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Parents Pricing Home NSC Mechanical Technology Welding and Metalwork Welding terms and symbols A steel ring must be manufactured using a 60 x 60 mm square steel bar
A steel ring must be manufactured using a 60 x 60 mm square steel bar - NSC Mechanical Technology Welding and Metalwork - Question 5 - 2021 - Paper 1 Question 5
View full question A steel ring must be manufactured using a 60 x 60 mm square steel bar. The ring has an outside diameter of 960 mm.
5.1.1 Calculate the mean diameter of the ring.
5... show full transcript
View marking scheme Worked Solution & Example Answer:A steel ring must be manufactured using a 60 x 60 mm square steel bar - NSC Mechanical Technology Welding and Metalwork - Question 5 - 2021 - Paper 1
5.1.1 Calculate the mean diameter of the ring. Only available for registered users.
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To calculate the mean diameter, use the formula:
e x t M e a n D i a m e t e r = e x t O u t s i d e D i a m e t e r − e x t P l a t e T h i c k n e s s ext{Mean Diameter} = ext{Outside Diameter} - ext{Plate Thickness} e x t M e an D iam e t er = e x t O u t s i d eD iam e t er − e x t Pl a t e T hi c kn ess Given that the outside diameter is 960 mm and the plate thickness is 60 mm:
e x t M e a n D i a m e t e r = 960 e x t m m − 60 e x t m m = 900 e x t m m ext{Mean Diameter} = 960 ext{ mm} - 60 ext{ mm} = 900 ext{ mm} e x t M e an D iam e t er = 960 e x t mm − 60 e x t mm = 900 e x t mm Thus, the mean diameter of the ring is 900 mm .
5.1.2 Calculate the mean circumference of the ring (round off answer to the nearest whole number). Only available for registered users.
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The mean circumference can be calculated using the formula:
e x t M e a n C i r c u m f e r e n c e = e x t π i m e s e x t M e a n D i a m e t e r ext{Mean Circumference} = ext{π} imes ext{Mean Diameter} e x t M e an C i rc u m f ere n ce = e x t π im ese x t M e an D iam e t er Substituting the mean diameter we calculated:
e x t M e a n C i r c u m f e r e n c e = e x t π i m e s 900 e x t m m e x t M e a n C i r c u m f e r e n c e e x t ≈ 2827.43 e x t m m ext{Mean Circumference} = ext{π} imes 900 ext{ mm} \
ext{Mean Circumference} \ ext{≈} 2827.43 ext{ mm} e x t M e an C i rc u m f ere n ce = e x t π im es 900 e x t mm e x t M e an C i rc u m f ere n ce e x t ≈ 2827.43 e x t mm Rounding off the answer to the nearest whole number gives us 2827 mm .
5.2.1 Spot weld Only available for registered users.
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The symbol for a spot weld is:
5.2.2 Projection weld Only available for registered users.
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The symbol for a projection weld is:
5.2.3 Seam weld Only available for registered users.
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The symbol for a seam weld is:
5.2.4 Foil seam weld Only available for registered users.
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The symbol for a foil seam weld is:
5.2.5 Flash or resistance weld Only available for registered users.
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The symbol for a flash or resistance weld is:
5.3 Identify the templates shown in FIGURES 5.3.1–5.3.3 below. Only available for registered users.
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The identified templates are as follows:
5.3.1 : Flange Template5.3.2 : Strip Template5.3.3 : Web Template
5.4 Name THREE hand tools that a template maker uses. Only available for registered users.
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Three hand tools that a template maker uses include:
Hand saws
Chisels
Measuring tape
5.5 Name TWO machines used in the template loft. Only available for registered users.
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Two machines used in the template loft are:
Circular saw
Drilling machine
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