Question

The solid cylindrical bar shown is subjected to a bending force F = 55kN, an axial load P = 8 kN, and a torsional load T = 30 kN-mm. The bar is made of AISI 1060 CD steel (Sy = 280 MPa, Sut = 330 MPa). Find the factor of safety based on Maximum Distortion Energy Theory.

Answer 1

The factor of Safety by Maximum Distortion Energy Theory is given as: N = 6.04

What is Maximum Distortion Energy Theory?

According to this idea, failure by yielding happens when the distortion energy per unit volume in a condition of combined stress equals that associated with yielding in a basic tension test at any location in the body.

What is the calculation for the above value?

Note that we are given the following:

A solid Steel bar subjected to a bending force of F = 55kN

Torsional load T = 30 kN-mm; and

Yield Strength (sy) = 280Mpa

Sut = 330 Mpa

The formula for Normal Stress is given as:

$\alpha _{x}$ = [UP/πd²] + [ (32* f * L)/πd³]

= [(4 * 8 * 10³)/π * (0.02)²] + [(32 * 55 *10³ * 0.1)/ π * (0.02)³]

= 25.46 + 7.003

$\alpha _{x}$= 32.5 Mpa

Tyz = (16T)/πd³

= (16 * 30)/ π * (0.02)³

= 19.098

Tyz = 19.098

Following from the above, sate of stress, we have:

α₁ = αₓ = 32.46 Mpa; and

α₂ = Tyz = 19.098 Mpa;

α₃ = -Tyz = -p19.098 Mpa;

Utilizing the formula given by the Maximum Distortion Energy Theory:

α₁² + α₂² + α₃² - (α₁α₂ + α₂α₃ + α₃α₁)

= Sy²/N²

→ 32.46² + 19.098² + 19.098² - ( 32.46* 19.098 - 19.098² -32.46 *  19.098)

= 280/N²

N = 6.04

Learn more about Maximum Distortion Energy Theory at:
brainly.com/question/22435706
#SPJ1