.1-48

• Pr=Pu=(1.2)(80)+(1.6)(100)= 256 k

• Mrx=Mux= (1.2)(10)+(1.6)(12)=31.2 ft-k

• Pc=φtFyAg=(0.9)(50)(8.85)=398.2 k

• Mcx=φbMpx=177 ft k from AISC 14th ed. Table 3-2, pp 3-26

• Pr/Pc=256/398.2= 0.643 > 0.2

• Must use AISC Eqn H1-1a

• Pr/Pc+(8/9) [(Mrx/Mcx)+(Mry/Mcy) =0.643+(8/9)(31.2/177)=0.800 < 1.00 OK

(c) Al Nasra 209

Solution - ASD • Pa=80+100= 180 k = Pr

• Mrx= Max= 10+12= 22 ft-k

• Pc=Pn/Ωc= FyAg/Ωt=(50)(8.85)/1.67=265 k

• Mcx= Max/Ωb=118 ft k form AISC Table 3-2

• Pr/Pc= 180/265 =0.679 > 0.2

• Must use AISC Eqn H1-1a

• Pr/Pc+(8/9) [(Mrx/Mcx)+(Mry/Mcy)= 0.679+(8/9)(22/48)=0.845 <1.00 OK

(c) Al Nasra

210

Example

• Repeat the previous problem if the member is also subjected to the service moments MDY= 10 ft-k and MLY= 5 ft-k

• Using a W14x30 ( A= 8.85 in2, φbMpx=177 ft-k, Mpx/Ωb =118 ft-k, AISC Table 3-2, Lp= 5.26 ft)

(c) Al Nasra 211

Solution - LRFD • Pr=Pu=(1.2)(80)+(1.6)(100)= 256 k • Mrx=Mux= (1.2)(10)+(1.6)(12)=31.2 ft-k • Mry= Muy=(1.2)(10)+(1.6)(5)=20 ft-k • Pc=φtFyAg=(0.9)(50)(8.85)=398.2 k • Mcx=φbMpx=177 ft k from AISC Table 3-2 • Mcy= φbMpy=33.7 k-ft • Pr/Pc=256/398.2= 0.643 > 0.2 • Must use AISC Eqn H1-1a • Pr/Pc+(8/9) [(Mrx/Mcx)+(Mry/Mcy)= • 256/398.2+(8/9)(31.2/172+20/33.7)=1.332 >1.00

NG

(c) Al Nasra 212

Solution - ASD • Pa=80+100= 180 k = Pr • Mrx= Max= 10+12= 22 ft-k • Mry==May=10+5=15 ft=k • Pc=Pm/Ωc= FyAg/Ωt=(50)(8.85)/1.67=265 k • Mcx= Max/Ωb=118 ft k form AISC Table 3-2 • Mcy=May/ Ωb=22.4 ft-k • Pr/Pc= 180/265 =0.679 > 0.2 • Must use AISC Eqn H1-1a • Pr/Pc+(8/9) [(Mrx/Mcx)+(Mry/Mcy)= • 180/256+(8/9)(22/118+15/22.4)= 1.44 >1.00 NG

(c) Al Nasra

213

First and Second Order Analysis • Pnt

• Pnt = axial compression force -1st order

• δ

• Mnt is the first order moment

• Mr=Mnt+Pntδ

• B1= magnification factor to account for the P-δ effect

(c) Al Nasra 214

Magnification Factors • B1= Cm/[1-α(Pr/Pe1)] ≥ 1.0

• α= 1.00 for LRFD and 1.60 for ASD

• Cm= 0.60 -0.4(M1/M2) Modification factor, single curvature

• M1/M2= ratio of the smaller moment to the larger moment at the end of the un-braced length in the plane of bending under consideration

• Pe1 = member’s Euler buckling strength

• Pe1= (π2EI)/(KL)2

(c) Al Nasra 215

• Pnt

• Δ

• B2= a magnification factor to account – for P-Δ effect

– Mr=Mlt+PntΔ

– Mlt=secondary moment due to sidesway

– Mr=B1Mnt+B2Mlt = final moment in a particular member

– Pr=Pnt+B2Plt = Final axial strength

– Pnt = axial compression – 1st order

(c) Al Nasra 216

• B2=1/[1-((αΣPnt)/(ΣPe2))]

• ΣPe2=Σ[(π2EI)/(K2L)2] , All column on that level

• K2L= effective length in the plane of bending, based on a sidesway buckling analysis

(c) Al Nasra 217

Moment Capacities • LRFD

– φbMn=Cb[φbMpx-BF(Lb-Lp)]≤φbMpx

For ASD Mn/Ωb=Cb[Mpx/Ωb-BF(Lb-Lp)] ≤Mpx/Ωb

BFs= bending factors from AISC Manual Cb= (12.5 Mmax/[12.5Mmax+3MA+4MB+3Mc]) Rm ≤3.0 Mmax = largest moment in an unbraced segment of the beam,

while MA, MB, and MC are, respectively, the moments at the ¼ point, ½ point, and ¾ in the segment

Rm= cross section mono-symmetry parameter =0.5 +2(Iyc/Iy)2

Rm=1 for doubly symmetric members Iyc = moment of inertia about y-axis referred to the compression

flange if single curvature bending is present

(c) Al Nasra 218

Example • A W10x54 pin-connected beam=column that is not

subjected to sidesway is 15 ft long. The load Pa= 180 k or PU= 300 K is applied to the column at its upper end, with an eccentricity of 2 in. so as to cause bending about the major axis of the section. Check the adequacy of the member if it consists of 50 ksi steel and is used in a braced frame so that Mltx=Mlty=0 . Ky=1.0 while Kx =1.0. Assume Cb=1.0 and Cmx=Cmy=0.85

• Using a W10x54 ( A=15.8 in2, Ix=303 in4, φbMpx=250 ft-k, Mpx/Ωb= 166 ft-k, Lp=9.04 ft, Lr=33.7 ft for LRFD BF= 3.74, for ASD BF =2.49)

(c) Al Nasra 219

LRFD Solution • Pr=Pu=300 k • Mr=Mu=300(2/12)=50 ft=k • KxLx=KyLy=(1.0)(15)=15 ft • Pc=φcPn=496 k • Pr/Pc=300/496=0.605 >0.2 • Must use AISC Eqn H1-1a • Cm=0.6-0.4M1/M2=0.6-0.4(-50/50)=1.0 • Pe1= (π2EIx)/(KxLx)

2 = (π2(29000)(303)/(12x15)2=2667 k • B1= Cm/[1-α(Pr/Pe1)] =1.0/[1-((1.0)(300)/2667)]=1.13 • Mrx= B1Mr=(1.13)(50)=56.5 ft k • Since Lb=15 >Lp< Lr

• Mcx= φbMnx=1.0[250-3.74(15-9.04)]=227.7 ft k • Pr/Pc+(8/9) [(Mrx/Mcx)+(Mry/Mcy)= 300/496+(8/9)(56.5/227.7)= 0.826

<1.0 • The section is satisfactory

(c) Al Nasra 220

ASD Solution • Pr=Pa=180 k • Mr=Ma=(180)(2/12)=30 ft k • KxLx=KyLy=1.0(15)=15 ft • Pc= Pn/Ωc=330 k • Pr/Pc= 180/330 =0.545 >0.2 • Must use AISC Eqn H1 1a • Cm=0.6-0.4M1/M2=0.6-0.4(-30/30)=1.0 • Pe1= (π2EIx)/(KxLx)

2 =2667 k • B1= Cm/[1-α(Pr/Pe1)] =1.0[1.0-((1.6)(180)/2667)]=1.12 • Mrx= 1.12(30)=33.6 ft k • Since Lb>Lp<Lr

• Mcx= φbMnx=1.0[166-2.49(15-9.04)]=151.2 ft k • Pr/Pc+(8/9) [(Mrx/Mcx)+(Mry/Mcy)=

180/330+(8/9)(33.6/151.2)=0.743<1.0 • The section is satisfactory

(c) Al Nasra 221

Example • A pin-connected W10x49 consisting of 50 ksi

steel is used as a beam-column. If sidesway is prevented and a load Pa= 200 k or Pu= 300 k is applied 2.0 in off center at both ends so to cause a single curvature bending about x axis, is the member satisfactory? Length =12 ft , Kx=1.4, Ky= 1.0

• Using a W10x49( A=14.1 in2, Ix=272 in4, φbMpx=227ft-k, Mpx/Ωb= 151 ft-k, Lp=8.97 ft, Lr=31.6 ft for LRFD BF= 3.67, for ASD BF =2.44)

(c) Al Nasra 222

LRFD - Solution • Pr=Pu= 300 k • Mrx=Mux =(2/12)(300)=50 ft k • KyLy=(1)(12)=12 ft Governs • KxLx=1.4(12)=16.8 • Pc=φcPn=513 k (AISC Table 4-1) • Pr/Pc=300/513=0.585>0.2 • Must use AISC Eqn H1-1a • Cm=0.6-0.4M1/M2=0.6-0.4(-50/50)=1.0 • Pe1= (π2EIx)/(KxLx)

2 = (π2(29000)(272)/(12x12)2=5606k • B1= Cm/[1-α(Pr/Pe1)] =1.0/[1-((1.0)(300)/5606]=1.06 • Mrx= B1Mr=(1.06)(50)=53ft k • Since Lb=12>Lp< Lr

• Mcx= φbMnx=1.0[227-3.67(12-8.97)]=215.9ft k • Pr/Pc+(8/9) [(Mrx/Mcx)+(Mry/Mcy)= 300/513+(8/9)((53/215.9)+0)= 0.803<1.0 • The section is satisfactory

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ASD Solution • Pr=Pa=200 k • Mrx=Max=(200)(2/12)=33.3 ft k • KxLx=KyLy=1.4(12)=16.8 ft • KyLy=(1)(12)= 12 , KxLx/(rx/ry)=16.8/1.71=9.82 • Pc= Pn/Ωc=341k (AISC Table 4-1) • Pr/Pc= 200/341=0.58>0.2 • Must use AISC Eqn H1 1a • Cm=0.6-0.4M1/M2=0.6-0.4(-33.3/33.3)=1.0 • Pe1= (π2EIx)/(KxLx)

2 =5606k • B1= Cm/[1-α(Pr/Pe1)] =1.0[1.0-((1.6)(200)/5606)]=1.06 • Mrx= 1.06(33.3)=35.3 ft k • Since Lb=12>Lp=8,97<Lr=31.1 • Mcx= φbMnx=1.0[151-2.44(12-8.97)]=143.6ft k • Pr/Pc+(8/9) [(Mrx/Mcx)+(Mry/Mcy)= 200/341+(8/9)((35.3/143.6)+0)=0.805<1.0 • The section is satisfactory

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Bolted Connection

Moayyad AL Nasra, PhD, PE

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Types of Bolts • Unfinished bolts, ordinary or common bolts.

Classified by ASTM as A307 bolts

• High-strength bolts. Made from medium carbon heat – treated steel and from alloy steel and have tensile strength two or more times those of ordinary bolts. Commonly available as A325, and A490.

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Advantages of High-Strength Bolts

• Smaller screws

• Fewer bolts needed

• Less training and experience

• Relatively less noisy

• Less cost

• Less hazards

• Better fatigue resistance

• Can be removed easily

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Methods for Fully Pre-tensioning High-Strength Bolts

• Turn-of-the-Nut: Bolts are brought to a snug-tight condition, then are given one-third to one full turn.

• Calibrated Wrench : Bolts are tightened with an impact wrench to the required torque putting the necessary tension in the bolts

• Direct Tension Indicator: Consists of a hardened washer that has protrusions on one face in the form of small arches. The arches will be flattened as a bolt is tightened. The amount of gap at any one time is a measure of the bolt tension

• Alternative Design Fasteners: Bolts with splined ends that extend beyond the threaded portion of the bolts, called twist-off bolts.

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Bearing-Type Connections – Loads Passing Through Center of Gravity of Connections

• Shear Strength: The shear strength of bolts are given in AISC Manual Table J3.2 PP16.1-120, 14th edition 2011.

• Bearing Strength: Based upon the strength of the parts being connected and the arrangements of the bolts. Expressions for the nominal bearing strength (Rn values) at bolts are provided in section J3.10 of the AISC, pp.16.1.127

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• A.) if deformation around bolts holes is a design consideration ( if we want deformation to be ≤ 0.25 in) then

• Rn= 1.2LctFu≤2.4dtFu (AISC 14th ed. Eqn J3-6a, pp. 16.1-127)

• If deformation around the bolt is not a design consideration ( deformation > 0.25 in)

• Rn= 1.5 LctFu ≤ 3.0 dtFu (AISC 14th ed.Eqn J3-6b)

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• B. For bolts used in connections with long-slotted holes, the slots being perpendicular to the forces

• Rn= 1.0 LctFu ≤ 2.0 dtFu (AISC 14th Eqn J3-6c)

• d= bolt diameter

• t= thickness of members bearing against the bolt

• Lc= clear distance between the edge of holes and the edge of the adjacent holes or edges of the material in the direction of the force

• Fu= minimum tensile strength of the connected material

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Example

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2

6

6

2

2 3 3 2 PL 1X 16

PL 1 X 16

• Determine the LRFD design tensile strength and the ASD allowable strength for the member shown, assuming bearing-type connection. Use A325 ¾ inch bolts, threads excluded from shear plane. Use AISC Specifications, standard size holes, members have clean mill-scale surfaces (Class A), Fy= 36 ksi and Fu= 58 ksi, deformation at service loads is a design consideration, and ignore block shear

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LRFD Solution • A.) Gross section Yielding of plates • Pn= FyAg= 36(1x16)=576 k, φt= 0.90 • φtPn=(0.90)(576) = 518.4 k • B.) Tensile Rupture Strength of Plates • An=16-3(3/4 +1/8)(1.0) =13.3 in2 • U=1.0 as all parts connected • Pn= FuAe=(58)(13.375)(1.)=775.75 k • Also φtPn=(0.75)(775.75)=581.8 k , φt= 0.75 • C.) Bearing Strength of Bolts • Lc= lesser of 2-(1/2)(3/4+1/8) or 3-(3/4+1/8)

=1.5625 in

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• Rn=1.2LctFu(no. of bolts)≤2.4dtFu (No. of bolts)

• Rn= (1.2)(1.5625)(1.0)(58)(9)=978.8 k > 2.4(3/4)(1)(58)(9)=939.6 k

• φPn=(0.75)(939.6)=704.7 k, , φ= 0.75

• D.) Shearing Strength of bolts

• Area of the bolt = 0.44 in2

• Fn = 68 ksi nominal shear stress, AISC Table J3.2, PP. 16.1-120 AISC 14th edition

• Number of bolts used in the connection = 9

• Rn=FnAb=68(0.44)(9)=269.3 k

• φRn= (0.75)(269.3)=202 k, φ= 0.75

• The strength is 202 k (c) Al Nasra

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ASD Solution • A.) Gross section Yielding of plates • Pn= FyAg= 36(1x16)=576 k, Ωt= 1.67 • Pn/ Ωt=(576) / 1.67 = 344.9 k • B.) Tensile Rupture Strength of Plates • An=16-3(3/4 +1/8)(1.0) =13.3 in2 • U=1.0 as all parts connected • Pn= FuAe=(58)(13.375)(1.)=775.75 k • Also Pn/ Ωt=(775.75)/2.0=387.9k , Ωt= 2.0 • C.) Bearing Strength of Bolts • Lc= lesser of 2-(1/2)(3/4+1/8) or 3-(3/4+1/8)

=1.5625 in

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• Rn=1.2LctFu(no. of bolts)≤2.4dtFu (No. of bolts)

• Rn= (1.2)(1.5625)(1.0)(58)(9)=978.8 k > 2.4(3/4)(1)(58)(9)=939.6 k

• Pn/ Ω=(939.6)/2.00=469.8k, Ω= 2.0

• D.) Shearing Strength of bolts

• Area of the bolt = 0.44 in2

• Fn = 68 ksi nominal shear stress, AISC Table J3.2, PP. 16.1-120 AISC 14th edition

• Number of bolts used in the connection = 9

• Rn=FnAb=68(0.44)(9)=269.3 k

• Rn/Ω= (269.3)/2.00=134.6 k, Ω= 2.0

• The strength is 134.6 k (c) Al Nasra

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