graphs

Question

6.3 The following readings were obtained in a tensile test on a mild steel bar: Load in kN 2,3 9,2 18,4 27,6 36,8 Extension in mm 0,0056 0,0246 0,0456 0,066 0,0896 Gauge length — 61 mm Original diameter of the bar . 13,3 mm 6.3.1 Draw the load -extension graph for the given values. 6.3.2 Determine Young's modulus of elasticity by means of the graph 6.3.3 Calculate the percentage reduction in area if the diameter of the rod was 7,32 mm at the fracture.

Answer 1

(6.3.1)

Let us draw the Load-Extension Graph.

Load and Extension table is given as follows.

Load	2.3	9.2	18.4	27.6	36.8
Extension	0.0056	0.0246	0.0456	0.066	0.0896

Graph.

Answer 2

(6.3.2)

Calculate Young's Modulus E

Load and Extension table is given as follows.

Load	2.3	9.2	18.4	27.6	36.8
Extension	0.0056	0.0246	0.0456	0.066	0.0896

Graph.

Young's Modulus E = Stress / Strain.

Let us consider two points from the table (0.0056, 2.3) and (0.0246,9.2)

E =(9.2K-2.3K)/(0.0246 - 0.0056)

E = 363,157.89 MPa

Therefore Young's Modulus = 365,157.89 MPa.

Answer 3

(6.3.3)

Find the percentage of reduction in area .

Original Diameter = 13.3

Area of the Tensile rod with diameter as 13.3 is

Then radius = (diameter/2) = 6.65

Area a1= πr²

Area a1 = π(6.65)²

Area a1 = 138.92mm²

Area of the Tensile rod with diameter as 7.32 is

Then radius = (diameter/2) = 3.615

Area a2= πr²

Area a2= π(3.615)²

Area a2= 41.05mm²

Percentage Reduction in area is (a1 - a2) / a1.

(a1 - a2) / a1 =  (138.92 - 41.05) / 138.92.

(a1 - a2) / a1 = 0.7045

Percentage Reduction = 70.45 %