7.1 Define the following terms: 7.1.1 A force 7.1.2 Forces in equilibrium 7.1.3 Resultant of a system of forces 7.2 A load of 40 kN causes a tensile stress of 20 MPa in a round brass bar - NSC Mechanical Technology Fitting and Machining - Question 7 - 2017 - Paper 1

Forces in equilibrium refer to a situation where the sum of all forces acting on a body is zero, resulting in a state of rest or uniform motion.

The resultant of a system of forces is the single force that represents the combined effect of all the individual forces acting on an object.

To calculate the diameter of the bar:

Given:

• Load (F) = 40 kN = 40,000 N
• Tensile Stress (σ) = 20 MPa = 20 × 10⁶ N/m²

Using the formula for stress:

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

Where:

• Area (A) = rac{ ext{π}D^2}{4}

Setting the equations equal:

20 imes 10^6 = rac{40,000}{rac{ ext{π}D^2}{4}}

Rearranging to solve for D:

D^2 = rac{40,000 imes 4}{20 imes 10^6 imes ext{π}}

To calculate the strain:

Using the formula:

ext{Strain} = rac{ ext{Change in Length}}{ ext{Original Length}}

We know:

• Tensile Stress = $20 imes 10^6$ N/m²
• Young's Modulus (E) = $90 imes 10^9$ N/m²

From the relation:

ext{Strain} = rac{ ext{Stress}}{E}

Calculating:

ext{Strain} = rac{20 imes 10^6}{90 imes 10^9} = 2.222 imes 10^{-4}

For the change in length using:

$ext{Change in Length} = ext{Strain} imes ext{Original Length}$

Substituting the values:

$ext{Change in Length} = (2.222 imes 10^{-4}) imes 800 ext{ mm} = 0.1776 ext{ mm}$

To find the resultant forces in Figure 7.3:

First, calculate the x and y components of the forces:

• X-component: $R_x = 280 + 300 imes ext{Cos}(30°) - 400 imes ext{Cos}(120°)$
• Y-component: $R_y = 300 imes ext{Sin}(30°) + 150 - 170$

Calculating:

• For $R_x$, substituting: $R_x = 280 + 300 imes 0.866 - 400 imes -0.5 = 133.9 ext{ N}$
• For $R_y$: $R_y = 150 + 150 - 170 = 133.9 ext{ N}$

Now, calculate the magnitude of the resultant:

$R = ext{sqrt}(R_x^2 + R_y^2)$

$R = ext{sqrt}(133.9^2 + 133.9^2) = 189.73 ext{ N}$

The direction is given by: an^{-1}rac{R_y}{R_x} provide direction in degrees.

To determine reactions at A and B in the given beam:

Assuming downward forces:

1. Sum of vertical forces must equal zero: $800 + 350 + (80 imes (6.2)) = R_A + R_B$
2. Moments about A to find R_B: $0 = (350 imes 6.2) + (80 imes 6.2 imes 0.85) - R_B imes 6.2$
3. Solve these simultaneous equations to find: $R_A = 693.96 ext{ N}$ and $R_B = 952.03 ext{ N}$