A curved beam is being forced open by a force in the center. Determine the fatig

Question

A curved beam is being forced open by a force in the center. Determine the fatigue factor of safety using the data below.

A curved beam that is made of 1018 steel (Sy 220 MPa and Sut 341 MPa) is being pried open with a force that fluctuates from -100 N to 1000 N. The main focus is to determine the factor of safety using the modified-goodman criteria on the inner wall. The outer diameter (OD) is approximately 120 mm while the inner diameter (ID) is approximately 100 mm; while the cross-sectional area is a circle. The curved beam will be cold-drawn and requires a reliability of 80%, while operating in a room at about 90 degrees C.

Explanation / Answer

given

Su=220MPA

Sy=341MPA

LMIN= 100N

LMAX=1000N

OD=120mm

ID=100mm

Operating temperatures =90 and 80 degree celsius

Modified Goodman equation

FOS = SU /(Savg + Sr* Kf)(Su/Se)

Smax= Lmax/A = 0.28MPA

Smin= Lmin/A = 0.028MPA

STEADY STRESS

Savg = (Smax +Smin)/2

Savg = (0.28+0.028)/2 = 0.38MPA

EVERSING STRESS

Srev = (Smax -Smin)/2

Srev = (0.28-0.028)/2 = 0.25MPA

Endurence stress = Se = KLOAD*KSIZE*KTEMP*KRELIABILITY*KD*S'e

                                                KTEMP= 1 for T<=450 degree celsius

                                                KRELIABILITY = A* Su0.265

                                                KD= 1/2.5 = 0.35

                                                   S'e= 0.5*Su = 0.5*341

                                                 1*0.75*1*0.016*0.35*0.5*341

                                           Se = 0.64MPA

FOS = 341/(0.38+(0.25*1))*(341/0.64)

FOS = 2.55

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