Problem 2 Fatigue strength of machine members 14



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Homework 2


Problem 2 Fatigue strength of machine members
14. Using conditions of the Problem No 1, it is necessary to calculate beam on fatigue strength and to determine dynamic safety factor kd for one operation regime of the system (e.g., for operation regime determined in point 1 or for regime determined in point 2B). Strength influence coefficients may be taken as follows:
=1,15 – notch-sensitivity index of static stress;
=1,23 – notch-sensitivity index of dynamic stress (for symmetric cycle);
= 0,84 – scale factor
=0,85 – coefficient of surface quality

Initial data for calculations
Material of the beam - steel St. 3: E = 2·1011 MPa; σt = 250 MPa; σ-1 = 200 MPa; σb = 420 MPa;
Main parameters of the double-T shape No 24: h = 24 cm, b = 11.5 cm, d = 0.56 cm, F(1) = 34.8 cm2, Ix(1) = 3460 cm4, Wx(1) = 289 cm3, μ = 27.3 kg/m.

Obtained results from Problem 1

maximal static stress- σst max = 35.9 MPa (N/mm2); maximal stress in dynamic regime σd max = 52.7 MPa.

Safety factor k can be found using the above given results:
k= = = 9.72 MPa

1) Parameters of the stress cycle:

Mean value of stress:


σm = σSt max = 35.9 MPa.

Amplitude of stress cycle:


σa = σd max − σSt max = 52.7 − 35.9 = 16.8MPa.

Graph of stress σ versus time t





16.8 MPa



2) Diagram of ultimate stress

Given coefficients:


* =1,15 – notch-sensitivity index of static stress;
* =1,23 – notch-sensitivity index of dynamic stress (for symmetric cycle);
* = 0,84 – scale factor
* =0,85 – coefficient of surface quality

Effective values of mean of stress cycle:


Effective values of amplitude of stress cycle:


Using mechanical characteristics of material (σt = 250 MPa; σ-1 = 200 MPa; σb = 420 MPa), approximate diagram of ultimate stresses is drawn on co-ordinate plane σa – σm (border lines on the diagram are red).

Point K with stresses and is marked on the diagram:

A straight line through points O and K is drawn (on the picture this line is blue). This line crosses the stress limit line in pont L, which corresponds to fatigue limit stress σR. Dynamic safety factor kd is determined graphically with the aid of diagram. Straight line segments OK and OL are measured:

OK=29.4 OL=106

, MPa


t
-1








m, MPa

Dynamic safety factor:

kd =



Conclusion Therefore, if real cyclicity of stresses is taken into account, value of safety factor is sufficiently reduced:
kd = 3.6 < k = 9.72
But even in the case of cyclic changing of stresses safety factor is bigger than 1 ( kd = 3.6 >1). This ensures for the beam a necessary reserve of strength to dynamic load.
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