Tension in members (Simple Welds)
1)Find the forces in the truss members required
2)Find L, w and t, then A , n1ann2A (S22)n3nd use to find A = A + A +ne <= An1 n2 n3 g
*Note that t should be subtracted to w to account for the weld
• A n1wt
• If L >= 2w An2 wt
2w>L>=w An2 0.50wt + 0.25Lt
W > L An2 0.75 Lt
• If L>= w An3 (1-x/L)wt where x is the distance from the weld to the
centroid
W > L An3 0.50 Lt
• Convert dimensions (1” = 25.4mm) in mm if required Use appropriate table to find
A gW: S8, S79, Angles: S25, S31, Channels: S37, ...)
For Complicated cross-section geometries use : A = A (1-neL) g
3) Find tension resistance
Use table pS2 to determine F = F vnd Fywith reupect to the date published
-3
T r φA Fgx y0 , where φ = 0.90
T r φ AuFne u0 , where φ = 0.75u
4) Maximum Slenderness
I
Check that : L/r <= 300, where r = radius of gyration = √ A , find both Rxand R any L
' 2
is the total length. (I =I+Ax ́ )
Tension in members (Bolted Connections)
1) Find the forces in the truss members
2) Find the required constants
F ynd F frum table pS3
A, d, b, t, w from the member table
3) Determine A (Sn6) for a number of fracture paths. The smallest controls the design *Note: Bolt hole size = bolt diameter + 2 mm for drilled holes
= bolt diameter + 4 mm for punched holes (or plasma
cutting)
A n1-1 w n = A gA boltsFor segments of failure path normal to the applied force
A n2-2A g (-nd +hs /4g)t For segments of failure path inclined to the applied force (pS21)
s is the spacing between holes parallel to the applied force
g is the spacing perpendicular to the applied force
4) Determine Tension Resistance (evaluate all these cases, lowest controls the design)
i) Failure occurs over gross area, T =rφA F g yhere φ = 0.90
ii) Shear lag controls, Tr= φ u ne,uwhere φ = 0u75 and see pS27 for A ne
iii) Block Shear failure (S17): Tr= φ uUA t n u.6A (F gvF y/2) u
U t 1.0 for symmetric blocks of failure patterns and concentric loading)
exceptions pS17
where A =gvross area in shear(ignore holes) and A =net nrea in
tension(consider holes)
iv) Plug shear failure: Tr= φ u0,6A gv +y )/u), note that A igvcalculated from sum of bolt
lines
5) Check Slenderness
Axial Compression
1) Find factored concentric load
2) Use appropriate table to find the values: F ,yA, R ,xR y d, b, t, w
3) Check Maximum width-to-thickness ratio:
bel b/2 200
= ≤ to avoid flange local elastic buckling or anything applicable
t t √F y
table 1 pS41
h ≤ 670
w F to avoid web local elastic buckling
√ y
kL I I
≤200 r = y∧r = x
4) Check Slenderness limits: r , where y √ √ x A , see pS46 for k Selection tables can only be used if the weak axis controls
5) Check that C>C
r f
−1
Cr=ϕ A g y1+λ2n)n , where ϕ=0.90 and
P
λ= y , where P yA F y (yield or squash load) and
√ PE
PE=Eulerbucklingload
P y A Fy kL Fy
λ= P = 2 = r 2
For Euler Flexural buckling: E π EA2 √π E Use the larg

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