Physics, 15.07.2019 14:30 hardwick744

Arod 150 cm long and of diameter 2.0 cm is subjected to an axial pull of 20 kn. if the modulus of elasticity of the material of the rod is 2 x 105 n/mm2 ; determine: (i) the stress, (ii) the strain, and (iii) the elongation of the, rod

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The stress, strain and modulus of elasticity are connected by an equation. Thus, the given question has the following answers:

(i) The stress is 2 KN/ $mm^{2}$

(ii) The strain is 0.01

(iii) The elongation is 15 mm

From the question, given: l = 1500 mm, D = 20 mm, F = 20 000 N, λ = 2 x $10^{5}$ N/ $mm^{2}$ .

Thus,

(i) Stress = $\frac{Force}{Area}$

Area of the rod = Area of a circle = $\pi$ $r^{2}$

where r is the radius.

radius, r = $\frac{diameter}{2}$

= $\frac{20}{2}$

r = 10 mm

so that;

Stress = $\frac{20000 }{10}$

= 2000

Stress = 2 KN/ $mm^{2}$

(ii) To calculate the strain;

modulus of elasticity, λ = $\frac{stress}{strain}$

2 x $10^{5}$ = $\frac{2000}{strain}$

strain = $\frac{2000}{200000}$

= 0.01

strain = 0.01

(iii) The elongation of the rod is also its extension, e.

So that;

strain = $\frac{extension}{length}$

extension = strain x length

= 0.01 x 1500

extension = 15 mm

Therefore: the stress is 2 KN/ $mm^{2}$ , the strain is 0.01, and the elongation is 15 mm.

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Answer from: jamyiaa93

Given:
rod of circular cross section is subjected to uniaxial tension.
Length, L=1500 mm
radius, r = 10 mm
E=2*10^5 N/mm^2
Force, F=20 kN = 20,000 N
[note: newton (unit) in abbreviation is written in upper case, as in N ]

From given above, area of cross section = π r^2 = 100 π =314 mm^2
(i) Stress,
σ
=force/area
= 20000 N / 314 mm^2
= 6366.2 N/mm^2
= 6370 N/mm^2 (to 3 significant figures)

(ii) Strain
ε
= ratio of extension / original length
= σ / E
= 6366.2 /(2*10^5)
= 0.03183
= 0.0318 (to three significant figures)

(iii) elongation
= ε * L
= 0.03183*1500 mm
= 47.746 mm
= 47.7 mm (to three significant figures)

Answer from: Quest

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Answer from: Quest

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