For the structure and loading system shown:B) If the resultant force and couple pair is replaced by a single resultant force, where does its line of action intersects a vertical line along OB, measured from O? ### Analysis of Forces and Resultant Movements on a Structure#### Problem Statement:The image shows a structural system subjected to various loads. The structure consists of two main members, arranged perpendicularly to each other.#### Description of the Structure:- **Horizontal Member (AB):** - Point B has an applied load of 300 lb at an angle of \( \cos^{-1}(\frac{3}{5}) \) from the horizontal. - Point A has a downward load of 50 lb, positioned 1 ft from point B. - Point A also has a leftward load of 100 lb. - The total length between points A and B is 2 ft, with point A located 1 ft from the left end of the horizontal member.- **Vertical Member (CD):** - Positioned vertically from point A to a base support at point O. - Point C has a downward load of 50 lb, located 1 ft from A down to C, with another 50 lb to the left at point O. - The length from A to C is 3 ft, with sections of 1 ft each from base O and to loading points.#### Given Task:B) Determine the location where a single resultant force, replacing the given forces and their moments, intersects a vertical line along the member OB, measured from point O.#### Graphical Notation and Dimensions:- **Coordinate Axes:** - The x-axis is horizontal, and the y-axis is vertical.- **Spatial Configuration:** - Points on the structure are labeled as A, B, C, and O with respective distances. - A rigid base at point O supports the vertical member.### Steps for Solution (Provided Explanations):1. **Resultant Force (FR):** - Sum all the vertical and horizontal forces to find the components of the resultant force \( \vec{F_R} \).2. **Moment (MR = MO):** - Calculate the moments caused by the distributed forces about point O to find the total moment \( M_O \).3. **Line of Action:** - Determine the point where the resultant force intersects the line OB using the relationship between net force, moments, and distances from point O to give the precise location.### Conclusion:This analysis enables the understanding of how complex force interactions can be represented as a single resultant force and moment, simplifying the

Given:- To find:- From the given question it is required to find the line of action of resultant…

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B) If the resultant force and couple pair is replaced by a single resultant force, where does its line of action intersects a vertical line along OB, measured from O?

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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Concept explainers

System of Forces

A force is an external influence or agent that changes the state of a system. According to Newton's second law, the force applied to a body is proportional to the change in momentum of a body. Hence, whenever a force is applied to a body, its momentum changes. If the body is at rest, under the action of force, its state of rest changes to motion. In other words, the magnitude of the momentum of the body changes from zero to a measurable value. Similarly, if the body possesses some initial velocity, it has some associated momentum, the effect of force can cause the body to halt or come to rest, that is, the momentum reduces to zero.

Question

For the structure and loading system shown:

B) If the resultant force and couple pair is replaced by a single resultant force, where does its line of action intersects a vertical line along OB, measured from O?

### Analysis of Forces and Resultant Movements on a Structure

#### Problem Statement:

The image shows a structural system subjected to various loads. The structure consists of two main members, arranged perpendicularly to each other.

#### Description of the Structure:

- **Horizontal Member (AB):**
- Point B has an applied load of 300 lb at an angle of \( \cos^{-1}(\frac{3}{5}) \) from the horizontal.
- Point A has a downward load of 50 lb, positioned 1 ft from point B.
- Point A also has a leftward load of 100 lb.
- The total length between points A and B is 2 ft, with point A located 1 ft from the left end of the horizontal member.
- **Vertical Member (CD):**
- Positioned vertically from point A to a base support at point O.
- Point C has a downward load of 50 lb, located 1 ft from A down to C, with another 50 lb to the left at point O.
- The length from A to C is 3 ft, with sections of 1 ft each from base O and to loading points.

#### Given Task:

B) Determine the location where a single resultant force, replacing the given forces and their moments, intersects a vertical line along the member OB, measured from point O.

#### Graphical Notation and Dimensions:

- **Coordinate Axes:**
- The x-axis is horizontal, and the y-axis is vertical.
- **Spatial Configuration:**
- Points on the structure are labeled as A, B, C, and O with respective distances.
- A rigid base at point O supports the vertical member.

### Steps for Solution (Provided Explanations):

1. **Resultant Force (FR):**
- Sum all the vertical and horizontal forces to find the components of the resultant force \( \vec{F_R} \).

2. **Moment (MR = MO):**
- Calculate the moments caused by the distributed forces about point O to find the total moment \( M_O \).

3. **Line of Action:**
- Determine the point where the resultant force intersects the line OB using the relationship between net force, moments, and distances from point O to give the precise location.

### Conclusion:

This analysis enables the understanding of how complex force interactions can be represented as a single resultant force and moment, simplifying the

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Replace the loading by an equivalent resultant force and couple moment acting at point 0. Solution Strategy:1. Break the distributed load into “primal” geometric shapes. This problem lends itself to analysis in 3 primal shapes.2. The areas below the load profiles will tell you the magnitudes of the equivalent concentrated loads.3. The position of the individual concentrated loads resulting from breaking up the given load profile will be located at the centroid of each primal.4. Once you have the equivalent concentrated loads and positions for the three primals you can find the resultant force and moment about ? in one of two ways:a. Find a single equivalent force and position (like you did for 11-W) and then relocate this single force to ?? using the techniques from topic 10.b. Use the techniques from topic 10 directly on the three equivalent concentrated loads you found in steps 2 & 3.