Question

A tensile test is carried out on a bar of mild steel of diameter 20 mm. The bar yields under a load of 80 kN. It reaches a maximum load of 150 kN, and breaks finally at a load of 70 kN. Find (i) the tensile stress at the yield point (1i) the ultimate tensile stress; (iii) the average stress at the breaking point, if the diameter of the fractured neck is 10mm

Answer 1

tensile stress at yield = 254 MPa

ultimate stress = 477 MPa

average stress = 892 MPa

Step-by-step explanation:

Given data in question

bar yields load = 80 kN

load maximum = 150 kN

load fail = 70 kN

diameter of steel (D) = 20 mm i.e. = 0.020 m

diameter of breaking point (d) = 10 mm i.e. 0.010 m

to find out

tensile stress at the yield point , ultimate tensile stress and average stress at the breaking point

solution

in 1st part we calculate tensile stress at the yield point by this formula

tensile stress at yield = yield load / area

tensile stress at yield = 80 ×10³ /
`\pi` /4 × D²

tensile stress at yield = 80 ×10³ /
`\pi` /4 × 0.020²

tensile stress at yield = 254 MPa

in 2nd part we calculate ultimate stress by given formula

ultimate stress = maximum load / area

ultimate stress = 150 ×10³ /
`\pi` /4 × 0.020²

ultimate stress = 477 MPa

In 3rd part we calculate average stress at breaking point by given formula

average stress = load fail / area

average stress = 70 ×10³ /
`\pi` /4 × d²

average stress = 70 ×10³ /
`\pi` /4 × 0.010²

average stress = 892 MPa