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Two circular rods, one steel and the other copper, are joined end to end. Each rod is 0.750 m long and 1.50 cm in diameter. The combination is subjected to a tensile force with magnitude 4000 N . For each rod, what are (a) the strain and (b) the elongation?

Short Answer

Expert verified

(a) Steel :1.110-4

Copper :2.110-4

(b) Steel :8.310-5m

Copper:1.610-4m

Step by step solution

01

Given information

Tensile force: F = 4000 N,

l0=0.75m

Area=(d24)=((1.5010-2m)24)=1.7710-4m2

02

Concept/Formula used

Y=l0Fl

Where, Y is Young鈥檚 modulus,l0is length of muscle, F is muscle force, A is cross-sectional area and localid="1668002448923" lis elongation.

03

(a) Strain Calculation

The strain is : ll0=FYA

For steel:Y=2.01011Pa

ll0=4000N(2.01011Pa)(1.7710-4m2)=1.110-4

For copper :Y=1.11011Pa

ll0=4000N(1.11011Pa)(1.7710-4m2)=2.110-4

04

Elongation Calculation

For steel:

(1.110-4)(0.75)=8.310-5m

For Copper:

(2.110-4)(0.75)=1.610-4m

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