University of Michigan

2009-01

ARCH 324 - Structures 2, Winter 2009

von Buelow, Peter Buelow, P. v. (2008, December 19). Structures 2. Retrieved from Open.Michigan - Educational

Resources Web site: https://open.umich.edu/education/tcaup/arch324-winter2009.

<http://hdl.handle.net/2027.42/64938>http://hdl.handle.net/2027.42/64938

Deep Blue deepblue.lib.umich.edu

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© 2009, Peter Von Buelow 

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University of Michigan, TCAUP Structures II Slide 2/9

Architecture 324 Structures II

Combined Stress

•  Axial vs. Eccentric Load •  Combined Stress •  Interaction Formulas

Source: Caspar David Friedrich from “Man und Frau den Mond betrachtend” 1830-35 Alte Nationalgalerie, Berlin

University of Michigan, TCAUP Structures II Slide 3/9

Axial Stress

•  Loads pass through the centroid of the section

•  Member is straight

•  Load less than buckling load

•  f = P/A

Source: University of Michigan, Department of Architecture

University of Michigan, TCAUP Structures II Slide 4/9

Eccentric Loads

•  Load offset from centroid

•  M = P e

•  Total load = P + M

Interaction formula

Source: University of Michigan, Department of Architecture

University of Michigan, TCAUP Structures II Slide 5/9

Combined Stress •  Stresses combine by superposition •  Values add or subtract by sign

Source: University of Michigan, Department of Architecture

University of Michigan, TCAUP Structures II Slide 6/9

Example

Determine external reactions

Source: University of Michigan, Department of Architecture

University of Michigan, TCAUP Structures II Slide 7/9

Example

Determine internal member forces: Axial and Flexural

Determine axial and flexural stresses

Use interaction formula to determine combined stresses at key locations (e.g. extreme fibers)

University of Michigan, TCAUP Structures II Slide 8/9

Rafters Flexure + Axial

P = 22.5 # A = 5.25 in2

fc = 22.5 / 5.25 = 4.28 psi

M = PL/4 = 39# x 96″ /4 = 936 ″- # c = 1.5″ /2 = 0.75″ Iy = 3.5(1.53)/12 = 0.984 in4 fb = 963 (0.75)/0.984 = 713.14

f = fc + fb = 717.4 psi

Or

M = PL/4 = 45# x 83.14″ /4 = 936 ″- # fb = 963 (0.75)/0.984 = 713.14

90°

60°

30°

45#

45#(sin60) = 39#

45#(cos60) = 22.5#

University of Michigan, TCAUP Structures II Slide 9/9

Other Examples

University of Michigan, TCAUP Structures II Slide 10/9

Second Order Stress “P Delta Effect”

1.  Eccentric load causes bending moment

2.  Bending moment causes deflection, Δ

3.  P x Δ causes additional moment

4.  The additional PΔ moment causes additional deflection, Δ1 Δ

P

University of Michigan, TCAUP Structures II Slide 11/19

Treehouse

•  Form Exploration

University of Michigan, TCAUP Structures II Slide 12/19

Treehouse

•  Details –  Pinned at ends –  Radius bent at joints –  ‘Continuous’ sections –  Welded seams

University of Michigan, TCAUP Structures II Slide 13/19

Treehouse

•  Details –  Pinned at ends –  Radius bent at joints –  ‘Continuous’ sections –  Welded seams

University of Michigan, TCAUP Structures II Slide 14/19

Treehouse

•  Erection –  3 sections –  Trucked to site –  Erected with mobile crane –  One day process

University of Michigan, TCAUP Structures II Slide 15/19

Treehouse

•  Erection –  3 sections –  Trucked to site –  Erected with mobile crane –  One day process

University of Michigan, TCAUP Structures II Slide 16/19

Treehouse

•  Finished

University of Michigan, TCAUP Structures II Slide 17/19

Treehouse

•  Finished

University of Michigan, TCAUP Structures II Slide 18/19

Treehouse

•  Finished

University of Michigan, TCAUP Structures II Slide 19/19

Treehouse

•  Finished