On the use of thermographic method to measure fatigue limits

Traditional fatigue limit measurements are expensive and time consuming, requiring a large number of specimens and a long time to be completed. An alternative approach is used in this work to obtain the fatigue limit of a cold drawn steel, in a rotating bending machine, by monitoring temperature variations induced by different load range levels in just a few specimens. Comparisons with Dixon’s traditional up-and-down method confirm that this thermographic approach is indeed a much more efficient, faster, and cheaper method to obtain fatigue limits.

Keywords
Fatigue; Fatigue limits; Thermographic method

Authorship

Carlos Filipe Cardoso Bandeira

Departamento de Engenharia Mecânica, Pontifícia Universidade Católica do Rio de Janeiro - PUC-Rio, Rio de Janeiro, RJ, Brasil. E-mail: fillypebandeira@hotmail.com, jtcastro@puc-rj.brPontifícia Universidade Católica do Rio de JaneiroBrazilRio de Janeiro, RJ, BrazilDepartamento de Engenharia Mecânica, Pontifícia Universidade Católica do Rio de Janeiro - PUC-Rio, Rio de Janeiro, RJ, Brasil. E-mail: fillypebandeira@hotmail.com, jtcastro@puc-rj.br

Programa de Pós-Graduação em Engenharia Mecânica e Tecnologia de Materiais, Centro Federal de Educação Tecnológica Celso Suckow da Fonseca - CEFET/RJ, Rio de Janeiro, RJ, Brasil. E-mail: paulo.kenedi@cefet-rj.brCentro Federal de Educação Tecnológica Celso Suckow da FonsecaBrazilRio de Janeiro, RJ, BrazilPrograma de Pós-Graduação em Engenharia Mecânica e Tecnologia de Materiais, Centro Federal de Educação Tecnológica Celso Suckow da Fonseca - CEFET/RJ, Rio de Janeiro, RJ, Brasil. E-mail: paulo.kenedi@cefet-rj.br

Paulo Pedro Kenedi ^**Corresponding author

Jaime Tupiassú Pinho de Castro

*Corresponding author

SCIMAGO INSTITUTIONS RANKINGS

Figures | Tables

Figures (12)
Tables (6)

C	Si	Mn	P	S	Cr	Ni	Mo	Al	Cu	Ti	Nb	V
0.226	0.114	0.510	0.020	0.0028	0.024	0.011	0.003	0.017	0.026	<0.001	<0.003	0.001

Specimen	$N$	Final condition	Specimen	$N$	Final condition
1	$6.0 \times 10^{6}$	Survival	14	$3.1 \times 10^{6}$	Failure
2	$5.5 \times 10^{6}$	Survival	15	$5.4 \times 10^{6}$	Survival
3	$5.3 \times 10^{6}$	Survival	16	$3.7 \times 10^{6}$	Failure
4	$2.0 \times 10^{6}$	Failure	17	$3.2 \times 10^{6}$	Failure
5	$1.0 \times 10^{7}$	Survival	18	$5.0 \times 10^{6}$	Survival
6	$8.1 \times 10^{5}$	Failure	19	$5.2 \times 10^{6}$	Survival
7	$3.8 \times 10^{6}$	Failure	20	$2.9 \times 10^{5}$	Failure
8	$5.0 \times 10^{6}$	Survival	21	$5.4 \times 10^{6}$	Survival
9	$2.7 \times 10^{6}$	Failure	22	$1.2 \times 10^{6}$	Failure
10	$5.7 \times 10^{6}$	Survival	23	$5.3 \times 10^{6}$	Survival
11	$5.9 \times 10^{6}$	Survival	24	$1.5 \times 10^{6}$	Failure
12	$9.7 \times 10^{5}$	Failure	25	$5.1 \times 10^{6}$	Survival
13	$6.1 \times 10^{6}$	Survival	-	-	-

SP	Fitting curve	$σ_{a} / S_{ut} (%) (\partial T / \partial N_{I} = 0)$
1	$\frac{\partial T}{\partial N_{I}} = 3.6 \frac{σ_{a}}{S_{ut}} - 168.2$	46.9
2	$\frac{\partial T}{\partial N_{I}} = 3.6 \frac{σ_{a}}{S_{ut}} - 169.5$	46.6
3	$\frac{\partial T}{\partial N_{I}} = 4.1 \frac{σ_{a}}{S_{ut}} - 186.4$	45.9
4	$\frac{\partial T}{\partial N_{I}} = 3.2 \frac{σ_{a}}{S_{ut}} - 144.5$	45.4

$μ$ (%)	$φ$ (%)	$S_{L_t h}^{'} / S_{ut}) (%)$	$S_{L_t h}^{'}$ (MPa)
46.2	0.7	46.2 ± 0.7	307.2 ± 4.7

$S_{y}$ (MPa)	$S_{ut}$ (MPa)	RA (%)
576.6	666.6	50.1

$μ$ (%)	$φ$ (%)	$S_{L_u d}^{'}$ (MPa)
46.3	1.1	308.9 ± 7.1