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2 k/ft Problem 7.46 Use the virtual work method to determine the horizontal deflection B at joint C of the frame shown. 10 ft 15 ft 30 ft El = constant E = 29,000 ksi 1 = 1.500 in.

2 k/ft Problem 7.46 Use the virtual work method to determine the horizontal deflection B at joint C of the frame shown. 10 ft 15 ft 30 ft El = constant E = 29,000 ksi 1 = 1.500 in.

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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**Problem 7.46** **Objective:** Use the virtual work method to determine the horizontal deflection at joint C of the frame shown. **Description:** The frame consists of three members. Member AB extends vertically 10 ft from the ground to the top of member BC. Member BC is a horizontal beam spanning 30 ft and carries a uniform distributed load of 2 kips/ft. Member CD is a vertical column extending 15 ft from the top of the beam to the ground. **Material Properties:** - The modulus of elasticity, \( E = 29,000 \) ksi. - The moment of inertia, \( I = 1,500 \) in\(^4\). **Conditions:** - The frame has fixed supports at points A and D. - All members are rigidly connected. **Diagram Explanation:** - The frame structure is depicted with three key points: A, B, C, and D. - The length of AB is 10 ft, BC is 30 ft, and CD is 15 ft. - The beam BC is subjected to a uniformly distributed load (UDL) of 2 kips/ft. - The supports at A and D are fixed, indicating no rotation or translation at those joints.
Transcribed Image Text:**Problem 7.46** **Objective:** Use the virtual work method to determine the horizontal deflection at joint C of the frame shown. **Description:** The frame consists of three members. Member AB extends vertically 10 ft from the ground to the top of member BC. Member BC is a horizontal beam spanning 30 ft and carries a uniform distributed load of 2 kips/ft. Member CD is a vertical column extending 15 ft from the top of the beam to the ground. **Material Properties:** - The modulus of elasticity, \( E = 29,000 \) ksi. - The moment of inertia, \( I = 1,500 \) in\(^4\). **Conditions:** - The frame has fixed supports at points A and D. - All members are rigidly connected. **Diagram Explanation:** - The frame structure is depicted with three key points: A, B, C, and D. - The length of AB is 10 ft, BC is 30 ft, and CD is 15 ft. - The beam BC is subjected to a uniformly distributed load (UDL) of 2 kips/ft. - The supports at A and D are fixed, indicating no rotation or translation at those joints.
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