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6.1 Define the following terms: 6.1.1 Stress 6.1.2 Strain 6.2 A steel bar experiences a stress of 250 MPa - NSC Technical Sciences - Question 6 - 2020 - Paper 1

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6.1-Define-the-following-terms:--6.1.1-Stress--6.1.2-Strain--6.2-A-steel-bar-experiences-a-stress-of-250-MPa-NSC Technical Sciences-Question 6-2020-Paper 1.png

6.1 Define the following terms: 6.1.1 Stress 6.1.2 Strain 6.2 A steel bar experiences a stress of 250 MPa. The modulus of elasticity is 190 GPa. The bar has a dia... show full transcript

Worked Solution & Example Answer:6.1 Define the following terms: 6.1.1 Stress 6.1.2 Strain 6.2 A steel bar experiences a stress of 250 MPa - NSC Technical Sciences - Question 6 - 2020 - Paper 1

Step 1

Define the following terms: Stress

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Answer

Stress is defined as the internal resistance offered by a material to deformation when subjected to an external load. It is mathematically expressed as:

Stress=ForceAreaStress = \frac{Force}{Area}

where Force is the applied load and Area is the cross-sectional area over which the load is distributed.

Step 2

Define the following terms: Strain

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Answer

Strain is the measure of deformation representing the displacement between particles in a material body. It is defined as the change in length divided by the original length and is expressed as:

Strain=ΔLL0Strain = \frac{\Delta L}{L_0}

where (\Delta L) is the change in length and (L_0) is the original length.

Step 3

Calculate the: Strain on the bar.

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Answer

To calculate the strain on the bar, we can use the formula relating stress and strain:

Strain=StressModulus of ElasticityStrain = \frac{Stress}{Modulus\ of\ Elasticity}

In this case:

  • Stress = 250 MPa = 250 \times 10^6 Pa
  • Modulus of Elasticity = 190 GPa = 190 \times 10^9 Pa

Substituting:

Strain=250×106190×109=0.001315790.00132Strain = \frac{250 \times 10^6}{190 \times 10^9} = 0.00131579 \approx 0.00132

Step 4

Calculate the: Force exerted on the bar.

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Answer

To calculate the force exerted on the bar, we can use:

Force=Stress×AreaForce = Stress \times Area

First, we need to find the cross-sectional area (A) of the bar:

A=π(d2)2=π(0.062)22.82743×103 m2A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.06}{2}\right)^2 \approx 2.82743 \times 10^{-3} \ m^2

Then we apply:

Force=250×106×2.82743×10370700 NForce = 250 \times 10^6 \times 2.82743 \times 10^{-3} \approx 70700 \ N

Step 5

What is the effect of an increase in temperature on the viscosity of a fluid?

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Answer

An increase in temperature typically results in a decrease in the viscosity of a fluid. This is because higher temperatures provide more energy to the molecules, allowing them to move more freely and overcome the intermolecular forces that contribute to viscosity.

Step 6

Define a perfectly plastic body.

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Answer

A perfectly plastic body is a type of material that undergoes permanent deformation without any increase in stress beyond a certain yield point. Once this yield point is reached, the material will deform indefinitely under constant stress without returning to its original shape.

Step 7

Give TWO examples of perfectly plastic bodies.

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Answer

  1. Clay
  2. Mud

Step 8

State Pascal's law in words.

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Answer

Pascal's law states that in a confined fluid at rest, any change in pressure applied at any point in the fluid is transmitted undiminished throughout the fluid in all directions.

Step 9

What is the minimum force that must be applied to lift the vehicle?

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Answer

Using Pascal's law, we can find the minimum force required:

F1A1=F2A2\frac{F_1}{A_1} = \frac{F_2}{A_2}

Given:

  • (F_2 = 20000 \ N, A_1 = 0.8 \ m^2, A_2 = 0.05 \ m^2)

Substituting in:

F1=F2×A1A2=20000×0.80.05=320000 NF_1 = F_2 \times \frac{A_1}{A_2} = 20000 \times \frac{0.8}{0.05} = 320000 \ N

Step 10

Define the thrust of a liquid.

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Answer

The thrust of a liquid refers to the axial force exerted by the liquid when it flows or is displaced. This force is a result of the pressure exerted by the liquid on the walls of the container or on an object within the liquid.

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