Structural Engineering Crane Loads 6 Crane Loads. 𝐶𝑜𝑙𝑢𝑚𝑛 𝑉 𝑉 𝐻 𝐻 𝐸𝑙𝑒𝑣𝑎𝑡𝑖𝑜𝑛 ➢ 𝑳𝒐𝒂𝒅𝒔 𝒓𝒆𝒔𝒖𝒍𝒕𝒊𝒏𝒈 𝒅𝒖𝒆 𝒕𝒐 𝑪𝒓𝒂𝒏𝒆 +25% 𝑜𝑓 𝑊ℎ𝑒𝑒𝑙 𝐿𝑜𝑎𝑑 𝑑𝑢𝑒 𝑡𝑜 𝐼𝑚𝑝𝑎𝑐𝑡 1) 𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 → 𝑂𝑤𝑛 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝐶𝑟𝑎𝑛𝑒 𝐵𝑟𝑖𝑑𝑔𝑒 + 𝐶𝑟𝑎𝑛𝑒 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 2) 𝐻𝑍 𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑆ℎ𝑜𝑐𝑘 → 10% 𝑜𝑓 𝑊ℎ𝑒𝑒𝑙 𝐿𝑜𝑎𝑑 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝐼𝑚𝑝𝑎𝑐𝑡 (𝑑𝑢𝑒 𝑡𝑜 𝐻𝑍 𝑀𝑜𝑣𝑒𝑚𝑒𝑛𝑡 𝑜𝑓 𝑇𝑟𝑜𝑙𝑙𝑒𝑦) 3) 𝐵𝑟𝑎𝑘𝑖𝑛𝑔 𝐹𝑜𝑟𝑐𝑒 → 1 𝑜𝑓 𝑊ℎ𝑒𝑒𝑙 𝐿𝑜𝑎𝑑 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝐼𝑚𝑝𝑎𝑐𝑡 7 𝑅𝑒𝑠𝑖𝑠𝑡𝑒𝑑 𝑏𝑦 𝑀𝑎𝑖𝑛 𝑆𝑦𝑠𝑡𝑒𝑚 𝑅𝑒𝑠𝑖𝑠𝑡𝑒𝑑 𝑏𝑦 𝑉𝐿 𝐵𝑟𝑎𝑐𝑖𝑛𝑔 6 Crane Loads. 0.5 m 1) 𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 B−1 B 0.5 m 6 Crane Loads. 1) 𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑂. 𝑊 𝑜𝑓 𝐵𝑟𝑖𝑑𝑔𝑒 0.2 → 0.3 𝑡/𝑚 𝐶𝑟𝑎𝑛𝑒 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝐹𝑜𝑟 𝑡ℎ𝑖𝑠 𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡𝑟𝑜𝑙𝑙𝑒𝑦, 𝐿𝑜𝑎𝑑 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙𝑙𝑦 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡𝑤𝑜 𝐶𝑇𝐺 ′ 𝑠 𝑎𝑛𝑑 𝑡ℎ𝑖𝑠 𝑖𝑠 𝑛𝑜𝑡 𝑡ℎ𝑒 𝑤𝑜𝑟𝑠𝑡 𝑐𝑎𝑠𝑒 𝑤𝑒 𝑠𝑒𝑒𝑘. 6 Crane Loads. 1) 𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑂. 𝑊 𝑜𝑓 𝐵𝑟𝑖𝑑𝑔𝑒 0.2 → 0.3 𝑡/𝑚 0.9 m 𝐶𝑟𝑎𝑛𝑒 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑀𝑖𝑛. 𝐿𝑜𝑎𝑑𝑠 𝑜𝑛 𝐿𝑒𝑓𝑡 𝐶𝑇𝐺 𝑀𝑎𝑥. 𝐿𝑜𝑎𝑑𝑠 𝑜𝑛 𝑅𝑖𝑔ℎ𝑡 𝐶𝑇𝐺 6 Crane Loads. 1) 𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑂. 𝑊 𝑜𝑓 𝐵𝑟𝑖𝑑𝑔𝑒 0.2 → 0.3 𝑡/𝑚 0.9 m 𝐶𝑟𝑎𝑛𝑒 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑀𝑖𝑛. 𝐿𝑜𝑎𝑑𝑠 𝑜𝑛 𝑅𝑖𝑔ℎ𝑡 𝐶𝑇𝐺 𝑀𝑎𝑥. 𝐿𝑜𝑎𝑑𝑠 𝑜𝑛 𝐿𝑒𝑓𝑡 𝐶𝑇𝐺 𝐶𝑟𝑎𝑛𝑒 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑂. 𝑊 𝑜𝑓 𝐵𝑟𝑖𝑑𝑔𝑒 0.2 → 0.3 𝑡/𝑚 0.9 m 𝑅𝑚𝑎𝑥. B−1 𝑅𝑚𝑖𝑛. 6 Crane Loads. S 1) 𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 S 𝐴 𝐴 6 Crane Loads. S 1) 𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝐴 S 𝑃𝑚𝑎𝑥. 𝐿𝑒𝑡 ′ 𝑠 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑡ℎ𝑖𝑠 𝐵𝑟𝑎𝑐𝑘𝑒𝑡 𝑃𝑚𝑎𝑥. 𝐴 𝑊𝑒 𝑛𝑒𝑒𝑑 𝑡𝑜 𝑝𝑙𝑎𝑐𝑒 𝑊ℎ𝑒𝑒𝑙 𝐿𝑜𝑎𝑑𝑠 𝑖𝑛 𝑜𝑟𝑑𝑒𝑟 𝑡𝑜 𝑔𝑒𝑡 𝑡ℎ𝑒 𝑚𝑎𝑥. 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑛 𝐵𝑟𝑎𝑐𝑘𝑒𝑡 𝑃𝑚𝑎𝑥. = 𝑅𝑚𝑎𝑥. 2 𝑃𝑚𝑖𝑛. = 𝑅𝑚𝑖𝑛. 2 6 Crane Loads. 1) 𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 S 𝐴 S 𝑃𝑚𝑎𝑥. 𝑃𝑚𝑎𝑥. 𝐿𝑒𝑡 ′ 𝑠 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑡ℎ𝑖𝑠 𝐵𝑟𝑎𝑐𝑘𝑒𝑡 𝐴 𝑉𝐷𝐿 𝑫𝑳 𝑉𝐷𝐿 𝑙𝑒𝑓𝑡 𝑉𝐷𝐿 𝑟𝑖𝑔ℎ𝑡 𝑉𝐷𝐿 = 𝑉𝐷𝐿 𝑙𝑒𝑓𝑡 + 𝑉𝐷𝐿 𝑟𝑖𝑔ℎ𝑡 = 0.2 × 𝑆 ×2 2 𝑡𝑜𝑛 𝑉𝑚𝑎𝑥 = 𝑉𝐷𝐿 + 1.25 𝑉𝐿𝐿 𝑃𝑚𝑎𝑥. = 𝑅𝑚𝑎𝑥. 2 𝑃𝑚𝑖𝑛. = 𝑅𝑚𝑖𝑛. 2 𝑉𝐿𝐿 𝑳𝑳 𝑉𝑚𝑖𝑛 = 𝑉𝐷𝐿 + 1.25 𝑉𝐿𝐿 𝑉𝐿𝐿 𝑙𝑒𝑓𝑡 = 0 𝑉𝐿𝐿 = 𝑉𝐿𝐿 𝑙𝑒𝑓𝑡 + 𝑉𝐿𝐿 𝑟𝑖𝑔ℎ𝑡 𝑉𝐿𝐿 𝑟𝑖𝑔ℎ𝑡 𝑉𝐼𝑚𝑝𝑎𝑐𝑡 = 𝑉𝐿𝐿 × 0.25 𝑉𝐿𝐿+𝐼𝑚𝑝𝑎𝑐𝑡 = 1.25𝑉𝐿𝐿 𝐷𝑢𝑒 𝑡𝑜 𝑃𝑚𝑖𝑛. 6 Crane Loads. 1) 𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑽𝒎𝒂𝒙 𝑽𝒎𝒊𝒏 𝑉𝑚𝑎𝑥 = 𝑉𝐷𝐿 + 1.25 𝑉𝐿𝐿 𝑉𝑚𝑖𝑛 = 𝑉𝐷𝐿 + 1.25 𝑉𝐿𝐿 𝐷𝑢𝑒 𝑡𝑜 𝑃𝑚𝑖𝑛. 6 Crane Loads. 2) 𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑊ℎ𝑒𝑒𝑙 1 𝑚𝑚 𝐶𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 𝐶𝑟𝑎𝑛𝑒 𝑅𝑎𝑖𝑙 𝐻𝑚𝑎𝑥 𝐶𝑇𝐺 ➢ 𝐷𝑢𝑒 𝑡𝑜 𝐻𝑍 𝑀𝑜𝑣𝑒𝑚𝑒𝑛𝑡 𝑜𝑓 𝑇𝑟𝑜𝑙𝑙𝑒𝑦 ➢ 1 𝑚𝑚 𝐶𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑊ℎ𝑒𝑒𝑙 𝑎𝑛𝑑 𝑅𝑎𝑖𝑙 𝐶𝑜𝑙𝑢𝑚𝑛 𝐵𝑟𝑎𝑐𝑘𝑒𝑡 𝐻𝑍 𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑆ℎ𝑜𝑐𝑘 𝑖𝑠 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑒𝑑 𝐻𝑚𝑎𝑥 = 0.1 𝑉𝐿𝐿 𝐷𝑢𝑒 𝑡𝑜 𝑃𝑚𝑎𝑥. 𝐻𝑚𝑖𝑛 = 0.1 𝑉𝐿𝐿 𝐷𝑢𝑒 𝑡𝑜 𝑃𝑚𝑖𝑛. 𝑾𝒊𝒕𝒉𝒐𝒖𝒕 𝑰𝒎𝒑𝒂𝒄𝒕 6 Crane Loads. 2) 𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑽𝒎𝒂𝒙 𝑽𝒎𝒊𝒏 𝑯𝒎𝒂𝒙 𝑯𝒎𝒊𝒏 𝑉𝑚𝑎𝑥 = 𝑉𝐷𝐿 + 1.25 𝑉𝐿𝐿 𝐻𝑚𝑎𝑥 = 0.1 𝑉𝐿𝐿 𝐷𝑢𝑒 𝑡𝑜 𝑃𝑚𝑎𝑥. 𝑉𝑚𝑖𝑛 = 𝑉𝐷𝐿 + 1.25 𝑉𝐿𝐿 𝐻𝑚𝑖𝑛 = 0.1 𝑉𝐿𝐿 𝐷𝑢𝑒 𝑡𝑜 𝑃𝑚𝑖𝑛. 𝐷𝑢𝑒 𝑡𝑜 𝑃𝑚𝑖𝑛. 6 Crane Loads. 𝐶𝑎𝑠𝑒𝑠 𝑜𝑓 𝐿𝑜𝑎𝑑𝑖𝑛𝑔 𝐹𝑜𝑟 𝐶𝑟𝑎𝑛𝑒 𝑃𝑟𝑖𝑚𝑎𝑟𝑦 𝐶𝑎𝑠𝑒𝑠 𝑆𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝐶𝑎𝑠𝑒𝑠 𝑉𝐿 𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑂𝑛𝑙𝑦 𝑉𝐿 + 𝐻𝑍 𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 𝟑 𝟏 𝑽𝒎𝒂𝒙 𝑽𝒎𝒊𝒏 𝟒 𝑽𝒎𝒂𝒙 𝑽𝒎𝒊𝒏 𝑽𝒎𝒂𝒙 𝑯𝒎𝒊𝒏 𝑯𝒎𝒂𝒙 𝑽𝒎𝒊𝒏 𝑯𝒎𝒊𝒏 𝑯𝒎𝒂𝒙 𝑪𝑹 𝑽𝑳 𝑳𝒆𝒇𝒕 𝑪𝑹 𝑯𝒁𝟏 𝑳𝒆𝒇𝒕 𝑪𝑹 𝑯𝒁𝟐 𝑳𝒆𝒇𝒕 𝟐 𝟓 𝟔 𝑽𝒎𝒊𝒏 𝑽𝒎𝒂𝒙 𝑽𝒎𝒊𝒏 𝑽𝒎𝒂𝒙 𝑯𝒎𝒊𝒏 𝑪𝑹 𝑽𝑳 𝑹𝒊𝒈𝒉𝒕 𝑯𝒎𝒂𝒙 𝑪𝑹 𝑯𝒁𝟏 𝑹𝒊𝒈𝒉𝒕 𝑽𝒎𝒊𝒏 𝑽𝒎𝒂𝒙 𝑯𝒎𝒊𝒏 𝑯𝒎𝒂𝒙 𝑪𝑹 𝑯𝒁𝟐 𝑹𝒊𝒈𝒉𝒕 6 Crane Loads. 𝑆𝑢𝑚𝑚𝑎𝑟𝑦 𝑜𝑓 𝐶𝑟𝑎𝑛𝑒 𝐿𝑜𝑎𝑑𝑠 1) 𝑆𝑜𝑙𝑣𝑒 𝐶𝑟𝑎𝑛𝑒 𝐵𝑟𝑖𝑑𝑔𝑒 𝑎𝑛𝑑 𝑔𝑒𝑡 𝑅𝑚𝑎𝑥. , 𝑅𝑚𝑖𝑛. 2) 𝐴𝑠𝑠𝑖𝑔𝑛 𝑅𝑚𝑎𝑥 𝑜𝑛 𝐶𝑇𝐺 𝑎𝑠 𝑡𝑤𝑜 𝑊ℎ𝑒𝑒𝑙 𝐿𝑜𝑎𝑑𝑠 𝑃𝑚𝑎𝑥 . 𝐺𝑒𝑡 𝑉𝐷𝐿 , 𝑉𝐿𝐿+𝐼𝑚𝑝𝑎𝑐𝑡 𝐺𝑒𝑡 𝑉𝑚𝑎𝑥 𝑉𝐷𝐿 = 𝑉𝐷𝐿 𝑙𝑒𝑓𝑡 + 𝑉𝐷𝐿 𝑟𝑖𝑔ℎ𝑡 = 𝑊ℎ𝑒𝑟𝑒 𝑃𝑚𝑎𝑥 = 0.2 × 𝑆 ×2 2 𝑅𝑚𝑎𝑥 2 𝑡𝑜𝑛 𝑉𝐿𝐿+𝐼𝑚𝑝𝑎𝑐𝑡 = 1.25 𝑉𝐿𝐿 𝑉𝑚𝑎𝑥 = 𝑉𝐷𝐿 + 1.25 𝑉𝐿𝐿 𝑈𝑠𝑖𝑛𝑔 𝑃𝑚𝑖𝑛 𝑅𝑒𝑝𝑒𝑎𝑡 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑠𝑡𝑒𝑝𝑠 𝑡𝑜 𝑔𝑒𝑡 𝑉𝑚𝑖𝑛 3) 𝐺𝑒𝑡 𝐻𝑍 𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑆ℎ𝑜𝑐𝑘 𝐻𝑚𝑎𝑥 = 0.1 𝑉𝐿𝐿 𝐷𝑢𝑒 𝑡𝑜 𝑃𝑚𝑎𝑥. 4) 𝐷𝑒𝑓𝑖𝑛𝑒 𝑎𝑛𝑑 𝐴𝑠𝑠𝑖𝑔𝑛 6 𝑐𝑟𝑎𝑛𝑒 𝑙𝑜𝑎𝑑 𝑐𝑎𝑠𝑒𝑠 𝐻𝑚𝑖𝑛 = 0.1 𝑉𝐿𝐿 𝐷𝑢𝑒 𝑡𝑜 𝑃𝑚𝑖𝑛. 𝑾𝒊𝒕𝒉𝒐𝒖𝒕 𝑰𝒎𝒑𝒂𝒄𝒕 6 Crane Loads. 𝑆𝑢𝑚𝑚𝑎𝑟𝑦 𝑜𝑓 𝐶𝑟𝑎𝑛𝑒 𝐿𝑜𝑎𝑑𝑠 𝐼𝑛𝑠𝑡𝑒𝑎𝑑 𝑜𝑓 𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝐶𝑟𝑎𝑛𝑒 𝐵𝑟𝑖𝑑𝑔𝑒 𝑡𝑜 𝑔𝑒𝑡 𝑃𝑚𝑎𝑥 , 𝑃𝑚𝑖𝑛 𝐶𝑟𝑎𝑛𝑒 𝑡𝑎𝑏𝑙𝑒𝑠 𝑐𝑎𝑛 𝑏𝑒 𝑢𝑠𝑒𝑑 𝑡𝑜 𝑔𝑒𝑡 𝑡ℎ𝑒𝑠𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑎𝑠 𝑓𝑜𝑙𝑙𝑜𝑤𝑠: 𝐺𝑖𝑣𝑒𝑛 𝑡ℎ𝑎𝑡: 𝐶𝑟𝑎𝑛𝑒 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 = 10 𝑡𝑜𝑛 𝑆𝑝𝑎𝑛 = 20 𝑚 𝑃𝑚𝑎𝑥 = 7450 𝐾𝑔 = 7.45 𝑡𝑜𝑛 𝑃𝑚𝑖𝑛 = 2000 𝐾𝑔 = 2 𝑡𝑜𝑛 𝐹𝑜𝑙𝑙𝑜𝑤 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑠𝑡𝑒𝑝𝑠 𝑡𝑜 𝑔𝑒𝑡 𝑉, 𝐻 𝑇ℎ𝑒𝑟𝑒 𝑖𝑠 𝑎𝑙𝑠𝑜 𝑎𝑛𝑜𝑡ℎ𝑒𝑟 𝑇𝑎𝑏𝑙𝑒 𝑡𝑜 𝑔𝑒𝑡 𝑊ℎ𝑒𝑒𝑙 𝑆𝑝𝑎𝑐𝑖𝑛𝑔 𝐷𝑜𝑤𝑛𝑙𝑜𝑎𝑑 𝐶𝑟𝑎𝑛𝑒 𝑇𝑎𝑏𝑙𝑒𝑠 𝐻𝑒𝑟𝑒 6 Crane Loads. 𝑊ℎ𝑎𝑡 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑡ℎ𝑖𝑟𝑑 𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡? 𝐵𝑟𝑎𝑘𝑖𝑛𝑔 𝐹𝑜𝑟𝑐𝑒 𝐵𝑟𝑎𝑘𝑖𝑛𝑔 𝐹𝑜𝑟𝑐𝑒 𝑎𝑡 𝑒𝑎𝑐ℎ 𝑤ℎ𝑒𝑒𝑙 = 𝑃𝑚𝑎𝑥 7 𝐶𝑜𝑙𝑢𝑚𝑛 𝑅𝑒𝑠𝑖𝑠𝑡𝑒𝑑 𝑏𝑦 𝑆𝑡𝑟𝑢𝑡 𝑢𝑠𝑒𝑑 𝑎𝑡 𝐶𝑇𝐺 𝐿𝑒𝑣𝑒𝑙