Question

Find the dimensions of a circular cross section steel bar subjected to tension by a force N 20000 N in two hypotheses: a) the maximum allowable stress is 150 N/mm2, b) the maximum acceptable strain is 0.0005. Take E 207GPa

Answer 1

Answer:

a). d = 13 mm

b). d = 16 mm

Explanation:

a). Given :

  Force = 20000 N

  Maximum stress, σ = 150 N/$mm^{2}$

Therefore, we know that that

σ = $\frac{Force}{area}$

150 = \frac{Force}{\frac{pi}{4}\times d^{2}}

150 = \frac{20000}{\frac{pi}{4}\times d^{2}}

$d^{2}$ = 169.76

d = 13.02 mm

d $\simeq$ 13 mm

b). Given :

   Strain, ε =  0.0005

   Young Modulus, E =  207 GPa

                                   = 207$\times$$10^{3}$ MPa

Therefore we know that, Stress σ = E$\times$ε

                                                         = 207$\times$$10^{3}$$\times$0.0005

                                                         = 103.5 N/$mm^{2}$

We know that  

σ = $\frac{Force}{Area}$

103.5 = $\frac{Force}{\frac{pi}{4}\times d^{2}}$

$d^{2}$ = 246.27

d = 15.69 mm

d $\simeq$16 mm

Answer 2

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 = $\pi$/4 × d²

put the area in stress equation and find out d

d² = 4×force / $\pi$ ×stress

d² = 4× 20000 / $\pi$ ×150

d = $\sqrt{ 4× 20000 / [tex]\pi$ ×150}[/tex]

d = 13 mm

and now we know starin = stress / E

same like stress we find d here

d = $\sqrt{ 4× 20000 / [tex]\pi$ ×0.0005×207×10³ }[/tex]

so d =  15.68 mm