Notes on AISC 13th(1)

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)


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Ag =

Ix =

rx =

Iy =ry =

J =

Cw =

Xo =

Yo =


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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


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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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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


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


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


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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


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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)


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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


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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


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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


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


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


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)


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5.860806


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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)]


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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


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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


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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

Notes on AISC 13th(1)

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