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:
Mc
m 
I
Where
My
x  
I
I = moment of inertia
y = distance from the N. Surface
c = max distance
(5.1)
(5.2)
Recalling, elastic section modulus, S = I/c,
M
m 
S
hence
(5.3)
For a rectangular cross-section beam,
1
S  bh 2
6
From Eq. (5.3),
(5.4)
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
 w
dx
(5.5)
Integrating Eq. (5.5) between points C and D
xD
VD  VC    wdx
xC
(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 :
x
( M  M )  M  V x  w x
0
2
1
M  V x  w ( x ) 2
2
M dM

x  0  x
dx
lim
or
M
1
 V   x
x
2

dM
V
dx
(5.7)
dM
V
dx
(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 max c
m 
I
M max
m 
S
(5.1’,5.3’)
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 d4y
 4
EI dx
V
d3y
 3
EI dx
M d2y
 2
EI dx
dy
 
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) n when x  a
{
0
when x  a
(5.15)
(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  dx   x  a 
  x  a  dx   x  a 
1
  x  a  dx  2  x  a 
1
  x  a  dx  3  x  a 
 x  a  2dx   x  a  1
1
0

0
1
1
2
2
3
1
 x  a  dx   x  a  4
4
3
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.
1
V ( x)  w0 a  w0 x  a
4
1
1
M ( x)  w0 ax  w0 x  a
4
2
(5.11)
2
(5.12)
x
1
M ( x)  M (0)   V ( x)dx   w0adx   w0 x  a dx
0
0 4
0
•After integration, and observing that M (0)  0 We obtain as before
x
x
1
1
M ( x)  w0 ax  w0 x  a
4
2
w( x)  w0 x  a
2
0
w( x)  0 for x< 0
(5.13)
x  a  ( x  a )0  1
0
w( x)  w0 for x  a
1
V  w0 a for x  0
4
x
x
0
0
0
V ( x)  V (0)    w( x)dx    w0 x  a dx
1
V ( x)  w0 a   w0 x  a
4
1
V ( x)  w0 a  w0 x  a
4
1

1
x  a dx 
xa
n 1
n
d
xa
dx
and
n
 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
1
1
V ( x)  P  P x  L
2
2
1
1
M ( x)  Px  P x  L
2
2
0
1
0
w( x)  w0 x  a  w0 x  b
0
•Load and Resistance Factor Design
•
 D M D   L M L   MU
Example 5.03