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For the circular arch, find the internal axial force at a location 5 m to the right of point A. Use the correct positive internal force sign convention. If it helps you, the radius of the arch is 12 m. 400 kN Hinge 220 kN –1.6 m 4.6 m B 8.5 m 5.8 m (A '3.5 m' 8.5 m 6 m 4.5 m

For the circular arch, find the internal axial force at a location 5 m to the right of point A. Use the correct positive internal force sign convention. If it helps you, the radius of the arch is 12 m. 400 kN Hinge 220 kN –1.6 m 4.6 m B 8.5 m 5.8 m (A '3.5 m' 8.5 m 6 m 4.5 m

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 Statement:** For the circular arch, find the internal axial force at a location 5 m to the right of point A. Use the correct positive internal force sign convention. If it helps you, the radius of the arch is 12 m. **Diagram Explanation:** - The diagram features a circular arch with two supports at points A and B. - There is a hinge located at the apex of the arch. - A downward force of 400 kN is applied vertically near point A. - An inward horizontal force of 220 kN is applied near point B. - The vertical distance from the ground to point A is 8.5 m. - The vertical distance at point B includes a height of 5.8 m with an extension of 4.6 m to the total height. - Horizontal distances are marked between points: - A to the first intermediate point is 3.5 m. - The first intermediate point to the hinge is 8.5 m. - The hinge to a second intermediate point is 6 m. - The second intermediate point to B is 4.5 m. **Note:** - The radial distance from the center of the arch to any point along it is 12 m. - The internal axial force at a location 5 m to the right of point A needs to be calculated, adhering to the positive internal force sign convention.
Transcribed Image Text:**Problem Statement:** For the circular arch, find the internal axial force at a location 5 m to the right of point A. Use the correct positive internal force sign convention. If it helps you, the radius of the arch is 12 m. **Diagram Explanation:** - The diagram features a circular arch with two supports at points A and B. - There is a hinge located at the apex of the arch. - A downward force of 400 kN is applied vertically near point A. - An inward horizontal force of 220 kN is applied near point B. - The vertical distance from the ground to point A is 8.5 m. - The vertical distance at point B includes a height of 5.8 m with an extension of 4.6 m to the total height. - Horizontal distances are marked between points: - A to the first intermediate point is 3.5 m. - The first intermediate point to the hinge is 8.5 m. - The hinge to a second intermediate point is 6 m. - The second intermediate point to B is 4.5 m. **Note:** - The radial distance from the center of the arch to any point along it is 12 m. - The internal axial force at a location 5 m to the right of point A needs to be calculated, adhering to the positive internal force sign convention.
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