Design and assessment of structures

Transcription

1 Design an assessment o structures Limit state esign requires the structure to satisy in two principal criterias: the ultimate limit state (ULS) an the serviceability limit state (SLS) In Europe, the Limit State Design is enorce by the Eurocoes. The ultimate limit state is reache when the applie stresses actually excee the allowable strength o the structure or structural elements it causes to ail or collapse o the structure. We use magniication actor to get higher the loa (= esign loa) an reuction actors to get lower strength o the structure ( = esign strenght). The serviceability limit state is the point where a structure can no longer be use or it's intene purpose (but it woul still be structurally all right). The tolerances or serviceability epen on the intene use o the structure ( large eormations). ultimate limit state we compare carrying capacity an esign internal orce serviceability limit state we compare eormation an limit given eormation / 59

2 Characteristic an esign loa F k - characteristic value o loa ( use or SLS) F - esign value o loa ( use or ULS) F F k.g g Coeicients o reliability or loa: permanent loa g G, changeable loa g Q EU Czech republic g G,35, (,35) g Q,50,50 2 / 59

3 Design an assessment o tensile beams ultimate limit state serviceability limit state R l l all inner normal orce in esign value (rom F ) R carrying capacity Δl eormation in axial loa = alongation or contraction loa is given in characteristic values Δl all = Δl lim - allowable (limit) given eormation 3 / 59

4 Strenght o material Strenght o material = resistace o structure R (carrying capacity) (on the contrary o the loa it gets lower in esign values): Material: steel Fe 430/S275 y... Stress at yiel limit (rom stress- strain iagram) = 275 MPa u... Stress at ultimate limit 430MPa k y yiel limit is the value or esigning accoring to ultimate limit state in elastic-plastic range Strenght o material:. characteristic value = k 2. esign value k g M g g M... Coeicients o reliability or material 4 / 59

5 Design an reliability assessment o bar expose to axial orces Design o carrying structure, req, req juste esign Dimensioning Reliability assessment o esign (Limit state o carrying capacity). R R For steel: = (yiel limit) k g M Realization 5 / 59

6 Tension an compression basics o asssessment R x,a F 3 F 2 F x -R x,a l 3 l 2 l F F ormal stress [Pa] x l From the last lesson: Deormation [m] ili E i F 2 F2 3 2 F3 i σ = const. F Cross sectional characteristic or tension an compression is area ssessment o members in stress: yk Yiel limit g ULS: SLS: allowable R M allowable req max ssessment o members in eormations : l real ikli E i i Require area req... 6 / 59

7 Example ULS + SLS Make the esign an assessment o the square section beam accoring to both limit states. F k =25 k, l=2,5m, yk =235MPa, l all = 3mm, γ G =.30, E=20GPa, γ M =.0 F l=2,5m pproach: ULS: SLS: Design + assessment: - Distribution o 5 - charakt = k 7 - choose higher a req roun up real kl I I II II 3 - a 6 - lall req areq 9 assessment: req req E a real 4 reqi a req I ULS: R > SLS: l max > l Results: =32,5k, a reqi =,8mm, k =25k, a II req = 9,96mm, a =2mm, R =33,84k, l=2mm. 7 / 59

8 Example 2 ULS (ultimate limit state) R b b. Determine stress in both ros rom characteristic values (proile I80 area rom tables o one I80: =0, mm 2 ). 2. Make the assessment accoring to ULS: yk = 50MPa, γ M =,00, γ G =,. a b γ Determination Two Equilibrium Conitions in the hinge a: 2 a F b a sin g cos g a b a b F F x z 0 : 0 : sing F cosg Conitions o solution: ) Hinges in noes 2) Loa in noes Then in ros just orces a = 2 m, b = 5 m, F k = 39,3 k, k 2k 02, k 94,3k 2 2,32k 03,68k F =43,23k c R c 8 / 59

9 Example 2 - ULS Diagram o orces along the ros F Distribution o the stress in the section - ormal stress is constant 2 2 σ x = / x Results: ormal stress σ x =34,9 MPa, σ x2 =24,5 MPa, R 3,55k 2, 32k 9 / 59

10 Example 3 ULS, SLS Make esign o the ro mae rom IP - proile accoring to ULS. Fe 430/S275, g,35 ( reqi ) Make esign o the ro also accoring to SLS, E=2,.0 5 MPa, Δl all = 20 mm ( req II ) Choose resulting esign an make the assessment accoring both limit states Calculate stress in the ro σ x, raw its istribution over the (σ x =const.) R az R or: az, k M M ib ib k 0 : Raz, k 4,5 gk 32,5 gk, : k 4,5 gk 32,5 gk, ssessment: R l ov l kl E 0m g k = 200k/m σ x = / 3,5 B R bx =0 R bz Results.: = 270k, reqi = 98,82mm 2, k =200,0k, req II =476,2mm 2, IP00, R = 29,5k, Δl = 8,98mm 0 / 59

11 Example 4 - SLS Steel bar o the circle cross section area sia a =6mm. E=2,.0 5 MPa. - etermine normal stress in all parts o a bar. - etermine the elongations o the bar Δl an compare with δ lim = 5mm (assessment accoring SLS) Don t orget to construct orces. Calculate in given characteristic values. b =,7 m c =, m P P 2 P 3 = 0,6 m P = 20 k b c σ x = / P 2 = 0 k P 3 = 20 k y Results: Δl =,376 mm < δ lim = 5 mm σ =7,9MPa, σ 2 =39,06MPa, σ 3 =78,3MPa all parts are tensile / 59