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Final essay SEA & INLAND WATERWAY TRANSPORT FIATA

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Tiêu đề Final Essay Sea & Inland Waterway Transport FIATA
Chuyên ngành Sea & Inland Waterway Transport
Thể loại Essay
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Số trang 2
Dung lượng 17,64 KB

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Trang 1

Task 1:1 20 boxes of Product A, each weighing 350 kg

10 crates of Product B, each weighing 500 kgDunnage material totaling 250 kg

15 pallets (120x110 cm) with each weighing 120 kgTotal weight of cargo and materials: 350 x 20 + 500 x 10 + 250 + 120 x15 = 14,050 kg

Tare weight of the container: 3,850 kgVGM = Tare weight + Net weight = 3,850 + 14,050 = 17,900 kg

2 The size of a 40-foot container is 1220 x 244 x 260 (cm).

The size of each pallet is 120 x 110 (cm).This container will load up to 20 pallets.Because this container is loaded with 15 pallets, 5 pallets should be added.The maximum gross weight for a 40-foot container is 30,480 kg

This container can still accommodate an additional 30,480 - 17,900 = 12,580kg

The total weight of 5 pallets which were added is 120 kg.The total remaining weight of the container is 12,580 – 120 = 12,460 kg

Option 1: Additional boxes of Product A that can be added without

exceeding the weight limit: 12,460/350 = 35.6 boxes.A maximum of 35 boxes of Product A can be added

Option 2: Additional crates of Product B that can be added without

exceeding the weight limit: 12,460/500 = 24.92 crates.A maximum of 24 crates of Product B can be added

Option 3: The number of boxes of Product A and crates of Product B that

can be added without exceeding the weight limits are 7 and 20 respectively:350 x 7 + 500 x 20 = 12,450 kg

The shipper should consider loading 5 pallets and up to 35 boxes of Product Aor up to 24 crates of Product B or 7 boxes of Product A and 20 crates of Product B

Task 2:

A 40-foot container (tare weight 3,800 kg, max gross weight 26,500 kg) is beingloaded with steel coils

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Each coil: 1.5 meters diameter, 0.5 meters width, density of 7,850 kg/m³Number of coils: 14

Coil Volume: V = π × r2 × h = π × (1.5/2)2 x 0.5 = 0.884 m3

Total Cargo Volume: Vtotal = 14 × 0.884 = 12.376 m3

Coil Weight: W = V x D = 0.884 x 7850 = 6939.4 kgTotal Cargo Weight: Wtotal = 14 x 6939.4 = 97151.6 kg

Estimated Gross Weight which we can load into that container: Gross Weight

= Tare Weight + Wtotal = 3800 + 97151.6 = 100951.6 kgMax payload = Max gross weight – Tare weight = 26,500 - 3,800 = 22,700 kgWe can load 3 steel coils Gross Weight = 3 x 6939.4 = 20818.2 kg

Ngày đăng: 01/09/2024, 16:44

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