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

the required diameter of the rod is 9.77 mm

Step-by-step 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 = (P)/(A)`

A is the area = πd²/4 = (3.14 × d²)/4

`S = (P)/(((3.14*d^(2) )/(4)) )`

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

`3.14*S*{d^(2)} = {4P}`

`{d^(2)} =(4P)/(3.14*S)`

`d= \sqrt{(4P)/(3.14*S) }`

Substituting the values, we get

`d= \sqrt{(4*3*10^(3) )/(3.14*40*10^(6) ) }`

`d= \sqrt{(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