3° Cuatrimestre Trabajo final
Enviado por juve09 • 3 de Abril de 2017 • Prácticas o problemas • 2.300 Palabras (10 Páginas) • 286 Visitas
[pic 1]
Construcción 1
3° Cuatrimestre
Trabajo final
- Arellano Ramírez Nicolás Fabián
Resistencia de materiales
Karla Kenia Arteaga barron
16/agosto/2015
σ X =-150 MPA σ Y= 120 MPA τ XY= 80 MPA [pic 2][pic 3][pic 4][pic 5]
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[pic 7]
R= 81.39[pic 8]
σ PROM= σ PROM= [pic 9][pic 10]
σ MIN= σ PROM – R [pic 11]
σ MIN= 15MPA-81.39 = -66.39 MPA[pic 13][pic 12]
σ MAX= σ PROM + R
σ MAX= 15MPA + 81.39 =96.39 MPA
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R= 81.39 σ MIN=66.39 MPA σ PROM= [pic 22][pic 23]
[pic 24][pic 25][pic 26]
σ X =-8 N/M2 σ Y= 6 N/M2 τ XY= 4 N/M2[pic 27][pic 28][pic 29][pic 30]
[pic 31]
[pic 32]
R=4.123[pic 33]
σ PROM= σ PROM= [pic 34][pic 35]
σ MIN= σ PROM – R [pic 36]
σ MIN= 1MPA-4.123 = -3.123 N/M2[pic 37]
σ MAX= σ PROM + R
σ MAX= 1MPA + 4.123 =5.123 N/M2
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[pic 44][pic 45]
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R=4.123
σ MIN=-3.123 N/M
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σ PROM= [pic 48]
[pic 49][pic 50][pic 51]
σ X =-412 N/M2 σ Y= 360 N/M2 τ XY=230 N/M2[pic 52][pic 53][pic 54][pic 55]
[pic 56]
[pic 57][pic 58]
R=449.33 [pic 59]
σ PROM= σ PROM= [pic 60][pic 61]
σ MIN= σ PROM – R [pic 62]
σ MIN=-386 N/M2-449.33= -63.33 N/M2[pic 63]
σ MAX= σ PROM + R[pic 64]
σ MAX= -386 N/M2 + 449.33 =835.33 N/M2
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[pic 71][pic 72]
R=449.33 σ MIN=-63.33 N/M2 835.33 N/M2 σ PROM [pic 73][pic 74]
[pic 75][pic 76][pic 77]
[pic 78]
σ X =-1900 MPA σ Y= 2400 MPA τ XY= 1300 MPA [pic 79][pic 80][pic 81]
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R= 2512.46[pic 85]
σ PROM= σ PROM= [pic 86][pic 87]
σ MIN= σ PROM – R [pic 88]
σ MIN= -2150MPA-2512.46 = -4662.46 MPA[pic 89]
σ MAX= σ PROM + R
...