stress

Question

6.2. a round bar with a diameter of 43mm and a length of 0.86m is machined to have a square area of 22mm sides over a length of 0.425m. the stress in the square section is not to exceed 421MPa. Calculate the ffg: 6.2.1. the stress in the round bar The total extension of the bar if Young’s modulus of elasticity is 108 GPa. 6.3. the following readings were obtained during a tensile test on a mild steel bar: Original diameter at the gauge length section is 26mm Gauge length is 62mm. 6.3.1. draw a stress-strain graph for the given values 6.3.2. determine Young’s modulus of elasticity with the aid of the graph

Answer 1

(6.2.1)

The diameter of the round bar = 43 mm

Radius of the round bar = 43/2 = 21.5mm

Length of the round bar = 0.86 m.

Area of the square bar = 22mm

Length of the square bar = 0.425m.

Stress of the Square bar = 421MPa.

We know that Stress = Force/Area.

Force = Stress * Area.

Force on the Round Bar = Force on the Square bar.

Where are the stress of the round bar and the Square bar and   are the area of the round bar and the Square bar respectively.

Area of the round bar = πr²

Area= π(21.5)²

Ar = 1452.2 mm².

Stress of the Round bar = 6.2 MPa.

Answer 2

(6.3)

The Young's Modulus E = 108 GPa.

Length of the bar = 0.86 m

Young's Modulus  = Stress / strain.

Force on the Round Bar = Force on the Square bar.

Area of the round bar = πr²

Area= π(21.5)²

Ar = 1452.2 mm².

Stress of the Round bar = 6.2 MPa. 

Strain = Extension / Length of the bar.

Young's Modulus =  Stress / ( Extension / Length)

dl = 4.9 * 10^-5

dl = 0.000049 m

dl = 0.049 mm

Total Extension of the bar = 0.049 mm.