Respuesta :
Answer:
See explanation
Explanation:
a)
MSS: σ1 = 100 MPa ; σ2 = 50 MPa ; σ3 = 0
n =350 /(100 - 0) = 3.5 Ans.
DE: σ' = 100² -100*(50)+50²) ^0.5 = 86.6 MPa
n = 350/86.6 = 4.04 Ans.
b) σA, σB = 100/2 ± √((100/2)² + (-75)²) = 140,
σ1 = 140, σ2= 0, σ3= -40 MPa
MSS: n = 350/(140-(-40)) =1.94 Ans.
DE: σ' = [100² + 3 * (-75)²] ^0.5 = 164 MPa
n= 350/164 = 2.13 Ans.
c) σA,σB = ((-50 -75) / 2) ± √(((-50 +75)/2)² + (-50)²)
σ1 = 0, σ2 = -11.0, σ3 = -114.0 MPa
MSS: n = 350/(0-(-1142.0)) = 3.07 Ans.
DE: σ' [(-50)² - (-50)(-75) + (-75)² + 3(-50)²]^0.5 = 105.0 MPa
n = 350/109.0 = 3.21 Ans.
d) σA,σB = (100 +20)/1 ± √(((100-20)/2)² + (-20)²)
σ1 = 104.7, σ2 = 15.3, σ3 = 0 MPa
MSS: n= 350/(140.7 - 0) = 3.34 Ans.
DE: σ' = [100² - 100(20) + 20² + 3 (-20)²]^0.5 = 98.0 MPa
n= 350/ 98.0 = 3.57 Ans.
It is the ratio of ultimate load and allowable load. the safety factor is always larger than one. The factor of safety values are MSST N= 3.5,MDET N=4.04
What is a factor of safety?
It is the ratio of ultimate load and allowable load. the safety factor is always larger than one.
The permissible stress is always smaller than the eventual failure stress, as the preceding calculation shows. As a result, the safety factor is always larger than one.
Safety Factor = Ultimate Load /Allowable Load
(a) (a) σx = 100 MPa, σy = 50 MPa (
MSST(Maximum shear stress theory)
[tex]\sigma _1 = 100 MPa \\\sigma_2 = 50 MPa \\ \sigma_3 = 0\\\\ \rm N =\frac{350}{100} \\ \rm N= 3.5[/tex]
MDET()Maximum distortion energy theory)
[tex]\sigma' = 100^2 -100\times (50)+50^2) ^0.5 = 86.6 MPa\\\\[/tex]
[tex]\rm N=\frac{350}{86.6} \\\\ \rm N= 4.04[/tex]
(b) σx = 100 MPa, τxy = –75 MPa
[tex]\sigma_A,,\sima_B = \frac{100}{2} \sqrt{ \frac{100}{2} + (-75)^2)} = 140,\\\\ \sigma_1 = 140, \sigma_2= 0, \sigma_3= -40 MPa[/tex]
MSST
[tex]n =\frac{350}{(140-(-40)} ) =1.94[/tex]
MDET
[tex]\sigma' = [100^2 + 3 \times (-75)^2] ^{0.5 } = 164 MPa\\\\n= \frac{350}{164} = 2.13[/tex]
(c) ) σx = –50 MPa, σy = –75 MPa, τxy = –50 MPa
[tex]\sigma_A, \sigma_B = \frac{100}{2} + \sqrt{( \frac{100}{2} ^2 + (-75)^2) = 140[/tex]
[tex]\sigma_1 = 140, \sigma_2= 0, \sigma_3= -40 MPa[/tex]
MSST
[tex]n =\frac{ 350}{-(1142-(-40))} =3.07[/tex]
MDET
[tex]\sigma'= [(-50)^2 - (-50)(-75) + (-75)^2 + 3(-50)^2]^{0.5} = 105.0 MPa[/tex]
[tex]\rm n=\frac{350}{109.0} \\\\ \rm n= 3.21[/tex]
(d) σx = 100 MPa, σy = 20 MPa, τxy = –20 MPa
[tex]\sigma_A,\sigma_B =\frac{100+200}{1} +\sqrt{(100-\frac{20}{2} )^2 + (-20)^2}[/tex]
[tex]\sigma_1 = 104.7, \sigma_2 = 15.3, \sigma_3 = 0 MPa[/tex]
MSST
[tex]\rm n= \frac{350}{140.7} \\ \\\rm n=3.34[/tex]
MDET
[tex]\sigma' = [100^2 - 100(20) + 20^2 + 3 (-20)^2]^0.5 = 98.0 MPa\\\\[/tex]
[tex]\rm n= \frac{350}{98.0} \\ \rm n=3.27[/tex]
Hence the value for the factor of safety for case 1 is MSST N= 3.5,MDET N=4.04
To learn more about factors of safety refer to the link;
https://brainly.com/question/16918369
