7.1 Four pulling forces of 100 N, 200 N, 300 N and 185 N are acting from the same acting point, as shown in FIGURE 7.1. Determine, by means of calculations, the magnitude and direction of the resultant of the system of forces in FIGURE 7.1. 7.2 A load of 40 kN causes a tensile stress of 20 MPa in a round brass bar. The original length of the bar is 300 mm. Young's modulus for brass is 90 GPa. Calculate the: 7.2.1 Diameter of the bar 7.2.2 Strain 7.2.3 Change in length 7.3 FIGURE 7.2 shows a uniform beam that is supported by two vertical supports, A and B. A uniformly distributed force of 80 N/m is exerted over the whole length of the beam. Determine, by means of calculations, the magnitudes of the reactions in supports A and B.

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7.1 Four pulling forces of 100 N, 200 N, 300 N and 185 N are acting from the same acting point, as shown in FIGURE 7.1 - NSC Mechanical Technology Welding and Metalwork - Question 7 - 2016 - Paper 1

Question 7

7.1-Four-pulling-forces-of-100-N,-200-N,-300-N-and-185-N-are-acting-from-the-same-acting-point,-as-shown-in-FIGURE-7.1-NSC Mechanical Technology Welding and Metalwork-Question 7-2016-Paper 1.png

7.1 Four pulling forces of 100 N, 200 N, 300 N and 185 N are acting from the same acting point, as shown in FIGURE 7.1. Determine, by means of calculations, the magn... show full transcript

Worked Solution & Example Answer:7.1 Four pulling forces of 100 N, 200 N, 300 N and 185 N are acting from the same acting point, as shown in FIGURE 7.1 - NSC Mechanical Technology Welding and Metalwork - Question 7 - 2016 - Paper 1

Step 1

Determine the resultant of the system of forces in FIGURE 7.1

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Answer

To find the resultant force, we need to calculate the horizontal and vertical components of each force. Using trigonometric functions:

Horizontal Components:
- For 200 N: $200 \cos(25^\circ)$
- For 300 N: $300 \cos(45^\circ)$
- For 185 N: $185$ (acting horizontally)
- For 100 N: $100$ (acting horizontally)
Thus, the total horizontal component (H) is: $H = -200 \cos(25^\circ) + 300 \cos(45^\circ) + 185 - 100$
Vertical Components:
- For 200 N: $200 \sin(25^\circ)$ (acting upwards)
- For 300 N: $300 \sin(45^\circ)$ (acting upwards)
- The vertical component of 100 N is simply $-100$ .
Hence, the total vertical component (V) is: $V = 200 \sin(25^\circ) + 300 \sin(45^\circ) - 100$
Calculate the magnitudes:
- $H = 215.87 N$ (calculated value)
- $V = 196.65 N$ (calculated value)
Resultant Force (R): $R = \sqrt{H^2 + V^2}$ $R = \sqrt{(215.87)^2 + (196.65)^2} = 292.01 N$
Direction (angle θ): $tan(θ) = \frac{V}{H}$ Thus, $heta = 42.33^\circ$ north of east.

Step 2

7.2.1 Diameter of the bar

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Given:

Load (F) = 40 kN = $40 \times 10^3$ N
Tensile stress ( $\sigma$ ) = 20 MPa = $20 \times 10^6$ Pa

Using the formula for tensile stress: $\sigma = \frac{F}{A}$ Where: $A$ is the cross-sectional area.

Thus, we rearrange the formula: $A = \frac{F}{\sigma} = \frac{40 \times 10^3}{20 \times 10^6} = 2 \times 10^{-3} \text{ m}^2$

The area of a circle is given by: $A = \frac{\pi D^2}{4}$

Setting the two area equations equal to each other: $\frac{\pi D^2}{4} = 2 \times 10^{-3} \Rightarrow D^2 = \frac{4 \times 2 \times 10^{-3}}{\pi}$ $D = \sqrt{\frac{4 \times 2 \times 10^{-3}}{\pi}} = 0.05046 \text{ m} \approx 50.46 \text{ mm}$

Step 3

7.2.2 Strain

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

Using: $\text{Strain} = \frac{\sigma}{E}$ Where:

Young's modulus (E) = 90 GPa = $90 \times 10^9$ Pa
Tensile stress ( $\sigma$ ) = 20 MPa = $20 \times 10^6$ Pa

Thus: $\text{Strain} = \frac{20 \times 10^6}{90 \times 10^9} = 0.22 \times 10^{-3}$

Step 4

7.2.3 Change in length

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The change in length ( $ΔL$ ) can be calculated using: $\text{Strain} = \frac{\Delta L}{L_0}$ Where:

$L_0 = 300 \text{ mm} = 0.3 \text{ m}$

Rearranging gives: $\Delta L = \text{Strain} \times L_0 = (0.22 \times 10^{-3}) \times 0.3 = 0.07 \text{ m}$ So, $ΔL \approx 0.07 ext{ m}$ .

Step 5

Determine the reactions in supports A and B using FIGURE 7.2

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Answer

To find the reactions at supports A and B, we can apply the equilibrium of moments.

Calculate Moments about B:
- Taking moments about B: $\sum RHM = \sum LHM$ $(A \times 4) + (300 \times 6) + (80 \times 3) = (550 \times 2)$
Solving for A: $A \approx 637.5 N$
Calculate Moments about A:
- Taking moments about A: $\sum LHM = \sum RHM$ $(B \times 8) + (300 \times 10) = (550 \times 2) + (80 \times 8)$
Solving for B gives: $B \approx 1012.5 N$

7.1 Four pulling forces of 100 N, 200 N, 300 N and 185 N are acting from the same acting point, as shown in FIGURE 7.1 - NSC Mechanical Technology Welding and Metalwork - Question 7 - 2016 - Paper 1

Worked Solution & Example Answer:7.1 Four pulling forces of 100 N, 200 N, 300 N and 185 N are acting from the same acting point, as shown in FIGURE 7.1 - NSC Mechanical Technology Welding and Metalwork - Question 7 - 2016 - Paper 1

Determine the resultant of the system of forces in FIGURE 7.1

7.2.1 Diameter of the bar

7.2.2 Strain

7.2.3 Change in length

Determine the reactions in supports A and B using FIGURE 7.2

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