NS18steeldesign 3 PDF
NS18steeldesign 3 PDF
NS18steeldesign 3 PDF
Steel Design
Notation:
a =name for width dimension db = nominal bolt diameter
A =name for area D = shorthand for dead load
Ab =area of a bolt DL = shorthand for dead load
Ae =effective net area found from the e = eccentricity
product of the net area An by the E = shorthand for earthquake load
shear lag factor U = modulus of elasticity
Ag = gross area, equal to the total area fc = axial compressive stress
ignoring any holes fb = bending stress
Agv = gross area subjected to shear for fp = bearing stress
block shear rupture fv = shear stress
An = net area, equal to the gross area fv-max= maximum shear stress
subtracting any holes, as is Anet fy = yield stress
Ant = net area subjected to tension for F = shorthand for fluid load
block shear rupture Fallow(able) = allowable stress
Anv = net area subjected to shear for block Fa = allowable axial (compressive) stress
shear rupture Fb = allowable bending stress
Aw = area of the web of a wide flange Fcr = flexural buckling stress
section Fe = elastic critical buckling stress
AISC = American Institute of Steel FEXX = yield strength of weld material
Construction Fn = nominal strength in LRFD
ASD = allowable stress design = nominal tension or shear strength of
b = name for a (base) width a bolt
= total width of material at a Fp = allowable bearing stress
horizontal section Ft = allowable tensile stress
= name for height dimension Fu = ultimate stress prior to failure
bf = width of the flange of a steel beam Fv = allowable shear stress
cross section Fy = yield strength
B1 = factor for determining Mu for Fyw = yield strength of web material
combined bending and compression F.S. = factor of safety
c = largest distance from the neutral g = gage spacing of staggered bolt
axis to the top or bottom edge of a holes
beam G = relative stiffness of columns to
c1 = coefficient for shear stress for a beams in a rigid connection, as is Ψ
rectangular bar in torsion h = name for a height
Cb = lateral torsional buckling hc = height of the web of a wide flange
modification factor for moment in steel section
ASD & LRFD steel beam design H = shorthand for lateral pressure load
Cc = column slenderness classification I = moment of inertia with respect to
constant for steel column design neutral axis bending
Cm = modification factor accounting for Itrial = moment of inertia of trial section
combined stress in steel design Ireq’d = moment of inertia required at
Cv = web shear coefficient limiting deflection
d = calculus symbol for differentiation Iy = moment of inertia about the y axis
= depth of a wide flange section J = polar moment of inertia
= nominal bolt diameter
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Steel Design
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Materials
American Society for Testing Materials (ASTM) is the organization responsible for material and
other standards related to manufacturing. Materials meeting their standards are guaranteed to
have the published strength and material properties for a designation.
A36 – carbon steel used for plates, angles Fy = 36 ksi, Fu = 58 ksi, E = 29,000 ksi
A572 – high strength low-alloy use for some beams Fy = 60 ksi, Fu = 75 ksi, E = 29,000 ksi
A992 – for building framing used for most beams Fy = 50 ksi, Fu = 65 ksi, E = 29,000 ksi
(A572 Grade 50 has the same properties as A992)
ASD Rn
Ra ≤
Ω
where Ra = required strength (dead or live; force, moment or stress)
Rn = nominal strength specified for ASD
Ω = safety factor
Factors of Safety are applied to the limit stresses for allowable stress values:
LRFD where Ru = Σγ i Ri
Ru ≤ φRn
where φ = resistance factor
γ = load factor for the type of load
R = load (dead or live; force, moment or stress)
Ru = factored load (moment or stress)
Rn = nominal load (ultimate capacity; force, moment or stress)
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The design strength, φRn , of each structural element or structural assembly must equal or exceed
the design strength based on the ASCE-7 (2010) combinations of factored nominal loads:
1.4D
1.2D + 1.6L + 0.5(Lr or S or R)
1.2D + 1.6(Lr or S or R) + (L or 0.5W)
1.2D + 1.0W + L + 0.5(Lr or S or R)
1.2D + 1.0E + L + 0.2S
0.9D + 1.0W
0.9D + 1.0E
Drawing V and M diagrams will show us the maximum values for design. Remember:
V = Σ(− w)dx dV dM
= −w =V
M = Σ(V )dx dx dx
For a prismatic member (constant cross section), the maximum normal stress will occur at the
maximum moment.
For a non-prismatic member, the stress varies with the cross section AND the moment.
Deflections
1 M ( x)
If the bending moment changes, M(x) across a beam of constant material and cross =
section then the curvature will change: R EI
1
EI ∫
The slope of the n.a. of a beam, θ, will be tangent to the radius of θ = slope = M ( x)dx
curvature, R:
1 1
The equation for deflection, y, along a beam is: y=
EI ∫ θdx =
EI ∫∫ M ( x)dx
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Elastic curve equations can be found in handbooks, textbooks, design manuals, etc...Computer
programs can be used as well. Elastic curve equations can be superimposed ONLY if the stresses
are in the elastic range.
The deflected shape is roughly the same shape flipped as the bending moment diagram but is
constrained by supports and geometry.
All building codes and design codes limit deflection for beam types and damage that could
happen based on service condition and severity.
y max ( x) = ∆ actual ≤ ∆ allowable = L
value
Lateral Buckling
With compression stresses in the top of a beam, a sudden “popping” or buckling can happen
even at low stresses. In order to prevent it, we need to brace it along the top, or laterally brace it,
or provide a bigger Iy.
Concentrated forces on a steel beam can cause the web to buckle (called web crippling). Web
stiffeners under the beam loads and bearing plates at the supports reduce that tendency. Web
stiffeners also prevent the web from shearing in plate girders.
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Pn (interior) = ( 5k + N )Fyw t w
where tw = thickness of the web
Fyw = yield strength of the web
N = bearing length
k = dimension to fillet found in beam section tables
φ = 1.00 (LRFD) Ω = 1.50 (ASD)
In order to determine the loads on a beam (or girder, joist, column, frame, foundation...) we can
start at the top of a structure and determine the tributary area that a load acts over and the beam
needs to support. Loads come from material weights, people, and the environment. This area is
assumed to be from half the distance to the next beam over to halfway to the next beam.
The reactions must be supported by the next lower structural element ad infinitum, to the ground.
For determining the flexural design strength, φb Mn , for resistance to pure bending (no axial
load) in most flexural members where the following conditions exist, a single calculation will
suffice:
Σγ i R i = M u ≤ φ b M n = 0 . 9 F y Z
f
Plastic Section Modulus
fy = 50ksi
Plastic behavior is characterized by a yield point and an
increase in strain with no increase in stress. E
1
ε
εy = 0.001724
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Atension = Acompression
Shape Factor:
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Va ≤ Vn / Ω or Vu ≤ φvVn
The nominal shear strength is dependent on the cross section shape. Case 1: With a thick or stiff
web, the shear stress is resisted by the web of a wide flange shape (with the exception of a
handful of W’s). Case 2: When the web is not stiff for doubly symmetric shapes, singly
symmetric shapes (like channels) (excluding round high strength steel shapes), inelastic web
buckling occurs. When the web is very slender, elastic web buckling occurs, reducing the
capacity even more:
E
Case 1) For h tw ≤ 2.24 Vn = 0.6Fyw Aw φv = 1.00 (LRFD) Ω = 1.50 (ASD)
Fy
where h equals the clear distance between flanges less the fillet or corner
radius for rolled shapes
Vn = nominal shear strength
Fyw = yield strength of the steel in the web
Aw = twd = area of the web
E
Case 2) For h tw > 2.24 Vn = 0.6Fyw AwCv φv = 0.9 (LRFD) Ω = 1.67 (ASD)
Fy
Beam design charts show available moment, Mn/Ω and φb Mn , for unbraced length, Lb, of the
compression flange in one-foot increments from 1 to 50 ft. for values of the bending coefficient
Cb = 1. For values of 1<Cb≤2.3, the required flexural strength Mu can be reduced by dividing it
by Cb. (Cb = 1 when the bending moment at any point within an unbraced length is larger than
that at both ends of the length. Cb of 1 is conservative and permitted to be used in any case.
When the free end is unbraced in a cantilever or overhang, Cb = 1. The full formula is provided
below.)
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Compact Sections
For a laterally braced compact section (one for which the plastic moment can be reached before
local buckling) only the limit state of yielding is applicable. For unbraced compact beams and
non-compact tees and double angles, only the limit states of yielding and lateral-torsional
buckling are applicable.
bf E h E
Compact sections meet the following criteria: ≤ 0.38 and c ≤ 3.76
2t f Fy tw Fy
where:
bf = flange width in inches
tf = flange thickness in inches
E = modulus of elasticity in ksi
Fy = minimum yield stress in ksi
hc = height of the web in inches
tw = web thickness in inches
where Mp = Mn = FyZx
and Cb is a modification factor for non-uniform moment diagrams where, when both ends of
the beam segment are braced:
12.5M max
Cb =
2.5M max + 3M A + 4 M B + 3M C
Mmax = absolute value of the maximum moment in the unbraced beam segment
MA = absolute value of the moment at the quarter point of the unbraced beam segment
MB = absolute value of the moment at the center point of the unbraced beam segment
MC = absolute value of the moment at the three quarter point of the unbraced beam
segment length.
Plots of the available moment for the unbraced length for wide flange sections are useful to find
sections to satisfy the design criteria of M a ≤ M n / Ω or M u ≤ φb M n . The maximum moment
that can be applied on a beam (taking self weight into account), Ma or Mu, can be plotted against
the unbraced length, Lb. The limiting length, Lp (fully plastic), is indicated by a solid dot (•),
while the limiting length, Lr (for lateral torsional buckling), is indicated by an open dot ().
Solid lines indicate the most economical, while dashed lines indicate there is a lighter section
that could be used. Cb, which is a lateral torsional buckling modification factor for non-zero
moments at the ends, is 1 for simply supported beams (0 moments at the ends). (see figure)
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Design Procedure
The intent is to find the most light weight member (which is economical) satisfying the section
modulus size.
1. Determine the unbraced length to choose the limit state (yielding, lateral torsional buckling
or more extreme) and the factor of safety and limiting moments. Determine the material.
2. Draw V & M, finding Vmax and Mmax.for unfactored loads (ASD, Va & Ma) or from factored
loads (LRFD, Vu & Mu)
3. Calculate Zreq’d when yielding is the limit state. This step is equivalent to determining if
M max M max M max Mu
fb = ≤ Fb , Z req' d ≥ = and Z req 'd ≥ to meet the design criteria that
S Fb Fy φ b Fy
Ω
Ma ≤ Mn / Ω or Mu ≤ φbMn
If the limit state is something other than yielding,
determine the nominal moment, Mn, or use plots of
available moment to unbraced length, Lb.
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P
7. Provide adequate bearing area at supports. This step is equivalent to determining if fp = ≤ Fp
A
is satisfied to meet the design criteria that Pa ≤ Pn / Ω or Pu ≤ φPn
Tρ T
8. Evaluate shear due to torsion fv = or ≤ Fv (circular section or rectangular)
J c1 ab2
9. Evaluate the deflection to determine if ∆maxLL ≤ ∆LL−allowed and/or ∆maxTotal ≤ ∆Totalallowed
**** note: when ∆calculated > ∆limit, Ireq’d can be found with: ∆too big
I req' d ≥ I trial
and Zreq’d will be satisfied for similar self weight ***** ∆lim it
Tables exist for the common loading situation of uniformly distributed load. The tables provide
the safe distributed loads based on bending and deflection limits. For specific clear spans, they
provide maximum total loads and live loads for a specific deflection limits. wequivalent L2
If the load is not uniform, an equivalent uniform load can be calculated M max =
from the maximum moment equation: 8
If the deflection limit needed is not that of the table,
the design live load must be adjusted. For example: L / 360 table limit
wadjusted = wll − have
L / 400 wanted
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Fa = 1 −
( )
Kl 2
r Fy
2Cc2 F.S.
with:
( ) ( )
Kl
5 3 r
Kl
r
3
F .S . = + −
3 8C c 8C c3
Pa ≤ Pn / Ω or Pu ≤ φc Pn where Pu = Σγ i Pi
γ is a load factor
P is a load type
φ is a resistance factor
Pn is the nominal load capacity (strength)
where : Ag is the cross section area and Fcr is the flexural buckling stress
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KL E
when ≤ 4.71 or ( Fe ≥ 0.44Fy ):
r Fy
Fy
Fcr = 0 .658 Fe F y
KL E
when > 4.71 or ( Fe < 0.44Fy ):
r Fy
Fcr = 0.877Fe
π 2E
where Fe is the elastic critical buckling stress: Fe =
(KL r ) 2
Design Aids
Tables exist for the value of the flexural buckling stress based on slenderness ratio. In addition,
tables are provided in the AISC Manual for Available Strength in Axial Compression based on
the effective length with respect to least radius of gyration, ry. If the critical effective length is
about the largest radius of gyration, rx, it can be turned into an effective length about the y axis
by dividing by the fraction rx/ry.
Sample AISC Table for Available Strength in Axial Compression
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3. Pick a trial section based on if we think r or A is going to govern the section size.
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4. Analyze the stresses and compare to allowable using the allowable stress method or
interaction formula for eccentric columns.
6. Change the section choice and go back to step 4. Repeat until the section meets the
stress criteria.
Pr P 8 Mx My Pu 8 M ux M uy
+ + + ≤ 1.0
M ny
For ≥ 0.2 : + ≤ 1.0
Pc Pn 9 M nx φc Pn 9 φb M nx φb M ny
Ω Ω Ω
(ASD) (LRFD)
Pr P Mx My Pu M ux M uy
For < 0. 2 : + + ≤ 1.0 + + ≤ 1 .0
Pc 2Pn M nx M ny 2φ c Pn φ b M nx φb M ny
Ω Ω Ω
(ASD) (LRFD)
where:
for compression φc = 0.90 (LRFD) Ω = 1.67 (ASD)
for bending φb = 0.90 (LRFD) Ω = 1.67 (ASD)
m
C
For a braced condition, the moment magnification factor B1 is determined by B1 = ≥ 1.0
1 − α( Pu Pe1 )
where Cm is a modification factor accounting for end conditions
When not subject to transverse loading between supports in plane of bending:
= 0.6 – 0.4 (M1/M2) where M1 and M2 are the end moments and M1<M2. M1/M2 is
positive when the member is bent in reverse curvature (moments the same
direction), negative when bent in single curvature.
When there is transverse loading between the two ends of a member:
= 0.85, members with restrained (fixed) ends
= 1.00, members with unrestrained ends π 2 EA
Pe1 =
α = 1.00 (LRFD), 1.60 (ASD) (Kl r )
2
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Connections must be able to transfer any axial force, shear, or moment from member to member
or from beam to column.
Connections for steel are typically high strength bolts and electric arc welds. Recommended
practice for ease of construction is to specified shop welding and field bolting.
1. tension yielding
2. shear yielding
3. bearing yielding
4. bending yielding due to eccentric loads
5. rupture
A307: similar in strength to A36 steel (also known as ordinary, common or unfinished
bolts)
A325: high strength bolts (Group A)
A490: high strength bolts (higher than A325) (Group B)
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Available shear values are given by bolt type, diameter, and loading (Single or Double shear) in
AISC manual tables. Available shear value for slip-critical connections are given for limit states
of serviceability or strength by bolt type, hole type (standard, short-slotted, long-slotted or
oversized), diameter, and loading. Available tension values are given by bolt type and diameter
in AISC manual tables.
Available bearing force values are given by bolt diameter, ultimate tensile strength, Fu, of the
connected part, and thickness of the connected part in AISC manual tables.
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The minimum edge desistance from the center of the outer most bolt to the edge of a member is
generally 1¾ times the bolt diameter for the sheared edge and 1¼ times the bolt diameter for the
rolled or gas cut edges.
The maximum edge distance should not exceed 12 times the thickness of thinner member or 6 in.
Standard bolt hole spacing is 3 in. with the minimum spacing of 2 2 3 times the diameter of the
bolt, db. Common edge distance from the center of last hole to the edge is 1¼ in..
The smallest effective are must be determined by subtracting the bolt hole areas. With staggered
holes, the shortest length must be evaluated.
A series of bolts can also transfer a portion of the tensile force, and some of the effective net
areas see reduced stress.
The effective net area, Ae, is determined from the net area, An, multiplied by a shear lag factor, U,
which depends on the element type and connection configuration. If a portion of a connected
member is not fully connected (like the leg of an angle), the unconnected part is not subject to the
full stress and the shear lag factor can range from 0.6 to 1.0: Ae = AnU
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1. yielding Rn = Fy Ag
φ = 0.90 (LRFD) Ω = 1.67 (ASD)
2. rupture Rn = Fu Ae
φ = 0.75 (LRFD) Ω = 2.00 (ASD)
Welded Connections
Weld sizes are limited by the size of the parts being put
together and are given in AISC manual table J2.4 along
with the allowable strength per length of fillet weld,
referred to as S.
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The minimum length of a fillet weld is 4 times the nominal size. If it is not, then the weld size
used for design is ¼ the length.
Coping is the term for cutting away part of the flange to connect a
beam to another beam using welded or bolted angles.
AISC provides tables that give bolt and angle available strength knowing number of bolts, bolt
type, bolt diameter, angle leg thickness, hole type and coping, and the wide flange beam being
connected. For the connections the limit-state of bolt shear, bolts bearing on the angles, shear
yielding of the angles, shear rupture of the angles, and block shear rupture of the angles, and bolt
bearing on the beam web are considered.
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In addition to resisting shear and tension in bolts and shear in welds, the connected materials
may be subjected to shear, bearing, tension, flexure and even prying action. Coping can
significantly reduce design strengths and may require web reinforcement. All the following
must be considered:
• shear yielding
• shear rupture
• block shear rupture -
failure of a block at a beam as a
result of shear and tension
• tension yielding
• tension rupture
• local web buckling
• lateral torsional buckling
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where:
Anv is the net area subjected to shear
Ant is the net area subjected to tension
Agv is the gross area subjected to shear
Ubs = 1.0 when the tensile stress is uniform (most cases)
= 0.5 when the tensile stress is non-uniform
Gusset Plates
Gusset plates are used for truss member connections where the geometry prevents the members
from coming together at the joint “point”. Members being joined are typically double angles.
Decking
Shaped, thin sheet-steel panels that span several joists or evenly spaced support behave as
continuous beams. Design tables consider a “1 unit” wide strip across the supports and
determine maximum bending moment and deflections in order to provide allowable loads
depending on the depth of the material.
The other structural use of decking is to construct what is called a diaphragm, which is a
horizontal unit tying the decking to the joists that resists forces parallel to the surface of the
diaphragm.
When decking supports a concrete topping or floor, the steel-concrete construction is called
composite.
Frame Columns
Because joints can rotate in frames, the effective length of the column in a frame is harder to
determine. The stiffness (EI/L) of each member in a joint determines how rigid or flexible it is.
To find k, the relative stiffness, G or Ψ, must be found for both ends, plotted on the alignment
charts, and connected by a line for braced and unbraced fames.
Σ EI l
c
G =Ψ =
EI
Σ l
b
where
E = modulus of elasticity for a member
I = moment of inertia of for a member
lc = length of the column from center to center
lb = length of the beam from center to center
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(unified ASD)
Fy = 50 ksi
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Example 2
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Example 3
For the same beam and loading of Example 1, select the most economical
beam using Load and Resistance Factor Design (LRFD) with the 18” depth
restriction. Assume the distributed load is dead load, and the point load is
live load. Fy = 50 ksi and E = 30x103 ksi
(1.6)20k = 32k
(1.2)1k/ft = 1.2k/ft
1. To find Vu-max and Mu-max, factor the loads, construct a new load diagram,
shear diagram and bending moment diagram.
32.8k 32.8k
3. Check the shear capacity to satisfy Vu ≤ φvVn: Aweb = dtw and d=17.99 in., tw = 0.355 in. for the W18x50
So 33.64k ≤ 191.6 k OK
4. Calculate the deflection from the unfactored loads, including the self-weight now because it is known,
and satisfy the deflection criteria of ∆LL≤∆LL-limit and ∆total≤∆total-limit. (This is identical to what is done in
Example 1.) Ix =800 in3 for the W18x50
∆total-limit = L/240 = 1.4 in., say ∆LL = L/360 = 0.93 in
So 1.26 in. ≤ 1.4 in., and 0.658 in. ≤ 0.93 in. OK ∴FINAL SELECTION IS W18x50
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Example 4
A steel beam with a 20 ft span is designed to be simply supported at the ends on columns and to carry a floor system
made with open-web steel joists at 4 ft on center. The joists span 28 feet and frame into the beam from one side
only and have a self weight of 8.5 lb/ft. Use A992 (grade 50) steel and select the most economical wide-flange
section for the beam with LRFD design. Floor loads are 50 psf LL and 14.5 psf DL.
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Example 5
Select a A992 W shape flexural member (Fy = 50 ksi, Fu = 65 ksi) for a beam with distributed loads of 825 lb/ft
(dead) and 1300 lb/ft (live) and a live point load at midspan of 3 k using the Available Moment tables. The beam is
simply supported, 20 feet long, and braced at the ends and midpoint only (Lb = 10 ft.) The beam is a roof beam for
an institution without plaster ceilings. (LRFD)
1.6(3k) =4.8k
SOLUTION: 10’ 10’
To use the Available Moment tables, the maximum moment required is plotted against
the unbraced length. The first solid line with capacity or unbraced length above what is
needed is the most economical.
w = 1.2(825 lb/ft)+1.6(1300 lb/ft) =3.07k/ft
DESIGN LOADS (load factors applied on figure):
Plotting 662 k-ft vs. 10 ft lands just on the capacity of the W21x83, but it is dashed (and not the most economical) AND we need to
consider the contribution of self weight to the total moment. Choose a trial section of W24 x 76. Include the new dead load:
Replot 680.2 k-ft vs. 10ft, which lands above the capacity of the W21x83. We can’t look up because the chart ends, but we can
look for that capacity with a longer unbraced length. This leads us to a W24 x 84 as the most economical. (With the additional self
weight of 84 - 76 lb/ft = 8 lb/ft, the increase in the factored moment is only 1.92 k-ft; therefore, it is still OK.)
Evaluate the deflection with respect to the limits of L/240 for live (unfactored) load and L/180 for total (unfactored) load:
L/240 = 1 in. and L/180 = 1.33 in.
10’ 10’
∆
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Example 6
Select the most economical joist for the 40 ft
grid structure with floors and a flat roof. The
roof loads are 10 lb/ft2 dead load and 20 lb/ft2
live load. The floor loads are 30 lb/ft2 dead
load 100 lb/ft2 live load. (Live load deflection
limit for the roof is L/240, while the floor is
L/360). Use the (LRFD) K and LH series
charts provided.
(Top values are maximum total factored load in lb/ft, while the lower (lighter) values
are maximum (unfactored) live load for a deflection of L/360)
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Example 6 (continued)
(Top values are maximum total factored load in lb/ft, while the lower (lighter) values
are maximum (unfactored) live load for a deflection of L/360)
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Example 7 (LRFD)
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Example 8
A floor with multiple bays is to be supported by open-web steel joists spaced at 3 ft. on center and spanning 30 ft.
having a dead load of 70 lb/ft2 and a live load of 100 lb/ft2. The joists are supported on joist girders spanning 30 ft.
with 3 ft.-long panel points (shown). Determine the member forces at the location shown in a horizontal chord and
the maximum force in a web member for an interior girder. Use factored loads. Assume a self weight for the open-
web joists of 12 lb/ft, and the self weight for the joist girder of 35 lb/ft.
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Example 9
A floor is to be supported by trusses spaced at 5 ft. on center and spanning 60 ft. having a dead load of 53 lb/ft2 and
a live load of 100 lb/ft2. With 3 ft.-long panel points, the depth is assumed to be 3 ft with a span-to-depth ratio of
20. With 6 ft.-long panel points, the depth is assumed to be 6 ft with a span-to-depth ratio of 10. Determine the
maximum force in a horizontal chord and the maximum force in a web member. Use factored loads. Assume a self
weight of 40 lb/ft.
tributary widths
area loads
Pdead Plive
wdead wlive
(=wdead ⋅ A) (=wlive ⋅ A)
(K) (K)
+ 0.14 = 3.49
+ 0.29 = 7.00
self weight 0.04 k/ft (distributed) 3 1.2Pdead = 1.2wdead ⋅ tributary width = 0.14 K
6 1.2Pdead = 1.2wdead ⋅ tributary width = 0.29 K
NOTE – end panels only have half the tributary width of interior panels
FBD 2 of cut just to the left of midspan FBD 3 of cut just to right D1 = 0.5(3.49 k)/0.707 = 2.5 k (minimum near midspan)
of left support
ΣFx =-C1 + T1 + D1⋅cos45° = 0 C1 = 174.5 k
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Example 12
Solution:
ASD:
1. Pa = 140 k + 420 k = 560 k
2. The effective length in the weak (y-y) axis is 15 ft, while the effective length in the strong (x-x) axis is 30 ft. (K = 1, KL = 1×30 ft).
To find kL/rx and kL/ry we can assume or choose values from the wide flange charts. ry’s range from 1 to 3 in., while rx’s range from
3 to 14 inches. Let’s try ry = 2 in and rx = 9 in. (something in the W21 range, say.)
kL/ry ≅ 15 ft(12 in/ft)/2 in. = 90 ⇐ GOVERNS (is larger)
kL/rx ≅ 30 ft(12 in/ft)/9 in. = 40
3. Find a section with sufficient area (which then will give us “real” values for rx and ry):
If Pa ≤ Pn/Ω, and Pn = Fcr A, we can find A ≥ PaΩ/Fcr with Ω = 1.67
The tables provided have φFcr, so we can get Fcr by dividing by φ = 0.9
φFcr for 90 is 24.9 ksi, Fcr = 24.9 ksi/0.9 = 27.67 ksi so A ≥ 560 k(1.67)/27.67 ksi = 33.8 in2
4. Choose a trial section, and find the effective lengths and associated available strength, Fcr :
Looking from the smallest sections, the W14’s are the first with a big enough area:
Try a W14 x 120 (A = 35.3 in2) with ry = 3.74 in and rx = 6.24 in.: kL/ry = 48.1 and kL/rx = 57.7 (GOVERNS)
φFcr for 58 is 35.2 ksi, Fcr = 39.1 ksi so A ≥ 560 k(1.67)/39.1 ksi = 23.9 in2
Choose a W14 x 90 (Choosing a W14 x 82 would make kL/rx = 59.5, and Areq’d = 24.3 in2, which is more than 24.1 in2!)
LRFD:
1. Pu = 1.2(140 k) + 1.6(420 k) = 840 k
2. The effective length in the weak (y-y) axis is 15 ft, while the effective length in the strong (x-x) axis is 30 ft. (K = 1, KL = 1×30 ft).
To find kL/rx and kL/ry we can assume or choose values from the wide flange charts. ry’s range from 1 to 3 in., while rx’s range from
3 to 14 inches. Let’s try ry = 2 in and rx = 9 in. (something in the W21 range, say.)
kL/ry ≅ 15 ft(12 in/ft)/2 in. = 90 ⇐ GOVERNS (is larger)
kL/rx ≅ 30 ft(12 in/ft)/9 in. = 40
3. Find a section with sufficient area (which then will give us “real” values for rx and ry):
If Pu ≤ φPn, and φPn = φFcr A, we can find A ≥ Pu/φFcr with φ = 0.9
φFcr for 90 is 24.9 ksi, so A ≥ 840 k/24.9 ksi = 33.7 in2
4. Choose a trial section, and find the effective lengths and associated available strength, φFcr :
Looking from the smallest sections, the W14’s are the first with a big enough area:
Try a W14 x 120 (A = 35.3 in2) with ry = 3.74 in and rx = 6.24 in.: kL/ry = 48.1 and kL/rx = 57.7 (GOVERNS)
φFcr for 58 is 35.2 ksi, so A ≥ 840 k/35.2 ksi = 23.9 in2
Choose a W14 x 90 (Choosing a W14 x 82 would make kL/rx = 59.5, and Areq’d = 24.3 in2, which is more than 24.1 in2!)
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Example 13
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Example 14
Investigate the accepatbility of a W16 x 67 used as a beam-column under the unfactored
loading shown in the figure. It is A992 steel (Fy = 50 ksi). Assume 25% of the load is dead
load with 75% live load.
SOLUTION:
DESIGN LOADS (shown on figure):
Axial load = 1.2(0.25)(350k)+1.6(0.75)(350k)=525k
Moment at joint = 1.2(0.25)(60 k-ft) + 1.6(0.75)(60 k-ft) = 90 k-ft
Determine column capacity and fraction to choose the appropriate interaction equation:
so use
There is no bending about the y axis, so that term will not have any values.
Determine the bending moment capacity in the x direction:
The unbraced length to use the full plastic moment (Lp) is listed as 8.69 ft, and we are
over that so of we don’t want to determine it from formula, we can find the beam in the
Available Moment vs. Unbraced Length tables. The value of φMn at Lb =15 ft is 422 k-ft. 525 k 525 k
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Example 15
Example 16
φS = 6.96 k/in
8 in + 6 in + 8 in = 22 in.
6.96 k/in = 153.1 k
72.9 k
φPn = φFyAg φ = 0.9
72.9 k 3.31 k/in.
0.9 x 36 k/in2 x 3/8” x 6”= 72.9 k
From Available Strength table, use 3/16” weld
φPn = 72.9 k
(φS = 4.18 k/in.)
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Example 17
7-1,
Fu = 65 ksi
Example 18
Verify the tensile strength of an L4 x 4 x ½, ASTM
36, with one line of (4) ½ in.-diameter bolts and
standard holes. The member carries a dead load of
20 kips and a live load of 60 kips in tension.
Assume that connection limit states do not govern,
and U = 0.869.
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Example 19
The steel used in the connection and beams is A992 with Fy = 50 ksi,
and Fu = 65 ksi. Using A490-N bolt material, determine the
maximum capacity of the connection based on shear in the bolts,
bearing in all materials and pick the number of bolts and angle length
(not staggered). Use A36 steel for the angles.
SOLUTION:
The maximum length the angles can be depends on how it fits between the top and bottom flange with some clearance allowed
for the fillet to the flange, and getting an air wrench in to tighten the bolts. This example uses 1” of clearance:
Available length = beam depth – both flange thicknesses – 1” clearance at top & 1” at bottom
= 21.62 in – 2(0.93 in) – 2(1 in) = 17.76 in.
With the spaced at 3 in. and 1 ¼ in. end lengths (each end), the maximum number of bolts can be determined:
Available length ≥ 1.25 in. + 1.25 in. + 3 in. x (number of bolts – 1)
number of bolts ≤ (17.76 in – 2.5 in. - (-3 in.))/3 in. = 6.1, so 6 bolts.
It is helpful to have the All-bolted Double-Angle
Connection Tables 10-1. They are available for ¾”, 7/8”,
and 1” bolt diameters and list angle thicknesses of ¼”,
5/16”, 3/8”, and ½”. Increasing the angle thickness is
likely to increase the angle strength, although the limit
states include shear yielding of the angles, shear rupture
of the angles, and block shear rupture of the angles.
Tables 10-1 (not all provided here) list a bolt and angle
available strength of 271 kips for the ¾” bolts, 296 kips
for the 7/8” bolts, and 281 kips for the 1” bolts. It
appears that increasing the bolt diameter to 1” will not
gain additional load. Use 7/8” bolts.
φRn = 367.8 kips for double shear of 7/8” bolts φRn = 296 kips for limit state in angles
We also need to evaluate bearing of bolts on the beam web, and column flange where there are bolt holes. Table 7-4 provides
available bearing strength for the material type, bolt diameter, hole type, and spacing per inch of material thicknesses.
a) Bearing for beam web: There are 6 bolt holes through the beam web. This is typically the critical bearing limit value
because there are two angle legs that resist bolt bearing and twice as many bolt holes to the column. The material is
A992 (Fu = 65 ksi), 0.58” thick, with 7/8” bolt diameters at 3 in. spacing.
φRn = 6 bolts⋅(102 k/bolt/inch)⋅(0.58 in) = 355.0 kips
b) Bearing for column flange: There are 12 bolt holes through the column. The material is A992 (Fu = 65 ksi), 0.615” thick, with
1” bolt diameters.
φRn = 12 bolts⋅(102 k/bolt/inch)⋅(0.615 in) = 752.8 kips
Although, the bearing in the beam web is the smallest at 355 kips, with the shear on the bolts even smaller at 324.6 kips,
the maximum capacity for the simple-shear connector is 296 kips limited by the critical capacity of the angles.
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Example 20
SOLUTION:
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Available Critical Stress, φcFcr, for Compression Members, ksi (Fy = 36 ksi and φc = 0.90)
KL/r φc Fcr KL/r φc Fcr KL/r φc Fcr KL/r φc Fcr KL/r φc Fcr
1 32.4 41 29.7 81 22.9 121 15.0 161 8.72
2 32.4 42 29.5 82 22.7 122 14.8 162 8.61
3 32.4 43 29.4 83 22.5 123 14.6 163 8.50
4 32.4 44 29.3 84 22.3 124 14.4 164 8.40
5 32.4 45 29.1 85 22.1 125 14.2 165 8.30
6 32.3 46 29.0 86 22.0 126 14.0 166 8.20
7 32.3 47 28.8 87 21.8 127 13.9 167 8.10
8 32.3 48 28.7 88 21.6 128 13.7 168 8.00
9 32.3 49 28.6 89 21.4 129 13.5 169 7.91
10 32.2 50 28.4 90 21.2 130 13.3 170 7.82
11 32.2 51 28.3 91 21.0 131 13.1 171 7.73
12 32.2 52 28.1 92 20.8 132 12.9 172 7.64
13 32.1 53 27.9 93 20.5 133 12.8 173 7.55
14 32.1 54 27.8 94 20.3 134 12.6 174 7.46
15 32.0 55 27.6 95 20.1 135 12.4 175 7.38
16 32.0 56 27.5 96 19.9 136 12.2 176 7.29
17 31.9 57 27.3 97 19.7 137 12.0 177 7.21
18 31.9 58 27.1 98 19.5 138 11.9 178 7.13
19 31.8 59 27.0 99 19.3 139 11.7 179 7.05
20 31.7 60 26.8 100 19.1 140 11.5 180 6.97
21 31.7 61 26.6 101 18.9 141 11.4 181 6.90
22 31.6 62 26.5 102 18.7 142 11.2 182 6.82
23 31.5 63 26.3 103 18.5 143 11.0 183 6.75
24 31.4 64 26.1 104 18.3 144 10.9 184 6.67
25 31.4 65 25.9 105 18.1 145 10.7 185 6.60
26 31.3 66 25.8 106 17.9 146 10.6 186 6.53
27 31.2 67 25.6 107 17.7 147 10.5 187 6.46
28 31.1 68 25.4 108 17.5 148 10.3 188 6.39
29 31.0 69 25.2 109 17.3 149 10.2 189 6.32
30 30.9 70 25.0 110 17.1 150 10.0 190 6.26
31 30.8 71 24.8 111 16.9 151 9.91 191 6.19
32 30.7 72 24.7 112 16.7 152 9.78 192 6.13
33 30.6 73 24.5 113 16.5 153 9.65 193 6.06
34 30.5 74 24.3 114 16.3 154 9.53 194 6.00
35 30.4 75 24.1 115 16.2 155 9.40 195 5.94
36 30.3 76 23.9 116 16.0 156 9.28 196 5.88
37 30.1 77 23.7 117 15.8 157 9.17 197 5.82
38 30.0 78 23.5 118 15.6 158 9.05 198 5.76
39 29.9 79 23.3 119 15.4 159 8.94 199 5.70
40 29.8 80 23.1 120 15.2 160 8.82 200 5.65
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Available Critical Stress, φcFcr, for Compression Members, ksi (Fy = 50 ksi and φc = 0.90)
KL/r φc Fcr KL/r φc Fcr KL/r φc Fcr KL/r φc Fcr KL/r φc Fcr
1 45.0 41 39.8 81 27.9 121 15.4 161 8.72
2 45.0 42 39.6 82 27.5 122 15.2 162 8.61
3 45.0 43 39.3 83 27.2 123 14.9 163 8.50
4 44.9 44 39.1 84 26.9 124 14.7 164 8.40
5 44.9 45 38.8 85 26.5 125 14.5 165 8.30
6 44.9 46 38.5 86 26.2 126 14.2 166 8.20
7 44.8 47 38.3 87 25.9 127 14.0 167 8.10
8 44.8 48 38.0 88 25.5 128 13.8 168 8.00
9 44.7 49 37.8 89 25.2 129 13.6 169 7.91
10 44.7 50 37.5 90 24.9 130 13.4 170 7.82
11 44.6 51 37.2 91 24.6 131 13.2 171 7.73
12 44.5 52 36.9 92 24.2 132 13.0 172 7.64
13 44.4 53 36.6 93 23.9 133 12.8 173 7.55
14 44.4 54 36.4 94 23.6 134 12.6 174 7.46
15 44.3 55 36.1 95 23.3 135 12.4 175 7.38
16 44.2 56 35.8 96 22.9 136 12.2 176 7.29
17 44.1 57 35.5 97 22.6 137 12.0 177 7.21
18 43.9 58 35.2 98 22.3 138 11.9 178 7.13
19 43.8 59 34.9 99 22.0 139 11.7 179 7.05
20 43.7 60 34.6 100 21.7 140 11.5 180 6.97
21 43.6 61 34.3 101 21.3 141 11.4 181 6.90
22 43.4 62 34.0 102 21.0 142 11.2 182 6.82
23 43.3 63 33.7 103 20.7 143 11.0 183 6.75
24 43.1 64 33.4 104 20.4 144 10.9 184 6.67
25 43.0 65 33.0 105 20.1 145 10.7 185 6.60
26 42.8 66 32.7 106 19.8 146 10.6 186 6.53
27 42.7 67 32.4 107 19.5 147 10.5 187 6.46
28 42.5 68 32.1 108 19.2 148 10.3 188 6.39
29 42.3 69 31.8 109 18.9 149 10.2 189 6.32
30 42.1 70 31.4 110 18.6 150 10.0 190 6.26
31 41.9 71 31.1 111 18.3 151 9.91 191 6.19
32 41.8 72 30.8 112 18.0 152 9.78 192 6.13
33 41.6 73 30.5 113 17.7 153 9.65 193 6.06
34 41.4 74 30.2 114 17.4 154 9.53 194 6.00
35 41.1 75 29.8 115 17.1 155 9.40 195 5.94
36 40.9 76 29.5 116 16.8 156 9.28 196 5.88
37 40.7 77 29.2 117 16.5 157 9.17 197 5.82
38 40.5 78 28.8 118 16.2 158 9.05 198 5.76
39 40.3 79 28.5 119 16.0 159 8.94 199 5.70
40 40.0 80 28.2 120 15.7 160 8.82 200 5.65
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ASD
(Unified) Allowable Stress or LRFD
LRFD Design?
Collect data: Fy, Fu, and Collect data: load factors, Fy,
safety factors Ω Fu, and equations for shear
capacity with φV
Find Vmax & Mmax from
constructing diagrams or
using beam chart formulas Find Vu & Mu from
constructing diagrams or
using beam chart formulas
with the factored loads
Find Zreq’d and pick a section
from a table with Zx greater or
equal to Zreq’d
Pick a steel section from a chart having
φbMn ≥ Mu for the known unbraced length
Determine ωself wt (last number in OR
name) or calculate ωself wt. using A find Zreq’d and pick a section from a table
found. Find Mmax-adj & Vmax-adj. with Zx greater or equal to Zreq’d
Yes
No
Is Mu ≤ φbMn
No
Is Vmax-adj ≤ (0.6FywAw)/Ω.?
pick a new section with a
larger web area
Is Vu ≤ φv(0.6FywAw)
Yes No
pick a section
Yes with a larger
Calculate ∆max (no load factors!)
using superpositioning and beam web area
chart equations with the Ix for the
section
∆too big
I req' d ≥ I trial
is ∆max ≤ ∆limits? ∆lim it
This may be both the limit for live load No
deflection and total load deflection.) pick a section with a larger Ix
Yes (DONE)
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