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goblinko [34]
3 years ago
5

A tension-force replaces what objects when creating a Free Body Diagram?

Physics
1 answer:
algol133 years ago
3 0

Answer:

Option C D and E

chains cables and ropes.

Explanation:

A free body diagram is a simple sketch showing the forces acting on objects in a physical system.

Usually tensile forces  are those forces that occur on objects that are pulled by a force. To get the correct answer choices we need to identify the objects that are most likely to be pulled by a force in any physical system. Such objects from our options include: chains cables and ropes.

This means that whenever these are acting in a physical system, they are mostly used in tension.

Consequently, whenever a free body diagram of such system is drawn, chains cables and ropes.are usually replaced by tension force arrows

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8-14 A Cu-30% Zn alloy tensile bar has a strain-hardening coefficient of 0.50. The bar, which has an initial diameter of 1 cm an
77julia77 [94]

Answer:

The true stress at true strain 0.05cm/cm is 80MPa

Explanation:

Given that

the strength coefficient is K

true strain is ε

strain hardening exponent is n

initial diameter of bar is d = 1cm, (10mm)

tensile force is F

engineering stress(S) = 120

the engineering stress(S) = \frac{F}{\frac{\pi }{4}(d^2) }

To find force (F) =

                    120 = \frac{F}{\frac{\pi }{4}(100^{2} )}

                     F = 120 * (π/4) * (100)

                     F = 9425N

Calculate the true strain  (ε) = In (l₀ / l₁)

where

l₀ =  initial length of the metallic bar = 3cm

l₁ = final length of metallic bar = 3.5cm

ε = In (3.5 / 3)

  = In 1.1667

  = 0.154cm/cm

Calculate the true stress (σ) at fracture point

          = \frac{F}{\frac{\pi }{4}(d^2) }}

tensile force is F and final diameter of bar is d₁ (d in the eqn)

Substitute 9425 N for F and 0.926 cm (9.26mm) for d₁ (d in the eqn)

σ = \frac{9425}{\frac{\pi }{4}(9.26^2) }}

   = 140MPa

To find the strength coefficient (K) of the material bar

K = \frac{140}{\sqrt{0.154} }

K = \frac{140}{0.3925}

   = 356.75MPa

To calculate the true stress σ true strain of 0.05cm/cm

K  = 356.75MPa

σ = 356.75(0.05)^0^.^5

  = 356.75 ( 0.2236)

  = 80MPa

The true stress at true strain 0.05cm/cm is 80MPa

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