THEME IS FIRST OCCURANCE OF YIELDING THE LIMIT?

Transcription

1 CIE309 : PLASTICITY THEME IS FIRST OCCURANCE OF YIELDING THE LIMIT? M M - N N + + σ = σ = + f f BENDING EXTENSION Ir J.W. Welleman page nr

2 0 kn Normal conditions during the life time WHAT HAPPENS DUE TO FURTHER LOADING? 00 kn Failure due to maximum load MATERIAL? CROSS SECTION? STRUCTURE? plastisc hinge 00 kn plastic hinge (ULTIMATE LOAD) (LIMIT STATE ANALYSIS) plastic hinge Analsis model with plastic hinges Ir J.W. Welleman page nr

3 RELEVANT MATERIAL PROPERTIES (STEEL) Experiment TEST RESULT bi-linear MODEL stress, σ [N/mm ] strength f t stress, σ [N/mm ] ielding hardening ield stress, f [N/mm ] Young s modulus, E [N/mm ] unloading brittle failure ductile ield stress, f [N/mm ] Young s modulus, E [N/mm ] brittle failure ductile strain =0, % vl = % t u =0 % plastic deformation 0, % % 0 % strain Ir J.W. Welleman page nr 3

4 (STRUCTURAL) STEEL name Yield stress Strength ield-strain limit-strain N/mm N/mm ( ) % ( u ) % S , 9 S , 6 S , 6 Young s modulus : E=, 0 5 N/mm = 0 GPa Ir J.W. Welleman page nr 4

5 ELASTIC OR PLASTIC ANALYSIS? Piece of Art situation m 3 cables S355 Cross6 section 60 mm mass : 5500 kg 4,0 m Question : Is this safe? 4,0 m Ir J.W. Welleman page nr 5

6 Piece of Art situation 3 cables S355 Cross section 60 mm mass : 5500 kg Question : Is this allowed? 4,0 m Ir J.W. Welleman page nr 6

7 ELASTIC LIMIT EA E 60 k = = = 60E l situation : F = A f = = 30, 0 N u e p Fp 30, 0 = u p = = =,39 mm k 5 60, 0 G = k u = 3 k u G 3 e e e e = 3 30, 0 = 90,3 0 N ( SAFE) (all cables simultaneous loaded up to capacit) situation : G e = 5 e e 3 e 3 e k u + k u + k u = k u = 50, 0 N (centre wire at 00% of capacit but outer two loaded at 33% of capacit) 3 3 Situation Situation load distribution per cable (LE) load distribution per cable (LE) 33% 33% 33% 0% 60% 0% elastic limit is failure load after elastic limit all additional load is taken b the outer two cables. The centre cable remains at 00% load and is not allowed to break before the outer two cables are loaded up to 00% of the capacit. (required ductilit) Failure occurs when all three cables are loaded up to their capacit. The outer cables will have an elongation of 3,39 = 7,6 mm Ir J.W. Welleman page nr 7

8 RESULT 90,3 situation situation, wire, and 3 F [kn] 55 kn 50, 30, margin = 35,3 kn γ F =,64 >,5 OK contribution of wire en 3 ielding of wire 4,77 mm = 0,33 % < % = OK,39 7,6 u [mm] Ir J.W. Welleman page nr 8

9 PRELIMINARY CONCLUSIONS PLASTIC ANALYSIS SHOWS ADDITIONAL LOAD BEARING CAPABILITIES FOR STATIC INDETERMINATE STRUCTURES AFTER THE ELASTIC LIMIT HAS BEEN REACHED DUE TO PLASTICITY THE STATIC SYSTEM CHANGES AND THUS A REDISTRIBUTION OF THE FORCES OCCURS IN CASE OF A STATIC INDETERMINATE STRUCTURE PLASTIC DEFORMATION CAPACITY IS ESSENTIAL IF THE ADDITONAL LOAD BEARING CAPACITY IS USED AND SHOULD BE CHECKED Ir J.W. Welleman page nr 9

10 BENDING ELASTICITY VERSUS PLASTICITY FULLY PLASTIC MOMENTCAPACITY SHAPE FACTOR EXAMPLES BEHAVIOUR OF THE CROSS SECTION MOMENT-CURVATURE PLASTIC ZONES IDEAL PLASTIC HINGE STRUCTURAL BEHAVIOUR (LIMIT STATE ANALYSIS) BEAMS FRAMES Ir J.W. Welleman page nr 0

11 BENDING R = κ dϕ Cross section I Linear strain distribution over the depth of the beam : ( z ) = κ z h h -axis κ neutrale lijn druk trek z-axis = κ h σ = E Cross section strain stress Ir J.W. Welleman page nr

12 - - κ f neutral axis f f neutral axis f STRESS AND STRAIN IN THE CROSS SECTION DUE TO INCREASING CURVATURE = κ h strain-diagram + σ = E stress-diagram κ = κ h strain-diagram σ + stress-diagram = E = f see figure 3.3 par 3.. (Dutch Book) - - κ 3 neutral axis + + κ 4 3 = κ 3 h σ 3 = f 4 = κ 4 h σ 4 = f strain-diagram stress-diagram strain-diagram stress-diagram Ir J.W. Welleman page nr

13 FULLY PLASTIC MOMENT h ½h compression N - ½h N H = f = 0 N A = ' bh = f N ½h tension N + M p = N. h = 4 bh f b cross section stress : σ = 5 f DEMAND : HORIZONTAL EQUILIBRIUM THUS : RESULTANT COMPRESSION = RESULTANTE TENSION Ir J.W. Welleman page nr 3

14 MOMENT CURVATURE DIAGRAM (M-κ diagram) moment M p plastic M e elasto-plastic elastic curvature κ M p 4 bh f SHAPE FACTOR : α = RECTANGLE: α = =, 5 M bh f e 6 Ir J.W. Welleman page nr 4

15 FULLY PLASTIC MOMENT NO Smmetrical section, load along the axis of smmetr homogeneous material NO Not part of these lectures load Determine the medium line such that : F compression = F tension Determine the cross sectional area : A total medium line splits the cross section in two equal parts : A compression = A tension =½A total F = F = comp tension A total f Determine the resultant compressive and tensile forces acting on the cross section and the internal lever (arm) z. axis of smmetr Full plastic moment capacit of the cross section becomes : M p =F tension z Ir J.W. Welleman page nr 5

16 EXAMPLES OF PLASTIC MOMENT 50 mm 50 mm 5 mm 5 mm 50 mm 350 mm 5 mm 300 mm d 360 mm 400 mm 3φ6 Concrete C30/37 f cd = 0 N/mm Steel B500 f d = 435 N/mm 50 mm 50 mm Ir J.W. Welleman page nr 6

17 SOLID CIRCULAR CROSS SECTIONS - elastic moment - median line - plastic moment - shape factor median line - + f z F p F p plastic moment 50 mm f M = F p p z Ir J.W. Welleman page nr 7

18 I-SECTION 50 mm 5 mm F 94,3 mm 5 mm 3,5 mm 50 mm F neutral axis median line,5 mm 5 mm F 3,5 mm 38,75 mm F 4 N = 0! M = sum of the moments 300 mm Ir J.W. Welleman page nr 8

19 CONCRETE BEAMS ( hxb=400x50 mm ) f cd = 0 N/mm druk x u ½x u N c = 50 0 x u d 0,9 h d=360 mm z=d-½x u trek N s = N f d = 435 N/mm steel : A s =603 mm b=50 mm strain stress forces H = 0 N = N x = c s u ( ) M = N z = A f d x p s s d u A f s b f d cd Ir J.W. Welleman page nr 9

20 SOME SHAPE FACTORS,0,5,7,7,,5,0,0 α ELASTO PLASTIC BEHAVIOUR α,0 IDEAL PLASTIC BEHAVIOUR κ κ Ir J.W. Welleman page nr 0