B. 1
BEAMS
Common shapes:
Three basic considerations in design:
i) Bending strength
⎫
⎬ → strength consideration
ii) Shear strength
⎭
iii) Deflection
→ serviceability consideration
Bending Strength: Moment Capacity
Bending strength of a beam depends upon laterally unsupported length
of the compression flange. Thus, whenever you design a beam, find first the
condition of lateral support to compression flange,
Because of the compressive force in the compressive flange, the beam
tendency to buckle lateral ("lateral-torsional buckling"), if there is no adequate
lateral support.
Think compression flange
as a compression member
Compression flange
may buckle laterally
(sidesway)
Tension flange
(no problem)
Bending Strength of a Laterally Stable Beam (laterally Supports)
f<F y
f=F y
f=F y
f=F y
Plastic
X
X
Elastic
Entirely
Plastic
Plastic
M<M y
M=M y
M y <M<M p
M=M p
Stress disterbution at different stages of loading
B. 2
Elastic formula for beam in bending is
M ⋅ Ymax M ⋅ c M
M ⋅Y
M
; f b. max =
=
=
=
fb =
I
I
I c S
I
I
S = elastic section of modulus =
c
I = moment of inertia about axis of bending
M
Thus f b. max =
within elastic limit
S
The yield moment (moment at yield stress is reached, i.e. f max = Fy ) is given
as:
M y = S ⋅ Fy
S = elastic section modulus
As more loads are applied be beyond the yield level, the beam enters
the plastic zone. When the entire section becomes fully plastic, the plastic
(ultimate) moment capacity is:
M p = Z ⋅ Fy
Z = plastic modulus
The ratio
Mp
My
= ε is as the shape factor
Z
ε > 1.0 (usually 1.09 to 1.20 )
S
Shape factor gives an indication of how much reserve strength we have
beyond the yield moment.
Example
Calculate M y , M p and ε for W14x43 (A-36). Strong axis bending.
Thus, ε =
Ix
⎫
= 62.7in 3 ⎪
62.7 × 36
d 2
= 188.1k '
⎬M y = yield moment =
12
3
⎪
Z x = 69.6in
⎭
69.6 × 36
M p = plastic moment =
= 208.8k '
12
69.6
ε = shape factor =
= 1.11
62.7
A Note on Plastic Section Modulus
Plastic moment capacity = Z ⋅ Fy
Sx =
Z = plastic section modulus about the axis of
bending
How to calculate Z
1. Find the plastic centroid or the equal area axis (axis about which
the area is equal)
2. Take moment of the areas about the "equal area axis"
3. The total moment of the areas = Z
B. 3
fy
1/2x8
Example
Calculate Z for the following sections
6"
3/8x10
Z = 2[12 × 8 × (6 − 0.25) + 83 × 5 × 2.5] = 55.4in 3
M p = Z ⋅ Fy
11"
equal area
axis
8 × 12 + 83 ⋅ Y = 12 × 6 × 83 (10 − Y )
Y = 3.76"
fy
Z = 8 × 12 × (3.67 + 0.25) + 83 × 3.67 × 1.833
+ 83 × 6.33 × 3.81 + 3 × (6.33 + 0.25)
= 15.68 + 2.25 + 7.55 + 19.74 = 45.49in
M p = Z ⋅ Fy
fy
1/2x8
Y
3
3/8x10
11"
1/2x6
fy
Compact and Noncompact Section
Depending on the width-thickness ratio of elements, sections are
classified as "compact” and "noncompact" (table B5.1, P.16.1 1-14). Compact
sections must have axis of symmetry. Compact elements have lower b t ratio
λ p than noncompact elements, λr . If laterally supported, a compact section
can attain full plastic moment, M p
Design Requirement of AISC
→ In this class, we shall consider only I and C as beam sections
→ Beams are loaded in the plane of parallel to a principal axis through the
shear center (no torsion)
Depending on the λ -values and the lateral supported to compression
flange, failure can be:
i) Full yielding
ii) Lateral-torsional buckling
iii) Local flange and web buckling
The item (iii) depends on λ (width to thickness ratio).
Flexural strength (moment capacity) = φb M n = 0.9 M n
M n = nominal moment capacity must be calculated according to
different cases
equal area
axis
B. 4
Case A: Beams having full lateral support
A.1 Compact sections ( b t ratio λ ≤ λ p )
Full supports are usually
provided by slabs
Failure by full yielding (beam can
reach M p )
M n = M p = Z ⋅ Fy ≤ 1.5M y (Eq. F1-1)
M n = Z ⋅ Fy ≤ 1.5M y a Limit set by AISC
The slab prevents lateral
buckling, if flange is inside the
slab or connected to it
(composite design)
A.1 Noncompact sections ( λ P ≤ λ ≤ λ r )
Major axis bending
⎡ λ − λp ⎤
M n = M p − (M p − M r )⎢
⎥ (Eq. A-F1-3) (P.16.1-96)
⎣⎢ λ r − λ p ⎦⎥
(Eq. F1-7)
M n = FL S x = (Fy − Fr )S x
Fr = comp. residual stress in flange
= 10ksi for rolled sections; 16.5ksi for welded sections
Case B: Beams not having full lateral support
Length designation:
l b = distance between lateral supports (bracing points) to the
compression flange. (known) ( l b = 0 ,if full lateral support)
l r = limiting laterally unbraced length for lateral-torsional
buckling. (given by equation)
l p = limiting laterally unbraced length for full plastic moment
capacity M p . (given by equation)
For C and I sections,
300ry
E
(eq.F1-4)
; Fyf = yield stress of the flange = Fy
l p = 1.76ry
=
Ftf
Fyf
(eq.F1-6)
(eq.F1-8)
(eq.F1-9)
lr =
ry x1
1 + 1 + x 2 FL2
FL
FL = (Fy − Fr )
Fr = comp. residual stress in flange
= 10ksi for rolled sections; 16.5ksi for welded sections
G = shear modulus = 11,200ksi
⎧
π EGJA
=
x
⎪ 1
E = modulus of elasticity
Sx
2
⎪
J = torsional constant; in 4
⎨
2
⎪ x = 4C w ⎛ S x ⎞
A = cross sectional area
⎜
⎟
⎪ 2
I
GJ
C w = warping constant; in 6
⎠
y ⎝
⎩
B. 5
B.1 Compact sections (λ ≤ λ p ) and lb ≤ l p
M n = M p = Z ⋅ Fy ≤ 1.5M y same as A.1 case
B.2 Compact sections (I & C) with l p < lb ≤ l r
(eq.F1-2)
(eq.F1-7)
(major axis bending)
⎡
⎛ lb − l p ⎞⎤
⎟⎥ ≤ M p
M n = C b ⎢ M p − (M p − M r )⎜
⎜
⎟
l
−
l
⎢⎣
⎝ r p ⎠⎥⎦
M n = (Fy − Fr )S x
Value of C b
12.5M max
2.5M max + 3M A + 4 M B + 3M C
AISC allows taking C b = 1.0 a (safe)
(eq.F1-3)
Cb =
C b can be taken from table 4-1 (P.5-35) for same cases. If not sure take
C b = 1.0 a (conservative)
Note that M n now will lie between M p and M r (i.e. between full plastic and
effective elastic moment capacity)
B.3 Noncompact sections (I & C) with l p < lb ≤ l r
(major axis bending)
λ p < λ ≤ λr (B 5.1)
M n = smaller of (eq.F1-2) above
⎡
⎛ λb − λ p ⎞ ⎤
⎟⎥
(eq.A-F1-3) Or, = C b ⎢ M p − (M p − M r )⎜
⎜
⎟⎥
−
λ
λ
p ⎠⎦
⎝ r
⎣⎢
B.4 I & C (compact or noncompact λ ) with l p > l r
(eq.F1-12)
(eq.F1-13)
(major axis bending)
M n = M cr ≤ M p
M cr = C b
π
lb
2
⎛ πE ⎞
⎟⎟ I y C w
EI y GJ + ⎜⎜
⎝ lb ⎠
C b S x x1 2
x12 x 2
(using x1 and x 2 values)
1+
=
2
l b ry
2(lb ry )
( M cr value independent of Fy )
Note: M n will now generally be less than M r
B. 6
Moment Capacity of Beams
(flow chart)
Two items must be known or determined first
a) Lateral support to compression flange. The value of lb
b) Compact or noncompact section in bending.
λ p and λr values in table B5.1
Full lateral support l b = 0
compact
Mn = M p
(Eq. F1-1)
Mn
noncompact
⎡ λ − λp ⎤
M n = M p − (M p − M r )⎢
⎥
⎣⎢ λ r − λ p ⎦⎥
(Eq.A-F1-3)
lb ≠ 0
compact
A-F1-3
lb ≤ l p
lb ≤ l p
l p < lb ≤ l r
l p < lb ≤ l r
l p > lr
(eq.F1-1)
noncompact
l p > lr
(eq.F1-2)
smaller of F1-2
or A.F1-3
(eq.F1-12)
M n = M cr ≤ M p
Moment Capacity Diagram
eq. F1-1
eq. F1-2
M u = φM n
φ = 0.9
eq. F1-12
Mn
(for a compact section)
(for noncompact section,
Diagram similar)
lr
lp
unbraced length,lb
B. 7
Procedure for Calculation of M n
1. Chick the section is compact or noncompact
2. Find the laterally unsupported (unbraced) length, lb .
3. Chick which formula applicable (item A.1, A.2 and B.1 to B. in notes)
4. Use formula to find M n
Finally, M u = φM n
( φ = 0.9 )
Note that
i) A vast majority of rolled W-shape are compact
ii) Property table gives the value of b 2t f and h t w (widththickness ratio) to check compactness.
(P.5-42+) (Table 5-3)
Use of Beam Tables
All essential properties of beam sections (W-shapes) are calculated
and tabulated for rapid design.
Following values are available:
l r , l p , φb M r and BF
BF =
φb ( M p − M r )
lr − l p
Example #1
Calculate the following and compare with those given in table (5-3) for
W16x40 ( Fy = 50ksi ).
φM p , l r , l p , φb M r and BF
Table
indicates
if section
is compact
⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎩
Check
For flange
compactness:
⎛
b
65
65 ⎞⎟
E
(table B 5.1)
=
≈ 9.2⎜ 0.38
=
⎜
2t f
Fy
Fy
Fy ⎟⎠
⎝
For web
⎛
h
640
640 ⎞⎟
E
=
≈ 91⎜ 3.76
=
⎜
tw
Fy
Fy
Fy ⎟⎠
⎝
For W16x40- b 2t f = 6.9 & h t w = 46.6 → elements compact
72.9 × 50
= 304k '
12
φM p = 0.9 × 304 = 274k '
M p = Z x ⋅ Fy =
lr =
ry x1
1 + 1 + x 2 FL2
FL
FL = Fy − Fr = 50 − 10 = 40ksi
1.57 × 1890
1 + 1 + 127000 × 10 −6 × 40 2 = 176.4" = 14.7'
40
300ry 300 × 1.57
=
= 66.7" = 5.55' ≈ 5.6'
lp =
Fy
50
lr =
M r = (Fy − Fr )S x = (50 − 10) × 64.7 12 = 21.6k '
B. 8
φM r = 0.9 × 216 = 194k '
φb ( M p − M r ) 274 − 194
=
lr − l p
14.7 − 5.55
Table values are the same
BF =
Example #2
Find the design capacity of W16x40 ( Fy = 50ksi ) if lateral support are at
i) 5 ft interval
ii) 10 ft interval
iii) 24 ft interval. C b = 1.0
i)
lb = 5'
Since lb < l p and compact section
M n = M p ; M u = φM p = 274k '
ii) lb = 10'
lb > l p = 5.6' but < l r = 14.7'
Mn
φM n
⎡
⎛ lb − l p ⎞⎤
⎟⎥
= C b ⎢ M p − (M p − M r )⎜
⎜
⎟
−
l
l
⎝ r p ⎠⎦⎥
⎣⎢
⎡
BF
(lb − l p )⎤⎥
= Cb ⎢M p −
φ
⎣
⎦
= C b φM p − BF (lb − l p )
[
]
= 1.0[274 − 8.74 × (10 − 5.6 )] = 236k '
iii) lb = 24'
lb > l r = 14.7'
M n = M cr
φM n
=
C b S x x1 2
x12 x 2
1+
2
l b ry
2(lb ry )
=
1.0 × 64.7 × 1890 2
1890 2 × 127000 × 10 −6
1+
2
24 × 12 1.57
2(24 × 12 1.57 )
= 1225k " = 102k '
= 0.9 × 102 = 91.8k '
Example #3
Select the lightest beam (W) to carry uniformly distributed load of
0.2k / ft and live load 0.8k / ft for a simple span of 20' . Full lateral support.
( Fy = 50ksi )
Add small self weight say → 50lb / ft = 0.05k / ft
wu = 1.2 wDL + 1.6 wLL = 1.2 × 0.25 + 1.6 × 0.8 = 1.58k / ft
wu l 2 1.58 × 20 2
=
= 79k '
8
8
Assume compact section
φM n = φM p (full lateral support, l b = 0 )
Mu =
φM p = φZ x ⋅ Fy = 79k '
B. 9
79 × 12
= 21.0"
0.9 × 50
Go to table 5-3, select a beam with bold section (explain). [bold
sections are the best]
W10x19 ( Z x = 21.6" )
W12x19 (better)? ( Z x = 24.7" )
• Beam should be checked for SHEAR
(to be covered later)
Required Z x =
Selection of Beams with Given lb
Pages 5-71 to 5-101 give the design moment capacity of beams for
given unbraced length lb . Fy = 50ksi and 36ksi and C b = 1.0 only.
0.9M n=M u
How to use charts
1. Find the value of M u from analysis, lb and C b .
2. If C b > 1.0 , find the required
moment capacity required
Calculated .M u
φM n =
( assuming
Cb
lateral-torsional buckling controls)
3. Enter chart with lb and M u .locate
the point.
4. Any point on the top RHS quadrant
of the vertical and horizontal lines
lp
through the point is OK. But the
best beam (lightest) is the nearest one with solid line.
all beams OK
here. select the
nearest one with
solid line
Example #1
Select a W-shape to carry factored moment of 640k ' with unbraced
length of 20' . Fy = 50ksi . Assume C b = 1.0 .
Enter page 5-89 with M u = φM n = 640k ' ( C b = 1.0 ) and lb = 20' . Nearest
line beam is W30x90 ( φM n = 692k ' )
Example #2
Select a W-shape to carry factored moment of 640k ' with unbraced
length of 20' . Fy = 50ksi . Assume C b = 1.5 .
640
= 427.0k '
1.5
Enter page 5-154 with φM n = 427.0k ' ( C b = 1.0 ) and lb = 20' . Nearest
line beam is W18x76 (solid line) ( φM n ≈ 490k ' )
φM n =
B. 10
Shear Strength
(P. 16.1-35, section F.2)
Web area Aw = d × t w ( d = over all depth, t w = web thickness)
Design shear strength = φ vVn = 0.9Vn
For
417
h
E
≤ 2.45
≤
tw
Fyw
Fyw
Fyw = yield stress of web
= Fy for rolled section
Vn = 0.6 Fyw ⋅ Aw → (F2.1)
For
418
Fyw
≤
h
523
≤
tw
Fyw
(
)
Vn = 0.6 Fyw ⋅ Aw 418 Fyw / (h t w ) → (F2.2)
→ For almost all rolled shapes, Eq. F2.1 is not applicable.
h
Note that → ratio is given in the properties tables
tw
⎧
⎪
⎨
⎪
→ φ vVn values are given in P.5-49
⎩
Example
Find shear strength of the beam W18x76. ( Fy = 50ksi ).
h
= 37.8 ; φvVn = 0.9 × 0.6 × Fy Aw
tw
= 0.9 × 0.6 × 50 × 18.2 × 0.425 = 902k
Design note: As steel beams have high shear strength, choose a beam
first based on M u and check for Vu .(Unless M u is small and
Vu is very high for special cases)
Example #1
Design the beam shown to carry the service loads as indicated .lateral
supports to the beam at A, B and C. ( Fy = 50ksi )
Factored load:
self wt say
DL=0.6k/ft
10k
wDL = 1.2 × (0.6 + 0.06 ) = 0.79k / ft
LL=1.0k/ft
wLL = 1.6 × 1.0 = 1.6k / ft
Also, 1.6 × 10 = 16k
Total U DL = wu = 1.6 × 0.79 = 2.39k / ft
15'
15'
B. 11
Vmax = 8.0 + 2.39 × 15 = 44k
16k
2.39k/ft
2
wl
PL
M max = u +
8
4
2.39 × 30 2 16 × 30
=
+
= 389k '
8
4
Design values:
M u = 398k '⎫
⎬
Vu = 44k ⎭
15'
15'
V1
V1
Assume C b = 1.0 (actual C b > 1.0 )
Required φM n ≥ 389k ' ; lb = 15' ; Fy = 50ksi
Section is W21x62 ( φM n = 406k ' )
Check shear:
Design shear strength = φ vVn
φ vVn = 0.9 × 0.6 × Fy Aw
M max
= 0.9 × 0.6 × 50 × 20.99 × 0.4 = 227 k > 44k (we can get this
value from P. 5-49)
Use W21x62
Example #2
Design the beam below ( Fy = 50ksi ), considering two cases:
a) Full lateral support
b) Lateral support at A, B, C and D.
Factored load = 1.6 × 16 = 25.6k
Assume self wt, say = 80lb / ft
Factored wu = 1.2 × 0.8 ≈ 0.1k / ft
From analysis,
0.1 × 40 2
M u = 25.6 × 15 +
8
= 384 + 20 = 404k ' (at mid span)
Vu = 25.6 + 2.0 = 27.6k
Moment at B
0.1 × 40 2
M u = 348 + 2 × 15 −
8
= 384 + 30 − 11 = 403k '
A
15'
16k
16k
B
C
10'
D
15'
40'
25.6k
25.6k
0.1k/ft
a) Full lateral support
Assuming compact section, φb M n = φb M b ≥ 404k '
From beam selection table, section is W21x50 ( φb M px = 413 ,
φ vVu = 213k -OK
B. 12
b) Lateral support at A, B, C and D
for safe design, assume C b = 1.0
i) Segment A-B & C-D : M u = 403k ' , lb = 15'
ii) Segment A-B & C-D : M u = 403k ' , lb = 10'
Thus, part A-B/C-D is critical.
From chart, with C b = 1.0 , section is W21x62
Shear check: φ vVn = 227k OK
Selection of Beam Having C b > 1.0
Notice that C b values affects the strength only for the case
where lb > l p . Moment capacity M n is directly proportional to C b .
Thus, whenever lb > l p , find the equivalent design moment by
dividing the actual required M u by exact C b , i.e.
equivalent design moment (with C b = 1.0 ) M u =
actual.M u
Cb
Example
For the previous example – use exact value of C b for (b) segment
A-B/C-D: M u = 403k ' , lb = 10' ; C b = 1.76 (table 5-1, P5-35)
Equivalent design moment M u with C b = 1.0 is
actual.M u 403
=
= 241k '
Cb
1.67
Segment B-C; M u = 403k ' , lb = 15' ; C b = 1.0
Assuming B-C controls, section is W21x55
With lb = 15' , capacity (with C b = 1.0 ) = 308k ' > 241k '
Mu =
Also, l p for W21x55 = 6.11' ; lb > l p
Shear OK.
Thus, W21x55 is OK.
B. 13
Beam Deflection
To insure serviceability. Deflection due to LL should be checked,
especially for larger span beams. Engineers (or codes) specify the allowable
limits on deflection, depending on the type of structure.
l
Allowable LL deflection of beams supporting a plastered ceiling =
360
Note: LL deflection to be checked under service load only (no load factor)
Example
Compute the deflection of the beam shown
δ max at midspan from AISC beam diagram and formulas
δ max
4
3
5wl
Pl
=
+
384 EI x 48EI x
6k
DL=3k/ft
12'
12'
6conversion
4748
⎡
⎤
1 ⎢ 5 × 3 × 24 × 12 × 12 × 12 6 × 24 × 12 × 12 × 12 ⎥
W21x62
+
δ max =
⎢
⎥
384
48
EI x ⎢
⎥
⎣
⎦
12 × 12 × 12 × 14688
=
= 0.658"
29000 × 1330
l
24 × 12
as allowable limit, the limit is:
= 0.8"
If we use
360
360
• AISC beam formulas (P.5-162+) gives δ values for many cases!
B. 14
Cover Plated Beams
(best design)
Material can be saved by adding cover plates to flange of W-shape.
However, there added cost of welding
In many situations where a suitable rolled section not available or
beam needed to be reinforced for extra strength, cover plates are used.
Example
Area Ap
tp
d
X
X
Moment
capacity
Area Ap
Mmax
Capacity provided
by cover plate
Moment capacity
of beam
For a cover plated beam as shown,
⎛ d tp ⎞
Z x = Z x + 2 A p ⎜⎜ + ⎟⎟ = Z x + A p (d + t p )
{
⎝2 2 ⎠
beam
length of
cover plate
or Z x ≈ Z x + A p ⋅ d ; if t p is small
Z x − Z x required .Z x − Z x of .beam
=
d
d
Thus, we have an idea of A p . Because of approximation, A p will
Thus, A p =
be slightly more.
Thus, A p ≈ A p calculated above.
Select size of plate, b p × t p = A p , keep width less than b f (flange
width), unless unavailable. Keep thickness low, but be sure that t f is
OK for welding! (this will be covered in welding compatibility) . calculate
the exact properties, once plate size is known.
Example
For simple beam shown, design the beam utilizing W18x46 which is
available in stock. Fy = 36ksi . Assume full lateral support.
Factored wu = 1.2 DL + 1.6 LL
= 1.2 × 1.2 + 1.6 × 2.8 = 5.92k / ft
5.29 × 20 2
= 296k '
8
= 5.92 × 20 2 = 59.2k
DL=1.2k/ft
LL=2.8k/ft
M max =
Vmax
20'
B. 15
Thus, required M u = 296k ' , Vu = 59.2k .
Section is compact; Z x =
Mu
296 × 12
=
= 109.6in 3
φb Fy 0.9 × 36
Z x of W18x46 = 90.7in 3 < 109.6in 3
Thus, cover plate is needed.
Z needed − Z x .beam 109.6 − 90.7
=
= 1.05in 2
Approx. A p = x
d
18.06
Try A p ≈ 1.2in 2
Select A p = 165 × 4 = 1.25in 2
⎛
⎞
⎜ 18.06 0.3125 ⎟
⎟ = 113.7in 3 > 109.6in 3 OK
Actual Z x = 90
+
.7 + 2 × 1{
.05 × ⎜
{
2
23⎟
⎜ 12
z x beam
Ap
12
⎜ d3
⎟
2
t
p 2
⎝
⎠
Shear strength of W18x46, φ vVn = 176k > 59.2k
Use W18x46 + cover plate of 165 × 4
Now we need to find the length of the cover plate
Length of cover plate
moment capacity of beam alone
0.9 × 90.7 × 36
φb M p =
= 245k '
12
(or from tables)
Up to 'x', the beam capacity is OK; but
after 'x', actual moment > beam capacity.
Therefore, we need a cover plate
59.2 x 2
Moment at x = M x = 59.2 x −
= 245
2
x1 = 5.85' & x 2 = 14.15'
need plate
296k'
x
5.92k/ft
20'
59.2k
Theoretical length of cover plate = x 2 − x1 = 8.30'
Development Length
Needed extra length to develop the strength of the plate. This extra
length at each end is called "development length". This should be added to
the theoretical length to get the actual length.
• Minimum extra length of each end = 2 × width of the plate
}
2× 4× 2
In this example, we need total length = 8.3 +
= 9.05'
12
Use 9'−4" length
twice. width
5 '- 4 "
9 '- 4 "
5 '- 4 "
W 18x46
20'
5 /1 6 x 4 p la t e
245k'
B. 16
Concentrated Force on Beams
(Ch.K, P.16.1-71, 3 &4)
Rn = maximum concentrated force that
can be applied to the beam
N = length of bearing
(concentrated force is applied through
a small bearing plate)
Rn
N
N
Two modes of failure, if beam is held against
end reaction
Sidesway flange moment at point application
of load:
Rn
a) Local web yielding
⎫
Rn smaller of
⎬
b) Web crippling
two values
⎭
a) Local web yielding
Web yielding strength = φRn = Rn ( φ = 1.0 )
i) Load applied at > d from end:
Rn
>d
k
N
tw
k
Fyw
Rn = (5k + N )Fyw t w → (K1-2)
ii) Load applied at ≤ d from end:
Rn = (2.5k + N )Fyw t w → (K1-3)
= yield stress of web
tw
= web thickness
k
= distance from outer flange to toe
N
= length of bearing
b) Web crippling
Web crippling strength = φ r Rn = 0.75 Rn ( φ r = 0.75 )
i) Load applied at ≥ d / 2 from end:
1.5
⎡
⎛ t w ⎞ ⎤ E ⋅ Fyw ⋅ t f
N
⎛
⎞
2
Rn = 0.8t w ⎢1 + 3⎜ ⎟⎜ ⎟ ⎥
→ (K1-4)
tw
⎢
⎝ d ⎠⎜⎝ t f ⎟⎠ ⎥
⎣
⎦
1.5
⎡
⎤
⎛ N ⎞⎛⎜ t w ⎞⎟ ⎥ Fyw ⋅ t f
2
⎢
= 136t w 1 + 3⎜ ⎟
tw
⎢
⎝ d ⎠⎜⎝ t f ⎟⎠ ⎥
⎣
⎦
ii) Load applied at < d / 2 from end:
a. For N d ≤ 0.2
⎡
⎛ N ⎞⎛ t
Rn = 0.4t ⎢1 + 3⎜ ⎟⎜ w
⎢
⎝ d ⎠⎜⎝ t f
⎣
2
w
⎞
⎟
⎟
⎠
1.5
⎤ E ⋅ F ⋅t
yw
f
⎥
→ (K1-5a)
tw
⎥
⎦
B. 17
b. For N d > 0.2
1.5
⎡
⎛ t w ⎞ ⎤ E ⋅ Fyw ⋅ t f
N
4
⎞
⎛
Rn = 0.4t ⎢1 + ⎜
→ (K1-5b)
− 0.2 ⎟⎜ ⎟ ⎥
tw
⎢ ⎝ d
⎠⎜⎝ t f ⎟⎠ ⎥
⎣
⎦
If the beam fails in local web yielding or web crippling, web is stiffened by
adding a pair of stiffener at the load point.
2
w
Stiffener
Not that equations K1-3, K1-5a and K1-5b are applicable for end reaction only
(load distance < d )
Example
Compute the required width of bearing plate for a concentrated force of
120kips (factored), when it acts on a beam W21x62 as
a) Interior force, and
W21x62
b) End reaction. Fy = 36ksi
⎧
a) Interior force: (load distance > d )
k = 1.12"
⎪
i) Local web yielding
⎪
d = 21.0"
Rn = (5k + N )Fyw t w = (5 × 1.12 + N ) × 36 × 0.4
⎪⎪
⎨
= 80.64 + 14.4 N
⎪
t w = 0.40"
120 = 80.64 + 14.4 N
⎪
N = 2.37"
⎪
t w = 0.615"
⎪⎩
ii) Web crippling
1.5
⎡
⎤ E ⋅ F ⋅t
⎛ N ⎞⎛ t ⎞
yw
f
φRn = 120 = φ 0.8t w2 ⎢1 + 3⎜ ⎟⎜⎜ w ⎟⎟ ⎥
tw
⎢
⎝ d ⎠⎝ t f ⎠ ⎥
⎣
⎦
φ
}
= 0.75 × (1.626 + 12.2 N )
No need for N for crippling
Use N = 2 34 " needed for local web yielding
b) End reaction: (load distance < d )
i) Local web yielding
Rn = (5k + N )Fyw t w = 42.32 + 14.4 N = 120
N = 5.53"
ii) Web crippling
Try N d > 0.2
1.5
⎡
⎛ t w ⎞ ⎤ E ⋅ Fyw ⋅ t f
N
4
⎛
⎞
Rn = 0.4t ⎢1 + ⎜
− 0.2 ⎟⎜ ⎟ ⎥
tw
⎢ ⎝ d
⎠⎜⎝ t f ⎟⎠ ⎥
⎣
⎦
φRn = 0.75 × (72.6 + 8.1N ) ⇒ N = 11" , N d > 0.2 ; assumption OK
Use N = 11" , controlled by web crippling
2
w
B. 18
Bending About Minor (Weak) Axis
For I-shapes and C2, there is no lateral-torsional buckling, when
bending occurs about the minor axis, the section can attain full plastic
moment
M n = Z y ⋅ Fy ; M u = φb M n = 0.9Z y ⋅ Fy
y
y
Bi-axial Bending of I Sections or Doubly Symmetric Sections
The following interaction equation can be used.
M uy
M ux
+
≤ 1.0
φb = 0.9
φb M nx φb M ny
M nx = capacity of beam in bending about x-axis
M ny = capacity of beam in bending about y-axis
M uy , M uy = design values of moment about x and y axes
For a selected beam, calculate M ux and M uy in usual way then use
interaction equation.
Example
Check the adequacy of a laterally unsupported simple span
(span = 16' ) beam of W14x48 ( Fy = 50ksi ). If calculated M ux = 200k '
and M uy = 25k '
From chart, with lb = 16' , C b = 1.0
Moment capacity of W14x48 is φb M nx = 323k '
19.6
Z y = 19.6in 3 ; φb M n = φb Z y ⋅ Fy = 0.9 ×
× 50 = 73.5k '
12
M uy
M ux
200 25
+
=
+
= 1.2 > 1.0
φb M nx φb M ny 232 73.5
Beam is unsafe
y