4. Flexural Members

Mapua Institute of Technology School of CE-EnSE-CEM STEEL & TIMBER DESIGN

FLEXURAL MEMBERS

Design of Flexural Members Classification of Steel Section: 1. Compact Section 2. Non – Compact Section 3 Slender Element Section 3. Stiffened Element – supported along two edges parallel to the direction of the compression force. Unstiffened Element – supported along one edge, parallel to the direction of the compression force. Compact Co pact Sect Section o bf < 170 (for compression flanges of I – Section and Channel Section) 2tf √Fy d < 1680 tw √Fy

(for webs in flexural compression)

Lecture Notes of Engr. Edgardo S. Cruz, MSCE

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Non – Compact Section bf < 250 (for compression flanges of I – Section and Channel Section) 2tf √Fy h < 1995 tw √Fy

(for webs in flexural compression)

Slender Element Section bf > 250 (for compression flanges of I – Section and Channel Section) 2tf √Fy h > 1995 tw √Fy

(for webs in flexural compression)

*NOTE: Compact Sections - ALL of its elements must be compact. For sections other than mentioned above, please see Table 5-1-Limiting Width Thickness Ratio for Compression Members (page 194 of Gillesania).

Allowable Bending Stress for I – Section & Channel Section Bent About Their Major Axis 1. Members with Compact Section with Lb < Lc ; Lb=braced length

(allowable bending stress in both tension & compression) 2. Members w/ Non – Compact Section a. For

except that their flanges are non – compact (excluding )). built – up members & members w/

Lecture Notes of Engr. Edgardo S. Cruz, MSCE

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Mapua Institute of Technology School of CE-EnSE-CEM STEEL & TIMBER DESIGN

b. For Built – Up Members w/

where: iff

,

if

c. For Members w/ Non – Compact Section not included above and d ,

3. Members w/ Compact or Non–Compact Section w/ a. Allowable bending stress in tension,

b Allowable bending stress in compression, b. compression

:

• When

• When

• For any value of

Lecture Notes of Engr. Edgardo S. Cruz, MSCE

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Mapua Institute of Technology School of CE-EnSE-CEM STEEL & TIMBER DESIGN

where:

Reversed Curvature

Single Curvature

(+)

Lecture Notes of Engr. Edgardo S. Cruz, MSCE

(-)

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Sample Problem 1: A simply supported beam has an I – Section made from 3 A36 steel plates welded together. Flanges are 450 x 20mm and web is 500 x 20mm.

W=?

L

Calculate the maximum uniform load “w” that the beam can carry. If the span of the beam is: a.) 4 m c.) 10 m b.) 6 m d.) 16 m

Lecture Notes of Engr. Edgardo S. Cruz, MSCE

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Mapua Institute of Technology School of CE-EnSE-CEM STEEL & TIMBER DESIGN

Solution: a.)

Check if the section is compact or not? Compact p flanges g :

;

Compact web :

Non – Compact flange

.: Section is Non – Compact

since the section is non-compact with LbLc a. b.

Lecture Notes of Engr. Edgardo S. Cruz, MSCE

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Mapua Institute of Technology School of CE-EnSE-CEM STEEL & TIMBER DESIGN

still under case 3.b., solving for Fbc

* Not satisfied! **

***

Not satisfied also because

For any value of

= 229.83 >0.6Fy

.: use

Lecture Notes of Engr. Edgardo S. Cruz, MSCE

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Mapua Institute of Technology School of CE-EnSE-CEM STEEL & TIMBER DESIGN

Thus applying flexure formula,

ans.

Sample Problem 2: Calculate the moment gradient multiplier

, for the ff. beams.

1.

2.

3.

Lecture Notes of Engr. Edgardo S. Cruz, MSCE

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Mapua Institute of Technology School of CE-EnSE-CEM STEEL & TIMBER DESIGN

4. single curvature

5. single curvature

Allowable Bending Stress, of I – Section, Solid Bars, Channel & Plates on Their Weaker Axis. •Members with Compact Sections

•Members with Non – Compact Sections

Shear Stress on Beams,

Lecture Notes of Engr. Edgardo S. Cruz, MSCE

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Mapua Institute of Technology School of CE-EnSE-CEM STEEL & TIMBER DESIGN

Allowable Shear Stress, *When

or

*When

or

where: when

when

when

when

where:

Sample Problem 3: From sample problem 1, check the adequacy of the beam against shear.

a. b. c. d.

Lecture Notes of Engr. Edgardo S. Cruz, MSCE

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Mapua Institute of Technology School of CE-EnSE-CEM STEEL & TIMBER DESIGN

Solution:

Seatwork 1: A floor system is supported by W18x158 A36 steel beams 10m long and spaced 3.50m on centers. The beams are simply supported at their ends and are laterally supported over their entire span. Allowable stress for bending and shear are 0.66Fy and 0.40Fy, respectively. Deflection of each beam should not exceed 1/360 of the span. overall depth= 500.9 mm web thickness= 20.6 mm section modulus= 5085.5x103 mm3 beam weight=235.0 kg/m 1. Which of the following most nearly gives the maximum floor load without exceeding the allowable bending stress of each beam. a. 11.08kPa b.18.37kPa c. 15.41kPa d. 19.03kPa 2. Which of the following most nearly gives the maximum floor load without exceeding the allowable shearing stress of each beam. a.57.83kPa b. 50.31kPa c. 68.49kPa d. 61.77kPa 3. Which of the following most nearly gives the maximum floor load without exceeding the allowable deflection of each beam. a. 15.53kPa b. 22.06kPa c. 19.82kPa d.14.87kPa

Lecture Notes of Engr. Edgardo S. Cruz, MSCE

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Mapua Institute of Technology School of CE-EnSE-CEM STEEL & TIMBER DESIGN

Design/Analysis of Purlins

Allowable Bending  Stress

Compact  Section

Non‐Compact  Section

Fbx

0.66Fy

0.6Fy

Fby

0.75Fy

0.6Fy

Deflection of Purlins,

Design/Analysis of Purlins with Sagrods and Tierods a. Sagrods at Midspan of Purlins

Lecture Notes of Engr. Edgardo S. Cruz, MSCE

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Mapua Institute of Technology School of CE-EnSE-CEM STEEL & TIMBER DESIGN

b. Sagrods at Third Points of Purlins

Sample Problem 4: A 28m Fink Truss are spaced at 6m o.c. The purlins are made up of W8X18 sections and are spaced at 1.96m on centers. Roof LL is 960N/m2 of roof surface and roof covering is assumed to be 720N/m2 of roof surface. Determine whether the purlins are adequate to carry such loadings. Sagrods & tierods are placed at midspan. A36 steel is used. tierod

sagrods 1.96

7m

28 m

Lecture Notes of Engr. Edgardo S. Cruz, MSCE

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Mapua Institute of Technology School of CE-EnSE-CEM STEEL & TIMBER DESIGN

W8x18 Purlins

Solution:

Lecture Notes of Engr. Edgardo S. Cruz, MSCE

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Mapua Institute of Technology School of CE-EnSE-CEM STEEL & TIMBER DESIGN

Subs. Subs

Substituting,

0.543 d ; Case b

R= P/2

tw = 14.7mm N= 600mm k = 31mm Fy=250Mpa d=550mm tf = 23.6mm

d = 0.55m ;

P = 1831.2525 KN ans. Web Crippling, Crippling x = 3 d = 0.55m 0 55m x > d/2 ; Case a

P = 2001.32 KN Pmax = 1831.2525 KN ans.

a. Nmin @ supports=? Web Yielding, x = 0 x < d ; Case a

d = 0.55m

; Nsup = 300 mm Web Crippling, x = 0 d = 0.55m x < d/2 ; Case b

Nsup = 510.34 mm .: Nmin @ support = 510.34mm ans.

Lecture Notes of Engr. Edgardo S. Cruz, MSCE

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