CIVE1143 – Structural Analysis Shear Force and Bending ......Sep 17, 2017 · Shear Force and...
Transcript of CIVE1143 – Structural Analysis Shear Force and Bending ......Sep 17, 2017 · Shear Force and...
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CIVE1143 – Structural Analysis
Shear Force and Bending Moment
Equations and Diagrams
Lecturers: Srikanth & Indu
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Revision: Member Internal Forces
To determinate the internal forces at a
point such as B, we need to imaginarily cut
the beam through the point into two
segments. Thus the internal loading will be
exposed on the free-body diagram of the
segment.
A B
1P2P
xA
yAAM
xA
yAA B
AM
N
V
M
N
V
MB
1P2P
OR
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Revision: Sign Convention
N N
N N
V V
V V
M M
M M
Positive axial force
(tension)Positive shear force Positive bending moment
N
V
M
N
V
M
xA
P
A BC
yAyB
B
yB
CNCV
CM
P
AxA
yA
CN
CV
CM
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Revision: Calculation of the Internal Forces
xA
P
A BC
yAyB
P
AxA
yA
CN
CV
CM
• Identify a section in the structure where
you wish to determine the values of the
internal forces.
• Consider the free body diagram from
this section to the end of the beam
obtained by “cutting” the body through
this section. You may consider either the
right or the left part of the beam obtained,
whichever gives the easier calculation.
•Use the sign convention defined earlier
to provide the positive direction of these
forces. Treat the 3 internal forces at the
section as unknowns.
• 3 unknown internal forces can be
determined by 3 equilibrium equations.
∑ = 0xF+ ∑ = 0yF+ ∑ = 0BM+
B
yB
CNCV
CM
OR
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V and M Functions
� Design of beam requires detailed knowledge of the
variations of V and M.
� Internal axial force N is generally not considered as
��The loads applied to a beam normally act perpendicular to the beThe loads applied to a beam normally act perpendicular to the beamam’’s s
axisaxis
��For design purpose, a beamFor design purpose, a beam’’s resistance to shear & bending is more s resistance to shear & bending is more
important than its ability to resist normal forceimportant than its ability to resist normal force
��An exception is when it is subjected to compressive axial force An exception is when it is subjected to compressive axial force where where
buckling may occurbuckling may occur
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V and M Functions
� To obtain the variation of V and M, it is necessary to section the beam at
an arbitrary distance x from one end of beam. For example
5m
x
10kN/m
A B
The variation range of x:
m50 ≤≤ x
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V and M Functions
5m
x
10kN/m
A B
The variation range of x:
m50 ≤≤ x
� The functions of V and M along the beam’s axis can be obtained by the
method of sections.
x
10kN/m
B
N
V
M
N
V
M
Sign Convention:
V
M
FBD:
(the right part)
The resultant force of distributed load is
equivalent to the area under the loading
diagram, and act through the geometric
center of this area.
xxarea 1010 =×=
10x
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V and M Functions
5m
x
10kN/m
A B
The variation range of x:
m50 ≤≤ x
� The functions of V and M along the beam’s axis can be obtained by the
method of sections.
x
10kN/m
B
V
M
FBD:
(the right part)
10x
:0∑ =M+
xV 10=
25xM −=
02
10 =×−−x
xM
Equilibrium Equations:
:0∑ =yF+ 010 =− xV
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V and M Diagrams
5m
x
10kN/m
A B
The variation range of x:
m50 ≤≤ x
� If the resulting functions of x are plotted, the graphs are termed the shear
force diagram and bending moment diagram.
xV 10=25xM −=
)pointat(0when Bx =
0010 =×=V
m5.2when =x
kN255.210 =×=V
)pointat(m5when Ax =
kN50510 =×=V
V
x 02.5m
25kN
5m
50kN
SFD:
Straight line
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V and M Diagrams
5m
x
10kN/m
A B
The variation range of x:
m50 ≤≤ x
� If the resulting functions of x are plotted, the graphs are termed the shear
force diagram and bending moment diagram.
xV 10=25xM −=
)pointat(0when Bx =
005 2=×−=M
m5.2when =x
kNm25.315.25 2−=×−=M
)pointat(m5when Ax =
kNm12555 2−=×−=M
V
x 05m
50kN
SFD:
BMD: x
M
02.5m
-125kNm
5m
Parabola(open downward)
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V and M Diagrams
5m
x
10kN/m
A B
� Bending moment diagram is
plotted on the compression side of
the beam.
V
x 05m
50kN
SFD:
BMD: x
M
0
-125kNm
5m
Parabola(open downward)
T
C
� UDL causes a linear change (a
straight line) to the SFD.
� UDL results in the BMD
assuming a parabolic curved shape.
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Example
6kN/m
A B
9m
Draw the shear force and bending moment
diagrams for the beam.
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Example
6kN/m
Draw the shear force and bending moment
diagrams for the beam.
Reaction Forces:
6kN/m
A B
9m
FBD:
The resultant force of distributed load is
equivalent to the area under the loading
diagram, and act through the geometric
center of this area.
kN272
m9kN/m6=
×=area
A B
Ay By
27kN
3m
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Example
6kN/m
Draw the shear force and bending moment
diagrams for the beam.
Reaction Forces:
6kN/m
A B
9m
FBD: A B
Ay By
27kN
3m
:0∑ =BM+
kN18=yB
kN9=yA
03279 =×+×− yA
:0∑ =yF+ 027 =+− yy BA
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Example
18kN
Draw the shear force and bending moment
diagrams for the beam.
V and M Functions:
6kN/m
A B
9m
FBD: Ax
9kN
9kN
x
N
V
M
N
V
M
Sign Convention:
V
M
Establish an arbitrary distance x from point A
where m90 ≤≤ x
?
6kN/m
?
x9m
96
? x=
3
2?
x=
2x3
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Example
18kN
Draw the shear force and bending moment
diagrams for the beam.
V and M Functions:
6kN/m
A B
9m
FBD: Ax
9kN
9kN
x
V
M
Establish an arbitrary distance x from point A
where m90 ≤≤ x
2x3
The resultant force of distributed load is
equivalent to the area under the loading
diagram, and act through the geometric
center of this area.
33
2
2
1 2xx
xarea =××=
x3
x2
3
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Example
18kN
Draw the shear force and bending moment
diagrams for the beam.
V and M Functions:
6kN/m
A B
9m
FBD: Ax
9kN
9kN
x
V
M
Establish an arbitrary distance x from point A
where m90 ≤≤ x
2x3
x3
x2
3
:0∑ =M+
39
2x
V −=
xx
M 99
3
+−=
033
92
=+×+×− Mxx
x
:0∑ =yF+ 03
92
=−− Vx
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Example
18kN
Draw the shear force and bending moment
diagrams for the beam.
V and M Diagrams:
6kN/m
A B
9m
SFD:
9kN
xm)90( ≤≤ x
39
2x
V −=
)pointat(0when Ax =
kN9=V
)pointat(m9when Bx =
kN18−=V
x
V
9kN
-18kN
,0when =V
03
92
=−x
m2.5=x
5.2m
0
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Example
18kN
Draw the shear force and bending moment
diagrams for the beam.
V and M Diagrams:
6kN/m
A B
9m
SFD:
9kN
xm)90( ≤≤ xx
xM 9
9
3
+−=
)pointat(0when Ax =
0=M
)pointat(m9when Bx =
0=M
x
V
9kN
-18kN
,m2.5when =x
kNm2.31max =M
5.2m
0
31.2kNm
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Discontinuous V and M Functions
�� The type or magnitude of the distributed load changesThe type or magnitude of the distributed load changes
�� Concentrated forces or couple moments are appliedConcentrated forces or couple moments are applied
In general, the internal shear force and bending moment functions will be
discontinuous or their slope will discontinuous at points where:
Because of this discontinuous property, V and M functions must be determined
for each segment of the beam located between two discontinuities of loading.
P
w
A B C D
3m 3m 4m
x1
x2
x3
3m0 1 ≤≤ x
Range of variations:
m63m 2 ≤≤ x
m106m 3 ≤≤ x
(from A to B)
(from B to C)
(from C to D)
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Discontinuous V and M Functions
�� The type or magnitude of the distributed load changesThe type or magnitude of the distributed load changes
�� Concentrated forces or couple moments are appliedConcentrated forces or couple moments are applied
In general, the internal shear force and bending moment functions will be
discontinuous or their slope will discontinuous at points where:
Because of this discontinuous property, V and M functions must be determined
for each segment of the beam located between two discontinuities of loading.
P
w
A B C D
3m 3m 4m
3m0 1 ≤≤ x
Range of variations:
m30 2 ≤≤ x
m40 3 ≤≤ x
(from A to B)
(from B to C)
(from C to D)
x1 x2 x3
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Example
Draw the shear force and bending moment diagrams for the beam.
6kN
A B C
2m 4m
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Example
Draw the shear force and bending moment diagrams for the beam.
6kN
A B C
2m 4m Reaction Force:
Ay Cy
:0∑ =CM+
kN2=yC
kN4=yA
0466 =×+×− yA
:0∑ =yF+ 06 =+− yy CA
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Example
Draw the shear force and bending moment diagrams for the beam.
6kN
A B C
2m 4m
V and M Functions:
4kN 2kN
For Segment AB, establish an arbitrary distance
x1 from point A where m20 1 ≤≤ x
x1
A
x14kN
V
M
N
V
M
N
V
M
Sign Convention:
FBD::0∑ =M+
14xM =
kN4=V
04 1 =+×− Mx
:0∑ =yF+ 04 =−V
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Example
Draw the shear force and bending moment diagrams for the beam.
6kN
A B C
2m 4m
V and M Functions:
4kN 2kN
For Segment AB, establish an arbitrary distance
x1 from point A where m20 1 ≤≤ x
x1
A
x24kN
V
M
N
V
M
N
V
M
Sign Convention:
FBD::0∑ =M+
14xM =
kN4=V
04 1 =+×− Mx
:0∑ =yF+ 04 =−V
For Segment BC, establish an arbitrary distance
x2 from point A where m6m2 2 ≤≤ x
x2
2m (x2 – 2)
:0∑ =M+
122 2 +−= xM
kN2−=V
0)2(64 22 =+−×+×− Mxx
:0∑ =yF+ 064 =−− V
6kN
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Example
Draw the shear force and bending moment diagrams for the beam.
6kN
A B C
2m 4m
V and M Functions:
4kN 2kN
For Segment AB m)20( 1 ≤≤ x
x1
4kN
14xM =kN4=V
For Segment BC m)6m2( 2 ≤≤ x
x2
122 2 +−= xMkN2−=V
0
V and M Diagrams:
SFD:
-2kN
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Example
Draw the shear force and bending moment diagrams for the beam.
6kN
A B C
2m 4m
V and M Functions:
4kN 2kN
For Segment AB m)20( 1 ≤≤ x
x1
4kN
14xM =kN4=V
For Segment BC m)6m2( 2 ≤≤ x
x2
122 2 +−= xMkN2−=V
0
V and M Diagrams:
SFD:
-2kN
BMD: 0
)pointat(m2when 1 Bx = kNm8=M
)pointat(m6when 1 Cx =
kNm8=M)pointat(m2when 2 Bx =
0=M
)pointat(0when 1 Ax = 0=M
8kNm
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Example – Another option
Draw the shear force and bending moment diagrams for the beam.
6kN
A B C
2m 4m
V and M Functions:
4kN 2kN
For Segment AB, establish an arbitrary distance
x1 from point A where m20 1 ≤≤ x
x1
FBD::0∑ =M+
14xM =
kN4=V
04 1 =+×− Mx
:0∑ =yF+ 04 =−V
For Segment BC, establish an arbitrary distance
x2 from point C where m4m0 2 ≤≤ x
x2
:0∑ =M+
22xM =
kN2−=V
02 2 =×+− xM
:0∑ =yF+ 02 =+V2kN
C
V
M
x2
N
V
M
N
V
M
Sign Convention:
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Example – Another option
Draw the shear force and bending moment diagrams for the beam.
6kN
A B C
2m 4m
V and M Functions:
4kN 2kN
For Segment AB m)20( 1 ≤≤ x
x1
4kN
14xM =kN4=V
For Segment BC m)40( 2 ≤≤ x
22xM =kN2−=V
0
V and M Diagrams:
SFD:
-2kN
x2
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Example – Another option
Draw the shear force and bending moment diagrams for the beam.
6kN
A B C
2m 4m
V and M Functions:
4kN 2kN
For Segment AB m)20( 1 ≤≤ x
x1
4kN
14xM =kN4=V
For Segment BC m)40( 2 ≤≤ x
22xM =kN2−=V
0
V and M Diagrams:
SFD:
-2kN
BMD: 0
)pointat(0when 1 Ax = 0=M
)pointat(m2when 1 Bx = kNm8=M
)pointat(m4when 1 Bx =
0=M)pointat(0when 2 Cx =
kNm8=M8kNm
x2
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Example
Draw the shear force and bending moment diagrams for the beam.
6kN
A B C
2m 4m
Features of SFD:
4kN 2kN
4kN
0SFD:
-2kN
BMD: 0
8kNm
SFD follows the force arrows – if a
load or reaction points upward, move
upwards on SFD.
SFD should be a horizontal line where
no load is applied such as segment AB
and segment BC separately.
SFD jumps at where a point load is
applied, and the step equal to the value
of the load.
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Example
Draw the shear force and bending moment diagrams for the beam.
6kN
A B C
2m 4m
Features of BMD:
4kN 2kN
4kN
0SFD:
-2kN
BMD: 0
8kNm
BMD follows the deformation shape
of the beam – plot BMD on the
compression side.
BMD should be a straight line where
no load is applied such as segment AB
and segment BC.
BMD has a sudden change in the slope
at where a point force is applied.
C
T
BMD always has a zero value at a end
pin or roller support, unless a moment
is applied at that point.
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Complicate Examples
F1
A B Cx1
x2
F2
D
x3
F1
A B Cx1
x2
F2
D
x3
F1
A B Cx1
x2
F2
D
x3
OR
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Summary
� Design of beam requires detailed knowledge of the variations of V
and M. Internal axial force N is generally not considered.
� To obtain the variation of V and M, it is necessary to section the
beam at an arbitrary distance x from one end of beam.
� If the resulting functions are plotted, the graphs are termed the
shear force diagram and bending moment diagram.
� Because of this discontinuous property, V and M functions must be
determined for each segment of the beam located between two
discontinuities of loading using x1, x2, ….(Note the variation range).
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CIVE1143 – Structural Analysis
Shear Force and Bending Moment Equations and Diagrams
Lecturers: Srikanth & Indu
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Revision: Discontinuous V and MFunctions
The type or magnitude of the distributed load changesThe type or magnitude of the distributed load changesConcentrated forces or couple moments are appliedConcentrated forces or couple moments are applied
In general, the internal shear force and bending moment functions will be discontinuous or their slope will discontinuous at points where:
Because of this discontinuous property, V and M functions must be determined for each segment of the beam located between two discontinuities of loading.
P
wA B C D
3m 3m 4m
x1
x2x3
3m0 1 ≤≤ x
Range of variations:
m63m 2 ≤≤ x
m106m 3 ≤≤ x
(from A to B)
(from B to C)
(from C to D)
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Revision: V and M Functions and DiagramsGeneral procedure for SFD and BMD
Determine all the Determine all the reactive forcesreactive forces and couple momentsand couple moments if necessary.if necessary.
Specify proper coordinators (Specify proper coordinators (xx11,, xx22,,……)) , draw the , draw the freefree--body diagramsbody diagramsfor corresponding segments.for corresponding segments.
Applying the equilibrium equations to obtain the Applying the equilibrium equations to obtain the shear force shear force functionsfunctions and and bending moment functionsbending moment functions for each segment (for each segment (VV((xx11), ), VV((xx22) ) …… and and MM((xx11), ), MM((xx22), ), ……). ).
Plot the Plot the shear force diagrams shear force diagrams andand bending moment diagramsbending moment diagramssegment by segment below the freesegment by segment below the free--body diagram of the structure. body diagram of the structure.
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Features of SFD and BMD
StepHorizontalApplied moment
CubicParabolaLinearly varying distributed load
ParabolaStraight lineUDL
KinkStepPoint load
Straight lineHorizontalNo load
Shape of BMDShape of SFDType of load
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Mathematical relationship between SFD and BMDConsider the small segment of beam subjected to only a uniform distributed load
w
V V+ΔV
M+ΔMM
Δx
wΔx Applying the force equation of equilibrium to segment, we have
:0∑ =yF+xwV Δ−=Δ
0)( =Δ+−Δ− VVxwV
wdxdV
−=
Dividing by Δx, and letting Δx → 0, we get
Slope of shear diagram = Negative of distributed load intensity
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Mathematical relationship between SFD and BMDConsider the small segment of beam subjected to only a uniform distributed load
w
V V+ΔV
M+ΔMM
Δx
wΔx
Applying the moment equation of equilibrium to segment, we have
:0∑ =M+
2)( xwxVM Δ−Δ=Δ
0)(2
=Δ++Δ
×Δ+−Δ− MMxxwMxV
wdxdV
−=
Dividing by Δx, and letting Δx → 0 and treating 2nd order terms as negligible, we get
Slope of shear diagram = Negative of distributed load intensity
Vdx
dM=
Slope of moment diagram = Shear
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Mathematical relationship between SFD and BMDConsider the small segment of beam subjected to only a uniform distributed load
w
V V+ΔV
M+ΔMM
Δx
wΔx wdxdV
−=
Slope of shear diagram = Negative of distributed load intensity
Vdx
dM=
Slope of moment diagram = Shear
Hence, the mathematical relationship among load, shear force and bending moment is
Load Shear Force
Bending Moment
Differentiate Differentiate
Integrate Integrate
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Mathematical relationship between SFD and BMDConsider the small segment of beam subjected to only a uniform distributed load
w
V V+ΔV
M+ΔMM
Δx
wΔx wdxdV
−=
Slope of shear diagram = Negative of distributed load intensity
Vdx
dM=
Slope of moment diagram = Shear
Location of the point on the beam where the bending moment is a maxium or minimum
M is achieves a maximum or minimum where
0=dx
dM
Vdx
dM=Since
the maximum or minimum M will occur where V=0
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Mathematical relationship between SFD and BMDConsider the small segment of beam subjected to only a single load
V V+ΔV
M+ΔMM
Δx
F Applying the force equation of equilibrium to segment, we have
:0∑ =yF+FV −=Δ
0)( =Δ+−− VVFV
The change in shear is negative, so that on the SFD the shear will jump downwards on the beam. Likewise, the jump in shear ΔV is upward when F acts upward.
Load SFD
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Mathematical relationship between SFD and BMDConsider the small segment of beam subjected to only a single moment
V V+ΔV
M+ΔMM
Δx
Applying the moment equation of equilibrium to segment (assuming Δx → 0), we have
0MM −=Δ
The change in moment is negative, ie. BMD will jump downward if M0 is anti-clockwise. Likewise, the jump in ΔM is upward when M0 is clockwise.
Load BMD
M0
:0∑ =M+ 0)(0 =Δ+++− MMMM
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Example 1
AB
CD
E
Draw shear force and moment diagrams for the beam.
10kN 10kN 6kN2m 2m 2m 2m
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Example 1Draw shear force and moment diagrams for the beam.
Reaction Forces:
AB
CD
E
10kN 10kN 6kN
By Dy
:0∑ =BM+
kN17=yB
kN9=yD
0664210210 =×−×+×−× yD
:0∑ =yF+061010 =−+−+− yy DB
AB
CD
E
10kN 10kN2m 2m 2m 2m 6kN
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Example 1Draw shear force and moment diagrams for the beam.
AB
CD
E
10kN 10kN2m 2m 2m 2m 6kN
9kN17kN
Shear Force Diagram:
SFD:
SFD follows the force arrows – if a load or reaction points upward, move upwards on SFD.
-10kN
0
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Example 1Draw shear force and moment diagrams for the beam.
AB
CD
E
10kN 10kN2m 2m 2m 2m 6kN
9kN17kN
Shear Force Diagram:
SFD:
SFD follows the force arrows – if a load or reaction points upward, move upwards on SFD.
-10kN
0
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Example 1Draw shear force and moment diagrams for the beam.
AB
CD
E
10kN 10kN2m 2m 2m 2m 6kN
9kN17kN
Shear Force Diagram:
SFD:
SFD follows the force arrows – if a load or reaction points upward, move upwards on SFD.
-10kN17kN0
7kN
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Example 1Draw shear force and moment diagrams for the beam.
AB
CD
E
10kN 10kN2m 2m 2m 2m 6kN
9kN17kN
Shear Force Diagram:
SFD:
SFD follows the force arrows – if a load or reaction points upward, move upwards on SFD.
-10kN
07kN
-3kN
6kN
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Example 1Draw shear force and moment diagrams for the beam.
AB
CD
E
10kN 10kN2m 2m 2m 2m 6kN
9kN17kN
Bending Moment Diagram:
SFD:
1. The moment of Zero at Point A.
-10kN
07kN
-3kN
6kN
BMD: 0
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Example 1Draw shear force and moment diagrams for the beam.
AB
CD
E
10kN 10kN2m 2m 2m 2m 6kN
9kN17kN
Bending Moment Diagram:
SFD:
1. The moment of Zero at Point A.
-10kN
07kN
-3kN
6kN
BMD: 0
2. The moment of Point B
kNm202)10(0 −=×−+=
MMM AB Δ+=
ΔM - the area of SFD.
-20kNm
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Example 1Draw shear force and moment diagrams for the beam.
AB
CD
E
10kN 10kN2m 2m 2m 2m 6kN
9kN17kN
Bending Moment Diagram:
SFD:
1. The moment of Zero at Point A.
-10kN
07kN
-3kN
6kN
BMD: 0
2. The moment of Point B
kNm62720 −=×+−=
MMM BC Δ+=
ΔM - the area of SFD.
-20kNm
3. The moment of Point C
-6kNm
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Example 1Draw shear force and moment diagrams for the beam.
AB
CD
E
10kN 10kN2m 2m 2m 2m 6kN
9kN17kN
Bending Moment Diagram:
SFD:
1. The moment of Zero at Point A.
-10kN
07kN
-3kN
6kN
BMD: 0
2. The moment of Point B
kNm122)3(6 −=×−+−=
MMM CD Δ+=
ΔM - the area of SFD.
-20kNm
3. The moment of Point C
-6kNm
4. The moment of Point D.
-12kNm
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Example 1Draw shear force and moment diagrams for the beam.
AB
CD
E
10kN 10kN2m 2m 2m 2m 6kN
9kN17kN
Bending Moment Diagram:
SFD:
1. The moment of Zero at Point A.
-10kN
07kN
-3kN
6kN
BMD: 0
2. The moment of Point B
02612 =×+−=MMM DE Δ+=
ΔM - the area of SFD.
-20kNm
3. The moment of Point C
-6kNm
4. The moment of Point D.
-12kNm
5. The moment of Point E.
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Example 2
A BC
20kN/m
4m 4m
Draw shear force and moment diagrams for the beam.
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Example 2
A BC
20kN/m
4m 4m
Draw shear force and moment diagrams for the beam.
Reaction Forces:
ByAy
80kN
:0∑ =BM+
kN60=yB
kN20=yA
02808 =×+×− yA
:0∑ =yF+080 =+− yy BA
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Example 2
A BC
20kN/m
4m 4m
Draw shear force and moment diagrams for the beam.
Shear Force Diagram:
60kN20kN
SFD: 0
20kN
SFD follows the force arrows – if a load or reaction points upward, move upwards on SFD.
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Example 2
A BC
20kN/m
4m 4m
Draw shear force and moment diagrams for the beam.
Shear Force Diagram:
60kN20kN
SFD: 0
20kN
SFD follows the force arrows – if a load or reaction points upward, move upwards on SFD.
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Example 2
A BC
20kN/m
4m 4m
Draw shear force and moment diagrams for the beam.
Shear Force Diagram:
60kN20kN
SFD: 0
20kN
SFD follows the force arrows – if a load or reaction points upward, move upwards on SFD.
The shear force at point B
kN604)20(20 −=×−+=
VVV CB Δ+=
ΔV - the area of UDL.-60kN
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Example 2
A BC
20kN/m
4m 4m
Draw shear force and moment diagrams for the beam.
Shear Force Diagram:
60kN20kN
SFD: 0
20kN
SFD follows the force arrows – if a load or reaction points upward, move upwards on SFD.
-60kN
x x
4m
20
80
48020 x
= m1=x
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Example 2
A BC
20kN/m
4m 4m
Draw shear force and moment diagrams for the beam.
Bending Moment Diagram:
60kN20kN
SFD: 0
20kN
-60kN
1m
BMD: 0
1. The moment of Zero at Point A.
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Example 2
A BC
20kN/m
4m 4m
Draw shear force and moment diagrams for the beam.
Bending Moment Diagram:
60kN20kN
SFD: 0
20kN
-60kN
1m
BMD: 0
1. The moment of Zero at Point A.
2. The moment of Point C
kNm804200 =×+=
MMM AC Δ+=
ΔM - the area of SFD.
80kNm
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Example 2
A BC
20kN/m
4m 4m
Draw shear force and moment diagrams for the beam.
Bending Moment Diagram:
60kN20kN
SFD: 0
20kN
-60kN
1m
BMD: 0
1. The moment of Zero at Point A.
2. The moment of Point C
kNm902
12080 =×
+=
MMM C Δ+=max
ΔM - the area of SFD.
3. The maximum moment
80kNm90kNm
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Example 2
A BC
20kN/m
4m 4m
Draw shear force and moment diagrams for the beam.
Bending Moment Diagram:
60kN20kN
SFD: 0
20kN
-60kN
1m
BMD: 0
1. The moment of Zero at Point A.
2. The moment of Point C
02/3)60(90 =×−+=
MMM B Δ+= max
ΔM - the area of SFD.
3. The maximum moment
80kNm90kNm
4. The moment of Point B