Page 315 - Mechanical Behavior of Materials

316 Chapter 7 Yielding and Fracture under Combined Stresses

From Eqs. 7.59 and 7.65, we obtain

σ
770 MPa
uc
¯ σ CM = MAX(C 12 , C 23 , C 31 ) = 260.4MPa, X CM = = = 2.96
¯ σ CM 260.4MPa
σ ut 214 MPa
¯ σ NP = MAX(σ 1 ,σ 2 ,σ 3 ) = 94.31 MPa, X NP = = = 2.27
¯ σ NP 94.31 MPa
Finally, from Eq. 7.66, the controlling safety factor is the smaller of the two:

X MM = MIN(X CM , X NP ) = 2.27 Ans.

(c) The additional compressive force causes a stress of
P −100,000 N
σ x = = =−141.47 MPa
A π(15 mm) 2
so the overall state of plane stress and the resulting principal normal stresses are now

σ x =−141.47, σ y = 0, τ xy = 94.31 MPa
σ 1 = 47.16, σ 2 =−188.63, σ 3 = 0MPa

The latter, with the same m, σ , and σ ut values as before, give the following from Eqs. 7.58,
uc
7.59, 7.65, and 7.66:
C 12 = 271.7, C 23 = 188.63, C 31 = 83.05 MPa

σ
uc
770 MPa
¯ σ CM = MAX(C 12 , C 23 , C 31 ) = 271.7MPa, X CM = = = 2.83
¯ σ CM 271.7MPa
σ ut 214 MPa
¯ σ NP = MAX(σ 1 ,σ 2 ,σ 3 ) = 47.16 MPa, X NP = = = 4.54
¯ σ NP 47.16 MPa

X MM = MIN(X CM , X NP ) = 2.83 Ans.

Discussion In (b), X NP is the smaller of the two safety factors, so that the maximum normal
stress component of the M-M failure criterion is controlling. But in (c), X CM is smaller, so the
C–M component is controlling.