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2011

2011. S 1. S 2. t yx. 4. 1. x. x. s nx. 4. 1. y. dy. 3. 2. s nx. t xy. 3. 2. p. t xy. ) 2. ) 2. (. (. s nx + s ny. s nx + s ny. s nx - s ny. s nx - s ny. +. -. + t 2. + t 2. 2 a. s 1 =. s 2 =. s 1. a. f. 2. 2. 2. 2. s nx. n. s 3. 2 g.

hadassah-haynes
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2011

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  1. 2011

  2. S1 S2 tyx 4 1 x x snx 4 1 y dy 3 2 snx txy 3 2 p txy )2 )2 ( ( s nx+ s ny s nx+ s ny s nx- s ny s nx- s ny + - + t2 + t2 2 a s1 = s2 = s1 a f 2 2 2 2 snx n s3 2 g 2 t xy tyx tan 2a = s nx- s ny p’ Tensiones Principales. Líneas Isostáticas.

  3. Ejes pricicipales de una sección Son los ejes que pasando por G el momento de inercia de la sección es máximo y mínimo, se demuestra que son perpendiculares entre si. Cuando en una sección existe un eje de simetría es un eje principal

  4. y z f Mf y z Ejes pricicipales de una sección. Línea neutra Flexión Recta: Mf coincide con eje principal Flexión Esviada: Mf no coincide con un eje principal + Línea neutra: no existe tensión normal. s = 0 - Mfz = Mf cos f Mfy = Mf sen f s = Mf ·z·sen f /Iy - Mf · y·cos f /Iz y/z = tag f· Iz /Iy Si Iz > Iy :La línea neutra se acerca a “y” o mínimo esfuerzo

  5. Secciones compuestas B b E2·B·h32/12 = E1·b·h32/12 E2 h2 b = B· E2·/E1 E1 E1 Mf/E·Iz

  6. f H H Flexión Hiperestática A q B Parábola : y = 4·f·x·(L-x) / L2 Longitud : Lf = L + 8/3 ·f 2 /L Alargamiento : d = Lf - L = 8/3 ·f 2 /L = H·L/S·E H = S·E ·8/3 ·(f /L) 2 s = H/S = E ·8/3 ·(f /L) 2

  7. MfA f A A C B B MfCB C M/ E·Iz A’ B’ Empotramientos elásticos fA = MfA /KA A Rigidez del empotramiento : KA =MfA /fA Permisividad del empotramiento : 1/KA = fA/ MfA fA = MB·L/6·E·Iz fB = MB·L/3·E·Iz KB =MB /fB = 3·E·Iz /L 1/KB = fB/ MB= L/3·E·Iz

  8. M M Asientos en vigas empotradas A B R·L = 2 M R D A M DAB = M·L2/6·E·Iz B M R DAB · 6·E·Iz/L2= M DAB · 12·E·Iz/L3= R

  9. D P C A L L L B P C A + + + - B HB RB Semipórtico IAB = IAC = Iz MC = HB·L + RB·2·L – P·L dHB = HB ·L3/ 3·E·Iz+ FA·L FA= HB·2·L2/E·Iz+ RB·4·L2/2·E·Iz-P·L2/2·E·Iz dVB = HB2·L3/E·Iz+ RB·4·L2/(2·E·Iz)·(2/3)·2·L– P·L2/(2·E·Iz)·(5/3)·L FA·2·E·Iz/L2 = HB·4+ RB·4 - P dHB ·6·E·Iz/L3 = 2·HB+ HB·12+ RB·12 – 3·P 0 = 14·HB+ 12 · RB– 3·P dVB·6·E·Iz/L3 = 0 =12HB + 16· RB– 5·P 36HB + 48· RB– 15·P = 56·HB+ RB·48 – 12·P HB= - 3/20·P dVB=0 dHB=0 RB = (17/40)·P RC = (23/40)·P MC = - 3/10·P·L

  10. P D C I I I L A B L Resolución de Pórtico

  11. P - + D C I - - - - I I L A B L SFH = 0 dVB=0 SFV = 0 dHB=0 SMF = 0 FB=0 Resolución de Pórtico FB=0 0 =(3·L·MB -HB·(L2/2+ L2+L2/2)-P·(L2/8+ L2/2)+RB·(L2/2+L2)/E·Iz 0 = (3·L·MB -HB·2·L2-P·5·L2/8+RB·3·L2/2)/E·Iz dVB=0 0 =(MB·(3L2/2) -HB·L3 -P·(29·L3/48) + RB·(4·L3/3))/E·Iz dHB=0 0 =(MB·(2·L2) -HB·(5/3·L3 )-P·(3·L3/8) + RB·L3)/E·Iz P·5·L/8 = 3·MB - HB·2·L + RB·3·L/2 P·29·L/48 = 3·MB/2 - HB·L+ RB·4·L/3 P·3·L/8 = 2· MB – 5·HB·L /3 + RB·L RB = P /2 HB = P /8 MB = P·L /24

  12. + - + P D C - L - - + + dVB=0 L dHB=0 A FB=0 Resolución de Pórtico SdVB=0 dVB1 = (MB·(L2/2+ L2))/E·Iz dVB2 = (-HB·(L2·L/2+L2/2 ·L))/E·Iz dVB3 = (-P·(L2/8·(L/2+2/3·L/2)+ L2/2·L))/E·Iz dVB4 = (RB·(L2/2·2/3·L+L2 ·L))/E·Iz B FB= 0 =(3·L·MB -HB·2·L2-P·5·L2/8+RB·3·L2/2)/E·Iz dVB= 0 =(MB·(3L2/2) -HB·L3 -P·(29·L3/48) + RB·(4·L3/3))/E·Iz

  13. + - + P D C - - L - - + + + dVB=0 L dHB=0 A FB=0 B Resolución de Pórtico SdHB=0 dVH1 =(MB·(2·L2/2+ L2))/E·Iz dVH2 =(-HB ·(2·L2/2 ·2/3·L+ L3))/E·Iz dVH3 =(-P·(L2/8·L+L2/2·L/2))/E·Iz dVH4 = (RB·(L2/2·L+L2 ·L/2)/E·Iz FB= 0 = (3·L·MB -HB·2·L2-P·5·L2/8+RB·3·L2/2)/E·Iz dVB= 0 =(MB·(3L2/2) -HB·L3 -P·(29·L3/48) + RB·(4·L3/3))/E·Iz dHB= 0 =(MB·(2·L2) -HB·(5/3·L3 )-P·(3·L3/8) + RB·L3)/E·Iz

  14. dB = 0 q B q B A C A C L L L L L RB q + + A C - - 2 · L = L’ R’B B A C 2 · L = L’ Viga hiperestática 3 apoyos N = 0 VB = +5/8· q·L VB = - 5/8· q·L MFB = - 1/8· q·L2 R’B = - 5/8· q·L’ RB = + 5/4· q·L RA = RC = + 3/8· q·L

  15. S2 S3 P P x C C E D E D S1 S4 L L x x A A B B L HA HB RA RB - + P + + L + - + dVA=0 B dHA=0 Pórtico RA = RB = P/2 HA = - HB 0 = dHA dHA1 = HA·L3/3·E·Iz dHA2/3 = HA·L3/E·Iz+ RA·L3/(2·E·Iz) – P·L3/(8·E·Iz) dHA4 = HA·L3/3·E·Iz+ RA·L3/(2·E·Iz) – P·L3/(4·E·Iz) dHA = 0 =5/3 HA+ RA – 3/8·P 5/3 HA= 1/8·P HA= 3/40·P

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