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8/3/2019 Notes on AISC 13th(1)
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CH Subject
9 Fatigue
12 Effective HSS design wall thickness t =
13 Net Area (Ref D3.3)
4 - Limiting b-t ratio (Table B4.1)
C 2 - 2nd Order Analysis Requirement- - Tension
3 3 Effective Net Area (Ref B3.13)
- - Compression
1 (a) W, S, HSS (w/o SLENDER ELEMENT)
(b) All others & Build-up members
3 - Flexural Buckling of members w/o SLENDER
ELEMENT (b/t ratio exceeds noncompact
limit)
4 - Torsional & Flexural-Torsional Buckling of
members w/o SLENDER ELEMENT,
otherwise E7 applies.
- Single Angle
(a) Connecting thru L or S legs to web of
planner elements (web, gusset)
(b) Connecting thru L or S legs to web of Box or
Space Trusses
(c) For conditions other than (a) & (b)
6 - Built-up Members7 - Members w/SLENDER ELEMENTS
Flexural Buckling is the prevalent failure
mode
B 3
E
5
D
Torsional buckling, and flexural buckling limit
state must be checked.
Limit applicable to all members w/o
SLENDER ELEMENT, otherwise E7 applies
See above note.
For angle connecting thru S-Leg, kl/r is to be
increased by a factor.
Applicable to singly symm, unsymmetric,
non-rolled double symm, and built-up
columns except SINGLE ANGLE [see E5]
Section Note
An
Design per App. 3
0.93 tw (ERW); tw (SAW)
Slenderness of element under consideration
of local buckling.
Ae = An*U (U - Shear Lag, see Table D3.1)
8/3/2019 Notes on AISC 13th(1)
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Ag =
Ix =
rx =
Iy =ry =
J =
Cw =
Xo =
Yo =
8/3/2019 Notes on AISC 13th(1)
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AISC 13th E.3 - Compression (Flexural Buckling) Strength of Column w/o Slender Elements
* Verify the column doesn't have slender sections before proceed. Otherwise, select
a column with stronger sections, or use provisions on E.7 to compute Pn.
Col ID :
Shape:
Pr = 450 kipsKx = 1
Ky = 1
Lx = 31 ft
Ly = 31 ft
Properties:
A = 35.1 in2
rx = 7.9 in
ry = 2.69 in
Fy = 50 ksi
KLx/rx = 47.1
Kly/ry = 138.3 KLy/ry Governs
KL/r = 138.3
Fe = 14.96 ksi Fe < 0.44 Fy, cal Fcr per EQ E3-3
Fcr = 13.12 ksi
Pn = 460.5 kips
Pa = 275.8 kips ASD
Pa = 414.5 Kips LRFD
W18x119
Frame Beam
8/3/2019 Notes on AISC 13th(1)
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AISC 13th E.5 - Single Angle Compression Members
E.5(a) - Conneted through long leg to individual members or gusset/chord of a planner truss.
(Note, in a truss, the adjacent web member shall be connected to the same side of
gusset/chord.)
Shape: Fy = 36 ksiL = 12 ft
Ag = 2.86 in2
Note:
rx = 1.23 in rx = radius of gyration about geometric axis parallel
L/rx = 117.1 to connected leg (either rx or ry on table)
Case 1: Per EQ E5-1 & E5-2
KL/r = 236.9
Case 2: Same as condition set above, but connected through the shorter leg:
bl = 4
bs = 4
rz = 0.779
bl/bs = 1.00
For bl/b/s < 1.7:
KL/r = 236.9
Case 1 or 2: KL/r = 200 Input value obtained above, but need not to exceed 200.
Fe = 7.16 ksi Fe >= 0.44 Fy, cal Fcr per EQ E3-2
Fcr = 6.28 ksi
Pn = 18 kips
Pa = 10.8 kips ASD
Pa = 16.2 Kips LRFD
L4x4x3/8
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8/3/2019 Notes on AISC 13th(1)
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AISC 13th E.3 - Flexural Buckling
Col ID :
Shape:
Fy = 50 ksi
Pr = 0 kips
Kx = 1Ky = 1
Lx = 16 ft
Ly = 16 ft
Properties:
A = 29.6 in2
rx = 3.1 in
ry = 3.02 in
KLx/rx = 61.9
Kly/ry = 63.6 KLy/ry Governs
KL/r = 63.6
Fe = 70.76 ksi Fe >= 0.44 Fy, cal Fcr per EQ E3-2
Fcr = 37.2 ksi
Pn = 1101.1 kips
Pa = 659.4 kips ASD (Pn/1.67)
Pa = 991 Kips LRFD (0.9*Pn)
0
WT 10.5x100.5
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AISC 13th E.4 (a) - Torsional & Flexural-Torsional Buckling - Double Angles
Col ID :
Shape:
Fy = 36 ksi
Slenderness of Compression Flange Leg (Table B4.1):
b = 6 int = 0.875 in
d/tw = 6.9 Flange leg is non-slender, proceed.
Pr = kips
Kx = 1
Ky = 1
Lx = 16 ft
Ly = 16 ft
Properties:
A = 16 in2
Taken from Table 4-10 (p. 4-118 thru 156)
rx = 1.1 in See above
ry = 2.96 in See above
Ix = 19.4 in4
Ix = Arx2
Iy = 140.2 in4
Iy = Ary2
J = 4.06 in4
2 x Value given on Table 1-7 (p. 1-40 thru 47)
xo = 0 in Distance from shear ctr to centroid
yo = 0.683 in Distance from shear ctr to centroid = y^ - tf/2
y^ = y' (LLBB) = x' (SLBB). See Table 1-7 for y', x'
r'o = 3.23 in Taken from Table 1-15 (p. 1-10 thru 107)
r'o2
= 10.433 in2
H = 0.956 Taken from Table 1-15 (p. 1-10 thru 107)
Kly/ry = 64.9
Fe = 67.95 ksi Fe >= 0.44 Fy, cal Fcr per EQ E3-2
Fcry = 28.84 ksi
Fcrz = 272.4 ksi EQ E4-3
Calculate Fcr:
Let m = (Fcry + Fcrz)/2H = 157.55 ksi
n = 4FcryFcrzH/(Fcry + Fcrz)2
= 0.331
Then, Fcr = m*[1 - (1 - n)1/2] = 28.69 ksi EQ E4-2
Pn = 459 kips
ASD (Pn/1.67) Pa = 274.9 kips Govn'd by Flexural (E.3)
LRFD (0.9*Pn) Pa = 413.1 Kips Govn'd by Flexural (E.3)
2L6x4x7/8 - 3/8" Seperator, SLBB
Note: The calculated Pa has a 1.5% margin of error compared to
8/3/2019 Notes on AISC 13th(1)
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AISC 13th E.3 - Flexural Buckling
Col ID :
Shape:
Fy = 36 ksi
Pr = 0 kips
Kx = 1Ky = 1
Lx = 16 ft
Ly = 16 ft
Properties:
A = 16 in2
rx = 1.1 in
ry = 2.96 in
KLx/rx = 174.5 KLx/rx Governs
Kly/ry = 64.9
KL/r = 174.5
Fe = 9.4 ksi Fe < 0.44 Fy, cal Fcr per EQ E3-3
Fcr = 8.24 ksi
Pn = 131.8 kips
Pa = 78.9 kips Governs ASD (Pn/1.67)
Pa = 118.7 Kips Governs LRFD (0.9*Pn)
0
2L6x4x7/8 - 3/8" Seperator, SLBB
8/3/2019 Notes on AISC 13th(1)
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F.10 - Flexural Strength of (Equal Leg) Single Angle without Continuous Restraint
F.10.1 - Yielding
Fy = 36 ksi
Sy = 1.5 in3
My = 54 in-kips
Mn = 81 in-kips
Ma (ASD) = 48.5 in-kips Does Not Govern
F.10.2 - Lateral Torsional Buckling
b = 4 in
t = 0.375 in
L = 56 in Length between Brace Points
Cb = 1
Me (a) = 311.1 in-kips Toe in Compression F10-4a
Me (b) = 1482.9 in-kips Toe in Tension F10-4b
Governing Me = 311.1 in-kips Min of Me(a) & Me(b)
8/3/2019 Notes on AISC 13th(1)
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A B.66Eb
4tCb/L
2 [1+0.78(Lt/b2)
2]1 2
Me(a) = A*(B-1) = 311.1 in-kips Toe in C 585.92 1.5309
Me(b) = A*(B+1) = 1482.9 in-kips Toe in T
Toe in C Toe in T
Mn (F10-2) = -18.4 -5558.5 in-kips
Mn (F10-3) = 77.4 91.6 in-kips
8/3/2019 Notes on AISC 13th(1)
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AISC 13th F - Flexural
Beam ID : Frame Beam
Fy = 36 ksi Lb = 4.167 ft = 50.004 in
Shape: C12x20.7 C
8/3/2019 Notes on AISC 13th(1)
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Lr
Inner most roo2nd root 2nd*1st root
1.735296193 1.653873 0.066012 145.616 12.13467
F2-1
Mn = Mp = 921.6 in-kips
(E/Fy)^0.5 = 28.38231
Lp = 39.81243 in
Lr = 145.616 in
LbLrMn (in-k) = 921.6 885.0156 921.6
F2-2 F2-3
(1) (2) (1)/(2) A B
Lb-Lp Lr-Lp (Lb/rts)^2 0.078*{..} (1+A)^0.5 Fcr
10.19 105.80 0.096325 2587.632 0.321543 1.149584 127.1563
F6-1 FyZy 1.6Fysy
Mn = 124.92 99.072 in-kips
A B
F6-2(b) Lenda Ter Mp-0.7FyS Mn-0.44583 81.576 161.289
F6-2(c) Fcr Mn
2315.008 3981.813
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AISC 13th F - Limiting Width-Thickness Ratios for Compression Elements
Case Case
1 3
Compact Noncomp Compact Noncomp
Fy .38(E/Fy)1/2
1.0(E/Fy)
1/2
Fy .56(E/Fy)1/2
36 10.79 28.38 36 NA 15.89
50 9.15 24.08 50 NA 13.49
6 5
Compact Noncomp Compact Noncomp
Fy .54(E/Fy)1/2
.91(E/Fy)1/2
Fy .38(E/Fy)1/2
1.0(E/Fy)1/2
36 15.33 25.83 36 NA 12.77
50 13.00 21.92 50 NA 10.84
7 8
Compact Noncomp Compact Noncomp
Fy .38(E/Fy)1/2
1.0(E/Fy)1/2
Fy .38(E/Fy)1/2
1.0(E/Fy)1/2
36 10.79 28.38 36 NA 21.29
50 9.15 24.08 50 NA 18.06
9 10
Compact Noncomp Compact Noncomp
Fy 3.76(E/Fy)1/2
5.7(E/Fy)1/2
Fy .38(E/Fy)1/2
1.0(E/Fy)1/2
36 106.72 161.78 36 NA 21.29
50 90.55 137.27 50 NA 18.06
Elements subjected to Flexural Elements under Axial Compression
Leg of Single Angle & All Other Unstiffened
Elements (except elements of built-up
shapes)
Web of W & C Shapes (h/tw) - h is the cleardistance between flanges, see B4.2(b) for built-up
shapes.
Web of W (h/tw) - see 9 for "h".
Flange of Tees Stem of Tees (d/t)
Flange of Rolled W & C Shapes Flange of Rolled W & C and Out-Standing
Leg of 2L in Continuous Contact
Leg of Single Angle and Out-Standing Leg of
2L with Seperators (see case 3 for out-
standing legs in continuous contact)
8/3/2019 Notes on AISC 13th(1)
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5.860806
8/3/2019 Notes on AISC 13th(1)
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1. WELDING: Minimum Base Metal Thickness Required to Match Weld Strength
(Manual p.9-5)
Case I - Weld on both sides of the connecting element
For Exx = 60 ksi
For Exx = 70 ksi
D - Number of weld size in 16th
Fu - Specified min tensile strength of the connecting element (see Manual T3-2)
Case II - Weld on one side of the connecting element
For Exx = 60 ksi
For Exx = 70 ksi
2. BOLTING: Strength Reduction for Flanges with Bolt Holes
(Manual F13.1, p.16.1-61)
Case I - for Fu*Afn >= Yt*Fy*Afg - The limit state of Tensile Rupture DOES NOT apply.
Afg = Gross Tension Flange Area = bf*tf D3.1
Afn = Net Tension Flange Area = tf*bf(net) D3.2Yt
1.0
1.1
For single row of holes across flange width:
dh = hole diameter
For flange with chain of of holes in longitudinal direction:
m = number of gage (across the width of flange)
s = longitudinal ctr-ctr spacing of bolt holes (pitch)
g = transverse ctr-ctr spacing of bolt holes (gage)
Case II - for Fu*Afn < Yt*Fy*Afg - The nominal flexural strength:
F13-1
Fr(LRFD) = 0.9
Fr(LRFD) = 0.9
t(min) = 2.65*(D/Fu)
t(min) = 3.09*(D/Fu)
t(min) = [0.6*Exx*0.707*(D/16)*2]/(0.6*Fu)
t(min) = 5.30*(D/Fu)
t(min) = 6.19*(D/Fu)
Fy/Fu 0.8
bf(net) = bf - n*dh
Mn = Fu*Sx*(Afn/Afg)
bf(net) = bf - n*dh - m*[s2/(4*g)]
8/3/2019 Notes on AISC 13th(1)
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Ch. D Effective Net Area (does not apply here but listed for information use)
For Tension Member:
D3-1
- U (Shear Lag Factor): T-D3.1 (see table for cases not shown)
For W, Tees -
U = 0.9
U = 0.85
U = 0.70
For Single Angles -
U = 0.70
U = 0.60
Ch. J.4-J.8 Net Area Associated Topics
1. Strength of Connected Elements in Tension Fr(LRFD) Fr(ASD)
J4-1 0.90 1.67
J4-2 0.75 2.00
Note: For bolted splice plate, Ae = An
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8. Concrete Bearing Strength:
On full Area J8-1
Less than full area J8-2
Fr(LRFD) Fr(ASD)
0.60 2.50
Ch. J.10 1. Flange Local Bending due to Tensile ForceNote: This check was originally developed out of concerns over weld rupture
(seperation due to deformation) between the connected beam-column
flanges, thus, it does not apply to bolted connection, nor flange in
compression. Also, it does not apply to moment end-plate and tee-stub
type of connections - see C.J10.1, p16.1-356 for references)
a. The applied load is located at a distance > = 10*tf from member end:
J10-1
b. The applied load is located at a distance < 10*tf from member end:
member end. Fr(LRFD) Fr(ASD)
see note on J.10.1 0.90 1.67
2. Web Local Yielding due to Concentrated (Compressive/Tensile) Force
a. The applied load is located at a distance > d from member end:
J10-2
b. The applied load is located at a distance = k for end beam reactions)
Note: A pair of stiffeners, or a doubler plate, may be provided to satisfy the
requirements.
3. Web Crippling due to Concentrated Compressive Force (see Work Sheet)
4. Web Side Sway Buckling due to lack of lateral restraint on both flanges at the
point of application of the concentrated force
5. Web Compression Buckling due to a pair of Concentrated Compressive Force
applied on both flanges
b. The applied load is located at a distance >= d/2 from member end:
J10-8b. The applied load is located at a distance < d/2 from member end:
see note on J.10.5 Fr(LRFD) Fr(ASD)
0.90 1.67
Note: A pair of stiffeners, or a doubler plate, extending the full depth of the web
may be provided to satisfy the requirements.
6. Web Panel Zone Shear due to double concentrated forces applied to one or
both flanges
Rn = 24*tw
3
*(E*Fy)
1/2
/h
Rn = 12*tw3*(E*Fy)
1/2/h
Rn = 6.25*tf
2
*Fyf/2
Pp =0.85*fc'*A1
Pp =0.85*fc'*A1*(A2/A1)1/2
8/3/2019 Notes on AISC 13th(1)
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8. Additional Stiffener Requirement for Concentrated Forces
9. Additional Doubler Plate Requirement for Concentrated Forces
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J10.3 Web Crippling due to Concentrated Compressive Force Work Sheet
Shape:
Fy = 50 ksi
d = 8 in
tw = 0.245 in
bf = 6.5 intf = 0.4 in
k1 = 0.5625 in
k = 0.79 in k = bf/2-k1
N = 1 in Use N >= k
- Loading case 1. The applied load is at a distance >= d/2 from member end.
- Loading case 2 & 3. The applied load is at a distance < d/2 from member end, but
- for N/d 0.2, input "3" in loading case input field below.
N/d = 0.13 (see loading cases 2 & 3 above for applications)
Loading Case: 1 (Input "1", "2" or "3" per conditions given above)
Rn = 42.3 kips J10-4, J10-5a, J10-5b
Ra = 31.7 kips LRFD (Fr = 0.75)
Ra = 21.1 kips ASD (Fr = 2.00)
W8x24
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tw2
N/d (tw/tf)1.5
(Efytf/tw)1/2
0.060025 0.125 0.479357 1538.619
LC 1 2 3
Rn 87.2 43.6 42.3
8/3/2019 Notes on AISC 13th(1)
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Max/Min Stresses due to Pure Bending (Based on Generalized Theory of Pure Bending)
Shape L6x4x1/2
Mx = 10 in-k (+) rotated about +X axis
My = 0 in-k (+) rotated about +y axis
A = 4.75 in2
Ix = 17.4 in4
Iy = 6.27 in4
Iz = 3.6 in4 (Leave Iz empty if X & Y are principal axes)
Tana = 0.44 rads
Iw = 20.07 in4
a = 23.75 degrees
Ixy = 6.07 in4
- Locate Nutral Axis:
Tan = 0.97 Rads
= 44.13 degrees (+) Counterclockwise from +X axis.
- Calculate Stresses at Points of Interests:
Points of
Interestsx (in) y (in) x (ksi)
A -0.987 -4.01 2.65 Dist. along X-axis (see sign convention)
B -0.987 1.99 -2.56 Dist. along Y-axis (see sign convention)
C 4.01 1.99 1.64
+ Tension
Iw = Ix+Iy-Iz
Ixy =1/2(Iw-Iz)sin(2a)
Tan =My*Ix + Mx*Iyz
Mx*Iy + My*Iyz
x =(My*Ix + Mx*Ixy)*x - (Mx*Iy + My*Ixy)*y
Ix*Iy - I2
xy
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Mx My Ix Iy Ixy x yx
A 10 0 17.4 6.27 6.07 -0.987 -4.01 2.65
B -0.987 1.99 -2.56
C 4.01 1.99 1.64