Arch324 w09 Lecture Unit5

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Architecture 324

Structures II

Composite Sections and Steel Beam Design

•Steel Beam Selection - ASD

•Composite Sections

•Analysis Method

University of Michigan, TCAUP Structures II Slide 2/21

Standard section shapes:

W – wide flangeS – american standard beamC – american standard channelL – angleWT or ST – structural TPipeStructural Tubing

Steel W-sections for beams and columns


Source: University of Michigan, Department of Architecture

Columns:

Closer to squareThicker web & flange

Beams:

Deeper sectionsFlange thicker than web





Columns:

Closer to squareThicker web & flange

Beams:

Deeper sectionsFlange thicker than web



CC:BY-SA Gregor_y (flickr) http://creativecommons.org/licenses/by-sa/2.0/

Steel Beams by ASD

Yield Stress Values• A36 Carbon Steel Fy = 36 ksi• A992 High Strength Fy = 50 ksi

Allowable Flexure Stress• Fb = 0.66 Fy • = Lc

– Compact Section

– Braced against LTB (l <Lc)• Fb = 0.60 Fy ° = Lu

– Compact or Not

– Lc < l < Lu• Fb < 0.60 Fy

– Compact or Not

– LTB failure mode (l >Lu)

Allowable Shear Stress• Fv = 0.40 Fy

– fv=V/(twd)University of Michigan, TCAUP Structures II Slide 6/21

Source: AISC, Manual of Steel Construction Allowable Stress Design, 9th ed. 1989

Section Modulus Table

• Calculate Required Moment • Assume Allowable Stress

– Fb = 0.66Fy = 24 ksi (A36)– Fb = 0.60Fy = 21.6 ksi (A36)

• Using the flexure equation, – set fb=Fb and solve for S

• Choose a section based on S from the table (D-35 and D-36)

– Bold faced sections are lighter– F’y is the stress up to which the

section is compact (•• is ok for all grades of Fy)

b

bb

FMS

FSM

IMcf


Source: Structural Principles, I. Engel 1984

1. Find the Section Modulus for the given section from the tables (D-35 and D-36).

2. Determine the maximum moment equation.

Example – Load Analysis of Steel Beam

Find Load w in KLF



3. Using the flexure equation, fb=Fb, solve for the moment, M.

4. Using the maximum moment equation, solve for the distributed loading, w.

Example – Load Analysis cont.

W30x116

w = 1.28 KLF



1. Use the maximum moment equation, and solve for the moment, M.

2. Use the flexure equation to solve for Sx.

Design of Steel Beam

Example



3. Choose a section based on Sx from the table (D35 and D36).

4. Most economical section is: W16 x 40Sx = 64.7 in3


Example


Source: I. Engel, Structural Principles, 1984

5. Add member self load to M and recheck Fb (we skip this step here)

6. Check shear stress:

Allowable Stress

Fv = 0.40 Fy

Actual Stressfv=V/(twd)

fv ≤ Fv


Example


6. Check Deflections

calculate actual deflection

compare to code limits

if the actual deflection

exceeds the code limit

a stiffer section is needed


Example

University of Michigan, TCAUP Structures II Slide 1335

Source: Standard Building Code, 1991

Composite Design

Steel W section with concrete slab “attached” by shear studs.

The slab acts as a wider and thicker compression flange.


Source: University of Michigan, Department of ArchitectureSource: University of Michigan, Department of Architecture

Effective Flange WidthSlab on both sides:(Least of the three)• Total width: ¼ of the beam span• Overhang: 8 x slab thickness• Overhang: ½ the clear distance to next

beam (i.e. the web on center spacing)

Slab on one side:(Least of the three)• Total width: 1/12 of the beam span• Overhang: 6 x slab thickness• Overhang: ½ the clear distance to next

beam




Analysis Procedure

1. Define effective flange width2. Calculate n = Ec/Es3. Transform Concrete width = n bc

4. Calculate Transformed Itr do NOT include concrete in tension

5. If load is known, calculate stress

or

6. If finding maximum load use allowable stresses. The lesser M will determine which material controls the section.

trconc

trsteel

InMcf

IMcf

ncIFM

cIFM

trconcc

trsteels


Given:• DLslab = 62.5 psf• DLbeam = 135 plf• n = 1/9• fsteel = 24 ksi ( Fy = 36 )• fconc = 1.35 ksi

For this example the floor capacity is found for two different floor systems:

1. Find capacity of steel section independent from slab

2. Find capacity of steel and slab as a composite

Non-composite vs. Composite Sections



• Find section modulus, Sx in chart.

• Assume an allowable stress, Fb.

• Determine the total moment capacity of the section, M.

• Subtract the DL moment to find the remaining LL moment.

• Calculate LL capacity in PSF.

Part 1 Non-composite Analysis



1. Determine effective width of slab.(using 90”y92”)

2. Find n=Ec/Es (1/9)

3. Draw transformed section(transform the concrete)

4. Calculate Transformed Ix:

• Locate neutral axis.

Part 2 - Composite Analysis




4. Calculate Transformed Ix: Use parallel axis theorem.

Ia=Ig+Ad2

Composite Analysis cont.



5. Calculate moment capacity for steel and concrete each assuming full allowable stress level.

6. Choose the smaller moment. It will control capacity.




7. Subtract the DL moment to find the remaining LL moment.

8. Calculate the LL in PSF based on the MLL.





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Arch324 w09 Lecture Unit5