30 k 3 k/ft Problem 7.39 Use the virtual work method to D 21 determine the rotation of joint D of the frame shown. 8 ft E = 2,000 ksi 40k 21 I= 10,000 in. B ft 16 ft FIG. P7.39, P7.40

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
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**Problem 7.39**

Use the virtual work method to determine the rotation of joint D of the frame shown.

**Diagram Explanation:**

- The frame is a vertical structure with three main joints: A, B, and C.
- Joint A is at the bottom left corner, joint B is at the center, and joint C is at the top.
- At joint B, there is a horizontal force of 40 kips acting to the right.
- The vertical column from A to C is divided into equal segments of 8 ft.
- From joint C, a horizontal beam extends to the right towards joint E, which is 24 ft long in total.
- At joint C, a vertical force of 30 kips is acting downward.
- Along the beam from joint C to joint E, there is a uniformly distributed load of 3 kips per foot over a span of 16 ft.
- The distance from joint C to point D is 8 ft (1L), and from point D to joint E is 16 ft (2L).
- Material properties are given: Modulus of Elasticity, \(E = 2,000 \text{ ksi}\), and Moment of Inertia, \(I = 10,000 \text{ in}^4\).

The problem involves calculating the rotation at joint D using the virtual work method.
Transcribed Image Text:**Problem 7.39** Use the virtual work method to determine the rotation of joint D of the frame shown. **Diagram Explanation:** - The frame is a vertical structure with three main joints: A, B, and C. - Joint A is at the bottom left corner, joint B is at the center, and joint C is at the top. - At joint B, there is a horizontal force of 40 kips acting to the right. - The vertical column from A to C is divided into equal segments of 8 ft. - From joint C, a horizontal beam extends to the right towards joint E, which is 24 ft long in total. - At joint C, a vertical force of 30 kips is acting downward. - Along the beam from joint C to joint E, there is a uniformly distributed load of 3 kips per foot over a span of 16 ft. - The distance from joint C to point D is 8 ft (1L), and from point D to joint E is 16 ft (2L). - Material properties are given: Modulus of Elasticity, \(E = 2,000 \text{ ksi}\), and Moment of Inertia, \(I = 10,000 \text{ in}^4\). The problem involves calculating the rotation at joint D using the virtual work method.
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