metal rod of length 10 cm and cross sectional area of 0.0002 m2 is under tensile

Question

metal rod of length 10 cm and cross sectional area of 0.0002 m2 is under tensile test. There are two loading conditions. In 1st loading condition the metal is pulled by a force of 10 N and as a result the length of the rod changes to 11cm. In the 2nd condition the metal is pulled by a force of 17.5 N but due to some problems the change in length cannot be measured. Assume the stresses due to the loads are in elastic region.

What are the stresses developed due to these two loading conditions? And what is the change in length for 2nd loading condition?

Explanation / Answer

Stress in 1st case = F/A = 10/0.0002 = 50000 Pa = 50 kPa

Stress in 2nd Case = F/A = 17.5/0.0002 = 87.5 kPa

50 k = Y * (11-10)/10 ( Y denotes young modulus)

87.5 k = Y*(l/10)

Dividing these two

  l/1 = 87.5/50

=> l = 1.75 cm

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.