Question

Two steel wires are connected together, end to end, and attached to a wall as shown below. The two wires have the same length and elastic modulus, but the ratio of the radius of the first wire to the radius of the second wire is 7 : 3. As the wires are initially the same length, the midpoint of the combination coincides with the connection point. An applied force then stretches the combination by 0.550 mm while the two wires stay connected together. After the wires are stretched, what is δ, the distance from the midpoint to the connection point?

Answer 1

Solution:

The thicker wire stretches less. Let's say it stretches "d" and the thinner wire stretches "D."

d + D = 0.550 mm and D = d*(7/3)² = 5.44d

so

d + 5.44d = 6.444 d = 0.550 mm

d = 0.0853 mm

D = 0.465 mm

Let's say each was originally length L. The total length is 2L + 0.550 mm, and so the midpoint is now L + 0.257 mm from each end.

But the connection point is only L + 0.0853 mm from one of the ends, so

= (0.257 - 0.0853)mm

= 0.1897 mm

= 0.2 mm.

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