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Great explanation for drawing the shear and moment diagrams for the beam
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The term beam refers to a slender bar that carries
transverse
loading
; that is, the applied force are perpendicular to the bar.
In a beam, the internal force system consist of a shear force and a bending moment acting on the cross section of the bar. Theshear force and the bending moment usually vary continuouslyalong the length of the beam.
The internal forces give rise to
two
kinds of stresses on a
transverse section of a beam: (1) normal stress that is caused bybending moment and (2) shear stress due to the shear force.
Knowing the distribution of the shear force and the bending moment in a beam is essential for the computation of stressesand deformations. Which will be investigated in subsequentchapters.
Figure 4.
Statically determinate beams
cantilever beam
is built into a rigid support at one end, with the
other end being free, as shown in Fig.4.1(b). The built-in supportprevents displacements as well as rotations of the end of the beam.
An
overhanging beam
, illustrated in Fig.4.1(c), is supported by a
pin and a roller support, with one or both ends of the beamextending beyond the supports.
The three types of beams are
statically determinate
because the
support reactions can be found from the equilibrium equations.
( )
g
g
( )
g
g
(c) 2003 Brooks/Cole Publishing / Thomson Learningā¢(c) 2003 Brooks/Cole Publishing / Thomson Learningā¢
concentrated load
, such as
in Fig. 4.1(a). In contrast a
distributed load
is applied over a finite area. If the distributed
load acts on a very narrow area, the load may be approximated bya
line load
The
intensity
w
of this loading is expressed as force per unit
length (lb/ft, N/m, etc.) The load distribution may be uniform, asshown in Fig.4.1(b), or it may vary with distance along the beam,as in Fig.4.1(c).
The weight of the beam is an example of distributed loading, butits magnitude is usually small compared to the loads applied tothe beam.
( )
g
g
( )
g
g
(c) 2003 Brooks/Cole Publishing / Thomson Learningā¢(c) 2003 Brooks/Cole Publishing / Thomson Learningā¢
4.3 Shear- Moment Equations and Shear-Moment Diagrams
Ā
The determination of the internal force system acting at a
given
section of a beam : draw a free-body diagram that expose theseforces and then compute the forces using equilibrium equations.
The goal of the beam analysis
determine the shear force
and
the bending moment
at
every
cross section of the beam.
To derive the expressions for
and
in terms of the distance x
measured along the beam. By plotting these expressions to scale,obtain the
shear force and bending moment diagrams
for the
beam.
The shear force and bending moment diagrams are convenientvisual references to the internal forces in a beam; in particular,they identify the maximum values of
and
a.
Sign conventions
(
)
k /
l
bli hi
/
h
i
(
)
k /
l
bli hi
/
h
i
Figure 4.3 Sign conventions for external loads; shear force,
and bending moment.
Determine the expressions for
and
from the equilibrium
equations obtainable from the FBD. These expressions, whichare usually functions of x, are the shear force and bendingmoment equations for the segment.
Plot the expressions for V and M for the segment. It is visually desirable to draw the
-diagram below the FBD of the entire
beam, and then draw the
-diagram.
The bending moment and shear force diagrams of the beam arecomposites of the
and
diagrams of the segments. These
diagrams are usually discontinuous, or have discontinuousslopes. At the end-points of the segments due to discontinuitiesin loading.
The simply supported beam in Fig. (a) carries two concentratedloads. (1) Derive the expressions for the shear force and the bendingmoment for each segment of the beam. (2) Sketch the shear forceand bending moment diagrams. Neglect the weight of the beam.Note that the support reactions at
and
have been computed and
are shown in Fig. (a).
Solution Part 1 The determination ofthe expressions for
and
for each of the
three beam segments(
, and
) is
explained below.
Segment AB
ļ¼xļ¼
5 m)
y
= +18-14 = +4 kN
Answer
E
= +18x-14(x-2) = 4x+28 kNĀ· m
Answer
Segment CD
(5 m
ļ¼xļ¼
7 m)
y
= +18-14-28 = -24 kN
Answer
G
= +18x-14(x-2) ā (x-5) = -24x+168 kNĀ· m
Answer
The simply supported beam in Fig. (a) is loaded by the clockwisecouple
0
at
. (1) Derive the shear and bending moment
equations. And (2) draw the shear force and bending momentdiagrams. Neglect the weight of the beam. The support reactions Aand C have been computed, and their values are shown in Fig. (a).^ SolutionPart 1^ Due to the presenceof the couple
0,
We
must analyzesegments
and
separately.
Segment AB (
0ļ¼xļ¼
Answer
Answer
Segment BC (
4ļ¼xļ¼L)
Answer
Answer 0
0
0
= ā ā ā + = ā
V
L C
F
y
L C
V
0
ā
=
ā
=
0
D
M
x
L C
M
0
ā
=
0
0
=
M
x
L C
0
0
0
= ā ā ā + = ā
V
L C
F
y
L C
V
0
ā
=
ā
=
0
E
M
0
0
0
=
ā
M
C
x
L C
0
0
C
x
L C
M
ā
=
The cantilever beam inFig.(a) carries a triangularload. The intensity ofwhich varies from zero atthe left end to 360 lb/ft atthe right end. In addition, a1000-lb upward verticalload acts at the free end ofthe beam. (1) Derive theshear force and bendingmoment equations. And(2) draw the shear forceand bending momentdiagrams. Neglect theweight of the beam.
Solution Note that the triangular load has beenreplaced by is resultant, which is theforce 0.5 (12) (360) = 2160 lb (areaunder the loading diagram) acting atthe centroid of the loading diagram.
Part 1 Σ
y
x
2
x
2
lb
Answer
C
x
x
2
x
x
x
3
lbĀ· ft
Answer
Because the loadingis continuous, thebeam does not haveto be divided intosegment.
w
x
= 360/12, or
w
x
lb/ft.