PSI - Issue 28
Christos F. Markides et al. / Procedia Structural Integrity 28 (2020) 710–719 Christos F. Markides and Stavros K. Kourkoulis / Structural Integrity Procedia 00 (2019) 000–000
715
6
Then substituting in Eqs.(3) and (4) from Eqs.(5) and (6) for φ ο ( ζ ) and ψ ο ( ζ ), and reverting to the variable z through Eq.(2), one obtains the solution of the problem as:
z z z z z z z z 1 2
iP
1 2 3 z z z z
z
log
z
4
2
2
2
R
3
4
(7)
z z z
2 R z z
1
1
z z z z
z z
n
log
log
z z z
k
k
k
k
k
k
2
2
2
2
1
2
R z z R z z R z z
R
k
1
k
k
k
k
z z z z
iP
z
z
z
z
z
1
2
log
1
2
3
4
2
3 z z z z z z z z z z z z z z z z R z z R z 4 1 2 3 4 2 2
1
z
n
(8)
log
log
k
k
k
k
k
k
2
2
2
1
z z z z z z
R z z z R z z R z z
k
1
k
k
k
k
k
k
2 R z z z z z
2 R z z z z z
2 R z z
1
z z
2
2
k
k
k
k
k
k
R
R
z z
k
k
k
k
2
2
2 z R z z R z z 2
z
2 R z z
2 R z z
k
k
k
k
3. The stress field in the flattened disc Stresses corresponding to the above complex potentials are given by the well-known Muskhelishvili’s (1963) formulae (single and double primes denote the first and second derivatives with respect to z , and denotes the real part): 2 4 ( ), ( ) ( ) ( ) ( ) i r r r z i z z e z z z (9) Combining Eqs.(7), (8) and (9), stresses are obtained at any point or along any locus of the truncated disc (apart the singular points z 1 , z 2 , z 3 , z 4 , , k k z z ), which, according to the assumptions adopted in the present study, approximate the respective ones developed in the flattened disc. For the sake of brevity only two loci, of strategic importance from the engineering point of view, are considered in this study, just to underline the potentialities of the solution introduced, and, also, to highlight some interesting aspects of the stress field developed in a FBD under uniform load along its straight edges. The first locus is the “loaded diameter”, i.e., the diameter along which the resultant P frame of the external load acts (coinciding with the direction of y -axis). The specific locus is, perhaps, the most interesting one, taking into account that fracture is expected to start at its central point (which is, in fact the center of the disc). Since the specific locus is an axis of symmetry, no shear stresses appear all along its length and therefore only normal stresses (transverse, σ θ , and radial, σ r ) are developed. In order to obtain a quantitative overview of the stress distribution along this locus, a FBD of radius R =50 mm and thickness t =10 mm, made of a material with Young’s modulus E =3.2 GPa and Poisson’s ratio ν =0.36, is compressed by equal point forces P resulting to an overall load P frame =20 kN. The central (loaded) angle was set equal to 30 o (cor responding to Θ =75 ο ). The stresses developed on the FBD are plotted in Fig.5a (employing Eqs.(9) for 0< r ≤0.9 R sin Θ , i.e., far enough from the singular point z k , k =( n +1)/2), in juxtaposition to the stresses, developed along the respective locus 0< r ≤ R in a standard (circular) BD of the same dimensions and under the same overall external load, distributed according to a parabolic manner (Markides and Kourkoulis 2012, Kourkoulis and Markides 2014). It is evident from Fig.5a that the magnitude of both σ θ and σ r developed in the FBD are lower from the respective ones in the standard BD. The differences are more or less imperceptible until half of the loaded radius (0< y <0.5 R ). From this point on, the differences start increasing abruptly. What is more, for y tending to its ultimate value, i.e., while the loaded edges are approached, the increasing tendency of the magnitude of both σ θ and σ r is reversed to a very low rate. Thus, while the stress field at the center of the FBD is almost identical to that of the standard BD (concerning both the magnitude of the stresses and, also, their σ θ / σ r ratio, which is again close to 3) things become dramatically different while one approaches the loaded boundaries. Indeed, while for the standard BD normal stresses with maximum value equal to about 115 MPa are developed for y ≡ r = R , the respective maximum values for the FBD for y ≡ r =0.9 R sin Θ are equal to only 70 MPa for σ r and 25 MPa for σ θ .
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