structural curriculum for construction management and architecture students 1 prepared by: ajay...

Structural Curriculum for Construction Management and Architecture Students

1

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

2

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


4

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.


Yielding occurs when stresses exceed the yield stress of the member.

5

1.5. General Concepts of Beam Design1. Yielding

☞

Figure 1.5.1. Stress-strain diagram

6


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.

7

Section X-X


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.

8


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

9

► 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.

10


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.

11

PNA

Zx= A1Y1 + A2Y2

= 12x0.5 + 12x0.5 = 6 + 6 = 12 in3

Zx =A1Y1 + A2Y2

= 12x1 + 12x1 = 12 + 12 = 24 in3


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.

12

Zx= A1Y1 + A2Y2

= 12x1.5 + 12x1.5 = 18 + 18 = 36 in3

Zx= A1Y1 + A2Y2

= 12x3 + 12x3 = 36 + 36 = 72 in3


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.

13

Zx= A1Y1 + A2Y2 + A3Y3 + A4Y4

= 8x4.5 + 4x2 + 4x2 + 8x4.5 = 36 + 8 + 8 + 36 = 88 in3


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


14


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


15


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.

16

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.

17

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

18


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

☞

Figure 1.5.12. Local buckling of beam section under compression.

19

► 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).


2. Local Buckling

AISC Specification Chapter F2 User note states:

20

►Yielding is the upper limit on strength►However, lateral-torsional buckling, based

on beam unbraced length, may control strength


3. Lateral Torsional Buckling

☞

21

►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



22

► 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.



☞

23

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.

24

► 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

25☞


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.

structural curriculum for construction management and architecture students 1 prepared by: ajay...

Documents

learning beam design

beam design activities

yielding figure

12x1 12x1

yield strength

yield stress

plastic neutral axis

a1y1 a2y2 a3y3 a4y4