ejercicio 6 de cs tarea 1
TRANSCRIPT
Datos de la probeta
Longitud inicial (Lo) 50Area (pi*r^2) 128.67963 #NAME?
Carga (N) Longitud (mm)0 50 0 0
4400 50.02 0.0200000000000031 0.000400000000000113200 50.06 0.0600000000000023 0.001222000 50.1 0.100000000000001 0.00230800 50.14 0.140000000000001 0.002833000 50.15 0.149999999999999 0.00334000 50.17 0.170000000000002 0.003435000 50.24 0.240000000000002 0.004836500 50.35 0.350000000000001 0.00736800 50.48 0.479999999999997 0.009599999999999935900 50.63 0.630000000000003 0.0126000000000001
b)
Esfuerzo (Mpa)= F/A0 0
0.0004000000000001 34.19344615771740.0012 102.580338473152
0.002 170.9672307885870.0028 239.354123104022
0.003 256.450846182881
∆l=lf - lo (mm) Deformacion= ∆l/Lo
Deformacion= ∆l/Lo
0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.00350
50
100
150
200
250
300
f(x) = 85483.6153942932 x
ԑ
σ
σ = 85484ԑ
E=85484MpaE=(85484)*(0,0000010000)= 85,5Gpa
C) te la mando por whatsap aunque no estoy segura de que sea asi
d)
d=((50,63-50)/(50))*100%d=1,26%
ductilidad(d)%= (e/L˳)*100%
Esfuerzo (Mpa)= F/A0
34.1934461577174102.580338473152170.967230788587239.354123104022256.450846182881264.222083945998271.993321709116283.650178353792285.981549682728278.987435695922
0 0.002 0.004 0.006 0.008 0.01 0.012 0.0140
50
100
150
200
250
300
350
ԑ
σ
0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.00350
50
100
150
200
250
300
f(x) = 85483.6153942932 x
ԑ
σ
0 0.002 0.004 0.006 0.008 0.01 0.012 0.0140
50
100
150
200
250
300
350
ԑ
σ