close

Вход

Забыли?

вход по аккаунту

код для вставкиСкачать
Chapter 5 Analysis and Design of
Beams for Bending
5.1 Introduction
-- Dealing with beams of different materials:
steel, aluminum, wood, etc.
-- Loading: transverse loads
 Concentrated loads
 Distributed loads
-- Supports
 Simply supported
 Cantilever Beam
 Overhanging
 Continuous
 Fixed Beam
A. Statically Determinate Beams
-- Problems can be solved using Equations of Equilibrium
B. Statically Indeterminate Beams
-- Problems cannot be solved using Eq. of Equilibrium
-- Must rely on additional deformation equations to solve
the problems.
FBDs are sometimes necessary:
FBDs are necessary tools to determine the internal
(1) shear force V – create internal shear stress; and
(2) Bending moment M – create normal stress
From Ch 4:
m 
Where
M c
I
x  
My
(5.1)
I
(5.2)
I = moment of inertia
y = distance from the N. Surface
c = max distance
Recalling, elastic section modulus, S = I/c,
m 
hence
M
(5.3)
S
For a rectangular cross-section beam,
S
1
bh
2
(5.4)
6
From Eq. (5.3),
max occurs at Mmax 
It is necessary to plot the V and M diagrams along the length
of a beam.

to know where Vmax or Mmax occurs!
5.2 Shear and Bending-Moment Diagrams
•Determining of V and M at selected
points of the beam
Sign Conventions
1.
The shear is positive (+) when external
forces acting on the beam tend to shear
off the beam at the point indicated in fig
5.7b
2. The bending moment is positive (+)
when the external forces acting on the
beam tend to bend the beam at the
point indicated in fig 5.7c
 Moment
5.3 Relations among Load, Shear
and Bending Moment
1. Relations between Load and Shear
  FY  0 :
V-(V+V )  wx  0
V   wx
Hence,
dV
dx
 w
(5.5)
Integrating Eq. (5.5) between points C and D
VD  VC   
xD
xC
wdx
(5.6)
VD – VC = area under load curve between C and D
1
(5.6’)
(5.5’)
2. Relations between Shear and Bending Moment

M C '  0 :
( M  M )  M  V x  w x
M  V x 
or
lim
x 0
M
x

dM
dx

M
x
dM
dx
1
2
V 
V
w ( x )
1
2
2
 x
(5.7)
x
2
0
dM
dx
V
(5.7)
xD
M D  M C   Vdx
xC
MD – MC = area under shear curve between points C and D
5.4 Design of Prismatic Beams for Bending
-- Design of a beam is controlled by |Mmax|
m 
M max c
I
m 
M max
(5.1’,5.3’)
S
Hence, the min allowable value of section modulus is:
Smin 
M
max
 all
(5.9)
Question: Where to cut? What are the rules?
Answer: whenever there is a discontinuity in the loading
conditions, there must be a cut.
Reminder:
The equations obtained through each cut are only valid
to that particular section, not to the entire beam.
5.5 Using Singularity Functions to Determine
Shear and Bending Moment in a Beam
Beam Constitutive Equations
w
4

d y
4
EI
dx
V
d y
3

3
EI
dx
M
d y
2

EI
dx

dy

2
dx
y
 f ( x)
Notes:
1.In this set of equations, +y is going upward
and +x is going to the right.
2. Everything going downward is “_”, and upward
is +. There is no exception.
3. There no necessity of changing sign for an y
integration or derivation.
Sign Conventions:
1. Force going in the +y direction is “+”
2. Moment CW is “+”
Rules for Singularity Functions
Rule #1:
xa
0
xa
n
1 when x  a
{
0 when x  a
( x  a ) when x  a
(5.15)
n
{
0
when x  a
(5.14)
Rule #2:
1. Distributed load w(x) is zero order: e.g. wo<x-a>o
2.
Pointed load P(x) is (-1) order: e.g. P<x-a>-1
3. Moment M is (-2) order: e.g. Mo<x-a>-2
Rule #3:

 xa
2
dx   x  a 
1

 xa
1
dx   x  a 
0

 x  a  dx   x  a 



0
1
 x  a  dx 
1
1
 xa
2
 xa
3
 xa
4
2
2
 x  a  dx 
1
3
3
 x  a  dx 
1
4
Rule #4:
1. Set up w = w(x) first, by including all forces, from the left
to the right of the beam.
2. Integrating w once to obtain V, w/o adding any constants.
3. Integrating V to obtain M, w/o adding any constants.
4. Integrating M to obtain EI , adding an integration
constant C1.
5. Integrating EI to obtain EIy, adding another constant
C2.
6. Using two boundary conditions to solve for C1 and C2.
V ( x) 
1
4
M ( x) 
1
4
w0 a  w0 x  a
1
w0 ax 
x
M ( x)  M (0)   V ( x)dx  
0
2
x
0
w0 x  a
1
4
(5.11)
2
(5.12)
x
w0 adx   w0 x  a dx
0
•After integration, and observing that M (0)  0 We obtain as before
M ( x) 
1
4
w0 ax 
1
2
w0 x  a
w( x)  w0 x  a
2
0
w( x)  0 for x< 0
(5.13)
xa
0
 ( x  a)  1
0
w( x)  w0 for x  a
1
V 
w0 a for x  0
4
x
0
x
V ( x)  V (0)    w( x) dx    w0 x  a dx
0
V ( x) 
1
4
V ( x) 
0
w0 a   w0 x  a
1
4
w0 a  w0 x  a
1

n
x  a dx 
d
dx
xa
n
1
n 1
and
xa
 n xa
n 1
n 1
for n  0 (5.16)
for n  1 (5.17)
Example 5.05
Sample Problem 5.9
5.6 Nonprismatic Beams
S
M
 all
V ( x) 
1
2
PP x
1
0
L
2
1
M ( x) 
1
2
Px  P x 
1
2
L
w( x)  w0 x  a
0
 w0 x  b
0
•Load and Resistance Factor Design
•
 D M D   L M L   MU
Example 5.03
1/--страниц
Пожаловаться на содержимое документа