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:
[tex]S = \frac{P}{A}[/tex]
A is the area = πd²/4 = (3.14 × d²)/4
[tex]S = \frac{P}{(\frac{3.14*d^{2} }{4}) }[/tex]
[tex]S = \frac{4P}{{3.14*d^{2} } }[/tex]
[tex]3.14*S*{d^{2}} = {4P}[/tex]
[tex]{d^{2}} =\frac{4P}{3.14*S}[/tex]
[tex]d= \sqrt{\frac{4P}{3.14*S} }[/tex]
Substituting the values, we get
[tex]d= \sqrt{\frac{4*3*10^{3} }{3.14*40*10^{6} } }[/tex]
[tex]d= \sqrt{\frac{12000 }{125600000 } }[/tex]
[tex]d= \sqrt{9.55*10^{-5} }[/tex]
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
