Download presentation

1
**Elastic Stresses in Unshored Composite Section**

The elastic stresses at any location shall be the sum of stresses caused by appropriate loads applied separately Steel beam Permanent loads applied before the slab has hardened, are carried by the steel section. Short-term composite section Transient loads (such as live loads) are assumed to be carried by short-term composite action. The short-term modular ratio, n, should be used. Long-term composite section. Permanent loads applied after the slab has been hardened are carried by the long-term composite section. The long-term modular ratio, 3n, should be used.

2
Elastic Stresses ( ) The procedure shown in this picture is only valid if the neutral axis is not in the concrete. Use iterations otherwise.

3
**Elastic Stresses (6.10.1.1) Effective Width (Interior)**

According to AASHTO-LRFD , the effective width for interior girders is to be taken as the smallest of: One quarter of the effective span length (span length in simply supported beams and distance between permanent load inflection points in continuous beams). Average center-to-center spacing. Twelve times the slab thickness plus the top flange width.

4
**Hybrid Sections 6.10.3, 6.10.1.10 The web yield strength must be:**

1.20 fyf ≥ fyw≥ 0.70 fyf and fyw≥ 36 ksi The hybrid girder reduction factor = Rh Where, b=2 Dn tw / Afn Dn = larger of distance from elastic NA to inside flange face Afn = flange area on the side of NA corresponding to Dn fn = yield stress corresponding to Afn

5
**Additional sections 6.10.1.4 – Variable web depth members**

– Stiffness – Flange stresses and bending moments – Minimum negative flexure concrete deck rft. – Net section fracture

6
**Web Bend-Buckling Resistance (6.10.1.9)**

For webs without longitudinal stiffeners, the nominal bend buckling resistance shall be taken as: When the section is composite and in positive flexure Rb=1.0 When the section has one or more longitudinal stiffeners, and D/tw≤ 0.95 (E k /Fyc)0.5 then Rb = 1.0 When 2Dc/tw ≤ 5.7 (E / Fyc)0.5 then Rb = 1.0

7
**Web Bend-Buckling Reduction (6.10.1.10)**

If the previous conditions are not met then:

8
**Calculating the depth Dc and Dcp (App. D6.3)**

For composite sections in positive flexure, the depth of the web in compression in the elastic range Dc, shall be the depth over which the algebraic sum of the stresses in the steel, the long-term composite and short term composite section is compressive In lieu, you can use

9
**Calculating the depth Dc and Dcp (App. D6.3)**

For composite sections in positive flexure, the depth of the web in compression at the plastic moment Dcp shall be taken as follows for the case of PNA in the web:

10
**6.10 I-shaped Steel Girder Design**

Proportioning the section (6.10.2) Webs without longitudinal stiffeners must be limited to D/tw ≤ 150 Webs with longitudinal stiffeners must be limited to D/tw≤ 300 Compression and tension flanges must be proportioned such that:

11
Section Behavior Moment Mp Compact My Noncompact Slender Curvature

12
**6.10 I-Shaped Steel Girder Design**

Strength limit state Composite sections in positive flexure ( ) Classified as compact section if: Flange yield stress (Fyf ) ≤ 70 ksi where, Dcp is the depth of the web in compression at the plastic moment Classified as non-compact section if requirement not met Compact section designed using Section Non-compact section designed using Section

13
**6.10.7 Flexural Resistance Composite Sections in Positive Flexure**

Compact sections At the strength limit state, the section must satisfy If Dp≤ 0.1 Dt , then Mn = Mp Otherwise, Mn = Mp(1.07 – 0.7 Dp/Dt) Where, Dp = distance from top of deck to the N.A. of the composite section at the plastic moment. Dt = total depth of composite section For continuous spans, Mn = 1.3 My. This limit allows for better design with respect to moment redistributions.

14
**6.10.7 Flexural Resistance Composite Sections in Positive Flexure**

Non-Compact sections ( ) At the strength limit state: The compression flange must satisfy fbu ≤ ff Fnc The tension flange must satisfy fbu + fl/3 ≤ ff Fnt Nominal flexural resistance Fnc = Rb Rh Fyc Nominal flexural resistance Fnt= Rh Fyt Where, Rb = web bend buckling reduction factor Rh = hybrid section reduction factor

15
**6.10.7 Flexural Resistance Composite Sections in Positive Flexure**

Ductility requirement. Compact and non-compact sections shall satisfy Dp ≤ 0.42 Dt This requirement intends to protect the concrete deck from premature crushing. The Dp/Dt ratio is lowered to 0.42 to ensure significant yielding of the bottom flange when the crushing strain is reached at the top of deck.

16
**6.10 I-Shaped Steel Girder Design**

Composite Sections in Negative Flexure and Non-composite Sections ( ) Sections with Fyf ≤ 70 ksi Web satisfies the non-compact slenderness limit Where, Dc = depth of web in compression in elastic range. Designed using provisions for compact or non-compact web section specified in App. A. Can be designed conservatively using Section 6.8 If you use 6.8, moment capacity limited to My If use App. A., get greater moment capacity than My

17
**6.10.8 Flexural Resistance Composite Sections in Negative Flexure and Non-Composite Section**

Discretely braced flanges in compression Discretely braced flanges in tension Continuously braced flanges: fbu≤ ff Rh Fyf Compression flange flexural resistance = Fnc shall be taken as the smaller of the local buckling resistance and the lateral torsional buckling resistance. Tension flange flexural resistance = Fnt = Rh Fyt

18
**Flange Local buckling or Lateral Torsional Buckling Resistance**

Fn or Mn Inelastic Buckling (Compact) Inelastic Buckling (non-compact) Fmax or Mmax Fyr or Mr Elastic Buckling (Slender) Lp Lr Lb lpf lrf lf

19
**6.10.8 Flexural Resistance Composite Sections in Negative Flexure and Non-Composite Section**

Fnc Compression flange flexural resistance – local buckling

20
**Fnc Compression flange flexural resistance Lateral torsional buckling**

21
**Lateral Torsional Buckling**

22
**Unstiffened Web Buckling in Shear**

D/tw Web plastification in shear Inelastic web buckling Elastic web buckling

23
**6.10.9 Shear Resistance – Unstiffened webs**

At the strength limit state, the webs must satisfy: Vu ≤ fv Vn Nominal resistance of unstiffened webs: Vn = Vcr = C Vp where, Vp = 0.58 Fyw D tw C = ratio of the shear buckling resistance to shear yield strength k = 5 for unstiffened webs

24
Tension Field Action Beam Action Tension Field Action D g d0

25
**6.10.9 Shear resistance – Stiffened Webs**

Members with stiffened webs have interior and end panels. The interior panels must be such that Without longitudinal stiffeners and with a transverse stiffener spacing (do) < 3D With one or more longitudinal stiffeners and transverse stiffener spacing (do) < 1.5 D The transverse stiffener distance for end panels with or without longitudinal stiffeners must be do < 1.5 D The nominal shear resistance of end panel is Vn = C (0.58 Fyw D tw) For this case – k is obtained using equation shown on next page and do = distance to stiffener

26
**Shear Resistance of Interior Panels of Stiffened Webs**

27
**Transverse Stiffener Spacing**

D End panel Interior panel

28
**Types of Stiffeners Transverse Longitudinal Intermediate D Bearing**

29
**6.10.11 Design of Stiffeners Transverse Intermediate Stiffeners**

Consist of plates of angles bolted or welded to either one or both sides of the web Transverse stiffeners may be used as connection plates for diaphragms or cross-frames When they are not used as connection plates, then they shall tight fit the compression flange, but need not be in bearing with tension flange When they are used as connection plates, they should be welded or bolted to both top and bottom flanges The distance between the end of the web-to-stiffener weld and the near edge of the adjacent web-to-flange weld shall not be less than 4 tw or more than 6 tw.

30
**Transverse Intermediate Stiffeners**

Single Plate Angle Double Plate Less than 4 tw or more than 6tw

31
Design of Stiffeners Projecting width of transverse stiffeners must satisfy: bt ≥ d/30 and bf/4 ≤ bt ≤ 16 tp The transverse stiffener’s moment of inertia must satisfy: It ≥ do tw3 J where, J = required ratio of the rigidity of one transverse stiffener to that of the web plate = 2.5 (D/do)2 – 2.0 ≥ 2.5 It = stiffener m.o.i. about edge in contact with web for single stiffeners and about mid thickness for pairs. Transverse stiffeners in web panels with longitudinal stiffeners must also satisfy:

32
Design of Stiffeners The stiffener strength must be greater than that required for TFA to develop. Therefore, the area requirement is: If this equation gives As negative, it means that the web alone is strong enough to develop the TFA forces. The stiffener must be proportions for m.o.i. and width alone

33
Design of Stiffeners Bearing Stiffeners must be placed on the web of built-up sections at all bearing locations. Either bearing stiffeners will be provided or the web will be checked for the limit states of: Web yielding – Art. D6.5.2 Web crippling – Art. D6.5.3 Bearing stiffeners will consist of one or more plates or angles welded or bolted to both sides of the web. The stiffeners will extend the full depth of the web and as closely as practical to the outer edges of the flanges. The stiffeners shall be either mille to bear against the flange or attached by full penetration welds.

34
Design of Stiffeners To prevent local buckling before yielding, the following should be satisfied. The factored bearing resistance for the fitted ends of bearing stiffeners shall be taken as: The axial resistance shall be determined per column provisions. The effective column length is 0.75D It is not D because of the restraint offered by the top and bottom flanges.

35
Design of Stiffeners Interior panel End panel D bt tp 9tw

36
**General Considerations**

Shear studs are needed to transfer the horizontal shear that is developed between the concrete slab and steel beam. AASHTO-LRFD requires that full transfer (i.e. full composite action) must be achieved. Shear studs are placed throughout both simple and continuous spans. Two limit states must be considered: fatigue and shear. Fatigue is discussed later.

37
**Strength of Shear Studs**

0.85 Cross-sectional are of the stud in square inches Minimum tensile strength of the stud (usually 60 ksi)

38
Placement A sufficient number of shear studs should be placed between a point of zero moment and adjacent points of maximum moment. It is permissible to evenly distribute the shear studs along the length they are needed in (between point of inflection and point of maximum moment), since the studs have the necessary ductility to accommodate the redistribution that will take place.

39
**Miscellaneous Rules Minimum length = 4 x stud diameter**

Minimum longitudinal spacing = 4 x stud diameter Minimum transverse spacing = 4 x stud diameter Maximum longitudinal spacing = 8 x slab thickness Minimum lateral cover = 1". Minimum vertical cover = 2”. Minimum penetration into deck = 2”

Similar presentations

Presentation is loading. Please wait....

OK

Reinforced Concrete Design

Reinforced Concrete Design

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on credit policy template Ppt on osteoporosis Ppt on life study of mathematician Ppt on obesity prevention grants Ppt on high level languages examples Ppt on mars one finalists Ppt on team building in hindi Ppt on civil engineering in india Ppt on story of human evolution Ppt on structure of chromosomes worksheet