Problem 3 Use the unit load method to compute the horizontal displacement at A. Take E =29,000 ksi. The moment of inertia of each frame member is shown in the figure. Assume that the support at D is a pin support, and a roller support at A. Ignore shear and axial deformations, i.e., assume that the deformations are due to bending only. 10 ft 2 k/ft IBC = 800 in.4 IAB = 500 in.4 ICD = 500 in.4 A -10 ft-
Problem 3 Use the unit load method to compute the horizontal displacement at A. Take E =29,000 ksi. The moment of inertia of each frame member is shown in the figure. Assume that the support at D is a pin support, and a roller support at A. Ignore shear and axial deformations, i.e., assume that the deformations are due to bending only. 10 ft 2 k/ft IBC = 800 in.4 IAB = 500 in.4 ICD = 500 in.4 A -10 ft-
Principles of Foundation Engineering (MindTap Course List)
8th Edition
ISBN:9781305081550
Author:Braja M. Das
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Chapter6: Vertical Stress Increase In Soil
Section: Chapter Questions
Problem 6.4P: Refer to Figure P6.4. A strip load of q = 900 lb/ft2 is applied over a width B = 36 ft. Determine...
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Use the unit load method to compute the horizontal displacement at A. Take E =29,000 ksi. The
moment of inertia of each frame member is shown in the figure. Assume that the support at D is a pin
support, and a roller support at A. Ignore shear and axial deformations, i.e., assume that the
deformations are due to bending only.
moment of inertia of each frame member is shown in the figure. Assume that the support at D is a pin
support, and a roller support at A. Ignore shear and axial deformations, i.e., assume that the
deformations are due to bending only.
Please show the actual steps and calculations/diagrams to solve. The last time I asked this question those were not provided.
Transcribed Image Text:Problem 3
Use the unit load method to compute the horizontal displacement at A. Take E =29,000 ksi. The
moment of inertia of each frame member is shown in the figure. Assume that the support at D is a pin
support, and a roller support at A. Ignore shear and axial deformations, i.e., assume that the
deformations are due to bending only.
10 ft
2 k/ft
IBC = 800 in.4
IAB = 500 in.4
ICD
= 500 in.4
A
-10 ft-
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