Given information: The total length of the beam is 14 ft (4.2 m). The downward load P, acting at 3 ft from the right support is 4,000 lb (17.8 kN). The uniformly distributed load wi is 6,000 lb (26.7 kN) acting throughout the span. The uniformly distributed load w2 is 2,000 lb (8.90 kN) acting along 8 ft from the right support. Calculation: Show the given beam as in Figure (1). 2,000 lb 4,000 lb (17.8 kN) 6,000 lb (8.90 kN) (26.7 kN) 6 ft (1.8 m) 5 ft (1.5 m) 3 ft (0.9 m) L= 14 ft (4.2 m) R1 R- Figure 1
Given information: The total length of the beam is 14 ft (4.2 m). The downward load P, acting at 3 ft from the right support is 4,000 lb (17.8 kN). The uniformly distributed load wi is 6,000 lb (26.7 kN) acting throughout the span. The uniformly distributed load w2 is 2,000 lb (8.90 kN) acting along 8 ft from the right support. Calculation: Show the given beam as in Figure (1). 2,000 lb 4,000 lb (17.8 kN) 6,000 lb (8.90 kN) (26.7 kN) 6 ft (1.8 m) 5 ft (1.5 m) 3 ft (0.9 m) L= 14 ft (4.2 m) R1 R- Figure 1
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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I'm having a diffucult time understanding the loads and reactive forces problem 1.7. How did the distributed load for I2 = 3ft and I3= 1 ft where did they get 3ft and 1 ft. I understand how they got I1= 7ft with a beam spaning 14 ft divide 2 = 7ft. I need an explanation on how and why I2 = 3ft and I3= 1 ft?
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