ENGG1300 FUNDAMENTAL MECHANICS
Bending Internal Force
Introduction
2
Introduction
3
Introduction
4
Internal Forces in Beams
• Internal forces on
cross-sections of
bending beams
m
F
A
B
m
FA
m
A
M
FS
m
m
F
FB
M
FS
m
The internal forces on cross-section
m-m can be simplified as shear
force and bending moment acting at
cross-sectional centroid.
B
5
Symmetric Bending
• Beam: a structural element which takes bending as the main form
of deformation.
• Symmetric bending: all external forces & moments acting in the
symmetric longitudinal plane (beam axis being bent from a straight
line to a curve within the symmetric longitudinal plane).
6
Types of Beams
Simply supported beam
Beam with an overhang
Cantilever beam
Compound beam
Types of Supports and Reactions
Fixed support
Hinged or pinned support
FRx
FRx
MR
FRy
Roller support
FRy
FRy
8
Determinacy of Beams
• Statically determinate beams
• Statically indeterminate beams
- Cantilever beam
- Simply supported cantilever beam
- Simply supported beam
- Continuous beam
- Overhanging beam
- Fixed beam
9
Sign Convention of shear Force and Bending Moment
• Shear force
- Positive: left up / right
down (clockwise loop)
FS
FS
FS
FS
• Bending moment
- Positive: upper half
under compression /
lower half under tension
(top concave / bottom
convex)
Sagging moment
M
M
Hogging moment
M
M
10
Example 4.1
• Find the shear force and bending moment at the left and right
cross-section at point C.
F
A
• Solution：
B
C
FA
a
a
F
FA = FB =
2
1. Reaction forces at the supports
2. Internal forces (assuming
positive) at the right crosssection of C.
F
FSCA = FA =
2
Fa
M CA = FA  a =
2
FB
FSCA
A
MCA
FA
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3. Internal forces (assuming
positive) at the left crosssection of C.
F
A
B
C
a
F
FSCB = − FB = −
2
Fa
M CB = FB  a =
2
a
FSCB
MCB
B
FB
Note: For a concentrated force acting at C, the change of bending
moment is zero while that of shear force is equal to the
magnitude of the concentrated force.
12
Example 4.2
• Find the shear force and bending moment at the left and right
Fa
cross-section at point C.
A
• Solution：
B
C
a
FA
a
FB
1. Reaction forces at the supports
F
FA = FB =
2
FSCA
2. Internal force at the right
cross-section of C.
F
FSCA = FA =
2
Fa
M CA = FA  a =
2
A
FA
MCA
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3. Internal force at the left
cross-section of C.
Fa
A
a
FSCB
F
= FB =
2
M CB
Fa
= − FB  a = −
2
B
C
a
FSCB
MCB
B
FB
Note: For a concentrated moment acting at C, the change of shear
force is zero while that of bending moment is equal to the
magnitude of the concentrated moment.
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Direct Calculation of Shear Force
• The shear force at a beam cross-section is equal to the net of the
external forces collected from either side of the cross-section.
• Upward net from left portion or downward net from right portion
results in positive shear force at the cross-section.
• All forces acting on a beam, including the reaction forces at the
supports, must be taken into consideration.
15
Example 4.3
• Directly calculate the shear force
at cross-section C without the
explicit use of method of section.
• Solution：
F1
F3
F4
q
A
B
C
FA
F2
FB
- Based on the left portion from cross-section C:
FSC = FA − F1 + F2 − F3
- Based on the right portion from cross-section C:
FSC = − FB + ql + F4
Note: in practice, we pick the portion with simpler loadings.
16
Direct Calculation of Bending Moment
• The bending moment at a beam cross-section is equal to the net of
the external moments with respect to the cross-section centroid,
collected from either side of the cross-section
• Clockwise net from left portion or counter clockwise net from right
portion results in positive bending moment.
• All forces acting on a beam, including the reaction forces at the
supports, must be taken into consideration.
17
Example 4.4
• Directly calculate the bending moment at cross-section C without
the explicit use of method of section.
• Solution：
F1
M1
A
- Based on the left portion from
cross-section C
FA
C
B
F2
FB
MC = FAd A − F1d1 − M1
- Based on the right portion from cross-section C
MC = − FB d B + F2d 2
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Diagram of Shear Forces & Bending Moments
• Equation of shear forces and bending moments
- Equation of shear forces: FS = FS(x)
- Equation of bending moments: M = M(x)
- x: denotes the position of cross-section.
• Diagram of shear forces and bending moments
- Abscissa: cross-section position (x)
- Ordinate: shear forces (FS) / bending moments (M)
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Example 4.5
• Draw the diagram of shear forces and bending moments
F
x1
A
FA
x2
B
C
a
a
FB
• Solution:
F
1. Reaction force at the supports: FA = FB =
2
2. Equation of internal forces
F
FS ( x1 ) = FA =
2
F
FS ( x2 ) = − FB = −
2
x1  [0, a)
x2  [0, a)
Positive: left & upward
Negative: right & upward
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FS
Fx1
x1  [0, a)
M ( x1 ) = FA x1 =
2
0.5F
0.5F
x1
x2
F
C
A
B
Positive: left & clockwise
Fx2
M ( x2 ) = FB x2 =
2
x2  [0, a)
FA
a
a
FB
Positive: right & counter clockwise
M
0.5 Fa
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Example 4.6
• Draw the diagram of shear forces and bending moments
x1
A
• Solution:
FA
C
a
1. Reaction forces at the supports
x2
Fa
a
B
FB
FS
0.5F
F
FA = FB =
2
2. Equations of internal forces
F
F
FS ( x1 ) = ; FS ( x2 ) =
2
2
Fx1
Fx2
M ( x1 ) =
; M ( x2 ) = −
2
2
-0.5Fa
M
0.5Fa
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Example 4.7
q
A
B
ql 2
FS
x
• Plot the diagram of shear
forces and bending moments
• Solution:
ql 2
l
• ql / 2
1. Reaction forces at the
supports
2. Equation of internal forces
ql
FS ( x) = − qx
2
•
−ql / 2
ql
x
M ( x) =
x − qx
2
2
2
•
M
q l2 / 8
3. Diagram of internal forces
q 
l  q l2
=− x−  +
2
2
8
(0  x  l )
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Relations among Loads, Shear Forces and Bending Moments
q( x)
q( x)
x
M (x )
M ( x ) + dM ( x )
dx
FS ( x)
FS ( x) + dFS ( x)
FS ( x ) + q ( x ) dx − FS ( x ) − dFS ( x ) = 0
1
M ( x ) + dM ( x ) − q ( x ) d 2 x − M ( x ) − FS ( x ) dx = 0
2
24
Relations among Loads, Shear Forces and Bending Moments
FS ( x ) + q ( x ) dx − FS ( x ) − dFS ( x ) = 0
x2
dFS ( x )
q ( x) =
,
FS2 − FS1 =  q ( x ) dx
dx
x1
1
M ( x ) + dM ( x ) − q ( x ) d 2 x − M ( x ) − FS ( x ) dx = 0
2
dM ( x )
FS ( x ) =
,
dx
dM 2 ( x )
q ( x) =
dx 2
M 2 − M1 =
x2
 F ( x ) dx
S
x1
Sign convention of q(x): up positive / down negative.
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Relations among Loads, Shear Forces and Bending Moments
• Without distributed load: q(x) = 0
dFS ( x )
q ( x) =
=0
dx
FS ( x ) = const
Note: the slope of the diagram of shear forces is zero.
dM ( x )
= FS ( x ) = const
dx
M ( x ) = ( const )  x + ( const )
Note: the slope of the diagram of bending moments is constant.
26
Relations among Loads, Shear Forces and Bending Moments
• With uniformly distributed load: q(x) = const
dFS ( x )
q ( x) =
= const
dx
FS ( x ) = ( const )  x + ( const )
Note: the slope of the diagram of shear forces is constant.
dM ( x )
= FS ( x )
dx
M ( x ) = ( const )  x 2 + ( const )  x + ( const )
Note: M(x) is a second order polynomial of x.
27
Relations among Loads, Shear Forces and Bending Moments
Loads
No load
(P=0 & q=0)
Uniformly distributed
q=const
Concentrated load
(P)
C
Diagram of shear forces
FS  0
q0
q0
Discontinuous Fs
C
Concentrated moment
(m)
C
FS  0
Diagram of bending moments
FS  0
FS  0
q0
q0
C
Continuous but not
differentiable M
Discontinuous M
Const
C
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Example 4.8
• Draw the diagram of shear forces and bending moments.
A
B
• Solution:
FA
ql
FA =
8
5ql
FB =
8
FS
q
l
FB 2
l
C
ql / 2
−ql / 8
− ql 2 / 8
M
29
Example 4.9
q
• Draw the diagram of
shear forces and bending
moments.
A
• Solution:
FA
3ql
FA =
8
ql
FB =
8
FS 3ql
8
B
l
2
l
2
x = •3l 8 •x = l 2
•
M
9ql
2
128
FB
−
ql
8
•
ql 2
16
30
Example 4.10
• Draw the diagram of
shear forces and bending
A
moments.
q
• Solution:
a
FA = qa
FB = qa
FA
FS
qa
B
a
−qa
M
qa 2 / 2
qa
a
FB
−qa 2 / 2
qa 2 / 2
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Example 4.11
• Draw the diagram of
shear forces and bending
moments.
q
qa 2 / 2
A
q
a
• Solution:
qa
FA =
2
FB = qa
2
FS
qa
B
FA
a
FB
qa / 2
−qa / 2
M
qa 2 / 8
qa 2 / 2
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