Class Notes (836,067)
CIVE 462 (1)
Lecture

# Steel notes 318.docx

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School
Department
Civil Engineering
Course
CIVE 462
Professor
Colin A Rogers
Semester
Fall

Description
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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