the loads for initial plastic deformation of a 0039 in diame

the loads for initial plastic deformation of a 0.039 in. diameter copper wire and a 0.027 in. diameter steel wire are 40 lb and 31 lb respectively. the maximum possible loads carried by the two wires ( copper and iron) are 66 lb and 55 lb respectively. The diameter at the fracture point of the copper wire is 0.026 in. The load at fracture was 49 lb.

A) which wire has the lowest yield strength?

B) will the ultimate strength of the copper be greater or less than the yield strength of the steel?

C) what is the breaking strength for copper?

D) what is the true stress at the point of fracture for the copper?

E) What is the reduction in area of the copper?

Solution

Given

Dia of cu wire= 0.039in= 0.039×2.54= 0.09906cm

Load of cu wire= 40lb= 40×9.81÷2.21=177.55N

Dia of stee wire= 0.027in= .027×2.54= 0.068

Area of steal wire = (÷4)× 0.068^2 = 3.63×10^-3 cm square

Load of steel wire = 31lb= 31×9.81÷2.21=137.6N

Area of cu wire =( ÷4)× .09906^2= 7.7 × 10^-3

Now yeild strength of cu wire= 177.55÷ (7.7×10^-3)=23.03KN/cm square

And yield strength ofsteal wire= 137.6÷ (3.63×10^-3)= 37.46KN/cm square

AnsA -from above discussion it is clear that cu wire has lowest strength.

AnsB - Max load carried by copper wire= 66= 66×9.81÷2.21=292.96N

Ultimate strength of cu wire= 38.07KN/cm square

Hence ultimate strength of cu wire is little more than that of yield strength of steal wire

And c- dia of fracture point wire of cu is = 0.026in= 0.026×2.54= 0.066cm

Than area = 3.42×10^-3

Hence braking strength= 292.96÷ (3.42×10^-3)=85.56KN/cm square

And d- true stores;

Here fracture strength = 217.2N

Hence true stress= 217.2÷{7.7×10^-3}=28.8KN/cm square