Question

Q3. (a) Define "Brittle Define "Brittle Materials and Ductile Materials in terms of Ductinty. (12 marks) (b) A solid metal bar of 30 mm tensile load of 120 kN. Find the stress, strain an Assume the Young's Modulus of the metal is 200 kN/mmº. Dar of 30 mm in diameter and 8 m in length is subjected to a ad of 120 kN. Find the stress strain and elongation in the matena. (9 marks) (c) A metal bar is subjected to a tensile load of 900 N and the stiffness o is 2 kN/mm. Find the extension of the bar by Hooke's Law. (4 marks) the bar by Hooke's Law.

Answer 1

Hi,

Hope you are doing well.

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PART (a):

Ductility: Ductility is the measure of ability of a material to undergo plastic deformation due to a deforming force, before breaking, or simply the ability to stretch.

Brittle materials are those materials with less ductility (easily breakable with a less amount of deforming force) whereas Ductile materials are those materials with high ductility, hence ductile materials can be hammered to thin sheets or plates.

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PART (b):

Given that:

Radius of the metal bar, R = 15 mm = 0.015 mm

Hence, Area of cross section is given by:

A = TR2 = 1 x (0.015 m)

.: A = 7.0686 x 10-4 m2

Tensile load, F = 120 kN = 120 x 10³ N

Length of the rod, L = 8 m

Young's modulus, Y = 200 kN/mm² = (200 x 10³) N /(10^-6m²) = 200 x 10⁹ N/m²

We know that Young's modulus is given by,

FL Y = AAL

Therefore, Change in length (Elongation) is given by:

(120 x 105N) × (8 m) (7.0686 x 10-4 m²)× (200 x 109 N/m2) AY

L.AL = 0.006791 m

Therefore, Stress is given by:

Force 120 x 103 N Stress = Area = 7.0686 x 10-4 m2

. Stress = 1.698 x 108 N/m2

Strain is given by:

Strain = 41 -0.006791 m 8 m

. Strain = 8.489 x 10-4

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PART (c):

According to Hooke's Law, we have:

F = kx

Given that,

Force, F = 900 N

Spring constant/ Hooke's constant, K = 2 kN/mm = 2000 x 10³ N/m

Therefore, Extension of the bar is given by:

900 N x = K = 2000 x 103 N/m

_.x = 4.5 x 10-4 m

OR

.x = 0.45 mm

(10^-3​​​​​​​ m = 1 mm)

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