Chapter 3 LOAD DETERMINATION 111

Fi η m 0.98
= vi =1 = 1 285.9 force
W gδ st s m
9.81 2 (0.00006 m ) ratio
s
(c ) 7 000
Fi = 1 285.9(9.81 N ) = 12 612 N
6 000
The variation in force ratio with changes in l / d ratio for a constant amount of 3
moving mass and a constant impact velocity (i.e., constant input energy) is shown in 5 000
Figure 3-20. As the l / d ratio is reduced, the rod becomes much stiffer and generates 4 000
much larger dynamic forces from the same impact energy. This clearly shows that
impact forces can be reduced by increasing the compliance of the impacted system. 3 000

3 For part (b), we will keep the l / d ratio constant at 10 and vary the ratio between 2 000
the moving mass and the rod mass over the range of 1 to 20. The files EX03-01B 1 000
on the CD-ROM calculate all values for a range of mass ratios. The results for a
mass ratio of 16.2 are the same as in part (a) above. Figure 3-21a shows that the 0
dynamic force ratio Fi / W varies inversely with the mass ratio. However, the value 0 10 20
of the dynamic force is increasing with mass ratio as shown in Figure 3-21b, because
l/d ratio
the static force W is also increasing with mass ratio.
force in
Newtons

70 000
3.9 BEAM LOADING
60 000
A beam is any element that carries loads transverse to its long axis and may carry loads
50 000
in the axial direction as well. A beam supported on pins or narrow supports at each end
is said to be simply supported, as shown in Figure 3-22a. A beam fixed at one end and 40 000
unsupported at the other is a cantilever beam (Figure 3-22b). A simply supported beam 30 000
that overhangs its supports at either end is an overhung beam (Figure 3-22c). If a beam
has more supports than are necessary to provide kinematic stability (i.e., make the ki- 20 000
nematic degree of freedom zero), then the beam is said to be overconstrained or inde- 10 000
terminate, as shown in Figure 3-22d. An indeterminate beam problem cannot be
solved for its loads using only equations 3.3 (p. 78). Other techniques are necessary. 0
This problem is addressed in the next chapter. 0 10 20
l/d ratio
Beams are typically analyzed as static devices, though vibrations and accelerations
can cause dynamic loading. A beam may carry loads in three dimensions, in which case
equations 3.3a apply. For the two-dimensional case, equations 3.3b suffice. The review FIGURE 3-20
examples used here are limited to 2-D cases for brevity. Dynamic Force and Force
Ratio as a Function of l/d
Ratio for the System in
Example 3-1
Shear and Moment
A beam may be loaded with some combination of distributed and/or concentrated forces
or moments as in Figure 3-22. The applied forces will create both shearing forces and
bending moments in the beam. A load analysis must find the magnitudes and spatial
distributions of these shear forces and bending moments on the beam. The shear forces
V and the moment M in a beam are related to the loading function q(x) by
dV d 2 M
q( x ) = = (3.16 a)
dx dx 2

Ch 03 4ed Final 111 10/18/09, 10:18 AM

 112 MACHINE DESIGN - An Integrated Approach

The loading function q(x) is typically known and the shear V and moment M distribu-
force
ratio
tions can be found by integrating equation 3.16a:
5 000 VB xB
4 500
4 000
∫VA
dV =
∫x A
qdx = VB − VA (3.16b)

3 3 500 Equation 3.16b shows that the difference in the shear forces between any two points, A
3 000 and B, is equal to the area under the graph of the loading function, equation 3.16a.
2 500
2 000 Integrating the relationship between shear and moment gives
1 500 MB xB
1 000
500
∫M A
dM =
∫x A
Vdx = M B − M A (3.16c)

0
showing that the difference in the moment between any two points, A and B, is equal
0 10 20
to the area under the graph of the shear function, equation 3.16b.
mass ratio
SIGN CONVENTION The usual (and arbitrary) sign convention used for beams is to con-
sider a moment positive if it causes the beam to deflect concave downward (as if to
force in
collect water). This puts the top surface in compression and the bottom surface in ten-
newtons
sion. The shear force is considered positive if it causes a clockwise rotation of the sec-
16 000 tion on which it acts. These conventions are shown in Figure 3-23 and result in positive
14 000 moments being created by negative applied loads. All the applied loads shown in Fig-
12 000 ure 3-22 are negative. For example, in Figure 3-22a, the distributed load magnitude
from a to l is q = –w.
10 000
8 000 EQUATION SOLUTION The solution of equations 3.3 (p. 78) and 3.16 for any beam prob-
6 000 lem may be carried out by one of several approaches. Sequential and graphical solu-
tions are described in many textbooks on statics and mechanics of materials. One
4 000
classical approach to these problems is to find the reactions on the beam using equa-
2 000 tions 3.3 and then draw the shear and moment diagrams using a graphical integration
0 approach combined with calculations for the significant values of the functions. This
0 10 20 approach has value from a pedagogical standpoint as it is easily followed, but it is cum-
bersome to implement. The approach most amenable to computer solution uses a class
mass ratio of mathematical functions called singularity functions to represent the loads on the
FIGURE 3-21
beam. We present the classical approach as a pedagogical reference and also introduce
the use of singularity functions which offer some computational advantages. While this
Dynamic Force and Force
Ratio as a Function of approach may be new to some students, when compared to the methods usually learned
Mass Ratio for the in other courses, it has significant advantages in computerizing the solution.
System in Example 3-1

Singularity Functions
Because the loads on beams typically consist of collections of discrete entities, such as
point loads or segments of distributed loads that can be discontinuous over the beam
length, it is difficult to represent these discrete functions with equations that are valid
over the entire continuum of beam length. A special class of functions called singu-
larity functions was invented to deal with these mathematical situations. Singularity
functions are often denoted by a binomial in angled brackets as shown in equations 3.17.
The first quantity in the brackets is the variable of interest, in our case x, the distance
along the beam length. The second quantity a is a user-defined parameter that denotes
where in x the singularity function either acts or begins to act. For example, for a point
 Chapter 3 LOAD DETERMINATION 113

y l y l
V V
a a F
w 〈x–a〉0 〈x–a〉–1

M1
x x positive shear 3

R1 R2 R1

(a) Simply supported beam with (b) Cantilever beam with M M

uniformly distributed loading concentrated loading

y l y l
positive moment
a w〈x–a〉1 b
a
w 〈x–a〉0 FIGURE 3-23
Mz Beam Sign Convention
x x
〈x–0〉–2
R1 R2 R1 R2 R3
(c) Overhung beam with moment (d) Statically indeterminate beam with
and linearly distributed loading uniformly distributed loading

FIGURE 3-22
Types of Beams and Beam Loadings

load, the quantity a represents the particular value of x at which the load acts (see Fig-
ure 3-22b). The definition of this singularity function, called the unit impulse or Dirac
delta function, is given in equation 3.17d. Note that all singularity functions involve
a conditional constraint. The unit impulse evaluates to ∞ if x = a and is 0 at any other
value of x. The unit step function or Heaviside step function (Eq. 3.17c) evaluates
to 0 for all values of x less than a and to 1 for all other x.

Since these functions are defined to evaluate to unity, multiplying them by a coef-
ficient creates any magnitude desired. Their application is shown in the following three
examples and is explained in the most detail in Example 3-2B. If a loading function
both starts and stops within the range of x desired, it needs two singularity functions to
describe it. The first defines the value of a1 at which the function begins to act and has
a positive or negative coefficient as appropriate to its direction. The second defines the
value a2 at which the function ceases to act and has a coefficient of the same magni-
tude but opposite sign as the first. These two functions will cancel beyond a2, making
the load zero. Such a case is shown in Example 4-6 in the next chapter.
Quadratically distributed loads can be represented by a unit parabolic function,
2
x−a (3.17a)
which is defined as 0 when x ≤ a, and equal to (x – a)2 when x > a.

Linearly distributed loads can be represented by a unit ramp function,

1
x−a (3.17b)
 114 MACHINE DESIGN - An Integrated Approach

which is defined as 0 when x ≤ a, and equal to (x – a) when x > a.

A uniformly distributed load over a portion of a beam can be represented math-
ematically by a unit step function,
0
x−a (3.17c)
3
which is defined as 0 when x < a, unity when x > a, and is undefined at x = a.

A concentrated force can be represented by the unit impulse function,

−1
x−a (3.17d )
which is defined as 0 when x < a, ∞ when x = a, and 0 when x > a. Its integral evalu-
ates to unity at a.

A concentrated moment can be represented by the unit doublet function,

−2
x−a (3.17e)

which is defined as 0 when x < a, indeterminate when x = a, and 0 when x > a. It gen-
erates a unit couple moment at a.
This process can be extended to obtain polynomial singularity functions of any order
<x – a>n to fit distributed loads of any shape. Four of the five singularity functions de-
scribed here are shown in Figure 3-22, as applied to various beam types. A computer
program is needed to evaluate these functions. Table 3-7 shows five of the singularity
functions implemented in a “BASIC-like” pseudocode. The For loops run the variable
x from zero to the beam length l. The If test determines whether x has reached the value
of a which is the location of the start of the singularity function. Depending on this test,
the value of y(x) is set either to zero or to the specified magnitude of the singularity
function. This type of code can easily be implemented in any computer language (e.g.,
C+, Fortran, BASIC) or in an equation solver (e.g., Mathcad, MATLAB, TK Solver,
EES).

The integrals of these singularity functions have special definitions that, in some
cases, defy common sense but nevertheless provide the desired mathematical results.
For example, the unit impulse function (Eq. 3.17d) is defined in the limit as having zero
width and infinite magnitude, yet its area (integral) is defined as equal to one (Eq. 3.18d).
(See reference 8 for a more complete discussion of singularity functions.) The integrals
of the singularity functions in equations 3.17 are defined as

x 3
2 x−a
∫ −∞
λ − a dλ =
3
(3.18a)

x 2
1 x−a
∫ −∞
λ − a dλ =
2
(3.18b)

x
0 1
∫ −∞
λ − a dλ = x − a (3.18c)

x
−1 0
∫ −∞
λ−a dλ = x − a (3.18d )
 Chapter 3 LOAD DETERMINATION 115

Table 3-7 Pseudocode to Evaluate Singularity Functions

‘ Pulse Singularity Function

For x = 0 to l
If ABS (x - a) < 0.0001 Then y(x) = magnitude, Else y(x) = 0*
Next x
3
‘ Step Singularity Function
For x = 0 to l
If x < a Then y(x) = 0, Else y(x) = magnitude
Next x
‘ Ramp Singularity Function
For x = 0 to l
If x < = a Then y(x) = 0, Else y(x) = magnitude * (x – a)
Next x
‘ Parabolic Singularity Function
For x = 0 to l
If x < = a Then y(x) = 0, Else y(x) = magnitude * (x – a)^2
Next x
‘ Cubic Singularity Function * Note: This routine does not
generate the infinite value of the
For x = 0 to l Dirac delta function. Rather, it
If x < = a Then y(x) = 0, Else y(x) = magnitude * (x – a)^3 generates the magnitude of a
point load applied at location a for
Next x use in plotting a beam’s loading
function.

x
∫
−2 −1
λ−a dλ = x − a (3.18e)
−∞

where λ is just an integration variable running from –∞ to x. These expressions can be

used to evaluate the shear and moment functions that result from any loading function
that is expressed as a combination of singularity functions.

EXAMPLE 3-2A
Shear and Moment Diagrams of a Simply Supported Beam
Using a Graphical Method

Problem: Determine and plot the shear and moment functions for the simply
supported beam with uniformly distributed load shown in Figure 3-22a.

Given: Beam length l = 10 in, and load location a = 4 in. The magnitude of the
uniform force distribution is w = 10 lb/in.

Assumptions: The weight of the beam is negligible compared to the applied load and
so can be ignored.

Solution: See Figures 3-22a and 3-24.