4. Flexural Members

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Mapua Institute of Technology School of CE-EnSE-CEM

STEEL & TIMBER DESIGN

Lecture Notes of Engr. Edgardo S. Cruz, MSCE engredgar2k@gmail.com

FLEXURAL MEMBERS

Design of Flexural Members

Classification of Steel Section:1. Compact Section2. Non – Compact Section3 Slender Element Section3. Slender Element Section

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 SectionCo pact Sect obf < 170 (for compression flanges of I – Section and Channel Section)

2tf √Fy

d < 1680tw √Fy (for webs in flexural compression)

Non – Compact Sectionbf < 250 (for compression flanges of I – Section and Channel Section) 2tf √Fy

h < 1995tw √Fy (for webs in flexural compression)

Slender Element Sectionbf > 250 (for compression flanges of I – Section and Channel Section) 2tf √Fy

h > 1995tw √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 (excludingbuilt – up members & members w/ ).built up members & members w/ )

b. For Built – Up Members w/

where:fif

c. For Members w/ Non – Compact Section not included aboved

3. Members w/ Compact or Non–Compact Section w/

a. Allowable bending stress in tension,

b Allowable bending stress in compression :b. Allowable bending stress in compression,

• When

• For any value of

where:

Reversed Curvature Single Curvature

(+) ( - )

Sample Problem 1:

A simply supported beam has an I – Section made from 3 A36 steel plateswelded together. Flanges are 450 x 20mm and web is 500 x 20mm.

Calculate the maximum uniform load “w” that the beam can carry. If the span of the beam is:

a.) 4 m c.) 10 mb.) 6 m d.) 16 m

Solution:

Check if the section is compact or not?

Compact flanges : ;p g

Compact web :

Non – Compact flange

.: Section is Non – Compact

since the section is non-compact with Lb<Lc.:equations on case 2.b. applies.

Thus applying flexure formula,

Substituting:

Sample Problem 1: Part (b)

Solution:

b.) , and section is non – compact!

Check case 2.c.

Case 3: members with compact or non compact section with Lb>Lc

Not satisfied!

still under case 3.b., solving for Fbc

For any value of

Not satisfied!

Not satisfied also because

.: use

= 229.83 >0.6Fy

Thus applying flexure formula,

Sample Problem 2:

Calculate the moment gradient multiplier , for the ff. beams.

single curvature4.

single curvature

•Members with Compact Sections

Allowable Bending Stress, of I – Section, Solid Bars, Channel &Plates on Their Weaker Axis.

•Members with Non – Compact Sections

Shear Stress on Beams,

Allowable Shear Stress,

*When or

where:

when when

where:

Sample Problem 3:

From sample problem 1, check the adequacy of the beam against shear.

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 mmweb thickness= 20.6 mmsection 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

Design/Analysis of Purlins

Allowable Bending Stress

Compact Section

Non‐Compact Section

Deflection of Purlins,

Fbx 0.66Fy 0.6Fy

Fby 0.75Fy 0.6Fy

Design/Analysis of Purlins with Sagrods and Tierods

a. Sagrods at Midspan of Purlins

b. Sagrods at Third Points of Purlins

Sample Problem 4:

A 28m Fink Truss are spaced at 6m o.c. The purlins aremade up of W8X18 sections and are spaced at 1.96m oncenters. Roof LL is 960N/m2 of roof surface and roofcovering is assumed to be 720N/m2 of roof surface.

tierod

sagrods

Determine whether the purlins are adequate to carry suchloadings. Sagrods & tierods are placed at midspan. A36steel is used.

7 m1.96

W8x18 PurlinsW8x18 Purlins

Solution:

SubsSubs.

Substituting,

0.543<1.0

Design of Sagrods:

From the previous problem, design the diameterof the sagrods needed using steel rod withFy=248Mpa. Sagrods are placed @ midspan ofeach purlins.

Tension on gross area,

Solution:

Design of Tierod:

From the previous problem, design the diameter of the tierod needed using steelrod with Fy=248Mpa.

Solution:

Tension on gross area,

WEBS UNDER CONCENTRATED LOAD

a. Web Yielding

b. Web Crippling

A. WEB YIELDING

N+2.5k Toe of fillet

where:N = length of bearing plate g g pk = distance of toe of fillet from top/bottom flange

( found in the steel manual)R = a beam reaction or any concentrated load

a. For loads applied at or near the end of member

Compressive stress =

0.66 Fy

b. For loads applied at a distance “x” greater than “d” from the end of member.

Compression Stress =

0.66 Fy

B. WEB CRIPPLING

SAMPLE PROBLEM 5

A simply supported beam carries a concentrated load at the center of the wide flange section having a web thickness of 14.7mm. The base plate placed directly on the load is used to prevent web yielding & crippling has a width “N” of 600mm, k= 31mm, Fy= 250 Mpa a Determine the max concentrated load that the base plate can carrya. Determine the max concentrated load that the base plate can carry.b. Determine the minimum length (N) of bearing plate to be placed under supports.

tw = 14.7mm N= 600mmk = 31mm Fy=250Mpad=550mmtf = 23.6mm

Solution:

R= P/2 R= P/2

a. Pmax @ midspan=?Web Yielding, x = 3 d = 0.55m

x > d ; Case b

P = 1831.2525 KN ans.

Web Crippling x = 3 d = 0 55mWeb Crippling, x = 3 d = 0.55mx > d/2 ; Case a

P = 2001.32 KN

Pmax = 1831.2525 KN ans.

a. Nmin @ supports=?

Web Yielding, x = 0 d = 0.55mx < d ; Case a

Web Crippling, x = 0 d = 0.55mx < d/2 ; Case b

Nsup = 300 mm

Nsup = 510.34 mm

.: Nmin @ support = 510.34mm ans.

4. Flexural Members

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