A load of 40 kN causes a compressive stress of 16 MPa in a square brass bar - NSC Technical Sciences - Question 5 - 2022 - Paper 1

To find the strain ($ext{e}$), we use the relationship between stress and strain:

$ext{stress} = E imes ext{strain}$

where E is the Young's modulus. Rearranging for strain gives:

ext{strain} ( ext{e}) = rac{ ext{stress}}{E} = rac{16 imes 10^6}{90 imes 10^9} = 1.78 imes 10^{-4}

The change in length ($ext{ΔL}$) can be calculated using:

$ext{ΔL} = L imes ext{strain}$

where L is the original length (300 mm = 0.3 m):

$ext{ΔL} = 0.3 imes 1.78 imes 10^{-4} = 0.0000534 ext{ m} = 0.0534 ext{ mm}$

To calculate the fluid pressure (P) in the hydraulic system, we use:

P = rac{F}{A}

where F is the load of 20 kN and A is the area of Piston B. First, we calculate the area (A) of Piston B:

For a circular area: A = rac{ ext{π} imes d^2}{4} = rac{ ext{π} imes (0.2)^2}{4} = 3.142 imes 10^{-2} ext{ m}^2

Now using the load (F): P = rac{20 imes 10^3}{3.142 imes 10^{-2}} = 636.54 imes 10^{3} ext{ Pa}

Using the relationship of pressures in a hydraulic system, since the pressure is the same:

P = rac{F}{A}

Using the force on Piston B:

• The area of Piston A is given as 1.96 × 10⁻³ m².

Thus: $F = P imes A = 636.54 imes 10^{3} imes 1.96 imes 10^{-3} = 1247.61 ext{ N}$