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Chapter 1: Q26E (page 359)

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 :1.1脳10-4

Copper :1.1脳10-4

(b) Steel :8.3脳10-5m

Copper:1.6脳10-4m

Step by step solution

01

Given information

Tensile force: F = 4000 N,

l0=0.75m

Area=(d24)=((1.50脳10-2m)24)=1.77脳10-4m2

02

Concept/Formula used

Y=l0F鈭�l

Where, Y is Young鈥檚 modulus, l0 is length of muscle, F is muscle force, A is cross-sectional area and 鈭�l is elongation.

03

(a) Strain Calculation

The strain is :鈭�ll0=FYA

For steel: Y=2.0脳1011Pa

鈭�ll0=4000N(2.0脳1011Pa)(1.77脳10-4m2)=1.1脳10-4

For copper :Y=1.1脳1011Pa

鈭�ll0=4000N(1.1脳1011Pa)(1.77脳10-4m2)=2.1脳10-4

04

Elongation Calculation

For steel:

(1.1脳10-4)(0.75)=8.3脳10-5m

For Copper:

(2.1脳10-4)(0.75)=1.6脳10-4m

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