Question

A 1.5-m-long aluminum rod must not stretch more than 1 mm and the normal stress must not exceed 40 MPa when the rod is subjected to a 3.0-kN axial load. Knowing that E = 70 GPa, determine the required diameter of the rod.

Answer 1

Answer:

the required diameter of the rod is 9.77 mm

Explanation:

Given:

Length = 1.5 m

Tension(P) = 3 kN = 3 × 10³ N

Maximum allowable stress(S) = 40 MPa = 40 × 10⁶ Pa

E = 70 GPa = 70 × 10⁹ Pa

δ = 1 mm = 1 × 10⁻³ m

The required diameter(d)  = ?

a) for stress

The stress equation is given by:

$S = \frac{P}{A}$

A is the area = πd²/4 = (3.14 × d²)/4

$S = \frac{P}{(\frac{3.14*d^{2} }{4}) }$

$S = \frac{4P}{{3.14*d^{2} } }$

$3.14*S*{d^{2}} = {4P}$

${d^{2}} =\frac{4P}{3.14*S}$

$d= \sqrt{\frac{4P}{3.14*S} }$

Substituting the values, we get

$d= \sqrt{\frac{4*3*10^{3} }{3.14*40*10^{6} } }$

$d= \sqrt{\frac{12000 }{125600000 } }$

$d= \sqrt{9.55*10^{-5} }$

d = (9.77 × 10⁻³) m

d = 9.77 mm

b) for deformation

δ = (P×L) / (A×E)

A = (P×L) / (E×δ) = (3000 × 1.5) / (1 × 10⁻³ × 70 × 10⁹) = 0.000063

d² = (4 × A) / π = (0.000063 × 4) / 3.14

d² = 0.0000819

d = 9.05 × 10⁻³ m = 9.05 mm

We use the larger value of diameter = 9.77 mm