Tie-Rod Combined Loading and Fatigue at a Thread Root
A 25 mm diameter A36 steel tie rod (Sy=250 MPa, Su=400 MPa) is part of a press linkage. It carries an axial tensile load that varies from 20 kN to 50 kN and a steady bending moment of 200 N·m at a threaded section. The thread root has a fatigue stress-concentration factor Kf=2.4. The endurance limit of the polished laboratory specimen is estimated as 0.5Su; Marin factors give Se=140 MPa. The designer's static check used only the maximum axial load and reported a factor of safety of 2.45 against yield. The maintenance team wants to know whether the rod can run 10^6 cycles or should be replaced with a larger diameter. A strain gauge at the plain shank reads 102 MPa at peak load, which the designer says 'proves the rod is fine.' The bending moment comes from a misaligned linkage and is not measured by the gauge.
1. Explain why the strain gauge reading of 102 MPa does not prove the rod is acceptable. Compute the bending stress at the plain shank and the combined peak stress.
Read the worked answer
The strain gauge at the plain shank measures only the axial component if it is aligned with the rod axis. Area A=π(25²)/4=490.9 mm². Axial peak σa=50,000/490.9=101.9 MPa, matching the gauge. But the rod also carries a steady 200 N·m bending moment. I=π(25⁴)/64=19,175 mm⁴, c=12.5 mm, so σb=Mc/I=200,000(12.5)/19,175=130.4 MPa. The combined peak on the tensile side is 101.9+130.4=232.3 MPa. The gauge reading therefore misses a large bending stress and cannot prove the rod is fine.
2. Recompute the static factor of safety against yield using the combined stress and comment on the designer's reported 2.45.
Read the worked answer
The designer's 2.45 is 250/101.9, an axial-only static check. Using the combined nominal peak, FoS=250/232.3=1.08 against yield. That is much lower and leaves little margin for tolerances, eccentricity, and material variability. At the thread root the local elastic stress would be higher; local yielding may redistribute some static load, but the low nominal factor means the design should not be accepted on the strength of the axial-only number. Recompute with the actual thread-root section and the appropriate static stress-concentration factor.
3. Using Kf=2.4, Se=140 MPa, Su=400 MPa, and the load range 20–50 kN, estimate the modified Goodman damage ratio at the thread root. Recommend whether to run, inspect, or replace/redesign.
Read the worked answer
At the thread root, the axial load varies 20–50 kN, so nominal axial mean=35,000/490.9=71.3 MPa and nominal axial alternating=15,000/490.9=30.6 MPa. Bending is steady at 130.4 MPa. Total mean≈201.7 MPa; alternating≈30.6 MPa. Apply Kf to the alternating part: σa=2.4(30.6)=73.4 MPa. Modified Goodman: σa/Se+σm/Su=73.4/140+201.7/400=0.524+0.504=1.03. A ratio above 1 predicts finite life before 10^6 cycles. Recommend not simply running to 10^6 cycles; inspect for cracks, reduce bending by aligning the linkage, increase diameter, or improve the thread root. Recheck with local stress/strain and a mean-stress correction if yielding occurs.