1. Used in buildings: A. Joists: Closely spaced beams - - PDF document

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1. Used in buildings: A. Joists: Closely spaced beams - - PDF document

Jan 21, 2023 250 likes •712 views

Analysis and Design of Beams

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SLIDE 1

1

  • Analysis and Design of Beams
  • Beams are members that are subjected to transverse
  • loads. (Chapter F in the specifications in the manual)
  • Types of beams:
  • 1. Used in buildings.
  • 2. Used in roofs and walls.
  • 3. Used in bridges.
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SLIDE 2

2

  • 1. Used in buildings:
  • A. Joists: Closely spaced beams supporting floors of

buildings.

  • B. Lintels: On top of doors and windows.
  • C. Spandrel beams: To support external walls.
  • 2. Used in roofs and walls:
  • A. Purlins: on roofs.
  • B. Girts: on walls.
  • 3. Used in bridges:
  • A. Stringers: beams running parallel to the roadway.
  • B. Floor beams: large beams which are perpendicular to

the roadway.

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SLIDE 3

3

  • Types of sections:
  • 1. Rolled steel sections:
  • W (most economical, used in this class), S, C, T, L.
  • 2. Built up sections:
  • Plate girder.
  • If h/tw < 5.70 (E/Fy)0.5 : It is considered beam.
  • If h/tw > 5.70 (E/Fy)0.5 : It is considered plate girder
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SLIDE 4

4

!!" #

  • 1. Elastic beam theory:
  • Plane section before bending remains plane after

bending.

  • !!"

#

  • 2. Increasing Load (Moment), reaching yielding:
  • Reaches yield stress and strain.
  • Still linear.
  • Sx : Section modulus (I/c)
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SLIDE 5

5

!!" #

  • 3. Increasing Load (Moment), beyond yielding:
  • Elastic zone (core).
  • Plastic zone.
  • !!"

#

  • 4. Plastic moment (Mp):
  • Strain and stress reach yielding every where.
  • Full Plastic zone, Pu, Mp.
  • Formation of plastic hinge.
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SLIDE 6

6

!!" #

  • !!"

#

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SLIDE 7

7

!!" #

  • $%
  • Example 1: A572, Grade 50. Determine:
  • 1. Yielding moment and elastic section modulus.
  • 2. Plastic moment and plastic section modulus.
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SLIDE 8

8

$%

  • 1. .
  • $%
  • 2. .
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SLIDE 9

9

$%

  • 2. .
  • $%
  • Example 2:
  • Determine plastic section modulus.
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SLIDE 10

10

$%

  • #!
  • 1. Lateral Torsional Buckling (LTB):
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SLIDE 11

11

#!

  • Mn = Mp (if the beam remains stable up to the fully

plastic condition)

  • #!
  • Lateral torsional buckling will not occur if the

compression flange of a member is braced or if twisting of the beam is prevented at frequent intervals.

  • Two Categories of lateral support:
  • 1. Continuous lateral support by embedment of the

compression flange in a concrete slab.

  • 2. Lateral support at short or long intervals.
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SLIDE 12

12

#!

  • 2. Classifications of Shapes :
  • A. Compact sections:
  • Capable of developing a fully plastic stress distribution

before buckling.

  • If λ < λp and the flange is continuously connected to

the web.

  • B. NonB Compact sections:
  • Yield stress can be reached in some but not all of the

elements.

  • If λp < λ < λr
  • #!
  • C. Slender sections:
  • If λ > λr
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SLIDE 13

13

#!%

  • Beams can fail by reaching Mp and become fully

plastic , or it can fail by:

  • 1. Lateral Torsional Buckling (LTB), elastic or inelastic.
  • 2. Flange Local Buckling (FLB), elastic or inelastic.
  • 3. Web Local Buckling (WLB), elastic or inelastic.
  • #!%
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SLIDE 14

14

#!%

  • 1. Plastic Behavior ( Zone 1):
  • Compact Shape.
  • Compression flange is continuously braced laterally or

lateral bracing are provided at short intervals, i.e. Lb > Lp

  • Where
  • , φb =0.9
  • #!%
  • For bending about minor axis (yBy); there is no (LTB)

in the doubly symmetrical sections.

  • Always zone 1.
  • Φb Mn = φb Mp

= φb Fy Zy < φb 1.6 Fy Sy

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SLIDE 15

15

$%&'()

  • Example 1:
  • W16 x 31, A992 steel.
  • Continuous Lateral Support.
  • Is the beam adequate in flexure?
  • $%&'()
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SLIDE 16

16

$%&'()

  • $%&'()
  • Example 2:
  • Fy = 50 ksi.
  • Continuous lateral support.
  • Design.
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SLIDE 17

17

$%&'()

  • Case 1 + Case 9 Table 3B23 in the manual.
  • $%&'()
  • Case 1 + Case 9 Table 3B23 in the manual.
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SLIDE 18

18

$%&'()

  • Case 1 + Case 9 Table 3B23 in the manual.
  • Wu = 1.2 (2) = 2.4 k /ft
  • Pu = 1.6 (24) = 38.4 k
  • Mu = (2.4)(30)2 / 8 + (38.4)(10) = 654 kBft
  • or :
  • Zreq. = (654)(12) / (0.9)(50) = 174.4 in3
  • Using Table 3B2, try W24 x 68, Zx = 177 > 174.4
  • $%&'()
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SLIDE 19

19

$%&'()

  • Wu = 1.2 (2+0.068) = 2.4816 k/ft
  • Pu = 38.4 k
  • Mu = (2.4816)(30)2 / 8 + (38.4)(10) = 663.18 kBft
  • Zreq. = (663.18)(12) / (0.9)(50)

= 176.8 < 177 (O.K)

  • Use W24 x 68
  • #!%
  • 2. Inelastic LTB (Zone 2):
  • The bracing is insufficient to permit the member to

reach a full plastic distribution.

  • Yield strain is reached in some but not all of its

compression elements. Lp < Lb < Lr

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SLIDE 20

20

#!%

  • #!%
  • Cw = Warping constant, in4
  • The moment capacity in zone (2) is:
  • Or : φb Mn = Cb [ φb Mp – BF (LbB Lp)] <φb Mp
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SLIDE 21

21

#!%

  • Cb = Modification factor for nonBuniform moment

diagrams, when both ends of the beam segment are braced.

  • #!%
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SLIDE 22

22

#!%

  • Cb is used to account for the effect of different

moment gradients on LTB.

  • Cb = 1.0 for Cantilever and over hangs.
  • #!%
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SLIDE 23

23

#!%

  • Example (Cb):
  • Bracing at ends
  • Mmax = PL/8
  • MA = 0
  • MB = PL/8
  • MC = 0
  • Cb = (12.5)(PL/8) / (2.5PL/8 + 0 + 4 PL/8 + 0)

= 1.92

  • $%&'*)
  • Example 1:
  • W12 x 30, A992 steel.
  • Lb = 10 ft
  • Cb = 1.0
  • Compute the flexural design strength.
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SLIDE 24

24

$%&'*)

  • $%&'*)
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SLIDE 25

25

$%&'*)

  • Mn = 1.0 [ (50)(43.1) – {(50)(43.1) –

(0.7)(50)(38.6)} [ (10B5.369) / (15.6B5.369)] = 1791.1 kipBin < Mp = 2155 kipBin

  • Φb Mn = (0.9)(1791.1)

= 1611.99 kipBin = 134.33 kipBft

  • $%&'*)
  • Example 2:
  • Mu = 290kBft.
  • Fy = 50 ksi
  • Lb = 10 ft
  • Cb = 1.0
  • Design.
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SLIDE 26

26

$%&'*)

Using Table 3B2

  • $%&'*)
  • Try W 18 x 40: (assume Zone 1)
  • Φb Mp = 294 kBft,
  • Lp = 4.49 ft,
  • BF = 13.3,
  • Lr = 13.1 ft
  • Zone 2 ( 4.49< 10 <13.1)
  • Φb Mn = 1.0 [ 294 – (13.3)(10B4.49)]

= 220.72 <290 (N.G)

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SLIDE 27

27

$%&'*)

  • Try W 21 x 44:
  • Φb Mp = 358 kBft,
  • Lp = 4.45 ft,
  • BF = 16.8,
  • Lr = 13 ft
  • Zone 2 ( 4.45< 10 <13.0)
  • Φb Mn = 1.0 [ 358 – (16.8)(10B4.45)]

= 264.76 <290 (N.G)

  • $%&'*)
  • Try W 21 x 48:
  • Φb Mp = 398 kBft,
  • Lp = 6.09 ft,
  • BF = 14.7,
  • Lr = 16.6 ft
  • Zone 2 ( 6.09< 10 <16.6)
  • Φb Mn = 1.0 [ 398 – (14.7)(10B6.09)]

= 340.52 > 290 (O.K)

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SLIDE 28

28

#!%

  • 3. Elastic LTB (Zone 3):

Lb > Lr

  • $%
  • Example 1: (Design by Charts)
  • A572, Grade 50.
  • Bracing at ends.
  • Design by charts.
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SLIDE 29

29

$%

  • From charts page 3B126:
  • $%
  • Example 2:
  • A992 steel.
  • Lateral bracing at the ends and at mid span.
  • 30% dead load, 70% live load.
  • Design.
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SLIDE 30

30

$%

  • $%
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SLIDE 31

31

$%

  • #!"%
  • Flange Local Buckling (FLB):
  • (NON –COMPACT)
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SLIDE 32

32

#!"%

  • Flange Local Buckling (FLB):
  • λ > λr (SLENDER)
  • $%&#!"

%)

  • Example:
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SLIDE 33

33

$%&#!" %)

  • $%&#!"

%)

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SLIDE 34

34

$%&#!" %)

  • $%
  • Bracing at A, B, C, and D.
  • Fy = 50 Ksi.
  • Is W 14 x 132 adequate in flexure?
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SLIDE 35

35

$%

  • Segment ABC:
  • Wu = 1.2 (3.5+0.132) + 1.6 (1.0) = 5.958 kip/ft
  • Segment CD:
  • Wu = 1.2 (0.132) = 0.1584 kip/ft
  • Pu = 1.6 (25) = 40 kips.
  • Mmax = 858 kipBft at 16.97 ft
  • Check Compactness (section is compact)
  • $%
  • W14 x 132 :
  • Lp = 13.3 ft, Lr = 56.0 ft
  • For segment BC :
  • Lb = 9 ft < Lp = 13.3 ft (Zone 1)
  • Φb Mn = φb Mp = 878 kipBft < Mu = 858 kipBft (O.K)
  • For segment CD:
  • Lp= 13.3 ft <Lb = 18 ft < Lr = 56 ft (Zone 2)
  • Cb = 12.5 (855.4) / (2.5)(855.4)+(3)(646.4)+

(4) (434.1)+ (3)(218.7) = 1.653

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SLIDE 36

36

$%

  • Φb Mn = 1.653 [ 878 –(7.7)( 18B13.3)]

= 1391.512 kipBft > φb Mp = 878 kipBft

  • Use φb Mp = 878 kipBft > Mu = 855 kipBft (O.K)
  • For segment AB :
  • Lb = 9 ft (Zone 1)
  • φb Mp > Mu (O.K)
  • Section is adequate in flexure
  • Shear stress :
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SLIDE 37

37

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SLIDE 38

38

slide-39
SLIDE 39

39

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SLIDE 40

40

$%&)

  • $%&)
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SLIDE 41

41

!

  • Example:
  • W: Service live

load.

  • $%&!)
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SLIDE 42

42

#$#

  • #$#
slide-43
SLIDE 43

43

$%&#$#)

  • $%&#$#)
slide-44
SLIDE 44

44

$%&#$#)

  • Is that point through which the loads must act if there

is to be no twisting or torsion of the beam.

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SLIDE 45

45

  • Location of shear center:
  • Location of shear center:

qv = VQ / I (Shear flow)

  • H h = V e
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SLIDE 46

46

$%&)

  • qv at B = VQ / I
  • I = (0.3)(9.4)3 / 12 +
  • [ (3)(0.3)3 / 12 +
  • (3)(0.3)(4.85)2}] (2)
  • = 63.12 in4
  • Q = (3B0.15)(0.3)(5B0.15)
  • = 4.146
  • qv at B = V (4.146) / 63.12
  • = 0.0657 V
  • $%&)
  • Total area H = (0.5)(0.0657 V)(2.85) = 0.0936 V

V e = H h V e = (0.0936 V) (9.7) e = 0.91 in (from center line

  • f web)