# 4.0 INTERNAL FORCES AND MOMENTS IN MEMBERS - PowerPoint PPT Presentation

PPT – 4.0 INTERNAL FORCES AND MOMENTS IN MEMBERS PowerPoint presentation | free to download - id: 55d721-NjcyM The Adobe Flash plugin is needed to view this content

Get the plugin now

View by Category
Title:

## 4.0 INTERNAL FORCES AND MOMENTS IN MEMBERS

Description:

### ... the shear force, Q, and the bending moment ... Example Sketch the shearing force and bending moment diagrams for the beam given 4 kN C D B A 1 m 1 m 2 m 0.5 ... – PowerPoint PPT presentation

Number of Views:188
Avg rating:3.0/5.0
Slides: 15
Provided by: Eng83
Category:
Tags:
Transcript and Presenter's Notes

Title: 4.0 INTERNAL FORCES AND MOMENTS IN MEMBERS

1
4.0 INTERNAL FORCES AND MOMENTS IN MEMBERS
4.1 Introduction 4.2 Idealization of Structural
Systems and Loads 4.3 Method of
Analysis 4.4 Differential Equailibrium
Relations 4.5 Examples
2
4.1. Introduction
• Structural members are subjected to many
different types of external forces. In the design
of these members, it is essential to compute the
stresses and deformations produced by the

The study of structural members
v
1. Idealization of the structural system and the
2. Computation of the external reactions
3. Determination of the internal forces
4. Calculation of the stresses and deformations
caused by these internal forces

v
3
4.2. Idealization of Structural Systems and Loads
4.2.1 Types of Supports
(a) and (b) can resist the force only in the
direction of line AB
(c) and (d) can only resist a force which s
perpendicular to the direction of movement of the
roller.
4
(2) Pin or Hinge Supports
(a) and (b) can resist a force acting in any
direction in the plane. Therefore, in general,
hinged or pinned supports have two unknowns
(b)
(a)
(3) Fixed, clamped, built-in or encastre supports
This type of support is capable of resisting a
force in any direction and is also capable of
resisting a moment or a couple.
5
A force is transmitted to a member through a
post, hanger or another member. In such cases the
load is applied over a very limited portion of
the member. Such forces are idealized as
concentrated forces and shown with a single
vector acting at a point.
In general, forces and loads are distributed over
a finite length or area of the member. These are
If the total force on length Dx is denoted as DF,
6
(No Transcript)
7
4.3. Method of Analysis
Internal forces in members are determined by
passing an imaginary section across the member at
the point concerned, and writing the equilibrium
equations for the free body either to left or
right of the cut.
8
Example
Compute the internal forces at point D for the
beam given.
At D (use right portion)
9
4.4. Differential Equilibrium Relations
is very small compared to the other two terms and
will be neglected

In the limit, when Dx approaches zero
These are basic differential equations relating
the external load, q(x), the shear force, Q, and
the bending moment, M.
10
Lets write this relations in integral form as
i and i1 are any two sections along the member
First equation shows that the change in shear
between any two sections is equal to minus the
area of the load diagram between these two
sections.
Second equation shows that the change in moment
between any two sections is equal to minus the
area of the shear diagram between these sections.
11
Sketch the shearing force and bending moment
diagrams for the beam given
Example
It is to be noted that because of the 4 kN load
at C, these values correct strictly just short of
C.
12
(No Transcript)
13
RA4 kN RB8 kN
Example
q can be written as
Therefore Q
RA
14
The maximum moment occurs where the shearing
force is zero, _at_ x1.9 m, and is given by
and for MB