1 Answer

Pdksock · Answer 1 · 2021-08-12T04:59:32+0000

Complete Question:

An AISI 1018 steel has a yield strength,
S_(y) = 295 MPa . Using the distortion-energy theory for the given state of plane stress,

$\sigma_(x) = 31 MPa$ ,
$\sigma_(y) = 41 MPa$ ,
T_(xy) = 45 MPa

(a) determine the factor of safety,

(b) plot the failure locus, the load line, and estimate the factor of safety by graphical measurement.

Answer:

a) Factor of safety = 3.42

b) See the plot failure locus in the file attached

Step-by-step explanation:

According to the maximum distortion energy theorem:

$\sigma_(1) ^(2) - \sigma_(1) \sigma_(2) + \sigma_(2) ^(2) \leq ((S_(y) )/(n)) ^(2)$ ..............(1)

n = Factor of safety

We have to calculate
$\sigma_(1) and \sigma_(2)$

$\sigma_(1) = (\sigma_(x)+ \sigma_(y) )/(2) + \sqrt{((\sigma_(x)-\sigma_(y) )/(2)) ^(2) + T_(xy) ^(2) }$

$\sigma_(1) = (\sigma_(x)+ \sigma_(y) )/(2) - \sqrt{((\sigma_(x)-\sigma_(y) )/(2)) ^(2) + T_(xy) ^(2) }$

$\sigma_(1) = (31+ 41 )/(2) + \sqrt{((31-41 )/(2)) ^(2) +45 ^(2) }$

$\sigma_(1) = 81.28 MPa$

$\sigma_(2) = (31+ 41 )/(2) - \sqrt{((31-41 )/(2)) ^(2) +45 ^(2) }$

$\sigma_(2) = -9.28 MPa$

Substituting these values into equation (1)

$81.28 ^(2) +(81.28*9.28) + (-9.28) ^(2) \leq ((295 )/(n)) ^(2)$

$n^(2) \geq 11.7$

$n \geq √(11.7) \\n \geq 3.42$

Please log in or register to add a comment.