Structural Curriculum for Construction Management and Architecture Students
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Prepared by:Ajay Shanker, Ph.D., P.E. Associate Professor Rinker School of Construction ManagementUniversity of Florida
1.5. General Concepts of Beam Design
Design Parameters:► Introduction to beam design
Resources:► AISC Steel Construction Manual
Learning Concepts and Objectives: ► Understand the main steps for beam design
Activities:► Learning beam design through AISC Steel Construction
Manual
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3
Mu ≤ ɸbMn
Mu = the required strength
Mn = the nominal flexural strength (the flexural capacity of the beam)
ɸb = the resistance factor = 0.9 for flexure
1.5. General Concepts of Beam Design
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Limit States to determine Mn
1. Yielding: yielding is the upper limit for all shapes.
2. Local buckling: buckling of elements before they are able to reach yield.
3. Lateral-torsional buckling: a combination of lateral buckling and twist.
1.5. General Concepts of Beam Design
Yielding occurs when stresses exceed the yield stress of the member.
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1.5. General Concepts of Beam Design1. Yielding
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Figure 1.5.1. Stress-strain diagram
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1.5. General Concepts of Beam Design
X
X
Section X-X
d
bFactored Uniform Load, wu
Span, L
2
8u
u
w LM
0.9u n nM M M
1. Yielding
Figure 1.5.2 (a). Simply supported beam with uniform loading.
Figure 1.5.2 (b). Section cut through
beam.
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Section X-X
1.5. General Concepts of Beam Design
F
y
Co
mp
ress
ion
Ten
sio
n
2
2 2 4
n
n y y
M Force LeverArm
bd d bdM F F
n y xM F Z
b
d/2
d/2
PNA
1. Yielding
F
y
C = FyAC
T = FyAT
d/4
d/4
Figure 1.5.3. Loading on section cut of beam.
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1.5. General Concepts of Beam Design
1. Yielding
x i iZ AY
Yi = The distance from the centroid of the Area, Ai, to the plastic neutral axis.
Mn = FyZx
Fy = Yield strength of the steelZx = Plastic Section Modulus
= the moment of the area about the PNA
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► How do we define the plastic neutral axis, PNA?
► Thus, the area below the PNA must equal the area above the PNA.
y T y C
T C
T C
F A F A
A A
1.5. General Concepts of Beam DesignPlastic Section Modulus, Zx
0Forces T CAC
b
C = FyAC
T = FyAT
PNA
AT
Figure 1.5.4. Balances forces on section cut of beam.
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1.5. General Concepts of Beam Design
1 1 2 2
2 2 2
2 4 2 4
8 8 4
x i iZ AY
A Y A Y
bd d bd d
bd bd bd
A1
Y1
Y2d/2
d/2d/4
d/4
b
Plastic Section Modulus, Zx
PNA
For a rectangle the PNA is in the middle.
A2
Figure 1.5.5. Calculating plastic section modulus.
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PNA
Zx= A1Y1 + A2Y2
= 12x0.5 + 12x0.5 = 6 + 6 = 12 in3
Zx =A1Y1 + A2Y2
= 12x1 + 12x1 = 12 + 12 = 24 in3
1.5. General Concepts of Beam DesignPlastic Section Modulus, Zx
1”
1”
12 in
A1=12 in2
A2=12 in2
Y2=1/2 in
Y1=1/2 in
2”
2”
6 in
Y2=1”Y1=1”
Section 1 Section 2
Figure 1.5.6 (a) and (b). Calculating plastic section modulus.
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Zx= A1Y1 + A2Y2
= 12x1.5 + 12x1.5 = 18 + 18 = 36 in3
Zx= A1Y1 + A2Y2
= 12x3 + 12x3 = 36 + 36 = 72 in3
1.5. General Concepts of Beam DesignPlastic Section Modulus, Zx
4 in
PNA
3 in
3 in
A1=12 in2
A2=12 in2
Y2=1.5 in
Y1=1.5 inY1=3 in
Y2=3 in
6 in
6 in
Section 3Section 4
Figure 1.5.7 (a) and (b). Calculating plastic section modulus.
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Zx= A1Y1 + A2Y2 + A3Y3 + A4Y4
= 8x4.5 + 4x2 + 4x2 + 8x4.5 = 36 + 8 + 8 + 36 = 88 in3
1.5. General Concepts of Beam DesignPlastic Section Modulus, Zx
8 in
PNA10 in
1 in
1 in
1 in
Y1=4.5 in
Y4=4.5 in
Y2=2 in
Y3=2 in
A1
A4
A2
A3Total Area = A1 + A2 + A3 + A4
= 8 + 4 + 4 + 8= 24 in2
Section 5
Figure 1.5.8. Calculating plastic section modulus.
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1.5. General Concepts of Beam DesignPlastic Section Modulus, Zx
PNA
10 in
10 in
1 in
1 in
1/2 in
Y1=4.5 inY2=2 in
Y3=2 inY4=4.5 in
Zx= A1Y1 + A2Y2 + A3Y3 + A4Y4
= 10x4.5 + 2x2 + 2x2 + 10x4.5 = 45 + 4 + 4 + 45 = 98 in3
Total Area = A1 + A2 + A3 + A4
= 10 + 2 + 2 + 10= 24 in2
A1
A4
A2
A3
Section 6
Figure 1.5.9. Calculating plastic section modulus.
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1.5. General Concepts of Beam Design
PNA
10 in
18 in
1 in
1 in
1/4 in
Y1=8.5 in
Y2=4 in
Y3=4 in Y4=8.5 in
Total Area = A1 + A2 + A3 + A4
= 10 + 2 + 2 + 10= 24 in2
A1=10 in2
A4
A2
A3
Zx = A1Y1 + A2Y2 + A3Y3 + A4Y4
= 10x8.5 + 2x4 + 2x4 + 10x8.5 = 85 + 8 + 8 + 85 = 186 in3
Section 7 Figure 1.5.10. Calculating plastic section modulus.
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Zx = 186 in3
Zx = 98 in3
Zx = 88 in3
Zx = 24 in3
Zx = 12 in3
Zx = 36 in3
Zx = 72 in3
1
2
3
4
5
7
6
All seven shapes have same area of 24 in2 but have increasing Zx and moment capacity. Shape 7 has (186/12 = 15.5 times) more moment capacity than Shape 1.
Figure 1.5.11. Comparing section modulus for different shapes of same area.
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Shape Wt/ft Zx, in3
W8x67 67 70.1
W10x68 68 85.3
W12x65 65 96.8
W14x68 68 126
W16x67 67 130
W18x65 65 133
W21x68 68 160
W24x68 68 177
Plastic Section Modulus of Different W-Shape of Approximately Same Weight
W8x67 W10x68 W12x65 W14x68 W16x67 W18x65 W21x68 W24x68 0
20
40
60
80
100
120
140
160
180Zx, in3
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1.5. General Concepts of Beam Design
M M
Compression, leads to possiblelocal buckling
Tension, no buckling
A
A
MM Flange Local Buckling (FLB)
A
A
MM Web Local Buckling (WLB)
A
A
2. Local Buckling
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Figure 1.5.12. Local buckling of beam section under compression.
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► If a shape is capable of reaching the plastic moment without local buckling it is said to be a compact shape.
► Most standard shapes are compact.
All current W, S, M, C and MC shapes except W21x48, W14x99, W14x90, W12x65, W10X12, W8x31, W8x10, W6x15, W6x9, W6x8.5, and M4x6 have compact flanges for Fy = 50 ksi (345 MPa); all current ASTM A6 W, S, M, HP, C and MC shapes have compact webs at Fy < 65 ksi (450 MPa).
1.5. General Concepts of Beam Design
2. Local Buckling
AISC Specification Chapter F2 User note states:
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►Yielding is the upper limit on strength►However, lateral-torsional buckling, based
on beam unbraced length, may control strength
1.5. General Concepts of Beam Design
3. Lateral Torsional Buckling
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►The compression portion of the bending member tries to behave like a column but can’t
►Tension region resists buckling down►Tension region also resists buckling laterally
►Thus, the shape twists as it buckles laterally
1.5. General Concepts of Beam Design
3. Lateral Torsional Buckling
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► To control lateral-torsional buckling the beam must be properly braced.► Intermediate points along the span may be
braced against translation and twisting.► The distance between braced points is referred
to as the unbraced length, Lb.
1.5. General Concepts of Beam Design
3. Lateral Torsional Buckling
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1.5. General Concepts of Beam DesignBeam Lateral Bracing Examples
Cross beam acts as a lateral brace since it will prevent lateral displacement of the girder’s compression flange.
Compression Flange, top
Tension Flange, bot.
Continuous concrete floor slab provides continuous bracing for the compression flange, Lb=0, no LTB.Concrete Slab
Lateral displacement of the bottom compression flange is prevented by the diagonal members (typically angles).
Compression Flange, top
Tension Flange, bot.
Tension Flange, top
Compression Flange, bot.
Figure 1.5.13. Beam Lateral Bracing Examples.
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► If Lb ≤ Lp , the limit state of yielding controls bending strength:
► If Lb > Lr the limit state of elastic lateral-torsional buckling controls.
► If Lp < Lb < Lr the limit state of inelastic lateral-torsional buckling controls.
n y xM F Z
1.5. General Concepts of Beam Design3. Lateral Torsional Buckling
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1.5. General Concepts of Beam Design
Lateral bracing points from secondary beams
Figure 1.5.14. Partial floor framing plan
LATERAL SUPPORT OF STEEL BEAMS
47 kip-ft
28.5 kip-ft
CASE II
W10X12
CASE IIICASE I
CASE II CASE IIICASE I
CASE I if lateral brace is spaced 0-2.75’
CASE II if lateral brace is spaced 2.75’-8’
CASE III if lateral brace is spaced more than 8’
W10X12
CASE I
Design Moment Strength = = 47 kip-ftNM
BRACING DISTANCE
BRACES
W10X12
b0 L 2.75'
CASE II
Design Moment Strength reduces as increasesbL
AT b 2.75 'L
AT b 'L 8
NM 47kip-ft
NM 28.5 kip-ft
Linear variation in & bL NM
W10X12b2.75' L 8'
LATERAL BRACES
bL
bL
AT b 'L 8
AT b 'L 18
NM 28.5
NM 8.25
Design Moment Strength reduces as increases
Should be avoided for load bearing floor beams.
kip-ft
kip-ft
CASE III
bL
bL 8'
Top flange embedded in concrete
bL 0
ADDITIONAL CASE I CONSTRUCTION DETAILS
bL 0
x
x
Y
Y
X-X
Y-Y
CONCRETE
W-SHAPESTEEL STUDS
Use 50 ksi and select a shape for a typical floor beam AB. Assume that the floor slab provides continuous lateral support. The maximum permissible live load deflection is L/180. The service dead loads consist of a 5-inch-thick reinforced-concrete floor slab (normal weight concrete), a partition load of 20 psf, and 10 psf to account for a suspended ceiling and mechanical equipment. The service live load is 60 psf.
yF
• Fy = 50ksi
• Case 1
•
• LIVE LOAD = 60psf
LL
• DEAD LOADS
1) 5” Slab
2) Partition = 20psf
3) Ceiling, HVAC = 10psf
GIRDER
GIRDER
(PRIMARY BEAMS)
not to exceed 180
L
Figure 1
DESIGN LOAD (kips/ft) on AB = wu x TRIBUTORY AREA
LENGTH OF BEAM
1.2x(62.5 20 10) 1.6(60) x[6x30]
30
[ ]x
1
1000= 1.242 kips/ft
12” of Slab = 150 psf6” of Slab = 75 psf1” of Slab = 12.5 psf5” of Slab = 62.5 psf(12.5 psf for every inch of concrete thickness)*5 x 12.5 = 62.5
wu = 1.242 kips/ft
Mu2
8uw L 21.242x30
8= = = 139.7 ft-kips CASE 1
139.7
STRENGTH OF W14 X 26
= 150.7 ft-kips
STRENGTH OF W16 X 26
= 165.7 ft-kips
Page # 3-127
Page # 1-22
Page # 1-23
Page # 1-20
SELECT W16X26
SHAPE AREA
W 14x26 7.69 245
W16x26 7.68 301
XI
Page # 1-21
EXTRA SELF WEIGHT MOMENT = 21.2x0.026x30
8= 3.3 ft-kips
MOMENT STRENGTH APPLIED MOMENT 165.7 139.7 + 3.3
W16X26 is OK
CODE 30x12
180
360
180ACTUAL
45
384
wl
EI= = = 2” =
Page # 3-211
w =(60)x(6x30)
30= 360 lb/ft
0.360
12
45x0.03x(30x12)
384x29,000x301
= 0.360 kips/ft
= 0.03 kips/in
= = 0.75” 2” OK
= 45
384
wl
EI;
.kips
in
4( )in
2.
kips
in
2in
in= in
4( )in
=
kips/in
Typical Copes for a shear connection of a large girder to column web.
Note that duct holes have to be strengthened by plates. Also, holes are at third point where shear & moment are not maximum.
Cantilever construction for projected balcony.
If shear studs are noticed on beams and column then those members have to be encased in concrete for increasing fire resistance of steel.
Details of web opening in steel girders for HVAC ducts.