Classes #11 & #12
Civil Engineering Materials – CIVE 2110
Bending
Fall 2010
Dr. Gupta
Dr. Pickett
1
ASSUMPTIONS OF BEAM BENDING THEORY






Beam Length is Much Larger Than
Beam Width or Depth.
so most of the deflection
is caused by bending,
very little deflection is caused by shear
Beam Deflections are small.
Beam is Perfectly Straight,
With a Constant Cross Section
1
(beam is prismatic).
Beam has a Plane of Symmetry.

Resultant of All Loads
acts in the Plane of Symmetry.
Beam has a Linear Stress-Strain
Relationship.
E




1




ASSUMPTIONS OF BEAM BENDING
THEORY
Beam Material is Homogeneous.
Beam Material is Isotropic.
Beam is Loaded ONLY by a
Moment about an axis
Perpendicular to the long axis of
Symmetry.
Thus Moment is CONSTANT
across the
Length of the Beam.
There is NO SHEAR.

d

ASSUMPTIONS OF BEAM BENDING
THEORY
Plane Sections Remain Plane.
No Warping
(no buckling, no rotation about vertical axis).
Motion is only in Vertical Plane.
Beam Cross Sections originally
Perpendicular to Longitudinal Axis
Remain Perpendicular.

d
BEAM BENDING THEORY

When a POSITVE moment is applied,
TOP of beam is in COMPRESSION
BOTTOM of beam is in TENSION.
NEUTRAL SURFACE:
- plane on which NO change
in LENGTH occurs.
BEAM BENDING THEORY

When a POSITVE moment is applied, (POSITIVE Bending)
TOP of beam is in COMPRESSION
BOTTOM of beam is in TENSION.
NEUTRAL SURFACE:
- plane on which
NO change
in LENGTH occurs.
Cross Sections
perpendicular to
Longitudinal axis
Rotate about the
NEUTRAL (Z) axis.
BEAM BENDING THEORY

When a POSITVE moment is applied, (POSITIVE Bending)
TOP of beam is in COMPRESSION
BOTTOM of beam is in TENSION.
NEUTRAL SURFACE:
- plane on which
NO change
in LENGTH occurs.
Cross Sections
perpendicular to
Longitudinal axis
Rotate about the
NEUTRAL (Z) axis.
BEAM BENDING
THEORY

For M = +
Any line segment, Δx :
- shortens,
if located
above Neutral Surface.
BEAM BENDING
THEORY

For M = +
Any line segment, Δx :
- does not change length,
if located
at Neutral Surface.
BEAM BENDING
THEORY

For M = +
Any line segment, Δx :
- lengthens,
if located
below Neutral Surface.
BEAM BENDING
THEORY
Re m em bering Norm al Strain  

Change in Length
s 's
 lim
s 0
Original Length
s
Before deform ation s  x
After deform ation x  [Tan( )]
Since deform ations are sm all
Tan( )   Radians
and
x  
and
s '  (   y )
(   y )  
Thus   lim
s 0

y
 

BEAM BENDING
THEORY
Thus Norm al Strain
in the Longitudinal Direction
varies linearily with  y
above or below
Neutral Surface
 
Thus
y

and  max 
c

 y
     max
 c
Flexural Bending Equation

We assumed:

Cross Sections remain constant

However,
do to the Poisson’s Effect;
there will be strains
in the 2 directions
perpendicular to the
Longitudinal Axis.
 y   x
and
 z   x
Axial
Compressive
Strain
Axial
Tensile
Strain

BEAM BENDING
THEORY
For material that is:




Homogeneous
Isotropic
Linear-Elastic
We can conclude for
STRESS, σ


E
thus   E
and  max  E max
 y
 y

and sin ce      max then   E    max 
 c
 c

then

 y
E  
 c
  max
 max  E
thus

x
 y
E  
 c


 y
 
 c
 y
 c
     max
 y   x

x =X E
and
 z   x
Y




Y

Y


BEAM BENDING
THEORY
For material that is:




Homogeneous
Isotropic
Linear-Elastic
We can conclude for
STRESS, σ

E

Compressive Strain
Tensile Strain
Compressive Stress
thus   E
 y
and sin ce      max
 c
 y
then
     max
 c
Tensile Stress





BEAM BENDING
THEORY
Internal Moment must resist
External Moment.
Internal Resisting Moment:
Caused by an Internal Force
resisting an External force

Neutral Axis
Compressive
Stress
Can find Neutral Axis by balance
of Forces:
Σ Internal Forces must = ZERO
Fx  0   dF    (dA)
A
A
 y
0      max (dA)
A
 c
0
 max
c

A
y (dA)
Tensile Stress

BEAM BENDING
THEORY
Can find Neutral Axis by balance
of Forces:
Σ Internal Forces must = ZERO

Neutral Axis
Neutral Axis = Centroidal Axis
Compressive
Stress
0 =1st Moment of Area
about Neutral Axis
0   y (dA)
A
Tensile Stress



BEAM BENDING
THEORY
Internal Moment must resist
External Moment.

M = (Lever Arm)x(Internal Force)
M = (Lever Arm)x(Stress x Area)
Compressive
Stress
M Z External  M Z Internal
M Z External  A y (dF )
M Z External  A y (dA)
M Z External
 y


  y   max  (dA)
A

 c

M Z External    max A y 2 (dA)
c
Tensile Stress
Neutral Axis



BEAM BENDING
THEORY
Internal Moment must resist
External Moment.

M = (Lever Arm)x(Internal Force)
M = (Lever Arm)x(Stress x Area)
Compressive
Stress
M Z External  M Z Internal
M Z External    max A y 2 (dA)
c
I   y 2 (dA)
A
I  2 nd
Mom ent of
M Z External  
 max
c
I
Area about Z
or  max  
axis
M Z external c
I
Tensile Stress
Neutral Axis


BEAM BENDING
THEORY
Flexural Bending Stress Equation:
For Stress in the Direction of the
Long Axis (X),
Compressive
At any location, Y, Stress
above or below the
Neutral Axis
Neutral Axis
x  
M Z external y
I
Tensile Stress
Beam Bending
2nd Moment of Area
Calculation
A Rectangular Cross Section
x
Y
2
dA 
A

h
2
x
b
2
b

h 2

2


h

2
Y dzdy   Y dyz 
2
2

h
2

h
2
b
2
b

2

h
2
 b  b 
  Y 2 dy       Y 2 dyb
 2  2   h
h

2

2
h
2
Y 3 
2
 b  Y dy  b  
h
3

h
2
h

2

b   h   h 
 
  
3  2   2 
3
3



2
b  h3
 
3 8
 h 3  b  h 3  h 3  b  h 3 
           
 8  3  8  8  3  4 
bh3
2
nd
Y
dA


I

2
Mom ent of Area about Neutral Axis
ZZ
A
12
PARALLEL AXIS THEOREM FOR 2nd MOMENTS OF AREA
2nd MOMENTS OF COMPOSITE AREAS
B & J 8th,9.6, 9.7
Z
Y2
I Z  I 1Z  I 2 Z  I 3 Z
3
3

 b1h13



b
h
b
h
2
2
2
3 3
2 2






 A1Y1   
 A2Y2   
 A3Y3 
 12
  12
  12

Z
QUESTION: Why are I-beams shaped like I
????????
PARALLEL AXIS THEOREM
ANSWER: In order to achieve maximum strength
(and least deflection) for
the least weight, by maximizing the
Second Moment of Area, I,
of a beam. This is achieved by
maximizing the distance between
beam material in the flanges and the
beam mid-height.
FOR 2nd MOMENTS OF AREA
For Example: Construct an I-beam from three pieces
of balsa wood,
With each piece of balsa wood
3”x3/8”x36”
I beam  I A  I B  I C  3.21_ in 4  0.844_ in 4  3.21_ in 4  7.26 _ in 4

3
IA 
BA H A
2
  AA  B A H A    y A 
12
3"   3" 

3
 8    3"  3"    1.5" 1  3" 


  
12
2 8
 8  

 0.013in 4  3.2in 4
IA 
IA
I A  3.21_ in 4
3


BB H B
2
  AB  B B H B    y B 
12
 3" 
3
   3"
  3" 

8
2
IB   
     3"  0  
12

  8 

IB 
I B  0.844in 4  0in 4
I B  0.844_ in 4
2



Sign Convention for Diagrams
MIntrnl=+
MIntrnl=+
Compression
Free End
Or
V=+
Free End
Or
Pinned End
Pinned End
V=-
Tension
Compression
MIntrnl=MIntrnl=-
MExtrnl=-
Tension
MIntrnl=-
Tension
Tension
Tension
MIntrnl=+
Tension
MIntrnl=-
MIntrnl=-
Compression
Fixed End
Compression MExtrnl=+
Fixed End
Steps for V and BM diagrams
1.Draw FBD
2.Obtain reactions:
M (@left support) to obtain reaction at right;
M (@Right support) to obtain reaction at left;
Check Fy = 0
3. Cut a section ;
Obtain internal F (or P), V, M at cut section ;
M, Fy, Fx
4. Record, draw internal F (or P), V, M on both sides of cut sections ;
- magnitude
- units
- direction on both sides of cut
BEAM END CONDITIONS
Roller Pin - Pin
Fixed - Free
Fixed - ?
VL=RLY
BEAM END CONDITIONS
Pin
VL=RLY
BEAM END CONDITIONS
Roller
VL=RLY
Pin