# Internal Forces in Structures - PowerPoint PPT Presentation

PPT – Internal Forces in Structures PowerPoint presentation | free to download - id: 789a19-OTg5N

The Adobe Flash plugin is needed to view this content

Get the plugin now

View by Category
Title:

## Internal Forces in Structures

Description:

### ... 21, 24, 26 & 35 Due Wednesday, November 13 Generate a shear / bending ... shear force M(x) moment Shear and Moment Diagrams using Sectioning Method V M V ... – PowerPoint PPT presentation

Number of Views:41
Avg rating:3.0/5.0
Slides: 15
Provided by: Timoth198
Category:
Tags:
Transcript and Presenter's Notes

Title: Internal Forces in Structures

1
ME 221 Statics Lecture 27 Section 7.4
2
Homework 9
• Chapter 5 problems
• 54, 56, 62, 64, 69, 71 73
• Due Today

3
Homework 10
• Chapter 7 problems
• 5, 8, 19, 21, 24, 26 35
• Due Wednesday, November 13

4
Last Lecture Internal Forces in Structures
• Reviewed internal/external forces
• Found internal forces
• Started shear moment diagrams

5
Shear and Moment Diagrams using Sectioning Method
• Generate a shear / bending diagram as follows

1. Find reaction forces
2. Take a section on each side of an applied
force or moment and inside a distributed load
(take a new section whenever there is a change in
the load or shape of the beam)
- draw a FBD and sum forces / moments
3. Repeat 2 along the length of the beam
V(x) shear force
M(x) moment
6
Sign Convention
Positive Shear and Positive Moment
7
Effect of External Forces
Positive Shear
Positive Moment
8
Relations Between w, V, and M
• In balancing forces, we can come up with
differential equations relating w, V, and M.
These are as follows

w(x)
MdM
V
M
Thus,
VdV
dx
This means you can integrate the shear diagram
to obtain the moment diagram.
9
Shear and Moment Diagrams Using Integration Method
w(x)
MdM
V
M
VdV
dx
Area under the load intensity diagram between xj
and x j1
10
Similarly
Area under the shear diagram between xj and x j1
11
P
P
MdM
V
M
C
VdV
3 - Concentrated Moment
MdM
V
M
C
VdV
12
Notes
• Concentrated force will cause a jump in the shear
diagram by an amount equal to the applied load
• Concentrated moment will cause a jump in the
moment diagram by an amount equal to the negative
of the applied moment

Connecting points in the shear and moment diagrams
xj
Negative increasing slope
xj1
xj
Positive decreasing slope
xj1
Negative decreasing slope
xj1
Positive increasing slope
xj1
xj
xj
13
Developing Shear and Bending Diagrams
1. Show FBD and statics for each section
2. Determine equation for V(x) and M(x)
3. Draw shear and bending diagrams indicating
linear or parabolic
4. Label end points of diagram as well as every
region endpoint
14
(No Transcript)