1. 3 Steel structures 3 Steel structures

2. If you have any doubts, you can check your textbook, pp. 198-199

3. Steel structures »Lesson 1 Steel profiles are divided in 4 classes depending on their tendency to become more or less plastic Class 1 Generally IPE and HE sections Class 2 Some UNP and L sections (rarely IPE and HE) Effective cross-section With bolt holes A net = A – A f, where: A = gross area of the profile A f =( ∅ ⋅ n) ⋅ t sections of type asections of type b where: s pitch of the holes p cross-centre distance between holes n number of holes t thickness of the steel sheet ∅ diameter of the holes With welded sections A net = A c + 0,5 ⋅ A nc, where: A c welded area A nc non welded area Class 3 Some UNP and L sections, parts of compressed flanges or webs of HE sections Class 4 Parts of compressed flanges or webs of HE sections

4. Steel structures »Lesson 2 Stretched elements Allowable stress design method N = normal stress A = net area of the profile Limit state design method N sd ≤ N t,Rd N sd = design normal stress N t,Rd = resistance of the profile equal to Extension

5. Steel structures »Lesson 3 Compressed elements Allowable stress design method N = normal stress A = net area of the profile Limit state design method N sd ≤ N c,Rd N sd = design normal stress N c,Rd = resistance of the profile equal to profiles classes 1, 2, 3 profiles class 4

6. Steel structures »Lesson 4 Test of stability Slenderness  = () Allowable stress design method Limit state design method classes 1, 2, 3 class 4 with

7. Steel structures »Lesson 5 Simple bending Allowable stress design method Limit state design method »Lesson 6 Biaxial bending Allowable stress design method Limit state design method where the resistant moments are calculated as in Lesson 5 classes 1, 2 class 3 class 4

8. Steel structures »Lesson 7 Deformability  c = initial camber displacement  1 = elastic displacement resulting from permanent loads only  2 = elastic displacement resulting from variable loads  tot =  1 +  2 = total displacement resulting from the loads  max =  tot –  c = maximum displacement Double test to carry out: 1)  max ≤ K · L 2)  2 ≤ K 2 · L where the coefficients K and K 2 depend on the type of structural element tested

9. Steel structures »Lesson 8 Shear stress Allowable stress design method S = static moment of the section with respect to the neutral axis s a = thickness of the web J n = moment of inertia with respect to the neutral axis Limit state design method A v = t w · h w se A f /A w ≤ 0,6 A v = A – 2bt f + (t w + 2r) ⋅ t f I-beams or H-beams in the plane of the web A v = A – 2bt f + (t w + r) ⋅ t f C-beams or U-beams A v = 0,9 ⋅ (a – bt f ) T-beams A v = A – h w ⋅ t w I-beams or H-beams in the plane of the flanges

10. Steel structures »Lesson 9 Bending and shear stress Allowable stress design method Von Mises stress Limit state design method Se V Ed > 0,5 V c,Rd (design shear resistance) is calculated in the following way f T = (1 – ) ⋅ f yk with Shear stress test Bending test

11. Steel structures »Lesson 10 Combined bending and compression Allowable stress design method Limit state design method according to EC03 We first calculate from which check