Statically indeterminate examples - axial loaded members, rod in torsion, members in bending

Transcription

1 Elsticity nd Plsticity Stticlly indeterminte exmples - xil loded memers, rod in torsion, memers in ending Deprtment of Structurl Mechnics Fculty of Civil Engineering, VSB - Technicl University Ostrv 1 / 8

2 Stticlly indeterminte structures Condition of solution: elstic (liner) ehviour of strin-stress digrm of mteril Stticlly indetermined prolems: numer of unknown vriles > numer of equilirium equtions Solution: numer of unknown vriles numer of equilirium + equtions numer of deformtion conditions Stticlly determined nd indetermined exmples / 8

3 Axilly loded memers 1. Fixed supported column on oth end. Rods 3. Nehomogenized r (steel pipe filled in y concret). 3 / 8

4 Exmple 1: Fixed supported column on oth ends R l Condition of solution: elstic (liner) ehviour of strin-stress digrm of mteril Unknown vriles in exmple: R ( N ) R ( ), N 1 F l 1 l Equilirium eqution: R 0 : R + R F 0 z N1 + N F 0 Deformtion eqution: R l 0 : l N1l1 Nl + E1. A1 E. A 1 + l 0 Stticlly determined nd indetermined exmples 4 / 8

5 Determine norml stress in oth prts, cross sections I140 nd I180, F650kN. N 1 R + N - -R Exmple 1 I 140 I 180 R F R l 1 1,5m l,5m We cn determine the unknows just from the one eqution deformtion condition. 1) 1x stticlly indetermined in the xil tsk ) Equilirium equtions (just xil tsk): F i,verticl 0 R + R F 0 3) Deformtion condition: l 0 From the digrm ohf norml forces : 1 R N R F N N l1 Nl R 1 ( ) 1 l R F l EA EA EA EA 1 l l + l 1 By sustituting into the deformtion condition: 1 0 R F l l A 1 A1 + l 1 A N 1 R N R - F ( -R) 5 / 8

6 Exmple 1 Determine norml stress in oth prts, cross sections I140 nd I180, F650kN. R R N1 R 338, 58kN N 1 + I 140 F l 1 1,5m N 311, 4kN Norml stress in the r: N - -R I 180 R l,5m N1 σ x1 186, 03MP A 1 N σ x 111, 6MP A 6 / 8

7 Exmple The r is loded y forces see the picture. A 1 3cm, A 10cm, E 1 E, F 1 0kN, F 45kN. Guess the direction of the rections nd digrm of norml forces Divide the r on prts, where there will e different vlue of stresses nd clculte them. x Results.: N 1 1,875kN N -18,15kN N 3 6,85kN R 6,85kN R 1,875kN F 1 F 1 0,6m 0,8m 0,4m σ 1 6,5MP σ -18,15MP σ 3 6,875MP 7 / 8

8 Exmple Determine norml stress in the r U100, which is under the temperture chnge T90 C. l6m, E, MP, α T 1, 10 5 [ ] o 1 C ΔT 1) 1x stticlly indetermined in the xil tsk ) Equilirium equtions (just xil tsk): F i,x 0 R R 0 R R N R R 3) Deformtion condition: 4) Norml stress in the r: N l 0 Nl l + αt T l EA N αt T EA 0 55,15 kn σ x N A 06,8MP We cn determine the unknows just from the one eqution deformtion condition. 8 / 8

9 Exmple 3 The r of the lenght l 1 m is etween two stiff wlls, the hole is 0, mm. Wht vlue of norml stress is in the rod, if the temperture chnge is +50 C? α T C -1, E 1, MP 0, mm l 1000 mm 1) Deformtion eqution: l 0, 10 Nl EA 3 m + αt T l 0, 10 ) Result: 3 σx 71,5MP 9 / 8

10 Exmple 4 Determine stress in the rods, if oth re from I 140. Conditions of solution: elstic ehviour of the rods, idelly stiff sl 1 g 10 knm -1 c l 1 m 3 m 4 m L 7 m 10 / 8

11 Exmple 4 Determine stress in the rods, if oth re from I 140. Conditions of solution: elstic ehviour of the rods, stiff ehviour of the em. 1) 1x stticlly indetermined ) Equilirium equtions : R R c F ix 0 R x 1 g l 1 m F iz 0 M i 0 3) Deformtion condition: 3 m 4 m R z L 7 m 11 / 8

12 Exmple 4 Determine stress in the rods, if oth re from I 140. (Equilirium equtions): F ix 0 F iz 0 M i 0 1 l 1 c l l 1 m l1 l + N l EA N 1 g N N1l 1 1 EA ( + ) ( unknowns forces N 1, N, choose n eqution from Equilirium equtions, which includes just N 1 N ) N 3) Deformtion condition: N 7 q 7 3,5 0 1 / 8

13 Exmple 5 Determine N in rods. I 450, 1m. Conditions of solution: elstic ehviour of the rods, idelly stiff sl c 1 3 l 1 m F 400 kn 13 / 8

14 Exmple 5 Determine norml forces in the rods. Cross sections re I ) 1x stticlly indetermined in the xil ts R R R c c N N 1 N N 3 N N 3 ) Equilirium conditions: l 1 m F ix 0 F iz 0 M i 0 R N 1, R N, R c N 3 F 400 kn 14 / 8

15 Exmple 5 Determine norml forces in the rods. Cross sections re I 450. Deformtion of construction: c 1 3 l 1 m l l 3 l 1 F 400 kn 15 / 8

16 Exmple 5 Determine norml forces in the rods. Cross sections re I ) Deformtion condition: l 1 c 1 3 l 1 m l l 3 (coordinte system) y x z l1 l3 l l3 4 (3 unknowns forces N 1, N, N 3, choose equtions from Equilirium equtions) Chosen equtions: F iz 0 M i 0 16 / 8

17 Exmple 5 Solution: 1 3 N 1 N N 3 F 400 kn results: N 1 450kN, N 1400kN, N 3 350kN y z c x l 1 m Deform. condition: l1 l3 l l3 4 l l1 + l3 N l/ean 1 l/ea+n 3 l/ea Equilirium equtions: F iz 0 -N 1 - N - N 3 + F d 0 M i M i 0..N +4..N 3.F d 0 17 / 8

18 Exmple 6 Nehomogenized r (steel pipe filled in y concret). Determine norml stress in steel nd concret. d 1 80 mm (externl dimeter), d 70 mm (internl dimeter). E 10GP, E cm 4GP. Conditions of solution: - elstic ehviour of mterils, - F ffects uniformly to the section F 11 kn 1) 1x internlly stticlly indetermined l 0,5 m 18 / 8

19 N N o + N B N N O - - Exmple 6 F ) Deformtion eqution: l S l C l 0,5 m N sl E s A s Ncl E A ( unknowns, we tke one eqution from equilirium equtions) 3) Equilirium equtions: c F i,verticl 0 N B F - R 0 R F -N - N S - N C (F R 0) c F + N S + N C 0 3) Stresses: R σ S N A S, σ S N A C C results: N S -81,65kN, N C -30,35kN, σ S -69,31MP, σ C -7,91MP 19 / 8

20 Exmple 7: Reinforced concrete column Condition of solution: Elstic (liner) ehviour of strin-stress digrm of mteril nd uniform ffect of lod to cross-section re F steel Unknown vriles in exmple: N s, N c Equilirium eqution: F N s + N c concrete Deformtion eqution: l s l c N s. l E. A s s Nc. l E. A c c Stticlly determined nd indetermined exmples 0 / 8

21 Stticlly indeterminte prolems in torsion Both fixed ended shft l1 l l Condition of solution: liner elstic ehviour of mteril M x, M x, c M x, M x, + T c M x, Unknowns: M x, ( M x,1 ), M x, ( M x, ) Equilirium equtions: M i 0 : x, M x, M x, + M x, c 0 Deformtion condition: ϕ i 0 : i 1 T.l G.I i i i t, i 0 Stticlly indeterminte prolems 1 / 8

22 Stticlly indeterminte prolems in ending Methods of computing of stticlly indeterminte endings ems: ) Fourth-order integrtion of the function of lod ) Force method c) Methods sed on energetic principles (Elsticity nd plsticity II.) Fourth-Order Integrtion of differentil eqution (from the function of lod) Schwedlers reltions w ϕ y w M y E. I y. w Vz E. I y. w q ( x)? z E. I. w y IV E. I. w E. I E. I IV y q x y y ( ). w q + C1 V ( x) z. w q + C1. x + C M ( x) y x E. J. w y q + C1. + C. x + C ( x) 3 3 x x E. I y. w q + C1. + C. + C. x + C 6 ( x) 3 4 Solution: 4 unknowns C 1 C, C3,, C 4 oundry conditions 4 Stticlly indeterminte prolems in ending Fourth-order Integrtion / 8

23 Stticl nd deformtion oundry conditions Type of the end (oundry) Deformtion Boundry Condition Stticl Boundry Condition Free end w 0 ϕ 0 M 0 w 0 V 0 w 0 Simply supported end Fixed end w 0 ϕ 0 w 0 ϕ 0 M 0 w 0 V 0 w 0 M 0 w 0 V 0 w 0 Stticlly indeterminte prolems in ending Fourth-order Integrtion 3 / 8

24 Fourth-Order Integrtion of differentil eqution Exmple 1 Determine the digrms of internl forces (V, M) t stticlly indeterminte em. Use differentil reltions. q l 4 / 8

25 Fourth-Order Integrtion of differentil eqution Boundry conditions: 5 / 8

26 Fourth-Order Integrtion of differentil eqution 6 / 8

27 Fourth-Order Integrtion of differentil eqution Exmple Determine the internl forces (V, M) t stticlly indeterminte em y the Fourth-order integrtion of differentil eqution. q l Equtions the sme s Exmple 1, oundry conditions different 7 / 8

28 Force method Principle of the Force method: q( x ) q z ϕ Deformtion conditions: 0 ϕ ϕ ϕ, q +, M Superposition 0 + q( x ) Superposition ϕ, q 0 ϕ, M 0 q z M y Designte one of the rections s redundnt nd eliminte it The redundnt rection is then treted s n unknown lod tht together with other lods must produce deformtions tht re comptile with the originl supports The deflection or ngulr rottion t the point where the support hs een eliminted is otined y computing seprtely - the deformtions cused y the given lods nd y the redundnt rection Results will e otined y superposition Stticlly indeterminte prolems in ending Force method 8 / 8

29 Exmple 1 Force method Determine the digrms of internl forces (V, M) t stticlly indeterminte em. Use the Force method (method of superposition). q l Deformtion condition: w,q + w,r 0 q l R 9 / 8

30 Exmple 1 Force method Deformtion condition: w,q + w,r 0 w,q w,r 4 q.l 8.EI R.l/ 3.EI 3 Determine other rections from equilirium conditions nd construct the digrms of internl forces. 30 / 8

31 Exmple Force method Determine the internl forces (V, M) t stticlly indeterminte em y the Force method. (redundnt rection is M ) q ϕ 0 Deformtion condition: ϕ,q + ϕ,m 0 q ϕ,m M ϕ,q 31 / 8

32 Exmple Force method q ϕ,q q.l 3 4.EI ϕ,q ϕ,m M ϕ,m M.l 3.EI Deformtion condition: Other rections from equilirium conditions 3 / 8