In this paper, the stress intensity factor (SIF) is computed in a circumferentially cracked round bar (CCRB) subjected to applied tensile load or to imposed axial displacement, considering that the resistant ligament is circular and exhibits certain eccentricity in relation to the cylinder axis. The computation was performed by means of the finite element method, using a 3‐dimensional model and the J‐integral, the analyzed variables being the diameter and the eccentricity of the circular ligament. Results show that the SIF value is higher at the deepest point of the crack. For a given ligament diameter, an increase of ligament eccentricity (in relation to the bar axis) raises the difference between the SIF values at those points of the crack associated with the maximum and minimum depth. Furthermore, the maximum SIF value is higher for applied tensile load than for imposed axial displacement, mainly for high eccentricities and small diameters of the ligament, as a consequence of a greater rotation of the bar.