mecanica-viga-v2
3
E.E.E ΣM A =0; (.95lb) (6in)- R 2 (16in) + (.55lb) (21in)=0 (16in) R 2 = (.95lb) (6in) + (.55lb) (21in) (16in) R 2 = 17.25lb/in R 2 = 17.25 lb / ¿ 16 ∈¿¿ = 1.07lb ΣFy=0 ; R1- .95lb + 1.07lb-.55lb=0 r1= .95lb-1.07lb+.55lb R1= .43lb 1era sección 0< x <6 X o ΣFy= .43 – v=0 v M V= .43 Σ M o−o =0 ; 0.43(X) – M =0 o M = 0.43 (x) R1=0.43lb 2da sección 6 < x < 16 ΣFy= .43 - .95 – v=0 V= .43 - .95 Σ M o−o =0 ; .43(x) – 0.95(x – 6) - M x 0 6 M 0 2.5 8
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Problema de mecánica de materiales
Transcript of mecanica-viga-v2
E.E.EΣM A=0; (.95lb) (6in)-R2(16in) + (.55lb) (21in)=0
(16in)R2= (.95lb) (6in) + (.55lb) (21in)
(16in)R2= 17.25lb/in
R2=17.25lb /¿16∈¿¿ = 1.07lb
ΣFy=0 ; R1- .95lb + 1.07lb-.55lb=0r1= .95lb-1.07lb+.55lbR1= .43lb
1era sección 0< x <6
X o ΣFy= .43 – v=0 v M V= .43
ΣM o−o=0; 0.43(X) – M =0 o M = 0.43 (x) R1=0.43lb2da sección 6 < x < 16 ΣFy= .43 - .95 – v=0 V= .43 - .95
Σ M o−o=0 ; .43(x) – 0.95(x – 6) - M
M = 0.43(x) – 0.95(x-6)
3ra sección 16 < x < 21 V ΣFy=0; 0.43 – 0.95 + 1.07 – v= 0 V = .43 – 0.95+ 1.07
ΣM o−o=0; .43 (x) – 0.95 (x-6) + 1.07 (x-16) – M=0 M = 0.43(x) – 0.95(x – 6) + 1.07(x – 16)
x 0 6M 0 2.58
x 6 16M 2.58 -2.62
x 16 21M -2.62 0
C (viga de Aluminio)
0.125 in 0.0625in = cy 1 in
Momento Máximo:
Momento de Inercia
I= bh ³12
= ¿¿
Esfuerzo máximo:
σ=Mmáx (Cy )
I=(2.58 lb∗¿)¿¿
