Calculo de esfuerzos

Páginas: 2 (452 palabras) Publicado: 29 de febrero de 2012
Calculo de esfuerzos en los pasadores
Modulo de rigidez G=78Gpa
Diámetro de los pasadores= 0.0635 m
Area de los pasadores= π*r² = π*(0.031m) ² = 3.16x10¯³m²∑MC=0
19.692KN-m + 0.4609 FAD = 0
∑F(x) =0
Cx + FAD *cos40° = 0 Cx=-32.62 KN
∑F(y) =0Cy + FAD *sen40° - 65.64 KN = 0 Cy= 92.71 KN
Fc=Cx² + Cy² Fc= (32.62KN) ²+(92.71KN) ² = 98.28 KN

Tc = Pc 2*A = 98.28KN2*(3.16x10¯3m2) = 15.55 MPa

Øc = Tc*LG*r = 15.55MPa*0.2m75 GPa*0.031 m = 1.30 x10¯³ rad
Øc = 0.07°
TD = PD 2*A = 42.588KN2*(3.16x10¯3m2) = 6.73 MPa
ØD = TD*LG*r = 6.73 MPa*0.2m75 GPa*0.031 m = 5.65 x10¯⁴ rad
ØD = 0.032°

- FAC *sen30° - FAD *sen40° = 0
- FAC *sen30° - 42.588KN*sen40° = 0 FAC =- 54.75 KN
∑F(x) =0
-FAB + FAC *cos30° + FAD *cos40° = 0
-FAB -54.75KN *cos30° + 42.588KN*cos40° = 0FAB =14.79 KN
TB = PB 2*A = 14.79KN2*(3.16x10¯3m2) = 2.34 MPa

ØB = TB*LG*r = 2.34MPa*0.2m75 GPa*0.031 m = 2.01 x10¯⁴ rad
ØB = 0.011°

TC = PC 2*A = 54.75KN2*(3.16x10¯3m2) =8.66MPa

ØC = TC*LG*r = 8.66MPa*0.2m75 GPa*0.031 m = 7.44 x10¯⁴ rad
ØC = 0.042°
TA = PAB + PAC + PAD 2*A = 14.794KN+54.75KN+42.08KN2*(3.16x10¯3m2) = 17.74MPa

ØA = TA*LG*r =17.74MPa*0.2m75 GPa*0.031 m = 1.52 x10¯³ rad
ØA = 0.087°

∑MG=0
(65.64KN *2.80m) - (FFE *cos80.90°*0.2m) + (FFE *sen80.90°*1.25m) = 0
183.792KN-m – 1.26m FAD = 0
FFE=183.792KN-m 1.26 m =145.66 KN∑F(x) =0
Gx + FFE *cos80.90° = 0 Gx= 23.037 KN
∑F(y) =0
Gy + FFE *sen80.90° - 65.64 KN = 0 Gy= 78.18 KN
FG=Gx² + Gy² FG= (23.037KN) ²+(78.18KN) ² = 81.50 KN
TG = PG 2*A =81.50KN2*(3.16x10¯3m2) = 12.89MPa
ØG = TG*LG*r = 12.89MPa*0.2m75 GPa*0.031 m = 1.10 x10¯³ rad
ØG = 0.06°
TE = PE 2*A = 145.66KN2*(3.16x10¯3m2) = 23.01 MPa
ØE = TE*LG*r = 23.01...

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