Section 3 – 13, Stress Concentration
It is essentially impossible to determine stresses accurately near connections or applied loads, because there are stress concentrations that depend on the specific characteristics of the connection or load. No two welds are identical. No two concentrated forces are identical.
During fatigue, which we will discuss later in the course, stress concentrations at the ends of tiny cracks (flaws) cause the material to yield at those locations with each application of the load, eventually resulting in a crack large enough to propagate itself at the speed of sound through the material.
Stress-concentration factors have been determined theoretically, using the theory of elasticity, for some simple geometries. Stress-concentration factors have been determined experimentally, primarily with photo-elastic methods, for many more-complicated geometries.
The results are tabulated, or presented graphically, in big books. If it is ever necessary to perform stress-concentration calculations, it is necessary to use one of those big books.
This figure determines the stress concentration factor for a centered, circular hole in a rectangular bar.
MAE3501 - 7 - 1

Determine

AVG , MAX , the average, normal stress on a section through the center of the hole. This is the section with the minimum cross-sectional area, and therefore the maximum average, normal stress.

AVG , MAX

( w

P
2 r ) t
(7.1) r
Determine the ratio w
. Use this ratio to determine K , the stress-concentration factor.
The maximum value of K , equal to 3.0
, occurs for a very small hole.

MAX

K

AVG , MAX
(7.2)
This figure does not tell you, but

MAX occurs at the sides of the hole.
Stress-concentration factors are obviously very important in some applications, but they are not very important at this point in your education. You only need to know that these factors exist, and that it is necessary to have a big book in order to perform quantitative calculations.
MAE3501 - 7 - 2

Beam Deflection
The following coordinate system is used to denote position on a beam.
Y
X
Z
During the analysis of bending in the X - Y plane, the following relationship was determined.

X
 
M
Z
( X ) Y
I
Z
(7.3)
In order to use equation (7.3) , it is necessary to measure Y from the neutral axis of the beam cross section, both before and after deflection of the beam. Another coordinate, parallel to
Y , but with a fixed origin, is required to represent the deflection of the beam. Denote this other coordinate by a lowercase, script

, completely unrelated to V , the shear force.


O
X
We will spend a significant amount of time learning methods to determine

( X ) , the d

( X ) vertical deflection, and , the slope, for beams with various supports and loads. Only d X deflections resulting from bending will be considered. For all reasonable beams, the shear deflections are negligible compared to the bending deflections.
Shigley uses Y to denote the deflection of a beam. This is “sloppy.”
MAE3501 - 7 - 3

In strength of materials, you developed the following relationship between bending moment and beam deflection.
d
2 
M ( X )
 d X
2
(7.4)
E I
3



1
 d X


2



2
Equation (7.4) is a truly scary relationship. It would appear that only adult beverages will permit us to maintain our sanity.
However, we are fortunate that beams are initially straight and horizontal. Even more fortunately, only small deflection of beams will be considered. d


1 d X
(7.5) d X


2

1 (7.6)
Therefore, the denominator in equation (7.4) can be accurately approximated with 1 , and the following relatively simple relationship results.
M ( X
E I
)
 d
2  d X
2
(7.7)
Equation (7.7) is the relationship that will be used to determine the deflection curve of a beam.
MAE3501 - 7 - 4

There are two alternative forms of this relationship. These alternative forms will be used later in this course, Differentiate equation (7.7) with respect to X , and replace d M ( X ) d X with the shear force, V ( X ) .
V ( X )

E I d
3  d X
3
(7.8)
Differentiate equation (7.8) with respect to X , and replace d V ( X ) with the distributed d X force, q ( X ) . q ( X )

E I d
4  dX
4
(7.9)
MAE3501 - 7 - 5

Example
Consider a cantilever beam, of length L , with a concentrated force at its end.

P
L
M ( X )
 
X
P ( L

X ) (E.1)
Substitute equation (E.1) into the governing differential equation for beam deflection.
E I d
2 
( X )

M ( X )

(E.2) d X
2

P L

P X
Anti-differentiate equation (E.2) with respect to X . Include a constant of integration.
E I d

( X )
 d X

P L X

P X
2
2

C
1
(E.3)
There is a boundary condition for the slope, d

( X )
. This will not be the case when a dX simply supported beam is analyzed.
d
 d X
( X

0 )

0 (E.4)
( E .
3 )

( E .
4 )
( E .
5 )
C

(
1

E .
0
3 )
(E.5)
E I d

( X )
 d X

P L X

P X
2
2
(E.6)
Anti-differentiate equation (E.6) with respect to X . Include a constant of integration.
MAE3501 - 7 - 6

E I

( X )
 
P L X
2

P X
3

C
2
(E.7)
2 6
There is also a boundary condition for deflection. When a simply supported beam is analyzed, there will be two boundary conditions for deflection.

( X

0 )

0 (E.8)
( E .
7 )

( E .
8 )
C
2

0 (E.9)
( E .
9 )

( E .
7 )
E I

( X )
 
P L X
2
2

P X
3
6
(E.10)
Solve equation (E.10) for

( X ) .

( X )
 
P L X
2
2 E I

P X
3
6 E I
(E.11)
Determine expressions for the slope and deflection at the end of the beam. d
 dX
( X

L )
 
P L
2
2 E I
(E.12)

( X

L )
 
P L
3
3 E I
(E.13)
MAE3501 - 7 - 7

Example
Consider a cantilever beam, of length L , with a concentrated moment at its end.

L
X
M ( X )
 
M
0
(E.1)
Substitute equation (E.11) into the governing differential equation for beam deflection.
E I d
2 
( X ) d X
2

M ( X )
 
M
0
(E.2)
Anti-differentiate equation (E.2) with respect to X . Include a constant of integration.
E I d

( X ) d X
 
M
0
X

C
1 d
 d X
( X
There is a boundary condition for the slope, d

( X )
. dX

0 )

0
(E.3)
(E.4)
( E .
3 )
( E .
5 )

( E .
4 )
C
1

0

( E .
3 )
(E.5)
E I d

( X ) d X
 
M
0
X (E.6)
Anti-differentiate equation (E.6) with respect to X . Include a constant of integration.
E I

( X )
 
M
0
X
2
2

C
2
(E.7)
MAE3501 - 7 - 8

There is also a boundary condition for deflection.

( X

0 )

0
( E .
7 )

( E .
8 )
C
2

0
( E .
9 )

( E .
7 )
E I

( X )
 
M
0
X
2
2
Solve equation (E.10) for

( X ) .

( X )
 
M
0
X
2
2 E I
Determine expressions for the slope and deflection at the end of the beam. d
 dX
( X

L )
 
M
0
E I
L

( X

L )
 
M
0
L
2
2 E I
(E.8)
(E.9)
(E.10)
(E.11)
(E.12)
(E.13)
MAE3501 - 7 - 9

Example length
Consider a cantilever beam, of length L , with a uniform distributed force over its entire q
0

X L
M ( X )
  q
0
( L

X )
2
2
  q
0
2
L
2
 q
0
L X
 q
0
2
X
2
(E.1)
Substitute equation
E I d
2 
( X ) d X
2
(E.1)

into the governing differential equation for beam deflection.
M ( X )
  q
0
L
2
2
 q
0
L X
 q
0
2
X
2
(E.2)
Anti-differentiate equation (E.2) with respect to X . Include a constant of integration.
E I d

( X ) d X
  q
0
L
2
X

2 q
0
L X
2
2
 q
0
X
3
6

C
1
(E.3)
There is a boundary condition for the slope, d

( X )
. dX d
 d X
( X

0 )

0 (E.4)
( E .
3 )
( E .
5 )

( E .
4 )
C
1

0

( E .
3 )
(E.5)
E I d

( X ) d X
  q
0
L
2
X

2 q
0
L X
2
2
 q
0
X
3
6
(E.6)
Anti-differentiate equation (E.6) with respect to X . Include a constant of integration.
MAE3501 - 7 - 10

E I

( X )
  q
0
L
2
X
2
 q
0
L X
3
 q
0
X
4 6 24
There is also a boundary condition for deflection.
4

C
2

( X

0 )

0
( E .
7 )

( E .
8 )
C
2

0
( E .
9 )

( E .
7 )
E I

( X )
  q
0
L
2
X
2
4
 q
0
L X
3
6
 q
0
X
4
24
Solve equation (E.10) for

( X ) .

( X )
  q
0
L
2
X
2
4 E I
 q
0
L X
3
6 E I
 q
0
X
4
24 E I
Determine expressions for the slope and deflection at the end of the beam. d
 dX
( X

L )
  q
0
L
3
6 E I

( X

L )
  q
0
L
4
8 E I
(E.7)
(E.8)
(E.9)
(E.10)
(E.11)
(E.12)
(E.13)
MAE3501 - 7 - 11

Combine the results of the last three examples.

denotes the slope, positive counterclockwise, the tangent of the angle between the beam and its initially straight line. The tangent of a small angle is equal to the angle, measured in radian.

denotes the deflection of the beam, positive upward.

END

END

M
0
L

E I
M
0
L
2
2 E I
L
P

P L
2
2 E I

P L
3
3 E I
 q
0
L
3
6 E I
 q
0
L
4
8 E I
There are many cantilever-beam problems in the real world, including the simple problems shown above. Commit these six relationships to memory, and there will be many occasions when they will save a great deal of time. It will be necessary to for you to use these six relationships throughout this course.
MAE3501 - 7 - 12

Example
This problem appears straightforward. a

P b
O
Determine the reactions. Consider a FBD of what is to the left of a section that is to the left of the concentrated force, P . Refer to this region, to the left of the concentrated force, as region 1 , and denote variables in this region with a subscript of 1 . a
P b
 b
X
V
1
( X )
Sum moments about X .
M
1
( X )

M
X

0

M
1
( X )

P b X a
 b
Solve for M
1
( X ) and substitute into the governing differential equation for beam deflection.
E I d
2 
1
( X ) d X
2

P b a

X b
(E.1)
(E.2)
Anti-differentiate equation (E.2) twice. It is not possible to solve for the first constant of integration because there is no boundary condition for the slope in region 1 .
E I d

1
( X ) d X

2
P b X
2
( a
 b )

C
1
(E.3)
E I

1
( X )

P b X
3
6 ( a
 b )

C
1
X

C
2
(E.4)
MAE3501 - 7 - 13

There is one boundary condition for deflection in region 1 .

1
( X

0 )

0
( E .
4 )

( E .
5 )
(E.5)
C
2

0 (E.6)
Examine the region to the right of the concentrated force. Refer to this region as region
2 , and denote variables in this region with a subscript of 2 .
Make a section to the right of the concentrated force, P , and consider a FBD of what is to the left of that section.
P a
X M
2
( X )
P b a
 b
V
2
( X )
Sum moments about X .

M
X

0

M
2
( X )


 a
P b
 b


X

P ( X
 a ) (E.7)
Solve for M
2
( X ) and substitute into the governing differential equation for beam deflection.
E I d
2 
2
( X ) d X
2

P a b X
 b

P X

P a (E.8)
Anti-differentiate equation (E.8) twice. It is not possible to solve for the first constant of integration because there is no boundary condition for the slope in region 2 .
E I d

2
( X ) d X

2
P b
( a
X
2
 b )

P X
2
2

P a X

C
3
(E.9)
E I

2
( X )

P b X
3
6 ( a
 b )

P X
3
6

P a X
2
2

C
3
X

C
4
(E.10)
MAE3501 - 7 - 14

Expressions have been developed for the deflection,

( X ) , and slope, d

( d X
X )
, in each of the two regions of the beam. These expressions contain three remaining unknown constants of integration. In order to determine these unknown constants, it is necessary to use the following three boundary conditions.

2
( X
 a
 b )

0

1
( X
 a )
 
2
( X
 a ) d

1
( X
 a )
 d X d

2
( X d X
 a )
This problem has become a can of worms. Many beam problems contain more than two regions and it is necessary to solve a large number of simultaneous equations to determine the constants of integration.
Homework
Commit the six relationships on page 12 of this lesson to memory. A quiz requiring you to demonstrate some or all of these relationships is an option.
MAE3501 - 7 - 15