5.1 Give TWO reasons why the template loft is separated from the workshop in most cases. 5.2 Describe the purpose of purlins as used on roof trusses. 5.3 A steel ring with an outside diameter of 880 mm must be fabricated from a 50 x 50 mm square steel bar. 5.3.1 Calculate the dimensions of the required material. 5.3.2 Make a neat sketch of the steel ring indicating the mean diameter, outside diameter and the thickness of the material. 5.4 Identify the following resistance weld symbols: 5.4.1 5.4.2 5.5 FIGURE 5.5 below shows a welding symbol. Label parts A–F on the welding symbol. FIGURE 5.5

NSC Mechanical Technology Welding and Metalwork Developments and templates: 5.1 Give TWO reasons why the tem

5.1 Give TWO reasons why the template loft is separated from the workshop in most cases - NSC Mechanical Technology Welding and Metalwork - Question 5 - 2018 - Paper 1

Question 5

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5.1 Give TWO reasons why the template loft is separated from the workshop in most cases. 5.2 Describe the purpose of purlins as used on roof trusses. 5.3 A steel r... show full transcript

Worked Solution & Example Answer:5.1 Give TWO reasons why the template loft is separated from the workshop in most cases - NSC Mechanical Technology Welding and Metalwork - Question 5 - 2018 - Paper 1

Step 1

Give TWO reasons why the template loft is separated from the workshop in most cases.

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Answer

The template loft is separated from the workshop because it is quieter, allowing for more focus on delicate work.
The lighting is better in the template loft, which enhances visibility for precision tasks.

Step 2

Describe the purpose of purlins as used on roof trusses.

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Answer

Purlins serve to support the roof covering and stabilize the trusses, ensuring structural integrity and preventing movement.

Step 3

Calculate the dimensions of the required material.

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Answer

To calculate the dimensions of the steel ring:

Mean Diameter:

Mean Diameter = Outside Diameter - Plate Thickness

Mean Diameter = 880 mm - 50 mm = 830 mm
Mean Circumference:

Mean Circumference = π × Mean Diameter

Mean Circumference = π × 830 mm ≈ 2607.52 mm
A total of 2608 mm of 50 x 50 mm square steel bar is needed to fabricate the ring.

Step 4

Make a neat sketch of the steel ring indicating the mean diameter, outside diameter and the thickness of the material.

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Answer

In the sketch, the dimensions should include:

Outside Diameter: 880 mm
Mean Diameter: 830 mm
Material Thickness: 50 mm

The ring should be illustrated as a concentric circle displaying these measurements.

Step 5

Identify the following resistance weld symbols:

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Answer

5.4.1 This symbol represents a Spot Weld.

5.4.2 This symbol indicates a Seam Weld.

Step 6

Label parts A–F on the welding symbol.

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Answer

In the welding symbol:

A: Tail
B: Weld symbol (Fillet weld)
C: Pitch of weld
D: Site weld
E: Arrow
F: All around

5.1 Give TWO reasons why the template loft is separated from the workshop in most cases - NSC Mechanical Technology Welding and Metalwork - Question 5 - 2018 - Paper 1

Worked Solution & Example Answer:5.1 Give TWO reasons why the template loft is separated from the workshop in most cases - NSC Mechanical Technology Welding and Metalwork - Question 5 - 2018 - Paper 1

Give TWO reasons why the template loft is separated from the workshop in most cases.

Describe the purpose of purlins as used on roof trusses.

Calculate the dimensions of the required material.

Make a neat sketch of the steel ring indicating the mean diameter, outside diameter and the thickness of the material.

Identify the following resistance weld symbols:

Label parts A–F on the welding symbol.

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