Buchanan, G. R. “Shear and Moment Diagrams”

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Buchanan, G. R. “Shear and Moment Diagrams”
The Engineering Handbook.
Ed. Richard C. Dorf
Boca Raton: CRC Press LLC, 2000























































© 1998 by CRC PRESS LLC
7
Shear and Moment Diagrams
7.1 Sign Convention
7.2 Shear and Moment Diagrams
7.3 Shear and Moment Equations
George R. Buchanan
Tennessee Technological University
Computations for shear force and bending moment are absolutely necessary for the successful
application of the theory and concepts presented in the previous chapters. The discussion presented
here is an extension of Chapter 4 and the reader should already be familiar with computations for
reactions. This chapter will concentrate on statically determinate beams. Statically indeterminate
beams and frames constitute an advanced topic and the reader is referred to the end of the chapter
for further information on the topic. Even though the problems that illustrate shear and moment
concepts appear as structural beams, the reader should be aware that the same concepts apply to
structural machine parts. A distinction should not be drawn between civil engineering and
mechanical engineering problems because the methods of analysis are the same.
7.1 Sign Convention
The sign convention for moment in a beam is based on the behavior of the loaded beam. The sign
convention for shear in a beam is dictated by the convenience of constructing a shear diagram
using a load diagram. The sign convention is illustrated in Fig. 7.1. The x axis corresponds to the
longitudinal axis of the beam and must be directed from left to right. This convention dictates that
shear and moment diagrams should be drawn from left to right. The direction of the positive y axis
will be assumed upward, and loads that act downward on the beam will be negative. Note that it is
not mandatory in shear and moment computations for positive y to be directed upward or even
defined since the sign convention is independent of the vertical axis. However, for the more
advanced topic of beam deflections positive y must be defined [Buchanan, 1988]. Positive
bending moment causes compression at the top of the beam and negative bending moment causes
compression at the bottom of the beam. Positive shear forces act downward on the positive face of
the free body as shown in Fig. 7.1.























































© 1998 by CRC PRESS LLC
7.2 Shear and Moment Diagrams
Two elementary differential equations govern the construction of shear and moment diagrams and
can be derived using a free-body diagram similar to Fig. 7.1, with w corresponding to a continuous
load acting along the length of the beam.
dV = w dx or
Z
V
2
V
1
dV =
Z
x
2
x
1
w dx (7:1)
dM = V dx or
Z
M
2
M
1
dM =
Z
x
2
x
1
V dx (7:2)
The differential equations show that the change in shear V between any two points
x
1
and
x
2
on a
beam is equal to the area of the load diagram between those same two points and, similarly, that
the change in bending moment M is equal to the area of the shear diagram. It follows that the slope
of a tangent drawn at any point on the moment diagram is given by
dM=dx
and corresponds to the
magnitude of V at that point. When the tangent has zero slope,
dM=dx = 0;
that corresponds to a
maximum or minimum moment and can be located by examining the shear diagram for a point
(x location) where
V = 0:
Locating the largest positive or negative bending moment is important
for properly designing beam structures when using the equations of the previous chapters.
Shear and moment diagrams (as opposed to shear and moment equations) offer the most efficient
method for analyzing beam structures for shear and moment when the beam loading can be
represented as concentrated loads or uniform continuous loads. An elementary example will
serve to illustrate the concept. Consider the simply supported beam of Fig. 7.2. There is a single
concentrated load with reactions as shown. The shear is obtained by directly plotting the load; the
sign convention of Fig 7.1 specifically allows for this. The reaction on the left is plotted upward as
the change in the shear at a point,
x = 0:
The area of the load diagram between
x = 0
and
x = a
is zero since the load is zero; it follows that the change in shear is zero and the shear diagram is a
Figure 7.1 Beam element showing positive shear force V and positive bending moment M.























































© 1998 by CRC PRESS LLC
straight horizontal line extending from the left end of the beam to the concentrated load P. The
load changes abruptly by an amount P downward and a corresponding change is noted on the
shear diagram. Positive shear is above the axis of the shear diagram. There is no change in load
between
x = a
and
x = L
and the shear remains constant. The reaction at the right end of the
beam is upward and the shear is plotted upward to return to zero. The beam is simply supported,
indicating that the moment must be zero at the supports. The change in moment between the left
support and the point where the load is applied,
x = a;
is equal to the area of the shear diagram or
a positive
Pab=L:
The variation in moment appears as a straight line (a line with constant slope)
connecting
M = 0
at
x = 0
with
M = Pab=L
at
x = a:
The area of the remaining portion of the
shear diagram is
¡Pab=L
and, when plotted on the moment diagram, returns the moment to zero
and satisfies the simply supported boundary condition.
Figure 7.2
Shear and moment diagrams for a simple beam with a concentrated load.
Equations (7.1) and (7.2) indicate that the load function is integrated to give the shear function,
and similarly the shear function is integrated to give the moment function. The diagrams are a
graphical illustration of the integration. An important point is that the order (power) of each























































© 1998 by CRC PRESS LLC
function increases by one as the analyst moves from load to shear to moment. In Fig. 7.2, note that
when the load function is zero (1) the corresponding shear function is a constant and (2) the
corresponding moment function is linear in x and is plus or minus corresponding to the sign of the
area of the shear diagram.
Consider, as a second example, the beam of Fig. 7.3, where a series of uniform loads and
concentrated loads is applied to a beam with an overhang. The reactions are computed and shown
in the figure. The shear diagram is plotted starting at the left end of the beam. Shear and moment
diagrams are always constructed from left to right because Eqs. (7.1) and (7.2) were derived in a
coordinate system that is positive from left to right. The area of the load between
x = 0
and
x = 1
m is
¡4
kN; this value is the change in shear, which is plotted as a sloping line. The corresponding
change in bending moment equals the area of the shear diagram,
(¡4 kN)(1 m)=2 = 2 kN ¢ m:
A
curve with continually changing negative slops between
x = 0
and
x = 1 m
us shown in Fig. 7.3.
The point of zero shear occurs in the beam section,
2 m ∙ x ∙ 4 m;
and is located using similar
triangles as shown in the space between the shear and moment diagrams of Fig. 7.3. A textbook on
mechanics of materials or structural analysis should have a complete discussion of the topic.
Again, refer to "Further Information."
Figure 7.3
Shear and moment diagrams for a beam with an overhang. The concentrated loads and
uniform loads illustrate the concept of maximum moment and corresponding zero
shear.























































© 1998 by CRC PRESS LLC
Concentrated loads and uniform loads lead to shear diagrams with areas that will always be
rectangles or triangles; the change in moment is easily computed. The uniformly varying load
shown in Fig. 7.4(e) sometimes occurs in practice; locating the point of maximum moment (point
of zero shear) for some boundary conditions is not so elementary when using geometrical
relationships. In such cases the use of shear and moment equations becomes a valuable analysis
tool.
Figure 7.4
Shear and moment reactions for free-fixed beams.
7.3 Shear and Moment Equations
Shear and moment equations are equations that represent the functions shown in Figs. 7.2 and 7.3.
As with any mathematical function, they must be referenced to a coordinate origin, usually the left
end of the beam. Shear and moment equations are piecewise continuous functions. Note that two
separate equations are required to describe the shear diagram of Fig. 7.2 and, similarly, four
equations are required to describe the shear diagram of
Fig. 7.3
. The same is true for the moment























































© 1998 by CRC PRESS LLC
diagram.
The following procedure can be used to write shear and moment equations for almost any beam
loading: (1) Choose a coordinate origin for the equation, usually, but not limited to, the left end of
the beam. (2) Pass a free-body cut through the beam section where the shear and moment
equations are to be written. (3) Choose the free body that contains the coordinate origin. (4)
Assume positive unknown shear and moment at the free-body cut using the sign convention
defined by Fig. 7.1. (5) View the free body as a free-fixed beam with the fixed end being at the
free-body cut and the beam extending toward the coordinate origin. (6) Statically solve for the
unknown shear and moment as if they were the reactions at the fixed end of any beam, that is,
§F = 0
and
§M = 0:
(7) Always sum moments at the free-body cut such that the unknown
shear passes through that point. An example should illustrate the concept.
A complete description of the beam of Fig. 7.3 would require four shear and moment equations.
Consider a free body of the first sectionthe uniformly loaded overhang. The free body is shown
in Fig. 7.5(a). Compare Fig. 7.4(b) with Fig. 7.5(a); L is merely replaced with x, the length of the
free-body section. The seven steps outlined above have been followed to give
V
x
= ¡wx = ¡4x kN;M
x
= ¡wx
2
=2 = ¡4x
2
=2 kN ¢ m;0 ∙ x ∙ 1 m (7:3)
The second segment
(1 m ∙ x ∙ 2 m)
is shown in Fig. 7.5(b) and can be compared with Figs.
7.4(c) and 7.4(a): merely replace L with an x, a with 1 m, P with 12 kN, and w with 4 kN.
V
x
= ¡wa +P = ¡(4 kN=m)(1 m) +12 kN; (7:4)
M
x
= ¡wa(x ¡a=2) +P(x ¡a)
= ¡(4)(1)(x ¡1=2) kN ¢ m+12(x ¡1) kN¢ m (7:5)
The general idea is that any shear or moment equation can be broken down into a series of
individual problems that always correspond to computing the reactions at the fixed end of a
free-fixed beam. Continuing with the shear and moment equations for the beam of Fig. 7.3, the
third section
(2 m ∙ x ∙ 4 m)
is shown in Fig. 7.5(c) and should be compared with Figs. 7.4(a),
(c), and (d). There are four terms in each equation since there are four separate loadings on the free
body.
V
x
= ¡(4 kN=m)(1 m) +12 kN
¡4 kN ¡(3 kN=m)(x ¡2 m);(7:6)
M
x
= ¡(4 kN=m)(1 m)(x ¡1 m=2) +(12 kN)(x ¡1 m) ¡(4 kN)(x ¡2 m)
¡(3 kN=m)(x ¡2 m)
2
=2 (7:7)
The last beam section
(4 m

x

5 m)
is shown in
Fig. 7.5
(d). There are five separate loadings























































© 1998 by CRC PRESS LLC
on the beam. The shear and moment equations are written again by merely summing forces and
moments on the free-body section.
V
x
= ¡(4 kN=m)(1 m) +12 kN
¡4 kN ¡(3 kN=m)(2 m) ¡6 kN;(7:8)
M
x
= ¡(4 kN=m)(1 m)(x ¡1 m=2) +(12 kN)(x ¡1 m) ¡(4 kN)(x ¡2 m)
¡(3 kN=m)(2 m)(x ¡3 m) ¡(6 kN)(x ¡4 m) (7:9)
The point of maximum moment corresponds to the point of zero shear in the third beam section.
The shear equation, Eq. (7.6), becomes
V
x
= 10 ¡3x:
Setting
V
x
to zero and solving for x gives
x = 3:33 m
. Substituting into the corresponding moment equation, Eq. (7.7), gives the maximum
moment as
8:67 kN ¢ m:
Figure 7.5
Free-body diagrams for the beam shown in Fig. 7.3.
Defining Terms
Beam deflections: A theory that is primarily based upon the moment behavior for a beam and
leads to a second-order differential equation that can be solved to give a mathematical
equation describing the deflection of the beam.
Concentrated load: A single load, with units of force, that can be assumed to act at a point on a
beam.























































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Free body: A section that is removed from a primary structural system and is assumed to be in
equilibrium mathematically.
Free-fixed beam: A beam that is free to rotate and deflect at one end but is completely clamped
or rigid at the other end (also known as a cantilever beam).
Maximum or minimum moment: The bending moment that usually governs the design and
analysis of beam structures.
Simply supported beam: A beam that is supported using a pin (hinge) at one end and a surface at
the other end, with freedom to move along the surface.
Statically determinate beam: A beam that can be analyzed for external reactions using only the
equations of engineering mechanics and statics.
Uniform continuous load: A distributed beam loading of constant magnitude, with units of force
per length, that acts continuously along a beam segment.
References
Buchanan, G. R. 1988. Shear and moment in beams, Chap. 5, and Deflection of beams, Chap. 10, in
Mechanics of Materials. Holt, Rinehart and Winston, New York.
Further Information
Hibbeler, R. C. 1985. Structural Analysis. Macmillan, New York. Chapter 3 contains a discussion
of shear and moment concepts. Chapters 8 and 9 cover fundamental concepts for analysis of
indeterminate beams.
McCormac, J. and Elling, R. E. 1988. Structural Analysis. Harper & Row, New York. Chapter 3
contains a discussion of shear and moment concepts. Chapters 10, 11, and 13 cover
fundamental concepts for analysis of indeterminate structures.
Gere, J. M. and Timoshenko, S. P. 1990. Mechanics of Materials, 3rd ed. PWS, Boston. Chapter 4
contains a discussion of shear and moment concepts. Beam deflections are covered in Chaps.
7, 8, and 10.
Nash, W. A. 1994. Theory and Problems of Strength of Materials, 3rd ed., McGraw-Hill, New
York. Numerous solved problems for shear and moment are given in Chap. 6























































© 1998 by CRC PRESS LLC