Respuesta :
Answer:
d = 13 mm
d = 15.68 mm
Explanation:
Given data
force = 20000 N
stress = 150 N/mm²
strain = 0.0005
E = 207 GPa
Solution
we know stress = force / area
so 0.0005 = 20000 / area
area = [tex]\pi[/tex]/4 × d²
put the area in stress equation and find out d
d² = 4×force / [tex]\pi[/tex] ×stress
d² = 4× 20000 / [tex]\pi[/tex] ×150
d = [tex]\sqrt{ 4× 20000 / [tex]\pi[/tex] ×150}[/tex]
d = 13 mm
and now we know starin = stress / E
same like stress we find d here
d = [tex]\sqrt{ 4× 20000 / [tex]\pi[/tex] ×0.0005×207×10³ }[/tex]
so d = 15.68 mm
Answer:
a). d = 13 mm
b). d = 16 mm
Explanation:
a). Given :
Force = 20000 N
Maximum stress, σ = 150 N/[tex]mm^{2}[/tex]
Therefore, we know that that
σ = [tex]\frac{Force}{area}[/tex]
150 = \frac{Force}{\frac{pi}{4}\times d^{2}}
150 = \frac{20000}{\frac{pi}{4}\times d^{2}}
[tex]d^{2}[/tex] = 169.76
d = 13.02 mm
d [tex]\simeq[/tex] 13 mm
b). Given :
Strain, ε = 0.0005
Young Modulus, E = 207 GPa
= 207[tex]\times[/tex][tex]10^{3}[/tex] MPa
Therefore we know that, Stress σ = E[tex]\times[/tex]ε
= 207[tex]\times[/tex][tex]10^{3}[/tex][tex]\times[/tex]0.0005
= 103.5 N/[tex]mm^{2}[/tex]
We know that
σ = [tex]\frac{Force}{Area}[/tex]
103.5 = [tex]\frac{Force}{\frac{pi}{4}\times d^{2}}[/tex]
[tex]d^{2}[/tex] = 246.27
d = 15.69 mm
d [tex]\simeq[/tex]16 mm
