MAE314 Solid Mechanics Fall, 2008
51
Dr. Yuan
Chapter 5. Equilibrium of Beams
5.1 Introduction
→
Three internal forces {axial, shear and bending moment}
Bar
transmits axial force.
Shaft
transmits torque.
Beam
transmits axial force, shear force, and bending moment.
The assumptions about the geometry of the beam:
1. The axis of the beam is straight
2. Crosssectional dimensions (such as height, h, and width, b, for the rectangular section) are
small compared to the length of the beam (i.e., h/L << 1 and b/L << 1). In addition, cross
sectional dimensions are either constant or vary slowly with position along the axis of the
beam.
The axis of the length L of the beam is taken to be the xaxis.
Three types of supports
Type of
Supports
Symbol Types of
translations
and rotations
that the
connection
allows
Types of forces that can be
developed at the connection
Types of forces
that can be
developed when
the support is
inclined
Fixed
(Clamped)
Pinned
Roller
90
o
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Dr. Yuan
pinned
R
Ax
R
A
y
A
two unknowns
roller
R
B
B
one unknown
clamped
R
Cx
C
R
Cy
three unknowns
In summary,
Pinned support: restrain the beam from any translational movements at the support
can resist a general force in any direction, but cannot resist the moment
about the support
Roller support: restrain the beam from any translational movement normal to the
supporting surface.
cannot resist moment and lateral force along the surface of support
Clamped support: restrain the beam from both translation and rotation.
can resist force in any direction and moment about the connecting end.
simply supported beam
cantilever beam (clamped
end)
beam with an overhang
Loads
Point Load
( Concentrated )
Point Moment
( Concentrated )
Uniformly Distributed Load
Ramp Load
Distributed Load
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Dr. Yuan
Statically determinate structure (The reaction forces can be obtained from
0=
∑
x
F
,
0=
∑
y
F
,
0=
∑
M
. Otherwise, the structure is called statically indeterminate structure.
Examples 5.1.1 Determine the reaction forces and moments at the supports
(A)
L
a
b
R
a
R
b
A
B
P
1
q
↑
F
y
∑
=
0
,
R
a
+
R
b
=
P
1
+
qb
(a)
M
B
∑
=
0
,
qb
b
2
+ P
1
(L − a) − R
a
L = 0
R
a
=
P
L
−
a
L
+
q
b
2
2
L
(a)
R
b
= P
1
+ qb − R
a
=
Pa
L
+ qb
2
L
− b
2
L
(B)
L
a
C
R
a
R
b
A
B
P
2
M
1
↑
F
y
∑
=
0
R
a
+
R
b
=
P
2
M
A
∑
=
0
−
P
2
a
+
R
b
L
+
M
1
=
0
=
=
R
b
=
P
2
L
a −
M
1
L
R
a
= R
b
− P
2
=
P
2
(
L
− a)
L
+
M
1
L
or,
∑
=
0
B
M
0)(
12
=
+
−
+
−
MaLPLR
a
L
M
L
aLP
R
a
12
)(
+
−
=
This subsection begins a study of the type of internal forces and moments generated in a
beam carrying an external force system that acts transversely to the axis of the beam. The
concepts of shear and bending moment in beams introduced here are of fundamental importance
to an understanding of the behavior of beams under load. They also provide the basis for
developing tools for designing beams.
Consider the loaded cantilever beam illustrated in the figure below. There are two
primary ways in which the beam might fail as a result of the applied load. One potential type of
failure is for the load to cause two contiguous parts of the beam to slide relative each other in a
direction parallel to their plane of contact. This is called a shear failure. The internal force
developed in the beam that is associated with this phenomenon is called an internal shear force.
This internal shear force is developed in response to the components of the external force system
that act transversely to the long axis of the beam and tend to cause the type of transverse sliding
indicated in Fig. (c). These internal shear forces resist or balance the net external shearing force
MAE314 Solid Mechanics Fall, 2008
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Dr. Yuan
tending to cause the sliding. Failure of the type illustrated occurs when the beam can no longer
provide an equilibrium internal force of this type.
The second possible failure mode of failure is that illustrated in Fig. (c). This type of
failure is obviously associated with the tendency of the transverse external force to cause part of
the beam to rotate, or bend. Since the free rotation cannot occur in the clamped end (unless a pin
support is present), an internal resisting or balancing moment equal and opposite to the applied
moment (the moment associated with the rotational tendencies of the external forces) must be
developed within the beam. Failure of the type shown occurs when the beam can no longer
provide a resisting moment equal to the applied moment.
At any section of the loaded beam, internal shear forces and moments are developed
simultaneously. If the beam is decomposed at this point into two parts, the forces and moments
developed internally serve to maintain the translational and rotational equilibrium of each part.
They also represent the internal actions and reactions of one part of the beam on the other part.
As will be seen later, one of the objectives of the structural design process is to create a structural
configuration capable of providing these internal shears and moments in an effective way and
with factors of safety sufficient to prevent shear and moment failure.
(a) Cantilever beam.
(
b) Internal forces and moments are
generated in the beam which serves
to maintain the equilibrium of
portions of the beam.
(c) Possible failure modes: When
the beam cannot provide these
internal forces or moments, failure
of the type indicated can occur.
Sign Convention of Shear Force and Bending Moment
Unfortunately there are more than one type of sign conventions used by different authors. A
given sign convention depends on the coordinate and the relations of the forces and moments
with the stresses.
sign convention for the textbook:
x
V
M
(+)
(+)
1. A positive shear force tends to deform the element by causing the righthand face to move
downward with respect to the lefthand face
2. A positive moment gives compression on the upper point of the beam, tension on the lower
part of the beam.
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Dr. Yuan
Positive shear forces, V, tend to drag the beam down on the right. Positive bending
moments, M, tend to curve the beam concave upward.
In the next chapter it will be shown that the shear force V and moment M are directly
related to the stress components
τ
xy
and
σ
x
respectively. Therefore, the evaluation of the strength
of the beam will depend strongly on the magnitudes of the shear forces and moments. For
analyzing the beams, it is very important to have a systematic way to determine the locations and
magnitudes of the maximum moment M and of the maximum shear V transmitted by a beam
under a given set of loading.
Before determining the shear forces and moments along the beam, consider the situation
regarding the shear transmitted at the end support such as shown in the following figure.
Left reaction
(a)
R
1
R
1
V
Positive shear
(b)
R
1
V
Negative shear
(c)
A freebody diagram of a thin slice of the beam together with the corresponding forces exerted
on the rest of the beam is shown in the figure b. Recalling the definition for positive shear, it is
clear that when the reaction is in the direction shown, the shear is equal to the value of the
reaction and is positive in sign. Similarly, the freebody diagram of figure c shows that the shear
is negative the value of the reaction when the reaction is oppositely directed. For a support at the
right end of the beam as shown in the following figure, the freebody diagrams of the figs. b and
c picture the cases where the shear is negative and positive, respectively.
Right reaction
(a)
R
2
R
2
V
Negative shear
(b)
Positive shear
(c)
V
R
2
Thus the shear at the ends of a beam can be determined by inspection on the basis of the
values and directions of the reactions. Note also that at an unsupported end such as the overhang
shown, the shear must be zero on the basis of equilibrium for the freebody diagram of a thin
slice of the beam, as shown in fig. b. The shear force an unloaded overhang at the left end of the
beam would also be zero.
Overhang
(a)
R
2
Zero shear
(b)
V=0
R
2
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Dr. Yuan
If a vertical force P and moment M
1
applied to a free end, then V = P, and M = M
1
.
M
1
M
1
P
V
M
P
Similarly at a right free edge, V = P, M = M
1
.
M
1
P
V
M
P
M
1
5.2 Equilibrium of Beams using Finite FreeBody Diagrams
Example 5.2.1.
Determine M and V.
L/4
A B
P
L/4 L/2
M
0
L/4
R
ax
B
P
L/4
L/2
M
0
A
x
R
ay
R
by
First step: determine reaction forces and moments.
(
∵
simplysupported ends
→
no moments)
→
F
x
=
0
∑
†=
⇒
R
ax
=
0
↑
F
y
=
0
∑
††
⇒
R
ay
+
R
批
=
P
(1)
M
A
=
0
∑
=††=†
⇒ −
L
4
P − M
0
+ R
by
L = 0
(2)
(1) and (2)
R
by
=
P
4
+
M
0
L
R
ay
= P − R
ax
=
3P
4
−
M
0
L
R
ax
= 0
(a)
0 <
x
?
L
/4
M
V
x
R
ay
↑
F
y
∑
=
0
,
R
ay
−
σ
=
0
,
V
=
R
ay
M
A
∑
=
0
,
−
x
V
+
M
=
0
,
M
=
x
V =
x
R
ay
V
(
x
) =
3
P
4
−
M
0
L
M
(
x
) = (
3
P
4
−
M
0
L
)
x
0 <
x
<
L
/4
(3)
(b)
L
/4 <
x
?
L
/2
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Dr. Yuan
M
V
L/4
R
ay
P
x
↑
F
y
∑
=
0,
R
ay
−
P
−
σ
=
0,
V
= R
ay
− P
M
A
∑
=
0
, −P
L
4
− xV + M = 0, M= xV + P
L
4
V(x) = −
P
4
−
M
0
L
M(x) = x(−
P
4
−
M
0
L
) +
PL
4
L
4
< x <
L
2
(4)
(c)
L
2
< x < L
x
L/4
L/4
R
ay
P
M
0
V
M
↑
F
y
∑
=
0
,
R
ay
−
P
−
σ
=
0
V
=
R
ay
−
P
M
A
∑
=
0
,
−P
L
4
− M
0
− xV + M = 0
M= P
L
4
+ M
0
+ xV
V(x) = −
P
4
−
M
0
L
M(x) =
PL
4
+ M
0
+ x(−
P
4
−
M
0
L
)
L
2
< x < L
(5)
(d) M, V at a small distance to the left of the middle of the beam
V(
L
2
) = −
P
4
−
M
0
L
←
=
敱渮
㐩e
→
M(
L
2
)
=
L
2
(
−
P
4
−
M
0
L
)
+
PL
4
=
PL
8
−
M
0
2
←
=
⡥⤠(M, V at a small distance to the right of the middle of the beam
V(
L
2
)
= −
P
4
−
M
0
L
←
=
敱渮
㔩e
→
M(
L
2
) =
PL
4
+ M
0
+
L
2
(−
P
4
−
M
0
L
) =
PL
8
+
M
0
2
←
=
卨敡爠f→牣r= ←a楮i⁴he慭e= r→ss⁴桥i摤汥≤潦⁴桥= 慭.⁂= 湤i↑朠g→←e湴↑捲敡獥猠
慬来扲慩捡汬y== M
0
across the middle of the beam.
Example 5.2.2
Determine M and V at x from the free end.
MAE314 Solid Mechanics Fall, 2008
58
Dr. Yuan
A
B
x
L
q
q
0
↑
F
y
∑
=
0,
−q
x
2
2
− V = 0
V = −q
x
2
2
= −q
0
x
2
2
L
(q
0
= qL)
←
†
A
x
q
qx
V
M
M
A
∑
=
0
−
2
x
3
(
q
x
2
2
) −
Vx
+
M
= 0
M=
qx
3
3
+ Vx = q
x
3
3
+ (−q
x
2
2
)x
= −q
x
3
6
= −
q
0
x
3
6
L
(q
0
= qL)
←
=
A
x
V
M
2
x
3
x
3
qx
2
2
V(x) = −
q
0
x
2
2L
M(x) = −
q
0
x
3
6
L
0≤ x ≤ L
x = 0,
V
=
M
=
〠
x = L,
V = −
q
0
L
2
,M = −
q
0
L
2
6
Example 5.2.3
Determine M and V at point D
3 m
=28 kNP
=6 kN/mq
A
D
B
C
R
a
R
b
5 m
8 m 2 m
First step: calculate reactions at the supports.
↑ F
y
∑
= 0
, R
a
− 28
−
㘨6
+
㈩
+
R
b
=
0
R
a
+
R
b
=
88 kN (1)
M
A
∑
= 0
,
−3(28) − 5[6
×
⠸
+
㈩2
+
㠨 R
b
)
=
0
=
=
−
84
−
㌰3
+
8
R
b
=
0
R
b
=
48 kN (2)
From (1)
R
a
=
88
−
㐸
=
㐰 歎
=
MAE314 Solid Mechanics Fall, 2008
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Dr. Yuan
28 kN
6 kN/m
A
D
B C
40 kN
48 kN
D
6 kN/m
V
M
M
V
(b)
(c)
(A) From (b)
↑ F
y
∑
=
0,
40
−
㈸
−
㘨㔩
−
V
=
0
=
=
V
=
−
18 kN
M
A
∑
=
0,
−
3(28)
−
5
2
(6
×
5)
−
5(
V
)
+ M =
0
M
=
5
V +
84
+
75
=
㔨
−
ㄸ1
+
ㄵ1
=
㘹 歎
⋅
←
=
←
=
⡂⤠)
V
,
M
can also be determined from (c)
↑ F
y
∑
=
0
,
V
−
6(5)
+
㐸
=
0
=
=
V
=
−
18 kN
M
D
∑
=
0
,
−M−
5
2
(6
×
5)
+
3(48)
=
0
M
=
−
75
+
ㄴ1
=
㘹 歎
⋅
←
=
←
=
5.3 Equilibrium Relationships Among Loads, Shear Force, and Bending Moment
(A) Distributed Load
(
In the textbook, p(x) = q(x)
)
x
x
d
x
q(x)
dx
A
M
V
M+ Md
V+ Vd
q(x)
↑ F
y
∑
=
0
,
V −
(
V
+
d
V
)
−
q
(
x
)d
x
=
0
=
=
≤
V
d
x
= −
q(x)
(5.3.1)
M
A
∑
=
0
,
−M− q
d
x
(
d
x
2
)
−
(
V +
d
V
)d
x + M+
d
M =
0
neglected
d
2
d
d
d
V
x
qV
x
M
++=
MAE314 Solid Mechanics Fall, 2008
510
Dr. Yuan
d
M
d
x
= V
(5.3.2)
d
V
d
x
= −
q
The rate of change of
V
w.r.t.
x =

q
Special cases:
(1)
q
= 0 on the portion of the beam,
V
= constant
(2)
q
=
c
( uniformly distributed),
V
= 
cx
+ const.
d
V
A
B
∫
= − q
d
x
A
B
∫
,
V
B
− V
A
= − q
d
x
A
B
∫
* Restriction of the above equation:
no concentrated load acts between
A
and
B
.
d
M
d
x
=
V
The rate change of
M
=
V
Special cases:
(1)
V
= 0 on the portion of the beam,
M
= constant
V
= 0 on some points,
M
may reach a maximum or minimum.
(2)
V
=
c
on the portion of the beam,
M
=
cx
+ constant
d
M
A
B
∫
= V
d
x
A
B
∫
,
M
B
− M
A
= V
d
x
A
B
∫
* Restriction of the above equation
no concentrated moment acts between
A
and
B
Note: concentrated loads are O.K. between
A
and
B
(B) Concentrated Load acting on the Element
dx
A
M
V
M+M
1
V+V
1
P
↑
F
y
∑
=
0
,
V
−
P
−
σ
−
σ
1
=
0
Ⱐ,†=
V
1
= −
P
M
A
∑
=
0
,
−M− P
d
x
2
−
(
V + V
1
)d
x + M+ M
1
=
0
M
1
=
(
P
2
+ V + V
1
)d
x
d
x
→
0,
M
1
=
0
=
C潮→汵li潮→
㨠
= ㄮ†τhe↑=愠a→湣↑↑瑲tte≤潡≤=慣瑳琠t=p→i↑琬⁴her攠潣c×牳↑=慢r異琠cha湧↑==sh敡爠f→牣敳e=
= ㈮†τhe↑=愠a→湣↑↑瑲慴a≤潡≤=慣瑳琠t=p→i↑琬⁴her攠楳→=cha↑ge==←→←e湴↑灡獳s湧↑瑨牯rgh=
瑨攠灯楮t.=
=
乯te㨠
= 䱥晴楤e=→f⁴=攠敬敭e↑琠
=
≤
M
d
x
=
V
Right side of the element
MAE314 Solid Mechanics Fall, 2008
511
Dr. Yuan
d(
M
+
M
1
)
d
x
=
d
M
d
x
= V + V
1
= V − P
d
M
d
x
=
V
−
P
⇒
When a concentrated load acts at a point
d
M
d
x
decreases by 
P
from the left of the load to the right of the load
(C) Concentrated moment acting on the element
dx
A
M
V
M+M
1
V+V
1
M
0
↑
F
y
∑
=
0,
⇒
V
1
=
0
=
M
A
∑
=
0
,
−
M
+
M
0
−
(
V
+
V
1
)
d
x
+
M
+
M
1
=
0
M
1
=
−
M
0
+
(
V
+
V
1
)
d
x
d
x
→
0,
M
1
=
−
M
0
Conclusions
:
1. The shear force does not change at the point of application of a concentrated moment.
2. There is an abrupt decrease in the bending moment due to the
M
0
5.4 Shear and Bendingmoment Diagrams
There is commonly a variation in the magnitude (and often the sense) of the shears and
moments present at different sections in a structure. As an aid in visualizing the distribution of
these shears and moments, the values thus found can be plotted graphically to produce what are
called
shear and moment diagrams
. Shear and moment diagrams are invaluable aids in the
analysis and design of structures, and a working knowledge of what they mean and how to draw
them quickly is of permanent importance.
Example 5.4.1
Determine shear and moment diagrams.
q
A B
a
R
a
R
b
x
L
b
c
(a)
First step: determine reactions
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Dr. Yuan
qb
A
B
R
a
R
b
L
a +
b
2
b
2
+ c
R
a
= qb
(
b
2
+ c
)/
L =
qb
L
(
b
2
+ c
),
R
b
= qb
(
a +
b
2
)/
L =
qb
L
(
a +
b
2
)
Divide into three regions:
(a)
0
<
x
?a
M
V
R
a
x
F
y
∑
=
0
,
V
=
R
a
M
A
∑
=
0
,
M
=
xV
=
R
a
x
V
=
R
a
(cons
t
.)
M= R
a
x
0 < x < a
(b)
a
<
x
<
a
+
b
M
V
V(x=a)
x
M(x=a)
a ax
q(xa)
F
y
∑
=
0
,
V
(
x
=
a)
−
q(
x
−
a)
−
V
=
0
=
=
V
=
V
(
x
=
a
)
−
q
(
x
−
a
)
=
R
a
−
q
(
x
− a
)
←
(1)
M
∑
at
x
=
0
−M
(
x = a
)
−
(
x − a
)
V
(
x = a
)
+
x
?
a
2
q
(
x − a
)
+ M=
0
M= R
a
a +
(
x − a
)
R
a
−
q
2
(
x − a
)
2
= R
a
x −
q
2
(
x − a
)
2
←
=
⡣⤠(
a
+
b
<
x
<
L
M
R
b
V
Lx
F
y
∑
=
0
,
V
=
−
R
b
M
at
x
∑
=
0
,
M
=
R
b
(
L
−
x
)
d
M
d
x
=
V
Maximum value of
M
≡
V
=
0
=
䙲潭=渮
ㄩ↑=
=
R
a
− q
(
x
−
a
)
=
〬=†==
x = a +
R
a
q
M
max
= R
a
(
a +
R
a
q
)
−
q
2
[
a +
R
a
q
− a
]
2
=
qb
8
L
2
(
b +
2
c
)(4
aL +
2
bc + b
2
)
MAE314 Solid Mechanics Fall, 2008
513
Dr. Yuan
R
b
V
0
(b)
R
a
M
max
M
0
(c)
Example 5.4.2
Determine shear and moment diagrams.
A
B
a
P
1
x
L
b
P
2
(a)
(1) 0<x<a
P
1
V
M
x
↑
F
y
∑
=
0
,
−
V
−
P
1
=
0
Ⱐ,†=
V
= −
P
1
M
at
x
=
x
∑
=
0
,
M
+
P
1
x
=
0
Ⱐ,†=
M
= −P
1
x
(2) a<x<L
P
1
P
2
M
x
V
↑
F
y
∑
=
0
,
−
P
1
−
P
2
−
V
=
0
Ⱐ,†=
V
= −P
1
− P
2
M
at
x
=
x
∑
=
0
,
M
+
P
1
x
+
P
2
(
x
− a) = 0
M
=
−
P
1
x
−
P
2
(
x
− a)
P
1
V
P
1
P
2
0
(b)
MAE314 Solid Mechanics Fall, 2008
514
Dr. Yuan
P
1
a
M
P
1
LP
2
b
0
(c)
Example 5.4.3
Determine shear and moment diagrams.
q
=1.0 k/ft
B
A
R
b
R
c
M
0
= 12.0 ftk
L/2=8 ft
b=4 ft
C
L/2=8 ft
(a)
Determine reaction forces at B and C.
M
B
∑
=
0
,
2
×
1.0
×
4
+
ㄲ
−
ㄶ
×
R
C
=
0
R
C
=
1.25
k
M
C
∑
=
0
,
(8 + 8+ 2)
×
ㄮ0
×
4
−
ㄶ
×
R
B
+
12
=
0
=
=
R
B
=
5.25 k
0
4.0
+1.25
V
(k)
0
8.0
+2.0
M
(ftk)
10.0
+
()
()
()
Example 5.4.4
Determine shear and moment diagrams.
MAE314 Solid Mechanics Fall, 2008
515
Dr. Yuan
x
L
A
q
M
b
R
b
B
Reactions:
↑
F
y
∑
=
0,
R
B
−
煌
=
0
,
R
B
=
煌
M
b
∑
=
0
,
−
M
B
+
L
2
(qL)
=
0
, M
B
=
qL
2
2
FBD
x
qx
V
M
F
y
∑
=
0
,
V
=
−
煸
M
at
x
=
x
∑
=
0,
M
= −
qx
2
2
V
0
qL
()
M
0
()
−
qL
2
2
Example 5.4.5
Determine: (1) M in AB; (2) M
max
in AB;
A
q
C
B
x
2b
2q
b
M
C
∑
=
0
,
R
a
=
2qb
3
M
D
∑
=
0
M
+
qx
2
2
−
2
qb
3
x
=
0
M
=
2qb
3
x
−
qx
2
2
d
M
d
x
=
2qb
3
−
qx
=
0
MAE314 Solid Mechanics Fall, 2008
516
Dr. Yuan
A
x
q
D
V
M
2
qb
3
x
=
2b
3
d
2
M
d
x
2
= −
q
<
0,max.
M
max
x =
2b
3
=
2qb
2
9
Review:
Sign convention of moment and shear force
d
V
d
x
= −
q(x)
(
In the textbook, p(x) = q(x)
)
d
M
d
x
= V
MAE314 Solid Mechanics Fall, 2008
517
Dr. Yuan
P
a
b
L
R
a
=
P
b
L
R
b
=
Pa
L
R
a
=
P
b
L
V
0
− R
b
= −
Pa
L
P
M
0
P
ab
L
M
0
a b
L
V
0
R
a
M
0
R
a
=
M
0
L
R
b
= −
M
0
L
R
b
M
0
L
M
0
−
M
0
b
L
M
0
a
L
q
L
V
0
M
0
qL
2
qL
2
qL
2
−
qL
2
qL
2
8
q
b
V
0
M
0
a c
R
a
=
qb
L
(c +
b
2
) R
b
=
qb
L
(a +
b
2
)
R
a
R
b
0
M
max
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