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3. The rigid rod ACDB is loaded by a uniform distributed load of intensity 3.5 kN/m and supported by a pin at A and a roller at B. Calculate the internal axial force F, the shear force V and the bending moment M at cross section D. 3.5kN/M 6 m 45% 3m

3. The rigid rod ACDB is loaded by a uniform distributed load of intensity 3.5 kN/m and supported by a pin at A and a roller at B. Calculate the internal axial force F, the shear force V and the bending moment M at cross section D. 3.5kN/M 6 m 45% 3m

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 Description **3. The Problem Statement:** The rigid rod \( ACDB \) is loaded by a uniform distributed load of intensity \( 3.5 \, \text{kN/m} \) and is supported by a pin at \( A \) and a roller at \( B \). The task is to calculate the internal axial force \( F \), the shear force \( V \), and the bending moment \( M \) at cross section \( D \). ## Diagram Explanation - **Diagram Layout:** - The rod \( ACDB \) forms a structure divided into three segments. - Segment \( AC \) is at an angle and measures \( 2 \, \text{m} \). - Segment \( CD \) runs horizontally for \( 6 \, \text{m} \). - Segment \( DB \) is horizontal, dividing into two \( 3 \, \text{m} \) parts. - **Support Points:** - Point \( A \) has a pin support allowing rotation but resisting translational forces in both vertical and horizontal directions. - Point \( B \) has a roller support allowing horizontal movement but not vertical. - **Load Application:** - A uniform distributed load of \( 3.5 \, \text{kN/m} \) acts over segment \( DB \). ## Objective The objective is to determine: 1. **Internal Axial Force \( F \):** - The internal axial force within the rod due to applied loading and reactions at supports. 2. **Shear Force \( V \):** - The internal shear force at cross section \( D \). 3. **Bending Moment \( M \):** - The moment at cross section \( D \) resulting from the applied load and reactions. In solving this problem, principles of static equilibrium, such as summation of forces and moments, will be applied to find the required internal forces and moments.
Transcribed Image Text:## Problem Description **3. The Problem Statement:** The rigid rod \( ACDB \) is loaded by a uniform distributed load of intensity \( 3.5 \, \text{kN/m} \) and is supported by a pin at \( A \) and a roller at \( B \). The task is to calculate the internal axial force \( F \), the shear force \( V \), and the bending moment \( M \) at cross section \( D \). ## Diagram Explanation - **Diagram Layout:** - The rod \( ACDB \) forms a structure divided into three segments. - Segment \( AC \) is at an angle and measures \( 2 \, \text{m} \). - Segment \( CD \) runs horizontally for \( 6 \, \text{m} \). - Segment \( DB \) is horizontal, dividing into two \( 3 \, \text{m} \) parts. - **Support Points:** - Point \( A \) has a pin support allowing rotation but resisting translational forces in both vertical and horizontal directions. - Point \( B \) has a roller support allowing horizontal movement but not vertical. - **Load Application:** - A uniform distributed load of \( 3.5 \, \text{kN/m} \) acts over segment \( DB \). ## Objective The objective is to determine: 1. **Internal Axial Force \( F \):** - The internal axial force within the rod due to applied loading and reactions at supports. 2. **Shear Force \( V \):** - The internal shear force at cross section \( D \). 3. **Bending Moment \( M \):** - The moment at cross section \( D \) resulting from the applied load and reactions. In solving this problem, principles of static equilibrium, such as summation of forces and moments, will be applied to find the required internal forces and moments.
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