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Analysis of Tension Members
Moayyad Al Nasra, Ph.D, PE
(c) Al Nasra
1
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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26
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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
