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Question

6.2. the ratio between the outside diameter and the inside diameter of a copper pipe is 3,2 and the length is 872mm, when the pipe is not subjected to any load. When a tensile load of 937kN is applied to the copper pipe. The length increases by 0,941mm. youngs modulus for elasticity of the copper pipe is 315 Gpa. Calculate the inside diameter and the outside diameter of the copper pipe. 6.3. the following readings were obtained during a tensile test on a test piece. load in kN : a-0, b-7,c-14,d-21,e-28,f-35 elongated in mm: a-0, b-0.0121, c-0.0243, d-0.0361, e-0.0483, f- 0,0818 original diameter: 24mm gauge length: 85mm 6.3.1. draw a stress-strain graph of the given values 6.3.2. determine Youngs modulus of elasticity with the aid of a graph for a test piece

Answer 1

(6.2)

The ratio of the inner and outer diameter is 3 : 2.

Outer Diameter of  copper pipe is 3x

Inner Diameter of copper pipe is 2x

Length of the copper pipe is 872 mm.

Tensile load = 937 kN

The Length increases by 0.941 mm.

Young's modulus E = 315 Gpa

We know that E = stress/strain

Stress = E * strain.

strain = change in the length / original length.

Strain = 0.941 / 872 = 1.079 * 10^-3

Stress = E * strain.

Stress = 315 * 10^9 8 * 1.079 * 10^-3

stress = 33.99 * 10^7 Pa.

Diameter of outer pipe is 3x then outer radius = 3x/2.

Area of the copper pipe = π (3x/2)²

Stress = Force / area.

Area = Force / stress

Area = (937 * 10 ³) / 33.99 * 10^7

π (3x/2)² = 2.756 * 10^-3

(3x/2)² = 8.77 * 10^-4

3x/2 = 0.029

x = 0.019

x = 19.7mm

The Outer diameter of the pipe is 3x = 59.1 mm.

The Inner diameter of the pipe is 2x = 39.4 mm.

Answer 2

(6.3)

Let us draw the Load-Extension Graph.

Load and Extension table is given as follows.

Load	0	7	14	21	28	35
Extension	0	0.0121	0.0243	0.0361	0.0483	0.0818

Graph

Answer 3

(6.3.1)

Let us draw the Load-Extension Graph.

Load and Extension table is given as follows.

Load	0	7	14	21	28	35
Extension	0	0.0121	0.0243	0.0361	0.0483	0.0818

The Formula for the stress.

Stress = Force/area.

Area of the Copper wire is πr²

Area = π(0.012)²

Area = 4.5 * 10^-4.

Therefore Let us draw the stress table

Load	0	7K	14K	21K	28K	35K
Stress	0	15.56M	31.11M	46.67M	62.22M	77.78M

The Formula for the strain.

Strain = change in length/original length.

Strain = Extension/Original Length

Original Length = 24mm.

Therefore Let us draw the strain table

Extension	0	0.0121	0.0243	0.0361	0.0483	0.0818
Strain	0	0.504	1.0125	1.5041	2.0125	3.408

Therefore let us draw Stress and Strain table.

Stress	0	15.56M	31.11M	46.67M	62.22M	77.78M
Strain	0	0.504	1.0125	1.5041	2.0125	3.408

Graph the stress - Strain table.

.

Therefore the stress and strain graph is drawn.

Answer 4

(6.3.2)

The Stress and Strain table is.

Stress	0	15.56 MPa	31.11 MPa	46.67 MPa	62.22 MPa	77.78 MPa
Strain	0	0.504	1.0125	1.5041	2.0125	3.408

Young's Modulus E = Stress/strain.

Let us consider a point (0.504,15.56M)

Then Young's Modulus = 15.56M / 0.504

E = 30.87 MPa

Therefore Young's Modulus E = 30.87 MPa.