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Akip/ft P=8 kip B 4' 60 61 ,2' 31

Akip/ft P=8 kip B 4' 60 61 ,2' 31

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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**Problems 1, 2, and 3:** Calculate the equivalent concentrated force of the distributed force diagram, draw the free body diagram, and calculate reactions for the beams at right. **Notes:** 1. In problem 1, there is an internal hinge at B and a simple support at C. 2. The simple support at B in problem 2 is inclined by 60° with respect to the horizontal. ### Diagram Descriptions **Diagram 1:** - This is a horizontal beam with a fixed support at point A. - There is an internal hinge at point B, located 15 feet from A. - A point load \(P = 10 \text{kips}\) is applied at angle 60° with the horizontal at 5 feet from A. - There is a uniformly distributed load of \(w = 1500 \text{ lbs/ft}\) acting for 10 feet between B and C. - A simple support is depicted at point C at the distal end of the beam. **Diagram 2:** - Shows a simply supported beam with its left end supported on a roller (point A) and the right end attached at an incline (point B). - There is a triangular distributed load starting from zero at point A and increasing up to \(w = 1 \text{ kip/ft}\) over a span of 6 feet. - An 8 kip point load is placed at a distance of 7 feet to the right of A. - The inclined support at B makes an angle of 60° with the horizontal. **Diagram 3:** - A beam with fixed support at point A and a simple support at point B. - A distributed load of \(6 \text{ kip/ft}\) acts over a length of 6 feet from A. - A linearly increasing distributed load from 0 to \(3 \text{ kip/ft}\) is applied over the next 4 feet. - A 10 kip point load is applied at an angle of 45° at point B. These diagrams provide the basis for calculating equivalent forces, drawing free body diagrams, and determining reactions at supports.
Transcribed Image Text:**Problems 1, 2, and 3:** Calculate the equivalent concentrated force of the distributed force diagram, draw the free body diagram, and calculate reactions for the beams at right. **Notes:** 1. In problem 1, there is an internal hinge at B and a simple support at C. 2. The simple support at B in problem 2 is inclined by 60° with respect to the horizontal. ### Diagram Descriptions **Diagram 1:** - This is a horizontal beam with a fixed support at point A. - There is an internal hinge at point B, located 15 feet from A. - A point load \(P = 10 \text{kips}\) is applied at angle 60° with the horizontal at 5 feet from A. - There is a uniformly distributed load of \(w = 1500 \text{ lbs/ft}\) acting for 10 feet between B and C. - A simple support is depicted at point C at the distal end of the beam. **Diagram 2:** - Shows a simply supported beam with its left end supported on a roller (point A) and the right end attached at an incline (point B). - There is a triangular distributed load starting from zero at point A and increasing up to \(w = 1 \text{ kip/ft}\) over a span of 6 feet. - An 8 kip point load is placed at a distance of 7 feet to the right of A. - The inclined support at B makes an angle of 60° with the horizontal. **Diagram 3:** - A beam with fixed support at point A and a simple support at point B. - A distributed load of \(6 \text{ kip/ft}\) acts over a length of 6 feet from A. - A linearly increasing distributed load from 0 to \(3 \text{ kip/ft}\) is applied over the next 4 feet. - A 10 kip point load is applied at an angle of 45° at point B. These diagrams provide the basis for calculating equivalent forces, drawing free body diagrams, and determining reactions at supports.
Expert Solution
Step 1

(2)

The equivalent concentrated force of the UVL is calculated as,

Equivalent concentrated force=12×1×6=3 kips

The free body diagram of the beam is shown below.

Civil Engineering homework question answer, step 1, image 1

Consider the moment equilibrium about point B.

RA×15=8×3+12×6×113×6+2+3RA=3 kips

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