The stress in section AB is __ Mpa. A) 22.5 B) 26.3 C) 28.8 D) 31.8 E) 44.2 F) 56.6 4. The stress in section BC is Mpa.

A) 22.5

B) 26.3

C) 28.8

D) 31.8

E) 44.2

F) 56.6 5.

While under the load, the strain in section AB is x 10 -6 mm/mm.

A) 455

B) 808

C) 64,320

D) 125,300

E) 450,000

F) 722,300

6. While under the load, the strain in section BC is x 10 -6 mm/mm.

A) 455

B) 808

C) 64,320

D) 125,300

E) 45,000

F) 722,300

7. The total elongation of the bar while under the applied load is _ mm.

A) 0.110

B) 0.322

C) 0.596

D) 12.5

E) 17.7

F) 18.3

8. Upon removal of the load, the permanent deformation of the bar is ____ mm.

A) 0.110

B) 0.322

C) 0.596

D) 12.5

E) 17.7

F) 18.3

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Omm2 Omm2 · Answer 1 · 2021-02-17T14:54:40+01:00

Answer:

The stress in section AB is 31.8 Mpa

The stress in section BC is 56.6 Mpa

Explanation:

P.S - The exact question is -

As given,

[tex]P_{A}[/tex] = 10 KN , [tex]P_{B}[/tex] = 10 KN

d = 20 m for AB

d = 15 m for BC

Area of AB = [tex]\frac{\pi }{4} (d)^{2} = \frac{\pi }{4} (20)^{2} = \frac{\pi }{4} (400) = 100\pi[/tex]

Now,

Stress in AB = [tex]\frac{P_{A} }{Area} = \frac{10 KN}{100\pi } = \frac{10. 1000}{100\pi } = \frac{100}{\pi }[/tex] = 31.8 Mpa

Now,

Area in BC = = [tex]\frac{\pi }{4} (d)^{2} = \frac{\pi }{4} (15)^{2} = \frac{\pi }{4} (225) = 56.25\pi[/tex]

Now,

Stress in BC = [tex]\frac{P_{B} }{Area} = \frac{10 KN}{56.25\pi } = \frac{10. 1000}{56.25\pi } = \frac{178.4}{\pi }[/tex] = 56.6 Mpa