Chapter 5 Analysis and Design of Beams for Bending
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Transcript of Chapter 5 Analysis and Design of Beams for Bending
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Chapter 5 Analysis and Design of
Beams for Bending
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-- Dealing with beams of different materials:
steel, aluminum, wood, etc.
-- Loading: transverse loads
Concentrated loads
Distributed loads
5.1 Introduction
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-- Supports Simply supported
Cantilever Beam
Overhanging
Continuous
Fixed Beam
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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:
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FBDs are necessary tools to determine the internal
(1) shear force V – create internal shear stress; and
(2) Bending moment M – create normal stress
Where I = moment of inertiay = distance from the N. Surfacec = max distance
m
M cI
xMyI
From Ch 4:
(5.1)
(5.2)
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Recalling, elastic section modulus, S = I/c,
m
MS
hence (5.3)
For a rectangular cross-section beam,
216
S bh (5.4)
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!
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5.2 Shear and Bending-Moment Diagrams•Determining of V and M at selected
points of the beam
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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
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
Moment
Sign Conventions
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5.3 Relations among Load, Shear and Bending Moment
0 0: V-(V+ )YF V w x
V w x
dV wdx
(5.5)Hence,
1. Relations between Load and Shear
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1
D
C
x
D C xV V wdx
(5.6’)
(5.5’)
Integrating Eq. (5.5) between points C and D
VD – VC = area under load curve between C and D
(5.6)
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2. Relations between Shear and Bending Moment
0' :C
M
12
M V xx
02
( ) xM M M V x w x
212
( )M V x w x
dM Vdx
(5.7)
or
0limx
M dMx dx
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D
C
x
D C xM M Vdx
dM Vdx
MD – MC = area under shear curve between points C and D
(5.7)
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5.4 Design of Prismatic Beams for Bending
max max m m
M c MI S
maxmin
all
MS
(5.1’,5.3’)
(5.9)
-- Design of a beam is controlled by |Mmax|
Hence, the min allowable value of section modulus is:
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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.
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5.5 Using Singularity Functions to Determine Shear and Bending
Moment in a Beam
4
4
3
3
2
2
( )
w d yEI dxV d yEI dxM d yEI dx
dydx
y f x
Beam Constitutive Equations
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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 “+”
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0 1 when {
0 when
x ax a
x a
( ) when {0 when
nn x a x a
x ax a
(5.15)
(5.14)
Rules for Singularity Functions
Rule #1:
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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
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Rule #3:
2 1
1 0
0 1
1 2
2 3
3 4
x a dx x a
x a dx x a
x a dx x a
1x a dx x a21x a dx x a31x a dx x a4
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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.
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0 01( )4
V x w a w x a
20 0
1 1( )4 2
M x w ax w x a
0 00 0 0
1( ) (0) ( )4
x x x
M x M V x dx w adx w x a dx
•After integration, and observing that (0) 0M We obtain as before
2
0 01 1( )4 2
M x w ax w x a
00( ) w x w x a
(5.11)
(5.12)
(5.13)
( ) 0 for x< 0w x
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0 0( ) 1 x a x a
0( ) for w x w x a
01 for 04
V w a x
0
00 0( ) (0) ( )
x xV x V w x dx w x a dx
1
0 01( )4
V x w a w x a
0 01( )4
V x w a w x a
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11 for n 01
n n
x a dx x an
1 for n 1 n nd x a n x a
dx
and
(5.16)
(5.17)
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Example 5.05
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Sample Problem 5.9
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5.6 Nonprismatic Beams
all
MS
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01 1( )2 2
V x P P x L
11 1( )2 2
M x Px P x L
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0 0
0 0( ) w x w x a w x b
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•Load and Resistance Factor Design• D D L L UM M M
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Example 5.03
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