5.1 A load causes a stress of 5,5 x 10^6 Pa in a round concrete bar, which has a diameter of 50 cm - NSC Technical Sciences - Question 5 - 2022 - Paper 1

To calculate the force on the bar, we use the formula for stress:

ext{Stress} = rac{F}{A}

Where:

• Stress (\sigma = 5,5 \times 10^6 , \text{Pa})
• Area (A = \frac{\pi d^2}{4} = \frac{\pi (0,5)^2}{4} = 1,96 \times 10^{-1} , m^2)

Rearranging the equation to find force:

$F = \sigma A = 5,5 \times 10^6 imes 1,96 \times 10^{-1} = 1,08 \times 10^6 \, N$

Strain is defined as the ratio of the change in length to the original length:

$\text{Strain} = \frac{\Delta L}{L}$

We can find strain using the relation for stress and Young's modulus:

$\text{Stress} = E \times \text{Strain}$

Thus:

$\text{Strain} = \frac{\text{Stress}}{E} = \frac{5,5 \times 10^6}{85 \times 10^9} = 6,47 \times 10^{-5}$

The change in length (\Delta L) can be calculated using the formula:

$\Delta L = \text{Strain} \times L$

Substituting the values:

$\Delta L = 6,47 \times 10^{-5} \times 3,5 = 2,26 \times 10^{-4} \, m$

Pascal's law states that in a continuous liquid in equilibrium, any change in pressure applied at any point in the liquid is transmitted undiminished to all other parts of the liquid.

Using Pascal's law, we know that:

$\frac{F_1}{A_1} = \frac{F_2}{A_2}$

Let:

• (F_1 = 40 , N)
• (A_1 = 4,8 \times 10^{-4} , m^2)
• (A_2 = 6,2 \times 10^{-2} , m^2)

Rearranging the equation:

$F_2 = F_1 \times \frac{A_2}{A_1} = 40 \times \frac{6,2 \times 10^{-2}}{4,8 \times 10^{-4}} = 5167 \times N = 5167 \, N$