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Using the double-integration method: a) Determine the equations of the elastic curve (deflection) and slope of the beam shown in Figure 4. Assume the x coordinate begins at the free end (point B). b) Given E = 29,000 ksi and I = 600 in, specify the slope and deflection at point B. Note: Use units of inches for lengths and kips for forces throughout your calculations. Hint: 2 k ft 1 k - 6 in 15 ft = 180 in 2 klf A EI 15 ft- Figure 4 B

Using the double-integration method: a) Determine the equations of the elastic curve (deflection) and slope of the beam shown in Figure 4. Assume the x coordinate begins at the free end (point B). b) Given E = 29,000 ksi and I = 600 in, specify the slope and deflection at point B. Note: Use units of inches for lengths and kips for forces throughout your calculations. Hint: 2 k ft 1 k - 6 in 15 ft = 180 in 2 klf A EI 15 ft- Figure 4 B

Engineering Fundamentals: An Introduction to Engineering (MindTap Course List)
Engineering Fundamentals: An Introduction to Engineering (MindTap Course List)
5th Edition
ISBN:9781305084766
Author:Saeed Moaveni
Publisher:Saeed Moaveni
Chapter14: Computational Engineering Tools Electronic Spreadsheets
Section: Chapter Questions
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Using the double-integration method: a) Determine the equations of the elastic curve (deflection) and slope of the beam shown in Figure 4. Assume the x coordinate begins at the free end (point B). b) Given E = 29,000 ksi and I = 600 in, specify the slope and deflection at point B. Note: Use units of inches for lengths and kips for forces throughout your calculations. Hint: 2 k ft 1 k - 6 in 15 ft = 180 in 2 klf A EI 15 ft- Figure 4 B
Transcribed Image Text:Using the double-integration method: a) Determine the equations of the elastic curve (deflection) and slope of the beam shown in Figure 4. Assume the x coordinate begins at the free end (point B). b) Given E = 29,000 ksi and I = 600 in, specify the slope and deflection at point B. Note: Use units of inches for lengths and kips for forces throughout your calculations. Hint: 2 k ft 1 k - 6 in 15 ft = 180 in 2 klf A EI 15 ft- Figure 4 B
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