•Download as PPTX, PDF•

0 likes•17 views

- The document presents the solution to a problem involving a beam with two point loads spaced 3m apart. It includes the influence line diagrams for shear force and bending moment. - At a section D located 6m from the left support, the maximum shear force is found to be 80kN when the 100kN load is on the section, and 160kN when the 200kN load is on the section. - The absolute maximum bending moment of 960kN-m occurs when the 200kN load is on the section. When the 100kN load is on the section, the bending moment is 720kN-m.

Report

Share

Report

Share

267579.ppt

The document discusses shear force and bending moment diagrams (SFD & BMD). It provides examples of calculating shear forces and bending moments at sections of beams under different loading conditions. Shear force is the sum of vertical forces acting on a section, while bending moment is the sum of moments acting on the beam. SFDs show how shear force varies along a beam's length, and BMDs show how bending moment varies. Points of contraflexure are where the bending moment changes sign in the BMD.

26-Sajid-Ahmed.pptx

The document discusses shear force and bending moment diagrams (SFD & BMD) for beams under different loading conditions. It provides examples of calculating reactions, shear forces, and bending moments at sections of beams with point loads and uniformly distributed loads. Shear forces are the sum of vertical forces on either side of a section, while bending moments are the sum of moments from forces on either side. SFDs show how shear force varies along a beam's length, while BMDs show how the bending moment varies. Points of contraflexure occur where the BMD changes sign.

shear force and bending moment diagram

The document discusses shear force and bending moment diagrams (SFD & BMD) for beams under different loading conditions. It provides examples of calculating reactions, shear forces, and bending moments at sections of simply supported and overhanging beams with point and distributed loads. Key steps include using the equations of equilibrium to find reactions, then determining shear forces and bending moments moving left to right along the beam. Points of contraflexure where the bending moment changes sign are also identified. Diagrams are drawn to illustrate the variation in shear force and bending moment.

SFD BMD.ppt

1. The document discusses shear force and bending moment diagrams (SFD & BMD) for beams under different loading conditions.
2. It provides an example of calculating the SFD and BMD for a simply supported beam with three point loads, finding the reactions, shear forces and bending moments at various sections.
3. A second example calculates the SFD and BMD for a double side overhanging beam with a udl and point loads, including locating points of contraflexure.

Shear force & bending moment.ppt

The document describes the calculation of shear forces and bending moments at different sections of beams. It provides an example problem where a simply supported beam is subjected to three point loads. The reactions, shear forces, and bending moments are calculated at various sections. Shear force and bending moment diagrams are drawn, which show a maximum bending moment of 29 Nm. Points of contraflexure, where the bending moment changes sign, are also identified.

Shear force and bending moment diagram

This document discusses shear force and bending moment diagrams (SFD & BMD) for beams under different loading conditions. It defines key terms like shear force, bending moment, sagging and hogging bending moments. It also describes the relationships between applied loads, shear forces and bending moments. Examples are provided to demonstrate how to draw SFDs and BMDs and calculate reactions, shear forces and bending moments at different sections of beams. Points of contraflexure, where the bending moment changes sign, are also identified.

AFD SFD BMD.ppt

The document discusses shear force and bending moments for a beam subjected to various loads. It provides an example of calculating the shear force and bending moment at a section of a simply supported beam under three point loads. The maximum bending moment is 29 Nm and occurs at section D. The maximum shear force magnitudes are 13.25 kN occurring at supports A and B. Points of contraflexure where the bending moment changes sign are also located.

Sfd bmd

The document discusses shear force and bending moments in beams. It provides examples of calculating the shear force and bending moment at a section for different loading conditions. Shear force is the sum of vertical forces to the left or right of a section, while bending moment is the sum of moments from forces left or right of the section. Shear force tries to shear a section, while bending moment bends it. Positive and negative signs are used to indicate the type of bending based on curvature.

267579.ppt

The document discusses shear force and bending moment diagrams (SFD & BMD). It provides examples of calculating shear forces and bending moments at sections of beams under different loading conditions. Shear force is the sum of vertical forces acting on a section, while bending moment is the sum of moments acting on the beam. SFDs show how shear force varies along a beam's length, and BMDs show how bending moment varies. Points of contraflexure are where the bending moment changes sign in the BMD.

26-Sajid-Ahmed.pptx

The document discusses shear force and bending moment diagrams (SFD & BMD) for beams under different loading conditions. It provides examples of calculating reactions, shear forces, and bending moments at sections of beams with point loads and uniformly distributed loads. Shear forces are the sum of vertical forces on either side of a section, while bending moments are the sum of moments from forces on either side. SFDs show how shear force varies along a beam's length, while BMDs show how the bending moment varies. Points of contraflexure occur where the BMD changes sign.

shear force and bending moment diagram

The document discusses shear force and bending moment diagrams (SFD & BMD) for beams under different loading conditions. It provides examples of calculating reactions, shear forces, and bending moments at sections of simply supported and overhanging beams with point and distributed loads. Key steps include using the equations of equilibrium to find reactions, then determining shear forces and bending moments moving left to right along the beam. Points of contraflexure where the bending moment changes sign are also identified. Diagrams are drawn to illustrate the variation in shear force and bending moment.

SFD BMD.ppt

1. The document discusses shear force and bending moment diagrams (SFD & BMD) for beams under different loading conditions.
2. It provides an example of calculating the SFD and BMD for a simply supported beam with three point loads, finding the reactions, shear forces and bending moments at various sections.
3. A second example calculates the SFD and BMD for a double side overhanging beam with a udl and point loads, including locating points of contraflexure.

Shear force & bending moment.ppt

The document describes the calculation of shear forces and bending moments at different sections of beams. It provides an example problem where a simply supported beam is subjected to three point loads. The reactions, shear forces, and bending moments are calculated at various sections. Shear force and bending moment diagrams are drawn, which show a maximum bending moment of 29 Nm. Points of contraflexure, where the bending moment changes sign, are also identified.

Shear force and bending moment diagram

This document discusses shear force and bending moment diagrams (SFD & BMD) for beams under different loading conditions. It defines key terms like shear force, bending moment, sagging and hogging bending moments. It also describes the relationships between applied loads, shear forces and bending moments. Examples are provided to demonstrate how to draw SFDs and BMDs and calculate reactions, shear forces and bending moments at different sections of beams. Points of contraflexure, where the bending moment changes sign, are also identified.

AFD SFD BMD.ppt

The document discusses shear force and bending moments for a beam subjected to various loads. It provides an example of calculating the shear force and bending moment at a section of a simply supported beam under three point loads. The maximum bending moment is 29 Nm and occurs at section D. The maximum shear force magnitudes are 13.25 kN occurring at supports A and B. Points of contraflexure where the bending moment changes sign are also located.

Sfd bmd

The document discusses shear force and bending moments in beams. It provides examples of calculating the shear force and bending moment at a section for different loading conditions. Shear force is the sum of vertical forces to the left or right of a section, while bending moment is the sum of moments from forces left or right of the section. Shear force tries to shear a section, while bending moment bends it. Positive and negative signs are used to indicate the type of bending based on curvature.

str.%20analysis.pptx

This document discusses shear force and bending moment calculations for beams subjected to different loading conditions. It provides examples of calculating the reactions, shear forces and bending moments at sections of beams. For one example beam, the maximum bending moment is found to occur at the section where the shear force is zero, located a distance of 2.24m from the left support. Points of contraflexure, where the bending moment changes sign, are also identified for some examples.

SFD & BMD Shear Force & Bending Moment Diagram

The document discusses shear force and bending moment in beams. It defines key terms like beam, transverse load, shear force, bending moment, and types of loads, supports and beams. It explains how to calculate and draw shear force and bending moment diagrams for different types of loads on beams including point loads, uniformly distributed loads, uniformly varying loads, and loads producing couples or overhangs. Sign conventions and the effect of reactions, loads and geometry on the shear force and bending moment diagrams are also covered.

Shear Force and Bending Moment Diagram

1. The document discusses shear force and bending moment diagrams. It defines shear force as a force that causes sliding, and bending moment as a force that causes rotation.
2. It provides an example of calculating the shear force and bending moment at a section for a simply supported beam with three point loads. The maximum shear force is 13.2 kN and the maximum bending moment is 39.2 kN-m.
3. The key steps to draw shear force and bending moment diagrams are outlined as calculating reactions, shear forces at sections, bending moments at sections, and then plotting the diagrams.

Mechanics.ppt

- The document discusses beams, shear force, and bending moment diagrams. It provides examples of calculating reactions, shear forces, and bending moments at different points along a beam under various loading conditions.
- Key steps include drawing free body diagrams, applying equations of equilibrium, and deriving equations to determine shear force and bending moment as a function of distance along the beam.
- Shear force and bending moment diagrams can then be plotted by inputting values of distance into the equations to show variations in shear force and bending moment along the length of the beam.

Mechanics.

- The document discusses beams, shear force, and bending moment diagrams. It provides examples of calculating reactions, shear forces, and bending moments at different points along a beam under various loading conditions.
- Key steps include drawing free body diagrams, applying equations of equilibrium, and deriving equations to determine shear force and bending moment as a function of distance along the beam.
- Shear force and bending moment diagrams can then be plotted by inputting values of distance into the equations to show variations in shear force and bending moment along the length of the beam.

SFD & BMD

In this presentation you will get knowledge about shear force and bending moment diagram and this topic very useful for civil as well as mechanical engineering department students.

unit 1 part 2 mechanics as AKTU syllabus first yr 2021

The document discusses beams, shear forces, bending moments, and provides examples of calculating shear force diagrams (SFD) and bending moment diagrams (BMD) for beams under different loading conditions. Key points:
- A beam is a structural element that is capable of withstanding load primarily by resisting bending.
- Shear force is the sum of all vertical forces acting on a beam section. Bending moment is the sum of moments of all forces acting on the beam section.
- SFD shows the variation of shear force along the beam length. BMD shows the variation of bending moment.
- Examples demonstrate how to calculate reactions, draw SFDs, and BMDs for beams with various

Shear Force and Bending moment Diagram

The document discusses shear force and bending moment in beams. It defines key terms like shear force, bending moment, and types of loads, supports and beams. It provides examples of different loading conditions and how to calculate and draw the shear force and bending moment diagrams for beams subjected to point loads, uniformly distributed loads, uniformly varying loads, couples and overhanging beams. The diagrams show the variations in shear force and bending moment, including locations of maximum and points of contraflexure where bending moment changes sign.

Shear and moment diagram

This document discusses shear and moment diagrams for beams. It provides examples of calculating reactions, shear, and bending moment at different points along simple beams subjected to various load cases including point loads and uniform loads. Key steps include determining reactions, drawing the shear diagram, and using the shear diagram to draw the moment diagram. The maximum shear and bending moment as well as points where shear and moment equal zero are identified. Overhanging ends can cause negative bending moments. Understanding shear and moment diagrams is important for beam design.

Mechanics of materials lecture 02 (nadim sir)

This document discusses shear force and bending moment in structural members. It defines shear force, normal force, and bending moment as the internal forces that develop in a beam due to applied loads. It presents methods for determining the shear force diagram and bending moment diagram of a beam based on the slope relationships between load, shear force, and bending moment. Several examples are worked through to demonstrate how to calculate and draw the shear force diagram and bending moment diagram for beams under different loading conditions.

Stucture Design-I(Bending moment and Shear force)

* Given: Maximum bending moment the beam can resist = 22 kNm
* Span of beam = 5 + 5 + 4 = 14 m
* Point loads = 1 + 2 + 1 = 4 kN
* To find: Maximum uniform load (w) the beam can carry
* Bending moment at center due to point loads
= (1 × 2.5) + (2 × 2.5) + (1 × 1.5) = 7.5 kNm
* Bending moment at center due to uniform load
= wl^2/8 = w × (14)^2/8 = w × 98 kNm
* Total bending moment should not exceed 22 kNm
7.5 + w

Design ppt

The document presents the design of a post-tensioned prestressed concrete tee beam and slab bridge deck. Key details include:
- The bridge will have an effective span of 30m and width of 7.5m with 600mm kerbs and 1.5m footpaths on each side.
- The project team will design the bridge to meet Class AA loading standards for a national highway.
- The bridge will have 4 main girders spaced at 2.5m intervals with a 250mm thick deck slab cast between them.
- The document outlines the design process for the interior slab panel, longitudinal girders, and calculation of design moments and shear forces. Properties of the main girder cross

10.01.03.029

1. The document discusses bending moment diagrams (BMD) which show the variation of bending moment along the length of a beam caused by applied loads.
2. It provides procedures for drawing shear force diagrams (SFD) and BMD for beams under different loading conditions such as concentrated loads, uniform distributed loads, and combinations of loads.
3. Key relationships discussed include: the slope of the shear diagram equaling the distributed load; the slope of the moment diagram equaling the shear; and the change in moment between two points equaling the area under the shear diagram between those points.

Strength example 1 5

The document contains solutions to examples from chapters in a strength of materials textbook, solving for things like beam reactions, stresses, deflections, shear and moment diagrams. It analyzes beams, braces, and steel strips under different loads and conditions to calculate values like forces, stresses, moments and radii of curvature. Diagrams are included to illustrate the shear and moment diagrams for one of the beam examples.

Sfd and bmd engineering science n4

1) The document discusses shear force diagrams and bending moment diagrams for loaded beams. It provides equations that govern loaded beams and defines clockwise and anticlockwise moments.
2) An example beam with different loads is used to calculate support reactions, draw the shear force diagram, and bending moment diagram. The shear force diagram shows vertical forces plotted against distance along the beam.
3) The bending moment diagram is created by calculating the area under the shear force diagram section by section, and plots bending moment against distance along the beam. This shows how bending moment changes along the beam.

Lecture--15, 16 (Deflection of beams).pptx

nbbnjhgvbbbbbbbbbbbbbbbbbbbbnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggg

Shear Force Diagram and its exampls

it contains the basic information about the shear force diagram which is the part of the Mechanics of solid. there many numerical solved and whivh will give you detaild idea in S.f.d.

Lecture-3-1.pptx

- Saint-Venant's principle states that the stress and strain distribution on a cross-section of a loaded material will be independent of the applied load if the cross-section is located away from the point of load application.
- The principle of superposition allows breaking down structures into individual load cases and adding their effects to determine the total stress, strain, or deflection.
- Statically indeterminate structures require additional compatibility equations relating deformations to solve for member forces.

deflection 4.ppt

- This chapter discusses the deflection of beams under lateral loading. Deflection is the displacement of points on the beam's axis from its original, unloaded position.
- The basic differential equation relates the bending moment to the second derivative of the deflection curve. This equation can be directly integrated or used with methods like Macaulay's to determine deflections.
- Macaulay's method involves integrating the loading equation rather than the bending moment equation directly. It is useful when the bending moment is difficult to obtain.
- Special cases like loads not starting from the beginning or ending at the beam's end are also discussed.

Deflection of Beams _Chapter 5_2019_01_17!10_37_41_PM.ppt

- This chapter discusses the deflection of beams under lateral loads. Deflection is the displacement of points on the beam's axis from its original, unloaded position.
- The basic differential equation relates the bending moment to the second derivative of the deflection curve. It can be directly integrated or used with methods like Macaulay's to determine deflections under different load cases.
- Macaulay's method allows integrating the loading function directly instead of first obtaining the bending moment. It is useful for concentrated or discontinuous loads.
- Special cases with loads not starting or ending at the beam's ends require modifying the integration limits.

Literature Review Basics and Understanding Reference Management.pptx

Three-day training on academic research focuses on analytical tools at United Technical College, supported by the University Grant Commission, Nepal. 24-26 May 2024

str.%20analysis.pptx

This document discusses shear force and bending moment calculations for beams subjected to different loading conditions. It provides examples of calculating the reactions, shear forces and bending moments at sections of beams. For one example beam, the maximum bending moment is found to occur at the section where the shear force is zero, located a distance of 2.24m from the left support. Points of contraflexure, where the bending moment changes sign, are also identified for some examples.

SFD & BMD Shear Force & Bending Moment Diagram

The document discusses shear force and bending moment in beams. It defines key terms like beam, transverse load, shear force, bending moment, and types of loads, supports and beams. It explains how to calculate and draw shear force and bending moment diagrams for different types of loads on beams including point loads, uniformly distributed loads, uniformly varying loads, and loads producing couples or overhangs. Sign conventions and the effect of reactions, loads and geometry on the shear force and bending moment diagrams are also covered.

Shear Force and Bending Moment Diagram

1. The document discusses shear force and bending moment diagrams. It defines shear force as a force that causes sliding, and bending moment as a force that causes rotation.
2. It provides an example of calculating the shear force and bending moment at a section for a simply supported beam with three point loads. The maximum shear force is 13.2 kN and the maximum bending moment is 39.2 kN-m.
3. The key steps to draw shear force and bending moment diagrams are outlined as calculating reactions, shear forces at sections, bending moments at sections, and then plotting the diagrams.

Mechanics.ppt

- The document discusses beams, shear force, and bending moment diagrams. It provides examples of calculating reactions, shear forces, and bending moments at different points along a beam under various loading conditions.
- Key steps include drawing free body diagrams, applying equations of equilibrium, and deriving equations to determine shear force and bending moment as a function of distance along the beam.
- Shear force and bending moment diagrams can then be plotted by inputting values of distance into the equations to show variations in shear force and bending moment along the length of the beam.

Mechanics. - The document discusses beams, shear force, and bending moment diagrams. It provides examples of calculating reactions, shear forces, and bending moments at different points along a beam under various loading conditions.
- Key steps include drawing free body diagrams, applying equations of equilibrium, and deriving equations to determine shear force and bending moment as a function of distance along the beam.
- Shear force and bending moment diagrams can then be plotted by inputting values of distance into the equations to show variations in shear force and bending moment along the length of the beam.

SFD & BMD

In this presentation you will get knowledge about shear force and bending moment diagram and this topic very useful for civil as well as mechanical engineering department students.

unit 1 part 2 mechanics as AKTU syllabus first yr 2021

The document discusses beams, shear forces, bending moments, and provides examples of calculating shear force diagrams (SFD) and bending moment diagrams (BMD) for beams under different loading conditions. Key points:
- A beam is a structural element that is capable of withstanding load primarily by resisting bending.
- Shear force is the sum of all vertical forces acting on a beam section. Bending moment is the sum of moments of all forces acting on the beam section.
- SFD shows the variation of shear force along the beam length. BMD shows the variation of bending moment.
- Examples demonstrate how to calculate reactions, draw SFDs, and BMDs for beams with various

Shear Force and Bending moment Diagram

The document discusses shear force and bending moment in beams. It defines key terms like shear force, bending moment, and types of loads, supports and beams. It provides examples of different loading conditions and how to calculate and draw the shear force and bending moment diagrams for beams subjected to point loads, uniformly distributed loads, uniformly varying loads, couples and overhanging beams. The diagrams show the variations in shear force and bending moment, including locations of maximum and points of contraflexure where bending moment changes sign.

Shear and moment diagram

This document discusses shear and moment diagrams for beams. It provides examples of calculating reactions, shear, and bending moment at different points along simple beams subjected to various load cases including point loads and uniform loads. Key steps include determining reactions, drawing the shear diagram, and using the shear diagram to draw the moment diagram. The maximum shear and bending moment as well as points where shear and moment equal zero are identified. Overhanging ends can cause negative bending moments. Understanding shear and moment diagrams is important for beam design.

Mechanics of materials lecture 02 (nadim sir)

This document discusses shear force and bending moment in structural members. It defines shear force, normal force, and bending moment as the internal forces that develop in a beam due to applied loads. It presents methods for determining the shear force diagram and bending moment diagram of a beam based on the slope relationships between load, shear force, and bending moment. Several examples are worked through to demonstrate how to calculate and draw the shear force diagram and bending moment diagram for beams under different loading conditions.

Stucture Design-I(Bending moment and Shear force)

* Given: Maximum bending moment the beam can resist = 22 kNm
* Span of beam = 5 + 5 + 4 = 14 m
* Point loads = 1 + 2 + 1 = 4 kN
* To find: Maximum uniform load (w) the beam can carry
* Bending moment at center due to point loads
= (1 × 2.5) + (2 × 2.5) + (1 × 1.5) = 7.5 kNm
* Bending moment at center due to uniform load
= wl^2/8 = w × (14)^2/8 = w × 98 kNm
* Total bending moment should not exceed 22 kNm
7.5 + w

Design ppt

The document presents the design of a post-tensioned prestressed concrete tee beam and slab bridge deck. Key details include:
- The bridge will have an effective span of 30m and width of 7.5m with 600mm kerbs and 1.5m footpaths on each side.
- The project team will design the bridge to meet Class AA loading standards for a national highway.
- The bridge will have 4 main girders spaced at 2.5m intervals with a 250mm thick deck slab cast between them.
- The document outlines the design process for the interior slab panel, longitudinal girders, and calculation of design moments and shear forces. Properties of the main girder cross

10.01.03.029

1. The document discusses bending moment diagrams (BMD) which show the variation of bending moment along the length of a beam caused by applied loads.
2. It provides procedures for drawing shear force diagrams (SFD) and BMD for beams under different loading conditions such as concentrated loads, uniform distributed loads, and combinations of loads.
3. Key relationships discussed include: the slope of the shear diagram equaling the distributed load; the slope of the moment diagram equaling the shear; and the change in moment between two points equaling the area under the shear diagram between those points.

Strength example 1 5

The document contains solutions to examples from chapters in a strength of materials textbook, solving for things like beam reactions, stresses, deflections, shear and moment diagrams. It analyzes beams, braces, and steel strips under different loads and conditions to calculate values like forces, stresses, moments and radii of curvature. Diagrams are included to illustrate the shear and moment diagrams for one of the beam examples.

Sfd and bmd engineering science n4

1) The document discusses shear force diagrams and bending moment diagrams for loaded beams. It provides equations that govern loaded beams and defines clockwise and anticlockwise moments.
2) An example beam with different loads is used to calculate support reactions, draw the shear force diagram, and bending moment diagram. The shear force diagram shows vertical forces plotted against distance along the beam.
3) The bending moment diagram is created by calculating the area under the shear force diagram section by section, and plots bending moment against distance along the beam. This shows how bending moment changes along the beam.

Lecture--15, 16 (Deflection of beams).pptx

nbbnjhgvbbbbbbbbbbbbbbbbbbbbnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggg

Shear Force Diagram and its exampls

it contains the basic information about the shear force diagram which is the part of the Mechanics of solid. there many numerical solved and whivh will give you detaild idea in S.f.d.

Lecture-3-1.pptx

- Saint-Venant's principle states that the stress and strain distribution on a cross-section of a loaded material will be independent of the applied load if the cross-section is located away from the point of load application.
- The principle of superposition allows breaking down structures into individual load cases and adding their effects to determine the total stress, strain, or deflection.
- Statically indeterminate structures require additional compatibility equations relating deformations to solve for member forces.

deflection 4.ppt

- This chapter discusses the deflection of beams under lateral loading. Deflection is the displacement of points on the beam's axis from its original, unloaded position.
- The basic differential equation relates the bending moment to the second derivative of the deflection curve. This equation can be directly integrated or used with methods like Macaulay's to determine deflections.
- Macaulay's method involves integrating the loading equation rather than the bending moment equation directly. It is useful when the bending moment is difficult to obtain.
- Special cases like loads not starting from the beginning or ending at the beam's end are also discussed.

Deflection of Beams _Chapter 5_2019_01_17!10_37_41_PM.ppt

- This chapter discusses the deflection of beams under lateral loads. Deflection is the displacement of points on the beam's axis from its original, unloaded position.
- The basic differential equation relates the bending moment to the second derivative of the deflection curve. It can be directly integrated or used with methods like Macaulay's to determine deflections under different load cases.
- Macaulay's method allows integrating the loading function directly instead of first obtaining the bending moment. It is useful for concentrated or discontinuous loads.
- Special cases with loads not starting or ending at the beam's ends require modifying the integration limits.

str.%20analysis.pptx

str.%20analysis.pptx

SFD & BMD Shear Force & Bending Moment Diagram

SFD & BMD Shear Force & Bending Moment Diagram

Shear Force and Bending Moment Diagram

Shear Force and Bending Moment Diagram

Mechanics.ppt

Mechanics.ppt

Mechanics.

Mechanics.

SFD & BMD

SFD & BMD

unit 1 part 2 mechanics as AKTU syllabus first yr 2021

unit 1 part 2 mechanics as AKTU syllabus first yr 2021

Shear Force and Bending moment Diagram

Shear Force and Bending moment Diagram

Shear and moment diagram

Shear and moment diagram

Mechanics of materials lecture 02 (nadim sir)

Mechanics of materials lecture 02 (nadim sir)

Stucture Design-I(Bending moment and Shear force)

Stucture Design-I(Bending moment and Shear force)

Design ppt

Design ppt

10.01.03.029

10.01.03.029

Strength example 1 5

Strength example 1 5

Sfd and bmd engineering science n4

Sfd and bmd engineering science n4

Lecture--15, 16 (Deflection of beams).pptx

Lecture--15, 16 (Deflection of beams).pptx

Shear Force Diagram and its exampls

Shear Force Diagram and its exampls

Lecture-3-1.pptx

Lecture-3-1.pptx

deflection 4.ppt

deflection 4.ppt

Deflection of Beams _Chapter 5_2019_01_17!10_37_41_PM.ppt

Deflection of Beams _Chapter 5_2019_01_17!10_37_41_PM.ppt

Literature Review Basics and Understanding Reference Management.pptx

Three-day training on academic research focuses on analytical tools at United Technical College, supported by the University Grant Commission, Nepal. 24-26 May 2024

BPV-GUI-01-Guide-for-ASME-Review-Teams-(General)-10-10-2023.pdf

Guia para el codigo ASME

Design and Analysis of Algorithms-DP,Backtracking,Graphs,B&B

Dynamic Programming
Backtracking
Techniques for Graphs
Branch and Bound

原版制作(unimelb毕业证书)墨尔本大学毕业证Offer一模一样

学校原件一模一样【微信：741003700 】《(unimelb毕业证书)墨尔本大学毕业证》【微信：741003700 】学位证，留信认证（真实可查，永久存档）原件一模一样纸张工艺/offer、雅思、外壳等材料/诚信可靠,可直接看成品样本，帮您解决无法毕业带来的各种难题！外壳，原版制作，诚信可靠，可直接看成品样本。行业标杆！精益求精，诚心合作，真诚制作！多年品质 ,按需精细制作，24小时接单,全套进口原装设备。十五年致力于帮助留学生解决难题，包您满意。
本公司拥有海外各大学样板无数，能完美还原。
1:1完美还原海外各大学毕业材料上的工艺：水印，阴影底纹，钢印LOGO烫金烫银，LOGO烫金烫银复合重叠。文字图案浮雕、激光镭射、紫外荧光、温感、复印防伪等防伪工艺。材料咨询办理、认证咨询办理请加学历顾问Q/微741003700
【主营项目】
一.毕业证【q微741003700】成绩单、使馆认证、教育部认证、雅思托福成绩单、学生卡等！
二.真实使馆公证(即留学回国人员证明,不成功不收费)
三.真实教育部学历学位认证（教育部存档！教育部留服网站永久可查）
四.办理各国各大学文凭(一对一专业服务,可全程监控跟踪进度)
如果您处于以下几种情况：
◇在校期间，因各种原因未能顺利毕业……拿不到官方毕业证【q/微741003700】
◇面对父母的压力，希望尽快拿到；
◇不清楚认证流程以及材料该如何准备；
◇回国时间很长，忘记办理；
◇回国马上就要找工作，办给用人单位看；
◇企事业单位必须要求办理的
◇需要报考公务员、购买免税车、落转户口
◇申请留学生创业基金
留信网认证的作用:
1:该专业认证可证明留学生真实身份
2:同时对留学生所学专业登记给予评定
3:国家专业人才认证中心颁发入库证书
4:这个认证书并且可以归档倒地方
5:凡事获得留信网入网的信息将会逐步更新到个人身份内，将在公安局网内查询个人身份证信息后，同步读取人才网入库信息
6:个人职称评审加20分
7:个人信誉贷款加10分
8:在国家人才网主办的国家网络招聘大会中纳入资料，供国家高端企业选择人才

22CYT12-Unit-V-E Waste and its Management.ppt

Introduction- e - waste – definition - sources of e-waste– hazardous substances in e-waste - effects of e-waste on environment and human health- need for e-waste management– e-waste handling rules - waste minimization techniques for managing e-waste – recycling of e-waste - disposal treatment methods of e- waste – mechanism of extraction of precious metal from leaching solution-global Scenario of E-waste – E-waste in India- case studies.

KuberTENes Birthday Bash Guadalajara - K8sGPT first impressions

K8sGPT is a tool that analyzes and diagnoses Kubernetes clusters. This presentation was used to share the requirements and dependencies to deploy K8sGPT in a local environment.

Generative AI leverages algorithms to create various forms of content

What is Generative AI?

introduction to solar energy for engineering.pdf

solar energy

一比一原版(IIT毕业证)伊利诺伊理工大学毕业证成绩单专业办理

IIT毕业证原版定制【微信：176555708】【伊利诺伊理工大学毕业证成绩单-学位证】【微信：176555708】（留信学历认证永久存档查询）采用学校原版纸张、特殊工艺完全按照原版一比一制作（包括：隐形水印，阴影底纹，钢印LOGO烫金烫银，LOGO烫金烫银复合重叠，文字图案浮雕，激光镭射，紫外荧光，温感，复印防伪）行业标杆！精益求精，诚心合作，真诚制作！多年品质 ,按需精细制作，24小时接单,全套进口原装设备，十五年致力于帮助留学生解决难题，业务范围有加拿大、英国、澳洲、韩国、美国、新加坡，新西兰等学历材料，包您满意。
◆◆◆◆◆ — — — — — — — — 【留学教育】留学归国服务中心 — — — — — -◆◆◆◆◆
【主营项目】
一.毕业证【微信：176555708】成绩单、使馆认证、教育部认证、雅思托福成绩单、学生卡等！
二.真实使馆公证(即留学回国人员证明,不成功不收费)
三.真实教育部学历学位认证（教育部存档！教育部留服网站永久可查）
四.办理各国各大学文凭(一对一专业服务,可全程监控跟踪进度)
如果您处于以下几种情况：
◇在校期间，因各种原因未能顺利毕业……拿不到官方毕业证【微信：176555708】
◇面对父母的压力，希望尽快拿到；
◇不清楚认证流程以及材料该如何准备；
◇回国时间很长，忘记办理；
◇回国马上就要找工作，办给用人单位看；
◇企事业单位必须要求办理的
◇需要报考公务员、购买免税车、落转户口
◇申请留学生创业基金
留信网认证的作用:
1:该专业认证可证明留学生真实身份
2:同时对留学生所学专业登记给予评定
3:国家专业人才认证中心颁发入库证书
4:这个认证书并且可以归档倒地方
5:凡事获得留信网入网的信息将会逐步更新到个人身份内，将在公安局网内查询个人身份证信息后，同步读取人才网入库信息
6:个人职称评审加20分
7:个人信誉贷款加10分→ 【关于价格问题（保证一手价格）
我们所定的价格是非常合理的，而且我们现在做得单子大多数都是代理和回头客户介绍的所以一般现在有新的单子 我给客户的都是第一手的代理价格，因为我想坦诚对待大家 不想跟大家在价格方面浪费时间
对于老客户或者被老客户介绍过来的朋友，我们都会适当给一些优惠。
8:在国家人才网主办的国家网络招聘大会中纳入资料，供国家高端企业选择人才
选择实体注册公司办理，更放心，更安全！我们的承诺：可来公司面谈，可签订合同，会陪同客户一起到教育部认证窗口递交认证材料，客户在教育部官方认证查询网站查询到认证通过结果后付款，不成功不收费！
学历顾问：微信：176555708

5214-1693458878915-Unit 6 2023 to 2024 academic year assignment (AutoRecovere...

Bigdata of technology

一比一原版(UMich毕业证)密歇根大学|安娜堡分校毕业证成绩单专业办理

UMich毕业证原版定制【微信：176555708】【密歇根大学|安娜堡分校毕业证成绩单-学位证】【微信：176555708】（留信学历认证永久存档查询）采用学校原版纸张、特殊工艺完全按照原版一比一制作（包括：隐形水印，阴影底纹，钢印LOGO烫金烫银，LOGO烫金烫银复合重叠，文字图案浮雕，激光镭射，紫外荧光，温感，复印防伪）行业标杆！精益求精，诚心合作，真诚制作！多年品质 ,按需精细制作，24小时接单,全套进口原装设备，十五年致力于帮助留学生解决难题，业务范围有加拿大、英国、澳洲、韩国、美国、新加坡，新西兰等学历材料，包您满意。
◆◆◆◆◆ — — — — — — — — 【留学教育】留学归国服务中心 — — — — — -◆◆◆◆◆
【主营项目】
一.毕业证【微信：176555708】成绩单、使馆认证、教育部认证、雅思托福成绩单、学生卡等！
二.真实使馆公证(即留学回国人员证明,不成功不收费)
三.真实教育部学历学位认证（教育部存档！教育部留服网站永久可查）
四.办理各国各大学文凭(一对一专业服务,可全程监控跟踪进度)
如果您处于以下几种情况：
◇在校期间，因各种原因未能顺利毕业……拿不到官方毕业证【微信：176555708】
◇面对父母的压力，希望尽快拿到；
◇不清楚认证流程以及材料该如何准备；
◇回国时间很长，忘记办理；
◇回国马上就要找工作，办给用人单位看；
◇企事业单位必须要求办理的
◇需要报考公务员、购买免税车、落转户口
◇申请留学生创业基金
留信网认证的作用:
1:该专业认证可证明留学生真实身份
2:同时对留学生所学专业登记给予评定
3:国家专业人才认证中心颁发入库证书
4:这个认证书并且可以归档倒地方
5:凡事获得留信网入网的信息将会逐步更新到个人身份内，将在公安局网内查询个人身份证信息后，同步读取人才网入库信息
6:个人职称评审加20分
7:个人信誉贷款加10分→ 【关于价格问题（保证一手价格）
我们所定的价格是非常合理的，而且我们现在做得单子大多数都是代理和回头客户介绍的所以一般现在有新的单子 我给客户的都是第一手的代理价格，因为我想坦诚对待大家 不想跟大家在价格方面浪费时间
对于老客户或者被老客户介绍过来的朋友，我们都会适当给一些优惠。
8:在国家人才网主办的国家网络招聘大会中纳入资料，供国家高端企业选择人才
选择实体注册公司办理，更放心，更安全！我们的承诺：可来公司面谈，可签订合同，会陪同客户一起到教育部认证窗口递交认证材料，客户在教育部官方认证查询网站查询到认证通过结果后付款，不成功不收费！
学历顾问：微信：176555708

DfMAy 2024 - key insights and contributions

We have compiled the most important slides from each speaker's presentation. This year’s compilation, available for free, captures the key insights and contributions shared during the DfMAy 2024 conference.

DEEP LEARNING FOR SMART GRID INTRUSION DETECTION: A HYBRID CNN-LSTM-BASED MODEL

As digital technology becomes more deeply embedded in power systems, protecting the communication
networks of Smart Grids (SG) has emerged as a critical concern. Distributed Network Protocol 3 (DNP3)
represents a multi-tiered application layer protocol extensively utilized in Supervisory Control and Data
Acquisition (SCADA)-based smart grids to facilitate real-time data gathering and control functionalities.
Robust Intrusion Detection Systems (IDS) are necessary for early threat detection and mitigation because
of the interconnection of these networks, which makes them vulnerable to a variety of cyberattacks. To
solve this issue, this paper develops a hybrid Deep Learning (DL) model specifically designed for intrusion
detection in smart grids. The proposed approach is a combination of the Convolutional Neural Network
(CNN) and the Long-Short-Term Memory algorithms (LSTM). We employed a recent intrusion detection
dataset (DNP3), which focuses on unauthorized commands and Denial of Service (DoS) cyberattacks, to
train and test our model. The results of our experiments show that our CNN-LSTM method is much better
at finding smart grid intrusions than other deep learning algorithms used for classification. In addition,
our proposed approach improves accuracy, precision, recall, and F1 score, achieving a high detection
accuracy rate of 99.50%.

Exception Handling notes in java exception

Java Exception handling

Advanced control scheme of doubly fed induction generator for wind turbine us...

This paper describes a speed control device for generating electrical energy on an electricity network based on the doubly fed induction generator (DFIG) used for wind power conversion systems. At first, a double-fed induction generator model was constructed. A control law is formulated to govern the flow of energy between the stator of a DFIG and the energy network using three types of controllers: proportional integral (PI), sliding mode controller (SMC) and second order sliding mode controller (SOSMC). Their different results in terms of power reference tracking, reaction to unexpected speed fluctuations, sensitivity to perturbations, and resilience against machine parameter alterations are compared. MATLAB/Simulink was used to conduct the simulations for the preceding study. Multiple simulations have shown very satisfying results, and the investigations demonstrate the efficacy and power-enhancing capabilities of the suggested control system.

RAT: Retrieval Augmented Thoughts Elicit Context-Aware Reasoning in Long-Hori...

RAT: Retrieval Augmented Thoughts Elicit Context-Aware Reasoning in Long-Horizon Generation

Harnessing WebAssembly for Real-time Stateless Streaming Pipelines

Traditionally, dealing with real-time data pipelines has involved significant overhead, even for straightforward tasks like data transformation or masking. However, in this talk, we’ll venture into the dynamic realm of WebAssembly (WASM) and discover how it can revolutionize the creation of stateless streaming pipelines within a Kafka (Redpanda) broker. These pipelines are adept at managing low-latency, high-data-volume scenarios.

Building Electrical System Design & Installation

Guide for Building Electrical System Design & Installation

14 Template Contractual Notice - EOT Application

EOT Application, Contractual Notice

Literature Review Basics and Understanding Reference Management.pptx

Literature Review Basics and Understanding Reference Management.pptx

Self-Control of Emotions by Slidesgo.pptx

Self-Control of Emotions by Slidesgo.pptx

BPV-GUI-01-Guide-for-ASME-Review-Teams-(General)-10-10-2023.pdf

BPV-GUI-01-Guide-for-ASME-Review-Teams-(General)-10-10-2023.pdf

Design and Analysis of Algorithms-DP,Backtracking,Graphs,B&B

Design and Analysis of Algorithms-DP,Backtracking,Graphs,B&B

原版制作(unimelb毕业证书)墨尔本大学毕业证Offer一模一样

原版制作(unimelb毕业证书)墨尔本大学毕业证Offer一模一样

22CYT12-Unit-V-E Waste and its Management.ppt

22CYT12-Unit-V-E Waste and its Management.ppt

KuberTENes Birthday Bash Guadalajara - K8sGPT first impressions

KuberTENes Birthday Bash Guadalajara - K8sGPT first impressions

Generative AI leverages algorithms to create various forms of content

Generative AI leverages algorithms to create various forms of content

introduction to solar energy for engineering.pdf

introduction to solar energy for engineering.pdf

一比一原版(IIT毕业证)伊利诺伊理工大学毕业证成绩单专业办理

一比一原版(IIT毕业证)伊利诺伊理工大学毕业证成绩单专业办理

5214-1693458878915-Unit 6 2023 to 2024 academic year assignment (AutoRecovere...

5214-1693458878915-Unit 6 2023 to 2024 academic year assignment (AutoRecovere...

一比一原版(UMich毕业证)密歇根大学|安娜堡分校毕业证成绩单专业办理

一比一原版(UMich毕业证)密歇根大学|安娜堡分校毕业证成绩单专业办理

DfMAy 2024 - key insights and contributions

DfMAy 2024 - key insights and contributions

DEEP LEARNING FOR SMART GRID INTRUSION DETECTION: A HYBRID CNN-LSTM-BASED MODEL

DEEP LEARNING FOR SMART GRID INTRUSION DETECTION: A HYBRID CNN-LSTM-BASED MODEL

Exception Handling notes in java exception

Exception Handling notes in java exception

Advanced control scheme of doubly fed induction generator for wind turbine us...

Advanced control scheme of doubly fed induction generator for wind turbine us...

RAT: Retrieval Augmented Thoughts Elicit Context-Aware Reasoning in Long-Hori...

RAT: Retrieval Augmented Thoughts Elicit Context-Aware Reasoning in Long-Hori...

Harnessing WebAssembly for Real-time Stateless Streaming Pipelines

Harnessing WebAssembly for Real-time Stateless Streaming Pipelines

Building Electrical System Design & Installation

Building Electrical System Design & Installation

14 Template Contractual Notice - EOT Application

14 Template Contractual Notice - EOT Application

- 1. Moving Loads and Influence Line Diagram Beam Mrs. S. M. Madhale Assistant professor Civil Engineering Department ATS’s SBGI Miraj
- 2. Solved Problem on BEAM ILD BY: Mrs. S. M. Madhale 2
- 3. • A Simple beam AB having span12m. And a 20Kn load is act at its center. Calculate the Reaction at support, maximum shear force and bending moment at a section of 3m from left support A. ILD BY: Mrs. S. M. Madhale 3
- 4. Reaction at support RA • Remove the support reaction RA and Lift end A in the direction of Ra using some amount of force F = 1 and End B is free to rotate. Hence due to force the beam will deflection as shown in Fig • Make ordinate at A equal to one. • Calculate the ordinate X1 • 1 12 = 𝑋1 9 • 𝑋1 = 9 12 = 0.75 • Reaction At support A = Load * Load factor X1 = 20 * 0.75 = 15 Kn ILD BY: Mrs. S. M. Madhale 4 A B 20 Kn 3 3 6 F RB 1 X1 3 9 12
- 5. Reaction at support RB • Remove the support reaction RB and Lift end B in the direction of Rb using some amount of force F = 1 and End A is free to rotate. Hence due to force the beam will deflection as shown in Fig • Make ordinate at B equal to one. • Calculate the ordinate X2 • 1 12 = 𝑋2 3 • 𝑋2 = 3 12 = 0.25 • Reaction At support A = Load * Load factor X2 = 20 * 0.25 = 5 Kn ILD BY: Mrs. S. M. Madhale 5 A B 20 Kn 3 3 6 F RA 1 X2 3 9 12
- 6. Shear Force at Section • Remove the ability to resist shear force the beam and providing rollers in vertical direction at section and apply unit deflection in the direction of shear force. • Make ordinate at section equal to one. • Calculate the ordinate X1 and X2 1 12 = 𝑋2 3 𝑋2 = 3 12 = 0.25 1 12 = 𝑋1 9 𝑋1 = 9 12 = 0.75 Shear force at section = [ Load * Load factor X1] + [ Load * Load factor X2] = { + [20 * 0.75] } + { - [20 * 0.25] } = +15 -5 = 10 Kn ILD BY: Mrs. S. M. Madhale 6 A B 20 Kn 3 3 6 3 9 F F X1 X2 1 1
- 7. Bending moment at Section • Remove the ability to resist moment by providing hinge at that section and Apply unit rotation equal to one. • Calculate the ordinate ∅, 𝜃 𝑎𝑛𝑑 𝛽 • 𝜃 = 1 = 𝜑 + 𝛽 = 𝑎 ∗𝑏 𝐿 = 3 ∗9 12 = 2.25 Bending Moment at section = [ Load * Load factor] = 20 * 2.25 = 45 Kn-m ILD BY: Mrs. S. M. Madhale 7 A B 20 Kn 3 3 6 a b 𝜃 ∅ 𝛽 a * b / L L
- 8. • Two point loads of 100 Kn and 200 Kn spaced 3 m apart cross a girder of span 15 m from left to right with 100 Kn load leading. Draw a ILD for shear force and bending moment and find the value of maximum shear and bending moment at a section D, 6 m from left support. Also, find out the absolute maximum bending moment due to given loading system ILD BY: Mrs. S. M. Madhale 8
- 9. • Here are 2 conditions where • 100Kn load lies on section • 200Kn load lies on section ILD BY: Mrs. S. M. Madhale 9 D 200 Kn 100 Kn B A 6 9 15
- 10. Reaction at support RA 100Kn load lies on section • Remove the support reaction RA and Lift end A in the direction of Ra using some amount of force F = 1 and End B is free to rotate. Hence due to force the beam will deflection as shown in Fig • Make ordinate at A equal to one. • Calculate the ordinate X1 and X2 • 1 15 = 𝑋1 12 1 15 = 𝑋2 9 • 𝑋1 = 12 15 = 0.8 𝑋2 = 9 15 = 0.6 • Reaction At support A = Load * Load factor X ={ Load * Load factor X1}+{ Load * Load factor X2} = {200 * 0.8} + {100 * 0.6} =160+60 =220 Kn ILD BY: Mrs. S. M. Madhale 10 A B 6 9 F RB 1 X1 6 9 15 3 200 Kn 100 Kn X2 3
- 11. Reaction at support RA 200Kn load lies on section • Remove the support reaction RA and Lift end A in the direction of Ra using some amount of force F = 1 and End B is free to rotate. Hence due to force the beam will deflection as shown in Fig • Make ordinate at A equal to one. • Calculate the ordinate X1 and X2 • 1 15 = 𝑋1 9 1 15 = 𝑋2 6 • 𝑋1 = 9 15 = 0.6 𝑋2 = 6 15 = 0.4 • Reaction At support A = Load * Load factor X ={ Load * Load factor X1}+{ Load * Load factor X2} = {200 * 0.6} + {100 * 0.4} =120+40 =160 Kn ILD BY: Mrs. S. M. Madhale 11 A B 6 9 F RB 1 X1 6 9 15 3 200 Kn 100 Kn X2 3 X2
- 12. Reaction at support RB 100Kn load lies on section • Remove the support reaction RB and Lift end B in the direction of Rb using some amount of force F = 1 and End A is free to rotate. Hence due to force the beam will deflection as shown in Fig • Make ordinate at B equal to one. • Calculate the ordinate X1 and X2 • 1 15 = 𝑋1 3 1 15 = 𝑋2 6 • 𝑋1 = 3 15 = 0.2 𝑋2 = 6 15 = 0.4 • Reaction At support B = Load * Load factor X ={ Load * Load factor X1}+{ Load * Load factor X2} = {200 * 0.2} + {100 * 0.4} =40+40 =80 Kn ILD BY: Mrs. S. M. Madhale 12 A B 6 9 F RA 1 X1 6 9 15 3 200 Kn 100 Kn X2 3
- 13. Reaction at support RB 200Kn load lies on section • Remove the support reaction RB and Lift end B in the direction of Rb using some amount of force F = 1 and End A is free to rotate. Hence due to force the beam will deflection as shown in Fig • Make ordinate at B equal to one. • Calculate the ordinate X1 and X2 • 1 15 = 𝑋2 9 1 15 = 𝑋1 6 • 𝑋1 = 9 15 = 0.6 𝑋2 = 6 15 = 0.4 • Reaction At support B = Load * Load factor X ={ Load * Load factor X1}+{ Load * Load factor X2} = {200 * 0.4} + {100 * 0.6} =80+60 =140 Kn ILD BY: Mrs. S. M. Madhale 13 A B 6 9 F RB 1 X1 6 9 15 3 200 Kn 100 Kn X2 3
- 14. Shear Force at Section 100Kn load lies on section • Remove the ability to resist shear force the beam and providing rollers in vertical direction at section and apply unit deflection in the direction of shear force. Make ordinate at section equal to one. • Calculate the ordinate X1 and X3 [for negative consider loads before section] • 1 15 = 𝑋1 6 𝑋1 = 6 15 = 0.4 • 1 15 = 𝑋3 3 𝑋3 = 3 15 = 0.2 Shear force at section = [ Load * Load factor X1] + [ Load * Load factor X3] = { - [100 * 0.4] } +{-[200*0.2]} = - 40 -40 = - 80 Kn ILD BY: Mrs. S. M. Madhale 14 A 6 9 1 X1 6 9 3 200 Kn 100 Kn X2 3 B 1 X3
- 15. Shear Force at Section 200Kn load lies on section • Remove the ability to resist shear force the beam and providing rollers in vertical direction at section and apply unit deflection in the direction of shear force. Make ordinate at section equal to one. • Calculate the ordinate X1 and X3 [for positive consider load after section] • 1 15 = 𝑋2 9 𝑋2 = 9 15 = 0.6 • 1 15 = 𝑋3 6 𝑋3 = 6 15 = 0.4 Shear force at section = [ Load * Load factor X3] + [ Load * Load factor X2] = { + [100 * 0.4] } +{+[200*0.6]} = + 40 + 120 = 160 Kn ILD BY: Mrs. S. M. Madhale 15 A 6 9 1 X1 6 9 3 200 Kn 100 Kn X2 3 B 1 X3
- 16. Bending moment at Section 100Kn load lies on section • Remove the ability to resist moment by providing hinge at that section and Apply unit rotation equal to one. 𝜃=1= 𝜑+ 𝛽 Calculate the ordinate ∅, 𝜃 𝑎𝑛𝑑 𝛽 • Calculate the ordinate X1 and X3 [for negative consider loads before section] (𝑎 ∗𝑏 )/𝐿 • 𝑋1 = 𝑎 ∗𝑏 𝐿 = 6 ∗9 15 = 3.6 • 𝑋2 = 3.6 6 ∗ 3 = 1.8 Bending moment at section = [ Load * Load factor X1] + [ Load * Load factor X3] = [100 * 3.6] +[200 * 1.8] = 360 + 480 = 720 Kn-m ILD BY: Mrs. S. M. Madhale 16 A 6 9 1 X1 6 9 200 Kn 100 Kn X2 3 B 1 3 3
- 17. Bending moment at Section 200Kn load lies on section • Remove the ability to resist moment by providing hinge at that section and Apply unit rotation equal to one. 𝜃=1= 𝜑+ 𝛽 Calculate the ordinate ∅, 𝜃 𝑎𝑛𝑑 𝛽 • Calculate the ordinate X1 and X3 [for negative consider loads before section] (𝑎 ∗𝑏 )/𝐿 • 𝑋1 = 𝑎 ∗𝑏 𝐿 = 6 ∗9 15 = 3.6 • 𝑋2 = 3.6 9 ∗ 6 = 2.4 Bending moment at section = [ Load * Load factor X1] + [ Load * Load factor X3] = [200 * 3.6] +[100 * 2.4] = 720 + 240 = 960 Kn-m ILD BY: Mrs. S. M. Madhale 17 A 6 9 1 X1 6 9 200 Kn 100 Kn X2 3 B 1 3 6
- 18. THANK YOU !!! ILD BY: Mrs. S. M. Madhale 18