12.07.2015 Views

EWPAA Structural Plywood and LVL Design Manual - Engineered ...

EWPAA Structural Plywood and LVL Design Manual - Engineered ...

EWPAA Structural Plywood and LVL Design Manual - Engineered ...

SHOW MORE
SHOW LESS
  • No tags were found...

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

F AFFFASS= (374.4 + 166.4)kN= 540.8kN (compression)= (+ 249.6 – 249.6)kN= 0 kNChoosing a point mid-way between C <strong>and</strong> B on the arch as shown in FIGURE 12.12 (a) results in the slopebeing:dydxdydx===x2 2 hL2 x15 x102301/ 3FIGURE 12.12: Exposed cross-section at mid-length <strong>and</strong> axial <strong>and</strong> shear force componentsFrom the equilibrium relationships the vertical (F VMS ) <strong>and</strong> horizontal (F HMS ) forces on the cross-sections ofthe free body diagrams are from:∑F = 0= 300 – F VMS – 10x15FVMS= 150kN∑FHR= 0= -450 + FHMSF = 450kNHMSResolving F vms <strong>and</strong> F HMS in the (t) <strong>and</strong> (n) directions results in the axial (F A ) <strong>and</strong> shear (F s ) componentsbeing:FA= (426.9 + 47.4)kNFA= 474.3kN(compression)FS= (142.3 − 142.3)kNF = 0kNSTo find the moment at any cross-section x from the arch centre, as shown in the free body diagram inFIGURE 12.13.167

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!