Advanced Steel Design with AISC LRFD

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Advanced Steel Designwith AISC LRFD
Dr. Jay Yoo Dr. Molly Hughes Department of Civil
Engineering Auburn University March, 2006
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EVOLUTION OF AISC DESIGN SPECIFICATION
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The first steel framed building, 1894-1895
Chicago Home Insurance, 10-story, demolished
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The Manhattan flatiron building, 1903
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The tallest structure
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Skyscrapers compared
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RELIABILITY IN DESIGN / LRFD DESIGN PHILOSOPHY
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Reliability in Design/LRFD Design Philosophy
Allowable Stress Design (ASD)
Rn Nominal Strength F.S. Factor Of
Safety Qni Load Effect
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Reliability in Design/LRFD Design Philosophy
Load and Resistance Factor Design (LRFD)
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Reliability in Design/LRFD Design Philosophy
Load and Resistance Factor Design (LRFD)
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Reliability in Design/LRFD Design Philosophy
Load and Resistance Factor Design (LRFD)
R
Q

• Structural behavior acceptable if R gt Q
• (strength gt load)
• Structural behavior unacceptable if R lt Q
• (strength lt load)

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Reliability in Design/LRFD Design Philosophy
Load and Resistance Factor Design (LRFD)

• Always some uncertainty in determining Q and R
• Examples
• Q Moment at given bridge section changes with
• changing traffic patterns
• R Concrete compressive strength may be
• greater than whats specified in design

13
Reliability in Design/LRFD Design Philosophy
Load and Resistance Factor Design (LRFD)

• Even for the most carefully designed and
  constructed structure there is a small but finite
  chance that Q gt R, i.e., that the limit state can
  be exceeded. A satisfactory structural design
  specification is one which minimizes this chance
  to an acceptably low level.
• -- Galambos, Load and Resistance Factor Design,
  Engineering Journal, AISC, Third Quarter, 1981.

14
Reliability in Design/LRFD Design Philosophy
Load and Resistance Factor Design (LRFD)
Probability Density
Q
R
Load Effect
Resistance
Probabilistic Description of Q and R

• Want to minimize overlap (probability of
  failure)
• So want curves to be as far apart as possible

15
Reliability in Design/LRFD Design Philosophy
Load and Resistance Factor Design (LRFD)
When R-Q lt1, limit state has been exceeded
(shaded area is probability that this will
happen). So we want to minimize this area.
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Reliability in Design/LRFD Design Philosophy
Load and Resistance Factor Design (LRFD)
ratio of mean to nominal value
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Reliability in Design/LRFD Design Philosophy
Load and Resistance Factor Design (LRFD)
Magnitude of positions the coordinates with
respect to the distribution curve. Larger is
better.
(R-Q)
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Reliability in Design/LRFD Design Philosophy
Load and Resistance Factor Design (LRFD)

• Based on statistical load and resistance data,
  on
• average (lots of variation), structural design
  in the US
• inherently has had

3.0 (gravity loads only)

• Target reliability value is currently 3.5.

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Reliability in Design/LRFD Design Philosophy
Load and Resistance Factor Design (LRFD)
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Reliability in Design/LRFD Design Philosophy
Load and Resistance Factor Design (LRFD)
-- Galambos, Load and Resistance Factor Design,
Engineering Journal, AISC, Third Quarter, 1981.
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Reliability in Design/LRFD Design Philosophy
Load and Resistance Factor Design (LRFD)
(ratio of mean to nominal value)
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Reliability in Design/LRFD Design Philosophy
Load and Resistance Factor Design (LRFD)
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Reliability in Design/LRFD Design Philosophy
Load and Resistance Factor Design (LRFD)

• Select reliability index (3.5)
• Select load factors
• Solve for resistance factors

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Reliability in Design/LRFD Design Philosophy
Load and Resistance Factor Design (LRFD)

• Load combinations have the general format

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Reliability in Design/LRFD Design Philosophy
Load and Resistance Factor Design (LRFD)
Example 1.2Dn 1.6 Sn 0.8Wn
Max. annual wind value
Max. snow load in the 75-yr lifespan of the
structure
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Reliability in Design/LRFD Design Philosophy
Load and Resistance Factor Design (LRFD)
Procedure designed to provide a more uniform
level of safety
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MINIMUM DESIGN LOADS
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ASCE Standard ASCE/SEI 7-05
Minimum Design Loads for Buildings and Other
Structures
2.3.2 Basic Combinations Structures,
components, and foundations shall be designed so
that their design strength equals or exceeds the
effects of the factored loads in the following
combinations
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ASCE Standard ASCE/SEI 7-05
Minimum Design Loads for Buildings and Other
Structures

• 1.4(DF)
• 1.2(DFT) 1.6(LH) 0.5(Lr or S or R)
• 1.2D 1.6(Lr or S or R) (L or 0.8W)
• 1.2D 1.6W L 0.5(Lr or S or R)
• 1.2D 1.0E L 0.2S
• 0.9D 1.6W 1.6H
• 0.9D 1.0E 1.6H

D dead load F loads due to fluids with well
defined pressures and maximum heights T
self-straining force L live load H load due
to lateral earth pressure, ground water
pressure, or pressure of bulk
materials Lr roof live load S snow load R
rain load W wind load E earthquake load
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ASCE Standard ASCE/SEI 7-05
Minimum Design Loads for Buildings and Other
Structures

• For design of structural steel members,
  pertinent combinations are

• 1.4D
• 1.2D 1.6L 0.5(Lr or S or R)
• 1.2D 1.6(Lr or S or R) (L or 0.8W)
• 1.2D 1.6W L 0.5(Lr or S or R)
• 1.2D 1.0E L 0.2S
• 0.9D (1.6W 1.0E)

Exception The load factor on L in combinations
(3), (4), and (5) is permitted to equal 0.5 for
all occupancies in which L0 (minimum uniformly
distributed live load) in Table 4-1 is less than
or equal to 100 psf, with the exception of
garages or areas occupied as places of public
assembly.
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ASCE Standard ASCE/SEI 7-05
Minimum Design Loads for Buildings and Other
Structures
"Each relevant strength limit state shall be
investigated. Effects of one or more loads not
acting shall be investigated. The most
unfavorable effects from both wind and earthquake
loads shall be investigated, where appropriate,
but they need not be considered to act
simultaneously."
2.3.3 Load Combinations Including Flood
Load 2.3.4 Load Cominations Including Atmospheric
Ice Loads
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Load Combination Example

• Given Compression member in the upper story of
  a building subjected to these
• service axial loads
• D 30 kips
• Lr 18 kips
• S 8 kips
• W -10 kips (uplift)
• R 5 kips
• Required Determine the controlling load
  combination and factored load

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• Solution

Load Combination 3 controls
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Load Combination Example

• Given Continuous steel girders spanning over
  interior columns support the roof of
• an industrial building. A typical
  interior girder elevation is shown below.
• Required Using load factors and load
  combinations from the AISC-LRFD
• Specification, calculate the
  maximum factored reactions and moments at
• the support at C.

-0.25 kips/ft (uplift)
WIND
1.0 kips/ft
SNOW
3.5 kips
2.5 kips
0.75 kips/ft
ROOF LIVE
3.0 kips
4.0 kips
1.0 kips/ft
DEAD
A
D
C
B
A
D
C
B
12'
0"
11'
-
0"
11'
-
0"
24'
-
0"
12'
-
0"
11'
-
11'
-
24'
-
0"
-
25'
-
0"
36'
-
0"
22'
-
0"
25'
-
0"
36'
-
0"
22'
-
0"
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Load Combination Example

• Solution

Reactions, Shear, and Moment Due to Unfactored
Wind Load

• Reactions, kips

• Shear, kips

• Moment, k-ft

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Load Combination Example

• Solution

Reactions, Shear, and Moment Due to Unfactored
Snow Load

• Reactions, kips

• Shear, kips

• Moment, k-ft

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Load Combination Example

• Solution

Reactions, Shear, and Moment Due to Unfactored
Roof Live Load

• Reactions, kips

• Shear, kips

• Moment, k-ft

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Load Combination Example

• Solution

Reactions, Shear, and Moment Due to Unfactored
Dead Load

• Reactions, kips

• Shear, kips

• Moment, k-ft

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Load Combination Example
Maximum Factored Reaction at C
Load Combination 3 controls
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Load Combination Example
Maximum Factored Moment at C
Load Combination 3 controls
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DESIGN EXAMPLES
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Design Flowcharts
Tension Member Selection
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Design Flowcharts
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Design Flowcharts
Tension Member Design (continued)
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Design Flowcharts
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Design Flowcharts
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Design Flowcharts
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Design Flowcharts
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Design Flowcharts
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Design Flowcharts
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Design Flowcharts
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Design Flowcharts
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Design Flowcharts
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Design Flowcharts
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Tension Member Analysis/Design

• Tension Limit States
• Excessive deformation, initiated by yielding
• Fracture

• Tension Failure Modes
• Gross Section Yielding
• Net Section Fracture
• Block Shear Failure

56
Tension Member Analysis/Design
or

• Pu factored axial tensile load
• resistance factor for tension
• Pn nominal tensile strength

• For yielding

• For fracture

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Tension Member Analysis/Design

• Gross Section Yielding

Ag gross area of cross section

• Net Section Fracture

Ae effective net area UAn
An net area of cross section
U shear lag reduction factor (AISC Table D3.1)
(For welded connections, Ae UAg , since An Ag)
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Tension Member Analysis/Design

• Shear Lag Reduction Factor, U

where
distance from centroid of connected area to the
plane of the connection

• If member has 2 symmetrically located planes of
  connection,

measured from the nearest one-half of the area
length of the connection in the direction of
the load

• For bolted connections, measure the largest
  center-to-center
• (c.c.) distance from one end of the line of
  bolts to the other

• For welded connections, measure from one end of
  the
• connection to the other (use the longest
  segment if there are
• segments of different lengths in the weld)

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Tension Member Analysis/Design

• Shear Lag Reduction Factor, U

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Tension Member Analysis/Design

• Shear Lag Reduction Factor, U

• Alternatives to "formula" U
• Table D3.1, AISC Manual, 13th Ed.
• Commentary B3, 3rd Ed.

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Tension Member Analysis/Design

• Block Shear (AISC J4.3)

a) When Fu Ant 0.6 Fu Anv (indicates shear
yield tension fracture)
AISC Eqn J4-3a
b) When FuAnt 0.6 Fu Anv (indicates shear
fracture tension yield)
AISC Eqn J4-3b

• Agt gross tension area
• Ant net tension area
• Agv gross shear area
• Anv net shear area

62
Tension Member Analysis Example

• Given W12 x 72, made of A242 steel, connected
  through the flanges at its
• ends to ?" thick plates with ?"
  diameter bolts as shown

• D 90 kips
• S 40 kips
• L 85 kips

• Find Determine the
• adequacy of the tension
• member, considering the
• three tension member
• failure modes of gross
• section yielding, net section
• fracture, and block shear.

63
Tension Member Analysis Example

• Solution

• Load Combinations
• 1. 1.4D 1.4(90) 126.0 kips
• 2. 1.2D 1.6L 0.5S 1.2(90) 1.6(85)
  0.5(40) 264.0 kips
• 3. 1.2D 1.6S 0.5L lt Case 2 since S lt L
• 4. 1.2D 1.6W 0.5L 0.5S lt Case 3 lt Case
  2
• 5. 1.2D 1.0E 0.5L 0.2S lt Case 4 lt Case
  3 lt Case 2
• 6. 0.9D (1.6W or 1.0E) lt All others

• Load Combination 2 controls ? Pu 264.0 kips

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Tension Member Analysis Example

• Gross Section Yielding

• A242 Steel Fy 50 ksi, Fu 70 ksi (Table
  2-1)
• W12 x 72 Ag 21.1 in2 , tf 0.670 in.

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Tension Member Analysis Example

• Net Section Fracture

? Use U 0.83
dimension taken from WT6 x 36
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Tension Member Analysis Example

• Block Shear

Agv 4(332.5)(0.670) 22.78 in2 Anv
22.78 4(2.5)(??)(0.670) 17.76 in2 Agt
4(2.75)(0.670) 7.37 in2 Ant 7.37
4(0.5)(??)(0.670) 6.37 in2
FuAnt (70)(6.37) 445.9 k 0.6FuAnv
(0.6)(70)(17.76) 745.9 k
Since 0.6FuAnv gt FuAnt ? Eqn J4-3b applies
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Tension Member Analysis Example

• Block Shear

FRn F(0.6FuAnv FyAgt) 0.75745.9(50)(7.37)
835.8 k Check Upper Limit F(0.6FuAnv
FuAnt) 0.75(745.9445.9) 893.9 k 835.8 k lt
893.9 k so the upper limit does not control
? FRn for block shear 835.8 kips
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Tension Member Analysis Example

• Gross Section Yielding FRn 949.5 kips
• Net Section Fracture FRn 831.6 kips
• Block Shear Failure FRn 835.8 kips

? FRn 831.6 kips gt Pu 264.0 kips
? Tension member is adequate for given loading
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Tension Member Design
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Tension Member Design Example

• Given Tension member shown connected between
  to plates with eight ¾"
• diameter bolts
• Service dead load 185 kips
• Service live load 40 kips
• Length 20 ft
• A36 Steel used for tension member

• Find Select an S-shape to resist the given
  loading.

• Solution

• Determine Pu. Only D and L are present, and D lt
  8L, so Load
• Combination 2 controls.

Pu 1.2D 1.6L 1.2(185) 1.6(40) 286.0 kips
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Tension Member Design Example

• Try S12 x 31.8 ? Ag 9.31 in2, rmin ry 1.00
  in., tw 0.350 in.

• Calculate Ae compare to (Ae)required.

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Tension Member Design Example

• An 9.31 2(¾?)(0.350) 8.70 in2

• Ae UAn
• U 0.70 (Table D3.1 W, M,S or HP shape or
  Tees cut from these shapes
• with web connected with 4 or
  more fasteners in the direction of
• loading)

• Ae 0.70(8.70) 6.09 in2 lt 6.57 in2 ? N.G.

73
Tension Member Design Example

• Try S12 x 35 ? Ag 10.2 in2, rmin ry 0.980
  in., tw 0.428 in.

• Calculate Ae compare to (Ae)required.

• An 10.2 2(¾?)(0.428) 9.45 in2

• Ae UAn 0.70(9.45) 6.62 in2 gt 6.57 in2 ?
  Okay

• Ag 10.2 in2 gt (Ag)required 8.83 in2
• Ae 6.62 in2 gt (Ae)required 6.57 in2
• rmin 0.980 in. gt (rmin)required 0.80 in.

? Use an S12 x 35 (check block shear after bolt
spacings have been determined)
74
Compression Member Analysis/Design
or

• Pu factored axial compressive load
• resistance factor for compression
• Pn nominal compressive strength

• For compression

(AISC Manual, 3rd Ed.)
75
Compression Member Analysis/Design
?

• Nominal compressive strength critical buckling
  stress times gross area

(AISC Eqn E2-4)
Nondimensional slenderness parameter
76
Compression Member Analysis/Design
If
? Inelastic buckling
(AISC Eqn E2-2)
If
? Elastic buckling
(AISC Eqn E2-3)

• Also, AISC recommends that kL/r remain less than
  200 for design.

77
Compression Member Analysis/Design
Local Buckling

• Design compressive strength given in AISC Eqns
  E2-2 and E2-3 must be
• reduced if element width-to-thickness ratio, ?,
  exceeds limiting w-to-t ratio for
• that shape, ?r, (? slender element) to account
  for possibility of local buckling

• AISC Table B5.1

78
Compression Member Analysis/Design
Local Buckling

• If ? gt ?r ? Go to Appendix B3 calculate
  reduction factor Q
• Eqns A-B5-3 through A-B5-10, depending on shape
• Compute
• If ?
  
• If ?
  
• Then

AISC Eqn A-B5-15
AISC Eqn A-B5-16
79
Compression Member Analysis/Design

• Tailored procedures for torsional and
  flexural-torsional buckling
• Torsional buckling doubly-symmetric cross
  sections with slender
• cross-sectional elements (e.g. cruciform)
• Flexural-torsional buckling unsymmetrical
  cross-sections with one
• axis of symmetry (e.g. channels, tees,
  double-angle shapes)
• or no axes of symmetry (e.g. unequal single-leg
  angles)
• AISC Section E3, 3rd Ed. ? double-angle and
  tee-shapes
• AISC Appendix E3, 3rd Ed. ? general approach,
  can be used for
• any unsymmetrical shape

80
Compression Member Design Example

• Given Factored axial compressive load 500
  kips
• KL effective length 18 ft

• Find Select the most economical W18 shape
  of A992 steel to resist the load.

• Solution W18 values are not given in the
  column tables, so can't just
• "look it up"

81
Compression Member Design Example

• Assume Fcr ? Fy (?)(50) 33 ksi

• Set

• Try a W18x65 (Ag 19.1 in2 gt 17.8 in2 ?OK)

? No local buckling of flange
82
Compression Member Design Example
? No local buckling of web
? Okay
83
Compression Member Design Example
? Use AISC Eqn E2-3 Elastic Buckling
? N.G.

• Try Fcr 15.37 ksi (value just calculated for
  W18x65)

84
Compression Member Design Example

• Try W18x143 (Ag 42.1 in2 gt 38.3 in2) ? Okay

? No local buckling of flange
? No local buckling of web
85
Compression Member Design Example
? Okay
? Use AISC Eqn E2-2 Inelastic Buckling
? Try again
86
Compression Member Design Example

• Try Fcr 25 ksi (between 33 ksi and 15.37 ksi)

• Try W18x86 (Ag 25.3 in2 gt 23.5 in2) ? Okay

? No local buckling of flange
? No local buckling of web
87
Compression Member Design Example
? Okay
? Use AISC Eqn E2-2 Inelastic Buckling

• Try next lighter section, W18x76 (h/tw 37.8 gt
  ?r 35.9)
• (We could use this section, but we'd have to
  reduce its design strength to
• account for the possibility of local buckling.)

88
Compression Member Design Example

• We should check this lighter section, because we
  just assumed the Fcr value
• that gave us our (Ag)req.
• If we'd chosen another Fcr value, we might have
  gotten a lower (Ag)req.
• Since we found a section that works, we should
  try the next lighter section to
• see if it works.
• If it does work, and the capacity is much
  greater than the required capacity,
• we could try the next lighter section after
  that.
• If it doesn't work, we know we've got the
  lightest one that will work.

89
Compression Member Design Example

• Try W18x71 (Ag 20.8 in2) (bf/2tf okay h/tw
  okay)

? Okay
? Use AISC Eqn E2-3 Elastic Buckling
? N.G.
Use a W18x86.
90
Compression Member Analysis/Design

• Note There's been a fair amount of
  modification to the treatment of
• compression members in the AISC
  Manual 13th edition.

• When

or
(AISC Eqn E3-2, 13th Ed.)

• When

or
(AISC Eqn E3-3, 13th Ed.)

• Slender elements, torsional, flexural-torsional
  issues dealt with in main body, Ch. E

91
Plastic Analysis/Design

• Participants, please turn to page 172 of your
  notes
• Click here to go to slide 344 .

92
Flexural Member Analysis
or

• Mu factored load bending moment
• (controlling load combination)
• resistance factor for flexure
• Mn nominal moment resistance

• For flexure

93
Flexural Member Analysis
Terminology

• Lb laterally unbraced length

• Mn nominal moment strength

• Mp fully plastic bending moment FyZ

• My yield moment, moment corresponding to onset
  of yielding at the
• extreme fiber from an elastic stress
  distribution FyS

• S elastic section modulus (based on x-sec
  geometry)

• Z plastic section modulus (based on x-sec
  geometry)

• Mr limiting buckling moment, Mcr, when ? ?r
  and Cb 1.0

• Cb bending coefficient based upon moment
  gradient

94
Flexural Member Analysis

• Summary of Analysis Procedure for Determining
  Nominal Moment Strength, Mn
• For I- and H-Shaped Sections And Channels Bent
  About The X-axis

• Determine whether the shape is compact,
  noncompact, or slender

Width-to-Thickness Parameters for Hot-rolled I-
and H-Shapes in Flexure
Element ? ?p ?r
Flange bf/2tf
Web h/tw
These values can also be used for channels,
except that ? bf/tf instead of bf/2tf.

• ? ? ?p ? Compact Shape
• ?p ? ? ? ?r ? Non-Compact Shape
• ? gt ?r ? Slender Shape

95
Flexural Member Analysis
2. If the shape is compact (? ? ?p)

• Check for lateral-torsional buckling (LTB) as
  follows

• If Lb ? Lp ? there is no LTB, and
• Mn Mp FyZx 1.5 My.
• (upper limit always satisfied for I- and H-shapes
  bent about their strong axis)

96
Flexural Member Analysis

• If Lp lt Lb ? Lr, there is inelastic LTB, and

G shear modulus 11200 ksi for structural
steel
J torsional constant (in4)
Cw warping constant (in6)
97
Flexural Member Analysis

• Cb bending coefficient
• 1.0 for constant (uniform) bending moment
  along the unbraced length

• For moment distributions along the beam other
  than constant moment

(AISC Eqn F1-3)

• Mmax absolute value of the max. moment w/in
  the unbraced length (including
• end points)
• MA abs. value of the moment at the quarter
  point of the unbraced length
• MB abs. value of the moment at the midpoint of
  the unbraced length
• MC abs. value of the moment at the ¾ point of
  the unbraced length

98
Flexural Member Analysis

• If Lb gt Lr, there is elastic LTB, and

Same as
99
Flexural Member Analysis

• If the shape is noncompact because of the flange,
  the web, or both, the
• nominal strength will be the smallest of the
  strengths corresponding to flange local buckling
  (FLB), web local buckling (WLB), and
  lateral-torsional buckling (LTB)

• a) Flange local buckling
• If ? ? ?p, there is no FLB.
• If ?p lt ? ? ?r, the flange is noncompact, and

Fr residual stress 10 ksi for hot-rolled
shapes
100
Flexural Member Analysis

• b) Web local buckling
• If ? ? ?p, there is no WLB.
• If ?p lt ? ? ?r, the web is noncompact, and

(values of ? are different from those for
flanges)
(also different than for flanges)
101
Flexural Member Analysis
c) Lateral-torsional buckling

• If Lb ? Lp, there is no LTB, and Mn Mp FyZ.
• If Lp lt Lb ? Lr, there is inelastic LTB, and

• If Lb gt Lr, there is elastic LTB, and

or
102
Flexural Member Analysis
Shear Strength for Flexural Members

• Vu factored load shear force
• (controlling load combination)
• resistance factor for shear
• Vn nominal shear resistance

• For shear

103
Flexural Member Analysis

• Web shear stress is much larger than flange
  shear stress, so the web will
• yield long before the flange
• Soshear yielding of the web is one of the shear
  limit states
• Shear yield stress is approximated as 60 of
  tensile yield stress (Fy)
• So the eqn for the stress in the web at failure
  for the web shear yielding limit
• state is approximated as fv Vn/Aw 0.60Fy
• Nominal strength corresponding to this limit
  state is then Vn 0.6FyAw
• (assuming there is no shear buckling of the web)

104
Flexural Member Analysis

• Shear buckling of the web depends on h/tw

• For

• Shear yielding of web (no web instability
  buckling)
• Vn 0.6 Fy Aw (AISC Eqn.
  F2-1)

2. For

• Inelastic web buckling can occur

(AISC Eqn. F2-2)
105
Flexural Member Analysis
3. For

• Elastic web buckling

(AISC Eqn. F2-3)
Aw area of the web (d)(tw)
d overall beam depth
4. For h/tw gt 260

• Web stiffeners required consult Appendix F2 (or
  Appendix G for
• plate girders)

106
Flexural Member Analysis
107
Flexural Member Design
General Design Procedure
1. Determine Mu.
Set it equal to the required design strength,
FbMn.

• Since the beam weight is part of the dead weight
  but isnt known yet, either
• 1) assume a beam weight, include it in the dead
  load, then, after picking
• section, recalculate the required section
  modulus and make sure the
• section you picked has one larger than what
  is required, or
• 2) ignore the beam weight initially, then,
  after you select a section,
• recalculate the factored load to include
  beam weight and make sure
• the section you picked can handle it.

108
Flexural Member Design
2. Select a shape that satisfies the strength
requirements

• by the usual procedure (assume shape,
  calculate design strength,
• compare to load, revise)
• or by using the beam design charts in Part 5 of
  the AISC manual, 3rd
• ed. (starting p. 5-71). There are two sets of
  charts one for W-shapes
• and one for C- and MC-shapes.
• (AISC Manual, 13th edition charts start on page
  3-96)

• 3. Check the shear strength.
• 4. Check the deflection.

109
Flexural Member Design Example

• Given Beam with loading shown
• Lateral support provided only at
  the vertical supports
• Allowable deflection due to dead
  and live loads is L/240.

• Find Select the best A992 steel W-shape
  bending is about strong axis.

• Solution Factored distributed load, wu 1.2D
  1.6L 1.2(2) 1.6(4) 8.8 k/ft

110
Flexural Member Design Example
Segment AB Segment DE Segment BC Segment CD ?
We'll only consider segments AB and BC
111
Flexural Member Design Example

• Segment AB

• Can determine Cb using moments computed at ¼
  points of span AB

• MA 530.6 k-ft
• MB 566.3 k-ft
• MC 106.9 k-ft
• Mmax .107wl2
• 847.4 k-ft

112
Flexural Member Design Example

• Segment BC

• Can determine Cb using moments computed at ¼
  points of span BC

• MA -39.7 k-ft
• MB 273.5 k-ft
• MC 91.0 k-ft
• Mmax .107wl2 847.4 k-ft

• Note AISC Manual 13th ed. limits Cb to 3.0 and
  includes an extra multiplier
• accounting for the cross-section's degree of
  symmetry

113
Flexural Member Design Example

• Both segments have the same Lb and must resist
  the same Mu.
• Cb is smaller for Segment AB, so it will have a
  lower strength (based on choice
• of a compact section), but it must resist the
  same Mu, so segment AB controls
• the design

• Go to beam design charts with Mu/Cb 847.4/1.68
  504.4 k-ft and Lb 30 ft

114
Flexural Member Design Example
115
Flexural Member Design Example

• Try a W14 x 99

• For Lb 30 ft, FbMn for a W14 x 99 539 k-ft
  (for Cb 1.0)
• For Cb 1.68, FbMn (539)(1.68) 905.5 k-ft
• But upper limit of FbMp' for W14 x 99 646 k-ft
• (W14 x 99 is noncompact, so use upper limit
  FbMp' )
• So, upper limit controls, and FbMn 646 k-ft lt
  Mu 847.4 k-ft ? N.G.

116
Flexural Member Design Example

• Try a W21 x 101

• For Lb 30 ft, FbMn for a W21 x 101 601 k-ft
  (for Cb 1.0)

• For Cb 1.68, FbMn (601)(1.68) 1009.7 k-ft

• Check upper limit of FbMp for W21 x 101 949
  k-ft

• Upper limit controls, so FbMn FbMp 949 k-ft
  gt Mu 847.4 k-ft ? Okay

117
Flexural Member Design Example

• Now check Mu including beam weight

• ?Mu 0.107(1.2)(0.101)(30)2 11.67 k-ft
• Mu 847.4 11.67 859.1 k-ft lt FbMn 949
  k-ft ? Okay for flexure

• Check shear

• Vu 0.607 wl 0.607(8.8)(30) 160.3 kips
• FvVn for W21 x 101 289 kips (Zx table)
• FvVn 289 kips gt Vu 160.3 kips ? Okay for
  shear

118
Flexural Member Design Example

• Check Deflection

• ?allowable L/240 (30 ft)(12 in/ft) / (240)
  1.50 in

• Note Use unfactored, service loads to compute
  deflection, a serviceability
• limit state

• ?max lt ?allowable ? W21 x 101 Okay for deflection

• Use a W21 x 101 for the given loading.

119
Plate Girder Analysis/Design

• Participants, please turn to page 109 of your
  notes
• Click here to go to slide 219.

120
Beam-Column Design

• If you assume Interaction Equation H1-1a governs
  (axial load dominant)

(AISC Eqn H1-1a)
Rewrite
Call this Equation 1
121
Beam-Column Design

• If you assume Interaction Equation H1-1b governs
  (bending load dominant)

(AISC Eqn H1-1b)
Rewrite
Call this Equation 2
122
Beam-Column Design
Design Procedure

• Select an average "m" value from Table 6-1 (or,
  if bending is dominant, select an average "b"
  value). If weak axis bending is included, also
  select an "n" value.
• Solve for b (or m, if you chose b) using Equation
  1 (or 2).
• Go to Table 6-2 select a shape that has b,m,n
  values close to the desired values.
• These values are based on the assumption that
  weak-axis buckling controls axial compressive
  strength and that Cb 1.0.
• Use b,m,n for the section selected to check
  interaction formulas (modified as Equations 1 and
  2).
• Repeat steps 3 and 4 as necessary.

123
Beam-Column Design

• Check assumptions
• If strong axis controls buckling, use
  to determine b from Table 6-2
• If Cb ? 1.0 ? must adjust "m" value

124
Beam-Column Design
Design of Beam-Columns in Unbraced Frames

• Almost always, B1 (amplification factor for
  non-sway moments) 1.0
• Assume B1 1.0 for preliminary design, then
  check to verify
• B2 (amplification factor for sway moments) is
  based on several quantities
• unknown until all columns in a frame have
  been chosen

(AISC Eqn C1-4)

• may not know ?oh (lateral drift) for preliminary
  design
• H sum of horizontal loads causing ?oh
• L story height

(AISC Eqn C1-5)

• SPe2 sum of Euler buckling loads for all
  columns in story
• can't know this until all columns in
  story are chosen

125
Beam-Column Design

• Two Methods for Evaluating B2 for Preliminary
  Design

• Assume B2 1.0. Select trial shape, then
  compute B2 from Eqn C1-5, assuming that
  is the same for all columns in the story
• Use a predetermined drift limit, , if
  available
• Use of a maximum permissible drift index is a
  serviceability requirement, so service lateral
  loads H should be used to compute .
  (Usually )

126
Beam-Column Design Example

• Given Single-story unbraced frame shown
  subjected to D, Lr, and W
• Approximate analysis results are
  shown
• All vertical loads are
  symmetrically placed
• (only contribute to non-sway moment, Mnt)
• Lateral load produces sway moments, Mlt
• Bending is about the strong axis, and each column
  is laterally braced at
• top and bottom
• Find Select a W14 shape for the columns
  using A992 steel
• Design for a drift index 1/400 based on service
  wind load

127
Beam-Column Design Example
128
Beam-Column Design Example

• Solution

• Load combinations involving D, Lr, W

2. 1.2D 0.5Lr
Pu 1.2(19) 0.5(33) 39.3 kips Mnt 1.2(79)
0.5(130) 159.8 kip-ft Mlt 0 kip-ft
3. 1.2D 1.6Lr 0.8W
Pu 1.2(19) 1.6(33) 0.8(-11 1.4) 67.92
kips Mnt 1.2(79) 1.6(130) 0.8(-46) 266.0
kip-ft Mlt 0.8(32) 25.6 kip-ft
129
Beam-Column Design Example
4. 1.2D 1.6W 0.5Lr
Pu 1.2(19) 1.6(-11 1.4) 0.5(33) 23.9
kips Mnt 1.2(79) 1.6(-46) 0.5(130) 86.2
kip-ft Mlt 1.6(32) 51.2 kip-ft

• Load Combination 3 controls (higher Pu and Mnt).
  Load Combination 4 has
• a higher Mlt, but will only control if the
  moment amplification factor for the
• sway case (B2, amplifying Mlt) is
  unrealistically large.

• Pu 67.92 kips
• Mnt 266.0 k-ft
• Mlt 25.6 k-ft

130
Beam-Column Design Example

• To select a trial shape, assume B1 1.0

2 columns in story
Therefore, Mu B1 Mnt B2 Mlt
(1.0)(266.0) 1.104(25.6)
294.3 k-ft
131
Beam-Column Design Example

• Frame member sizes aren't known, so can't use
  alignment charts to get K
• Table C-C2.1 Case (f) is closest to present
  end conditions
• Kx 2.0 for sidesway case
• Kx 1.0 for braced case
• Since columns are laterally braced (in
  out-of-plane direction), use Ky 1.0

132
Beam-Column Design Example

• Bending moment looks dominant ? use Equation 2

• For W14 with KL Lb 18 ft from Table 6-1 ? b
  0.624 x 10-3

? m 2.956 x 10-3
133
Beam-Column Design Example

• Try a W14 x 61, Ag 17.9 in2
• b 2.32 x 10-3
• m 2.76 x 10-3
• m is smaller than desired, b is larger ?
  hopefully will compensate

• Determine critical effective length for
  compression ? use sway condition
• 
  (worst case
  scenario)

• Therefore KL 18 ft., as assumed when using
  Table 6-1 to choose b

134
Beam-Column Design Example

• Verify that Equation 2 controls

• Therefore Interaction Equation 2 controls, as
  assumed

• For bending, consider braced condition first

• Cm coefficient dependent on column curvature
  caused by applied
• moments (only applies to braced
  condition)

• Pe1 elastic Euler buckling load for braced
  frame

135
Beam-Column Design Example

• No transverse loads, so

• M1 smaller moment at end of unbraced length
• M2 larger moment at end of unbraced length
• Ratio is positive for members bent in reverse
  curvature
• Ratio is negative for members bent in single
  curvature

Use K and r values for axis of bending
136
Beam-Column Design Example

• Therefore, B1 1.0, as assumed.

• Since B1 1.0 is same as assumed, and B2 won't
  change

Mu B1 Mnt B2 Mlt (1.0)(266.0)
1.104(25.6) 294.3 k-ft
(as previously calculated)
137
Beam-Column Design Example

• Must modify "m" to account for Cb ? 1.0 (since
  Tables tabulated with Cb 1.0)

• Case (g) applies ? Cb 1.67

• From Table 6-2
• FbMp 383 k-ft lt 537.8 k-ft

• Therefore

FbMn FbMp 383 k-ft
138
Beam-Column Design Example

• Now can check Interaction Equation 2

• Okay, W14 x 61 will work, but another may get
  closer to 1.0

139
Beam-Column Design Example

• Try W14 x 53, Ag 15.6 in2
• b 3.81 x 10-3
• m 3.58 x 10-3

• Determine critical effective length for
  compression ? use sway condition
• 
  (worst case
  scenario)

• Therefore KL 18 ft., as assumed when using
  Table 6-1 to choose b

140
Beam-Column Design Example

• Verify that Equation 2 controls

• Therefore Interaction Equation 1 controls

• For bending, consider braced condition first

141
Beam-Column Design Example

• Therefore, B1 1.0, as assumed.

• Since B1 1.0 is same as assumed, and B2 won't
  change

Mu B1 Mnt B2 Mlt (1.0)(266.0)
1.104(25.6) 294.3 k-ft
(as previously calculated)
142
Beam-Column Design Example

• Must modify "m" to account for Cb ? 1.0

• Case (g) applies ? Cb 1.67

• From Table 6-2
• FbMp 327 k-ft lt 414.6 k-ft

• Therefore

FbMn FbMp 327 k-ft
143
Beam-Column Design Example

• Check Interaction Equation 1

? N.G.

• Use a W14 x 61 for the given loading.

144
Design of Gable Frame

• Participants, please turn to page 73 of your
  notes.
• Click here to go to slide 146.

145
STRUCTURAL ANALYSIS
146
AISC Publication
147
DESIGN EXAMPLES (CONTINUED)
148
Design of Gable Frame
Examples Of Rigid Frames Designed For Industrial
Buildings
149
Design of Gable Frame
Typical Gable Frames and Building Terminology
150
Design of Gable Frame
Design Methods
The rigid frames are classified as Type FR
(fully restrained) in Article A.2. Connections
are assumed to be sufficient stiff to maintain
the angles between intersecting members. Thus,
two design methods are applicable LRFD using an
elastic analysis and plastic design using a
plastic analysis. In the elastic analysis, the
critical negative bending moment caused by
gravity loads may be reduced by 10 taking the
advantage of partial moment redistribution if
other requirements are met. The elastic analysis
of frames can be performed using any design aids
available (see AISC publication by Griffiths) or
by computer programs (see MFRAM).
151
Design of Gable Frame
Design Methods
Design by plastic analysis is a valid design
method for Type FR structures. Superposition of
statically indeterminate moments is not valid in
plastic analysis. The rafters and columns of a
single story rigid frame are stressed in bending
and axial compression. A member loaded in this
manner was treated in beam-columns in CIVL 4650.
However, if the axial forces are small compared
with the bending moments, they may be neglected.
Failure modes of a wide flange member includes
(1) inelastic instability (excessive deflection
in the plane of bending), (2) lateral-torsional
buckling (deflection plus twisting out of the
plane of bending), and (3) local buckling. The
shearing stresses within the connection between
the rafter and the column and the stability of
the haunched compression flanges should be
carefully checked.
152
Design of Gable Frame
Moment Redistribution
In order to take advantage of the 10 moment
redistribution, the frame must satisfy the
Article A.5. Moment redistribution is not
permitted for loads other than gravity loads such
as wind or other horizontal components. No such
limitation is placed on the plastic design
method. This makes the plastic design method
advantageous for tall frames or frames which must
support large horizontal loads provided that
sufficient bracing is available.
153
Design of Gable Frame
154
Design of Gable Frame
LRFD
1. Design Loads

• Bay dimension 24 feet
• Location somewhere in U.S. Midwest
• ASCE Standard 7-05, Minimum Design Loads for
  Buildings and Other Structures,
• used for minimum design load determination
• Assume A36 steel

155
Design of Gable Frame
LRFD
1. Design Loads
Roof roofing 4 psf (AISC
17-22) insulation 3 psf (AISC
17-22) metal deck 2 psf (supplier
brochure) Miscellaneous 4 psf purlins
3 psf (assumed) 16
psf snow load 20 psf (2 feet
accumulation is assumed)
156
Design of Gable Frame
LRFD
1. Design Loads
Rigid Frame roof 16x24x12.37/12/1000 0.395
frame 0.155 (assumed),
span/depth40, for span 120', d3'
(W36x155)
0.550 k/ft snow 20x24/1000
0.480 (ASCE7-05, page 10, Article
4.91, min roof LL1220 psf)
157
Design of Gable Frame
LRFD
1. Design Loads
Wind wind pressure
psf
k/ft
overturning moment
k-ft
eave thrust
kips, statically equivalent load
158
Design of Gable Frame
LRFD
1. Design Loads
Design of Purlins

• As per Fig. (a), purlins placed at 6' c to c on
  24' simple span.
• Purlin top flange assumed to be continuously
  braced against lateral-torsional
• buckling by tack welding of corrugated metal
  decking.

159
Design of Gable Frame
LRFD
1. Design Loads
Design of Purlins
dead load 16x6/1000 0.096 k/ft live load
20x6x12/12.37/1000 0.116 k/ft (the assumed live
load is for the horizontal projection)
k/ft
k/ft
k/ft
k/ft
160
Design of Gable Frame
(Article 2.3.2, ASCE 7-05, Basic Load Combination
3)
161
Design of Gable Frame

• Try W10x12

• Flange b/t3.96/2/.219.4lt

? Okay

• Web

? Okay
(Say), hence, compact
? Okay
162
Design of Gable Frame

• Roof ponding conditions are not checked here as
  it appears satisfactory.

• The simple span assumption for the purlins is
  conservative because of the
• actual support condition shown in Fig. (f)

• Check the assumed load versus actual
• purlin load 12/62 psf vs. 3, no need to
• revise the design.

163
Design of Gable Frame
Rigid Frame Analysis

• Loading on the Gable Frame

• Dead load 0.550 k/ft
• Live load from the roof 20x24
  0.480 k/ft

• For the analysis of the frame for the vertical
  load, see page 18 (Case 1),
• AISC publication.

164
Design of Gable Frame

• Assume

then
kips due to DL
kips due to LL
165
Design of Gable Frame

• For the analysis of the frame for the horizontal
  load, see page 18 (Case 6),
• AISC publication.

• b1 as the eave thrust is applied at the eave.

kips at leeward
kips at windward

• Refer to MFRAM outputs.

166
Design of Gable Frame
CIVL5650-6650GABLE FRAME INDUSTRIAL BUILDING
STRUCTURE DATA
BW NOM DOF NOJ NOR NORJ E 6 6
17 7 4 2 0.29000E05 BLANK COMMON ARRAY
527 COORDINATES OF JOINTS - STRUCTURE COORDINATE
SYSTEM JOINT X Y 1 0.000
0.000 2 0.000 240.000 3 591.360
387.840 4 720.000 420.000 5 848.640
387.840 6 1440.000 240.000 7 1440.000
0.000 MEMBER DESIGNATIONS AND PROPERTIES MEMBER
NE FE AREA INERTIA LENGTH DIR
COS DIR SIN NOSEG 1 1 2 34.700
5900.000 240.000 0.0000 1.0000 0 2
2 3 34.700 5900.000 609.560 0.9701
0.2425 0 3 3 4 34.700
5900.000 132.599 0.9701 0.2425 0 4
4 5 34.700 5900.000 132.599 0.9701
-0.2425 0 5 5 6 34.700
5900.000 609.560 0.9701 -0.2425 0 6
6 7 34.700 5900.000 240.000 0.0000
-1.0000 0 JOINT RESTRAINT JOINT X-RSTRT
Y-RSTRT Z-RSTRT 1 1 1 0 7
1 1 0
167
Design of Gable Frame
LOADING NO. 1 LOAD DATA NLJ NLM 0
4 TOTAL LOADING CASES IN EACH MEMBER MEMBER
LOADING 2 1 NN LOAD ID PCOS PSIN
DIS TO LOAD CONC LOAD UNIF LOAD CONC MOMT
1 2 0.2425 0.9701 1.0000
0.0000 0.0445 0.0000 MEMBER LOADING
3 1 NN LOAD ID PCOS PSIN DIS TO
LOAD CONC LOAD UNIF LOAD CONC MOMT 1
2 0.2425 0.9701 1.0000 0.0000
0.0445 0.0000 MEMBER LOADING 4 1
NN LOAD ID PCOS PSIN DIS TO LOAD CONC
LOAD UNIF LOAD CONC MOMT 1 2 -0.2425
0.9701 1.0000 0.0000 0.0445
0.0000 MEMBER LOADING 5 1 NN LOAD ID
PCOS PSIN DIS TO LOAD CONC LOAD UNIF
LOAD CONC MOMT 1 2 -0.2425 0.9701
1.0000 0.0000 0.0445 0.0000
168
Design of Gable Frame
ACTIONS AT END OF RESTRAINED MEMBERS DUE TO
LOADS MEMBER NE THRUST NE SHEAR NE MOMENT
FE THRUST FE SHEAR FE MOMENT 2
0.3287E01 0.1315E02 0.1296E04 0.3287E01
0.1315E02 -0.1296E04 3 0.7150E00
0.2860E01 0.6132E02 0.7150E00 0.2860E01
-0.6132E02 4 -0.7150E00 0.2860E01
0.6132E02 -0.7150E00 0.2860E01 -0.6132E02
5 -0.3287E01 0.1315E02 0.1296E04 -0.3287E01
0.1315E02 -0.1296E04 JOINT DISPLACEMENTS -
STRUCTURE COORDINATE JOINT X-DISPL
Y-DISPL Z-ROTN 1 0.000000
0.000000 0.003498 2 -0.522115
-0.007870 -0.000470 3 -0.023996
-2.069238 -0.001729 4 0.000000
-2.178035 0.000000 5 0.023996
-2.069238 0.001729 6 0.522115
-0.007870 0.000470 7 0.000000
0.000000 -0.003498 MEMBER END ACTIONS IN
MEMBER COORDINATES MEMBER NE THRUST NE SHEAR
NE MOMENT FE THRUST FE SHEAR FE MOMENT
1 0.3300E02 -0.2357E02 -0.8304E-12 -0.3300E02
0.2357E02 -0.5657E04 2 0.3087E02
0.2630E02 0.5657E04 -0.2430E02 -0.2998E-02
0.2359E04 3 0.2430E02 0.2998E-02
-0.2359E04 -0.2287E02 0.5717E01 0.1980E04
4 0.2287E02 0.5717E01 -0.1980E04
-0.2430E02 0.2998E-02 0.2359E04 5
0.2430E02 -0.2998E-02 -0.2359E04 -0.3087E02
0.2630E02 -0.5657E04 6 0.3300E02
0.2357E02 0.5657E04 -0.3300E02 -0.2357E02
-0.2883E-11 REACTIONS - STRUCTURE
COORDINATE JOINT X-REACT Y-REACT
Z-REACT 1 23.571639 33.000000
0.000000 7 -23.571639 33.000000
0.000000
169
Design of Gable Frame
LOADING NO. 2 LOAD DATA NLJ NLM 0
3 TOTAL LOADING CASES IN EACH MEMBER MEMBER
LOADING 1 1 NN LOAD ID PCOS PSIN
DIS TO LOAD CONC LOAD UNIF LOAD CONC MOMT
1 2 0.0000 1.0000 1.0000
0.0000 0.0400 0.0000 MEMBER LOADING
2 1 NN LOAD ID PCOS PSIN DIS TO
LOAD CONC LOAD UNIF LOAD CONC MOMT 1
2 -0.9701 0.2425 1.0000 0.0000
0.0097 0.0000 MEMBER LOADING 3 1
NN LOAD ID PCOS PSIN DIS TO LOAD CONC
LOAD UNIF LOAD CONC MOMT 1 2 -0.9701
0.2425 1.0000 0.0000 0.0097
0.0000 ACTIONS AT END OF RESTRAINED MEMBERS
DUE TO LOADS MEMBER NE THRUST NE SHEAR NE
MOMENT FE THRUST FE SHEAR FE MOMENT 1
0.0000E00 0.4800E01 0.1920E03 0.0000E00
0.4800E01 -0.1920E03 2 -0.2869E01
0.7171E00 0.1767E02 -0.2869E01 0.7171E00
-0.1767E02 3 -0.6240E00 0.1560E00
0.8362E00 -0.6240E00 0.1560E00 -0.8362E00
170
Design of Gable Frame
JOINT DISPLACEMENTS - STRUCTURE COORDINATE JOINT
X-DISPL Y-DISPL Z-ROTN 1
0.000000 0.000000 -0.003577 2
0.731418 0.000584 -0.002123 3
0.732305 -0.003220 0.001187 4
0.693879 0.148530 0.001115 5
0.724424 0.273650 0.000807 6
0.652489 -0.000584 -0.002162 7
0.000000 0.000000 -0.002997 MEMBER
END ACTIONS IN MEMBER COORDINATES MEMBER NE
THRUST NE SHEAR NE MOMENT FE THRUST FE
SHEAR FE MOMENT 1 -0.2450E01 0.1184E02
0.1984E-11 0.2450E01 -0.2238E01 0.1689E04
2 -0.2765E01 -0.1834E01 -0.1689E04 -0.2972E01
0.3268E01 0.1339E03 3 0.2972E01
-0.3268E01 -0.1339E03 -0.4220E01 0.3580E01
-0.3202E03 4 0.5408E01 -0.1173E01
0.3202E03 -0.5408E01 0.1173E01 -0.4757E03
5 0.5408E01 -0.1173E01 0.4757E03 -0.5408E01
0.1173E01 -0.1191E04 6 0.2450E01
0.4962E01 0.1191E04 -0.2450E01 -0.4962E01
-0.5826E-12 REACTIONS - STRUCTURE
COORDINATE JOINT X-REACT Y-REACT
Z-REACT 1 -11.837691 -2.450000
0.000000 7 -4.962309 2.450000
0.000000
171
Design of Gable Frame
LOADING NO. 3 LOAD DATA NLJ NLM 1
0 ACTIONS APPLIED AT JOINTS JOINT X-ACTION
Y-ACTION Z-ACTION 2
14.700 0.000 0.000 JOINT
DISPLACEMENTS - STRUCTURE COORDINATE JOINT
X-DISPL Y-DISPL Z-ROTN 1
0.000000 0.000000 -0.004165 2
0.873207 0.000584 -0.002586 3
0.856308 0.056771 0.001451 4
0.810894 0.235945 0.001274 5
0.844672 0.374184 0.000853 6
0.747385 -0.000584 -0.002517 7
0.000000 0.000000 -0.003413 MEMBER END
ACTIONS IN MEMBER COORDINATES MEMBER NE THRUST
NE SHEAR NE MOMENT FE THRUST FE SHEAR FE
MOMENT 1 -0.2450E01 0.9379E01 -0.6919E-12
0.2450E01 -0.9379E01 0.2251E04 2
0.4568E01 -0.3667E01 -0.2251E04 -0.4568E01
0.3667E01 0.1533E02 3 0.4568E01
-0.3667E01 -0.1533E02 -0.4568E01 0.3667E01
-0.4710E03 4 0.5757E01 -0.1086E01
0.4710E03 -0.5757E01 0.1086E01 -0.6150E03
5 0.5757E01 -0.1086E01 0.6150E03 -0.5757E01
0.1086E01 -0.1277E04 6 0.2450E01
0.5321E01 0.1277E04 -0.2450E01 -0.5321E01
-0.1518E-11 REACTIONS - STRUCTURE
COORDINATE JOINT X-REACT Y-REACT
Z-REACT 1 -9.378637 -2.450000
0.000000 7 -5.321363 2.450000
0.000000
172
Design of Gable Frame
Bending Moment Diagram Under Gravity Loading
173
Design of Gable Frame
Statics
174
Design of Gable Frame
Thrust and Shear of Rafter, Moment Redistribution
175
Design of Gable Frame

• Examination of the basic load combinations given
  by Article 2.3.2, ASCE 7-05
• reveals that the load combination 3

or the load combination 4
appears to be most critical.
176
Design of Gable Frame

• Since the superposition method is valid for the
  linear elastic analysis, values
• from the MFRAM outputs can be used to generate
  Table 1.
• Moment redistribution will be performed for the
  gravity loading assuming the
• gable frame will be classified as compact
  sections as per AISC Table B5.1

177
Design of Gable Frame
Table 1. Summary of Analysis
Table 1 indicates that the load combination 3
governs the design.
178
Design of Gable Frame
Rafter design

• Since the rafter bracing detail is as shown in
  Fig. (f), the rafter is assumed to
• be braced at every 6 feet against both flexure
  and compression.

• Try W36x135, A39.7 in2 ,

Eq. (H1-1b) controls.
179
Design of Gable Frame
? compact section
Hence
? Okay
180
Design of Gable Frame

• Try W33x118, A34.7 in2

Eq. (H1-1b) controls.
181
Design of Gable Frame
? compact section
Hence
Say ok (1 overstressed)
182
Design of Gable Frame
Column design

• The column is subjected to the same moment at
  the knee, 1086.9 k-ft, and
• almost the same thrust, 83.7 kips.
• However, the bracing interval is assumed to be
  7.5 ft (the girt spacing)
• If the side of the building is open for overhang
  sliding doors, etc., then the
• bracing length may be considered to be 20
  ft.
• Check for both bracing intervals.

183
Design of Gable Frame

• For

• Try W33x118, A34.7 in2,

• Determine the effective length factor with
  respect to the strong axis
• (sidesway not prevented).

Read off
184
Design of Gable Frame
? controls
Eq. (H1-1b) controls.
185
Design of Gable Frame

• compact
• section

Hence
Say ok (1 overstressed)
186
Design of Gable Frame

• Check for

Eq. (H1-1b) controls.
187
Design of Gable Frame

• not compact
• section

188
Design of Gable Frame
Say NG (4 overstressed)
189
Design of Gable Frame

• Try W36x135, A39.7 in2

Eq. (H1-1b) controls.
190
Design of Gable Frame

• not compact
• section

191
Design of Gable Frame
Hence
? Okay
192
Design of Gable Frame

• Use W36x135 for both rafters and columns.
• (If tapered members are permitted, a
  considerably smaller size member can be
• made acceptable.)
• (Cost analysis material vs. fabrication.)

193
Design of Gable Frame
Girt Design

• Horizontal girt
• Vertical girt
• Tributary area
• Sag rod
• Minor axis bending
• Connection details

194
Design of Gable Frame
Column Base Plate

• Effect of boundary conditions on
• Frame moments
• Anchor bolt pattern
• Grade beam
• Isolated footing
• Spread footing
• Wire mesh

195
Design of Gable Frame
Wind Bracing
"Bracing is expensive, and should not be
overdone. Just what is necessary in a given
situation is, to a considerable extent, a
question of judgment that depends on the
engineer's qualitative evaluation. Forces in the
members of the bracing system are rarely large
enough to control their design. Therefore,
except for whatever guideline is offered by
slenderness ratio limitations, their size is
largely a question of judgment."
Page 617, Design of Steel Structures, Gaylord and
Gaylord, 2nd ed., McGraw-Hill, 1972.
196
Design of Gable Frame
Part Side Wall

• See text