Question

2) An aluminum specimen with a defeet as shown having a thickness of 2 mm width 22 mm, the defeet has a hole of 4.S mm radius, the crack length is 12.5 mm from tip to tip the length of the specimen was 65 mm , this specimen is subjected to a tensile fore up to 6200 N, it fractured at s4s0 Find the ultimate stress. 10 marks

Strength of material

Answer 1

formula od ultimate strength

Ultimate Force F resisting cross A

here F = 5450 N

A = resiting area

the maximum stress will be at the least area resisting the force it also depends on the direction of the tensile force

or 10 marks 10 m condition 2 condition 1

for condition 1

resisting area = (22 - 12.5)*2 = 19 mm²

thus

Ultimate Force F resisting cross A

s307.368 MPa 2 5840 = 307368 19

for condition 2

resisting area = (22 - 2*4.8)*2 = 110.8 mm²

thus

Ultimate Force F resisting cross A

110.852.707_V 52.07 MPa

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